Mechanical Design of Process Systems-Vol 1 (Piping & Pressure Vessels)_2 كتاب موائع...

Mgchanica! Designof Process Systems

VolumelPiping and

Pressure Vessels

A.Keith Escoe

Gulf Publishing CompanyBook Division

Houston, London, Paris, Tokyo

Mechanical Dsign

of hocess Sy$erns

Volume IPiping ard hesun \bsels

Copyright O 1986 by Gulf Publishing Compann Houston,'Ibxas.All rights reserved. Printed in the United States of America. Thisbook, or parts thereof, may not be reproduc€d in any form withoutpermission of the publisher.

Library of Congrcss Cataloging-in-Publication DataEscoe, A. Keith.

Mechanical design of process systems.Bibliography: p.Includes index.1. Chemical plants-Design and consbuction.

L Title.TP155.5.E83 1986 [email protected]' 8l

ISBN G87201-562-9 (Vol. 1)ISBN G 87201-565-3 (Vol. 2)

85-22005

IY

Foreword ...,....viiby John J. McKetta

Preface .. , .... ... ix

Chapter 1

Piping Fluid Mechanics ........... 1

Basic Equations, INon-Newtonian Fluids, 5

Velocity Heads, 8Pipe Flow Geometries, 22Comoressible Flow. 25Piping Fluid Mechanics Problem Formulation, 25Example 1-1: Friction Pressure Drop for a

Hydrocarbon Gas-Steam Mixture in a Pipe, 27Example 1-2: Frictional Ptessure Drop for a Hot

Oil System of a Process Thnk, 33Example 1-3: Friction Pressure Drop for a Waste

Heat Recovery System, 42Example 1-4: Pressure Drop in Relief Valve

Piping System, 43Notation, 45References, 45

Chapter 2The Engineering Mechanics of Piping .,...47

Piping Criteria, 47Primary and Secondary Stresses, 49

Allowable stress Range for Secondary Stresses.

Flexibility and Stiffness of Piping Systems, 52Stiffness Method Advantages. FlexibilityMethod Advantages.

Stiffness Method and Large Piping, 58Flexibility Method of Piping Mechanics. PipeLoops.

PiDe Restraints and Anchors. 68- Pipe Lug Supports. Spfing Supports. ExpansionJoints. Pre-stressed Piping.

Contents

Fluid Forces Exerted on Piping Systems, 81

Extraneous Piping Loads, 83Example 2-l: Applying the Stiffness Method to a

Modular Skid-Mounted Gas LiquefactionFacility,88

Example 2-2: Applying the Flexibility Method toa Steam Turbine Exhaust Line, 95

Example 2-3: Flexibility Analysis for Hot OilPiping,96

Example 2-42 Lug Design, 98Example 2-5: Relief Valve Piping System, 99Example 2-61 Wind-Induced Vibrations of

Piping, 100Notation, 101

References, 101

Chapter 3Heat Transfer in Piping and Equipment ... 103

Jacketed Pipe versus Traced Pipe, 103

Tracing Piping Systems, 106Traced Piping without Heat Tmnsfer Cement.Traced Piping with Heat Transfer Cement.Condensate Return. Jacketed Pipe. Vessel andEquipment Traced Systems.

Heat Transfer in Residual Systems, 132Heat Transfer through Cylindrical Shells.Residual Heat Transfer through Pipe Shoes.

Example 3-1: Steam Tracing Design, 136Example 3-2: Hot Oil Tracing Design, 137Example 3-3: Jacketed Pipe Design, 139Example 3-4: Thermal Evaluation of a Process

Thnk, 140Example 3-5: Thermal Design of a Process

Tank, 142Internal Baffle Plates Film Coefficient. FilmCoefficient External to Baffles-ForcedConvection. Heat Duty of Internal VesselPlates. Outside Heat Transfer Jacket Plates.Heat Duty of Jacket Plates Clamped to BottomVessel Head. Total Heat Duty of Tank.

Example 3-6: Transient and Static Heat TransferDesign, 148

Static Heat Transfer Analysis. Total HeatRemoval. Water Required for Cooling.Transient Hear Transfer Analysis.

Example 3-7: Heat Transfer through VesselSkirts, 152

Example 3-E: Residual Heat Transfer, 154Example 3-9: Heat Transfer through Pipe Shoe,

156Notation, 156References, 157

Chapter 4The Engineering Mechanics of PressureVessels ... . ..... 159

Designing for Internal Pressure, 159Designing for External Pressure, 160Design of Horizontal Pressure Vessels, 166

Longitudinal Bending Stresses. Location ofSaddle Supports. Wear Plate Design. ZickStiffening Rings.

Steel Saddle Plate Design, 174Saddle Bearing Plate Thickness, 180Design of Self-Supported Vertical Vessels, 180Minimum Shell Thickness Reouired for

Combined Loads, 181Support Skirt Design, 183Anchor Bolts, 184Base Plate Thickness Design, 186Compression Ring and Gusset Plate Design, 189Anchor Bolt Torque, 189Whd Aralysis of Towers, 190

r'\'ind Design Speeds. Wind-Induced Moments.$ ind-Induced Deflections of Towers.l ind-Induced Vibrations on Tall Towers.O\aling. Criteda for Vibration Analysis.

Seismic Design of Tall Towers, 209\anical Distribution of Shear Forces.

Tower Shell Discontinuities and Conical Sections,1t i

Exanple {-l: Wear Plate Requirement Analysis,215

Example 12: Mechanical Design of ProcessColumn. 215

Sectron lt{omenls of Inertial lbwer SectionStress Calcularions. Skirt and Base PlateDesign- Section Centroids. Vortex-Inducedvibrarion. Equivalent Diameter Approachversus -{\S[ A58.1- 1982.

Example 4-3: Seismic Analysis of a VerticalTower, 237

Example 44: Vibration Analysis for Tower withLarge Vortex-Induced Displacements, 241

Moments of Inertia. Wind Deflections.Example 4-5: Saddle Plate Analysis of a

Horizontal Vessel, 249Saddle Plate Buckling Analysis. HorizontalReaction Force on Saddle.

Notation,252References,254

Appendix APartial Volumes and Pressure VesselCafcufations .....25s

Partial Volumes of Spherically Dished Heads,256

Partial Volumes of Elliptical Heads, 257Partial Volumes of Torispherical Heads, 259Internal Pressure ASME Formulations with

Outside Dimensions, 261Internal Pressure ASME Formulations with Inside

Dimensions,262

Appendix BNational Wind Design Standards ......... 265

Criteria for Determining Wind Speed, 265Wind Speed Relationships, 266ANSI A58.1-1982 Wind Cateeories. 267

Appendix CProperties of Pipe. ,.....271

Insulation Weight Factors, 278Weights of Piping Materials, 279

Appendix DConversion Factors ..... . 303

Alphabetical Conversion Factors, 304Synchronous Speeds, 31 1

Temperature Conversion. 3l 2Altitude and Atmospheric Pressures, 313Pressure Conversion Chart, 314

Index .. . .... . ... 315

vl

The engineer who understands the impact of process

design decisions on mechanical design details is in a po-

sition to save his client or his company a lot of money.That is because the test of any process design is in how

cost-effectively it yields the desired product, and how"cost" generally translates to "equipment": How muchwill the process require? How long will it last? Howmuch energy will it consume per unit of product?

In this two-volume work on Mechanical Design ofProcess Systems, A. K. Escoe has performed a monu-mental service for mechanical design engineers and

chemical process engineers alike. It is presented in sucha manner that even the neophyte engineer can grasp itsfull value. He has produced an in-depth review of theway in which process design specifications are inter-preted into precise equipment designs. Perhaps mostvaluable of all are the extensive worked examplesthroughout the text, of actual designs that have been suc-cessfully executed in the field.

The piping system is the central nervous system of a

fluid flow orocess. and the author has treated this with

Foreword

proper respect in two excellent chapters on fluid me-chanics and the engineering mechanics of piping.

The chapter on heat transfer in vessels and piping il-lustrates lucidly the interrelationship between process

and mechanical design. Every engineer working with in-dustrial process systems will benefit from reading thischaDter.

Although the author has made a herculean effort incovering the mechanical design of pressure vessels, heat

exchangers, rotating equipment, and bins, silos and

stacks, it is true that there are omissions. It is hoped that,as the author hints in his preface, a future volume mightbe added covering multiphase flow, specific cogenera-tion processes, turbines, and detailed piping dynamics.

Still, at this writing these two volumes comprise an

outstanding practical reference for chemical and me-chanical engineers and a detailed instructional manualfor students.

I recommend these volumes highly for each design en-gineer's professional library.

John J. McKexa. Ph.D., P.E.

Joe C. Waher Professor of Chemical EngineeringUniversitY of Texas ' Austin

vii

Dedication

To the memory of my beloved parents, Aubrey H. Es-coe and Odessa Davies Escoe; and to. the dedicated engi-neer, Dr. Judith Arlene Resnik, U.S. astronaut aboardthe ill-fated space shuttle Chnllenger (Flight 5l-L).

v|ll

d{ ry,'

This book's purpose is to show how to apply mechani-cal engineering concepts to process system design. Pro-cess systems are common to a wide variety of industriesincluding petrochemical processing, food and pharma-ceutical manufacturing, power generation (including co-generation), ship building, and even the aerospace indus-try. The book is based on years of proven, successfulpractice, and almost all of the examples described arefrom process systems now in operation.

While practicality is probably its key asset, this firstvolume contains a unique collection ofvaluable informa-tion, such as velocity head data; comparison ofthe flexi-bility and stiffness methods of pipe stress analyses; anal-ysis of heat transfer through pipe supports and vesselskirts; a comprehensive method on the design of hori-zontal vessel saddles as well as a method to determinewhen wear plates are required; detailed static and dy-namic methods of tower design considering wind gusts,vortex-induced vibration and seismic analysis of towers;and a comparative synopsis of the various national windcooes.

Topics include.d in the text are considered to be thosetypically encountered in engineering practice. There-fore, because most mechanical systems involve single-phase flow, two-phase flow is not covered. Because ofits ubiquitous coverage in the literature, flange design isalso excluded in this presentation. Since all of the majorpressure vessel codes thoroughly discuss and illustratethe phenomenon of external pressure, this subject is onlymentioned briefly.

This book is not intended to be a substitute or a re-placement of any accepted code or standard. The readeris strongly encouraged to consult and be knowledgeableof any accepted standard or code that may govern. It is

heface to Volume I

felt that this book is a valuable supplement to any stan-dard or code used.

The book is slanted toward the practices of the ASMEvessel and piping codes. In one area of vessel design theBritish Standard is favored because it nrovides excellenttechnical information on Zick rings. The book is writtento be useful regardless of which code or standard is used.The intent is not to be heavily prejudiced toward anystandard, but to discuss the issue-engineering. If onefeels that a certain standard or code should be men-tione.d, please keep in mind that there are others whomay be using different standards and it is impossible todiscuss all of them.

The reader's academic level is assumed to be a bache-lor of science degree in mechanical engineering, but en-gineers with bachelor of science degrees in civil, chemi-cal, electrical, or other engineering disciplines shouldhave little difficulty with the book, provided, of course,that they have received adequate academic training orexperience.

Junior or senior undergraduate engineering studentsshould find the book a useful introduction to the applica-tion of mechanical engineering to process systems. Pro-fessors should find the book a helpful reference (and asource for potential exam problems), as well as a practi-cal textbook for junior-, senior-, or graduateJevelcourses in the mechanical, civil, or chemical engineeringfields. The book can also be used to supplement an intro-ductory level textbook.

The French philosopher Voltaire once said, "Commonsense is not very common," and unfortunately, this issometimes the case in engineering. Common sense is of-ten the by-product of experience, and while both are es-sential to sound engineering practice, neither can be

--*

ix

learned from books alone. It is one ofthis book's eoats tounite these three elements of "book learning," c6mmonsense, and experience to give the novice a better grasp ofengineering principles and procedures, and serve as apractical design reference for the veteran engineer.

Finally, I wish to thank Dr. John J. McKetta, professorof chemical engineering at the University of Texas atAustin, who had many helpful comments, suggestions,

and words of encouragement. I also wish to thank otherengineering faculty members at the University of Texasat Austin for their comments. I must exDress thanks toLarry D. Briggs for reviewing some ialculations inChapter 4; and last, but certainly not least, I wish to ex-press gratitude to William J. Lowe and Timothy W. Calkof Gulf Publishing Company, whose hard work and pa-tience made this book oossible.

A. Keith Escoe, PE.

. { j&ir,,

The study of fluid energy in piping systems is a com-prehensive subject that could in itself fill countless vol-umes. This chapter is primarily concerned witl fluid en-ergy dissipated as friction resulting in a head loss.Although this topic is popularly known in industry as

"hydraulics," the term "piping fluid mechanics" is usedhere to avoid confusion.

BASIC EOUATIONS

The basic equation of fluid mechanics, originally de-rived by Daniel Bernoulli in 1738, evolved from theprinciple of conservation of energy:

Pr - Pz = V,t=- vrt + (y. _yr;€1pp 28" c"

where subscripts I and 2 refer to flow upstream (afterthe flow process) and downstream (before the flow pro-cess), respectively, and

Pt - Pz : change in pressure headp

Vt^- V' : change in velocity head (kinetic energy)29"

:dz

,]V r ,{E -r- ,llr. + ,1ll-29" g" ^

(Yr - Yr) I(l-l)

p

where

cm (kg)

P:

8":

dY:

density, lb./ft3 or g./cm3pressure, lb/ft2 or kg/cm2velocity, ftlsec or cm/secconversion constant, 32. 17 (ft-lb./sec2lbr)gravitational acceleration : 32.2 fllsecz,cm/sec2; g/9" : 1.0height above datum, ft, cmdifferential between height above datum andreference point, ft, cmhead loss, friction loss, or frictional pressuredrop, ft-lbr/Ib., cm-kg6/g.energy added by mechanical devices, e.g.pumps, ft-lb/Ib., cm-kg/g.energy extracted by mechanical devices, e.g.turbines, ftlb6/1b., cm-kg/g.

F:

He:

HE:

Rewriting Equation l-1 along a fluid streamline betweenpoints 1 and 2 with steady, incompressible flow and nomechanical energy added or extracted results in

Piping Fluid Mechanics

(r-2)

= change in static head (potential energy)

F : friction 1o* in !JlQ,

The following are expressions of the Bernoulli equa-tion when applied to various incompressible and com-pressible flow conditions:

Incompressible flow-p, - P. v,2 - v.2

P zE" gc

Compre s sib le -i s othermal f low -

FJn H : X[ _[*l [*l] + (zz - z,

+F+HA+HE

2 Mechanical Design of Process Systems

Compre s s ib le -adiabati c f low -

H [1 [' - (,*J'.-"'] : xl' -FJ^ [*J]+(22-z)+F+HA+HE

. /o \* /p\where l- | : l:l : general gas law

\Prl \rrl

k : .specific heat ratio (adiabatic coefficient),t- lt-

Cp : sPecific heat at constant pressure,Btu/lb.-'F

C, : specific heat at constant volume, Btu/lb--"F

Equation 1-2 is the analytical expression that states apressure loss is caused by a change in velocity head,

static head, and ftiction head. The most cofirmon unitsare "feet of head." lb. and lbr do not cancel out and theexpression is exactly "energy (ft-lb) per pound ofmass."

In most industrial fluid problems, Equation 1-2 is

cumbersome to use, because the friction loss is the pa-

rameter most often desired. The friction loss is the workdone by the fluid in overcoming viscous resistance. Thisfriction loss can only rarely be analytically derived and is

determined by empirical data developed through experi-mental testins .

Forcing a fluid through a pipe component requires en-ergy. This energy is expended by shear forces that de-velop between the pipe wall and the fluid, and to a lesserextent among the fluid elements themselves. These shearforces are opposed to fluid flow and require excess en-ergy to overcome. Figure 1-l shows a simple version ofthis phenomenon and illustrates how shear stresses in-crease in the radial direction away from the pipe centerline and are maximum within the boundary layer next tothe wall. Friction energy loss is a resuit of these shearstresses next to the pipe wall. Excess loss in energy oc-curs because of local turbulence and changes in the di-rection and speed of flow. As a fluid changes direction,energy is expended because of a change in momentum.The methods used to determine energy loss caused bywall friction are essentially the same, where the pipecomponent is treated as a straight piece of pipe. How-ever, the methods used to determine energy loss causedby change in momentum differ, and a couple are de-scribed as follows.

Equivalent Length

In this approach to determining energy loss caused bya change in fluid momentum, a piping component is ex-tended a theoretical length that would yield the same en-ergy loss as the actual component. This length is calledthe "equivalent length" because it is that length requiredto obtain the same amount of friction pressure drop as

the piping component alone. The major problem with

dvoy

x+c

rf> --

---[, . 9e a"] or1'1

Figure 1-1. Shear stresses in fully developed flow. Shown here are imaginary fluid elements "slipping" over one another.

this method is that the equivalent length for a pipe com-ponent varies with the Reynolds number, roughness,size, and geometry of the pipe. All these par.rmetersmust be analyzed in using this method.

Velocity Head llethod

Since the excess head loss is mostly attributed to fluidturbulence, the velocity head method is widely acceptedand is replacing the equivalent length method in fluidcalculations. Throughout this book, the velocity head ap-proach will be used.

The velocity head is the amount of kinetic energy in a

fluid, Y2l2g". This quantity may be represented by theamount of potential energy required to accelerate a fluidto a given velocity. Consider a tank holding a fluid with apipe entrance shown in Figure 1-2. We draw a streamlinefrom point 1 of the fluid surface to point 2 at the pipeentrance. Applying Equation 1-2 at point 1 we obtain thefollowins:

1=p

And applying Equation 1-2 at point 2 we have

Pr-P2_Pr_V22PP2g"

in which the change in fluid pressure between points Iand,2 is Y ] l2g, or one velocity head. A pressure gaugemounted on the pipe entrance would record the differ-ence of pressure of one velocity head. This term is ac-counted for in Equation 1-2 by Y y2 - Y2212g..

Analyzing a simple conversion from potential to ki-netic energy is an elementary procedure, as demon-strated. After the fluid passes through the pipe entrance

Piping Fluid Mechanics 3

into the piping system, the factor F in Equation 1-2 be-comes the desired parameter. This friction loss is thework done by the fluid in overcoming viscous resistanceand loss attributed to turbulence. The parameter F iscomposed of two components, pipe wall friction andlosses for the various pipe fittings, pipe entrances, pipeexits, and fluid obstructions that contribute to a loss influid energy. These latter losses are described in terms ofvelocity heads, K;. In solving for F in Equation 1-2, wefirst obtain pressure loss attributed to pipe wall friction,represented by

-AP. =.: eyll]' 2e. \d/(1-3)

By adding values of velocity head losses to Equation 1-3,we obtain the lollowing for any piping system:

t". \ .,,- aP, : ILL + )-r,l4I

\u I .6c(l-4)

where flld is the dependent pipe friction of the pipe ofdiameter d over the length L, and DK; the summation ofvelocity head losses. Equation l-4 provides the frictionpressure drop in a pipe for a steady-state incompressiblefluid of fully developed flow with a flat veiocity profile.Examples of this equation are given after the terms inEquation 14 are further explained.

The term (flld) (pV2l2g") expresses the amount of en-ergy loss attributed to shear forces at the pipe wall and isbased on experimental evidence. It is a function of thepipe component length and diameter and the velocity ofthe fluid. Writing the relationship for friction pressuredrop as a result of pipe wall friction results in

- [L pV']'p' - t+qd 2i- ' '-J'

where Fp, : i.i"aion torr, priL : length of pipe, in.d : corroded inside diameter, in.

The other terms are explained with Equation 1-1.Equation l-5 may be expressed in various forms. To ex-press flow rate in gpm (w) and d in inches use

FPf : 0.000217 fLW/d5 (l-5a)

Equation l-5 is the most commonly used relationshipand is known as the Fanning equation. Dividing theequation by p/144 yields feet of friction loss rather thanpsl.

The reader is cautioned in applying the friction factorf, because it is not always defined as above and some au-

\,, g

Figure 1-2. Storage tank.


thors use 4f1 in place of f. If such factors are used, par-ticular attention should be paid to the specific frictionfactor chart used.

The friction factor f is dependent upon the dimension-less term expressing the roughness of the pipe (E/D,where E is the depth of the pipe) and the dimensionlessReynolds number Nr" : dpV/M, where l1, is the absoluteviscosity of the fluid, inJb1-sec/ftz. The Reynolds num-ber is the single most important parameter in fluid me-chanics because it establishes flow regimes and dynamicsirnilarity. The relationship between the friction factor f,the pipe roughness, and the Reynolds number is shownin the classic relationship given by Moody in Figure 1-3.

Figure l-3 may be presented in a more convenientform as shown in Figure 1-4, where the relative rough-ness of the pipe is based on a single value of roughness.This value of roughness must be an average value esti-mated to simplii/ the problem. The figures presentedherein are the best available until more reliable friction

factor data can be obtained and better understoodthrough use of new methods for measuring roughness.

Figure 1-3 is broken into three flow regimes-laminar, transition from laminar to turbulent, and turbu-lent. The Reynolds numbers establishing these zones are2,100 for laminar, 2,100 to 3,000 for transition zone,and 3,000 or more for turbulent

The basis for Figure 1-3 is the classic Colebrook equa-tron

| ^. Idd 2.51 I

r1r, : -. to8ro [- " **,rpi (l -6a)

for (3,000 to 4,000) < NR" < 108

For laminar flow the friction factor is determined by thesimple expression

"64Nn.

\2 -q-s9l

(1-6b)

.09

.08

.07

.06

^ .04

:

=- .03

.05

.04

.01

.o?

.0t5

.01

.008

.006a

oo4 :003 :002

.0015 :^^, '-0008 -

.0006

.01

.009

.008 ? 3 4 56 I z J 4 56 8 rot 2 3 4 56 Blo5 2 3 4 56 € to7

R?ynotds Nunber n" = f r, '. -If* , o i' n., ,' ir *4r = ffFigure 1-3. Moody friction factors. (Repdnted from Pipe Friction Manual, @ 1954 by Hydraulic Institute. Data from L. F.Moody, Frioion Faaors for Pipe Flow, permission of ASME.)

#( -8u


Pipe Diafleier, in Inch€s -,/

Figure 1-4. Relative roughness of pipe materials and friction factors for complete turbulence. (Courtesy of Crane Company [5].Data from L. F Moody, Friction Factors for Pipe Flow permission of ASME.)

,=

.

Equation 1-6a, which describes the friction factor forturbulent flow in pipe of any roughness, is a simple addition of the Prandtl solution for smooth pipe and the vonKarman solution for rough pipe. The relationship holdsfor the transition between rough and smooth pipe.

To solve Equation 1-6a for the friction factor f an itera-tive analysis is required because the function is nonho-mogeneous and inseparable. There are several empiricalrelations of f expressed as an independent separate func-tion of f G/d, NR.), but with today's micro-computersEquation l-6 can be solved more accurately and expe-diently with iteration.

Dimensional forms of Equation 1-4 are presented inTable 1-1 [1], where the equation is conveniently shownin various units that are used to solve fluid pressure lossproblems.

NON.NEWTONIAN FLUIDS

The Colebrook equation holds for fluids whose flowproperties are dependent on the fluid viscosity. Thesefluids consist of all gases, liquids, and solutions of lowmolecular weieht and are known as Newonian fluids. In

Pipe oiameter, in Fe€t -/)

-rll

Mechanical Design of Process Systems

Table 1-1Dimensional Forms Used With Equation 1-4 [11

= Plessure Ol?p, ne r, ana

Row rate -APr or pHr Lw {*-r^,.i * + g nvz IrNr">2,ooo'r:[2 loglqQ27etD+(?/NR"o)],'

\ rr 'l pt)- pD pD2 e

" /!& * ",. 'l uoQ' cQP dQ pv'

- \ D ' -L\J D4 pD D2 .Units and constants

Conventlonal units Metric units

-AP(H'w(Q)

D

psllb/hln.in.ft

lb/ft3cpftlspsi

2.799x10-7t2

6.3160.05093

9,266

psl(gprn)

tn.tn.ft

lb/ft3cpftlspsr

1.801x 10-5t2

50.660.40859,266

(ft)tb/hln.ln.ft

lb/ft3cpftlsft

4.031 x l0-512

6.3160.0509364 1aY ^

(f9

ln.in.ft

lb/ft3cpft/sft

2.593 x 10-312

50.660.4085

64.35 x p

inHrO[60"F](acfm)

In.in.ft

tb/f13cp

ft/minin. H2O0.02792

t2379.0183.3

1.204 x 106

barkg/smmrnmm

kglm3mPa-s(cp)

m/sbar

8.106 x 1061,000

1.273 x 1061.273 x106

2xttr

bar(L/s)mmmmm

kg/m3mPa-s(cp)

m/sbar

8.1061,0001.2731.2732xlo5

(m)(m3/s)

mmm

kg/m3Pa-sm/sm

0.08265I

1.2731.273

19.61xp

Pakg/smm

eI

p

HVabcde

mkg/m3Pa-sm/sPa

0.8106I

r.2731.273

2a,b,c,d,e

Df

HIK

a== pipe diameter= \lbisbach friction factor= frictional head loss- number of velocity heads

piF lengthfrictional pressure dropvolumetric flowrateReynolds numbervelocity

= velocity head

= pipe roughness: fluid viscosity= fluid density

HV

Newtonian fluids the viscosity alone defines the rheolos_ical behavior.

Non-Newtonian fluids are those in which the viscosityalone does not define their rheological behavior. Sucirfluids are solutions composed of solid particles that ex_pand. Clay and very dense slurries are examples of non_)iewronian fluids. The flow properties of suih fluids area function of the particle characteristics, e.g., size andflexibility and thermal expansion.

Purely viscous non-Newtonian fluids are classifiedinto dree categories: time-dependent and time-indepen_dent and viscoelastic. A time-dependent fluid displaysslo*' changes in rheological properties, such as thixbtr-o_pic fluids that exhibit reversible structural chanses. Sev-eral ty,pes ofcrude oil fit inro this category. Anoiher rypeof tinre{ependent non-Newtonian fiuid is rheooecticfluids- Under constant sustained shear, these fluidi. rateof srrucrural deformation exceeds the rate of structuraldecav. One such category of fluids is polvester.Rheqectic fluids are less common than thixotrooic flu-ids.

_ Time-independent fluids that are purely viscous are

classified as. pseudoplastic, dilatant, Bingham, and yield-pseudoplastic fluids. ln pseudoplastic fluids an intinites-imal shear stress will initiate motion and the ratio ofshear stress with velocity decreases with increasins ve_locity gradient. This type of fluid is encountered in iolu_tions or- suspensions of fine particles that form looselybounded aggregates that can break down or reform witLan increase or decrease in shear rate. Such solutions areaqueous dispersions of polyvinyl acetate and of an acrv_clic copolymer: aqueous solutions of sodium carboxy_methyl cellulose, and of ammonium polymethacrylatl;and an aqueous suspension of limestone.

In dilatant fluids an infinitesimal shear stress will startmotion and the ratio of shear stress to velocity increasesas the velocity is increased. A dilatant fluid ij character-ized by an increase in volume of a fixed amount of dis-persion, such as wet sand, when subiected to a deforma_tion that alters the interparticli distances oI itsconstituents from their minimum-size confisuration.Such fluids are titanium dioxide particles in waier or su-

':bl&,,

crose solution. Dilatant fluids are much rarer thanpseudoplastic fluids.

ln Bingham fluids a finite shearing stress is required toinitiate motion and there is a linear relationship betweenthe shearing stress-after motion impends-and the ve-locity gradient. Such fluids include thickened hydrocar-bon greases, certain asphalts, water suspensions of clay,fly ash, finely divided minerals, quartz, sewage sludge,and point systems.

Yield-pseudoplastic fluids are similar to Bingham flu-ids, but the relationship between the excess shearingstress after motion impends and velocity gradient is non-linear. Fluids in this category are defined by their rheo-grams, where relationships between the shear stress andrate of shear exhibit a geometric convexity to the shearstress axis. Such fluids are many clay-water and similarsuspensions and aqueous solutions of carboxypoly-methylene (carbopol).

Viscoelastic fluids make up the last category of non-Newtonian fluids. The term "viscoelastic fluid" is ap-plied to the most general of fluids-those that exhibit thecharacteristic of partial elastic recovery of the fluidstructure. Whenever a viscoelastic fluid is subiected to arapid change in deformation, elastic recoil oi stress re-laxation occurs. Many solutions exhibit viscoelasticproperties under appropriate conditions-molten poly-mers, which are highly elastic; and solutions of long-charged molecules, such as polyethylene oxide and poly-acrylamides. Processes such as coagulation, oil-wellfracturing, and high-capacity pipelines rely on polymericadditives to cause pressure drops. Viscoelastic fluids ex-hibit the "Weissenberg effect," which is caused by nor-mal stresses and produces unusual phenomena, such asthe tendency of the fluid to climb up a shaft rotating inthe fluid.

For any time-independent non-Newtonian fluid, Met-zer and Reed [2] have developed the following general-ized Reynolds number fraction:

_ D' U2-np(1-7)

"l

: pipe ID, ft: average bulk velocity, ftlsec: density, lb,/fC: generalized viscosity coefficient, lb./ft

sec : gc c 8n-r (see Table 1-1): experimentally determined flow constant, //g"for a Newtonian fluid

n : empirical constant that is a function ofnon-Newtonian behavior (flow behaviorindex), 1.0 for Newtonian fluids

For n : 1.0 and C : p/g", Equation 1-7 reduces toNp" = Du p/p for Newtonian fluids. For 2,100 < NR"


( 100,000 the following empirical relations can be usedfor determinins the friction factor:

(Ni") b"

where bn : 0.0019498 (n)-45"n = (7.8958 x l0-7) (a") 182.1321

Typical values for "y and n are given in Table 1-2 [3].Values for "y and n not available in literature must be de-termined by viscosimeter measurements.

Figure 1-5 shows the rheological classification of non-Newtonian mixtures that behave as single-phase flow.The reader is urged to refer to Govier [4] for further in-formation on non-Newtonian fluid or other complexmixtures. Usually, the mechanical design of process sys-tems does not involve non-Newtonian fluids, but knowl-edge of them and their peculiarities is a must if the needanses.

SINGLE PHASEMULTI.PHASE

TRUE HOMOGENEOUS

=

9

I

PLASTIC C OILAIAI.II

Figure 1-5. Rheological classification of complex mixturesthat behave as single phase fluids [4].

N*"

where DUp^l

Mechanical Design of Proces: Slstems

Tabte 1-2Rheological Constants tor Some Typical Non-Newtonian Fluids* 13I

Rheological Constants Rheological Constanis

23.3% Illinois yellow clay in water

0.67 % carboxy -methyf cellulose(CMC) in water

1.5% CMC in water3.0% CMC in water33% \me water10% napalm in kerosene4% paper pulp in water54.3% cement rock in water

of Fluid18.6% solids, Mississippi clay in

water

14.3 7o clay in water2l .2% clay ln \nater25.0% clay in water31.9% clay in water36.8% clay in water40.4% clay in water23% Iime in water

0.022 0.105

0.350 0.03440.335 0.08550.185 0.2M0.251 0.4140.1'16 1.070.132 2.300. 178 1.04

ol Fluid n0.229

o.7160.5540.5660.1710.5200.5750.153

n0.863

0.1210.9202.800.9831. 186.130.331

* Reproduced by permission: A. B. Metzner and J. C. Reed, AICHE Jownal, l,434 (1955\.

VELOCITY HEADS

Returning to Equation 1-4, let's focus on the termEKi. This term represents the excess velocity heads lostin fluid motion due to fluid turbulence caused by localturbulence at the pipe wall and change in flow direction.The latter is the greatest contributor to the DKi term.When a fluid strikes a surface and chanses flow direc-tion. it loses momentum and. therelore. Jnergy. Consid-ering the 90' elbow in Figure l-6, we see that the fluidchanges direction from the x to the y direction and im-parts reactions Fx and Fy, each a function of the pressureand velocity of the fluid. End conditions of the elbow de-termine some of the velocity head loss, that is, where thefitting is a "smooth elbow" or a "screwed elbow." Asmooth elbow is one that is either flansed or welded tothe pipe such that a smooth internal srirface is encoun-

Figure 1-6. Reactionsflow.

tered by the flow. In a screwed elbow there are abruptchanges in the wall causing local turbulence and henieincreased velocity head loss.

Analytical determination of velocity heads can only beaccomplished in a few simplified cases. The values forvelocity heads must be determined and verified empiri-cally. Comprehensive listings of such velocity head (K)values are given in Figures 1-7 t5l, 1-8 [5], 1-9 [6], and1-10. Using these values in Equation 1-4, you can ana-lyze most cases of friction pressure drop for pipe under24 inches in diameter. For pipe with diameter greaterthan 24 inches, an additional analysis must be made insolving for the velocity head term. This method, pre-sented by Hooper [7] is called the "two-K method."

TWO.K METHOD

As explained previously, the value of K does not de-pend on the roughness of the fitting or the fitting size,but rather on the Reynolds number and the geometry ofthe fitting. The published data for single K values applyto fully-developed turbulent flow and K is independentof N*. when N^. is well into the turbulent zone. As Nq.approaches 1,000, the value of K increases. When Na"< 1,000, the value ofK becomes inversely proportionalto NR". In large diameter pipe ( > 24 in.) the value of NRemust be carefully considered if values of 1,000 or lessare encountered. The two-K method accounts for this de-pendency in the following equatron:

K : K1/Np" + K- (1 + lid)

where K1 : K for the fitting of NR" : IK- : K for a large fitting of NR" : oo

d : internal diameter of attached pipe, in.

(1-8)

kxt continued page 22

on an elbow induced by a change of

:i. a.t'- ;;t: :::a*a;=-:;i{ilif/r td

nt-*":m


Represenlolive Resisfonce Coeflicients (K) for Volves ond Fittings

PIPE FRICTION DATA FOR CTEAN CO'\AMERCIAL STEET PIPE

WITH FIOW IN ZONE OF COMPIETE TURBUTENCE

Nominol Size t/^n 3/q" I Y4" 1Y2" 2V2.3" 8.10" 12-16" t8-24"FriclionFocfor ( fr.)

.o27 .o25 .o23 .o22 .021 .0r9 .018 .o17 .01 6 .01 5 .014 .01 3 .ot2

FORMULAS FOR CALCULATING "K" FACTORS'FOR VALVES AND FITTINGS WITH REDUCED PORT

Kz=

K.Kr=

O - Formula z , Formula 4

lA,, K,- o r !sin i(r - trt + (r - E )2

tJ'|

k.Kr= j.n - 0(Formulaz 'Formula+) uhen d = r8o"

K,+O [o : (, - g') + (t - 9')']

o.s(in9(, - o'rt/\2/^1rz= Ba

: tr

Kz= "iG-p)\f"#a4

/.\2 6(sin+(I - P)'?

K, _ ,__]____184 =

|, a2\2

Krlf

K,E

KrR4

Kr=SO *Formr.rla I + Formula l

^ d,r

lJ \2 -az 12\ 1I" _\d,J _au

Subscript I dennes dimensionsand coefncients with reference tothe smaller diameter.Subscript 2 refers to the larger

Kr +sin3[o.a 0 - P\ +2.6 (t - 02)2)

K"=

SUDDEN AND GRADUAI. CONTRACTION

0 < +5".........K, - Formula r

45" <e< r8oo...Kr = Formula z

SUDDEN AND GRADUAT ENI.ARGEMENT

0. 4to. . .. . . . ..K2 - Formula 3

45o<0< r8o-. . .Kr = Formula 4

(Courtesy Crane Company [5].)

E;l

Figure 1-7A. Selected Crane Company velocity head values.


GATE VAIVESWedge Disc, Double Disc, or Plug Type

lf . ts = r,0 = o. . . . . . . . . . . . . K' : 8 -/rP< r and 0 < 45o ........K2: Formula 5

B< r and 45"<0< r8oo...Kz - Formula 6

GTOBE AND ANGTE VAIVES

All globe ancl angle valves,t hcthcr rcducccl scat or throttled,

Ii: 13 < r. . .l(2: Formula 7

SWING CHECK VATVES

Et#-

JLI I.-+ -ffa-r-

TL€ fNr,-

r-L-r FI-/f FK: rcof7

Minimum pipe velocity(fps) for full disc lift

=)5vv

K:sof,Minimum pipe velocity

(fps) for full disc lift-a8!V

If: B: r...Kr=6oo/z9. r.. .K, = Irormula 7

Minimum pipc r tlocitr itp.; ior full .lisc Iifr: F p2 \,/ vIf: B:r...Kt=l+ofr

lf A-t.. Kr=sjfr

If : A=r...lit=riofr If: B:r...Kr:S5fr

LIFT CHECK VAIVES

lf: 9 = r...K,: ;s frB < r. . . K, = Irormula 7

Minimum pipe velocicy (fps) for fr-rll disc lift: t4o B|V V

TIITING DISC CHECK VALVES

rE

tr

l'-F

i

Sizes zto 8'...K:Sizes ro to t+'...K:Sizes 16 to 18". . .K =

Minimum pipe velocity(fps) for full clisc lift -

l--4-lVI z++ll l-

LFtr-IF

Figure 1-78. Selected Crane Company velocity head values. (Courtesy Crane Company [5].)

li:il- -:::i::

8t'-n*"

+.[

lf d < 45. ora ( 22.50

Kr = 2.6(1 - B'?)2 sin e

ll 45" < 0 < 1800 or 22.50

K'=(1 -0"f. td,- d,l 0d=arcslnl- l=-

\21 /2Figure 1-8. Calculated Crane K-valuesdiffusers are tabulated in Table l-6.

40

830:E.4 20

llolPiping Fluid Mechanics 11

t.o 2.o 3.O 4.O 5.o 6.0

VBLoCITY-FPllxl03

SHADED AREA UNDER CURVE IS CRITERIA FOR UULTIPLYING

CORRECTION FACTOR, A,BY lHE VALUE OF It FOR EACH

FITTINC AND PIPE COI4PONENT .

CORRECTION FACTOR TABLB

Figure 1-9. Correction factor.

<o(90o

for concentric conical

1.0

T./1Dl ( ,l_l_ \,-,/

TWO.MITERED ELBOW

0Figure 1-10A. Velocity heads for change of flow [6].

'*


n =number ot mitersor segments

1

mitered ell

1Eo 644+I

smooth ell

.5 t.o 1.5R/o

Figure 1-108. Velocity heads for change of flow [6].

h*":ns

'4

2<.- <*3-V-

+s"(o<go.

an = az/og

Figure l-10C. Velocity heads for change of flow [6].



on= 9/ogFigure 1-10D. Velocity heads for change of flow [6].


\sri\7tAl| | I Ai= A2: A3

aR = or /ag

Figure 1-10E. Velocity heads for change of flow [6].


on = or/ag

Figure 1-10F. Velocity heads for change of flow [6].

k-...n*

'*

OR: O1/o3

Figure 1-10G. Velocity heads for change of flow [6].


18 Mechanical Desisn of Process Svstems

o*= 02/o3

2-----> -_)> 3

Figure 1-10H. Velocity heads for change of flow [6].

k*--=*

. ---)t -->2--v'.

OR= 01/O3

Figure 1-101. Velocity heads for change of flow l6l.


' 'drF'


2--+ ----)3-llt-

Kzg

on= oz/o,

Figure 1-10J. Velocity heads for change of flow [6].

it - 'I[||,,'

or: o.' /o3

Figure 1-10K. Velocity heads for change of flow [6].



Small pipe fittings have more surface roughness andabrupt changes in cross sections, making Kl insignificantat values of Nr" ) 10,000. For this reason, the newCrane method is recommended for pipe diameters 24 in.and less. Comparison of the methods for elbows is de-picted in Figure 1-11. Table 1-3 lists Kr and K- values.

The two-K method is preferred over the equivalentlength method because in large, multi-alloy sysiems theequivalent length method could predict losses 300% toohigh, resulting in oversized pumps and equipment. Withlaminar flow, the equivalent length method predicts headlosses too low. Also, in the equivalent length method,every equivalent length has a specific friction factor as-sociated with it, because the equivalent leneths are de-rived from the expression L. = K D/t.

The Hydraulic Institute's widely used K-factors aregood for l-in. to 8-in. pipe, but result in errors in largerpiping.

The disadvantage of the two-K method is it is limitedto the number of values of K1 and K- available, shownin Thble 1-3. For other fittings, approximations must bemade from data in Table 1-3.

PIPE FLOW GEOMETRIES

In using Equation 1-4 the geometry of the flow areamust be considered if the area is noncircular. In calculat-ing the Reynolds number and the diameter for a noncir-cular cross section, the hydraulic radius is applied:

R11 : hydraulic radius : cross-sectional flow area

wetted perimeter

This relationship applies to noncircular cross sectionsflowing full or partially full, oval, rectangulat etc., butnot to extremely narrow shapes, such as annular or elon-gated openings, where the width is small relative to thelength. In such cases the value of Rs is approximatelyone-half the width of the passage.

The value of 4RH is substituted for d in Equation 1-4.

**" : r,1*,

Thble l-4 provides hydraulic radii for various crosssections.

Table |-3Constants for the Two-K Method I7l

Filting Type Kl K-Standard (R/D : l),screwedStandard (R/D : 1),flanged/weldedLong-radius(R/D : 1.5), alltypqs 800 0.20

800

800

0.40

0.25

90" 1-Weld(90") I,000 1.152-Weld(45') 800 0.353-Weld(30') 800 0.304-Weld(22t/2") 800 o.275-Weld(18") 800 0.25

0.20

0.150.250. 15

Miteredelbows(R/D : l 5)

ElbowsStandard (R/D : l),all typesLong-radius(R/D : 1.5), all typesMitered, 1 weld, 45"Mitered, 2 weld, 22rlz"

500

500500500

180'

Standard (R/D : 1),screwedStandard (R/D : l),flanged/weldedLong radius(R/D : 1.s), alltypes

1,000 0.60

1,000 0.3s

1,000 0.30Standard, screwed

Used Long-radius,screwedas Standard, flansed or

etbow welded

500 0.70800 0.40

800 0.80

TeesSlub-in-type branch 1,000 1.00

0.100.500.00

Run- Screwedthrough Flanged or weldedtee Stub-in-type branch

200150100

Gate,

ball,

plug

0.10

0.15

0.25

Full line size,p: 1.0Reduced trim,p:0.eReduced trim,p=0.8

300

500

1.000Globe, standardGlobe, angle or Y-typeDiaphragm, dam typeButterfly

1,500 4.001,000 2.001,000 2.00

800 0.25Lift

Check SwingTilting-disk

2,000 10.001,500 1.501,000 0.50

Note: Use R/D = 1.5 values for R/D : 5 pipe bends, 45' ro 180'.Use appropriate tee values for flow through crosses.

< - -8l

Screwed 9d ell

Regular

DLongradiusr\

}J

Flanged 90oell

Longradius

t\\_t

Square-odged inlet [24]

Inward ptojocting pipe

20

l--*H K=o.b

/(=1.0

0.00' 0.50.02 0.280.04

o.240.06

0.r 50_10

0.090.15&up I o.o+

For (, see table 'Sharp{dged

Figure 1-11. Velocity head values for common piping components [1]. (Reprintedneering, @ 1978, by McGraw-Hill, Inc., New York.)

'd

Screwed tee

Lineflow

K10.8

rfi-I|lHK

Branchflow

1

D0.3

IT

FlangEd tee

o.4Branchflow 1

0.8

Contraction

Enlargem€nt

(:; /,*-1---lJK

p!"..ffio'?FidG,i\

o.4p

ffi'=.err-r,rf(fia-r]

p= lmall diamete./larse dismererK based on velocity in slnatter pipe

based on main pipe for orifice. l40j


Globe valve

Screwod

Flanged

10

60.3

10

4

Gate valve

Screwed

K o.2

0.10.3 0.6 I

D

Regular screwed 45oell

0.6

X 0.40.3

0.3 0.5

Lonq radius flanged 4soell

0.3

K

0.1

Hydraulic inltitote [24]

.........,...... crane It2l[,liller. based on water at 6 ft^ 132]

by special permission from Chemical Engi-

0


Swing chock valve

Screwed

Ev

n tLtLtt- _

flilj-ll-lH;ETvoictl dimh'ioo3 9

Nomimr LDr LEtize {in.l {in.l u tL

1/2 0.520 0.250314 0.750 0.275

r t.000 0.187r1n 1.500 0.200

2 2.O@ 0,1613 2.962 0.1434 4.000 0.081

4

.T

,I

0.o.12

0.08

0.04t04 106

Feynolds number, /VF"0.6 '1 24

K

FlangedHead loss in conical diffuse6

tr9l4 6 10

1.2

1.0

5 o.e

Angle valve

Scre$red

Head loss in circular mite6

147l

6l--4l-Ir,t-

,I0.3

llIl1

P, des.

0.6

63Flanged 1

Screw€d return bend

Pf0g cock valve lr9l Buttertly valve ll9l

Flanged return bend

F nninqfricrlon lactor for fllribl. m6.tl hc. Il4l

K= 1.2 (1 -cos0

0

to'2o'lo"60"

0.050.29r.5611.3

206.0

0

5"to'20"+0"6o'

o.24o.521.5410.8

I18.0

d is angle between pipe axit and llapper plate

Figure l-11. Continued.

d k anste ber@en pipe axis and plus cock axis

Table 1-4\ralues ot hydlaulic radius (RH) for various Cross

SectionsCross Section


where Y 1214a12 : 0 for incompressible flow sincea?- @, the term ar2 multiplied by the fluid density pl isthe bulk compressibility modulus of the fluid and givesthe pressure change for the fractional change in density.Values of the bulk compressibility modulus for varioussubstances are given in Appendix A. The term a1 is thevelocity of sound waves propagated in a compressiblemedium.

As Equation 1-9 shows, the velocity of the fluid is

compared to the fluid velocity of sound in the term V1'l4a12. If this ratio is small, compressibility effects can beignored because the error is the difference between thisterm and unity. This analysis is valid only for barotropicfluids, which are typical of most industrial applicationsinvolving flow of gas through a nozzle and the flow ofwater in conduits or over obstacles.

Compressibility effects of a fluid are small when thefluid velocity is small, compared to the fluid sonic ve-locity. If V1/al is equal to 0.3, the error in the velocity isless than I % when using the incompressible assumption.For ambient air, this limitation corresponds to a velocityof 300 ftlsec without causing significant error.

The phenomenon ofnonsteady flow is somewhat morecomplex than that of steady flow. The acceleration or de-celeration of liquid particles immersed in a two-phasesolution is one such example. The time required for thenonsteady phenomenon to occur is compared to the timereouired for a sound wave to traverse the flow in whichsubstantial differences in velocity occur. If the time dif-ferences are small, then the incompressible Bernoulliequation (Equation 1-2) may be applied.

PIPING FLUID MECHANICS PBOBLEMFORMULATION

To solve piping fluid problems a firm understanding ofbasic equations and units is essential. The units should becarefully defined and used throughout the calculations.Thble 1-5 presents reasonable velocities for various ser-vices used in mechanical systems. These velocities areonly guides intended to give the reader foresight for trialvalues and are only for mechanical systems; for suchvalues of chemical processes the reader should consultchemical engineering sources.

Viscosity

Widely misunderstood and often improperly applied,viscosity is perhaps the most recondite of all the proper-ties associated with fluid mechanics. However, a clearconception of this physical property is critical to the suc-

RH

L-

COMPRESSIBLE FLOW

The preceding analysis assumes steady and incom-:::ssible fluid flow. This is a reasonable assumption:jce most liquids are steady flow, but frequently the as-.-nption is valid for gases. Because some liquids and all:-ies are compressible, a criterion is needed to deter--,ne what percent of error is incurred assuming constant:::1Slty.

-\n estimate of the error can easily be made for a baro-::pic fluid-a fluid whose density is a function of pres-.-::. Sabersky and Acusta [8] have shown that for a

: .:id stream of velocity Vr, static pressure R and density: mat

IN/E+=0.153 Di

(1-9)


Table 1-5Reasonable Velocities

Liquids Feet per Second

Service water mainsGeneral service water pipingBoiler feed water pipingHeat medium oilsLubricating oils

GasesLow-pressure steam heating

and process pipingLow-pressure steam mainsHigh-pressure steam mainsSteam engine and pump pipingSteam turbine pipingNatural gasAir, 0 to 30 psigForced draft ductsInduced-draft fluesChimneys and stacksVentilating ducts

cessful design of hydraulic systems and rotating equip-ment that transport fluids (see Chapter 6).

Viscosity is the property of a fluid to resist flow. Con-sider how much more freely and easily gasoline pours

from a container than does black strap molasses. In fluidmechanics terminology, the heavier, bulkier nature ofmolasses is caused by the fluid's high shear stresses.

These high shear stresses make the molasses very resis-tant to flow.

The fundamental measurement of flow resistance is

the dynamic or absolute viscosity. In the cgs (centimeter-gram-second) system of units the basic unit of viscosityis the poise, which is equal to one hundred centipoises,(For a detailed explanation ofhow absolute viscosities offluids are determined, the reader should consult a basictext on fluid mechanics.) The centipoise (cp) is now thestandard unit of absolute viscosity, but because otherunits are still used, as illustrated in some of the examplesscattered throughout this book, methods for convertingto and from centipoises are provided. With the centi-poise, one must be careful in using the English system ofunits when converting to the kinematic viscosity. Illus-trating this conversion we have the following:

p = absolute viscosity, centipoise

Thus, ifone has a fluid such as a fuel oil (see Example6-1), which for a given temperature has an absolute vis-cosity of 139.53 centipoise, we calculate the kinematicviscosity, z, in the English system of units as follows:

p = 139.53 cp at 90'Fw : 54.725lb^/tr

/o.oooozol\ lb,-rT, :2. iz, ij!-.(rry )r)cP |1 r ., /= lbrsecr

(s4.72s) EfP

ft2r : 0.0017 :sec

ft20.0017 -::-

2to54 to l06 to 132to62to6

15 to 7070 to 165

165 to 400100 to 150150 to 3301007040 to 6030 to 50357

centistoke : 159.261 centistokes

0.00001076e ilsec

Since the kinematic viscosity is a function of the fluiddensity, the above value is only valid at the specifiedtemperature of 90'F. In the metric system the kinematicviscosity can be obtained by dividing the absolute viscos-ity by the specific gravity. This is only for the metic sys-

tem of units.It is a common mistake in using the Englishsystem of units to compute the kinematic viscosity by di-viding the absolute viscosity by the specific gravity ofthe fluid. Equipment manufacturers often use other unitsof viscosity. One ofthe most widely used units is the Sec-

onds Saybolt Universal (SSU). This unit represents thenumber of seconds required for sixty cubic centimetersof liquid at a constant temperature to flow through a cali-brated orifice. For liquids of high viscosity a larger ori-fice is used and the unit applied is termed Seconds Say-bolt Furol (SSF). It is customary to specify these units ofviscosity at standard temperatures.

The following are formulas for converting SSU's and

SSF'S to centistokes. Below the value of 32 the SSU is

undefined and below the value of 25 the SSF is unde-fined. Throughout this book, the centipoise and the cen-tistoke are the standard units of absolute and kinematicviscosity, respectively. Where the need arises, the centis-toke is converted to SSU.

SSU to centistokes

= i.0 lb-sec/ft2 = 478.7 poise = 4.787 centipoisez : kinematic viscosity, centistokQ2

: & 8", for the English system of units

fr-lhWnefe gc = JZ.|t aa---------a

lDrSeC'w = mass density of the fluid, lb./ft3

t : Seconds Saybolt Universal,/ : centistokes

For32(t(99,

For t ) 100,

6.2261 - P7 :t

o.zzu - !1 :t

*( ----L'


with COz in steam. The properties of the mixture are as

follows:SSF to centistokes

: : Seconds Saybolt Furol

Rrr 25 ( t ( 39,

5rt>40,

p = 0.01322 cpp = 0.085 lb/ft3e : 0.0015 in.

lR42.24r - -:- : v

t60

2.16t - -:-- = vt

P : 10 PSig

EXAMPLE l.l: FRIGTION PRESSURE DROPFOR A HYDROCARBON GAS.STEAM

MIXTURE IN A PIPE

An amine still reboiler boils off a hydrocarbon gas-

{eam mixture that flows in a 3M ss line connecting the:eboiler with an amine still tower. It is desired to deter-rine the maximum pressure drop in the line as shown inFieure 1-12. The sas is a small tract of amine immersed

The velocity head approach is used in determining thefriction pressure drop. The line shown in Figure 1-12 iscoming off two nozzles on top of the reboiler and merg-ing at a tee before entering the amine still tower. The di-mensions shown are identical with both sides coming offthe reboiler nozzles (exemplified by the word "TYR"meaning for both sides). To solve this problem, we mustapply Equation l-4. To use this equation, we divide theconnecting pipe into three components (see Figures1-13-l-15)-an 18-in. f portion with W = 25,291 lblhr a24-in. d portion with W : 25,291lblhr; and a24-in. d with W : 50,582 lb/hr, Equation 1-4 is applied toeach portion and the pressure drop for each is added to

Figure 1-12. Amine still reboiler hydraulics.

d


FLUID ANALYSIS FOR SINGLE PHAS€ FLOW - GAS OR LIOUID

coNFtq!84[!9X. LINE NUMBER IE"6 PoRTt oN - tr'6 scttEOULE tcs PAGE _OF _

L,, (z'.- a"\ r{z'-j"\ + (r'-et)

y- rerzr\H (=#l)(o.,s$$ f"*",i,"'(##)

, 1'- z" = 1, 1s.7 p7

= +i,+81#

SERVICE EiE;G'dr;BrE-vEEcrw REYNoLDS No= ov tFoR sERvtcE = Ki= No oF ver-ocrri ems; x =.ov/,

DEpENDENT ptpE FRtcrtoN = rt6K VALUES ILD = 5OK, D=rNStDE D|AMFTEFtfrll

FOR COMPONENTS:

)*,

t6" t R 90' ELL =

o,7 80o,079o, oza

PIPE ENTRANCE .?1.x ta" u FFUSER (cs/{rRrc). f=

lL= o,ot32? cpp = o,O85 Lb/cu tly= 1, I L'l I'€ = O,OOI5 in

D: 17' 50 in.[p= o, O?9 psiv = *9,18'1 1y ".NpE= 690,49/| = o,otl

tt/nr

) *=,..,u

Q=W= 2sz9t

Figure 1-13. Fluid analysis for single phase flow-gas or liquid.


FLUID ANALYSIS FOR SINGLE PHASE FLOW - GAS OR

coNFtquRAI!9!L LINE NUMBER

L,.= (i'-o') + (1'-t tv/1i']1 + (rg'' rr"\

..r- raszgr\E(#c'"J -' - (" ""n #n"cr.z\ i,,'' Gfi],,)

= 2+'- o 'tli = 2+076 ++

21,++2 *

SERVICEFEASOMBLE VELOCITY

REYNOLDS NO= DVM'^.,2

FOR SERVICE =KI=NO OF VELOCITY HEADSiK= .ov72DEPENDENT prpe rRrctroH = fl

K vALUEs [LD= soK,D=tNStDE DIAMFTERlft]l

FOR COMPONENTS:

)*,rwo 9d LR ELt s - K- o,1?o

lL= o,ot 322 cp

P = o.o?s Lb/cu lly = 21.C78 t'6 = OOOI5 1n.

/=-p= - 23,5 ;n.

[p= o,O29 1siy = 21a12 191"".

o.oll€t+,t17.r25

Q=

-gpm

w= ZS29l tyn

J

Figure'l-14. Fluid analysis for single phase flow-gas or liquid.


FLUID ANALYSIS FOR SINGLE PHASE FLOW - cAS OH LtOUtD

CONFI6URAT1ON LINE NUMBER

lt Z''tt" = 2,9l.1 k

Sh= "4q <2."

= r7,s84 *rs'srz'tiF (=za!=aifo,oa,iS h*l\ru.(ffi

K vALUEs ILD= 50K,D=tNstDE DTAMFTER tftllFOF COMPONENTS:

)*,coMatNtNG FLOV'IPIPE ExtT =

TEE,

lL= O.ol32? coP= O.O?S Lb/cu trt= ?-91'7 n'e= o,OOl5 in

04<!r- --:::_tn.Ap= oQ6, psiV = 5'1, 884 .11rl"""NRE= l.o2A,35+. espt= otol3Q=

--gpm

111r= So Sg2 .L/rrr

t2

)*= 2,2

Figure 1-15. Fluid analysis for single phase flow-gas or liquid.

p : (6.72 x 10-4)(0.01322) : (8.384 x l0 6)-]!LIt-sec

the other portions to give the total frictional pressuredrop for the line (For velocity head value of concentricconical diffusers, the reader is referred to Thble 1-6).The calculations are as follows.

Lr : 6.167 ft


Table 1-6K-Values tor Concentric Conical Diffusers

d2(in.) d1(in.) L(in.) d(deg) Kr

tlc 0.546 0.302ls 0.546 0.423

2.469 2.067 3.500 3.292 0.013

3.068 0.9571V+ 3.068 1.278 3.500 14.816 0.454

Sch80tl2 x

=DVP:p

lr1:50)r, 1+r.+tzr a ro.oasl k\ tz I sec rr- 3lt 0.742 0.423 1.500 6.104 0.126

(8.884 x 10-6) ;lb'n-sec

tlz 0.742 0.546 1.500 3.746 0.036

4s 0.957 0.423 2.000 7.672 0.2251lz 0.957 0.546 2.000 5.898 Q.1213lc 0.957 0.'742 2.000 3.081 0.022

Sch803lc x

: 690,491.450

Sch80

1X

,fl-05 - _2 ron,^ [{g * _?r_]" - [3'7 NR"(f)"'l

Let f : 0.014

'0.014)-05 : 8.452 : -2 logr0 [(2.317 x 10 5)

+ (3.072 x 10-5)l : -8.537f - 0.014

lPf =

-ro, : I

* , ,rol"--'l

( 1-6a)

fr lh

S€C'lD6

( o. o8r H ei . 44zf !(, * *-)fr lh

S€C'lD1

1.500 0.957 2.5W 6.235 0.099lUq 1.500 1.278 2.500 2.545 0.009

lc 2.067 0.742 3.000 12.758 0.4362.067 0.957 3.000 10.661 0.297

1.278 0.546 2.000 10.545 0.318

r.278 0.742 2.000 7.701 0.1531.278 0.957 2.000 4.603 0.040

lt/z 2.469 1.500 3.500 7.957 0.143

3lq 1.500 0.742 2.500 8.720 0.225

ll/c 2.067 1.278 3.000 7.556 0.131

llz 2.067 1.500 3.000 5.423 0.055

1.500 0.546 2.500 10.999 0.373

2.469 0.957 3.500 12.474 0.406Itlq 2.469 1.278 3.500 9.796 0.237

Sch80

I r/4 xtlz

Sch80

1tl2 x

(?. r'.l# (l -4)

Sch40

2x(0.014X6. 167X12)

(17.50)

lP1, : 9-929 O.t

L. : 24.078 tt

Similarly,

\R" : 514,177.125 and f = 0.014

_\p. - [(0.014X24.07s) + o.72olt (23.s0) I

,o.oss, ft (4e.487), g H*) Sch40

ztlz x

Sch40 1

3 x lyz 3.068 1.500 3.500 12.944 0.3373.068 2.067 3.500 8.221 0.11 I

Sch40 lV+ 3.548 1.278 4.000 16.484 0.559

2tl2 3.068 2.467 3.500 4.9@ 0.028

lUz 3.548 1.500 4.000 14.833 0.4493t/2 x 2 3.548 2.067 4.000 10.668 0.210

zth 3.548 2.469 4.000 7.151 0.0933.548 3.068 4.000 3.440 0.010

c--,-..,&


Table 1-6 continued Table 1-6 continued

4 x 2tl2 2.469 4.026 4.000 11.223 0.197

stdwt8

srdwt8Sch

402

stdwt

Size dr(in.) dl(in.) L(in.) d(deg) Kr

Sch40 ltlz 1.500 4.026 4.000 18.406 0.609

2.067 4.026 4.000 r4.r74 0.345

d2(in.) d10n.) L(in.)

7.981 r7.2s0 15.000 17.997 0.49610 10.020 r7.2s0 15.000 13.946 0.275

18 x 12 11.938 17.250 15.000 10.199 0.12514 13.250 17.250 15.000 7.662 0.058

16 15.250 r7.250 1s.000 3.823 0.008

7.981 19.25Q

10 10.020 19.250

12 11.938 19.250 20.000 10.533 0.18020x 14 13.250 t9.250 20.000 8.627 0.10816 15.250 19.250 20.000 s.739 0.036l8 17.250 19.250 20.000 2.866 0.005

l0 10.020 2t.25012 1 1.938 2r.25014 1,3.250 21,.250 20.000 11.537 0.19422x 16 15.250 21.250 20.000 8.627 0.09218 r7 .250 2r.250 20.000 5.739 0.030?0 19.250 2r.250 20.000 2.866 0.004

10 10.020 23.25012 11 .938 23.25014 13.250 23.250

5x

3.068 4.026 4.000 6.878 0.055

31lz 3.548 4.026 4.000 3.425 0.088

2.067 s.u'l 5.000 17.338 0.537

Ztlz 2.469 5.047 5.000 14.940 0.388

3.068 5.047 5.000 1r.4r4 0.205

3t/z 3.548 5.U7 5.000 8.621 0.100

4.026 5.U7 5.000 5.860 0.035

Sch40 Ztlz

6 x 3tlz 3.548 6.065 5.500 13.228 0.257

2.469 6.065 5.s00 19.08

3.068 6.065 5.500 15.810

4.026 6.065 5.500 10.682 0.151

5.047 6.065 5.500 5.310 0.023

3.068 7.981 6.000 24.168 0.726

3t/z 3.548 7.981 6.000 21.680 0.618

4.026 7.981 6.000 19.243 0.4765.V7 7.98t 6.000 r4.r52 0.229

6.065 7.98t 6.000 9.188 Q.O74

4.A6 10.020 7.000 25.350 0.703

5.M7 10.020 7.000 20.807 0.514

6.065 10.020 7.000 16.409 0.295

7.981 10.020 7.000 0.051

5.047 11.938 8.000 25.511 0.674

16 r5.2s0 23.250 20.000 1t.537 0.169

18 17.250 23.250 20.000 8.627 0.079

20 19.2s0 23.250 20.000 5.739 0.026

12 r1 .938 25.25014 13.250 25 .250

Sch403

8x stdwt

24xSch40

l0xSrdwr

Sch40

12x6.065 11.938 8.000 2r.535

16 15.250 25.250

26 x t8 r'1 .250 25 .250 24 .W0 9 .s94 0 .123

20 19.2sO 2s.2s0 24.000 7.181 0.057

22 2t.250 25.250 24.0W 4.780 0.018

24 23.250 2s.2s0 24.000 2.388 0.003

20 19.2s0 29.250 24.W0 12.025 0.17430x

24 23.2s0 29.250 24.000 7.181 0.04426 25.2s0 29.250 24.000 4.780 0.014

Sch40

14x

7.98t 11.938 8.000 14.319 0.197

10 10.020 11.938 8.000 6.885 0.027

13.250 13.000 t6.042 0.449

7.981 13.250 13.000 1r.692 0.21410 10.020 13.250 13.000 7.136 0.059

12 rr.938 13.250 13.000 2.892 0.005

6.065 15.250 14.000 19.150 0.604

7.98r 15.250 14.000 r5.U7 0.356

16 x 10 10.020 15.250 14.000 10.765 0.157

12 rr.938 15.250 14.000 6.793 0.046

stdWr 14

16

l8stdwt6

14 13.250 15.250 14.000 4.096 0.011 28 27 .2s0 29.250 24.WO 0.002

Table 1-6 continued

Size dr(in.) d1(in,) L(in.) @(deg) K1

26 25.250 33.250 24.000 9.594 0.078

30 29.250 33.250 24.000 4.780 0.011

32 31 .250 33 .250 24 .NO 0.001

24 23.250 35.250 24.W0 14.478 0.207


APt, : 6.961 O.'

Total Friction Pressure Drop for Line : AP

APl = APr, * APr, * AP;, : (0.029 + 0.005 + 0.061)pst

AP1 : 6.695 O"'

EXAMPLE t-2: FRIGTIONAL PRESSUREDROP FOR A HOT OIL SYSTEII OF A

PROCESS TANK

A pressure vessel storage tank contains 6,000 gallonsof filler coating that must be maintained at 370'F to beused in the manufacture of roofing products. To maintainthe coating mixture at the required temperature, externaljacket coils are placed on the outside shell and bottomhead as well as four internal coils inside the tank with anagitator. The tank is depicted in Figure 1-16 and the hot

33

Srd\\i 16

Std\\'t 16

Srd1\l

18

2022

_\ x 24 23.250 33.250 24.Un p.025 0.141

18

20

16x 26 25.250 35.250 24.000 12.025 0.128

30 29.250 35.250 24.000 7.181 0.032

32 3t .250 35 .250 24 .000 4.780 0.010

34 33.250 3s.250 24.000 2.388 0.001

24 23.250 41.250 24.000 22.024 0.4s426 25 .250 41 .250 24.000 19 .47 | 0.33930 29.250 41.250 24.000 14.478 0.161

32 31.250 4r.250 24.000 12.025 0.09834 33.250 41.250 24.000 9.s94 0.053

36 35.250 41.250 24.000 7.181 0.024

:P,. = 0.005 psi

'-. : 2.917 ft

\r. : 1,028,354.250 and f = 0.013

rP.- : I't + 2.201(0.013x2.917)

(23.s0)

(o.oss) H (s4.884F

-i1Fl",ooJr,.r r' ft lb.-\J-.-/ --7-a-

S€C'lD1 Figure 1-16. Process surge tank. (Courtesy of Tranter, Inc.)


oil system in Figures l-17, l-18, and 1-19. It is desiredto determine how much frictional pressure drop will beincurred for the entire tank so that pump sizes may beselected.

The tank is divided into two systems-the hot oil sup-ply system and the hot oil return system. Each systemconnects to the three components-the four internal coilsinside the tank, the outside shell jacket coils, and thejacket coils connected to the bottom head-and each ofthe three components must be analyzed separately.

Bottom llead System

A 2-in. pipe header supplies hot oil to the six inletjacket nozzles and returns hot oil from six outlet iacketnozzles. The supply nozzles are designated by an S andthe return nozzles by an R. We will analyze the supplysystem. The piping system is divided up into "stations,"which are points designating flow change due to separat-ing fluid. Each line following a station must be analyzedseparately because the flow rate decreases after the flowseparates in the tee. We will consider the pressure dropfrom point A to B, since that path involves more stationsand the maximum amount of pressure drop.

Bottom Head Hot Oil Supply

I Hot Oil Entrance from the 2-in. Header and FlowThrough Station 1. (Figure 1-U):

Y z.s't,,

6 gpm ,l/ 30 spm

"1,tI

: 36 gpm: 26.5 in., p = 0.15 cp, e : 0.0018:2.067, p = 58.7 lb/ft3: (6.72 x 10-4)(0.150) : 1.008 x 10-5

lb./ft-sec

: 3.442 ft.lsec

g=11h"x31a"C=11h"x1,D=2"xEa"E=2" x1"F=2'x1112"

For QLop

(36) sd (___u, ){_1.'"' min \7.479 gal/ \60 sec

Figure 1-17. Process surge tank bottom head coils.

Pipiry Fluid Mechanics

___.1

Figure 1-18, hocess surge tank-shell coiis Qooking south).


rl

0l l-{t \t I-\| \l

NIt\l

0l I{t \t I-\| \l

|\l| \lt\tt\_l I

-@

tlI t^J r3-q-q.\

7sm

Figure 1-19. Process surge tank-shell coils (ooking north).

fm&----*

Nn"


L : 4.0 ft

.io\ sar { I rc \ltgiv _ '--' rnin \7.479.9aU \60 sec/ : 2.869 ft/sec

ilrt I(J.JJbtrn.'t-......_l\144 in.'/l

37

_DVp_"IP (t.oo8 x lo ) -tD'

n-sec

:3,452,9\0

?8\o"*" n osD*

l{q!q'!)-z roe,o ffi +I

(1o s

: 0.040

K-factors

: 287 ,7U

-2 logp1

(fl0.5Pipe entrance

Branch flow tee

\- .-

K : 0.78

K = 0.46

+ (-0.78)

0.78 + 1. I 90" LR ellFlow-lhru tee

:K::K: 0.570

0.910

1.480

$rr : 0.186 psi

- Hot Oil Fbw from Station 1 to Station 2:

aP,: (dLL. D")#

l830

_ [to.o+oxz.zoelt rzr * ,.r6al

[ (2.067) I

or,^ _ [ro.o+orr+.orrtzt *,.orol'' I Q.067) I

.-^-. lb _^ ^-^.. fc I I ft, I()d. /) = {r.6oe]' , l.filn' sec' \r+4 ln.'/

lrql rr ft lb'-'--'-'sec2 lbr

APr, = 0.126 n.t

Z Hot Oil Fl.ow from Station 2 to Station 3:

Nn" :

..^. sal I 1ftr lltlxl-l-ll'--' min V.a79 ga| \I min60 sec : 2.837 ftlsec

12.036)in., I t tt'.1\144 in.J18gpm

., -Q,--*-Q,-

(ril!.lo o.ru r' rss.zr P\r2 l sec fi'th

(1.008 x 10 4) -j:'n-sec

22t,657

2.51 Ia,qsz,srol(D\

12 67lr,,z.sost a,sa.;, I

DVp _ \ 12 / sec

(1.008 x 10 4) --l!'n-sec

[/o.oora\ Il\-0. rzz i z.st I

[ 3r - e8lJwtfi{-'

rh rr2 / rfi2 \/5R?\::11 M)\2:L | -"

I'-' ,ftr.- -. ss62 \144 in.2/

,rrta j:b:l'-sec'lDf

:0.040

K-factors

I


1

(flo s

[/o oore\ I

l\ a r:+ 1 z.st I-z 'oc'o [- * rnrsvltf

1

1ft,,^, ca1 / I ft3 \/1 min\t"'

'oi" \2.+zs s"il \60 '*/ = 4.457 ftlsec

:0.M2

K-values

o 2-in. x ltlz-in. LR ell

/ t e'z \(0.864)in.,tfr1n}/

t#)o,0.0", A or.a *

(1.008 x 1o-) .lb'ft-sec

226,889.525

lb''4 Il\ o0s7 / 2.51 |-r rocro

[ 3i * tz26seo]flri

Nn. =

. - o "' [(9

t' - e'i],u, : 1u9'

1

(o0 5

:0.051

Ir rr o\2= l- -^"1 = 0.607

\2.M7 |

= P4 : 0.368

* _ 0.8 sin [(5.423)(l - 0.607)] _ o.o'(0.368)

. Flow-thru tee

1)Q*:18-:0.667-K:0.53

^R - - - r.v

Dr: o.oar + 0.53 : 0.611

or,, : [,oq?!u1'uo]t't, * 0.u,,]

.-^ -. lb .^ ^^-.. n, / r ft, \(J6. /) ftr

{r.6J /)' .*r- h44 i" j-l

z1zz.z1 !!y

Sec' lDf

APr, : 0.063 Ott

Z Hot Al Flow from Station 3 to Station 4:

6 gpm,l

t!' h Ir1 l2x't I

---A - | +6 -oFm12 spm I

@

K-values

. lr/2-in. x f-in. reducer

,. _ 0.8 sin [c(l - p,)] _--- et- - 0.8 sin [(6.23s)(0.593)10.166

. Flow-thru tee

. o, 6 -_ ArQ*: * = _:_ = 0.5lAn: _Q: t2 A3

Dr:0.:rt+0.87:1.181

= 1.0=K :0.87

*r:I + r. 1811(0.051x1.0x12)

(r .049)

,<o rr lb r,,, ,.t', fC / t t' \sec \t+a 1"31

fr-lh

S€C'-lD6

= O.222 psr


E Hot Oil Flow from Station 4 to Exit B:

L:2ft

rh rr2/ rcz\(58.7)

ft; (3.612F

**- |rfri".,I

tcrrr!jq.-S€C'-lD;

APq = 0.405 Ott

Total friction pressure drop from entrance A to exit B:

Path (Figure 1-17) APl psi

@ Entrance A thru branch flow tee 0.186 psi

@ Flow from station I to station 2 0.126 psi

@ Flow from statioo 2 to station 3 0.063 psi

@ Flow from station 3 to station 4 0.222 psi

@ Flow from statioD 4 to exit B O.zlO5 psi

1.002 psi

from entrance A to exit B

Z Shzll mils-Soutft ride (Figure 1-18)

Station 1-

35spm

. l-in. x 3/a-in. reducer

,. _ 0.8 sin [a(l - F1] ______7-_

3.612 ftlser

0.8 sin [(3.081X1 - 0.601I(0.361)

2o.7o" =:.r : : 0.167Qr 42

Kr = -0.032Header entrance = K = 0.78

station 1 : K = -0.03K : O.75

Q: : 42 epm'L = 10ft;/ = 0.15cp;d = 2.067in.p:58.7 lb/ft3;6 = 9.9613V = 4.016 ff/sec; f : 0.020; Nr, : 402,829APr : 9.195 *'

sal / 1ft3 \/r -i"\/4\ -

t-t t-l\"/ min \7.479 ga| \60 sec/ _

(0.533)in.'(+)

{o't'lo,r.utl a rsa.zr I\12l 'sec' 'tt' : A4.449

Dot, = 1.002 psi = Total frictional pressure drop

1^(o0.5

= 0.055

I-values

: 0.0'[8

. 3-90' LR ells

K : 3(0.025X30)

Pi1r exit + 1q :: 2.25O

Station 2-1.0

35 gpm

28 gpm

E* : o.o+s + z.zso + r.ooo = 3.298

_ l(0.055x2.0)(12)- t to^szat&r, + 3.2esl


n. 1Qr: ::] = .-:0.200Qr JJKaz = -0'03L = 5in. = 0.417 ft;d = 2.O67 in.Qa : 35 epmY : 3.346 tusec; f : 0.020; Nn : 335,691AP, : g.gg1 n"t

Station 3-

Q3 = 28 gPm

Q:.1 :o.zsQr 28L = 10ftd : 2.O67 rn.v : 2.677 ft/sec; K32 : -0.036Nr":268,553f : 0.020APr : 6.952 n.'

Station 4-

21 gpm

14gpm

Q:1:o.rgrQ: 2l&z = -0.030L:5in.:0.417ft; d = 2.067 ln.Qr : 21 epmV = 2.008 ff/sec; f : 0.021Nn' = 201'415AP+ : g.ggt O.t

Station 5-

9:1:o.sooQr 14L: 10ftKrz : 0.015v = 1.339 ft/sec;,f : 0.021Nn":134,276APs : 9.914 O.t

Friction pressure drop from station 6 to coil entrance

o^:Qt=roQr

&r = 1.28For 2-in. x 1tlz-in. reducer, K = O-129

lrlz-in. x l-in. reduceq K : 0.311l-in. x 3/a-in. reducer, K : 0.048

r/+-in. plug valve, K : 18(0.025) : 0.4502-l1lz-in.90" LR ells, K = 2(30)(0.021\ : 1.2@

1-1-in. 90" LR ell, K : (30X0.023) = 0.6901-3l+-in. 90. LR ell, K : (30X0.025) : 0.750

Exit into coil, K : 1.0Q = 7 gPm; L : 7 -25 f7

F.-!K = 5.168. For 2-in. { pipe, d : 2.067 in.

L:7in.:0.583ftK : 1.049V = 0.669 fl:/sec

3

ITspm

I

I2

Nr":67,138f:0.023

AP : 0.0M psi

o For lrlz-in. d Pipe, d : !.610 in.L:3ftK : 1.571V : 1.103 ftlsec; Nx" : 86,195;

f : 0.023AP : 0.016 psi

o For 1-in. { pipe, d : 1.049 in.L:ZftK = 0.738V : 2.599 ftlsec; Nq" : 132,292',

f : 0.024AP : 0.055 psi

. For 3/a-in. { pipe, d : 0.824 inL :2ftK : 2.2OY : 4.Zll ft/sec; Nx" : 168'416;

f:0.025AP = 0.330 pst

Toral frictional pressure drop from station 1 to bottom shell.nil is!s\llP = 0.195 psi + 0.001 psi + 0.052 psi + 0.001 psi

\-- i-!-Z \-!- --/-

station 1 station 2 station 3 station 4

+ 0.014 psi + (0.004 + 0.016 + 0.055+ 0.330) psi

station 6

I-t" : O.OOS psi : Total Drop for Shetl Coil on SouthSide

- Shell coils-Nonh side (Figure 1-18)

Sration 1-

21spm


n.QR::r = 0.333

\J3Kr: -0.030L: 10ftQr : 21 gpmp:58.7 lblff; p :0.015 cp; e:0.0018;

d : 2.067 \n.V = 2.008 tusec; Nr" : 201,415t f : 0.021AP1 : g.g3g O.'

Station 2 -

o, : Q' : o.5ool L = roftQr

K:z = 0.015Q3 : 14 gpmv : 1.339 ft/sec; N3" : 134,2'16: f : 0.021AP : 0.014 psi

Friction pressure drop from station 2 to coil entrance:

n.o" : lll = 1.0: K,, = 1.28r)^

For 2-in. x 1-in. reducer, K : 2.5381-in. x :/q-in. reducer, K = 0.048

3/4-in. plug valve, K : 0.450Exit into coil, K : 1.0

For l-in. 90" LR ell, K = 0.690For l-in. { pipe, d : 1.049 in.

4'l

station 5

3

+

I;'il]I

r2

d

14 gpm

For 2-i'].0 pipe:Q:7gpm,L: llftK:2.538V : 0.669 tusec

Nn":67'138.184f : 0.023

Ap : 0.0t1psi

For3A-in. opipe:d : 0.8241 L : 2 ft: K =V : 4.211 tusec

Nn":168'416f : 0.025

Ap : 0.245 psi

B{SS STAC i(

42'6 670 9a

4zV srD


Q : 7 gPm, L : 4fr,K : 0.738

V = 2.599f1/sec; f : 0.024Ns.:132'292

AP : 0.079

1.450

EXAIIPLE l-3: FRIGTION PBESSURE DROPFOR A WASTE HEAT RECOVERY SYSTEM

A gas turbine manufacturer specifies that the maxi-mum back pressure on the unit used in this system be 10in. of water pressure, therefore, the waste heat recoverysystem should be designed so that the frictional pressuredrop does not exceed 10 in. of water. The system isshown in Figure l-20.

Z Turbine exhaust dntq for outside air ot 6l)'F

Temp. ofexhaust gas : 795"F; V = 131 fusec

th .ntt : 0.0759 .':. \0.4132) --l= = 0.03 t cpIt-hr lb/n-hr

L : l2O ft; D

: O.OOOO+Z (commercial steel)

e = 0.00015Total frictional pressure drop from station 1 to bottom shellcoil issrl-r AP = 0.030 psi + 0.014 psi

\-!-\-.-/

station 1 station 2

+ (0.011 + 0.079 + 0.245)psi

station 3

srL AP = 0.3'19 psi : Toml Drop for Shell Coil on NorthSide

Maximum friction pressure drop in supply system is in-curred at bottom head coil line with AP : 1.002 psi.

N"" : VDP, a = 0.0759 lb {--!t '|

t' ft-hr \3,600 sec/

: 2.108 x 10-5 -lbft-sec

Nn:fl 1t.0) a(4t.25)i". I | ft I {0.031) !sec \12 in./ ftJ

(2.108 x 10 5) -.!!-rr-sec

= 662,224

Figure 1-20. Waste heat recovery system.

From Equation l-6a,

, I n o!goo38j-:* : -2 log,o 10.00001I + :-:ri t rttot /

f:0.0130

D = 41.25 in. = 3.438 ft

.. fL (0.0130)(120) _ ^,r..4D 3.438

From Equation 1-8,

t : K1/Np. + K- (l + '/d)

ior straight pipe, 42-in. { section

.. fL (0.0130)(120 - 30)

d 3.438

ii-\'alues

\:lr es and Fittings


For 42-in. d portion,

K.*r : 0.770 + 3.161 + 0.340 : 4.271From equation l-4,

op = ILL* rr)ey\d - I2e,

[(0.0130)(I20) , "-,1ur = t- 1- +,zt1lL (3.44) I

to.oiU -19, (r 3r .oo,12

ilrt" sec'

fr fr22\32.2) ". (144)

--:-sec' ln.'

th ff2(0.031r: (47.458), j ,It- sec'

fr fr22132.2) : 1144)

=sec' ln.'

AP : 0.271 psi

For l0 ft x 4 ft x 42-Lrr. { transition piece,

D:68.571 in.: L = 4.0ft: K:0.615

Kro, - [<o.ot:ot<+.ot * o u,rlnKr K- nK- | 5.714 I

tsrtterfly valveR.un-thru tee

0.25 0.250.50 0.50

0.75

I 800 800r 150 150

950 AP = 0.005 psi

qso / r\( _ -" | (0.75) ll I _l = 0.770662.224 \ 4t.251

Itirer K-values

in H,O ar 62.FAP = 0.005 osi t27.912t '::::::L' t psi: 0.140 in. H2O

AP thru heating coils : 2 in. H:O

Total AP = 7.564 tn. + 0.140 in. + 2.000 in.

OI

AP : 9.704 in. HrO < l0 in. allowed

l .0000.1611.0001.000

3. 161

i.De entrance at turbine nozzlerl-in.dx30-in.d.lrste heat recovery unit entry ductl-rck exit

:rr a rectangular duct,

i.=ab/2(a+b)

:-'r round pipe,

i. = Di4

EXAMPLE I-4 PRESSURE DROP IN RELIEFVALVE PIPING SYSTEM

Relief valve piping systems are designed to have mini-mum pressure drop. In this application the plant rulesstipulate that the pressure drop will not exceed 3 % of thevalve set pressure. The system is to have two valves,shown in Figure 1-21.

The relieving fluid is Freon 114 and the flow rate is W:243,755lblhr. First we compute the velocity heads, orK-values.

_ 2(10)ft (4)fr _ < ",, r,l0ft+4fta+b

68.571 in. : Equivalent circular diameter

6-in. tee

K : 60 ft = 60(0.015) : 0.9006-in. x 4-in. swage nipple = 0 : 60'

6:d,:4.@6=o.ao+' d2 6.065

From Figure 1-7,

0.5(l - 0.,141)(sin 30')05

Mechanical Desisn of Process Svstems

+Figure 1-21. Relief valve piping system.

The total pressure drop for 6-in. and 4-in. lines : Pr

Pr : 3.869 psi + 5.935 psi : 9.804 psiSet pressure : 205 psi

%^P:#:4.8vo>3vo

Consider moving 6-in. x 4-in. swages above gate valvesand making 90" LR and gate valve 6-in., as shown in Figure1-22. Recomputing the K-values we have

6-in. 90' LR ell, K : 30 fi : 30(0.015) : 0.4s06-in. gatevalve, K : 8ft : 8(0.015) : 0.120

Entrance, K : 0.780Tee, K : 0.900

6-in. 90' LR ell, K : 0.4506-in. gate valve, K = 0.120

6-in. x 4-in. swage, K : 0.019

D*: z.zas

For 6-in. @ line from entrance through swage, Lo : 10 ft

AP = 4.869 psi: Ns. : l4,g3l,g2gV : 37.N2 ft/sec; f = 0.01741

Vo AP :4869 : 0.024 = 2.470 <37o205

The pressure drop in the system in Figure 1-22 doesnot exceed 37o \p to the relief valve as the plant rulesrequire, thus, Figve l-22 is the final configuration.Latet in example 2-5, we will examine the structural in-tegrity of the system.

Flgure 1-22. Relief valve piping system.

K: = 1.0190.194

4-in. d 90' LR elbow,k : 30 ft : 30(0.015) : 0.450

4-in. gate valve,

B : I : 1.0 +r( = 8ft : 8(o.ol5) : o.l2o

Entrance, K : 0.78

6-in. d line from entrance thru swage, Lo : 5 ft\-rLtK : 0.'78 + 0.90 + 1.019 : 2.699AP : 3.869 psi; Np" : 14,931,929V : 37.002 tusec; f : 0.01741

4-in. line from swage to relief valve, L4 : 5 ft

Dr = o.oso+o.l2o = 0.570AP = 5.935 psi; NRe =22.494.325V :83.973 ftlsec: f = 0.01913

NOTATION

al : sonic velocity of sound waves in compressiblemedium, ft/sec

a. : rheological variable, dimensionlessAR : ratio of branch area to header area, dimensionlessb. : rheological variable, dimensionlessc : experimentally determined flow constant where

c : plE" for a Newtonian fluidd = inside diameter (lD) of pipe. in.D : inside diameter (ID) of pipe, ftf : friction factor, dimensionlessF : head loss, friction ioss or frictional pressure

droo. ft(.br) . cm(kgr)'. lb. g.

g : gravitational acceleration constant, 32.2 ftlseczlcm/sec-

g" : English system conversion factor, 32.17

lbtHa: energy added by mechanical devices, e.g.

pumps, ft(lb)/Ib.", cm(kg)/g.He: energy extracted by mechanical devices, e.g.

turbines, f(lbr)nb*, cm(kg)/g.k = specific heat ratio (adiabatic coefficient), Co/C,K : velocity head, (ft)(lb)/lb*

K- : velocity head for a large fitting at Np" = oL = length of pipe or piping component, in.n' = rheological variable, dimensionless

Nr" : Reynolds number, dimensionlessP : pressure, l!lt9, kgrlcrfi

Rn : hydraulic radius, ft, in'u = average bulk velocity, ftlsecv : velocity, lblt(, kgrlcrfiY : height above datum, ft, cm

The Engineering Mechanics of Piping 45

Greek Symbols

d : angle, degrees

6 : ratio of smaller diameter of pipe fitting to largerdiameter

"y = generalized viscosity coefficient lb'/(ft)Gec)e : absolute roughness or effective height of pipe wall

irregularities, ftp = absolute (dynamic) viscosity, centipoisey = kinematic viscosity, centistokescr : angle, degrees

REFERENCES

1. Simpson, L. L. and Weirick, M. L., "DesigningPlant Piping," Chem. Eng., April 3, 1978.

2. Metzer, A. B. and Reed, N. C. A.l.Ch.E. Jownal,vol. 1, no.434, A.S.M.E., New York, 1955.

3. Rase, H. F., Piping Design for Process Plazts, JohnWiiey, New York, 1963.

4. Govier, G. W. and Aziz, K., The Flow of ComplexMixtures in Pipes, Robert E. Krieger Publishing Co.,Huntington, New York, 1977.

5. Crane Co., Technical Paper No. 410 Flow of Fluids,Crane Co., New York. 1981.

6. SMACNA, HVAC Duct System Design, SMACNA,Vienna, Virginia 1981.

7. Hooper, B., "The Two-K Method Predicts HeadLosses in Pipe Fittings," Chem. Eng., Ang. 24,1981.

8. Sabersky, R. H. and Acosta, A. J. Fluid Flow-AFirst Course in Fluid. Mechanics, The MacMillanCompany, New York, 1964.

The Engineering Mechanics of Piping

Static and dynamic analyses require clear and precisec.efinition of terms-their misuse can often lead to mis-:-;nderstandings, a problem the engineer greatly appreci-

"tes. The application of engineering mechanics to piping:s mainly referred to in industry as "pipe stress analy--.rs." However, the term is not comprehensive enoughrecause engineers are usually more concerned about:orces and moments exerted on equipment than stress.Cerrainly, stress is a concern and is discussed along withrtier phenomena in the chapter.

-{nother popular term used in industry is "piping flex--:ility analysis." The word flexibility can pose a prob--em because in the stiffness method of analysis it is actu--:lh the structural stiffness of pipe supports, rather than:ieribility, that is important. For this reason the term-'piping flexibility analysis" is avoided.

-\ piping component is any constituent part of a piping.-, stem, of any finite length of pipe-a valve, flange, el-:\.\\\'. pump, or anything else within the piping system.llping is supported for various reasons-an obvious one

-rng to counteract the force of gravity-and to begin to-:rderstand the applications we must start with some ba-a.: concepts.

Consider a piping component as shown in Figure 2-1.i{:re we have a three-dimensional axis system with the: rmponent-a short length of straight pipe-subjected to: rrces and moments about each axis. The forces and mo-:.nts are considered as vector quantities and often ex-::essed in terms of resultant vectors. For convenience:: u'ill express resultant vectors in terms of a resultant' :.ror operator defined as follows:

; \.\.2) : ,tll*-Z (2-1)

-je resultant force and moments change in magnitude-J direction along the length of the piping system.

These forces and moments are controlled by structuralsupports attached to the piping using pipe supporrs rocontrol forces and moments in the pipe and attachingcomponents bring up two fundamental concepts-stiff-ness and f lexibility-which are discussed later in thischaDter.

PIPING CRITERIA

In analyzing piping mechanics, the following parame-ters must be considered:

1. The appropriate code that applies to the system.2. The design pressure and temperature.3. The type of material.4. The pipe size and wall thickness of each pipe com-

ponent.5. The piping geometry including movements of an-

chors and restraints.6. The allowable stresses for the desisn conditions set

by the appropriate code.7. Limitations of forces and moments on equipment

nozzles set by API, NEMA, or the equipment man-ulacturers.

8. Metallurgical considerations, such as protectingmaterial from critical temperatures, like carbonsteel below its transition temperature.

For any piping system, these criteria must be consid-ered and satisfied. While it is sufficient to analyze a pip-ing system, it is not always necessary. For example, asystem having only two terminal points and pipe of uniform size does not require a formal analysis if the fol-lowing approximate criterion is satisfied:

47

o?w""


Figure 2-1. An element in a pipe wall is subjected to four

SIrESSES.

onlv code that is different from the ASME codes is the

Geiman DIN code, where the basis of yield is different'

The code basis and theories of yield are discussed later'

Reeardless of what ASME codes are used, the user is

cauti6ned that the codes are written by ASME to be

euidelines and not design handbooks. The intent of the

lodes is merely to set minimum rules and procedures for

desrgn. This does not include operation ofplants' Opera-

tionil problems are not intended to be governed by

ASME codes. Such problems as bowing of the pipe and

geysering are considered operational and are not consrd-

ered as design Phenomena.Pioine codes-are not the only ones with which the de-

sien'eniineer should be familiar. It would be expedient

utia n.t-pru if he or she is familiar with ASME Section

vnl Di;. I and II. Also, the AISC (American Institute ofSteel Construction\ Manual of Steel Construction is

mandatory in the design of structural supports-a re-

quirement that will be obvious later.'The reader will notice a stark contrast between the

ASME and AISC philosophies of codes' The AISC Man-

ual of Steel Constiuaion is intended to be a design hand-

book and is considered as such. AISC' unlike ASME'

covers all industries of steel construction, from the

buildine of tall office buildings to major chemical plants '

Unlike ASME, the AISC codes give a commenta'ry on

what bases are used in formulating the code and why

these bases were used' It cannot be emphasized too much

that engineering mechanics crosses mechanical and civil

ensindrine aisiiplines as known in the United States A

kniwleaee"of some of both is necessary to understand

the overill perspective of piping mechanics'

ln satisfuing Step 6 in the list of criteria, once the ap-

propriate iodi is selected. the system must be analyzed

io ditermine if any portion of the system exceeds the al-

lowable stress given by the code' The allowable stress

"i;;; br the cide is' in the ASME and most foreign

EoA... Uu."a on ttte maximu-m shear stress theory of fail-

"i". itrit tft"ory is based orfthe fact that a material yields

when the maxihum sheir stress equals the yield stress'

This theory is in good agreement with experimental data

rnO"i .,"ufv stati and iatigue stress conditions and for

this reason has been adoPted'--iince tnowledge of thi different theories of yield is

noi dir""tly pertiient to industrial applications of pip-ing

r."ft-i.., the reader is referred to Faires Ill for further

discussion. It is pertinent to note what stresses are re-

cuired bv the codes in analyzing piping systems'' An element of pipe wall subjected to four stresses ls

shown in Figure 2:i. The pipe is under internal pressure

and the four stresses are as follows:

oL : longitudinal stress

oc : circumferential or hooP stress

(2-2)

where D, = outside diameter (OD) of piPe. in' (mm)

y = resultant of total displacement strains to be- absorbed by the piping system. in (mm)

L = developed length of line axis between

anchors, ft (m)U : anchor distance (length of straight line joining

anchors), ft (m)C : 0.03 for U.S. units

: 208.3 for SI units, in Parentheses

Usually. however. the piping sysrem has either more

rhan two terminal points or not all of the previous cnterla

are met, and a formal analysis is required'After the first five criteria are considered the next and

foremost factor to consider is Step 6-the allowable

stress of the pipe. To determine this, one must reter to

the appropriaie code that governs the piping system-' The

following are codes applicable to industrial piplng ln the

United States:

ASME 83l. I -Power Piping-governs pipingin the Power industries (e'g''high-Pressure steam lines)

ASME B31.3-Ct emical Plant and PetroleumRefinerY PiPing-governs PiP-ing sYstems used in the chemical

and Petroleum industrY

ASME B31.4-Zt qiid Petroleum TransportationPiPing SYstems

ASME 831.5-RdiEeration Plnins . ^.ASME 83l .8-C,as Transmission and ulstrlou-

tion PiPing SYstems

ASME Section |II-Nuclear PiPing'

Most foreign codes are similar to the ASME (Ameri-

can Society of Mechanical Engineers) codes' particu-

larlv as fai as the theoretical basis is concerned' lne

rR : radial stressJr : shear or torsional stress

The longitudinal stress is the sum of the followingdlree components:

l. Bending stress induced by thermal expansion.For straight pipe:

MoB : ,7 Q-3)

LM

For curved pipe:

M.oe: _ | (2-4)LM

2. Bending stress induced by the weight of the pipe.(This stress should not be a consideration ifthe pip-ing is properly supported and will not be consid-ered in this analysis.)

,1. Longitudinal stress induced by internal pressure.

The Engineedng Mechanics of Piping 49

Direct shear stress is negligible and is not consideredwhen caused by the piping temperature, because localyielding or "creep" reduces the stress at piping compo-nents. Local strain hardening restricts the local yieldingand prevents the material from rupturing. This phenom-enon of locai yielding reducing stress is termed "self-springing," and has the same or similar effect as cold orhot springing. The operating stress ("operating" is usedbecause it can be either hot or cold) diminishes withtime. This change in stress is compensated for by the al-lowable stress range, which is the sum of the operatingand down condition stresses and remains practically con-stant for one cycle. This sum is obtained as follows:

ot : f 0.25 o" i O.25 oe) (2-11)

f : stress range reduction factor for cyclic condition

Total no. of full temp.cycles over expected life

< 7,000< 14,000< 22,000< 45,000< 100,000

Expansion stress, caused by thermal expansion, mustnot exceed the allowable stress range, oo, and is defined:

oe=[@s)2+4(o)2] (2-tz)

The piping codes further state that the sum of the lon-gitudinal stresses caused by pressure, weight, and othersustained loadings shall not exceed op. This also in-cludes the longitudinal stress caused by internal pres-sure, op, defined above.

When torsional stress becomes significant, as in manymultiplane systems, the resultant fiber stress, or com-bined stress, is determined by the following:

t = llor+ op * [4(o1), * (o1 - op)r]05] (2-13 )

PBIMARY AND SECONDARY STRESSES

These two concepts are very important in analyzingpiping mechanics problems. A more detailed discussionof the various types of primary stresses is given in Chap-ter 4. The reader is encouraged to review Chapter 4 foran understanding of pressure vessels, as well as thischapter for help in solving piping mechanics problems.

Secondary stresses are called self-limiting or self-equilibrating because as they increase in magnitude, lo-

"' = Pi (2-s)

,- = P(D - 2Py)

2tE

3ecause both longitudinal stress caused by internal pres-.-re and bending stress act in the same direction,

:- = oBL + oP (2-6)

lle circumferential or hoop stress is caused primarily by::ernal pressure. Thus.

(2-'7)

i ,: thin-walled cylinders op is negligible. However, for--::k-walled pipe, the following relationship may be

--'d for determining the radial stress:

_ _ r,2P, - rozPo _ rozri2(Pi - Po)- -il -l- - Gt:'5r (2-6)

ic shere external pressure, P6 - 0, we have

,_ rilP; _ rotr,.P,

h: - ri2 (ro, - ar,)r

l::.ional or shear stress is

-TlZm

(2-9)

(2-10)

| ::ie torsion is generated in a multiplane system.


cal yielding causes local deformation which in turn re-duces the stresses. Self-springing is an example of thisohenomenon.-

Primary stresses are not selflimiting because as they

increase, local yielding does not reduce them. One ex-ample of primary stress is internal pressure. Under suffi-cient pressure a pipe will undergo local yielding and de-

form, but the stress will not diminish and the pipe walldeformations will be excessive and unacceptable. Forthis reason, it is necessary to assign lower allowablestress limits to primary stresses than to secondarystresses. This fact is extremely important, as prlmaryand secondary stresses are evaluated differently' and

have different allowable limits. It must be remembered

that piping and vessel codes give allowable stresses onlyfor primary sresses. Secondary stresses must be as-

signed allowable limits as shown in the following discus-

sion.

ALLOWABLE STFESS RANGE FORSECOI{DARY STRESSES

The most important secondary stresses are those in-duced by thermal expansion (or contraction) and surface

discontinuities, the latter being more relevant to vessels.

The most widely used approach in designing equipment'vessels, and piping is to keep the induced stresses in the

elastic range. In the case of ductile materials, the elastic

range is well defined by the minimum yield point. Duc-

tilehaterials are often used in piping systems subjected

to loads that induce secondary stresses. Materials that do

not have a well defined minimum yield point are de-

signed on the basis of their ultimate yield strength, whichis the maximum tensile load divided by the originalcross-sectional area of the specimen. The minimum yieldpoint is the tensile load required to develop permanent

deformation in the material. Materials that do not have a

well defined minimum yield point are generally not used

in piping systems subjected to extreme temperatures and

presiures. 'Thus,

this discussion applies to those materi-

als with minimum yield Points.Consider the stress-strain curve shown in Figure 2-2'

Here the metal specimen is loaded to point A and then

unloaded. Because point A is the minimum yield point'

no deformation occurs because the material is still in the

elastic range. Now, consider Figure 2-3 where the mate-

rial is loaded beyond point A' Because the minimumyield point is exceeded, plastic deformation sets in thatpermanently deforms the material to point B. When the

specimen is unloaded, er is the amount of permanent de-

formation, denoted by point C. Point B' is the theoreti-

cal stress point if the material had not deformed to point

B. Figure 2-4 shows a case where a specimen is loaded

e ---:>

Figure 2-2. Stress-strain curve.

Figure 2-4. Stress-shan curve.

inro the plastic region. For complete plastic deformationto occur, the entire area ofthe pipe wall must exceed theminimum yield point. This would not be acceptable inpractice because of permanent deformation and the pos-sibility of rupture.

There are acceptable cases where the loads will fail be-tween Figure 2-2 and Figve 2-3. This condition isshown on Figtre 2-4, where part of the pipe wall is inthe elastic range and the other part is on the plastic re-gion. For cases where the portion in the plastic range issmall compared with the portion in the elastic range, theamount of permanent plastic deformation is impercepti-ble. For this reason, the distance between points A and Bm Figure 2-4 is small compared to Figure 2-3 becausethe portion of material in the elastic range limits theamount of permanent deformation . Thus , when the spec-


imen is unloaded, residual stresses are developed thatcause reverse yielding when the material exceeds thecompressive yield point. This is shown graphically inFigure 2-5. The specimen is loaded to point A and an ex-cessive load deforms it to point B. At point B, part of thematerial is in the plastic range and the other portion is inthe elastic range. When the specimen is unloaded, thestresses in the material go into compression shown atpoint C. Residual stresses caused by the combination ofmaterial in the elastic and plastic regions make part ofthe material exceed the compressive yield point and thespecimen deforms from point C to point D. Upon appli-cation of the same initial tensile load, the material isloaded to point E. Point E is larger in value and, thus, tothe right of point A, because the initial loading of part ofthe specimen into the plastic range causes strain harden-

B

Itlll,l,l

STBAIN =>

Frgure 2-5. Stress-strain curve.


ing and, thus, increases the minimum yield point of thematerial. As excessive loads are applied, the minimumyield point E is exceeded and the material deforms topoint F. As the material is unloaded again the initial pro-cess repeats itself and the stresses in the material move topoint G and then to point H as the compressive yieldpoint is exceeded.

Point Q represents the stress in the loaded condition af-ter several loading cycles, and point P represents thestress in the unloaded condition. It is possible that no sig-nificant plastic deformation will occur after many loadcycles. However, should stress values of Q and P exceedthe fatigue limit of the material, small cracks will propa-gate throughout the strain-hardened material. After thesmall cracks appear, further cyclic loading will result inbrittle fracture failure. The stress magnitude P resultsfrom the specimen being unloaded when the load condi-tion, point Q, is reached. Thus, since Q is the tensilestress opposite to the compressive stress P in the paral-lelogram OB'QR the sides OB' and QP are equal iniength. Therefore, Q : 0.5 B'. Fracture by strain hard-ening will not occur if the theoretical tensile stress B'does not exceed twice the minimum yield stress of pointA, and the magnitude of Q does not exceed the ultimateyield strength of point A.

When a ductile material, that is a material with a de-fined minirnum yield point, is subjected to repeated load-ing, a certain behavior occurs. When a component, suchas a nozzle on a header pipe, is repeatedly loaded andunloaded, the strain hardening makes the materialstronger from load cycle to load cycle. As the materialbecomes harder, it is better able to resist yield. However,the maximum point at which this repeated loading cyclecan occur is 2oyp. The stress o : 2ovp is the limit ofthemaximum stress range. This process is called elasticshakedown. that is. the material "shakes down" to anelastic response, and undergoes deformations or strainsinduced by loads beyond the minimum yield point of thematerial.

It must be noted that at elevated temperatures the valueof 2oyp can be altered by hydrogen embrittlement. Car-bon steel exposed to hydrogen at elevated temperaturescan fail during elastic shakedown because the hydrogencombines with the carbon causing embrittlement.

The relationship between the maximum stress rangeand the initial yield point can be expressed as

o1,s 3 Zoyp (2-1,4)

where MR : maximum local stress range not producingfatigue failure, psi

YP : initial yield point of the matedal at theoperatrng temperature, psl

This analysis indicates that the allowable stress shouldbe based on the yield point rather than ultimate strength.

The material's ability to revert into compression andlimit itself to the amount of permanent plastic deforma-tion is termed "shake down." The material "shakingdown" limits the amount of deformation and, thus, hasan elastic response.

From this discussion, we see tlat there is a range ofallowable stresses available. Direct membrane stressesare limited by oy, bending stress is limited by l.5oy, anda limited, one-time permanent deformation from A to Boccurring from secondary stresses is limited by 2oy.

Table 2-l gives recommended values for design allow-able stresses. As shown in ASME Section VIII, DivisionI, paragraph UA-5e, different stress levels for differentstress categories are acceptable.

FLEXIBILITY AND STIFFNESS OF PIPINGSYSTEMS

There are two basic approaches to piping mechanics-flexibility and stiffness. The former approach is morecommon and easier to understand. Piping mechanics(more popularly known as "pipe stress") is often re-ferred to as "flexibility analysis," but it will become ob-vious in the following discussion that such a term is notcomplete.

In the flexibility approach, the piping configuration ismade more flexible by using loops that allow the pipe to

Table 2-1Allowable Stresses'

Pressure Component Design Conditionsl. Internal pressure . ....... oA2. Internal Dressure

plus therinal loading . ... . 1.25 (oa * op)3. Temporary mechanical

overload . ..... l.33oa < oy4. Hydrotest . . ... oo X hydrotest factor

Non-pressure Components Design Conditions1. Pipe supports and

connections otherthan bolts .. ... 1.330a

2. Bolting ....... Per AISC Manual of SteelConstruction considerablesavinss in material can beincuried if high strengthbols are utilized, such asSA-193-87. FollowinsAISC guidelines in n6npressure components willresult in prudenteconomical desisn.

' Courtes) of American Socier) o[ Mechanica] Engineers

displace itself, resulting in lower stresses, forces, andmoments in the system. This method is often the mostdesirable when relatively inexpensive piping material isused (pipe elbows can be very expensive in alloy piping)and space is available for the loop(s).

However, the stiffness method becomes quite impor-tant when the flexibility method is neither practicai noreconomical. When limited space reduces piping flexibil-ity or makes it irnpossible or undesirable to use flexibil-ity loops, restraining the piping using the stiffness ofpipe supports becomes the alternative. This approach isgaining popularity with the increased use of modular de-signs of petrochemical plants, offshore platforms, andother industrial facilities.

The following is a summary of the advantages of bothmethods:

St iffne s s Me thod Ady anta g e s

l. Requires less pipe fittings and is thus more eco-nomical than flexibility method, because pipe re-straints required are far less expensive than thenumber of fittings they replace. In alloy pipingthese savings are enormous.

2. Requires far less space for piping, such as in modu-Iar skid-mounted plants, offshore platforms, andships.

3. Method is safer because in case ofa failure, such asa leak in a weld crack, the pipe restraints can (andhave) kept systems from blowing apart.

4. Piping and system is more resistant to dynamicloads, such as vibration and seismic shock loads.

Flexibility Method Advantages

1. Utilizes simpler pipe supports, and requires lesspiping engineering skill.

2. Is more desirable in noncritical systems, e.g. ex-haust and flare lines.

3. Many solutions do not require a computer. Theproblems can be solved manually.

To better understand these two methods of piping me--'hanics, it is necessary to examine some basics of struc-::rral analysis.

Stiffness is the amount of force or moment reouired to:ioduce unit displacement. either translational or angu--.lr movement. The simplest concept of stiffness is to::nagine using X pounds to compress a spring one inch.Thus, the spring stiffness is in terms of pounds per inch.This simple example illustrates translational stiffness.Rotational stiffness can be thought of in a similar manneras a spring that resists rotational movement, foot-poundsrer unit degree of movement.


A piping element has six degrees of freedom, three intranslation and three in rotation, as shown in Figure 2-6.The amount of force or moment required to produce unitdisplacement in each degree of freedom at points allalong the piping element is described mathematically as

the stiffness matrix. K. which is defined as

P:KU

where we have an elastic element subjected to a set of nforces and moments

(2-ts)

the corresponding displacement of each P1 is describedby the matrix

(2-r6)

Therefore, the stiffness matrix can be expressed as

(2-17)

which can be in pounds per inch or foot pounds per de-gree. The relationship

(2-18)

is defined as the compliance or flexibility matrix and canbe in inches per pounds or degrees per foot-pounds.Thus, the stiffness K ofa system is the inverse of the sys-tem compliance or flexibility, C, that is, the piping sys-tem becomes more flexible, or less stiff than its initialconfiguration .

The system stiffness matrix, K, is made up of elementsthat are either direct stiffness or indirect stiffness compo-nents. The direct stiffness component K;; is the value ofstiffness at the point i when the displacement U1 is pro-duced by a force or moment P acting in the direction ofU1. The indirect stiffness Kij is the value of stiffness atthe point j, with the displacement Uj acting in the direc-tion ofj, due to a force or moment at another point i inthe direction of i. The indirect stiffness can also bethought of as relative stiffness-those stiffness values in-duced by forces and moments in the system other thanthe point in consideration. It is the combined grouping ofthe complete direct and indirect stiffness values thatform what is called the "stiffness matrix." Each directand indirect stiffness is considered in the matrix when allother matrix components are zero. Such as the systemdescribed in the followins:

p

U

IIP


P: tmi HniHtrHl iil]lH ft: e ft [: *11-u:J

Q-1e)

translational stiffness for a beam element fixed on one

end and pinned at the other end is

-- 3EIn't : t_3

For the 4-in. PiPe,

3(29 x 106) ${r.zr) in.o

K4 : ------GD3 ini

For the 10-in. PiPe,

3(29 x lo) |rtoo.s) ln."

Kro = - (48)' in '

The force required to move the 4-in. pipe l/+ in' is

lhF : (5.687.66t .11 (0.25) in. = 1.421.92 lb

To generate the same amount of force in a 10-in' pipe the

same length would have to move

|,421.92 lb : 0.011 in.

zo,qsl.+o !ln.

: S,Oal.OO Pm.where the values K11, Kzz, Fv:z, K44, K55, and K66 are

known as direct stiffness values and all the other compo-

n"rrt. u." known as indirect stiffness values' Each value

of U represents a unit displacement. The components ol

the stifiness matrix are ditermined by the nature (a'tial

force, bending moment, shear force) of the force or mo-

ment inducin! unit displacement U at or arvay froT the

point in que.iion. To eifectively see how these stiffness

io-pon"nt, ur. utilized in practical applications'.we willconsider each type of force or moment rnduclng ols-

olacements, thai is, each component of the P matrix cor-

iesponding to each value of the U matrix' Table 2-2 lists

in"'airect'uatues of stiffness induced by direct and indi-

rect loadings shown in Figure 2-6. For analytic deriva-

tions, the rlader is referred to Przemieniecki [2]'To illustrate how these concepts apply to piping me-

chanics, let us consider both a 4-in. schedule 40 pipe and

a 10-in. schedule 40 pipe shown in Figure 2-'7 He,re we

are considering two pipe spool pieces subjected to. a

force F shown. Referring to Table 2-2' we see that me

= r26,497.40Y1n.

Figure 2-6. Pipe element.


Table 2-2Stittness Properties ot Piping Elements

*"t t"

AFKtt=Kr:?

r,,:r,,:4EL

K:r:IQr:K5r=Iqr=0

Kzz=Kqz:6tr=lQr=0

-(*--u

Kr::Kzs:0

_- tzElor, = il-+ e)L3

,, -lzBlôr

: 11 1oy rr

Kla:K2a:0

,. l2Er^44:(l+o)L:

,. - tzEr^" : ( rJi) IJ-

f

K53 : rqi = {1 + O)trKy : Koq : ,=---.---= .-=-

6EI

rl'>'*- ->xA\ll,k \.tr\

K.. f.r

-""/ l--------9)'\-1._________J /,

/ P t"t--(.4-=-Y/^-.lffil

"( o-------4\YIt<----_T+

1+o)Ll

(2 - a)EIKes = Kso =

,/ ^.-_____________

T uTT T/f-l

&::ree:d#r,

-6FIKrs=K:r=illo)U

(4 + O)EIr\55 : 466 : .-);-- :-=:L(r r 9,

Note: In all cases

- lzBrGALI

and

-- cJK = -: lorsronal sunness

L

L(l + O)


g

FOR ONE END PINNED AND THE OTHER FIXED

K,, = -3!.L-' l1+olL'

a"q scHeoure

,r,10Q

K1o )) K4

SCHEDULE

\

40

Figure 2-7. Comparative stiffness.

The force required to move a 4-in schedule 40

0.070 in. is

thFq: (5.687.66) I tO.OUOr in. = 398.14 lb'

In other words, if the pipe itself moved because of ther-

mal expansion and theie was a restraint of a given spring

"onrt-t ."tttuioing the movement, the 10-in' pipe would

onlv have to rnonJ0.0l I in. to exert ihe same force as

the'4-in. pipe moving r/+ in' Thus' the l0-in pipe is

-ore tltun 2i ti-es stiffer than the 4-in' pipe, which is a

.igrifi"-t point because it indicates that the larger the

oiloine. the'less it must move to exert excessive forces

Iria rio."ntt on nozzle connections and pipe supports'

fi". ,fti, example it is obvious that the larger the pîp-

i*. tt'" *t"ut"t itte stiffness' Piping designers often fail

toiealizJ that larger piping does not have to move very

much to generate -greit

ioads . This basic fact is important

in ttt" OE ign of-pipe supports, particularly using the

stiffness apProach.- iarrvine^the analysis further. consider the two piping

"oniit,itu,iont shown in Figure 2-8 This situation is

similir to Figue 2-7 in that one end is fixed and the

other pinnedi'e., both systems have the same boundary

"""Jit'i"".. The segment-B-C is flexible enough to bend

with enough rotatidnal flexibility to consider tut "ld^::a pinned j6int. lf the temperature ofthe piping is -luu'f, the segment B-C moves

Ma: (-1.75)do,o'ft = -o.o70in'

pipe

Figure 2-8. Pipe size makes a significant difference in nozzle

loadins.s.

The force required to move a l0-in. schedule 40 pipe0.070 in. is

Fto = oz6,4s7.40r I to.ozol in. : 8,854.82 rbtn.

l ielding a moment of

vro: (8,854.82)(4) : 35,4r9.27tt-lbl2 : l7,7o9.64ft-tb

at the nozzles A and B.The 4-in. force of 398.14 lb would nroduce a moment of

\'r1 : (3e8.14) ! : na.ze uv

at nozzles A and B.It is clear that the 10-in. pipe would exert moments

u ell above the allowable moments for most rotating andstationary equipment. To reduce the loading at the noz-zle, the engineer is faced with two options-make thepiping configuration more flexible or restrain the piping.To fabricate the piping configuration to within a toler-ance of 0.070 in. would be well beyond the practicalrange of any fabricating shop.

First, we will analyze a case where space is premiumand there is not enough room to make the piping moreflexible. This requires using piping restraints to transferloads from the pipe to structural steel or concrete. Con-sider the piping system in Figure 2-9, where two alumi-num heat exchangers are piped parallel to one another.Here we use the fewest 90' elbows needed to give thesvstem enough flexibility to stay within the maximum al-iowable stress range for the material at the given temper-ature. Piping restraints are then placed close to the heat3\changers to transfer loads from the pipe to the steel in-stead of the nozzle of the exchanger.

Now, we analyze the component that makes the system$ork-the pipe restraint at the equipment nozzle. The:estraint's function is to transfer forces and moments ex-erted by the pipe to the structural steel below, simultane-ously allowing the equipment to move freely. This re-quires a more careful design of the piping restraint, as.\e are expecting it to do more.

In this example the piping restraints must allow the ex-Jhangers to move upward as shown in Figure 2-9. A re-straint that resists moments by transferring the moments:iom the pipe to the steel is termed a moment restrainingsupport (MRS). Different types of MRS supports areshown in Figure 2-10. An MRS can vary from a boiledplate connection shown in Figure 2-10A to a sophisti-iated type in Figure 2-10C. MRS restraints' sophistica-:ion is a function of how much rotation is resisted andiow much translational movement is allowed. The most


Aluminumexchangerllange

Figure 2-9. An MRS support-restraint designed to reduceforces and moments on an aluminum olate-fin heat exchanser.

simple MRS restraint is the anchor, where the pipe itselfor a pipe attachment is welded down to structural steel orimmersed in concrete. In that case, it is resisting threedegrees of freedom in translation and three degrees offreedom in rotation. In most applications, the momentsat nozzle connections can become excessive, and it is of-ten desirable to resist rotation in one. two- or tlree axeswhile allowins translational movement. Resistine rota-

Exchanger


C Fesl€inl = KTX, XRX. KRY KFz

Figure 2-10. Various designs of moment restraint supports(MRs)-arrows indicate direction of allowed movement.

tion along three axes is, if not impossible, wholly im-practical. An MRS allowing two degrees of freedom intranslation and resisting three degrees of rotation is quitecomplicated, although practical, very useful, and eco-nomical when the situation warrants. In designing suchrestraints Teflon and other materials with very low fric-tion coefficients are desirable. Care must be made in as-suring that such material selected can witlstand theforces and moments being resisted. If the material usedis not resistant to shear, cold flow will result, leading touneven surfaces and an improperly functioning restraint.

In the engineering of MRS restraints, the principlesdiscussed previously must continuously be applied. Nosupport or restraint can be expected to be infinitely rigidalong the degrees of freedom that are being restrained.Placing MRS devices in front of equipment nozzles willnot stop all loading exerted by the piping, because all re-straints have a corresponding stifftress value for each degof freedom, either lbs/in. for translation or ft-lbs/deg forrotation. The engineer must also understand what as-sumptions are being made by the piping stress programbeing applied. Almost all computerized pipe stress pack-ages consider an anchor as six springs, three resistingtranslational forces of 10e lbs/in. and three resistine rota-tional forces of l0e ft-lbs/deg. There is no infinitel! rigidanchor in nature, but 10e lbs/in. is sufficient to be calledan anchor in almost all applications.

In modular plant design it is often desirable for the en-gineer to enter the actual stiffness of any anchor or re-straint to obtain an accurate model of the piping systembeing analyzed on the computer.

STIFFNESS METHOD AI{D LABGE PIPING

Large piping is rnore difficult to restrain than smallpiping because of the surface to be restrained. The terms"large" and "small" are quantified in the following dis-cussion. The most common complication of restraininglarge piping is the phenomenon of shear flow, which oc-curs longitudinally and circumferentially. As illustratedin Figure 2-1 I , longitudinal shear flow transfers bendingmoments and shear forces to the equipment nozzle.

In modular construction longitudinal shear flow doesnot become a problem until one starts using l0-in. pipeand larger. Shear flow can be resisted to some degree bymaking the attachment pipe size or structural membersize close to that of the pipe, but is most often impracti-cal. What is often desirable is to mount an MRS on oppo-site ends ofthe pipe, either top and bottom or offto bothsides, depending on what space is available. In piping30-in. and larger MRS restraints must be attached onfour sides for the MRS effect to be effective. In pipe di---=4

-=a'/

Restrainl - KTX, KTY,KRX, KRY KRZ


Nozzle flange

I

_> Hequrres

Uniaxial longitudinal shear flownozzle tlange

D Requires

Biaxial longitudinal shear flowaround points A and B

meters 8-in. and smaller, attaching an MRS on one sideis sufficient for most modular construction.

Circumferential shear flow, on the other hand, is not alactor in most installations because torsion is very effec-lively transferred to the structural steel by the MRS re-sralnt.

Using piping restraints to transfer loads to structuraliteel or concrete to lower loads at equipment nozzles is'becoming quite popular and more widespread because itis more economical in modular skid design. Also, whereerpensive piping materials are used, the stiffness methodcan help reduce the number ofelbows used for flexibilityend, thus, reduce the cost of the job because restraintsand supports are far cheaper than piping elbows.

FLEXIBILITY IIETHOD OF PIPINGHECHANICS

In non-modular skid construction (block-mountedplants) and areas where there is ample space to place

Restraining pipe with MRS at

-normally required with pipe12" d and over

Restraining pipe with MRS at A, B, C & D

-normally required with pipe sizes-30 " d and over

equipment, it is often more economical and desirable todesign the piping to be flexible enough to reduce load-ings on supports and equipment nozzles. For pipe racks,long headers, etc. this method is the only practical ap-proach to solving piping mechanics problems. Tools usedin this approach include such well known devices andtechniques as piping loops, cut short and cut 1ong, andexpansion joints.

PIPE LOOPS

The most common types of pipe loops used today are"U" shapes, "2" shapes, and "L" shapes. Curves forthese shapes showing stresses plotted against the loop di-mensions are shown in Figures 2-12 and the equationsare as follows:

F1 = A1B -ll- tu, t" : in.o

AandBsizes

Figure 2-11. Longitudinal shear flow-a phenomenon of large pipe.


roltI,I

8l

,lRy

6

5

it

3

2

1

tofI

"II

"l_l

Ry

6

1

3

2

I

Figure 2-12A. UJoop with equal legs'


1o

Ry

4

3

1

ro

9

I

7

Ry

5

4

'|

to 12 14 16 1A 20 22 24Ay

Figure 2-128. Uloop with one leg twice the other leg.


4

2

1

10

9

a

7

6Ry

Figure 2-128 (continued). UJoop with one leg twice the other leg'

180 ^ z&

to

I

a

7

RY

5

1

3

2

I

Figure 2-12C. UJoop with one leg three times the other leg'

6Ry

5


Fv

"* o, ooo

10 t2 tO t"o, 22242a303234

Figwe 2-12C (continued). U-loop with one leg three times the other leg.


to

I

I

7

Rv

4

3

2

10

I

a

6Ry

4

2

I

15 2O 1, 25

Figure 2-12D. Uloop with one leg four times the other leg.

tao oo 22o

,o o, 25


'to

I

a

2oo 300 400 500 500 700 800

Figure 2-12D (continued). UJoop with one leg four times the other leg.

a = -i1 n=*

Ab

Figure 2-12E. UJoop: "2" configuration.

6

Ry

4


a

7

6

Ry

4

3

2

1

a

Ry

4

1

10 20 30 40 50 60 m 80 gOAv IOO llo l2O l3O lr|o 15O 160 17O t8O

40o A, so

Figure 2-12E (continued). U-loop: "2" configuration.


"=E

Figure 2-12F. U-loop: "L" configuration.


Figure 2-12F (continued). U-loop: "L" configuration.

Fv: AvB{lb,

oo: A"B P osi. L - fr. D = rn.-L

, , Thermal movement (in./100 ft)Eo172.800

SIF = 1.0 (Verified by computer stress analysis)

Loops such as circle bends, double offsets, and othergeometrics involving completed circular geometryshould be avoided because they are impractical, expen-sive, and unappealing to clients due to their complexity.If excessive looping is required, the stiffness methodshould be used to produce a practical, economical solu-tion. The use of both the flexibility and stiffness ap-proaches in areas, where applicable, can yield very at-tractive and appealing piping designs.

In pipe racks, the "U" shape loop is invariably themost practical shape to use because of its space efficiency. "U" loops are normally spaced together (i.e.,lines running together on a pipe rack are, where practi-cal, looped together as shown in Figve2-13). It is desir-able to guide the pipe on each side of the loop and at ev-ery other support thereafter as shown in Figure 2-14.Make sure the first guide is far enough from the loop toavoid jamming problems. Usually, this distance is twice

the bend radius of an elbow of the pipe size being used.If you cannot put piping guides on the pipe coming downfrom the loop, then put them on the inside ofthe loop as

shown in Figure 2-14.Other configurations, such as "2" and "L" shapes,

are used in the normal routing of piping systems. It mustbe remembered that when these shapes are anchored onopposite ends, the ratio of the shortest leg to the longestshould fall in the range of 1.0 to 10.0 to avoid over-stressing the pipe. When analyzing the shapes by com-puter, any ratio can be used, but usually the aforemen-tioned range is valid for most applications.

PIPE RESTRAINTS AND ANCHORS

Pipe restraints are used to counter forces of gravity,wind, earthquake, vibration, and other dynamic forcessuch as water hammer. The most common type is thegravity support, which merely restrains the force ofgravity. A piping restraint can act in one or all degrees offreedom. As discussed previously, there are no restraintsthat are infinitely rigid-each has its own spring rate ineach degree of freedom of translation and rotation. Evenan "absolute" restraint has in each desree of freedom a

rranslational stiffness of tOq lbs/in. aidior a rotational

Figure 2-13. Optimum grouping of UJoops.

Figure 2-14. Guides are necessary for controlling movementr: loops.

Frgure 2-15. Conceptualization of system stiffness. Each- -::rponent of the system-pipe, pipe supports, concrete, and.::--has translational and rotational values ofstiffness (matri--.: ilbout each axis. These values can be modeled into the sys-'.- as springs,


stiffness of 10e ft-lbs/deg. Such a restraint that restrains apipe in all degrees of freedom is termed an anchor.

Piping guides are restraints that counter movement inone or several directions but allow total freedom ofmovement in one or more directions. Total freedom isdefined as a stiffness value of zero. An anchor, by defini-tion, has some value of stiffness in every degree of free-dom, even though the anchor itself can move. The move-ment occurs while the anchor is still resisting movementat a certain stiffness in each degree of freedom. Thus,the term "sliding anchor" in place ofa pipe guide is er-roneous, because guides have a value of zero stiffness inone or more degrees of freedom. An anchor can restrainmovement, although it may move. It is important to becognizant of restraint terminology to avoid unnecessaryconfusion.

The stiffness of a support is not only a function of therestraint material, but also a function of the structuralsteel or concrete to which it is attached. Even thoushvery stiff in compression. concrete is not infinitely stifi.As shown in Figure 2-15, the pipe restraint has a stiff-ness value K,, the concrete a stiffness value of K6, andthe soil a value of IG. Because Ka ) ) Ks, the concretecan sink or move in the soil if the concrete support is de-signed correctly or if subsidence occurs. Movementscaused by soil conditions should be the responsibility ofthe piping engineer as well as the civil/structural engi-neer. The latter is responsible for limiting such move-ments as much as possible, and the piping engineer is re-sponsible for entering these movements in the stresscomputer run or manual calculations.

It was mentioned earlier that for a pipe restraint to beconsidered absolute in one direction it must restrain onebillion pounds per inch of translation and one billionpounds per degree of rotation. However, very few piperestraints in nature are so rigid (an anchor being a re-straint in three degrees of translation and three degreesof rotation). If the actual flexibility of the restraint ismodeled into the pipe stress analysis, more realistic reac-tions and moments are obtained. In the case of nressurevessels much work has been done in determining realis-tic spring constants for nozzles. For application to rotat-ing equipment, the reader is discouraged from usingthese spring constants, especially on equipment made ofbrittle material such as cast iron. Also, these spring con-stants are to be used only for ductile materials. Nozzleloadings should be based on either manufacturer recom-mendations or applicable standards. For further detailsand discussion of nozzle loadings on rotating equipmentsee Chapter 6.

To treat a restraint with elastic end conditions, only ro-tations are considered significant. Deformations inducedby radial force and other translations are ignored, be-cause their influence is insisnificant.

Line smallesl in size aod has leastlhermal movement is placed oninside

Lrne tnat has greatesl lhermalmovement and targesl size is placedon oulsade to allow lor movement

..PU

where KMeF

IDy

kf


The basic relationship for rotational deformation of Angle of Twist

nozzle ends is applying Equation 2-17 as

(2-r7)Longitudinal :

Circumferential

- MD"K, ..H, : " - tracransl

, MD"IC. ..:H: " - {fadl?llSlEI

where C1 : 0'09 for in Plane bendingC. : O.2'l for out of Plane bendingD = diameter of vessel or pipe header' in.

Dg = diameter of branch' in.E = modulus ofelasticity. lb/in 2

I : moment of inerria of branch. in.a

KL : longitudinal flexibility factor

K : circumferential flexibility factorM : apPlied moment, in -lbsOr : longitudinal angle of twist, radians

O. : circumferential angle of twist, radians

t : wall thickness of vessel or pipe header' in.tB = wall thickness of branch, in

In-plane bending refers to longitudinal bending in Ihe

pipe header or vessel in the plane formed by the intersec-

iion of the branch and vessel or pipe header centerlines'

Out-of-plane bending refers to circumferential be.nding

in a plane perpendicular to the vessel or pipe header. di-

ameter. These rotational spring rates are necessary wnen

the stiffness of an anchor must be considered in pipe

sfess analySis.

PIPE LUG SUPPORTS

These are about the most common pipe suppo(ts' The

lug can provide a means for spring hangers or simple

clevis-rod hangers. As simple as these supports are' a

failure by one could result in loss of property or lives

Thus, their simplicity should not allow one to take them

for granted thinking that any design will suffice'Tie following method is based on the Bijlaard analysis

discussed by wichman et al. [3]' Consider a pipe sub-

iected to a load P (lbs), as shown in Figure 2-16 The lug

Figure 2-16. Pipe lug support for a pipe with internal pres-

suie-primary and secondary stresses must be added'

M " ler I-

=

-l-le 180 [DNkr I

: KRX or KRY, ft-lb/deg= moment, ft.lb: angle of rotation, deg: modulus of elasticity of vessel metal at

ambient temperature, Psl: moment of inertia of vessel rLozzle, in.a

= diameter of vessel nozzle, in.: flexibility factor, referred to in piping codes

as "k"

The flexibility factor, kr, is a parameter that has had

several formulations over the years. One widely used

variant was that proposed by the "Oak Ridge ORNLPhase 3 Report- 1 15-3-1966 ." Since this document was

oublished in 1966, several revisions have been made'

the current ASME Section III Division I code gives de-

tailed discussions on the flexibility factor. If one is de-

siening piping for nuclear systems. then that person

str-oula only consult that code. Outside the nuclear indus-

try the piping engineer rarely knows all the parameters-

that are necessary to compute the flexibility factor ofSection III. Also, the piping engineer in nomuclear

work rarely knows which vendor will supply the piping

components, thereby making many Section III parame-

ters unknown.Therefore, the more elementary "ORNLI" factors are

Dresented here, because they produce lower values for

[, which, in turn, produce higher, more conservative

values of K. These factors are as follows:

Flexibility Factor

c","o" (|iLongitudinal = K.:

Circumferential : K" :

Rotational Spring Rate

r-onsitudinal = *.: #or*[ry*)

tD

^. ^ lr\:-,,8"8 \T/

Circumferential : R"

^C^L,RU' 2Rna 2Ru L

rr pq

< | .u : [, - i(- 'ff),,

connection is free of moments because the pin connec-tion at the lug hole allows the pipe to twist in all direc-tions. The usual oversight in designing a lug support isnot considering the primary and secondary stresses,which must be added together and compared to the mini-mum tensile strength of the pipe material. First, we willdiscuss the Bijlaard method, which is only concernedwith secondary stresses.

The pipe and 1ug geometries determine the attachmentparametet B, and the shell parameter, k, by

r ff> r,, : [' - ](ui - ')(' - *,)]


Figure 2-17. Membrane force, Na,/P/R-, induced by radialload P [3].

Table 2-3Radial Load P

(B,B)os Q-20)

- ",]

(Btp)o: (2-21)

where K1 and K2 are determined from Table 2-3. For cir-cumferential stress, od, the circumferential membraneand circumferential bending stresses are determined by

f" - (f' .l

( f.l : circumferenrial membrane stressr \P/RJ \R.V

T = H (9 - .,,".u'r.,ential bending stress

The membrane force, No/(P/R.), is determined fromFigure 2-17 or Figure 2-18, and the bending moment,\1"/P, is determined from Figure 2-19 or F\gure 2-2O.

Stress concentration factors must be accounted for inthe surface discontinuity between the rectangular surfacerfthe lug and the circular surface ofthe pipe. The mem-lrane stress concentration factor for Dure tension or:omoression is determined bv

/ \0.65

:< : r+l t I

\5 6w/

MdN,Nd

KrK,

0.911.48

1.68r.2

1.760.88

t.2r.25

Table 2-4Recommended Minimum Weld Sizes for Plates(2-22)

:rd the concentration factor for bendine stress is deter-Sned by

Thickness t ol Thicker Minimum Size, w' ot FilletPlare welded (in.) weld (in.)

(2-23)

; here w, the weld size, is given in Thble 2-4 for various:-ite sizes. These values for w are only recommended

=,*(-t'\"\9.4wi

t<Y2tlz1t13/c3l+<t1lt/zlrlz<t<211c2t/c<t<6

6<t

3/ro

51rc

3/z

!25/s

72

100

Mechanical Design of Process Syslems

Figure 2-18. Membrane force, N-6/P/R., induced by radialload P [3].

and the engineer should use whatever sizes are actuallyto be used in practice.

The total circumferential stress, ox, is determined byusing these factors in the following equation:

+ : H (9) : '"'r"'o''"r bending stress

The total longitudinal stress is thus found by addingthe two stresses,

(2-2s)

The longitudinal stress and circumferential stress rep-resent the secondary stresses in the pipe wall. Thesestresses must be added to the primary stress which, in thecase of internal pressure in the pipe, is the pressurestress. The pressure stress is determined by

,, N" ,, 6M,o"T-ô *

I€tEz l:\ P"GD)

OD: - . DSI'2t Q-26)

*: "(9 **,(9 (2-24)

The longitudinal stress, ox, is determined in a similarway. The membrane force, N*/(P/R.), and the bendingmoment, M*/P, are determined from Figure 2 -17 or 2-18and Figure 2-21 or Figure 2-22, respectively.

These parameters are used to determine the longitudi-nal membrane and circumferential bendins stresses.where

\ : (\) (i'l = r-onnituainal membrane stressr \P/RJ \R.V

Thus, the total stress for each secondary stress is as fol-lows:

oT = q6+ op

oT:qx+op

Q-2',1)

Q-28)

where o1 < 2oa 1 oy

Often, with large piping, a simple lug will be overlystressed because of localized stresses at the lug-pipe con-nection. When the lug attachment dimension, c, becomessmall to the pipe radius, a clamp is normally put aroundthe entire pipe with the lug attaching to the top of theclamp. This reduces localized stresses at the pipe wall byadding extra metal. This same principle applies to vesselnozzle reinforcement, which is discussed in Chapter 8.

SPRING SUPPORTS

These supports provide loading to a pipe that has un-dergone displacement. Simple supports are no longeruseful if the pipe raises off and loads are transfered toother supports or fragile equipment nozzles. To ensuresupport for the pipe while it moves, a moving support isdesired. The most practical device to fill this require-ment is the spring.

Springs come in two basic categories-variablesprings and constant springs. The former, which is by farthe most common, provides loading to a pipe at a fixedspring rate, lb/in., but the amount of force to deflect thespring varies with the amount of deflection. This forceversus spring rate is a linear relationship and is the rea-son for a "variable" spring. The constant spring is a


MxT.ol

Mx-T

p.30 .35 .40 45 .50 Figure 2-21. Bending moment, M"/P, induced by radial

load P [3].

Flgwe 2-22. Bending moment, M"/l induced by ra-dial load P [3].

spring that will provide the same spring rate for anyforce great enough to cause initial deflection. Constantsprings are used in critical installations where forces ordeflections induced on the piping system are critical.These springs are considerably more expensive than thevariable types and are usually avoided by piping engi-neers when not needed.

Constant springs provide constant supporting force forthe pipe throughout its full range of contraction and ex-pansion. As shown in Figwe 2-23, this constant supportmechanism consists of a helical coil spring working inconjunction with a bell crank lever in such a manner thatrhe spring force times its distance to the lever pivot is al-ways equal to the pipe load times its distance to the lowerpivot. Thus, the constant spring is used where it is notdesirable for piping loads to be transferred to connectingequipment or other supports.

Variable springs are used where a variation in pipingloads can be tolerated. As an example, consider the fol-Iowing example shown in Figtre 2-24. The spring isabove the pipe and is attached to it with a rod and clevis.This arrangement is called a spring hanger. As seen inFigure 2-24A, the spring supports the weight of the pipeand insulation. As the pipe heats up and expands it


moves upward. The amount of deflection, A, relates tothe amount of excessive force as

F" : AK' lbwhere K : spdng constant of spring, lb/in.

A = deflection, in.

(2-29)

It is common practice to calibrate the hanger in such amanner that when the piping is at its operating (hot orcold) condition, the supporting force of the spring isequal to the weight of the pipe. This means that the maxi-mum variation in supporting force occurs when the pipeis in the down condition, when primary stresses are non-existent because of no internal pressure. Therefore, inthe cold position, the suppo ing force of the spring is

F:F"+WP

where WP = pipe and insulation weight

(2-30)

To reduce the amount of variability, it is desirable touse the smallest type of variable spring provided that thedeflections will not exceed those of the spring range.Typical spring sizes and ranges are shown in Table 2-5.

--F _

Flgure 2-23. A constant load spring support provides constant.rpport loading in critical situations.

(A) F=WtCold Position

In thls case, hot = operating condition, cold = down condition

Figwe 2-24. The "cold" and "hot" loading positions ofa var-iable spring hanger.

(B)

Hot Condition


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It is common practice to utilize the smallest spring sizepossible.

In critical and large systems, spring loadings should beeYaluated by computer analysis. Often, in large systems,piping movements are not intuitively obvious and errorscan result because the entire system must be evaluated ifa correct analysis is to be obtained. In most systems,hand calculations are far too cumbersome.

Occasionally, springs are used as moment resisting de-vices, as shown in Figure 2-25. In such an application,the spring preloads the pipe in a specific direction. Asthe pipe expands or contracts, the spring counters therbrce created by the movement and, thus, reduces themoment at an end connection. Such a system in normalpractice usually works in the operating mode but whenihe system shuts down the spring overloads the piece ofequipment protected in the operating condition. Thus, ifsuch a scheme is used, care must be taken to ensure thatCre protected items are safe in both the operating andJown conditions. These schemes can be avoided by use..'i MRS devices where space does not warrant pipingt-lexibility.

lrcation of spring supports is of critical importance.\\-hile springs should be placed where they will be most3fficient, often such locations are undesirable from tleitructural engineer's viewpoint. The piping engineer:hould always be cognizant of available structural steelLrr concrete and loads to be placed on structures. Mostprings are supported from above at either mid spans or

at elbows. Many times it is desirable to support the pipetiom below. When using this type of spring, one must be.autious of pipe movement, as excessive movement ( >


l/a in.) will cause such a spring to jam, as shown in Fig-rre 2-26A. To avoid jamming, a guided load column isused to prevent such a problem (Figure 2-268).

Springs are often used to support equipment to reducenozzle loadings, which are discussed in Chapter 6.

EXPA]ISION JOINTS

These devices accommodate movement in pipingcaused by temperature changes. Such items range fromspecial slip joints that only allow movement in the axialdirection to corrugated bellows joints that can be de-

signed to accommodate movement in several directions.It is the latter type that we will concentrate on, as theyare by far the most numerous and complicated of expan-sion joints.

Corrugated bellows expansion joints have a bad repu-tation with some users because of ignorance. Many bel-lows expansion joints have been incorrecdy specifiedand the consequences attributed to the device itself. Thisis unfortunate because this device is invaluable when ei-ther re-routing the line is impossible or cold spring orother alternatives are not available. The surest way toavoid problems with bellows expansion joints is to havethe piping (stress) engineer specify the unit and to pro-cure the unit from a reputable manufacturer.

The bellows expansion joint is like the MRS devicediscussed earlier because the more the unit is required toaccomplish the more complex is its configuration. Thesimplest corrugated bellows expansion joint is the single

\lvla = momenl generated bYmovement at Pl A

) t" = moment generated by spring

)L2Nozzle

Direction oI Pt A movement

Figure 2-25. Utilizing a spring to counter a moment generated by piping is appropriate only when the spring movement, Ms, does

ilot overload the nozzle or overstress the piping system when the latter is in the down condition and there is no movement at A.This condition is required after the operating condition is met.


Figure 2-26'. (A) Enough piping movement will cock load flange andjam spring. Note: arrows indicate direction ofmovement.(B) A guide load column shown here will prevent situation in 1a--;. ttreie arJvarious designs for guide load columns, but for pipemovement greater than t/+" one should consider a column with rollers or Teflon on top;f the ioad flange.

bellows type shown in Figtxe 2-27 A. This specific jointis shown with flanges welded on each end, but is avail-able from manufacturers with pipe spool pieces on eachend to enable the unit to be directly welded into a line.The piping engineer should try to utilize this type ofjointwhenever possible because of economy and simplicity ofoperation. The single bellows is free to move in all de-grees of freedom except about the longitudinal cen-terline. In fact. no bellows expansion joint can accom-modate torsion and any tendency for the pipe to exert ahigh torsional moment could seriously damage thejoint.

External restraints are placed on the joint to restrictmovement in one or more degrees of freedom. Such de-vices are tie rods or hinges that restrict movement orpressure thrust. Figve 2-278-E are examples of jointsthat are so restricted. Following the same nomenclatureshown in Figure 2-10, we consider each joint in a three-dimensional axis system. KTX, KTY, and KTZ are thetranslational stiffness values lb/in., about the X. Y. andZ axes, respectively. KRX, KRY, and KRZ are therotational stiffness values, ft-lb/deg, about the X, Y, andZ axes, respectively. For all bellows joints,KRZ: (1.0 x 10') ft-lb/deg, as previously srared.

ln Figure2-27 A, we have finite values for KTX, KTy,KTZ, KRX, and KRI the joint is free to translate aboutthree axes and rotate about two axes. The bellows does

not support its own weight so this joint would not be de-sirable where each end exceeds the maximum amount ofpipe span shown, as calculated by the following equa-tion:

L:0.131

E : modulus of elasticity of pipe material, psiI = moment of inertia of pipe, in.aP : design pressure (psig)A : bellows effective area. in.2K : axial bellows stiffness (KTZ in F\gure 2-21)

The maximum length of unsupported pipe implies thatthe unit itself is within this length. Preferably, the joinr isclose to one support or nozzle to avoid excessive deflec-tron.

Thejoint in Figure 2-278 has values ofKTX and KRyand KTY : KTZ : 10e lbs/in. and KRX = KRZ = 10elbs/deg. This means that the joint is free to translate inthe X-direction and free to rotate in the y-direction and isrigid in all other directions. This type ofjoint is called a"hinged" joint and is self-supporting in the y-directionshown in Figure 2-27 . Placing high vertical loads on ajoint must be approved by the manufacturer.

::

aa

The joint in Figure 2-27C has values of KRX and KRYirut absolute values of KRZ, KTX, KTY, and KTZ. Nor-nally, these joints are used in pairs to allow rotationlbout two axes similar to swivel or ball joints and notellowing any translation. This action is seen in Figure 2-t8.

The unit in Figure 2-27D is a pressure balanced uni-r ersal joint. It is free to move about all degrees of move-:nent except KRZ and is restricted by tie rods that bal-.rnce pressure thrust. This type of joint is very commonin engine exhaust systems.

Figrre 2-278 depicts one of the most complicated ex-pansion designs-an in{ine pressure balanced expansion'oint. This joint eliminates pressure thrust, is self-sup-rorting, and does not require a change in the piping sys-:em to install. It is desirable where structural supports.ire not available and a joint is needed because flexibilityrs required of the piping.


Pressure thrust is the amount of force generated by in-ternal pressure and is simply internal pressure timesminimum bellows radius area (PA), lbs. This force canbecome quite high as the pipe size and the internal pres-sure increase. In many applications, the piping itself isanchored and the joint is allowed to compress when thethermal compression force exceeds the pressure thrustforce. As seen in Figure 2-28, when movement in theform of lateral translation is desired (KTZ and KTY), tierods are used to restrain the joint in the axial direction(KTZ: @). If tie rods are being used to overcomepressure thrust, then any equipment flanged to the jointshould be able to withstand the load reouired to over-come pressure thrust. Generally. tie rods are only used topermit lateral movement.

Bellows expansion joints can be restrained and com-bined in pairs or trios to perform certain tasks. It must beemphasized that just because a joint is free to move in

A

)

{:,

c

n

,-T-.Fz

Ftgure 2-27. Types of bellows expansion joints: (A) flanged-flanged end simple bellows joint; (B) hinge bellows expansion::nt: (C) gimbal bellows expansion joint; (D) pressure balanced bellows expansion joint; (E) "inJine" pressure balanced self-

'-:oorting bellows expansion joint. (Courtesy of Pathway Bellows, Inc.)

.%^

Figure 2-28. Generally the use of tie rods is to allow only lateral movement. (Courtesy of Pathway Bellows, Inc.)


-LATERAL MOVEMENT HOT

directions KTX, KTY, KTZ, KRX, and KRY does notmean that the corresponding stiffness values are small.As internal pressure and pipe size increase the values ofKTX, KTY, KTZ, KRX, and KRY increase, because thebellows wall thickness increases to resist increased inter-nal pressure. The bellows can be a single wall construc-tion (single ply) or multiple wall construction (multi-ply)and the stiffness values vary with each manufacturer.

Some people erroneously think that the purpose of us-ing bellows expansion joints is to make the pipe stressanalysis unnecessary. Such is definitely not the case, be-cause values of stiffness in each direction must be en-tered in each computer stress run so that it can be vei-fied that the displacement and piping loads are notexcessive to the equipment nozzles. As shown in Figure2-28 a pipe can either be properly guided or anchored,and such restrictions should be modeled into the com-puter stress analysis.

The piping engineer is encouraged to refer to the Stan-dards of the Expansion Joint Manufacturers Association(EJMA) t4l in accessing piping layouts when using bel-lows expansion joints. Also, it is desirable to specify thejoint such that the manufacturer is required to meetEJMA requirements. One should follow EJMA guide-lines and requirements, and include modeling restraintsand stiffness values in computer stress analysis to verifythat attached equipment is protected. Expansion jointsare not cataloged items to be bought at random but rathersophisticated pieces of equipment that must be engi-neered into the piping system. With this approach, theuser should not expedence any problems with bellowsexpansion joints.

lA: lnt€rmedlaleAnchorPG: Planar Guid6TUEJ: Tied Univorsal Expansion Joint

PRESTRESSED PIPING

Piping systems are sometimes prestressed to reduceanchor and restraint forces and moments. This prestress-ing of the pipe is best known as cold springing, but isalso called "cut short," meaning that the pipe is cut shorta percentage of the amount of thermal expansion ex-pected. The opposite is true in cold systems where thepipe contracts, so the pipe is fabricated extra long, withthe extra length being a percentage of the amount of ther-mal conEaction expected. This procedure is best knownas "cut long." Some refer to cut long as "hot springing,"which may cause confusion because it is not as popularas the term cold springing and to some it means hotforming, which hds nothing to do with fabricating thepipe extra long.

"Credit" may not be taken for prestressing the pipe incomputing the stress in the piping system. Several pipingcodes are specific about this and, if the piping is over theallowable stress range, one cannot cut short or long tolower the stress. However, credit may be taken for an-

chor and restraint reaction forces and moments.The procedure of cutting short or long involves a per-

centage of thermal movement. The whole purpose of theprestressing process is to balance the forces and move-ments between the down and operating conditions. Thus,cutting short or long 1007o (i.e., cutting short or long theexact amount of thermal movement) is normally notdone. Exceeding 100% is not recommended and doesn'tmake good sense. Normally, the amount cut is 50% and

should not exceed 66% of the thermal movement' Thereactions, R6 and Rp in the operating and down condi-

n":lr-?*l&- \ 3 lEo

Ro: XR

---: <4where

tions, respectively, are obtained from the reactions R de-rived from calculalions based on the modulus of elastic-ity at ambient temperature, 8". The relationships are asfollows:

ot

IR" : 11 --twhichever is greater, and with the additional conditionthat

1.0

X : cold (or hot) spring factor ranging from zeroto one, one being 100% cold or hot spring

E : computed expansion stress. psiEp : modulus of elasticity in the down condition,

psiEe : modulus of elasticity in the operating

condition, psiR : maximum reaction for full thermal movement

based on Ep which is the most severecondition. lb- or in.-lb

Rp : maximum reaction in down condition, lb orin.lb

Ro : maximum reaction in operating condition, lbor in.-lb

These formulations are not necessary nor desirablewhen computerized stress runs are made. All reactionsthat result from prestressing the pipe are much more ac-curately made by a computer. However, one is not al-ways privileged to use a computet especially at remotesites, so these formulations will yield conservative ap-proximations to feactions.

The biggest legitimate objection to prestressing thepipe is that often it is simply not done by the pipe fabrica-tor or construction workers. The orocess is often diffi-cult, especially in large pipe, and is unpopular with fab-rication personnel. When schedules get tight and peoplefall behind on the schedule, there is a tendency to over-look prestressing the pipe. To avoid such a problem,some large engineering companies issue cold spring re-ports that are signed off by inspectors. However, suchreports get lost fairly easily, unless a rigid system is im-plemented to treat them as control documents. There iscertainly nothing wrong with prestressing the pipe, ex-cept maybe a little extra paperwork.


Bellows expansion joints should be avoided if a moreeconomical and practical method is available for provid-ing flexibility oi restraint to the pipe. ln many ipplica-tions, only the bellows expansionjoint will suffice, e.g.,movement and vibration in straight runs of pipe at ele-vated temperatures between different pieces of equip-ment can only be compensated by bellows joints. How-ever, as the joint becomes more sophisticated andthereby more expensive, other alternatives should beconsidered. Such alternatives lie in either the flexibilityor stiffness methods Dreviouslv discussed.

FLUID FORCES EXERTED ON PIPINGSYSTEilS

When fluids move in a piping system, they import en-ergy to the system when they are forced to change direc-tion by the pipe. In other words, it requires energy tochange the direction of a moving f luid . This fundamentalfact is known as the impulse-momentum principle, ex-nressed as:

l)l ph : Mv, - Mv,\-]

(2-31)

This states that the change in momentum in a system re-mains constant during the exchange of momentum be-tween two or more masses of the system. Applying theequation to that of a pipe elbow shown in Figwe 2-29,we apply the principle to obtain:

Mvxr+DFxxt:MVy,

Mvyr+DFyxt=Mvy2

].

b

(2-32)

(z-33)

where t = 1 for unit timeforce in horizontal direction exerted by thebend on the flowing fluid, lbforce in vertical direction exerted by thebend on the flowing fluid, lbhorizontal velocity component at bend inlet,ftlsecvertical velocity component at bend inlet,ftlsechorizontal velocity component of bendoudet, ft/secvertical velocity component of bend oudet,ft/secWgi g" : fluid massweight of fluid in bend, lb,local acceleration due to grayity,approximately 32.2 ftl serzdimensional consiant : 32.17 lb-ft/lb1sec2

M=

6-

the analysis of chemical rocket engines is suitable for es-timating reaction forces. These calculations in such ananalysis agree with those reactions comDuted bv othermethods and have been found to be slightly conseivative.The method presented by Hesse [5] is desirable becauseof its simplicity and accuracy, and knowledge of the pro-cess fluid is limited onty to the specific heat ratio, k, andthe molecular weight, M. The derivation and explanationof the formulation is given by Hesse [5].

Consider the nozzle shown in Figure 2-30. The reac-tion force developed by a fluid exiting the nozzle is givenby the following:

Figure 2-29. Pipe reactions induced by change of momentumof fluid flowing through elbow.

When applying Equations 2-32 and 2-33 to reliefvalves, the fluid dynamics of nozzles must be consid-ered. The dynamics and thermodynamics of fluid motionthrough nozzles is a very involved subject and ratherthan investigate the various theoretical methods in thisbook, we will only investigate the various results anddiscuss their merits.

Relief valves can exert enormous forces when fluidsexit the nozzles. Often, the fluid exits the nozzle atspeeds exceeding Mach 1. Numerous private companies,as well as the ASME and API, have developed proce-dures to approximate such fluid forces. The ASMEB31.1 gives a method for computing the reaction forcesexerted by relief valves. The main drawback to thismethod is that it applies to steam only, because Code831.1 governs only power piping. Steam is one of themost comprehensively defined substances, with all prop-erties well known and published, but such is not the casewith many chemical processes. The 831.1 method re-quires that the properties of the substance be we de-fined, to the point of being rather cumbersome to use.The ideal method would require the fewest number ofphysical properties, but still provide the necessary data.One such method that is very easy to use is the ApI for-mulation in API 520 Part II, paragraph 2.4, which isused for gases or vapors. This formula loses accuracy asthe flow rate approaches Mach 1, so another method isdesired for predicting reactions at all flow rates in pro-cesses that have poorly defined properties.

The aerospace industry has done much research in thestudy of fluid dynamics and thermodynamics of nozzleflow. Because relief valves operate in a closed system,

[/ "'.2 \/ " \,-,F = \cac"A,P. I l= ,l l-i ,l-t\K-l/\K+u

',['where \

Ca=

lo.t

I *+tn-P.l (2-34)

nozzle correction factor : l/2(1 + cos o) - 1.0 for most relief valvesnozzle discharge coefficient, which0.97 < Cd < 1.15, normally Cd > 1.0specific heat ratio : CplC"

c = nozzle inlet sectiont = throat-where critical condition existe = nozzle exit sectionef = gffrat;rr exit section-where exhaust

gas pressure first equals ambientpressure, Po

l/r,/

-'.--7l\-

tFigure 2-30. The relief valve mecharusm.


cv: (rJ05 = 0.95 to 0.98nozzle adiabatic efficiencynozzle exit pressure, psiaambient pressure : 14.7 psianozzle exit area, in.2nozzle throat area, in.2

critical pressure : P, l---- I '\K -f l/

critical pressure ratio, determined tromFigure 2-31molecular weight of fluid, lb/molegas flow rate, lb/sec, whereG:

(2-34a)

these exDressrons mto Equation 2-34, we

G = C.A,p" [r- {_z\:ll- '[Rr.\k + u I

T" : critical temperaiure : ffir,, 't

.T,

Substitutinghave

/r. - r\I + l-.--1l M2 for adiabatic process

F : 1 028A,P" [-+ (#= [' - (,:)*]]"'

+ A"(P" - P,) (2-35)

Equation 2-35 assumes that the flow is isentropic andin addition to relief valves, includes turbines, compres-sors, jet engines, rockets, injectors, ejectors, and atomi-zers. Most nozzles used in current applications are eitherconvergent or convergent-divergent, also known as De-Laval or Level nozzles. Convergent-divergent nozzlesare used for high pressure ratios and supersonic flow,and convergent nozzles for low pressure ratios and sub-sonic flow. Thus, relief valves are Level types that canhandle high pressure flow.

Critical pressure of gas occurs at the point where thefluid velocity becomes Mach 1. This pressure is ob-tained at the minimum area of the nozzle and this mini-mum cross section is called the nozzle thrcat in the De-Laval nozzle. In the convergent nozzle, the cross sectionof minimum area is the exit section. The critical velocitycan be expressed in terms of the inlet temperature be-

cause the process is considered adiabatic making the to-tal temperature constant. Thus,

t)lrc = l-l r,. oF

\K + r,t

and

/"_,..-,f 5

vc=V.=(skRrJo5=|\#',f (2-36)

Reaction forces produced by relief valves can becomequite enormous and should not be overlooked. A struc-tural failure of a relief system could well result in a ca-tastrophe.

EXTRANEOUS PIPING LOADS

Vibration can be a real hazard in piping systems. Usu-ally, vibration problems that occur with piping have twosources-pulsations generated by reciprocating equip-ment and wind. Pulsation shock phenomena on rotatingequipment is briefly discussed in Chapter 6. The phe-nomenon of wind-induced vibrations on piping along talltowers is discussed here.

Wind-induced vibration is caused by vortex sheddingon the cylindrical surface of the pipe, and becomes aproblem with piping more than about thirty feet long.Vortex shedding usually occurs with piping that runs upalong the height of a vertical tower.

Analyzing and solving vortex shedding vibration prob-lems can best be handled by applying certain principlesthat include dimensionless parameters and experimentaldata. Sophisticated digital computer models are possible,and recently, vortex streets have been simulated withflow patterns around piping and structures. Such com-puter simulations are rigorous and expensive, so withcurrent software they are impractical to use for all pipingthat may be exposed to wind.

Several proposals have been made concerning vortex-induced vibrations around cylinders, but perhaps themost straightforward is the work by Belvins [6]. He de-veloped a dynamic model for vortex-induced vibrationusing random vibration theory. The theoretical basis is arepresentative spanwise correlation and cylinder ampli-tude is presented as a function of the vortex forces.When the state of resonance exists, the amplitude of thecorrelated lift force on the cylinder is represented as acontinuous function of cylinder amplitude. Also, at reso-nance, the spanwise correlation of the vortex force ispresented as a function of the characteristic correlationlength. This model is limited to the resonance of a singiemode with vortex shedding and a Reynolds number in


1009080

70

60

o"A,

10

8

7

6

Figure 2-31. Critical pressure ratroversus area ratio for various fluid spe-cific heat ratios (k).

80 lOO 2OO /rcO 600 80O loOO'clPa

the rate of200 ( Nn" ( 200,000, where a well-formedvortex stre€t exists.

Figtre 2-32 shows flow regimes of fluid flow acrossstationary circular cylinders, and illustrates how the vor-tex streets tend to separate as the flow velocity increases.Between the range 300 NR" < 300,000 thi region iscalled subcritical because as Nx" approaches 300,000,the boundary layer becomes completely turbulent andthe vortex shedding effect is lost. One parameter used inanalyzing vortex phenomena is the Strouhal number(s),

which is the numeric constant between the resonant fre-quency of vortex shedding (f) and the cylinder diameter(D), divided by the free stream velocity (V). This is ana-lytically written as

^ f.DQ-37)

For circular cylinders, the Strouhal versus Reynoldsnumber is shown in Figure 2-33. In a structure, the ob-

,,4.* 3,'o..' i. <35, o.L^YE i BAS UIiOEROOXE

-u/////t e' IYi?Y,",li,t lt6%i!13i*,,*.*[i'"-W voirEx srREtr rs.PP.REnt

3,5rro'< R. < cO t?lR€'€SJA8LISITTEIT OF Tt1€ TURAU.IENI IORIET SIREE' II]AI

'ASEvrD€tT rx 3oo< i.? 3r|o:Ih'S IIME TIE SOUNDIRY LAYER

lno THE uAr(€

Figure 2-32. Fluid flow regimes across circular cylinder l7l.

REVIIOLDS U E€h! R.

Figure 2-33. Strouhal-Reynolds number function for circularcylinders [7].


The response of a right circular cylinder at resonancewith vortex shedding is a function of the following:

Damping : (2zs)2 6.

where 6. : reduced damping

and

_ 2m(2rl)pW

(z-38)

| _ energy dissipated per cycle" 4?r(total energy of structure)

Mode shape : VW: i for a rigid cylinder)Aspect ratio : LlD, L = length between spans

The amplitude Ay/D can be approximated by loadingthe pipe with a uniform wind load and using the maxi-mum deflection as Ay. This can be used in Figure 2-34to estimate the damping at resonance for a given aspectratio. This damping is then compared with the naturalfrequency of the piping. The natural frequency of thepipe, especially for complex geometries, is computed bymodal extraction computer analysis or any other dy-namic computer software that computes the natural fre-quencies of piping systems. For short straight spans, thenatural frequency can be determined by comparing val-ues obtained from Table 2-6 and with the resonancedamping frequency in Figure 2-34.

In practice, the greatest problem with vortex sheddingoccurs on tall vertical towers when pipe four inches andsmaller is uninsulated and left hanging without support.It has been found that once insulation is applied to thepipe resonance vanishes.

The following simple guidelines will enable you toavoid the vast majority of wind-reduced vibrations:

l. Increase the flexural stiffness of the pipe so that itscritical velocity is above the range of moderatewinds.

2 . Use damping devices to restdct the amplitude of vi-bration.

3. Reduce the effective length of the member by usingintermediate struts.

4. Attach spoilers to the pipe to disrupt the flow nearthe tower surface; this impedes the formation ofvortices and thereby eliminates the cause of vibra-tions.

5. Span the piping supports at uneven intervals to pre-vent a periodic wave function from developing.

The analysis of wind-induced vibrations on tall verti-cal vessels is discussed in Chapter 4.

ject of design is to avoid resonance. If the inverse of theShouhal number < 1, where f is the natural frequency ofthe structure, then resonances with vortex shedding fromthe first, second, and third harmonics are avoided. Thiscan be accomplished by adding mass, such as insulation,and putting pipe support spacing at uneven intervals. Ifpipe supports are spanned evenly, periodic wave motionscan form, resulting in resonance.

Etl_g RcGrMa OF urrsEpARArEo FLO{.

voRTrcas rfi rnE w^(€.

Eg-5-E!-l-3aq

lOO<i.t 3r|ol

'rwo nEGrMEs lr wHrcH vofiTEr

PERTOOTCT'IY COVEFI|ED l'r LOWR. RII{CE BY I/(E

PERIOOICITY GOVERNEO II{ tiIOHR. RING'

'Y VOR'IEI

IRAXSIIIOI NINGE TO IUiBU.

voRrEx sritEt rs Frrl'tl


Table 2-6Natural Frequency ot vibration ot Beam Elements

Concentraled Load on RelatlvelyLight Beam or Spring

ffi

Uniform Load on Beam Supportedat Ends

Unitorm Load on Cantilever Beam

. / \0.5r -

t l9l'- t\-Di f: (3.55XD)-0 5 f : (3.89) (D) 0.5

that vibrate with itf : natural frequency of vibration. cycles per secondD : maximum static deflection of member under its own weight plus any weights

or r,rer r.re = r stn r3l s2/p#

Figure 2-34. Damping, d (dimensionless), versus amplitude, Ay/D (dimensionless).



EXAilPLE 2.1: APPLYIilc THE STIFFNESS]UIETHOD TO A IIODULAH SKID.IIOUIITED

GAS LIQUEFAGTION FACILITY

Figure 2-35 depicts the preliminary piping design of agas liquifaction plant mounted on a skid module. Spaceis severely limited, as the equipment and piping are lim-ited by the structural steel skid supports, so such devicesas piping loops are unthinkable. Expansion joints are notallowed by the client, because high-pressure hydrocar-bon gas is highly combustible and an expansion joint fail-ure would mean certain gross property damage and pos-sible loss of human lives. Therefore, the piping engineermust utilize the stiffness of pipe supports to transferloads from the piping to the structural steel rather than tothe equipment nozzles. This transfer of loads is not total,but enough to guarantee that the equipment nozzles load-ings will not exceed allowable levels.

For the stiffness method to work, the piping configura-tion must be flexible enough for the piping itself to bewithin allowable stress limits set by the applicable code.This is the first significant criterion, because if the pip-ing exceeds the allowable stress range in any part of thegeometry, the system design is faulty. Conversely, thepiping system can be well within the stress range and theequipment nozzles still be overloaded. Thus, the pipingitself must have a certain amount of flexibility to bewithin code allowables. The piprng supports must bestiff enough to protect equipment nozzles from excessiveloads. Here our case has been stated; adding additionalflexibility is not acceptable.

From computer calculations the original configurationin Figure 2-35 is found to be overstressed and the expan-sion stress exceeds the ASME 831.3 allowable stressrange provided in Equation 2-4 for 3O4 SS pipe. There-fore, the piping must be changed to bring the maximumstress within the accepted stress range. This analysis in-cludes the nozzle movements shown in the figure. Eachnozzle is considered as an anchor.

Figure 2-36 shows the final configuration after severaliterations are made to determine what configurationwould best suit the structural limitations set by the mod-ule skid. This configuration is found to have a maximumallowable well within the stress range of ASME B3 1 . 3 .

To achieve this acceptable stess, a limited amount offlexibility must be added to the system. Thus, regardlesswhich method is used-flexibility or stifftiess-a certainamount of flexibility is required to make the piping sys-tem operate properly.

Once we have obtained the minimum flexible configu-ration required, we now focus our attention to the equip-ment nozzles. To consider this question, we must distin-

guish between the types of equipment. The heatexchangers HE-A and B shown in Figures 2-35 and2-36are aluminum plate exchangers, and the cold separatorand power gas volume tank are made of reasonablythick-walled stainless steel. Thus, the critical items arethe aluminum heat exchangers. The line between pointsl0 and 25 in Figure 2-35 must be cut because the relativeZ-movement between these points overloads the nozzlesat points 5 and 30, creating a very high y-momenr and Z-moment, because the pipe wants to move in the -Z and*Y directions. These movements can be accommodatedby using certain structural devices, such as shown inFigure 2-37. Even though flexibility has been added tothe system to get the piping within the allowable stressrange, the equipment nozzles are still overloaded by ex-cessive moments above the X, Y, and Z axes-M;q, My,and Mz.

To counter the movements of the piping at the nozzlesnumbered 5 and 30 at HE-B and A, respectively, variablesprings are placed to support the pipe while allowing thepipe to move at the same time. One spring is placed atpoint 20 with a simple Y support added at point 56.These additional supports help reduce the moments atnozzles 5 and 30, but not enough. So, we must add MRSrestraints (see Figures 2-9 ard 2-10) in pipe members 5-l0 and 30-35. Each MRS is designed to allow nozzles 5and 30 to move upward but to transfer moments M;,Mv, and Mz from the pipe to the structual steel below.Also, each MRS allows pipe members 5-10 and 30-35freedom to nove along the axis so that we have the fol-lowing restraints at each MRS: KTZ, KRX, KRY, andKRZ of Figure 2-38 (see Figure 2-10). Thus, we haveone translational and three rotational restraints, eachwith a stifftress value K in lb/in. or ft-lb/deg. The pipeand exchanger are free to translate along the X and Yaxes.

One can readily see that the MRS restraints must allownozzles ar points 5 and 30 to move upward, as the ex-changers are bolted down to structural steel higher up onthe units. Restraining the nozzles from moving upwardwould anchor the unit at the nozzles and at the supportpoint causing the exchangers to rupture. Pipe members5-10 and 30-35 must be allowed to move along the x-axisfor thermal expansion.

We now have the conceptual model of what the solu-tion looks like and the next step is to finalize the details.The MRS restraints are resisting forces and momentsshown in Figure 2-38. It is necessary to design the re-straints such that each has enough stiffness to transfer theloads to the steel and protect the nozzles at points 5 and30. We will now compute the support stifftress valuesKTZ, KRX, KRY, and KRZ. Once these values are de-termined, they can be input back into the computer runand verified to be sufficient for the nozzles.


\,zr/\/W\\\\\\\\\ \\\ \'"\ \p\ 't 'a\--l 2,r^

o

ao

o

'4

q

o_-47

7^

, 4e.t 2l/o

o.,

z

tsJ

"32:1

e-:x $-

F

o

aoctNo

.DIL

9

z


o

',',

@ GnTNNELL slzE a rYPE D

@ venncr suPPoRT(sEE F|GURE 2-9)

@ GR|NIELL stzE 10 rYPE D SPR|NG

OHAr{GER suPpoRT


Figure 2-37. Plan view of location of MRS supports.

TO ENSURE PFOIEC'IoN

OF CAF8ON STEE L COSPONENTq

OISIRIB(IIION OVER LE GIH

\\ --- rtrslrL^r

--' +

STSE SOPMRT PIPE

Flgure 2-38. Two-axial translation-free multiple moment restraint support(BIAX-MMRS). Arrows indicate d!rections of freedom of movement.

€IIDE DIATES ASSEMELY }IOUSING

-_4

Mechanical Systems

-tb)o lb

ioo ft-rb

-x' ;> [ rr-r-owso roTRANSLATE

tt !x otngcrtots

For Torsion

T : 12,800 ft-lb

For Shear

._ _ 4(R2 + Rr + 12) . I"igit'1 :"it jotffi''. Tc,r:a:

(12,800)ft-rb (r., J (*, )3(R + r) ---- 4oJ0 in.o

4(10.976 + 9.545 + 8.297) = 1.975 in.316.194)

: 12,563 psi

For Tensile Stress

" : p_ : 1.20 t! = r+t prt- A 8.40 in.'

Shear DistributionAt Point A-

O : 22.83"For a circular .thin-walled cross-sectron'

Q : 2R'zt cos o

I-T--q

I

A_-$-Q.?.'_-* r,

rl4lc,

"=+:+i:(r,roonb (ry) in.'?(r.e75) in.

(40.49) in.a (0.432) in.

- n 16'-'\-625 5.76 I+

4(0.432) cos (22 83") = 7 '635

rs : 521-648 psi-max at neutral axis

For Bending

Mx : -(1,100 lbx3.0) ft - 700 ft-lb : -4,000 ftlb

Mz = -9,500 ft-lb

Mn : R(Mx, M, : (-4,000)" + (-9'500)tlo 5

: 10,308 ftib

/.^. \10,308 ft-lb lY

rn'l

M \ln/o =; = ----r,j/ti.-= = 1o,114Psi

(1,100) (7.635)= 480.143 psi

(40.49) (0.432)

At Point B-

Q = 2t9.588t (0.432) cos (67.166') = 3.215

_ ( 1,100) (3.215) _ 202.182 psl' (40.49) (0.432)

At Point c-A:?)R'lo=r= 480.143 psi


r=202.182+12,563

7 = 12,765.182 psi

Stress Elements

Point A is the most critical point.

,-, _ ox - oy * [/o* - o"'l', ",

1ot

2 -t\ 2 I ^Yl rov-0

lo,2s7 [h0,257\':+

-t

f2 -l\ 2 l

o : 19,143.674 psi

ro.ms ir-ru ll?:" )\ft/

l_ rr----.lr ____+

<-- ll c ll ----+ 6r=uspsi<-rL_-Jr+

16.81 in.3

Shear

Point A, Q :2(\8.59'17) (0.322) cos (22.83) : 11.039

(l.100) ( l 1.039)' (12.5) (0.322\

o = 7.358 psi + I43 psi = 7.501 psi

r = rs + z1 : 9,138 psi + 520 psi : 9,658 psi

7.501 [h.sor\' ^ -- .lo'6 = -|1J! - ll '--'l + {9-658)rl2 [\21 ]

o : l4,lll.\57 psi < 17,000 psi allowable stress

Therefore, use 8-in. Sch 40 A-312 TP3G4 SS pipe for the3-ft pipe spool piece.

From these calculations, we see that the minimum pipesize for the MRS is an 8-in. Sch 40, 3.0 feet in height.The stiffness values for an 8-in. Sch 40 pipe are as fol-lows:

From Thble 2-1, we have

KTx = l2EI' .

(1 + O)LJ

8.625r:

-:

+.JlJln.2

r 4 ?1?: = := : 0.120 << | - <D :0L JI'.U

12(2s.0 x 109 g (72.5) in.4

<- rfi, -;. 6r:1o,jr4+i/tg

.-- ll A ll --+ oi: 1o,2s7 psi<- {l lt __}-r'

l-, = r. + rr = 480,143 + 12,553

' = 13,043.143 psi

* lf--lf -

o:1oJ14 -143{_ lt e __-f<--ll - ll -----) O=seflpsi

r = 13,043.143 psi

lot(13,043.14311

For ,4312 GR TP 304SS, o"11 : 17,000 psi <19,143.614 psi allowable stress, so, try 8-in. d Sch 40.

Bending

KTX :(36.0)3 in.3

: 540,766.5 lb/in.

KRX -_ KRZ :(36.0) in.

: 233,611,1ll.l in.Jb/deg

4(2s.0 x 106) * (zz.s) in.o

fr-lh= 19.467 .s92.6 -:*qeg

106)

--1 (72.5) in.4

ln-'

o : 7,358 psi

Torsionin

( | 2,800)ft-lb (12) ^ t4.313)in.It' (72.5)in.3

Tensile Stress

l,200 lbo : 870GJ : r+r Psr

= 9.138 psi

or KRX : KRZ

2(29.0 xKRY = (36.0) in.

: 116,805,55.6 in.Jb/deg

or KRY : 9,133,196.3 ft-lbldeg

94 Mechanical Design of process Systems

Entering these stiffness values into the computer run.we see thar lhe nozzle loads fall very sharply it points 5and 30. Further reduction in loads can be obtained bvadding springs abo',e the MRS restraints to counter ;negative moment above the Z-axis. Using springs abovethese supports is not always necessary, but in this casethey are required because of the large vertical movementof points 5 and 30. A weight run should be made to ver-ify that the springs do not ovedoad the nozzles durinsshur-down.

The MRS restraints vary in design and are conceptu-ally shown in Figure 2-10 and Figure 2-39. These iup-ports are made ol interlocking sliding plates wirh eaihsliding surface coated with high-strength Teflon. Theprecise details of such supports vary and are customizedfor each application.

Looking to other parts of the piping system, we noticethat nozzle 75 on the cold separator has a high momentabout the negative x-axis. This moment is attributed tothe aluminum exchangers (HE-A and B) moving upwardand the cold separator shrinking downward. Becausespace is premium and we are "locked-in" and can't addany more flexible piping, we add a spring at elbow 65pulling downward to counter the exces5ive neqative x-moment at nozzle 15. The spring is sized ro b6 accept-able for operating and shut-down modes.

Table 2-7 lists the forces and moments at each equip-ment nozzle.

Upon reviewing Table 2-7 , you will notice the dispar-ity in nozzle loadings. The aluminum heat exchangers,HE-A and B, have lower loads, especially moments,than does the cold separator or power gas volume tank.This is because each has acceptable loadings that are dif-ferent. The cold separator is made of 23la-in. plate stain-less sreel. which makes rhe loads shown easilv acceot-able. {The method of determining whether such-loads ireacceptable on pressure vessels is discussed in Chapter 8.)Such loads would be very unacceptable lor the aluminumheat exchangers because aluminum cannol withstandnearly as great a load as steel and is not very elastic,

therefore making such a unit sensitive to external loads.Always be careful when subjecting rotating equipment orvessels made of light material to excessive nozzle loads.

- In the final analysis the pipe loadings transferred by

the MRS to the steel must be considered by the structuralengineer. who must design the loundation accordingly.Sometimes it is necessary to model the stiffness of thesteel foundation members when nozzle loadines becomecritical.

Figure 2-39. The BIAX-MMRS installed and in operation arolant facilitv.

Table 2-7Equipment Nozzle Forces and Moments

Heat Exchanger AHeat Exchanger BProcess Vessel AProcess Vessel B

-210.5 210.5-553.4 553.44501.9 8217.63163.0 3306.8

t44.7279.O126.2

- 38.5

-255.9-624.6

299.2-2437 .8

00

854.494.6

293.9684. I914.O

2440.0

00

-6175.0877 .1

EXAIIPLE 2-2: APPLYING THEFLEXIBILITY IIETHOD TO A STEAiI

TURBINE EXHAUST LINE

A client has added a steam turbine to a chemical plantand has piped up the turbine with make-shift parts and

existing pipe, plus a newly purchased bellows expansionjoint. When the turbine technicians determine they can-not cold align the turbine with the exhaust piping, the cli-ent decides that the piping must be rerouted, but requestsan evaluation of the system, which is shown in Figure240.

The system is modeled with a computer software pack-age, and the results indicate that a moment about the y-axis in the magnitude of 31,000 ft/lbs is exerted on theturbine exhaust nozzle under operating conditions. Sucha load is well above any turbine allowable. The reactions


along the other axes are moderate and the problem ofalignment must be solved. The extremely high y-momentis caused by the thermal expansion of the pipe memberextending along the z-axis from point 95 to point 145 al-most Ze in. With this expansion along the positive z-axis,the pipe rotates about the positive y-axis from point 20through the expansion joint at point 45 to the elbow atooint 75. This torsion is transmitted to the turbine nozzleit point 5. Thus, the adjustable base elbow support atpoint 31 is entirely useless in resisting this vertical mo-ment and the expansion joint at point 45 transmits all ofthe torsion motion to the turbine nozzle at point 5.

An earlier section discussed the fact that these jointsare totally rigid in torsion-a moment about the axis isparallel to the longitudinal axis, which in this example is

the y-axis. In fact, with the vertical moment as great as

31,000 ftlbs the expansion joint at point 45 will eitherbe destroyed or have a short service life because the bel-

lilri

iii

d-o.^* .u""'"'

Figure 2-40.psia.

Original piping configuration of 20-in. 0 steam line for turbine exhaust: temperature : 300"4 pressure : 16 Hg


lows are not designed to resist such high torsional mo-

ments. Thus, the diagnosis is to avoid the high torsion

and stop the .8-in. movement at point 135. To do this- ec-

onomically with minimum alteration to the piping, a bel-

lows expansion joint is added at point 123 and.the shoe

on the dummy leg is stopped in the *z direction (i'e''movement in the 1z direction is stopped, and the vessel

nozzle at point 85 is protected by the joint at point 123 'An expansion joint is sized based on the manufacturer's

standard dimensions for a 20-in. pipe and the joint stiff-ness values are as follows:

KTX : KTY - 1,500 lb/in.KRX = KRY : 200 in.-lb/degKTZ : l2O lblin.

These values are provided by the joint manufacturer'

The problem of turbine alignment is directly related to

the inabilitv of the turbine technicians to adjust the pipe

because of the pipe's inflexibility, which is caused by the

suided base elbow at point 3l . The base elbow support is

ieplaced by a spring depicted in Figure 2-41 and mod-

eled into the compuier siress program. This mn is made

with the added ixpansion joint at point 123 and. the

spring at point 3. ihe following results were obtained

from the computer run:

SPRIN6

Turbine nozzle (Point 5)-

Fx : 46.51bs, Fy : -530.8Ib, Fz : 343'9 lb, Fr: 634.l lb

Mx : 1,198.4 ftlb, MY : 1,978'2 ft-lb' Mz

= 745.3 ft-lb, Mr : 2,430.0 ftlb

Vessel nozzle (Point 85)-

F" = -46.4Ib, Fv = -3,311.8 lb' Fz: -3,348.5 lb, Fr : 4,709 9lb

Mx : 5,968.7 ft-lb, MY : 9,742 0 ft-lb ' Mz: 5,0?6.0 ft-lb, MR = 12,501.9 ft-lb

The loadings at the turbine nozzle are acceptable' (The

basis for conaluding this is discussed in Chapter 6') The

reactions at point 85 seem excessive and would be for a

steam turbin;, but considedng the vessel is five feet in

diameter and made of 3-in. plate, these loads are not ex-

cessive. Pressure vessel nozzle loading analysis is cov-

ered in Chapter 4, but one can deduce that pressure ves-

sel nozzles tan withstand much greater loads than most

tvDes of equipment.''The svstim is implemented and in two days the turbine

will be fired up and operating well. The concluding re-

marks are that the expansion joint at point 45 is accom-

plishing nothing and the capital expended for its pur-

"hu." riu. wa$;d' ln fact, it would not hurt to move the

unit, but this is not necessiuy since the high torsional

moment has been eliminated.The expansion joint at point 123 was specified and

ourchased for those stiffniss values previously listed'

the final configuration is shown in Figve 2-42

EXAilPLE 2'3: FLEXIBILITY AIIALYSISFOR HOT OIL PIPING

A olant in a remote area of Brazil has an emergency

need for a hot oil system. The plant manager has deter-

mined that a 3-inch Schedule 40 pipe is to be u-sed' based

on plant requirements and available pipe trom local

ioui.".. w" it" to design the piping and ensure it willnot be overstressed. There are no electronic computers

available anywhere near the plant and all calculations

must be made without a stress program'

For a hot oil header extending over some distance the

flexibility approach is the practical method in this appli-

"ution. fit" iiit" it to operaG at 550'F at 50 psig For-aj-

in. Sch. 40 pipe, d :3.50 in. A layout is n-rade o{ the

system and i preliminary loop is shown in Figure 2-43'

The piPe is ASTM A-53B PiPe.

I' Y 1':l .7

V

Figure 2-41. Sketch of spring that replaces base elbow^sup-

oon: installed load :713 lb. operatlng load = /uJ rD'

ipring : 300 lb/in.-": I = 3o : 8.oo: Rv

3.7 5:!:I:z.so

La4


-ttt t*f- I

,-tt^\.

97

tN:z Dlf,Ectro{s

Figute 2-42. Final piping configuration of 20-in. 0steam line for turbine exhaust: temperature :300"F, pressure = 16 Hg psia.

From Figure 2-12A, Ah = 220

B : (4. I l) (_2.9._q x ltr) _ 689.8r72,800

From Table 2-1 and ASME 831.3 the allowable stress is

oe : 1.25(20,000) + 0.25(18,100) : 29,525.0psi

The maximum bending stress is

,"_'-wJ_\

on = A,B [q] : tzzol toss.at tr.sol- \L/ 30.0

: 17,7M.9 psi

This is based on L = 30 ft between guides. Solving forL', the distance between anchors. we have

r^ : rr7.704.9\lL-'l' \L/

(29.525t (30\

" =-njaf =5u.urt

The available steel in the plant in the area the hot oilheader is to be run is spanned 4.5 m or 14.76 ft, makingtlre anchor points spaced at 18 m or 59 ft. Thus, L' =59.0 ft. We change l< : Ia: 6.0 + Rx = & : 5.0'At:98

(98) (689.8) (3.50)oB : ---10 j-r:-l : 7.886.7 psi < o^

Solving for total length L',

_ Qe,s25) (30)1t2.3 ft

7,886.7

112.3 fr15 ft (between supports)

: 7.5 supports


Figure 2-43. (A) Initial piping configuration; (B) final piping configuration installed and operating.

Therefore, place a loop 6 ft x 6 ft (arbitrary dimen-sions) every seven supports. One could increase L' bymaking the loop larger (increasing Ia and La'1, but spacelimitations in this application prevent it. See Figure2-43. The shess intensification factors (SID in the codewere made equal to one because computer stress runshave verified that the curves are conservative enoush tomake SIF : 1.0.

EXAMPLE 2.4: LUG DESIGN

Referring to Figure 2-16, a lug is to be designed for apipe with the following parameters:

Pipe - 8-in. { Sch. 40, C.S. 5A-1068T : 350'F, Pr = 500 psigP : 2,000 lbLet C : 3/s in., L : 2tlz in., and the lug has l/z-in. dholeR. : mean radius, in.From piping properties in App€ndix

R5?5+?qRtR.: ---- ; --'= 4.t52in.

t = corroded pipe wall thickness. rn.

0.188 in.; cr : : l.z) ln.

n ft> t,u : [' - i(,! -')rr - r,r] rB,B,ro,

L,

c-'2

rrff< r.u = [' -i(' -u,t),' - nr ]

to,o,lo'

wlth ; > L,tt2

B : 0.19 for Nd -* K, : 1.48p : 0.15 for N; - Kz : 1.200 : 0.11 for Md -- K, : 0.88B : 0.16 for M" - Kz : 1.25

Nd - Fisu.e 2-17

P/RM

Md -. Fisure 2-19P-

or,ao _ lrnrOl lotl _ ," ,", {6) (2,0q0) = E.*,.5.j2-i- : t-P l[-i,l

: tu t:t A.yzt"

Ci.rcumferential Stress, o{

"t=*"Y*",ryod : (1.38) (2,842.30) + (1.36) (rs,o4s.72)

: 24,309.65 lb

Letw : t/+-in. weld,

I t l'" I n q.r.r loo5K" :' +

ls.uwl =, * lr:rroC = r.38

[-, I

.. , [-1l" .lz1o.zzztlo"x': t + lr,+w-l = t- l*rcrr:l = r'36

t-rilLongitudinal Stress, o;

Nx -. Fizure 2-18Pi R-

!1r - Fisure 2-2rP-

*.:llolEl =,r't [P/R,l [R,tl

:3,739.87

6M-. [ur,-l [rp] ta\t-.-^ - r..^r r-. | - /n ,....-r'2'000)tz tPltt']l \0.322),

= 14,467.03

-. NY .. 6M"ox:K"itooti

qx : (1.38) (3,739.87) + (1.36) (14,467.03)

:24,836.19

Primary or pressur" ,1r"a, : o. = I2t


_ (500) (7,981)

2(O.322)

: 6,196.43 psi

Total stress : o-t : oO + or : 3g,5g6.Ot O.t

or : ox + oP : 31,032.62 Psi

Using greatest value of o1,

o7 12oa = 2(20,000) : 40,000 psi > 31,033 psi

Thus, a lug with these dimensions is acceptable.

Bolt shearing stress : 78

_ 2,000

Oolt or pin area) r(0.5)24

10,186 < oB Oolt allowable)

T": 294"F = 754"R

M : 170.9

*' : Eql[I = ,' r,t [P/Rml [R,tl

(2,000)

(4.1s2) (o.322) = 2,842.30

(2,000)

(4.rs2) (0.322)

os : 25,000 psi

The distance from the lug hole centerline to the lugedge is to be a minimum of AISC Table 1.16.5.1, p.5-51.

Weld Size

- P 2.000f,:-: -'-" : 1.600' ZvL 2(0.25) (2.50)

f*:0.707Eoe:E = Joint efficiencyf* : 14,140.0 psi

Weldsize : *: ftf*

: 0.113 + r/c-in. weld is acceptable.

EXAIIPLE 2.5: RELIEF VALVE PIPINGSYSTEM

Examine Example 1-4 relief valve system for externalloadings induced by valve discharge. The gas propertiesare as follows:

k:c1c":1.451N : 243,755 lb/hr

Ar = 28.89 in.2

[r"ln'l__:lIMl


Cc:

I l ^ \2451105

0.1443 (1.15) lrr.+srt | ' l* |'t v.45u I

I zs+ \n'\r?ot

:0.055

From Figure 2-31 or from the following we determine P"

and P" as

p-: G - 61 710 = 42.6t3' ccA, (0.055) (28.89)

D =14

F : (0.99X1.1s)(0.98)(28.89)(42.613)

[zrr.+sr,l_r )#[, _ l_ryq)l-,",,1"[ 0.45r v.4stl I \42.613/ I I

+(28.89) (4)

F : 2,385.879 lb

Reaction moment at the vessel nozzle is

MR : (2,385.879) lb (8.5) ft : 20,279.9'12 ft-lb

The reactions at the vessel nozzle are discussed in Chap-ter 4, along with external loadings on vessels.

EXAIIPLE 2.6: WIND-INDUGEDVIBRATIONS OF PIPING

A l-in. @ Schedule 40 pipe is to run up a process

tower, and it is necessary to determine what span inter-vals are needed to avoid vibration resonance caused byvortex shedding induced by wind external to the pipe.Piping designers have the line supported at 3O-foot evenintervals.

The first problem with the layout are the even intervalsfor the supports. Piping spans subjected to vibmtionshould be in uneven intervals to prevent sine wave oscil-lations that would be symmetric and pedodic, and thusself-destructive.

Experimental data from Blevins[6] support the follow-ing formulations:

I : 0.15 damping factor for small pipe (4-in. d<)

I : 0.025 for large pipe (>6-in. d)

A o.o7 c [^ .^ o.7z lo'5

D (6. + 1.9)52 [-'-- (6, + 1.9)Sl

L : 30 ft, l-in. { Sch. 40 pipe

(2-39)

For a uniformly loaded member with simple supports,

- 5WL3

384EI

where I : 0.0874 in.a; W : 1.68 lb/ft (30)ft

= 50.40 lb

6 : 12.569 in.

From Table 2-5,

(2-38)

- 3.55r =

-

: l.trut cvcles/sec(12'569)u r

. 4mzf'pD

For air at 60'F,

p : 0.076 lb./ft3

th0.140 i-

.= tn :0.004&ft ll.

'1) ) _:L-- - sec2

Air velocity under investigation : 25 milhr : 36.65 ttlsec

6, : 13.037; Nn : 2.54 x 1ff

From Figure 2-33,

S : 0.18

Damping = tul#I" : tr.utt

rnus, fi : o.tzo

Values from Figure 2-34 indicate we are close to reso-ftrnce, as we are within an L/D ratio of 5 and L/D = 30.Thus, we should experience resonance at 25 mph for thel-in. S Sch 40 bare pipe. The line should have more sup-ports added at uneven intervals closer than 30 ft and theprevious analysis repeated for a range of wind velocities .

Such a problem can be approached with a computer pro-gram based on experimental data.

As is obvious, vortex shedding vibrations is still a sub-_jective phenomenon based on empirical data, but this ex-ample should assist one in protecting piping surroundedby vortices.

I{OTATION

A : area, in.2C = compliance, in./lb or deg/ft-lbD : diameter, in.

Ep : modulus of elasticity in downcondition, psi

E" : modulus of elasticity in operat-ing condition, psi

= force, lbsmodulus of rigidity, psimoment of inertia, ft'polar moment of inertia, ftastiffness, either translational(lb/in.) or rotational (ft-lb/deg)stress concentration factor forbendingstress concentration factor forpure tension or compressiontranslational stiffness along X-axis, lb/in.rotational stiffness about X-axis, ft-lb/degtranslational stiffness along Y-axis, lb/in.rotational stiffness about Y-axis, ft-lb/degtranslational stiffness along Z-axis, lb/in.rotational stiffness abovt Z-axis, ft-lb/deglength, in.moment, ft-lbforce (lb) or moment (ft-lb) instiffness matrixinternal pressure, psigexternal pressure, psiginternal pressure evaluated atradius R, psigreaction, lb

Heat Transfer in Piping and Equipment 101

RD : reaction in down (non-operat-ing) condition, lb

R(x, y, z) :

OT

oy

T

F,M + X, +Y,+Z

4M-X, -Y,-Z

vector resultant operatorinside radius, in.outside radius, in.torsion, ftlbthickness, in.displacement, in.weight of fluid, lb.weld size, in.section modulus, in.3section modulus of mean sec-tion radius, in.3bending stress, psicircumferential stress, psilongitudinal stress, psipressure stress, psiradial stress, psitorsional stress, psiyield stress, psishear stress, psiforces or moments acting onlyin +X, +Y, or *Z direction,respectively

firoTt

Uw

Zz^

OR

oc

OL

o"OR

K=K:

KTX :

KRX :

KTY =

KRY :

KTZ :

KRZ :

FGIJ

K

: forces or moments acting on.lyin -X, -l or -Z direction,respectively

REFERENCES

Faires, V. M., Design of Machine Elements, TheMacmillan Company, New York, 1965.Przemieniecki, I.5., Theory of Matrix StructuralAnalysis, McGraw-Hill Book Co., New York, 1968.Wichman, K. R., Hopper, A. G., Mershon, J. L.,Welding Research Council Bulletin 107, LocalStresses in Spherical and Cylindrical Shells Due toExternal Loadings, Welding Research Council, NewYork, 1979.Expansion Joint Manufacturers Association, Inc.,Standards of the Expansion Joint Manufacturers As-sociation, Inc., New York.Hesse, W. J., Mumford, Jr., N. V. 5., Jet Propul-sion for Aerospace Applications , Second Edition, Pit-man Publishing Corporation, New York, 1964.Blevins, R. D., Flow Induced Vibration, van Nos-trand Reinhold Company, New York, 1977.Lienhard, J. H., "Synopsis of Lift, Drag and VortexFrequency Data For Rigid Circular Cylinders,"Washington State University, College of EngineeringResearch Division Bulletin 300, 1966.

2.

5.

l.

1

L:M=P=

Pi=P"=Pn:

6.

7.

R:

Heat Transfer in Piping and Equipment

Providing thermal energy to process systems andmaintaining desired temperatures are key responsibilitiesof mechanical design. Although they border on chemicalengineering, the concern here is with the mechanical as-p€cts of process systems, and not with the processesthemselves. (Chapters 2 and 4 illustrate how mechanicaldesign borders civil engineering in a similar manner.)

Process systems require thermal energy for variousreasons, and the most common are to accelerate chemi-cal reactions; to heat products and services so the prod-ucts remain liquid and do not clog piping or equipment,such as with asphalt and roofing materials, viscous fueloils, and syrups; and to cool products and services, forexample to protect epoxy from polymerizing.

In piping systems there are three ways to transfer heatto the process service-tubular tracers mounted exter-nally to the pipe, jacketing the process pipe with a largerpipe forming an annulus in which the heat transfer fluidflows, and electrically tracing the pipe. We will discussthe first two types of transfer systems.

JACKETED PIPE VERSUS TRACED PIPE

The difference between traced pipe and jacketed pipeis obviously the heat transfer area available on each. Thetwo types are depicted in Figures 3-1a and b. Jacketedsystems offer more heat transfer area, but are expensiveand can be difficult to maintain. One common nroblem iscracks that develop from the thermal stresses that are in-curred. Such cracks, which are difficult to locate and re-pair, can cause the heat transfer and process fluids tomix, which can have catastrophic resulis. However. thedisadvantages ofjacketed pipe must be weighed with theeconomics of adding tracers. A proven guideline is touse jacketed pipe for process fluids with viscosities of

5,000 centipoises or more. Such high-viscosity fluids arequite common with coating mixes used in manufacturingroofing tiles. Tracing such viscous mixtures with severaltracers has proven to be so inferior to jacketed pipe thatthe disadvantages ofjacketed systems are offset. With aviscosify of 4,000 centipoises, one should consider jack-eted pipe.

Most jacketed pipe is limited in commercially avail-able sizes. Normally 8-in. by 10-in. is the largest size

front vi6w

Figure 3-1A. Traced pipe.

103


Figure 3-1B. Jacketed pipe. (Courtesy of Parks-Cramer Co.)


DIMENSIONS

COMMON TO ALL 150 LB.' 3OO LB:

stzEI

tPsotPs u

TNPI .oD

Holes

BC RF K OD

Holeg

BC RF KNo. Dia. No. Dla,

Y2t1Y13/tt1Yz

1t21Yarz

1l2x2l2

2x3

3x4

4x6

6xE

8x10

Y2

,l

'| \/+

1L/2

2

3

4

6

8

1r/a

1Y2

2

, L/^

3

4

I10

2.56

2.56

3.44

4.69

4.44

4.31

4.31

4.88

Vq

%

1

1

1

1

1Y2

4.25

5.00

6.00

6.00

7.00

7.50

9.00

11.00

'13.50

16.00

I

0.62

0.62

0.75

0.75

0.75

0.75

0.75

0.88

0.88

1.00

3.12

3.88

4.75

6.00

7.50

9.50

11.75

14.25

2.00

2.84

3.62

3.62

4.12

s.00

6.19

8.50

10.62

12.75

0.75

0.75

0.75

0.62

0.69

0.75

0.94

1.00

1.12

1.19

4.88

6.12

6.50

7.50

8.25

'10.00

12.50

15.00

17.50

6

I8

12

12

16

0.75

0.E8

0.75

0.75

0.8E

0.68

0.8E

0.88

'L00

3.50

4.50

5.00

5.00

5.EE

6.62

7.88

10.62

13.00

'15.25

2.OO

2.88

3.62

3.62

4.12

5.00

6.'l9

8.50

10.62

12.75

0.88

0.88

0.6E

0.88

1.00

1.12

1.25

1.44

1.62

1.88

.Flanges of higher pressure class and other facings available.

Figure 3-lB. Continued.

Figure 3-1C. Expansion joints for jacketed pipe. (Courtesy of Parks-Cramer Co.)


DIMENSIONS150 LB., DUCTILE IRON. STEEL

All dimensions in inches (ins.) unless otherwise noted.'Flanges ol higher pressure class and other facings available.

Figure 3-1C. Continued.

carried in stock, but larger sizes can be specially fabri-cated. When a jacketed system is selected, a carefulstress analysis should be made to ensure that the systemis not overstressed. (Chapter 2 covers such stress analy-ses.)

TRACIilG PIPING SYSTEMS

When process fluids have low to moderate values ofviscosity 1g 4,500 cp), it is best to trace them with tubescontaining hot or cold fluids. The tracing can be donewith or without heat transfer cement around the tracertubes (Figure 3-2). We will consider two methods foranalyzing both systems.

Usually, steam or hot oil is used to trace systems. Hotoil is used when the fluid to be traced is hotter than satu-rated steam at typical operating pressures, which wouldbe about 350'F and above. Hot oil systems are some-

what simpler than steam-traced systems, because steamtraps and condensate return lines are unnecessary. How-ever, hot oil can be expensive and if there is ample auxil-iary steam available for tracing, steam is favorable formoderate- to low-temperature systems. When there ismuch piping to be traced, steam at the available tempera-ture and pressure may condense into hot water beforetracing the entire system. For these situations, only hotoil can be used. Thus, hot oil is used in tracing applica-tions where steam is either not practical or not available.

There are many types of hot oils marketed by variouschemical companies as heat transfer fluids.

It is most desirable and should be mandatory to useheat transfer cement in tracing tubes on process piping,because it provides more heat transfer area. Heat trans-fer cements are available in all major countries and insome of the larger Third World countries. Howeverthere are times of expediency in which traced systemsmust be installed without the cement.

FLANGE DIMENSIOIiISi

srzEIDtPs

OD Holea

BC RF xA

U

T

TIPTlns. mm. No. Dla- lns. mm.

1Y1r2

1t/2f,Y2

M3x4

4x6

6x8

8x10

1Y4

1Y2

2

I

6.00

7.@

7.50

9.00

11.00

13.50

16.00

152

178

190

28279

3/$

406

4

4

8

6

8

12

o.75

0.75

0.75

0.88

0.88

1.00

4.75

6.00

7.50

9.50

'tl.75

14.25

3.62

4.12

5.00

6.19

8.50

10.62

12.75

o.75

0_88

1.00

't,12

't.19

1.38

24.OO

25.00

25.38

26.00

26.50

27.U

u.25

610

635

645

660

673

470

3.44

4.69

4.44

4.31

4.31

4.8

3/t

3/q

'I

'|

,l

1

1Y1

Traced Piping Without Heat TransferCement

The modes of heat transfer in a system without heatkansfer cement are natural convection through the airspace inside the insulation, and to a much lesser extent,direct radiation between the tracer and pipe or equip-ment. Since the tracer tube and pipe surface have verylittle surface contact, conduction is minimal. Any effectof film resistance to heat transfer between the air space

outside the insulation and the inside insulation surface isnegligible.

The procedure for tracer design without heat transfercement is outlined in the foliowing steps (see Figure 3-3for parameters) :

l. Assume a value of air space temperature equal toor greater than the minimum temperature of theprocess temperature inside the pipe.

Figure 3-2, Various traced pipe configurations: (A) singletraced pipe, with tracer under pipe, with heat transfer cement(HTC); (B) process pipe with two tracers with HTC; (C) onetracer on top ofprocess pipe with HTC; (D) process pipe withthree tracers with HTC; (E) jacketed pipe.

Heat Transfer in Piping and Equipment 1O7

Tr = aclual insulationthickness

Figure 3-3. Traced pipe with one tracer under bottom withoutHTC.

2. Estimate the natural convection coefficient, h",from the tracer to the air space from Figure 3-4.

3. Calculate the equivalent cylindrical insulationthickness, T", as

(3- l)

4. Determine the outside film coefficient of the insu-lation to atmosphere, h., from Figure 3-5 and cal-culate Uo from the following:

1

u.(3-2)

Di : inside diameter of pipe insulation, ftho = outside film coefficient from insulation to

atmosphere, Btu/hr-ft2- "Fki : thermal conductivity of insulation,

Btu/hr-ftz-'FT" = equivalent thickness of cylindrical insulation, ftT, : actual insulation thickness, ftUo: overall heat transfer coefficient from the air

space to the atmosphere, Btu/hr-ft'-"F

5. Formulate a heat balance around the air sDace.solving for the temperature of the air space. q:

'" : {q'r=l)'" {9': =t)'\ 2l\Di I

T]kt h.

a: (v,xA")(t" - L)(hJ(AJ(n)(tt - t")

(3-3)(34)

Q: ) (r)(Qr) (3-5)

where Ao : outside insulation surface area, ft2lftA, : outside surface area of tracer tube, ftzlfth, : convection film coefficient from tracer or heat

transfer cement (HTC) to air space,Btu/hr-ft -'F


b 1.o

.c o.9

o.a

o.7

oo,ouTstDE D|aMETER OF CyL|NDER ltNcHESl

h: NAIURAL CONVECTTON FtLM COEFFICTENT

. ^ . .o.25

h. o.5lt-l

9

a:7

6

5

Figure 3-4, Natural convection onhorizontal cylinders.

Figure 3-5. Heat transfer outsidehorizontal pipes.

35710 152030 50

OUTSIOE DIAMETER OF INSULATION IINCHESI

ho=COMBINED OIJTSIDE HEAT TRANSFER FILM COEFFTCIENT

Qr : heat transfer per lineal foot from air space toatrnosphere. Btu/hr-ft

Qz : heat transfer per lineal foot from tracer to airspace, Btu/hr-ft

L : temperature of outside air, oF

ti = temperature of tracer fluid, oF

"y : safety factor; 1.3 for piping systems withoutHTC, 2.0 for piping systems with HTC, 1.5for vessels without HTC, 2.5 for vessels withHTC

lf ta > ti, then the system is adequate. The maximumspacing of tracer tubes for cylindrical vessels is calcu-lated in the same manner except that a flat plate approxi-mation (T. = t) is used to compute the heat losses, or Qvalues.

Traced Piping Wlth Heat Transfer Cement

One mode of heat transfer in a system with heat trans-fer cement is conduction from the tracer tube through thepipe or vessel wall to the point of the wall most distantfrom the tracer. The thermal distribution of such a sys-tem is shown in Figure 3-6. The other mode of heat

transfer is the natural convection from the tracel and thepipe or vessel wall to the air space. Thus, the air spacetemperature is lower than the minimum process pipe or


vessel wall temperature. The contribution of radiationfrom the tracer and pipe or vessel to the inside wall oftheinsulation is negligible, as is the film resistance to heattransfer on the inside insulation wall.

The procedure for tracer design with heat transfer ce-

ment is as follows:

1. Determine the scheme of tracers to be applied us-ing Figure 3-7.

2. Calculate the metal wall area (equals wall thick-ness) A*; the outside surface area of insulation, Ao,the outside surface area of pipe, Ao, and the outsidesurface area of tracer tube or heat tfansfer cement(Hrc).

3 . Assume a value of the minimum pipe wall tempera-ture, to, equal to or greater than the minimum pro-

8.

cess f luid temp€rature.Assume a value of air space temperature, ta.

Estimate the natural convection coefficient, h",from the HTC to air space.Calculate T" using Equation 3-1.Determine the outside film coefficient of the insu-lation to the atmosphere, h., from Figure 3-5 andcalculate Uo from Equation 3-2.Calculate the average pipe wall temperature tp andestimate t}le natural convection coefficient from thepipe or vessel to air space, ho, from Figure 3-4.

4.5.

6.7.

=

o

COLO SURFACE TEMPERAIURE It2I F

Figure 3-6. Heat transfer by radiatlon. ',{s-J[(!e"" I-e#9]


Formulate a heat balance around the pipe or vesselwall and air space and perform an iteration analysissolving for t" and te with the following steps:

Figure 3-7, Temperature distribution of a two tracer system.(Courtesy of Thermon Manufacturing Co. t

temperature of air space, 'Fambient temperature,'Flength of heat flow through metal, ftpipe temperature at point nearest tracer, 'Fpipe temperature at point farthest from tracer..F

Likewise for traced systems with HTC, for traced ves-sels, the maximum tracer tube spacing for traced vesselsis calculated by the same procedure, except that the flatplate approximation (Te : t) is used to compute the heatlosses, or Q values.

Condensate Return

Steam differs from hot oil in that condensate is formedby loss of heat energy. During energy shortages, the useof condensate return lines is normally justified. Consid-ering the use of 1/2-in. tracers, normally a l-in. conden-sate subheader will handle condensate from 2-8 tracers,a ltlz-in. header from 9-20 tracers, and a 2-in. sub-header from 21-50 tracers. With a condensate collectionand return system the steam supply pressure should be atleast 100 psig. Even though these rules of thumb are welltested in field practice, the reader is encouraged to calcu-

9.

Qr :Qz:

Q+)Qz+

where

(u.)(,\)(r" - t")(hJ(At)(tt - t")(hPxApxtp * t")

(3-6)(3-7)(3-8)

(3-e)

(3-10)(3-l l)

Qn : (2Xq) (I*)o.u,, * ,,,'

(r)(Q:)Q:)Qr

Am : cross-sectional pipe wall area (equals pipethickness), ft,/ft

Ap = outside surface area of pipe, ft /ftho = convection film coefficient from pipe to air

space, Btu/hr-ft2-oFk : thermal conductivity of vessel shell material,

Btu/hr-ft2-'Fn, : number of tracers, dimensionless

Q: = heat transf€r per lineal foot from pipe to airspace, Btu/hr-ft

Q4 : heat transfer per lineal foot from tracer topipe, Btu/hr-ft

Qr: Qq+Qz

For systems without HTc,

Qr:Q:

late the condensate load for his particular needs. Con-sider the following analysis:

Total heat loss from steam tracer : QrFor systems with HTC (by adding Equations 3-'7 and 3-9),

Heat Transfer in Piping and Equipment 11 1

where hr8 : enthalpy of vaporization (also called latentheat of vaporization), Btu/lb

The steam in the tracer is assumed to transfer energyas heat for a given mass of steam under constant pres-sure. A typical condensate return system is shown inFigure 3-8.

When collecting condensate, care must be taken toprevent water harnmer caused by the mixing of conden-sate at different temperatures and pressures. To preventwater hammer in condensate systems, spargers and

steam separation kegs should be considered.To size the condensate return lines, as well as the trac-

ers themselves, use the methods presented in Chapter Ifor line sizing. In systems where a large quantity of con-densate is formed by steam flashing, a condensate returnpump may be required. Normally, condensate returnpumps are the horizontal centrifugal type. Pumps andtheir applications are presented in Chapter 6.

(3-12)

(34)

(3-13)

The steam in the tracer is assumed to enter the systemas saturated steam at an initial temperature or pressure.Considering the amount ofheat loss over a given temper-ature range, the condensate load from n tracers on a

given process pipe is

,ir : $, rumnnfc

E

STEAM SUB

HOR.

COND.HOR.

SEEDETAIL A(TYP.)

SEE FIGURE 3

COND HOR.

SEE

OETAIL A(wP.)

OETAIL-

Figure 3-8. Condensate return header in tmcersystem. (Courtesy of Thermon Manufacturingco.)


Jacketed Plpe

Figure 3-9 illustrates details ofjacketed pipe. Forjack-eted systems, it is customary to assume a temperaturedrop over a given length of pipe for hot oil. In applica,tions of hot oil heating a viscous fluid such as asphalt,100'F drop per 100 ft, or I 'F per foot, is quite common.If one is not familiar with a given service, then a heatbalance must be made, like those done for tracers, How-ever, using a temperature drop over a given lenglh ofpipe simplifies the analysis and has been proven in prac-tice, because all examples cited are from actual, success-ful operating systems. The following steps illustrate onesuch method of designing jacketed pipe:

l. Compute the overall heat transfer coefficient, U,by the following relation:

for Dy'D" > 0.2

For an annulus, the hydraulic radius, Rs,

Dr-i - d.o

in which

D:4Rs

._ NC"

and thus

' - Nru"k.D

Rs (3-18)

(3-19)

(3-20)

(3-2r)

u = Er*r+ln(ry'r'* ll-r[h': kzr h

where N*" : Y%tt

hr: : r.86(NrJ"'(N",) "LP) ($ k$'"Nr" : o.o2o Nr;'Nr (oeJ

,1- rooo o socror.t

9d LR

PE(rcESS IINE

(3-14)

(3-r5)

(3- l6)

2. After solving for the overail heat transfer coeffi-cient, determine the amount of heat transfer fromthe relation

q : UA(LMTD) (3-22)

PrJ:1r2, 2r3,3t4,/rt 6linl

where A : outside area of inner tube. ft2. and the LMTD0_17\ is based on the assumed rate of heat loss per

unit lengrh of pipe.

ptt= t'ga,&1,4,€

Figure 3-9. Standard fabrication details for jacketed piping.


END OF JACKE'T DETAIL

PROCESS LINE IPI

JACKETIJI

CUIOE BAi OETAII- BANS PLACEO EVERY rEN FEEi OFPR@ESS IINE

PrJlinl GLJtoE BAR Srz€linlz'z

-le'9r-

z t" i,u31 4

-

r4r15 -2 In to€4t6

-t

rrYi6-r in tong

v2t. ss n BING

PFOCESS LITIE

rrNE srzE prJ li.L

ll

Figure 3-9, Continued.


The LMTD is solved using the following formula: To facilitate manual calculations refer to Figure 3-10 [1].The concept of the logarithmic mean temperature differ-ence is widely explained in most basic engineering text-books, so its explanation will not be presented here. Thereader is referred to Kern [2] or Ludwig [3] for a formaldescription of the significance of the LMTD.

3. Once the amount of heat transferred is determinedfrom Equation 3-22. assuming a given temperature

_ (GTTp) - (UrTp)

, /crro\]n l-l\rjrTD /

Chart for SolvingMTD Formula

,\,rn _ GTTD.LTTD-"- _ GTTDt-o8e LrrD

/1-r1r

where LMTD : logarithmic mean temperarure differenceGTTD : greater terminal temperature differenceLTTD : lesser terminal temDerature difference

^[email protected],*.rk@'

to,4

tm90

80

70

60

50

40

tllo

t

i5

EF

E

J

Greater Terminal Temperature Difference

Figure 3-10. LMTD chart. (O 1978 by Tubular Exchanger Manufacturers Association. Repdnted by permission.)

where rir

At

drop, the amount of flow rate ofthe heating fluid isdetermined by

q : rirCoAt (3-24)

: hot fluid flow rate, lb/hr: specific heat of hot fluid, Btu/lb-"F: hot fluid temperature drop

From this formulation we determine the flow rate re-quired.

4. Using Figure 3-11, the amount of pressure drop inthe annulus is determined and added to the Dressuredrop in the whole sysrem (which includes the pip-ing connecting the annuli). The pressure drop forthe piping other than the annuli is determined byusing the methods presented in Chapter 1. Chapter6 shows how to select and size the pumps to handlefluids that usually require jacketed services, suchas hot oil.

Once the flow rate is determined, the hydraulic analy-sis made, and the pressure drop judged adequate for thesize ofpumps selected, the jacketed system details can bedesigned.

Typical jacketed piping components are depicted inFigure 3-9. In extensively jacketed systems, valves canbe procured that have jacketed spaces built in. Thesetypes of valves are recommended for services wherejacketed pipe is required (p > 5,000 cp). Some of thesevalves are shown in Fieure 3-9.

Vessel and Equipment Traced Systems

Systems that require piping to be either traced or jack-eted likewise require similar components for vessels.The complexity of traced components depends on theviscosity of the process fluids being handled. For high-viscosity, non-Newtonian fluids special items must beadded to vessels, such as agitators that are composed ofblades and usually powered by electric motors. Thereare many reasons to use agitators, and one of the mostcommon is to keep suspended particles in a non-Newto-nian fluid evenly distributed to prevent particle settle-ment on the tank bottoms.

There are two basic types of heating and cooling de-vices used for vessels-internal and external iackets thatfit on the inside and outside of the vessel, respectively.These jacket types are shown in Figure 3-12. The hotfluid (normally steam or hot oil) enters one side of thecoil and flows through the baffle (inside the tank) or

Heat Transfer in Piping and Equipment 1 15

jacket (outside the tank) and exits through another sideheating the vessel's contents. This can be seen in Figure)-tz-

Before we analyze in detail these various components,we must first look at the overall heat reouirements of thevessel to determine how much heating surface is re-quired. The controlling criterion in determining theamount of heating panel surface area of a vessel is thetransient state, i.e., how much surface area is required toheat a given mass of fluid of specified properties to aspecified temperature within a specified time. Figure3-13 illustrates a control mass inside a oressure vessel.Consider two transient boundary conditions in the ves-sel-the fluid resting at steady state and the fluid movingthrough the tank at a given mass flow rate. Thus, the fol-lowing two criteria must be established before the heattransfer area required for a process vessel can be deter-mmeo:

1. A vessel shown in Figure 3-13(a) contains a staticfluid of X gallons at an initial temperature, Y'F.How many degrees of temperature per hour willthe fluid mass rise for a given surface area ofclamped-on jacketed coils?

2. Using clamped-on jacketed coils shown in Figure3-13@), how many degrees of temperature perhour will be transferred to a given mass of fluid ofdefined properties flowing through the vessel at aconstant mass flow rate with an initial temoeratureof Y'F

These two criteria are established bv considerins thefollowing relationships:

Q : mcpat

and

Q : UA(LMTD)

Equating Equation 3-25 to 3-26, we obtain

At : UA(LMTD)mCp

(3-2s)

(3-26)

(3-27)

The U value, or overall heat transfer coefficient, iscalculated on the basis of whether the panel of heat trac-ing tubes are clamped on outside the vessel or located in-side the vessel. These overall heat transfer U values aredetermined through extensive laboratory tests and accu-mulated field experience. The U value used in calcula-tions should be that recommended by the heat transferpanel manufacturer, as various panel designs are avail-able and the calculation of the U value analvticallv can


rh" x1" s,ch.40 Jackeled Pipecurves based on 100 l€et ol jacketod pipe (tiw 2010:lengrhs) and inclsde live 1" o.D. x.065" wal tubing jump-overs plus entrancs and exit losses. Warer @ 60'F. (16"C,)

A P-inchesoluater I

tt;

I

=

Flgure 3-11A. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)

't


,a" x1Y2" Sch.40 Jacketed Pipecurues based on 1oo reet or racceled p'pe (rrv;20 o_lengths, ano Include

'ive 3d" O o r 06s' wal lubing iump_

overs plus €ntrance and €xit losses. Water @ 60'F O6'C)

dl

=o

Figure 3-118. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)

1 18 Mechanical Design of Process Systems

1" x 2" Sch.40 Jackeled Pipecuryes based on 100 leel ol jack€ied pipe (livs 20{"rengrhs) and

'nclude rive t4" o.D. x .065" wall lubrng jump_

overs plus entrance and exii losses. water @ 60'F- {16'c.}

I

f,

fi

{.

E

JI

=oJ

g A P- inches of water

Figure 3-1|C. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co')


1Y4" x2" Sch.40 Jacketed PipeCutues based on 100 leet ol jacketed ppe (lve 20:0"lenqths) and include live 74'O.D. x .065'walrlubrnq jumpovers plus entance and exit Losses. Wate. @ 60"F. (16'C.)

d A P-psig <'

A P - inches ol water 3

ut

=o

Figure 3-11D. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)


1r/a" x 2V" Sch. 4O Jacketed PipeCurves based on 100

'eet ot jackered pipe (Uve 2010'

lengrl^sl dnd r clude rive r;'O D. ! .065" wdl tJbrng jump-overs pr!s entrance and exl rosses waler @ 60,F. (16.c )

t

'to

=J

AP-psig

Figure 3-11E. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)

Heat Transfer in Piping and Equipment 12'l

1Y2" x2t/2" Sch.40 Jacketed PipeCurues bas6d on 100 teei ol jackered pipe (live 20!0"lenglhs) and incrude nve ya" O O x obs" wrll ruo,rg tuhp-overs plls enrra.ce a.d ext losses. water @ 60"F. (16'c)

lD

I

=

Figure 3-11F. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)

122 Mechanical Design of Process Svstems

2" x 3" Sch. 40 Jackeled PipeCurves based on 100 leel ol iackeldd pipe (iive 2010"lenolnsl and nclude live 1" O.D. (.065'sall rubnq_Lmp-overs plus entrance 3nd exit rosses water @ 60.F. 06"c.)

A P - inches ot water

E

6

=J

Figure 3-11G. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)


3" x 4" Sch.40 Jackeled Pipecurves based on 100 leer ol jackeled pipe (tive 20'0"l6n9rhsj and Include l,!e I" o.D ^ 064 rali rlbr.g tuhp.overs plus enl.ance and exil losses. Water @ 60'F. (16'C )

d A P - psig o

I

=

Figure 3-11H. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)


3" x 5" Sch. 40 Jackeled PipeCLNes based on 100 leel ol iackeled pipe (live 2010"lengihs) a.d include live 1" O.D. x.065" walllubing jump-overs plus enirance and exil losses. Waler @ 60'F. (16'C.)

A P - inches ol waler q

=oJ

Figure 3-1 11. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)


4" x 6" Sch. 40 Jacketed PipeCurues based on 100 leei oi jacketed pipe (uve 20'4"lengrhsl and ,ruluoe lve I' O D ^ 065" $Ell rlbrnq tunp.overs pius entrance and exn losses. Water @ 60'F. (16'C.)

A P - inches ol wate. q

E

=o

Figure 3-11J. Engineering data pressure drop through jacket. (Courtesy of parks-Cramer Co.)


6" x 8" Sch. 40 Jackeled Pipecurues based on 100 ieer oi iackered pipe (nv€ 20a0'lenglhs) and inclLde nva I O o r 065'war' ruo ng Lmpovers plus enlrance and exil losses water @ 60"F. {16"c.)

AP-psig

E,o-(ra

=a,

lo

=

A P - inches ol water q

t-

,LENT 7

7.,)aa:

o(:-:,-:

jo(

7AM

'::tN/ \R

::::l::::l:

Park :s-l ,fal ne::::l::.1:.::1.::t,:

r:JACKETET N G SYST :l\4S t COMPONENTS

l,ill ' 1 ,'1,,,,1,,

Figure 3-11K. Engineering data pressure drop through jacket. (Courtesy of Parks-Cramer Co.)

=,(r.

=o,

8" x 10" Sch. 40 Jackeled Pipecu.ves based on 100 ieet ol tackeled prpe (rve 2oro,leng l-sl aro incrLde trr- t" o D. r 06r" do r_b'.q tu-p.overs pL!s enlrance and exrr tosses. water @ 60.F {16.c )

AP-psig


c

@

=o

A P - inches ol water q

-/;;JT

!r:::r'T FTOW i,?,/..:

i:,1-,:

h-0b, .1-=: F=

t1=

/. / I l::::F= t.-

t:l v)1

7/.

/

ul

LAMINARA I

Pe

:l

ks-l fame':r:i::::l

I:::l't

I,l

i. coMP

:;::l ::l...1..1

ONEI {TS

!l :

:::.[].. .t.._. . .1.-.

Figure 3-1 1 L. Engineering data pressure drop through jacket. (Courtesy of parks_Cramer Co. )


Moderate bracins formediurn agitation -coodi-tions. All 5rac6 are fromvessel wall and no circum-Iereotial rings are used.

Brrc$ 6ay be weldcd otbolted, Hemocd edgc pcrmounting lug detail, page,{4. desirable.

Flexible hoses desirableher€ w-hen possible aodwnen lorces are Severe.Also particularlv imoorcant, foi altematirig heitingano cooltnq conorttons.

Speial bracing.for heavyaertauon conorarons,

Figure 3-12A. Heat transfer internal plate or panel baffles inside a vessel. (Courtesy of Tranter, Inc.)

Heat Transt'er in Piping and Equipment 129

C=2%" MlN.

II

A

3RD. S€T OF CHANNETS

USED WHEN B DIM.EXCEEDS 7T'

OVER,AI.I WIDTHI(NO. OF PLATECOTL-|) (CJ.3',]

Figure 3-128. Schematic depicting how heat transfer panel plates heat up or cool down process fluid in tank. (Courtesy of Tran-ter. Inc.)

Figure 3-12C. Heat transfer panel plates designed to fit oncurved surfaces. (Courtesy of Tranter, Inc.)

,/ w*\ \/ _-.--\r\1 \'ffiFt*,(?,''l \-*'/\t e\ l[ [ l\ l\

D ----vHEADER SIZE

L-3S Luqs (typ)

ffi N


L 35 Llgs (typ)Cuslomer shoutd instatl

al iifre ol instartairon.

This sketch shows tineconlact pfovrded by the 1?,,

Figure 3-12C (continued). Heat transfer panel plates designedto fit on curved surfaces. (Courtesv of Tranter. Inc.)

lead to erroneous results if the exact details of the Daneldesign are not known. Most panel manufacturers hesitateto reveal such detailed features, so the U values on thelow side are recommended for situations where the panelmanulacturer does not have a recommended U value.With high-viscosity fluids such as tar and asphalt at tem-peratures of 300-700'F. a good U value lor lnternal baf-fle panels is 9.60 and for external clamped-on panels avalue of 4.00 is reaSonable.

After a U value has been selected, Equation 3-26 issolved, revealing the net temperature change per hour.

The second criterion involves the mass flow rate ofthesame fluid through the vessel. To estabiish this criterion,Equations 3-25 and 3-26 are solved tosether to deter-mine the temperature rise. The analysis o-f both criteria isgraphically illustrated later in Example 34.

Once both transient conditions I and 2 are satisfied bvthe selected heat transfer area, the detailed design of thebaffle panels (both external clamp-on and inteinal) canbe designed. This is best shown by example and done soin Example 3-5.

Further applications of Equations 3-25 and 3-26 aregiven in Example 3-6. In this example a material han-dling problem is analyzed in which both steady state andtransient heat transfer conditions are considered. Afterreviewing Examples 3-5 and 3-6, the reader is en-couraged to always consider transient conditions of heattransfer in similar situations. Transient criteria, as re-vealed, usually govern to a large degree.

iq-i-

Figure 3-12D. Vessels with typical external heat transfer plate panels. (Courtesy of Tranter, Inc.)


Figure 3-12D. Continued.

Figure 3-13. Two schemes in which the heat transferred must be considered: (A) conrol rnass scheme; (B) control volumescheme.

-'-- -*-ifluidx gallonsv'F

oF/min


HEAT TRANSFER IN RESIDUAL SVSTETyIS

lleat Transfer Through Gylindrical Shells

Heat transfer through pipe supports, vessel skirts, andempty branch piping connections to hot or cold headerscan cpuse critical stress problems as well as damage toequipment. Excessive thermal deflections can result inunacceptable loads on rotating equipment and vesselnozzles. In cryogenic service, vessel skirts can fail bybrittle fracture if the transition temperature point be-tween alloy steels and carbon steel is not considered.

This section discusses the analysis procedures for ana-lyzing heat transfer in such residual components as ves-sel skirts and pipe supports. The methods used have beentested with empirical data and have been used for severalyears in design practice. For derivations to the followingmethod on heat transfer through cylinders, the reader isreferred to the author's paper [4].

Vessel skirts are normally insulated on the inside andoutside surfaces as depicted in Figure 3-14. In cryogenicapplications, there are many reasons why a heat transferanalysis of the skirt is desirable. The primary reason isthe one previously cited-to protect carbon steel compo-nents from fracture failure. Another reason involves ec-onomics-a tall skirt made of alloy steel is much moreexpensive than a similar skirt made mostly of carbonsteel. Also, we will see how the skirt can actually deflectas a result of this heat exchange.

Consider the skirt in Figure 3-14. The vessel is at ei-ther an elevated temperature or a cold temperature deno-ted at the shell-skirt juncture as t.. Thermal conduction

and convection are the controlling modes of heat trans-fer. The convection can either be considered as naturalor free convection, or in the case of wind, forced con-vection. It has been found that using the free convectioncoefficient is the most desirable in many cases, sincevessels are normally surrounded by other equipment andstructures, making free convection more applicable.

Assume that the temperature inside the skirt is thesame as ambient temperature and wind chill factors arenot present. Air seepage under the skirt and open aper-tures on the shell allow for equilibrium to be establishedwith the outside temperature.

The first step is to determine the free convection filmcoefficient for the outside surface of the oressure vesselskirt insulation. In normal conditions. the air temoera-ture inside the vessel skirt. ti. is assumed five degreeslower than the outside ambient, ts. The free convectionfilm coefficient is found by iteration using the followingequatrons:

,, - [r, ln(r2lrr) , qlnG!lt2) ,

"--[ k- - l!-r+ ln (r+/r:) * 1 l-'k* -

-h.Jta = (Ua/ha 5)(! - t5) + t5

No, = [d3lgB I At | (3,600)2]tp2

NN" : C(NG.NP.)-

hl.s : G"i,NN,)/d

(3-28)

(3-29)

(3-30)

(3-31)

(3-32)

.f

@l = insulation

flll = metal

Figure 3-14. Vessel skirt insulation detail. Sometimes the in-side insulation is left off.

For free convection of cross flow around cvlinders.the following constants hold [5]:

l0 < Nc.Np. < 10e, C = O.525, m : tlc

lOe < Nc,Np, < 10", C = O.129,m : )13

These relationships are valid for applications for therefining, petrochemical, and gas processing industries.

Now, for a cylinder with insulation on both sides, weuse the final value of ha-5 after performing iterationsfrom Equations 3-28 through 3-32 in the following equa-trons:

^ /2"r,.,\ [ r l I

" - \U,q,/ tltt tr"'rrl - l" (t ,ttj

r 1-3'll

/ ^ \[z - ll:+l lroho,ttn . t5) - k2',

l\KmAny' [ [ ,, , l]L

tl" {r./rr) - I" (t/'i)lll

(3-34)

z : ztQ (3-35)

Substituting these parameters into the foliowing equa-tion, we obtain the temperature distribution down theskirt length:

2(t. - Zt e-oo 5

=. , *

"zrauJ

The difference between the process temperature insidethe vessel and the outside ambient temperature is themain driving force of heat transfer. It is analogous toelectrical EMF driving force or the potential energy ofheight differential from which a fluid is dropped andturned into kinetic energy.

The degree in significance of convection is inverselyproportional to the insulation thickness. The air aroundthe outside insulation surface is in a state of local turbu-lence and for this reason the variance of the Grashofnumber down the outside insulation wall is insignificant.Experimental measurements confirm this fact. Thereader will see in Examples 3-7 and 3-8 how to apply thismethod to vessel skirts.

Piping that is supported by piping sections is treated ina similar manner to vessel skirts. Such piping supportsare shown in Figure 3-15 in which the pipe supports andbranch lines are subject to thermal gradients from a hotor cold process header. Figure 3-15a shows a stub pieceused as a piping header support. The temperature gradi-ent through the stub piece must be analyzed to determineif the Teflon slide beneath the base plate will be pro-tected from the elevated temperature inside the process

(3-36)


Fragile piece ol equipment U

Figure3-15. (A) Stub piece used as header support: (B) pro-cess line is connected to a turboexpander. The line is sup-ported by a short section of pipe welded to a base plate; (C)branch line from a header (hot or cold) connected through ashut-off valve to a ftagile piece of process equipment.

header. If the process header is in cryogenic service, thestub piece must be analyzed to assure the design engineerthat the carbon steel structural members are adequatelyprotected from temperatures below the transition tem-perature.

Shown in Figure 3-l5b is a common situation in whicha process line connected to a turboexpander is supportedby a section of pipe welded to a base plate. If the pipestub deflects enough (shown by 61), the thermal deflec-

Pipe being analyz€d

-pip€ slub pi6ce


tion could induce a sufficient bending moment on theturbine to cause serious mechanical damage.

Figure 3-15c shows a branch line running from a hotor cold pipe header to a fragile piece of equipment. Eventhough the valve on the branch line is closed, the residualtemperature distribution through the branch line may beenough to cause the pipe to deflect and damage theequlpment.

Referring to Figure 3-16, the procedure for determin-ing the temperature distribution through the emptybranch pipe or pipe support is similar to the case of avessel skirt. First, solve for the free convection film co-efficient on the exterior surface of the pipe insulation. Todo this, use the equation for the overall heat transfer co-efficient:

ur:

t', : l=l (r, - to) + to

No, = [d37,gB( lAt l) (3,600),]/rr,

Nr" = C(Nc,NpJ'

(3-38)

where c and m are determined as previously for skirts

h,j = (k"r.Nr")/d (3-32)

tj : (u3/h;) (ti - t.) + t" (3-29)

Atj : t3 - ti < 2'F (3-39)

Once Atj criterion is met, we can proceed with the fi-nal iterative value for the film coefficient, h.. With thisfinal value. we solve lor the parameters Q, Z. and Z asfollows:

^ 2nk1

;"{$(3-40)

1,, ," (;) ,, ," (,:)

, ]

\ k- - r( -n;// r- \ r

(3-37) t l--^J lrrtr"tt: - t't -

t

kJ,---tTrn lll

\r,

(3-41)

Once Q and Z are known, we solve for the tempera-ture distribution with

o

(3-30)

(3-31) tx2(t. - Zte 'ao

5 ==-++LI + e2"au '

; 300'F

dia.,sch.40, cs

3-in. calcium silicate

Figure 3-16. Empty branch pipe with one end uniformly sub-jected to three temperatures.

(3-36)

You will notice that the form of the final solution.Equation 3-36, is the same for the skirt problem with in-sulation on the inside and outside shell surfaces as thepipe problem with insulation on only the outside surface.The difference in the solutions is because of the bound-ary conditions, i.e., a cylinder with insulation on bothinside and outside surfaces versus a cylinder with just in-sulation on the outside surface alone. The solutions to thebasic differential equations are affected by these differ-ences in boundary conditions. For further information onthis subject, the reader is referred to the author's paper

t4l.For cases of tapered vessel skirts the cylinder section

can be approximated by using an average diameter. Thisapproximation is very close to actual results becauseskirts should not taper more than 15" (see Chapter 4).

As a consequence of heat transfer along vessel skirtsand pipe connections, thermal deflections will occur.The deflection equations are the same regardless of wharcase is considered, whether it is a shell with insulation onthe inside and outside surfaces or a shell with only exter-nal insulation. The values of Q and Z are determinedfrom the appropriate equations of each respective case.

ts = 900'F;

The thermal deflection equations are dependent on thetype of material considered since the coefficient of ther-mal conductivity is the governing property of the partic-ular material being considered. Thking a differential ele-ment of a shell, we solve for the amount of thermaldeflection by

dL : @(t)t(x) dx (342)

Since the temperature varies over the shell length, we in-regrate Equation 3-42 to obtain the total deflection, 6, as


(2.Oss x ro ,7]Ll0o

(2.055 x l0-3) 1t. * ZY leuoorr - 1;

135

+

+l06eo5(1 + e2lao

5)

(1.06 x

'Isech (LQo 5) tanh(LQ0 5) + arctan [sinh(LQ0 5)]lr106

(_l8(t,

- z),l0-6)

o= jar- = JL crrrr(xr dx (3-41)

.. [sech(LQos.; tanh(LQo 1 +'t--- -------l arctan [sinhtLQo'y]l

-l

- tr "..] - |+ 4Zi,- Zt lî. arcran rerq" )l + ZrLl (3-45)

tv-' ) )

Like Equation 3-44, Equation 3-45 can be adequatelyhandled using only the first three terms. The use of theseequations will be demonstrated in the examples.

Residual Heat Transfer Through PipeShoes

Heat transfer through plate surfaces is much simplerthan more complex surfaces, because they can be han-dled with one-dimensional equations that are simple touse. Based on Figure 3-17, we consider the heat balancedown throush the shoe as follows:

- 87r, - 2f / e:Loo s

i,a"-- \l + etLao

5i

+ 42t. -2, [;L..,* r.'o"r] * z,r] B-441

For practical applications in the refining, petrochemi-cal, and gas processing industries, sufficient accuracymay be obtained by omitting the last term beginning with(6.536 x 10-?) in the calculations.

Similarly, for stainless steel, the thermal deflectionequation is as follows:

6., , 2& - Z) [8.96 + (4.1 I x 10 )Z] arctan (eLQ0 5)

2[5.89 + (2.496 x l0 ,22)& - Z) arctan (elo0 5;

The function, c(t), is the coefficient of thermal expan-sion for the particular material being considered. Valuesof the thermal expansion were curve fitted over a largerange of temperature and a relation in terms of tempera-ture was obtained for various materials. The function fort(x) is obtained from Equation 3-36 and is substitutedwith d(t) in Equation 3-43. Then, the product of a(t)t(x)is integrated over a length L and we obtain the thermaldeflection function for each particular material. For car-bon steel, the expanded thermal deflection equation is asfollows:

106e0.5

+(2.496 x lo-)(t, - Z)2(e2l-ao5 - 1)

(l09Qo 5(l + e2LQo 5)

(2.496 x ro ,z'Lt0-

(6.536,l0-?)

{rr,, - zr

(109

(106)Qo 5 Figure 3-17. Pipe supported on a shoe.


/Heat conducted rhrough\ - (Heat loss by convection from'l

\ shoe to base plate / - \ shoe to outside air l

Writing in equation form, we have for one-dimensional,steady state flow:

k.A, l:l = hJp(ar) (3-46)

For the conduction process, At : ti - tp

For the convection process, At : tp - to

Substituting into Equation 346, we have

Ue. (D' l=hoAp(tp-r.)

Solving for to, we have

. _ k-A.t, + hoApl-to .F'n- 1L.a. + trrl"I-r '^

where A. : (P x length of shoe) x 2, in.2Ao = Base width \ length of shoe. in

'z

h^ = free convection coefficient for shoe to air,

Btu/hr-ft2-'Fk : thermal conductivity of shoe material,

Btu/hr-fc"FL : shoe height, in.

Like the analysis for cylinders, the free convection co-

efficient, h., can be substituted with a forced convectioncoefficient. However, most pipe shoes are surrounded byenough obstructions to prevent a direct wind from blow-ing on the shoe for any length of time. Figures 3-18a and

3-18b show thermal gradients for various simple pipe

supports.

EXAMPLE 3-l: STEAM TRACING DESIGN

Determine the steam tracing requirements for an S-in.

Schedule 40 gas-vapor line with a minimum process tem-

perature of 140'F. The piping insulation is 2rlz in. cal-iium silicate, 9-inch nominal IPS. The system is to be

designed for an ambient temperature of 0"F and a 15

mph wind. The tracing medium will be 150 psig steam,

aid tlz-in. copper tubing without HTC will be used fortracing.

We first try using two tracers running alongside bot-tom of process pipe. Calculating the areas we have the

following:

A" = 3.63 ft

(3-47)

.-t = go"F

sos'r8o3'F

-

8O3"F

'-888'F

-

8sa"F

Figure 3-18A. Thermal gradient through pipe clamp, clevis,and supporting rod.

A' : 0.131 ft'?

From Figure 3-4, ht = 2.5Calculating T",

h {D' * 2t) = o.+rs

\Di /

/r rr\r. = l=l 0.41e = 0.256 ft

kr : (2.0X0.04) : 0.08

From Figure 3-5, h" : 4.5 (assuming At : 50'F)

U. = 0.292

Formulating a heat balance for t}re system we have thefollowing:

Qr (ah space to atm) : (0.292)(3.63)(140) : 148

Btu/hrQ2 (tracer to air space) : (2.5)(0.131)(2)Q26) : 147

Btu/hf

The assumed number of tracers is inadequate for 1 :1.3.

u.

i


I.=9OO"F= PROCESS FLUID TEMPERATUBE

Trying three tracers, we have

Qr : 148 Btu/hrQ: = 221 Btu/hr

Since Qz > (f)Qr, the system is adequate using threetracers.

EXAIIIPLE 3-2: HOT OIL TRACING DESIGN

A 3-in. schedule header contains asphalt which is to bemaintained at least to 445'F. The 3-in. header is to betraced with hot oil (Ce : 0.50 Btu/lb-'R p = 58.7 lblft3 at 475"F). Determine the size and number of hot oiltracers required to maintain the asphalt at a minimumtemperature of 450'F. For asphalt, Cp : 0.368 Btu/lb-"F at 500"F.

For most applications, l/z-in. copper tubing is the stan-dard size for tracing operations. We select a l/z-in. 18BWG gauge steel tube, At : 0.131 ftlft, k^ : 27.5Btu-ft/hr-ft2-'n First we will try one tracer,

Di : 3.50 + 0.50 : 4.00 in. : 0.333 ft

'{"1 ltll!'1-'-

*---l

Figure 3-188. Thermal gradient through PiPe clamP support.

D. : 8.00 in. : 0.667 ft;D" + 2ti :8.0 + 2(2.0) = 12.0in. : 1.0ft

,n [o.ooz + zro.roull = 0.203 rt[ 0.667 I

0.667 + 2(0.167)

A" = 2.095 ft'?lft; Ap = 0.916 ft2lftl' At : 0.131 ft'?lft

A- : o 216 : o.ol8 fr: t' - j1^ 12 2n,

- 0'690 ft : 0.345 ft2(1.0)

hr" = 0.33[ffiu'.*r,,J: r't,tr" : hr" + eh,:3.992 + (0.90)(1.185) : 5.059

1-0203+ I = u^:0.449u" 0.1 5.059

Now performing a heat balance we have

t, = 350'F and tn, : 490"F. Using Equations 3-6 through3-11 with 70'F ambient,

rrt___Iolo

i-1I

H_-

n-lI

Iol o

3

n, _ 0., (sso -:sojo" = z.z:o\ 0.s I

/ssn - rso\o 25hp:0.5(""ffiJ :r.375

9z - t2.236t(0.t31x550 - 350) =

qr = ( 1.375X0.916)(520 - 350) =

t)1 slqa = Qt (ffi/ ro ott'tsso - +ro'


qt : (0.449)(2.095)(350 - 70) : 263.383 Btu/hr-ft qt : (o.449)Q.095X345 - 70) : 258.680 Btu/hr-ft

q2 : Q.25)(O.131X550 - 345) : 60.4248tts/hr-ft

q3: (1.383X0.916)(497 .50 - 345.00) : 193.191Btu/hr-ft

tr1 slq4 = (2t l_:i_:- l{0.018X550 - 445) : 301.304 Bruihr_ft

\u.J+)/

q4 < 2q3 - No balance

Since we have reached the minimum desirable tempera-tures for q and to, it is clear that the system will not bal-ance using one Uz-in. tracer. Therefore, we will use twot/z-inch tracers. Referring to Figure 3-2b we consider thefollowing:

D; : 4.645 in. : 0.387 ft

D" = 8.645 in. : 0.720 ft

ti:2in. :0.167ft

_ Di * 2tr , iDr + 2t,l'1" = _-_ lnl |:0.224' 2 \D, l

A" : 2r(0.360) : 2.262 ft'?lft

58.583 Btu/hr-ft

214.1l5 Btu/hr-ft

: 172.174 Btu'lhr-ft

jt < 2qt - No balance

Consider t" = 300"F and te2 : 450oF.

h, : 0.5 l))u - ruul : 2.364' '- \ 0.5 I -'--

hp : 0.5 (::L roo)"' : r.+s+

qt : (0.449)(2.095X300 - 70) : 216.351 Btu/hr-ft

qz : Q.364)(0.131X550 - 300) : 77.421 Bttlhr-ft

q3 : ( 1.454)(0.916Xs50 - 300) : 266.373 Bttlhr-ft

t)1 \\q^ = t2) l-::-:l (0.018)(550 - 450) = 286.957 Btu/hr-ft

\u.J4)/

qq 129t = No balance

consider t" : 350'F and to, : 456"p

qt = (0.449)(2.09s)(350 - 70) : 263.383 Btu/hr-ft

q2 = Q.236)(0.131)(550 - 350) : 58.583 Btu/hr-ft

q3 = (1.375X0.916)(500 - 350) : 188.925 Btu/hr-ft

I 'tt s\9" = Qt 1:::--: -1(0.018X550 - 450) = 286.957 Btu/hr-ft

\u.J+),/

jq < 2q3 = No balance

Consider t" = 345'F and te2 : 445'F

r,, = os(try#, +10" = 2.25

h,":033[sffiffi,,t:,,,,h- : hr. + eh, : 1.996 + (0.90X1.185) : 3.063

| -0'224 * -l = U^ = 0.390u" 0.1 3.063

t^ : 2tr(2.323) - 2(0.886)(2.323) : 10.479 in.: 0.873 ft

r, : t' = 0 873 = o.ztsft2n, 2(2)

A, = 0.13t fP/fl: A. = 0.018 ft: Ap = 0.916 ft?/ft

consider t" : 350'F and to, : 490'F h, : 2.236ihp : 1'375

q : (0.390)(2.262)(350 -'t0) : 247.010 Btu/hr-ft

q2 : Q.236)(0.131X550 - 350) : 58.583 Btu/hr-ft

q3 - ( 1.175 )(0.916X520 - 350) : 214.ll5 Btuihr-ft

/rt.\9a = t2\Qt l '1j".

| (0.018X550 - 490) : 544.954 Btu,/h.:\u.z 1d/

r,": o.s({q::g)'" = 1.383


p=938.08 Co at 500'F; k:0.1 #h

139

t1'7 alq. - (2x2) l^'-:j" l(0.018)(550 - 500) = 454.128 Btu/hr-ft\u.l r6/

qr>2qandq2+q3>qr

Therefore, the system is balanced.For 100 ft of pipe,

: (100)f(454.128) Btu/hr-ft = 45,412.8 Btu/hr

: 22,706.422 Btu/hr for each tracer

_ 22,706.422 Btulhr

Consider L : 350'F and tp2 : 500'F

9r

9:

(0.390X2.262X350 - 70) : 247.010 Btu/hr-ft

(2.236X0.131)(550 - 3s0) : 58.583 Btu/hr-ft

(1.375)(0.916)(525 - 3s0) : 220.4r3 Bt'tlhr-ft

m

1.0

ra ra ln (ry'rr) , 1

ht z kz,: h: +

For film coating in inner tube,

Btuo = 95.909 lb/ftr: C- : 0.34 -

lb-'F

Nn"

Nn" = 3.242 <<

For laminar flow,

hrz : 1.86(NnJr"(N,)'' (P)'i3 H [:)."

V = I 5o -,"tut

(+, J F*r' J (,-r', J Hg)V = 3.781 ft/sec

^, _ VDP

(j.781) I t4.026)in. Ull '- lb 160 secl

sec 1,, ,n.7 '" ""ttr \ h, /

b'rL\{e38 08)ce l ill

\ *p i

2,100 + Laminar flow

we have [2]

N".:f/, , -,lo \

(e38.08)ce l+l to.:+r,oB-tu,

[o/nr Btu- ^ hr-fr-'F

: 7 ,654.733

/ rr'\D - 4 026

'" t,-rJ = 0 336: L = loo rt

hr., = 1.86(3.242)r r (7.654.733)r ' (H) '

(3-16)

In most instances, the ratio plpw = 1.0. Thus, we have

'(#)(10)#.Ff',, = 2.415-T-

nr-rt'- - f

For hot oil in annulus, Di/D. : 0.664 > 0.2, and Perry

[6] recommends

9r

or,

q

: 908.257 lb/hr(0.5) Btu/lb-'F (50)'F

-- t ,r, .i, \(c08.2s7) I lz.+s 4l lll . '- nr \ rr / \ou mln/

Q = -, _-- = t.ezy gpm

(58.D*

Thus, we see that two l/2-in. tracers containing hot oilflowing at I .929 gpm is adequate to maintain the asphaltat a minimum temperature of 450'F. The next step in thedesign is to do a hydraulic analysis using the principlesofChapter 1 and size pumps to handle the hot oil. (Chap-ter 6 discusses how to select the pumps required to dis-tribute the oil in the system.)

EXAMPLE 3-3: JACKETED PIPE DESIGN

A jacketed pipe shown in Figure 3-1b is to be ana-lyzed. The process fluid to be heated is a film coatingmixture used in the manufacture of roofins tiles.

NN" : 0.020 ^9,'*"1,,(*]'

^, - /rCo"p, - k

140 Mechanical Design of Process Sysrems

For the annulus,

Rs . hydraulic 1361r, = 3 033 --

2 250 .- 0.392 in.

D : 4Rn : 1.566 in. = 0.131 ft

For hot oil flowing at 0.5 gpm

Nr" : (4o.ro7r ]9:99: - ),0n.24r'' (0.1s)( L566)

Now,

q : UA(LMTD)

Rr''q : 0.75t "* ( 7.80)ftr(72.135).F

It-nr- "ts

q -- 6,381.625 lI ,, ,n" heat transfer required-hr

q : ricpat

For hot oil, At = toH - toc : 100'F and,

.' - ..r. Btu

' th-'FRtrl

6,381.625 =.nrm:-:

0.5 Btu ooo).Flb-'F'

Now,

th

ftr

= 0.271 gpm required

For t/z gpm,

lh Rr"I = (235.428) _: (0.5) _= (t00)oF- hr lb-'F '

or

q: I1,771.400 Btu/hr

Thus, 0.5 gpm is a sufficient flow rate to transfer the re-quired heat to the film coating mixture.

EXAMPLE 3.4: THERMAL EVALUATION OFA PROCESS TAilK

A coating surge tank contains 6,000 gal of fill coatingmix (see Figure 3-19). Two problems musr be solved: (a)how many degrees per hour can be obtained from aclamped-on jacketed system, when the fill coating mix-ture is static; (b) how many degrees per hour can be ob-tained from a clamped-on jacketed system, when the fill

127.$21!hr

Nr" = 0.020(5,01 r.24D0.8(z.s3s)'t3 (0.*U*)*' : ze.r2r

1^

Err'(29.121){0.071) ntu

_ NN,k hr-ft'/-"F/ftD 0.131 ft

rs.rs: Btu

hr-fC-"F

rr : |13.033) +- L 2.4rs

, 1 l-'- 15r$l

,0.,r,a" |i2.+ rorR-r,'l

,0.r, Iu_ '\ co / lb-'F

n n7r Btu" "' ' hr-fC'F/fr

(3.033) ln (2.2so/2.0r3)

Ri,r

ft']-hr-'F

A : heat transfer area : outside surface area of innertube

A - 1.178 fPlft : 117.800 ft, for 100 ft ofpipe

In hot oil applications it is common to assume that the hotoil decreases in temperature 100'F per 10 feet in jack-eted and traced systems. For the film coating mixture,

tcn : 500'F and t. : 459'P

For hot oil,

ton : 550'F and to. : 459'P

LMTD : 72.135"F

@0O gal

f.1.537ol

12.82 lb/gal

cP= o'g+

COATING MIXTUR€ ATTEMPERATURE t

Figure 3-19. Coating surge tank.

coating mixture is flowing through the tank at 150 gpmar 360"F.

From Figure 3-10, we have

LMTD : :104.869'F, [sso - :ool'" t4so - 4ool

with Q : mceAt

and Q : UA(LMTD)

, UA(LMTD)DLp

thm - {6.000)gal(12.82)

= : 76.920.00 lb

gal

Using heat transfer panels shown in Figure 3-12 wecompute the toial available heat transfer area as follows:


Flanged and dished head = (92X0.8)= 73.60 ftz

shell = (379.347X0.8)= 303.478 ft'?

4-internal heat transfer panels : 4(107)(12)/1,14: 35.667 ft2

The overall heat transfer coefficient, U-value, sup-plied by the panel manufacturer for applications to thefill coating mix is as follows:

Process Conditions (as determined by process engineersor client for desired capability of tank):Initial temperature of coating mixture = 360'FFinal temperature of coating mixture = 400'FFor internal panels, U : 9.52 Btu/hr-ft2-'FFor external clamp-on jacketing, U : 4.00 Btu/hr-ft2-"F

Substituting into the previous equation for At we have

^, _ (9.52x35.667)(LMTD) + (4.0X377.078XLMTD)-' - (?6,110 ooxo 34l

at:7.410'F/hr

Referring to Figure 3-19 we can now determine howmany degrees per hour the fill coating mix will rise usingexternal clamp-on jacketing on surfaces of the flangedand dished head, the vessel shell, and four internal pan-els just considered:

min oal lbQ : (60) : " (150) ":' (12.82)

-- hr 'min 'gal

x 0.34 Btu (t - 360)'Flb-'F'Q : 39,229.20(t - 360) Btu/hr

Now,

UA(LMTD) : 39,229.20(t - 360)

(9.s2)(35.667)(LMrD) + (4.0)(377 .U 8)(LMrD): 39,229.20(t' - 360)

1,847.862(LMTD) = 39,229.2O(t' - 360)

LMTD : 2L.230(t' - 360)

Now,

(550-360)-(450-r')

-

| 1550 - 3601'"t450-tl

COATING MIXTURE

(550-360)-(450-4oo)

21.23(t' - 360)


Solving for t,

(21.23r' - 1 ,642.80)

wP: wetted Perimeter : (--. Jo*t*

WP:

A : cross-sectional f low area : 0,68 I in.'? : 4a(W)

where W: a:0.4125 in.W : effective heat transfer length (see page 145)

A:2Yr+2wY

A = 2(O.4125)'? + 2(O.412r'1 : 0.681 in.2

^ A 0.681 in.'l'' WP -1.158 in.

D : 4RH : 0.862 in.

The equivalent circular cross-sectional area = r(0.431)'?: 0.584 in.'?

The hot oil properties are as follows:

, _ 550"F - 450'F 500'F {since we anticipale

in the plates)

Velocity of hot oil through baffles : 7.913 ft/sec

p : 58.7 lb/fc

k : 0.071 Btu/hr-ftl"F/ft

Cp : o.5o Btu/lb-'F

/^ . .. ," . \

a = 0.15 c- lz + rD/rl-nrl

- 0.3b0 tb/lr-hr' '\ 1Co I

The maximum pressure drop permitted through the in-ternal baffles, which are connected in parallel, is nor-mally 10 psi, thus

* - VDpl\Re -

-l-

(7.e*) r-t (0.862) in. {-.1. l' l,rr.r, g (lql. .'*)sec \ll ln./ rr- \ I nr /

0.360 Ib

ft-hrNr" = 333'661

lh Rr(0.360) ,j: (0.501 :j:

't <l<Btu

lu.v/lr-

(110.394t', - 39,841.956)ln (450 - r')

Letting y _ (21.23t' * "t ,642.80)(l 10.394t' - 39,841.956)

we have 1.0 : ln (450 - t'y

Or el.0 : eln (450-r')Y

in which 2.718 = (450 - t')Y

After several iterations, t' : 366.12'F

Thus, the temperature rise is

at : 366.12'F - 360"F : 6.12'F

The amount of heat required for the system is

Q : UA(LMTD)

Q : (9.52)(3s.667)(LMrD) + (4.0X377.078)(LMrD)

OI

Q = 1,847.862(LMTD)

Now,

LMTD = 1550 - 390) (450 - _366,12) = 1t9.789.F. lsso -:oo Iln |

-l[4s0 - 366.12l

in which Q : 1,847 .862(129.7 89) : 239,832. 162 Btulhr is the heat transferred to the coating mix.

EXAMPLE 3.5: THERMAL DESIGN OFPROCESS TANK

The coating surge tank of Example 3-4 is now ana-lyzed for detailed heat transfer requirements. The flowrates through the various types of heat transfer jacketplates are desired.

lntelna! Baffle Plates Film Goefficient

Some of the plates used are shown in Figure 3-12.Looking at Figures 3-20, 3-21, and 3-22 we determinethe hydraulic radius as follows:

N"': f


From Kern [],

- : = ro.o27r ([) ,t-.,., ,N,,,' ' (uJ"'"

BtuI /tl-

r. = {0.027) 'hr-ft2-'F/f

ro.aozrin. llrt I\ l2 in.i

x (333,661f 8 (2.535)1/r (1.0)

K23: 29 BTU/hr-t12-"F/fl

ri :0.44in =O.O37 ttr. =0.545 in = O.O45 tt

thn

1. . = qsl R? Btu is the film coefficient

hr_fd_.F inside the jacket baffleslocated inside the tank

Film Coefficient Exte?nal toBaff les-Forced Convection

The coating mixture inside the tank is in a state offlowacross the baffle plates made possible by agitator bladespowered by an electric motor. From Perry [61.

- i r\ir-:r.r.o\." /c-J''rtt = U.U9 l-ll "I l!l\DJ \ p / \k/

'h9'^ ("il'kl'ni = number of internal baffle plates : 4

For coating mix, 10r < NR" < 2 x 106

The properties of the mix are as follows:

p : 95.909 lb/ft3

A = 6,000 Co : 14,400 lb/ft-hr

k = 0.1 Btu/hr-ft-'F

Other properties related to the internal baffle plate ap-proximated as a string of cylinders with diameters equalto four times the hydraulic radius of the trapezoid platesections shown in Figures 3-20 and 3-21.

\6 = angular velocity of agitator, revolutions/hr

/-^ \\. - 56 11 lgel'nl : 3360 revrhr

min \ lhr /

Lp : diameter of agitator plate : 65 in. : 5.42 ft

Dj = 10.0 ft

D" = 0.862 in.

Figure 3-20. Cross section of panel plate tube approximatedas a cylinder whose surface area is equivalent to the heat trans-fer of contact area.(3-48)

HEATING AREA: A.

t= STRAIGHT SECTION BETWEEN CHANNELS

Figure 3-21. String of tubes.

Actual Internal Baftle Hot Oil Plate

Approximated as Circular Tubes

^ [(o r4){ r4.4oo)1,

, {lJ2l

, l]),'[ (0.I) l 110.0/ \41

h:.+ : 18.334 Btu/hr-ft2-.F for all four baffle plates

144 Mechanical Design of Process Systens

k : 0.071 Btu/hr-ft2-.F/ft

h,q - 0 0e (oo r-)

['s +:r't:'log.o

'qs eoqJ'^'

(1.0)

Now, to solve for the overall heat transfer coefficient,U, we must develop the appropriate equation.

Let Ao = the overall area of plate : LW

L = length of plate, W : width of plate

Referring to Figures 3-20 and 3-21 we have

e : UAa (tr - ta) for baffle plate

Ar : LW : wetted area : 2oD rL

where W : werred perimeter (WP)

srLJ | = '2 D" + r' . n = number of flow circuits

To account for the residual heat transferred through theplate connecting the hexagon tubes we consider r' to bethe equivalent radius of a cylinder that is the total surfacearea of the baffle plate separating the hexagon tubes.Now,

D, : ,o,u, distance between channels (Frgure 3-21)

Zor' = DtIt2tr

From above,

\.rq = UA. {f I - t1) - 2trlJ lzr 1t, - ta1

2T(tr - ta) n

r" r" ln (r./r)"-T-

--ffinl-2 Kt I

in which

or

lT -

[i^] o^ * E'l I u * r" rn tr,rr,r - _r lt\2/ - 2ur I lr;h1 , k, , h, J

for these baffle plates, n = 7, for which

,, 7o" -it\ --l

[(jJto to:' . i:]

I o art| - "' +[(0.326X949.883)

(0.045t ln (0.431 /0.32b) I18.334

u = 29.492 Btu

hr-ftr-'F

The baffle plate area for all four baffles is determinedfrom the baffle plate manufacturer's dimensions, as fol-lows:

Length of channel per baffle = 51.'123 ft

A = surface area - 2210.431)i" {.:-tf) rsl.72t rr)\rz ln./

A : lt.672 fe

For all four baffle plates,

A = 4(tr.672) ft2 : 46.690 ft2

Heat Duty of lnternal Baffle Plates

For hot oil the anticipated temperature through eachplate is 100'R as stated earlier, thus

q: UAAT

Rr',q - (29.492J, "-1:1, (46.690)ft, (100).Fnr-It'- -f

q : 137,698,1a8 f.!nr

Outside Heat Transfel Jaeket Plates

In the case of external jacket plates, the heat transferparameters are based on the dimension, I shown in

IL

U:D,Iq*r"lntr./r't * II

[r,hr-z k: : hr J

-tgure 3-22, because it is this surface that is in contact

^ rth the vessel wall. Consequently we can analyze the

: xfiguration in Figure 3-22 as a tube with circumfer-::re'e of W. Hence, we have the following:

i;r1 = Y7 : 1.375 in.

.rhere rr : tube inside radius, in.

:iom above,

| 175: = ; = 0.21C in. or D, : 9.433 in.

The tube equivalent flow rate for the length W mapped:nto a circle or radius rr is by the continuity condition ofiluid mechanics,

w= 1.375 in

w = eflective heat transfer area

For approximalion, analyze the ligure as a tube with a

circumference = w2Tr j = 'l.375in which,rz = O.219 in., or D2 = 0.438 in.

Equivalent llow,

13 1341/|) 111= 6.448 fusec

Thus for the equivalent tube,rz = 0.219 in.rs = 0.219 + 0.109 = 0.328 in.r+ = 0.328 + 0.375 = 0.703 in.


At = tube cross flow areawhere

At = r(0.219)1 : 0.151 in.':

Velocity of hot oil through outside plates : V2

Yz : 3.134 ftlsec (determined from process data)

internal cross flow area of baffle plate : 0.62 ln.2 (fromplate manufacturer's data)

(3.134) I (0.62) in.,secvr-

o.Lsrin: -

=

vDp

6.448 :-sec

c,

h where, tq 5

Djk

LP

N,

Equivalent Cylinder

Figure 3-22. Panel total flow cross section. Contact length wis mapped into an equivalent circular tube whose circuinfer-ence equals w.

I'r(6.448) a t0.438r in. I lll'l ,s8 7, 'g i{ryleclsec \12 in.i in.1 \ | hr /

(0.360) lLtt-hr

Nn" = 138,150.85

Solving for the over-all heat transfer coefficient, U, wehave

1.,1o. : PCo = 2.535k

h,, : (0.027) (H1.ol ,rrr,rro r5)0s (2.535),/, (1.0)

h',, = 77.260 Btu- hr-ft2-'F

Perry [6] gives the correlation for heat transfer for jack-eted walls to the agitated liquid as follows:

'' : "(;)t')1'(9^k)' (3-49)

film coefficient at vessel wall (see Figure 3-22)inside diameter of the vessel, ft = 10.0 ft0.1 Btu/hr-ft-'Fdiameter of agitator = 5.42 ftangular velocity, or rotation of agitator

3,360 rev/hr95,909 lb./fC


h,,. = 14.060 Btu' hr-fC-'F

Thus,

II _

14,400 lb/fi-hrspeciflc heat of coating mixviscosity at bulk temperature, lb/ft-hrviscosity at wall temperature, lb/ft-hrsee Table 3-l

.- t (0.703 )u=l +l(0.2t9)(71 .26)

(0.703 ) ln r0.703/0.328)

(0.73) ln (0.328/0.219)

I lrII14.06-lLaboratory tests were made on the coating mix and the

results showed that p6lp* : 0.65. Since the coating mixis a non-Newtonian fluid, it is strongly recommendedthat the physical properties be deterrnined by a qualifiedIaboratorl,. the ratio p5lp* should reyer be assumed to beI .0 for a non-Newtonian fluid without laboratory tests offluid samples.

For a disk. flarblade turbine agitator we find valuesfor a. b, and M from Thble 3-1 as follows:

a:0.54,b = 2A,M:0.14

since40 ( l38,l5l < 3 x 105

substituting above values into Equation 3-49 we have thefollowing:

(14,400)

U : 8.141 Btu

hr-ftr-'F

From manufacturer's drawings, the shell jacket plareheat transfer area, A, is

A : 37,043.82 in.'?

Now,

A' : area of channels in all nine jacket shell platesclamped-on to outside of shell

A' : 257 .249 ftl

Ar = 100"F for hot oil-coaring mix servicc

Rr' 'q - {8.14lr . :i-_r25t.24o1ft (100)'Fnr-It'- - t

or

Rr,'q = 209 .414.44 --:

nr

Heat Duty of Jacket Plates Clamped toBottom Vessel Head

The bottom head panel sections are depicted in Figurel-17. In Chapter 1, Example 1-2 we analyzed the hy-draulics for the hot oil flow through the panels mountedon the tank. From this analysis we determined the fol-lowing velocities required to obtain l0 psi pressure dropthrough the panels:

innerpanel : V : 7.315 fi/secouterpanel : V : 5.237 ft/sec

Heat Duty for Bottom Head Inner Panels

Similar to the shell panel plates above, we must com-pute the equivalent tube diameter and equivalent veloc-ity. As determined above the equivalent radius is

rr : 0.219 in.

,*,(,*)[ (s.42t(3,360.0X9s.909

l-,

lto

:ltr+,+ool]"' ,o.ur',o,o

Table 3-1Values of Constants for Equation 3-49

AgitatorFeynolds Number

RangeDisk, flat-blade

turbinePitched-blade

turblnePropellerPaddle

Helical ribbon

0.54 2lz 0.14

0.53 2lt 0.240.54 ,/3 0.140.36 2h 0.21

0.633 Vz 0. l8

40<NR.<3x105

80<NR"<200Nn" = 20003oo<NR"<3

x 105

8<NR.<105

S.lce the bottom baffles have the same flow area as the.rell plates, the cross flow area of the equivalent tube is

i = n(0.219) in.2 : 0.151 in.2

I re cross flow area inside the plate channel is found::om the manufacturer's catalog to be 0.31 in.2. Since:e equivalent tube circumference is equal to the contact::mension, w, as above we must compute the equivalent..locity. Thus

-quivalent velocity = Y. =(7.315) -:' (0.31)in.2

sec

0.151 in.'z

\-


q - uA at = (8.s90). lr)-- 1r.+r+)n, lroo;"r' hr-ftr-'F

in which

Btu

h,for both two inner plates

322,453.78

Nn.

-.. . : (0.027)

:..:152.2ll

l9J11l r:zz.+s:.28)0" {2.535,r 'r { 1.0)\u.4J6/

Btuhr-fP-"F

The vessel-side film coefficient is the same as for the

'hell panels,

i.r. . : t+.gt+ Btu- hr-ftr-'F

Thus,

, - | (0.703) (0.73) ln {0.70Ji0.J281' t(02lrr(15, rr1- - m-* (0.703) ln (0.703/0.328) * I

I29 14.914)

Eli, 'r, :8.590. :i^=nr-tt'-'t

Let A : heat transfer area of bottom head plates fromrhe manufacturer's data the flow path length is 388.231in. for one half of the head, hence,

1 = (18$.231) in. (1.375) in. (2) : 1,067.635 in.2

OI

A : '1.414 ft2 : area of the two inner head plates

q : heat duty

Heat Duty for Bottom Outer Panels

(5.237rt0.31 rtcutvatent \ etoclt\ - lu. /f I lt/sec0. 151

/r\{ 10.75l,(0.438r l-l r58.7 rr3.600 )

0.160

/^ ^--\h,2 - {0.027) lfffl rz.lo..lss.:41)0b r2.s35) '(1.01'-

\0.438/

h,, = 1t6.303 , :tu^=nr-lt'-'f

Similarly as for inner panels,

f'n, : .l.l+,SlO --!$-nr-rI'--t

Thus,

r (t.trr:, rn lo 34'l

,, | 0.703 _ \0.219/" tO) 9,( t 16:0l, 29

/^ -^^\ro.703 r ln lu /urI

, \0.328/, I I'- 2s ' r+.st+)

u : o.lg+ Bt'hr-ftr- "F

A : heat transfer area

A = 4(1,014.389) in. (1.375) in. = 5,579.140 in.2

OI

A = 38.'144 ft2 for all four outer panels

q : UA At = (6.394)(38.7 44)( 100) = 24,772.333 ryhr


Total Heat Duty of Tank

At maximum flow rates the total heat duty is as fol-lows:

q = 137,698.148 209,414.44

internalpanels

+ 6,368.11 + 24,772.333

Thus, the minimum hot oil flow rate in pipe header sup-plying the total hot oil to surge tank is 10 gpm, the actualflow rate is 16 gpm.

EXAIIPLE 3.6: TRANSIENT AND STATICHEAT TRANSFER DESIGN

Roofing shingles are made by adding asphalt, fillermaterial, granules, talc and adhesives to a plastic-glasssheet, which is the basic component of the roof shingle.The process is shown schematically in Figure 3-23.Granules are added last, after adhesives and talc. Thesheet must be cooled so that workers can handle it withgloves. Cooling is accomplished by water sprays, circu-lating water through the rollers, and using radiant heattransfer to the surroundings. The sheet, once cooled tothe desired temperature, is cut into specified dimensionsby mechanical cutters and then packaged into boxes forshipment.

There are two aspects to this problem-static heattransfer and transient heat transfer. First, we solve thestatic conditions and then the transient case to determinehow fast the sheet can be cooled with the coolins svstemdesigned in the static case.

Static Heat Transfer Analysis

The static criteria to be determined are as follows:

1. Specific heat of the composite sheet (Table 3-2).

2. Mean temperature of the sheet leaving the granulesection flhble 3-3).

Table 3-2Specitic Heat of Composite Sheet Leaving Granule

ComponentComponent wt., lb 0/o by wt. Cp o/o ol Cp

Btu ,

hr

Btuhr

shell sidepanels

Btuhr

Btuhr

q = 378,253.631 P!!hr

From Example 3-4 the total heat duty required is

q-t = 23g,832.rc28:nr

Now,

q: rh cp At

Elr"373.253.631 "-

Rrrl{0.50) :t: ( l00toF

tb-'F

two nnerpanels on

bottom head

7,s65.073 Pnr

th4.796.640 :nr

58.7.]9'n'

= 10.187 gpm

four outerpanels on

bottom head

flnln

Elr'r239 ,832.162 :::

nt

(0.50) Btu { 100).F

tb-'F

7,s65.073 l!nt

th4.796.64

nr

4.", : /r.+a gur\

\ri/5s.7 !ftr

: 16.067 gpm 0.2o.40.2170.20o.20.50

o.395'710.96888.217 46.03484.01280.3188

29.9483

Glass matAsphaltFillerGranulesThlcAdhesives

6.30 L978587.32 2',7.4220

120.58 3"t.868296.08 30.1740

6. I I 1.92002.03 0.6375

318,42 10000

)q q4R IC, = t;0- : 0.299 Btuilb-'F

f 7.48 gall

\- r/

Glass.A.sphaltFillerGranulesTalcAdhesives

80'F400'F400'F

80"F80'F

400'F

Table 3-3Mean Temperature ol Sheet

Leaving Granule Section

Q = mcp At for each componentTemp. of component

Component prior to mixing Cp Eo by wt.


3. Heat to be removed from sheet and amount of wa-ter required.

a. t- : 313.63'F (from Table 3-3)At:313.63 -212 = 101.63"FSpecific heat of sheet : 0.299 Btu/lb-'Fweight of sheet : 0.9375lb/ft2

Thus, the amount of heat to be removed persquare foot is

mCpAt = (0.9375X0.299)(101.63) = 29.49 .t 2tBtu/ft'?

b. Sensible heat loss through rollers : 3.0 x 106

Btu/hr

c. Heat loss through forced convection and radia-tion of heat passing through air medium is deter-mined as follows:

At : 313.63 - 90 : 223.63"F = temperaturedifference between sheet and ambient air

0.2 1.97850.4 2'7.4220.22 37 .86820.2 30.17398o.2 r.920.5 0.63"75

(1.978s)(0.2X80 - t^) + (27.422)(0.4)(400 - t,)+ (37 .8682)(0.22)(400 - t.)

= (30.17398X0.2)(t. - 80) + (1.92)(0.2Xt- - 80) +(0.6375X0.5)(r. - 80) 31.656 - 0.396 t, + 4387.520- 10.969 t^ + 3332.402 - 8.331 t.

: 6.035 t. - 482.',784 + 0.384 t. - 30.720 + 0.319 t.- 25.50

where t. : 313.633'F

FINALCOMPONENT

t

]|I

Figure 3-23. Process of manufacturing roof shingles.

'150 Mechanical Design of Process Systems

For convection,

Q : hAAt

For flowing air, h,in = 2 Btu/hrlft2/.Fh."^ : 50 Btu/hr/ftrl.F

Use h. = 23 Btu/hrtfP l"F

/6R {rA ";:-'(l) = s.708 fr2t2

For radiation,

For 600 shingles/hr (or 144,000 Btu/hr) the heat removalwould be : (144,000X0.9375)(O.299)(313.63 - 125)

: 7.614 mm Btu/hr

Thus. Total heat > Heat removalremoval requtled

and the cooling system is adequate.

For vaporization, q : 29 Btu/ft,

Qv : 104,400.00 Btu/min

For water, h1, : 1,000 Btu/bb

At 600 fpm, we have,

1r)4 400;*; - 104.40 lbi min - amounl of water required

Thus,

104.40 gmtn = 12.518 gpmgpm =th8.34:

gal

Thus, the water pump to be used is to be sized for 13gpm at a terminal exit pressure of 200 psi.

Transient Heat Transfer Analysas

This method is based on the Fourier analysis of un-steady-state heat conduction. The following assumptionsare made:

l. The composite sheet is approximated by a materialof average conductivity.

2. The sheet is infinitely long and is an isotropic rigidsolid.

Figure 3-23 shows a view of the roofing slab. Assum-ing that the material is a composite sheet approximatedby an integral sheet of average properties, the tempera-ture distribution is at all times symmetric about the mid-plane of the slab, thus x = 0 at the center of the sheet.From Fourier's law of conduction,

AIQ = -k-

dX

The heat transfer across x = 0 is zero and at the mid-plane of x : 0 the sheet behaves as a perfect insulator-an adiabatic surface. Consequently, the solution to thisproblem applies to a slab that is perfectly insulated at one

| ,lh, = F.Fo lo(tr" - t':")l

t (tr - tz) J

Fe:1.01 F": e = 0.90; o

h. : (0.90)(1.0)

: 0.173 x 10-8

h, : 1.857 Btu/hr/ft,/"F

Total Heat Removal

hr : h" + h. : 23.000 + 1.857 = 24.857 Btuthttfet"F

Q = h1A(At) : (24.8s7X5.708) (223.63) = 3r,729.464Btu/hr

Water Required for Cooling

Let Qv : heat removed by vaporization

The heat removed for a sheet 6 feet wide moving at 100ftlmin is

Qv : (100)(6) frrlmin (29) Btu/ft2 : 17,400 Btu/min

For a sheet velocity of 600 ftlmin,

Qv = 104,400.00 Btu/min

Qv : 6,264,000.00 Btu/hr

Total heat removal : 6,264,000 + 31,'729.464+ 3,000,000

Q = 9,295 ,729 .464 Btulhr


Iace, initially at a known temperature, to, and then ex-posed on one face to a fluid at a constant temperature, tr.

Temperature of the sheet = 314'F : t"

Gmperature of the spray water : 90'F : ti

Here we are spraying water on the sheet and we wishto determine the time required for the sheet to reach125'F.

0.:314-9O=224"F

0,L=o = 125 -90:35"F

l =-=0.156c. | ,=r 224

Thickness of sheet : 3/ro in. in which L : :/:zrn. : 0.094 in.

For water,Surface coefficient (worst condition) = 300 Btu/hr-ftl"F

k = o'30 : o.otthL (300X0.094)

Np. : Fourier number

From Figure 3-24, NF" : 0.90 : d7

L-

k 0.30 : 1.070 ft'?lhrp Cp (0.9375)(0.299)

0.90(0.094),r =

-

- U.UO/ hr = 0.2146 min1 .070

or r = 26.'756 = 2'7 sec for 90'water

Approximate length of sheet exposed to nozzlesplay : 150 ftVelocity of sheet : V,

For V, : 600 ft/min,

35 = 0.25 min = 15 sec150 ft

600 ftlmin

For V. : 400 ftlrnin,

150 ft : 0.375 min = 22.5 sec400 frlmin

The length of the cooling section and the velocity of thesheet are both fixe.d. The only parameter not fixed is thetemperature of the water.

Thus,

r: NroL2 : o.oo417ct

I

m/L2

Figure 3-24. Heisler's main chart for the infinite slab [7].

(*).,,-,, : oo,o


(Nr o),"o a (o oo4l7j( | o7o'

0.505' (0.094)'

For a Fourier number of 0.505,

Let t" : lequirad water temperature

l)5 - r

Jl+ - t,\

in which t* = -11.86"F and is well below the freezingtemperature of water.

Thus, for a cooling section of 150 feet long, the sheetmoving at 600 ftlmin cannot be cooled to 125'F sincethe theoretical value of t* is below freezing.

At V, = 499 Lr-tn,r = O.3"75 min : 0.0063 hr

(0.0063X 1.070){Nro),.u.r = .: ^: --j--- = 0.7568

(u.u94r

ln\l;l = 0.180

lts - r

314 - t*

and

t* : 83.51'F for a sheet velocity of 400 ftlmin

Thus, the sheet can be reasonably cooled while mov-ing at 400 ftlmin. If a velocity of 600 ftlmin is desired,additional water sprays must be added. However, onemust balance the sheet velocity against the cutting ma-chines and workers' capability to handle the additionalmaterial. It is found in most roof shingle plants that 400ftlmin is an optimum velocity. As demonstrated, thetransient heat transfer analysis is mandatory in evaluat-ing a system.

EXAMPLE 3-7: HEAT TRANSFERTHROUGH VESSEL SKIRTS

Calculate the temperature distribution down the lengthof a vessel skirt. The vessel contains a cold process fluidthat varies in temperature because of cyclic process con-ditions. Three operating temperatures are to be ana-

lyzed, -200"F, - 100'F, and -50'F. The skirt is madeof Type 3(X stainless steel and is insulated on the insideand outside as shown in Figure 3-28. The insulation issized for the most extreme process temperature that thevessel will be exposed to, -200"F. Data used in the ex-ample are given in Figure 3-25.

First, determine the natural convection film coeffi-cient for the skirt. The temperature inside the skirt, ti, isassumed to be five degrees lower than the ambient tem-peratute, t5.

uo : [', 'L(no,r

.,. O*4, * u'i,r:.',n) * iJ-'I r l-,= [7.r1s

+ -hJ

Assume h4,5 = 0.275

u4:0.093

4 = 55'FG = 60"Ft1 = 3048-in. :2.573lt/, = 367,b-in. = 3.073 ftf3 = 37%-in. - 3.135 ft/4 = 435//6-in. : 3.635 ft

A^ = (tt - r,2) = l.z0glt,F - 1(460+ 60) = 0.00'19231 = 0.07633 lb/tt3p = 0.04339 lbfit h

k*, = 0.01466 Btu/h ft "Fkyz 8.0 Blu/h ft.Fka t = kg-'q = 0.14 Btu/h ft'FNP' = 0'712

Figure 3-25. Cryogenic pressure vessel with internal and ex-ternal insulation on the skirt.


,. = [*J (tr - ts) + ts

/o.osl\ . -:l l{-))+bU\0.275i

: s8.31'F

L-t5:58.31 - 60 = -1.69"F

\o, = [d37,gB( | Atl)(3,6o0)2]/p,

l(7 .27)3(0.07 6T)' Q2.2)(0.00r923)(r.69)(3,6tJ0)21(0.04339f

Na,:1,613,'720,723

where d : 2te : 213.6tt) = 7.27 ft

Nc,Np, = (1,613,720,723)(0.7 12) : 1,14g,9tt,ttt> loe

C = 0.129 andm : r/:

N", : C(Nq,Np,)-

= 0.129(l,148,969,155)r/3 : 135. 1 1

n.1 = (fqi, I.") - (0.014661(l35.ll) : 0.2125' d 7.27

IOOqIIti - l#l (-s) -60: s8.29"F

: 58.31 - 58.29 = 0.02

:4.275

For a cylinder with insulation on both sides,

#[",('[4 - 'f5) - -,

['-h

- t, ll

'(;-.,Jll

ra-1nl.2gsr l'3 63sx0 27sx-1 69) - 0 14

t

I sa.: r 55.00

,,1. Iturr\-|ln t-t ln t-l[ \3.13s/ \3.073/

Z:65.1O2"F1ft2

z:? =6s'1o2 - 57.781'FQ 1.126'7

For t. : -299'P

2(t. - Zle*oo 5

I + e2*qo 5

t. _ (-515.56)(2.89)-+ 57.78.1

1 + (2.89y'

Similarly for:

t, : - 100'F,

t. _ ( -515.56X2.89)- + 57.781" r + (2.89)"

and for t, : -50"F

r _ (-215.56x2.89). + 5? ?Rll + (2.89t^

Figure 3-26 shows these distribution curves.The axial deflection of the skirt will now be calculated

using the first three terms of the stainless steel deflectionequation (Equation 3-45). The hyperbolic terms in theequation are not necessary when the steel temperature isgreater than -300'F or less than 1000'F,

^ _ 2(t. - Z)ls.sa + (4.i1 x 10-3)Zl arctan ielo0 5;

*._

(2.055 x l0-3\z2L-f

100

(2.055 x l0-rxr" - Z)2(e2oto' - l)

+z

.:[ffH][dil.*-.19]

I z "ro.

r+r I I r= t_t t_

[(8Xl.2oe, li . /z.sz:\ t

[l '" \'otr-/ |

! rl

';T#"'JQ : r.1267 ft'? 106 eo

5(1 * e2lao )


-200 -160 120 _80 _40 0 m 40 60Temperatur€, .F

Figure 3-26. Temperature distribution for the three shell-skirtJunctron temperatures.

EXAMPLE 3.8: RESIDUAL HEAT TRANSFER

A section of carbon steel process pipe is shown in Fig-ure 3-15c. Three conditions will be analyzed for processfluids at 900"F.600'F and 300"F. the basic analysis isthe same as used in Example 3-7 beginning with theiteration procedure to find the natural convection filmcoefficient. Note that it is assumed that the temperatureinside the empty pipe header, t1, is 130'F and that theambient temperature. t.. is 60'F

U: : [(r: ln (r1r')/k.) + 13 ln (r3lrr)/ki + '/h"]: (0.s26 In (0.26710.2527)t25) + 0.526 ln(0.526/0.276)10.027 + r/h.l r

U3 : [12.565 +'/h"] '

Let h" : 1.0 Btu/hr-ft2-'F

Ut : 0 '0137

t: = (U:/h")ftr - to) + t,,

= (0.07371 1.0)(130 - 60) + 60 : 65.16.F

At=t:-L=5.16'F

}'16, : [d3e,gB( lAt l)(3,600t]/r.,: (1.052t(0.0763r2(O.O01923)(32.2)x (3,600F(s. l6)l/(0.0433eF

: 14,920,198.65

Nc,N", : (14,920, 198.65X0.7 12) : 10,623,181.44

NNu : C(Nc.NPr).

where,

Let

0.525andm:r/+0.525 (10,623,181.44)\ra = 29.97(k"i,/d)NN" : (o.ot466n.052) 29.570.4t77(U3/h;)(ti -t.)+to = (0.073110.4177)(130 - 60)+60'72.35"Ft3 t3' = 65.16 - 72.35 : -7.19'Ftoo large, try another tdal valve for hn.

h" : 0.49 Btu/hr-ft-'FV: rt(.r2.565 + r/0.49): 0.0687

Btu/hr-ft -'F

(1.06 \ to o) [0,, _ 2,,

106 t","

, [sech rLQ05) tanh (LQ05) t arcran [.,inh rLeOstfllrElt.,82rt, - 2f / ezroo: 1,a'- \l + ela t

- - tr --l- 4zlt. - Zt l;'". arcran reLo"'tltQ"' )

For t. : -200"F. Q = 1.1267,2: 6.5

6,, : 0.00701 + 0.00004 + 0.00013

6., : 0.00718 fr : 0.08616 in.

.I L'L

: 57.781 and L

From Figure 3-26 we see that for the worst case ofts =-200'F that -20"F is obtained at x : 1.75 ft. At about2.0 ft and below, the skirt could be of carbon steel con-struction and considerable material savinss could be ob-tained.

: (.0.0684'710.49) 70 + 60 : 69.781'F:9.781"F:28,279,559.99: 20 ,r35 ,046 .7 |

= 0.4901 Btu/hr-ft-"F= (0.0684710.490r) 70 + 60 = 69.779"F: 69.'781 - 69.779 = 0.002'F < 0.1

t3

AtNc,

Nc.Np,Nr,h"'\,

At:'

= 0.49 Btu/hr-ft-'F= 2?rki/[kMAM ln (ry'ra)]= 2r (0.027 ) I [25 (0.0387 ) In

(0.52610.27 6)l : O.272 fr 2

= | [2rl(k.A.)][r3h.(t3 - t") - kit3/ln(ry'rJl I: I l21rt[25 (0.0387) [0.s26(0.49)(69.78- 60) -0.027 (69.78)/ln (0.52610.276)1

1: | -2.607 | : 2.60'1"F1ft'?: zlQ : 9.587'F= t2 (i. - Zt.'oo1lt 1-.:roo51; + Z ^-: iZ t,, - 9.587)e''0 "'7'o

5/11 + e2r'02?210s)l

+ 9.587


(r4 .233)(3s4 .3s2)(1 .497)(0.521 x ltr)

(2.496 x 10-3)(125 ,565 .623Xr82.902)(0.521 x 109(184.902)

+ 0.0008 - (6.s36 x 10-') [558,494,713.0+ 2706.95 + 2000884.26

+'74,115,250.451

6". : 0.0155 ft : 0.1860 in. axial deflection

This example shows that residual heat through a closedbranch line can be significant enough to cause thermalmovements, which can result in high stresses. Thesethermal deflections are particularly important whenspace is limited and the piping system has little flexibil-Ity.

X distanc€, ll

Figure 3-27. Temperature of a branch pipe connected to aheader through a closed valve plotted from the pipe to valveconnection every six inches for a distance of five feet in Figure3-15C.

155

h

aa

z

4Ztx

For t, : 900'F,

I.780.83 (1.313)' ^ -.| :

-

r q \87'' tl + ( 1.313)'?1

For t. : 699'P

1.180.83 0.313)-r, = . + 9.587' lr + (l.3l3y'l

For t, : 399'P

580.83 (l.3l3l + 9.587ll+ (1.313),.1 -'

Curves depicting t, are shown in Figure 3-27. UnlikeExample 3-7, the slopes of the curves change much less,almost approaching straight lines.

Axial temperature gradients along a section of pipingproduce thermal deflections. The pipe support will nowbe analyzed for thermal deflections.

The surface temperature, ts, of the branch pipe at thepoint of the contact with the header is 600'F. The aver-age temperature inside the pipe may be calculated fromthe 600"F curve in Figure 3-27 which shows a tempera-ture at a distance x of five feet to be 294'F.

t; = (600 +294)t2:447"F

Through the process of iteration, h. : 0.68 Btuihr-ft-'F at the average internal temperature of 447"F. Thiswas obtained using the natural convection iteration tech-nique described in Example 3-7.

Using the same techniques, Q = 0.2719 ft 2, Z :66.7916'Flftz, and Z : 245.6476"F.

To calcuiate the axial deflection, substitute these val-ues into the expanded thermal deflection equation forcarbon steel, Equation 3-39. Note: Values for the arctanused in the equation must be calculated in radians. Cal-culate the arctan m degrees and convert to radians inwhich the relationship is 2zr radians : 360 degrees.Using equation 3-39,

-E

E

E


EXAMPLE 3.9: HEAT TRANSFERTIIROUGH PIPE SHOE

A l2-in process header shown in Figure 3-28 is sup-ported by a shoe 14-in. long. The process fluid is at750"F and it is desired to determine the temperature ofthe bottom of the shoe base plate where Teflon ismounted to accommodate pipe movement. The Tefloncannot withstand a temperature in excess of 400'F.Referring to Figure 3-28 and using Equarion 3-47 wehave

, k.A.r, + hoAplto .D'r' (k,A. + h""AI)

h" : 3.0 Btu/hr-ftl'F for carbon steel in still air

k,,, : 26.0 Btu/hr-fr-'F

L : 10.0 in.

A. = (0.375X14) : 5.25 in.2

Ap : (8.0)(14) : ll2 in.,

L:90'F

t y'lgscu ao 5"cAlcruM srLrCATEINSULATION

/- to =9oo F

L= I oin

Rtrr /< rs ;- z\126.0), ": r.-l irr r750) "F . (3.0)

nr-rr--r \ t++ In. /ffi[126

0) hr-r' 'c (1ffi] rP + (3 0]

#" (,-,-ttt{,"J r,'(,i)r, rm,'-

P=0.375in

BASE }IIDTH

=8in

Figure 3-28. Heat transfer through pipe shoe.

Bru /nz in.,\ ^ . /ro.o\ ^ l_ t_t tt. t_t ttlhr-ft'z-'F \144 in.J " \ t2 / 'J

h = 306.303'F

Thus, the Teflon on the base is adequately protecte.The amount of heat loss through the shoe base plate :.

q = h"Ap (tp - t")

(r.0) , l:u - frrz in :) ,,' {rob.J'3 - e').F'- -'hr-ftr-"F

\r++ in.J " '-

= 504.706 Btu/hr

NOTATION

A- = area of metal in pressure vessel shell or pi5ft2

Ao : outside surface area of insulation, ftzlftAo : outside surface area of pipe, ft'?/ftAt = outside surface area of tracer tube or HTC

ft2lfrspecific heat, Btu/lb-'Foutside diameter of a pressure vessel, ftdiameter, ft, in.inside diameter of pipe insulation, ft, in.outside diameter, in.inside of outer ring of annulus, in.outside diameter of inner ring of annulus. ::inside diameter of tracer tube, in.acceleration of gravity, 32.2 ftlsec2natural convection coefficient at OD of :::piping insulation, Btu/hr-ft2-'F

ho' : corrected value for h", Btu/hr-ft'?-'Fho = convection coefficient, pipe to air space, B:-

hr-ftr- o Fht : convection coefficient, tracer or HTC to

"..:space, Btui hr-ft2-'F

h4-s : convection coefficient between the outsj.:vessel insulation and ambient air, Btu/hr-::-

A_D:D; :

D,-h-

\5' = corrected convection coefficient, Btu/hr-ft2-

ki = insulation conductivity, Btu/hr-ft-'Fki = thermal conductivity ofair inside empty pipe,

Btu/hr-ft-"Fk, : thermal conductivity of vessel skirt or pipe,

Btu/hr-ft-'FL = length of branch pipe, ft

N6. : Grashof number, dimensionlessN51. : Nusselt number, dimensionlessNp. : Prandl number, dimensionlessNp" : Reynolds number, dimensionless

Q : heat transfer factor, ft 2

Qr : heat transfer from air space to atmosphere,Btuihr

Qz : heat transfer from tracer to air space, Btu/hrQ: : heat transfer from pipe to air space, Btu/hrQ+ : heat transfer from tracer to pipe, Btu/hrt" : air space temperature, oF

t; : process fluid temperature, 'Ftj : air temperature inside the vessel skirt, pipe

support or branch pipe, 'Ft, : surface temperature of the branch pipe at con-

tact point with the header, or operating tem-perature in a pressure vessel, "F

tj( : temperalure at distance x along the vesselskirt, pipe support or branch pipe, 'F

to, t5 = ambient temperature, oF

t3, t3' : temperature and corrected temperature at ODof the pressure vessel insulation, 'F

t4, t4' : temperature and corrected temperature at ODof the pressure vessel insulation, 'F

At : tr - t in piping example, ta - t5 in vesselskirt exarnple, 'F

At' : t4 - ta' in vessel skirt example, "FAtj' : t3 - t3' in piping example, "F

U3 : overall heat transfer coefficient at OD of pipeinsulation, Btu/hr-ft'?-'F

U+ = overall heat transfer coefficient at 14. Btu/hr-ftr-'F


X : distance of plotted temperature points alongthe vessel skirt or piping, ft

Z : heat transfer factor, 'F/ft2Z = heat transfer factor, ZiQ, 'F

Greek Symbols

0 : volumeric coefficient of thermal expansion,,IK

6.,, 6,, : axial deflection of carbon or stainless steelskirt or pipe, in.

? : safety factor for traced pipep : absolute viscosity, lbift-hrp : densiry, lb/ft3

REFERENCES

1. Tubular Exchanger Manufacturers Association, Stan-dards of the Tubular Manufacturers AssociationQEMA), sixth edition, New York, N.Y, 1978.

2. Kern, Donald, Process Heat Tiansfer, McGraw-HillBook Company, 1950.

3. Ludwig, Ernest E., Applied Process Design forChemical and Petochemical Plazls, volume 3, sec-ond edition, Gulf Publishing Company, Houston,Texas, 1983.

4. Escoe, A. Keith, "Heat Transfer in Vessels and Pip-ing," Hydrocarbon Processing," January, 1983, vol.62, no. l, Gulf Publishing Company, Houston,Texas.

5. Chapman, Allen 8., Heat Transfer, third edition,Macmillan Publishing Company, New York, 1974.

6. Perry, Robert H. and Don Green, Perry's ChemicalEngineers' Handbook, sixth edition, McGraw-HillBook Company, New York, 1984.

7. Heisler, M. P., "Temperature Charts for Inductionand Constant Temperature Heating," Transactions ofthe A.S.M.E., vol. 69 (1947), pp.227-236.

The specifuing, design, and construction of pressurecontaining vessels varies all over the globe. Eachadopted code that has been used for any significantlength of time has proven to be workable because its usehas resulted in safe, economic designs. The main differ-ences in codes are the theories of yield that are used fordetermining maximum allowable stresses, material spec-ifications. and basic procedures.

With increasing international competition and cycliceconomic conditions, there is a growing need to empha-size economics and familiarity of foreign codes, andavoid unnecessary overdesign that relies on only one setof codes and standards. This chapter emphasizes the op-timization of economics and safety. If you choose to beconservative in your design, you can be; however, if youare bidding in a highly competitive market, you can usethese methods to produce a safe, economical design.

International competition and economic condltionshave caused engineers to restructure their thinking that agood design uses only enough material that produces asafe and economical product. Thus, this chapter's philos-ophy is to optimize engineering design within code rules,whatever the code. Overly conservative design that re-sults in excessive material use becomes unproductiveand expensive when one is competing in the world mar-ket today.

A thorough treatment of vessel engineering and itsconcomitant aspects of static and dynamic phenomenawould fill several volumes. To present this broad subjectwith clarity. various physical phenomena are briefly dis-cussed and references are made to sources that give de-tailed theoretical explanations. lt is not this boo-k's pur-pose to give a trearise of static and dynamic problems.but rather descriptions of proven practices. The theoryof these problems is always available, but proven solu-tlons are not-hence, the reason for this book.

The Engineering Mechanics ofPressure Vessels

The first problem you face in designing a vessel con-taining pressure is how to physically make the compo-nents and assemble them. In the petroleum refining in-dustry (CPl-Chemical Process Industry) and alliedindustries, the most practical and economical method iswelding. We will refer to welding later in more detail,but first we will look at the vessel from a pure engineer-ing viewpoint assuming perfect welds with given effi-ciencies. Some have proposed bonding pressure vesselstogether with glue, as is done with aircraft components.The main disadvantages to bonding are

1. Clean surfaces are required for assembly.2. Glues that exhibit high tensile and compressive

strengths are very expensive.3. Chemical bonding, especially in thick-walled ves-

sels, takes much longer than any welding process.

Another form of assembly that has been even more se-riously considered than bonding is threading componentsand screwing them in place. Even though this may ap-pear to be simple, the process becomes enormously ex-pensive with large diameters. Thus, welding is the mostpractical and economical means of assembling pressurevessels for the foreseeable future.

DESIGNING FOR INTERNAL PRESSURE

The two factors that must be considered in the desisnfor internaf pressure are crr??ponent thickness and quatiryof weds. Before either of these two factors can be ad-dressed, you must know what the vessel is to contain.This chapter only considers gases and liquids. Vessels,

159


silos and bins containing solids are discussed in Chapter

In the design for liquids under pressure, the most se-vere condition of coincident pressure and temperatureexpected in operation must be considered in computingshell thickness. This is fairly universal in codes through-out the world. The intent ofthe statement is that the mostfrequently occurring liquid level should be considered.For example, if a vessel is filled to a certain level "A'75% of the time and a higher level *8" 25% of the time,level "A' should be used for design purposes. The nor-mal liquid level to be used for vessel design and its quan-titative value should be determined by the process engi-neer. For upset conditions each code allows an increasein allowable stresses under temporary conditions, andyou should consult whichever code is to be used for exactamounts allowed. It is recommended that a value of 30psig or 10% be added to the operating pressure for de-sign pressure. This practice varies with each companythroughout the industry.

Once the internal pressure is determined it must be de-cided how the vessel is to be welded. The factors affect-ing this decision are as follows:

l. Size of vessel-whether rolled plate or seamlesspipe is used.

2. The toxic nature of the fluid to be contained.3. The economics of fabrication as to whether a full

joint efficiency is necessary.

One can appreciate the degree of types of welds re-quired for a vessel. A slug catchel which acts as a scrub-ber handling a non-toxic substance, does not require thesame caution as a vessel containing cyanide gas.

The quality of a weld joint is determined by a radio-graphic inspection. Full radiography includes a completeX-ray inspection (1OO% for butt weld and 907o for sin-gle-welded butt joint) and spot radiography implies 85 %for buttjoints. See Thble 4-1 for maximum allowable ef-ficiencies for arc and gas welded joints. The reader isstrongly urged to consult whatever code happens to gov-ern. Listed in Thble 4-2 are the joint efficiencies for thevarious welded combinations for pressure vessels underASME Section VIII, Division I[1].

Any discussion on designing for internal pressuremust include maximum allowable working pressure,which is the maximum gauge pressure permissibie at thetop of the completed vessel in its operating position for a

designated temperature. This pressure, MAVr'P, is nor-mally specified on two conditions*new and cold (ambi-ent) (NAC), and design. "New and cold" implies theMAWP for a new vessel (non-corroded) at atmosphericcondition, and "design" implies the vessel corroded at

design temperature and pressure. The value of theMAWP at the two conditions gives the exact range oftemperature and pressure that the vessel can withstand ifthe owner decides to use it in another application. Thereader is cautioned to consult his respective code on thepractice of using a vessel for another application. Thefollowing example illustrates how the MAVr'P is applied:

An ASME Section VIII Division I vessel is made ofSA 240-304 SS, design pressure : 500 psig, designtemperature : 150'F. The vessel has a shell thicknessof 1.00 in. and a ioint efficiencv of 1.0.

MAWP (NAC) =(18,800) x (1.00) x (1.00.)

(21.00)+(0.6)x(1.00)

870.4 psig

(18,300) x (1.00) x (1.00)

(21.00)+(0.6)x(1.00)

847.2 psig

The 18,300 psi is obtained by linear interpolation ofthe allowable stress values in Table UHA-23 of theASME Code.

The vessel owner knows the maximum allowable pres-sure for the shell at the new and cold condition as well asthe design condition. It is a common practice to limit theMAWP by the head or shell and not by the flanges oropenings, only the MAVr'P is determined by the flangesor openings when the vessel is to be reapplied in anotherapplication or a design oversight is made.

Finally, in computing the minimum thickness of theshell or head, mechanical allowances must be consid-ered. In the manufacture of heads, the metal is thinnedon forming the section (a forgery process). This formingallowance must be considered when the nominal thick-ness is specified. When a minimum thickness is specifiedto the head manufacturer, the forming allowance is notconsidered because it is the manufacturer's responsibilityto ensure the minimum thickness.

DESIGNING FOR EXTERNAL PRESSURE

The design for external pressure of vessels is fairlystandard in the ASME and codes of other nations. Theprocedures for determining minimum shell thickness,spacing, and section properties of stiffening rings arestraightforward and simple. Because there is much pub-lished material on external pressure design, the subject isnot discussed here. The reader is ursed to consult theoressure vessel code to be used.

MAWP (Design) :

The Engineering Mechanics of Pressure Vessels 161

Table 4-1Maximum Allowable Joint Efficiencies for Arc and Gas Welded Joints [11

Degree ot Examination

Type ol JoinlDescriptionNo. Limitations

(a)Fully

Radio"graphed

(b) (c)Spot Not Spot

Examined Examined(l) Butt joints as attained by dou-

ble-welding or by other meanswhich will obtain the samequality of deposited weld metalon the inside and outside weldsurfaces to agree with the re-quirements of UW-35. Weldsusing metal backing stripswhich remain in place are ex-cluded.Single-welded butt joint withbacking strip other than thoseincluded under (l).Single-welded butt joint withoutuse of backing strip.

Double tull fillet lap joint

Single firll fillet lap joins withplug welds conforming to UW-t7

Single tull fillet lap joints with-out plug welds

(a) None except as in (b) below(b) Butt weld with one plate off-set-for circumferential joints only,Circumferential joints only, notover 5/a in. thick and not over 24in. outside diameterl-ongitudinal joints not over 3/8 in.thick. Circumferential joints notover s/r in. thick(a) Circumferential joints for at-tachment of heads not over 24 in.outside diameter to shells not overt/2 in. thick(b) Circumferential joints for the at-tachment to shells ofjackets notover s/a in. in nominal thicknesswhere the distance from the centerof the plug weld to the edge of theplate is not less than 1r/2 times thediameter of the hole for the plug.(a) For the attachment of headsconvex to pressure to shells notover s/e in. required thickness. onlywith use of fillet weld on inside ofshell; or (b) for attachment of headshaving pressure on either side, toshells not oyer 24 in. inside diame-ter and not over t/+ in. requiredthickness with fillet weld on outsideof head flange only.

None 1.00

0.90

0.85

0.80

0.70

0.65

0.60

0.55

0.50

0.45

(2)

(3)

(4)

(s)

(6)


Table 4-2Joint Elficiencies for Arc and Gas Welded Joints per ASME

Asterisk (+) denotes which joint type governs. Illustration of weld joint locations Typical of Categories A, B,C. and D-see ASME Section VIII Division I.

Welded Head (Non-Hemispherical)-Welded ShellHead Thk. Calcu. Shell Thickness Calculations

Joint Types H, C, and LT1 = Type 1 Joinl (ASME UW-12)T2 = Type 2 Joint (ASME UW-12)

RadiographType

L- C- H' T1

E. Cir. Stress E. Long Stress

T1 T2

0.90

0.80

Spot 0.85

I .00

L00

1.00

Spot

None

Spot

Spot Spot 0.85Spot None

None Full IIIII

I

I

I

I

III

I

100

None Spot

Spot

None NoneFull Full 1.00

Spot Spot 0.85None

1.00

100II

II

III

I

I

I

I

I

I

I

IIII

V

SpotSpot

Spot

0.85 0.80

Spot Spot

Spot Spot Spot 0.85spot Spot

Spot None FullSpot None Spot

None None

None Full FullNone Full Spot

None FullNone Part

1.00

0.85

1.00

* L Weld soverns in circumferential stress calculations.

'FL^ E-^r-^^-r-^ i'[echanics of Pressure Vessels 163rrI LrrSrrr!!rur6 rr.

Table 4-2 continuedWelded Head (Non.Hemispherical)-Welded ShellHead Thk. Calcu. Shell Thickness Calculations

FadiographType

L' C' H.

E. Cir. Stress E. Long Slress

Tl f2None Part SpotNone Part NoneNone Spot FullNone Spot SpotNone Spot NoneNone None FullNone None SpotNone None None O.70

Welded Head (Hemispherical)-Welded Shell

0.85 0.80

IIrlYI

0.85 0.80iii!ii0.70 0.650.70 0.6s

RadiographTYPEJoint

Spot Spot NoneSpot None FullSpot None Spot

None Full Full

Head Thickness CalculationsShell Thickness Calculations

H=Tl H=Tl H=T2 H=T2 E. Cir. StressTt f2

E. Long StressT1 T2L- C' H' C=Tl C=T2 C=Tl C=T2 Vo

Full Full Full 1.0 0.90 0.90 0.90FUU Full Spot t {FullFullNoneiiiiFrrll snÎFrrll !ll!' "" _____i!::_____i_Full Spot Spot O.ps 0 p0 0.90 0.p0Full Spot NoneFull None Full

Spot Spot Spot 0..85 0.80 100 0.800.850.85 0.80 100 100

None Full SpotNone Full NoneNone Spot FullNone Spot Spot 0.85None Spot NoneNone None FullNone None

0.800.80

* L Weld governs in circumferential stress calculations.

Spot Full Full 1.00 0.90Spot Full SDot { |

I I

tSpot Full None i r:------------- _--------=--:-

|spot Spot tu i

None None None 0.70

164 Mechanical Design of Process Svsterrrs

Table 4-2 continued

Shell Thickness CalculationsRadiograph

Type

Full Spot

Full None

Spot FullSpot Spot

Spot NoneNone Full

H=T1C=Tl

H=T1C=r2

H=T2 H=T2C = Tl C=T2 o/o o^n

E. Cif. StressT1

E. Long StressT1 12

1.00 0.90Full Full 0.90 0.90 0.90l 00

0.70

0.85 0.80I .00 0.90

0.85 0.801.00 0.90

None Spot

SeamlessHead Thick. Calcu.

Head-Welded ShellShell Thickness Calculations

RadiographTYPE

E. Cir. Stress

1.00 0.90

E. Long Stress

T1 T2

1.00 0.90

0.85 0.80

1.00 0.90

0.85 0.80

I .00 0.900.85 0.800.70 0.65

Full Full 1.00 0.90Full Spot

Full None0.85 0.80

Spot Full L00 0.90+I

I

0.80iII

Ii

Spot Spot

Spot None0.85 0.80

None Full 1.00 0.90None Spot 0.85 0.80None None 0.70 0.65 0.65* C weld governs on head and longitudinal stress calculations.* L Weld governs on shell circumferencial stress calculations.

Seamless Head-Seamless ShellHead Thickness Calculations Shell Thickness Calculations

E. Long Stress

0.85 0.800.80

! C Weld go\ern5 ior head dnd longnudinal slre,s calculalion\.

The Engineering Mechanics of Pressure Vessels

Table 4-2 continuedSeamless (Non-Hemispherical) Head-Seamless Shell

Head Thick. Calcu. Shell Thickness Calculations

165

E. Cir.Stress

E. Long Stress

FullFart

Spot

\one

1.00 0.90

1.0 0.85 0.80

0.70 0.65

Seamless (Non-Hemispherical) Head-Welded ShellShell Thickness Calculations

RadiographTYPC

c'Head Thlck.Calculatlons E. Cir. Stress E. Long Stress

1.00FullFull 0.90100

85

Full Part1.00 0.90

Full SpotFull NoneSpot FullSpot PartSpot Spot

Spot NoneNoneNone PartNone Spot

None None

Welded (Non-Hemispherical)-Seamless ShellHead Thick. Calculations Shell Thickness Calculations

RadiographType E-Tl 12-

1.00 0.90 1.00 0.90Part

Full SpotFull None

I

II

I

0.80r00

85

II

I

II

I

I700.

i

t65

Full

0.65 0.

E. Cir.Stress

FullFullFull

100

-6)

100

----:-6)

III

I

II

80II

v

65

0.

-U.

0.85 0.80

Spot FullSpot PartSpot Spot

Spot NoneNoneNone PartNone SpotNone None 0.70* H Weld governs in head calculations.+ C Weld governs in loogitudinal sfress calculations.

100

85

1.00 0.90

0.85 0.80

Full 1.00 0.90

0.85 0.800.70 0.65


DESIGN OF HORIZONTAL PRESSUREVESSELS

The analysis of horizontal pressure vessels convergeson the design for internal pressure and vessel supports.This chapter only considers metal, cylindrical vessels,and focuses on the supports of horizontal pressure ves-sels.

L. P. Zick [2] of the Chicago Bridge and Iron Com-pany developed the method of analyzing supports forhorizontal cylindrical shells in 1951. We will not derivethe method, but rather summarize it in a seneral discus-sion along with guidelines and useful praciices thar makethe design of such items more straightforward.

Horizontal vessels should be desisned to withstandinternal and external pressures. and support reactionsproduced by the vessel weight and additional loads fromladders, platforms, piping, etc. Zick [2] showed thatsupporting horizontal vessels by more than two saddlesis not only inefficient, but incurs additional undesirableproblems. Figures 4-1 and 4-2 illustrate a horizontal ves-sel supported by two saddles.

LONGITUDINAL BEITIDING STRESSES

A horizontal vessel supported on two beams is thesame as a beam overhanging two supports. The maxi-mum longitudinal bending stresses occur at the supportsthemselves and at the center of the vessel, as shown inFigure 4-3. Zick [2) and Brownell and Young [3] give adetailed derivation of the equations for longitudinalbending stresses at the saddle and at mid-span. This anal-ysis is summarized in the following:

At Saddle

qr = longitudinal bending stress at saddleoa1 allowable stress in tension. psio"r = B = allowable stress in compression. psi

For tension,

01 : Eoan + op

where E : welding joint efficiencyop = pressure stress, psi

The allowable stress for compression is based on the ac-cepted formula for buckling of short cylindrical col-umns, which is

r = mean radius, ft

Figure 4-1. Horizontal vessels aresupported on saddles. The saddlescan be supported on concrete piersshown in Figure 4-2.

ts

or the allowable stress in compression is

o1 < Bl2

where r =

/"\i,\[,\-rt \;i f - (,1,'*,(i)]

D

radius of cylindrical shell, in.thickness of cylindrical shell, in.modulus of elasticity of shell, psiB factor in the ASME vessel code, psi

/'\- -- - '\,\l "

---,..T.,- .

>\.----r<]''..- 9--7

Figure 4-2. Horizontal vessel with saddles on concrete piers.


":-t*-[..(,

Referring to Figure 4-3, oy occurs at either lBl6 + 0l2ldegrees or zero degrees at the shell acting in the longitu-dinal direction. This only applies to unstiffened shells.The vessel must meet the allowable with or without ores-sure.

2A : arc, in radians, of unstiffened shell in plane saddleeffective against bending

At Mid-span

o : longitudinal bending stress at midspan

The longitudinal bending stress at midspan has the same

(4-3)

Thneential Shear Stress

l. For shell stiffened by ring in the plane of the saddle,

*'- "41,AL ll*ll

3L IJ

+

+

II

l-

where A, H, L, Q, r, and t. are defined in Figure 4-3.CA = corrosion allowance, in.

0 : angle of contact of saddle with shell, degrees(Figure 4-1)

* l<a \4 = _al1 +:Ol

180 \r2 l

nN

(4-r)

lA-) |

(4-4)_ _ (0.r8)Q /r- - ze - u\"r"- - rrr-tAr\ L-H /

o3 < 0.08ou1

\--il \

- --T-.-11 I

l-/

| | zT\-[llll/r\rr.,tt/ | \lllffi-["' Y V-t-+Figure 4-3. Bending moment diagram for a horizontal vessel developed by Zick l2l.


;&p

Figure 4-3. Continued.


I

I

I

Figure 4-3. Continued.

17O Mechanical Design of Process Systems

z

z

6 o.os-

==ozzUJ

)t-zgJ

,'u o'o2-

=E

1.O

RATIO A/r

L Unstiffened shell with>R)

',r'here

saddles awav from head (A

'( :in," I ,o-r,\7r-@+slnqcosd/

, sin 0R

.. 1

1t

sin 6-[ cos P

1 = r,# (; . rq], g & B in degrees

.ltu\\-/

. - r(#)'+ 2cos2B(4-12)

Q lsin o[ "- sinocoso \lr(rr-C{)1" \"-" + tin ""oso/l

06 is the same as 05 and also is located at (19/20)0. Withrhe shell stiffened by the head, then

o6 3 0.8o"1

Circumferential Stress at Horn of Saddle

For shell stiffened by head the maximum circumferentialstress at horn of saddle is,

IfL>8R,o7: 4( - CA) (b + 1.56(r(r - CA))u)

-,'QIu = p.s)2(r - CA)'?

-a@_ 12&QR _ (4_lo)

L(t - CAf

oe : ring compressive stress in shell over the saddle

This stress is located at O : 7r

-a

a /L-H -2A'l''v- - 11-ctr \ L+H /

(4-6)

r! : tangential shear stress, located at an angle of B/20

,j. Shell stiffened by head,

15 = shear stress in head

This stress occurs only when t}re shell is stiffened by thehead and when the head is located less than one shell ra-dius from the saddle. The rnaximum shear is located atan angle of (l9l2O)P as shown.

-1. Shear stress in shell,

q [.in rI o- rino.oro \l"6 - r(t5-CA)tn \" - " +.sindcose/ I


If A/r>1, then K6: O.42Z2e-a ot710 (4-11)where d : degIf A/r < 0.5, then K6 : K3/4Otherwise, use Figure 4-3.

0sin02

^/^\:srnpcosp, i .lslnpla \B I

(4-7) o, : circumferential comDressive stress

This stress is located at the horn of the saddle If o; (outr, it is not necessary to take credit for the wear plate.

(4-S) o7 < l'50 ou1

Additional Stress in Head When Used as a Stiffener

3Q I sin']" I"s - 3'-1u - 6e,1 Lr .o, + sin " cos "l

(4- 13)

08 S oall

Wear Plates-Ring Compression in Shell Over Saddle

(t - CAXb + 1.56(r(t - Ca;101

{ '1':'" }o"(0.5o,\7r-q+slnacosq/If L < 8R, o :

(4-14)


This stress is compressive and acts in a radial directionbetween the saddle and shell. The limitation of this stressIS

0.5ori"ra

where oyi"rd : the yield strength of the saddle material(metal or concrete)

Location ot Saddle Supports

For thin wall vessels with large diameters, it is desir-able to locate the saddles close to the head, where A =ID/4, using the stiffness of the head. Although arbitraryon what a thin shell is, and Zick [2] does not define theterm, a shell is generally regarded as ',thin" when D/t> 100, where D : shell diameter and t : shell thick-ness. For shells where D/t < 100 and the distance fromthe head tangent to head tangent is rather large (approxi-mately L/r > 10), the saddles are best spaced when thelongitudinal bending stress at rhe saddle, or, equals thelongitudinal bending stress at midspan. o2. Undei no cir-cumstances should the distance from the saddle centerline to the head tangent, A, exceed 0.25L.

A listing of allowable stress criteria is siven in Thble4-3. Each of the previously menlioned stress valuesshould be evaluated with this table and the appropriatecode.

Wear Plate Deslgn

One of the first things to consider when designing ahorizontal vessel is the need for wear plates. Too oftenthese plates are "auromatically" included with nolhought given to their necessity in each application. Wearplates involve material and labor expense and are a wasteif not needed.

Wear plates are not required if two criteria are met:The circumferential stress at the horn of the saddle mustbe less than 1.5 times the allowable stress, and the ringcompression stress in the shell over the saddle musr biless than one half the minimum yield strength. These cri-teria can be written as follows:o1 1 7.5 o^x

oe ( 0.5 o, 6n

Table 4-4 shows minimum allowable shell thicknessesrequired for horizontal vessels without wear plates. Thevalues are based on using a fluid 1.75 times the weightof watet and the metal has a minimum yield of 30,000psi and an allowable stress of 17,500 psi. For vessels inseismic regions wear plates should always be used tominimize stress concentrations at saddle plate-shell junc-ture.

Longitudinal Bending StressDPot T op : o' i, a o4 E. where E = joint efficiency

02 +op = 02

Tangential Shear Stressq4 < 0.806 < 0.8

Circumferential Stress at Horn Saddle

o7 < 1.5 o"1

Circumferential Stress at Bottom of Shetl

oe q 0.5 (or1-i") * Compressive Yield

Zick Stlffenang Rings

When the Zick stresses in a vessel become excessiveand the location of the saddles no longer is a factor be-cause the stresses are below the allowable stress, thentwo options are available-increase the vessel wall thick-ness or add stiffening rings. Almost always it is more de-sirable to add stiffening rings because it is cheaper to adda few rings than go to a larger size shell thickness, par-ticularly with expensive alloys. Also, if the vessel is sub-jected to external pressure , the Zick rings can act as ex-ternal pressure stiffening rings as well as Zick rings.

Referring to Figure 4-3, if two Zick stiffening ringsare located on each side of the saddle, then

Ln,n : l.Jb Vfl, It

Lr* : r, ft

The stress in the ring is

_ _ -KuQ KuQr ,- nA- n7,

Table 4-3Allowable Stress Values

ff 'o'' r

oall03'o5'

ir

Where Z

z

I*- "

= l-/c for ring in the plane of the saddle,tn.'

: I,-,/d at saddle horn at tip or flange ofstiffener ring, in.3

: moment of inertia of stiffening ring aboutaxis x-x, in.a (includes wear platethickness if one is used)

: cross-sectional area of stiffening ring, in.2: number of stiffener rings per saddle: mean vessel radius, in.: previously defined

nr

K6


Table 4-4Minimum Shell Thickness Required lor Horizontal vessels Without Wear Plates

175

lD (in.)

Not€s1. The above table is based on the following:

a. vessel is tully loaded with a fluid of specific gmvity of 1.5.b. The ratio of the shell outside radius, R., to shell thickness, t, is R-lt > '72.

c. vessel weight is computed with not€ (a.) and hemispherical heads.d. Vessel material has the following properties:

d,i" y,.rd = 30,000 psi and o.rr* = 17,500 psi2- In seismic zones 3 and 4 wear Dlates should be used.

78 84 90 96 102 108 114 120 132 144 156

8r/s in.

r/r in. I

I_r-9/ro in.

5/r in.

l0t2

ll/rc in.

lllro in.

rYrt in.

t4l6l820

30

?ro in.

r/+ in.

?/x in. I in.

40t2 tn,

Is/r6 in.50

60 78 in. I in.65 lYro in.


In compression, oro is negative,

oleAB(0.5o.,

In tension, o,6 is positive,

o'e * oo ( o.1 [tension]

where B = ASME compressive stress (see ASME Section8 Division l)

o", : compressive yield stress (see ASME Section gDivision 2)

op : internal pressure stress (includes wear platethickness if one is used)

In defining the parameter K7, it must be noted that theZick stiffening rings can fit on either the inside or out-side of the vessel shell. Many clients object ro the ringsbeing external to the vessel surface because of aesthetici.However, after insulation is applied, the rings are nolonger visible. We will consider rhe rings in both ways.The constant K7 is defined as follows:

For a ring in the plane of the saddle-

Kr:

strengthened with stiffener or web plates. but often toomany are used. which increases laboi and material costs.In the past, saddle plates have been purposely over-de-signed to guard against uncertainty. This is no longer re-quired, since literature on flat plate theory has increasedwith mounting experimental data. One such organizationthat has engaged in extensive research is U.S. Steel [4].

Figure 4-4 shows a typical saddle configuration for ahorizontal vessel. Section A-A shows that only an effec-tive portion of the member will resist compression.shear. and bending loads because when rhe member isloaded, the outside fibers ofthe web plates and the centerof the saddle plate -shown by rhe sh;ded areas in Figure4-4-go into the plastic range. The rest of the plate areais still in the elasric range because of residual stressesthat were created by non-uniform heating during rollingor welding. Presently, this "effective" area can be deter-mined only by experiment. Equations 4-9,4-10, and 4-14 are used in saddle design as follows:

b":KL

+ 1.00.340, 00.303, 0

0.250,0

: 120": 150': 180.

(4-15 )

where b" = effective width, in.K; = plate buckling coefficienr for either

compression, shear, bending, or acombination of these loads (see Figures 4-5,4-6, 4-7, 4-8,4-9, and 4-10).

i : c, b, s, or a combination of these characters,where K. : plate buckling coefficient for compression,

dimensionlessKb : plate buckling coefficient for bending,

dimensionlessKs : plate buckling coefficient for sheaq

dimensionless

We now have

For rings adjacent to saddle-For internal rings,

\: -1.0

Kr:: 120": 150': 180'

For external rings,

X = -1.0

Kz:: 120.:150'= 180'

STEEL SADDLE PLATE DESIGN

Once the shell conditions have been met. the saddleplates must be analyzed. The main phenomenon encoun-tered with saddle plates is local buckling with the platesundergoing bending, compression, shear, or any combi-nation of these loadings. Normally, saddle plates are

-a4(t - CA)[b" + 1.56{rrr - CA))o']

1.)K._ ;:--. rf L > 8R

o4(t - CA) tb" + 1.56(r(t - CA) f I

-.lt*'gl.,rrL < 8RL(r - LA)'

(o.271, 0

l.0.2r9, 0

[0.140, 0

(0.27 | , 010.2t9, 0

{0. r+0, o

(.4-9)

(4-10)

4ft - cA) tb. + 1.56G(t - CA)f 5l

ol

acos

sln1I

(\7t- cos -l'( (4-r4)


d"t/lllffl

b

f - b"-:lI t---------1 I

\r .,-.-lN -

I

'.-lF 1 |

'-ff l"'l*"*lJL.._"1 I-.T----sections A-A and B-8,

shaded areas are in theplastic range.

elevation view

Figure 4-4. Horizontal vessel saddle support detail.

Figures 4-5 to 4-10 are courtesy of United States Steel Corporation.USSC makes no warranties, express or implied, and no warranty as

to the merchantability, fitness fot any particular purpose, or accuracy

of the information contained in any material reproduced herein fromits Steel Design Manual. In the event of any liability arising out ofthepublication of such material herein, consequential damages arc ex-cluded.

Figure 4-5. Buckling coefficients for flat plates under uni-form compression. (Courtesy of U.S. Steel [4].)

EE

j

--'t------------- - --'l-----------_---i cAS€ 3 F ---l CISE 4I --r r- -- F-

I

l

ri\\\'

i\

_.1-...---------.1=-l casE 5 F

---- loaoEo EDGES FtXE0

LOADED EDGES\ SIMPLY SUPPORTEO

\\.ta.'a

5

z.

I


Figure 4-7. Buckling coefficients(Courtesy of U.S. Steel [4].)

tlT-- .]

\l t7/

3H.

jr:=-2l3r,r,.1 \

\-V {, = 1/3r, Y-tK

F----EH "=o y

Fry]=!l r, = r/3f I E/\t_-_____tr/

LOAD ING

{PU8E BENOING)

5.00

2.OO

't.00

0.50

I r r-r____-_r_: I,

F= f: = f, t= rp,,.. "îiil.*. -... --.....,--jtoN) 4.0

.VALUES GIVEN AAE BASED ON PLATES HAVING LOADED EDGESS{I\4PLY SUPPORTED AND ARE CONSERVATIVE FOR PLATESHAVING LOADED EDGES FIXED.

aaTro oFEENDING STBESS

-TOU\IFOR\,I MJNII\,4UI\,IBUCTI.NGCOLfFICITNI.'I,COVPBLSSION

ST-8ESS, UNLOAOED EDGES UNLOADED

-t!ft.

Sll\,4PLY SUPPonTED EDGES FtxED

Figure 4-6. Buckling coefficients for flatplates under compression and bending. (Cour_tesy of U.S. Steel [4].)

ni

j

LONC EDGES FIXED,SHONTED6ES SIMPLY STIPPOfi TEO

for flat plates in shear.

Figure 4-8. Buckling. coefficients for stiffened plates underuiform compression (one longitudinal stiffener at mid-point).,Courtesy of U.S. Steel [4].)

1.O 1.2 1.4

34

30

2A

26

24

22

20

18

t6

12

5 35 40 45 50 55

0.6 0.8 1.0t

14

13

12

tl

I

4 6 810 12 14 16 18

NONDIMENSIONAL PABAMETER, O

Figure 4-9. Buclding coefficients for stiffened plates underuniform compression (two longitudinal stiffeners at thirdpoints). (Courtesy of U.S. Steel [4].)


2.O

2.2

I

i

z

trUooz=f

F

2.8

3.0

I

j

:o(,

=

-

NONDIIUENSIONAL PAsAMETER. d


0.6 0.8 1.0 1.2 1_4 1.6 1.8 2.O

2.8

3.0

IFzqno

j

F

If no web plates are used then b" : t,. It is very com-mon_ for engineers and designers to use the we6 platewidth, b, instead of b". This is wrong. The only time b": b is when t, = b, as is true for a solid concrete sad-dle. With steel this never happens, as values of b can beas great as 24 in. and obraining plate that thick is impos_sible (ar least on this Dlanet).

Values ofb" depend upon K, and t,. Since the value of t.is known, the real independent variable in Equation 4-15is K,. Once again referring to Figure 4-4. we analyze thesaddle configuration for end (boundary) conditions. Sec-tion B-C is considered fixed-fixed in Fisure 4-5. since itjs stiffened by sections A-B and C-D. S=ections A-B andC-D are considered fixed-free since the outer web Dlateis not stiffened by another section. The fixed-free condi-tion is the most critical because it is more susceDtible tobuckling. and rhus ha: a lower value oi the plaie buck-ling coefficient than the fixed-fixed case. [t is interestinsto note that the plate buckJing coefficient for uniforri

Figure 4-10. Buckling coefficients for stiffened plates underuniform c_ompression (three Iongitudinal stiffeneis at quarterpoints). (Courtesy of U.S. Steel [4].)

compression for the fixed-free case, when multiplied byt., yields approximately the effective width, b", that isused for residual stress. In other words, if a member isknown or suspected to have residual stress and is sub-jected to compression, bending, shear, or a combination.the plate buckling coefficienr is equal [o rhe effecr:vewidth that is determined by the residual stress crirerion,which is as follows:

ti- d,t,(4-16)ldit.+2r*(b-l)l

The general equation in which the saddle plate stress dis-tribution is defined is as follows:

NONDII\,4ENSIONAL PARA]\IETER, d

K, zr2 Eo{:

--l

\,rztr - 4ld'1"\\/

(4-17 a)

shere di :

Substituting the elastic buckling stress in Equation-1-17a into

saddle plate length normal to vertical axis ofstiffener (web) plate, shown in Figure 4-4.modulus of elasticity, psiPoisson's ratioeffective saddle width, in. withsaddle plate thickness, in.


thickness of saddle plate, in.effective width of saddle plate that isperpendicular to the web plate, in.

d.=d,(0.25+0.91\)

\:lll\dJ

, o*2or:6y-71o.,

/, \2,. JolJ(r - /") oy'l-l

= or- ;; ;; "' ,l[ o,, 2 o.l2\7-D

gives the relationship of the plate buckling stress in theinelastic range. This equation is based on the conserva-rive assumption that a plate will always buckle before theyield stress is reached. However, U.S. Steel [4] statesthat plates will deform plastically without buckling be-cause of strain hardening. This process is similar to the"elastic shakedown" described in Chapter 2.

In most applications, as already cited, saddle plates arereinforced with stiffener plates. A simplified analysiscan be made to design saddles by using

Fs: n(A, + 2b"t.)o. (4-18)

where Fs : buckling load for compressive loading, LBqA, : section area of stiffener, in.2n : number of stiffeners

o, : maximum unit load the stiffener can carry as

a column, psi

Horizontal Reaction on Saddle

As shown in Figure 4-i 1, the load Q has a horizontalcomponent exerted on the horns of the saddle. The sad-dle must be designed to p{event the horns of the saddlefrom separating. To accomplish this make sure that theminimum cross-sectional area at the lowest point on thesaddle can resist the horizontal force component. Thisforce is as follows:

- ^h * cos 0 - 0.5 sin'z0l

l't-lJ+sInPcosP.l(4- l9)

(4-17b)

The effective cross section to resist the horizontal force

is As, shown in Figure 4-11 and calculated as follows:

iRlAe : l;l t,

where R : outside vessel radius

t1

Figure 4-11. The load distribution on a saddle.

R/3

0a;


SADDLE BEARING PLATE THICKNESS

.Designing bearing plates for saddles requires knowingwhat type of foundation the vessel will rest on. For con-crete the following analysis applies.

Consider a bearing plate with the dimensions shown inFigure 4-12. From ACI Standard 318-77 par. 10.16.1.the allowable bearing strength on concrete is

Table 4-5Bearing Plate Thickness Values tor Various Saddle

Loads

DESIGN OF SELF.SUPPORTED VERTICALVESSELS

Today's tall, cylindrical process towers are self-sup-porting, i.e., they are supported by a cylindrical or coni-cal shell (skirt) with a large base ring attached to a con-

, (,A :1:-3,L,,: L-1,1" r4-1rl\Ar l-"Using a factor of safety of 1.6, Equation 4-21 becomes

Using a minimum yield strength of 30,000 psi we havethe allowable stress for bending, per AISC recommenda-tron,

o.n : 0.66 o, : 0.66(30,000) : 19,800 psi

Thus, the maximum stress in the bearing plate is

riq\/bjM ,\2r"1\tl

op =; = -," lil

\6i

(4-))\

,: I Qt" lo'.,n.\24.600 Lrl

A-) 7l

where b : Lz

Q : load on saddle, lbs

BP: bearing pressure = Q ao'LrLz

Table 4-5 depicts values of bearing plate thickness forvarious saddle loads.

Ar=LrLzAz=LoL+

/,r \o s

o = @10.85 'J e,r (3

For bearing on concrete (ACI 9.3.2.e)

:0.70: 3000 psi

in which

o : (0.70)(0.85)(3,000)41

,' : *=r, r rs.63 A, (eir2Mtl!)"'

(4-20)

L1 L, Q.ax (tbs) t (in.) Bp (psi) o,(psi)17 4 2,858 0.165 42.029 108.852.56326 4 5,043 0.178 48.490 162.100.69430 4 8.103 0.2t0 67.525 t85.744.85733 4.25 .13l 0.241 79.365 213,447.89336 4.25 16,007 0.277 1U.62r 232,042.32439 5.'75 20,418 0.350 91.050 320.269.t3r42 5.75 25,387 0.3?6 lo5.t22 344.024.23345 5.75 33,523 0.4t7 129.557 367.1-1.7.37548 5.75 40,154 0.442 145.486 39t.528.914

s4 5.75 s9,549 0.508 r9r.784 439,028.224s7 5.75 68,777 0.531 209.846 462,77638260 s.75 84,203 0.573 244.067 486.523.,73663 5.75 101.759 0.6t4 280.908 5t0,270.39966 5.7s 114,664 0.637 302.145 534.016.46369 6.75 t28.417 0.715 275.721 637,918..16372 6.15 143,003 0.738 294.245 665.0.1s.97376 6.75 174,748 0.794 340.639 701,285.2,7584 6.75 210,035 0.828 370.432 773.70r.87395 6.75 2s0,290 0.850 390.316 873.271.364

Figure 4-12. Bearing plate dimensions.

,'rete foundation or steel structure with anchor boltsembedded in the concrete or steel. Normally, a vertical\'essel must be at least thirty feet tall to be classified as a"tower." This height is used because thirty feet is the oldfirst wind-zone demarcation in code use. However,smaller vertical vessels are governed by the same designcriteria, but are not usually referred to as towers.

The various phenomena that affect towers in normaloperation make their design complex and worthy of ex-perienced engineering personnel. Therefore, towersshould never be taken lightly by any design office, be-cause a failure could result in massive loss of materialand possibly lives.

MINIMUM SHELL THICKNESS REOUIREDFOR COIIBINED LOADS

High-speed electronic computers now provide de-tailed, exact solutions to complex mathematical prob-Iems, and so have replaced the "strong arm" approxima-tions of yesterday. An example is solving the equationsof the moments of inertia and section modulus. Beforethe advance of computers, the following expressionswere used to quicken computations on a slide rule or asmall electronic calculator:


the internal stress in the circumferentiai direction is com-bined with the bending and tensile (or compressive)stresses. Writing this expression we obtain,

(4-27)

where Z : section modulus of the shell cross section, in.lA : area of the shell cross section, in.2

Substituting Equations 4-24, 4-25, and 4-26 into Equa-tior 4-27 produces

"= -(.*J'H-(x)

I = nR3t; exact: I: #,o"" D'a)

Z : rRztt exact = z = ,a (gd ar'"'D

.I

A = ?Irt; exact: A: - D,)l,o"

(4-24)

| 4-15 |

(.4-26)

Using R as the mean radius minimizes the error andusing R as the outside radius results in considerable er-ror. Solving for the thickness or stress with the exact for-mulations involves iterative analysis, which is a key at-tribute of today's computers.

The minimum shell thickness required for internal orexternal pressure alone is often not sufficient for additional stresses induced by bending moments and weightloads. Bending stress is a result of static wind, dynamicwind gusts, vibration or seismic response spectra. In de-sign the engineer takes the largest bending moments in-duced by one of the following: wind, vibration or seis-mlc.

Referring to Figure 4-13, we analyze the stress ele-ment depicted. The maximum stress resulting from inter-nal pressure occurs along the x-axis, i.e., the hoop stressis twice the longitudinal stress. Wind, vibration, and/orseismic forces cause the shell to bend about the z-axis, so

, : * (ryf - (":zlur.. _ "i , h-

iLr, _ "J @-zB)

ipo\ / to o^lra \- - \4,/= \nrtO" -t DitrDr D,:t/

,( 2w \= \",tnt * Ot1

(4-29)

or

/po\ / to o"t'.1 \' - \+r/- \norD,-r D,rrD"r - D,2r/

,( 2w \= \""fr5 * t1

(4-30)

Referring again to Figure 4-13, we summarize the fol-lowing:

1. For the tension or windward side,

/po\ / ro o"v \' \ +t / \rrr D" - D,tt D"2 - D,2t /

lzwl- \"(D" + DJ/

or

. /.o\ I 16 D.M \' - \+, /- \ro{D" --J,DJ + D-l-,/

/zw\- \"r(DJ-D)/2. For the compression or leeward side,

/po\ i ro o.r,a )

\4tl \Tt(D" + D,XD.r I D,r)/

(4-31)

t4-1)\

lzwi- \"(D" + D)//4-11)

l- l--lo3lt ll,ld"dl^'l

' ,. f---l /

- -\'l < l<-lIIl

clF=ll ll'stN

l= x 'E'

1t--l /--r'l tr l<1_ -I

î^s-^ L- > E-o-l I ciNls" lci r o o"

\ | lr *--41 < lr>- -ll

I

/lol<r-rlI

i.;tkir

o

3 ielo ",*Rl+--i- ;l*,ll lll- !l < l<Lflti-t'-i1" Rlr

f x ^r^ i i.- - l'' o"-'l= ' I ' o- +

' {i ll I'-rdo[<r-lt ti--l-

x !:J t€

I

E"O,o^, i rine*il= "-1- ' f---------1r )-\1 <lK-il lr

t-lI olx o- td

35xo i -t-r r u N ^lFl_ "_12 il+

.Jt I/- +;l ql I:+it Ir..-t-''_Y\. pl*

182 Mechanical Design of Process Sysrems

ou,

U'

o-3ttGtq)o

q)

IL

tt,lo

th

atq)o

tttoC'

EObp!'<fr.u 3

o)

o.

a,lorl)

E(/,

o

o-

o

a9or l:!: (1,

U'|,,l

3. For vacuum vessels the maximum stress occurs onthe compressive side, such that Equations 4-29 and 4-30 become

OT

. _ /ro\ / ro o.rra \' - \4"/- \""(DJDJ(DJ +-3/lr*\- \"r(D" + DJ/

l2wl- \"(D. + DJ/and

. /po \ / ro o"r',r \' - \4")-F',D.+ qnD"' i-DI/_l 2wl

\"t(DJ DJ/


the cost-plus contractors seek to standardize designs anduse lower pressure vessel code allowables. As with wearplates on horizontal vessels, most lump-sum contractorswould elect to omit them whenever possible to save ma-terial and labor costs. This phiiosophy is becoming in-creasingly popular with recent economic upheavals andincreasing international competition.

Types of skirt supports are shown in Figure 4-14. Fig-ure 4-14b shows the most common and desirable skirt,since the shear is eliminated by the type of attachment.This type is used primarily on short vertical vessels.

The skirt is designed to resist loadings caused by bend-ing and the tower weight. Writing the expression that de-scribes this we obtain

(4-34)

(4-35)

(4-36)

Substituting Equations 4-25 and 4-26, as before, we ob-tain

16MD. 2W

7r't(D" + Dr(D"2 + Di) irr(Do + D,)(4-37)

(4-38)SUPPORT SKIRT DESIGN

The design ofvessel skirts is one area in which design-ers disagree philosophically. Lump-sum contractors seekto use higher allowables and thus less material, whereas

l6MD" , 2W5tD"'+ D-5 "'@. + D)no(D" +

Once again, Equations 4-37 and 4-38 must be solved byiteration. Normally, these equations do not govern theskirt thickness, as the reaction of external bolting and

IAINIGHf CIRCULAF CYLINDBICAL SKIBT

t,/l\l lJt tE, i

16I EXTEiNAL LAPPING SKIRI

Figure 4-14. Skirt designs.


compression rings is not considered. The stresses in theskirt shell that result in compressive loading on the com-pression ring and bolting chair can be quite high in appli-cations where external chairs shown in Fisure 4-15 ireused. See Brownell and Young [3], for a derivarion ofthereaction expression. The skirt thickness required to re_sist the reaction of external chairs or comp;ession ringfor a chair of the type in Figure 4-15 is determined aifollows:

Fi : see belowW : yessel weight, lbW. : operating weight, lbWE = empty weight, lbN = number of anchor bolts

The minimum initial bolt load required to maintain com-pression between the base plate and compression ring ex-ist when o" : 0. Thus, using Equation 4-40 and substi-tuting o. : 0 we have

mF.

B

: skirt thickness, in.: radius of skirt, in.: bolt spacing, in. or 28 in Figure 4-15= uplift bolt load, lb: radial distance from outside of skirt to bolt

(4-3e)

(4-40)

(4-41)

(4-42)No,

" = H). [oJ - (**J

where A" : f,to"t - o,r;

I,=#(D"4-D,4)

where BC : bolt circle diameter, rn.ou : allowable anchor bolt stress, psiM : in.lb

Equation 4-42 is one of the major differences in design-ing a tower under a lump-sum contract versus cost-plus.Most cost-plus designers use vessel code allowable stressvalues that are based on a factor of safety of 4:1. Thislarge a factor of safety is intended for components con-taining pressure. Thus, using vessel code allowablestresses for bolts leads to large anchor bolts, which is un-desirable because more concrete is required and largerbolts are much harder to torque, requiring biggerwrenches and being more susceptible to galling. To keepanchor bolt sizes down follow AISC euidelines for bolt-ing- since anchor bolts are purely stirctural in nature.

Table 4-6 provides the allowable stresses for boltinsper AISC l5l. Type 4325 bols and ASTM Al93-87high-strength bolts are used in most applications. A307bolts are used where bolt loadings are not large and thebolt size need not be massive. When bolt sizes get large(231a to 3 in.) or it is desirable to reduce the bolt size.then Al93-B7 or A325 bolts are used. One can see fromTable 4-6 that A325 has more than twice the allowablestress value as A307 bolts. The extra cost of the hieh-strength material will still be less than rhe addirional c6n-crete and labor costs associated with a larger bolt. Cer-tainly. if one pays more for high-strength stleel, he shouldbe permitted to use the larger allowable, as given byAISC. Normally, 40,000 psi is used with A325 and 193-87 bolts.

The spacing of anchor bolts is another critical parame-ter. Spacing the anchor bolts too close to one another

,=176[#r]",,,where t

r

circle, in., Figure 4-15G11 : gusset height, in.

Equation 4-39 is normaily the controlling criteria for askirt with external chairs. Howeveq for a skirt with orwithout external chairs, Equation 4-38 must be satisfied.

ANCHOR BOLTS

Anchor bolts are one of the most important aspects oftower design, and, unfortunately, are often not taken se-riously enough. Consequently, many problems related totowers during construction or operation can be linked toanchor bolts.

_ Wind and seismic loads are dynamic and result in cy-

clic loading of the anchor bolts. For this reason, I willonly present the method for analyzing preloaded anchorbolts. Initial preload is significant since pre-torquing thebolts reduces the variable stress range the bolts experi-ence during cyclic loading. The tower weight and boltload allow firm contact between the compression ringand concrete or steel such that the support base rotatesabout the neutral axis of the contact area, as shown inFigure 4-16. Referring to this figure we see that under amoment M at the base plate-concrete juncture the maxi-mum and minimum stress is

- 8M D. W,f,: -

- ____:' N(D", + D,1 N

The required bolt area is

[*")- *'


IT

IilJ IILr-*-l

k----il--+l

1I

t[liillF-r:-'-_i

NOTES:all dimenslons in inches BTHK to be evaluated by

eouations 4-57 or 4-60 all welds to be size "t"

f*- "---|

r?lzA,l

BOLTSIZEABCBH CHLLJMNPGH

'I

'l tl''l1la

1q8

1112

11la

2

2tl+

21h

23lq

31lz 'l4a 14q 1112 2

3{+ 148 17k '1518 21k

4221s1c211241lq 21la 2118 148 2sla

4tlz 21lc 21la 2 2sl+

5 2112 21lz 2th 31lq

51lz 24c 23lc 21h 31lz

Sslc 3 24c 23lt 34a

61lc31l+33411861lz 3rlz 3 Srla 4112

7 3glc 31lc 31lz 4gla

6 9e 51lz 5 31la 3 31lz

6 Ye 53lq 51lc 31lz 3 3{+

6 4t 6 5112 33t+ 3qo 4

7 4q 61lq 53lq 4 33lq 4318

8 3lq 6112 6 41lq 41la 4112

9 1 7 6tlz 4eh 4alc 43lq

10 1 7112 7 Stlq 51lz 6

12 1 Telc 7112 53lq 5718 6112

13 11/e 81lc 74c 64a 6 7'14 11lq 8112 I 61+ 6{e 71lz

16'l1la98r/+777slq

4e 31lt9e 33/e

4e 33lq

llz 41le

112 41lt5lB 53lc 51lz

3l+ 53lc

4e 64a

1 71lz

11k 8

74s3731rc31'tlq th 3

7112 1lq 3

81|q49 5h6 5

'12 {e 6

13 {e 7

144rc816 1lz 9

18 llz 10

Figure 4-15. Typical designs and dimensions of chair and base plates


ANCHOB BOLT

FOUNDATATION

MAXIMUMCOMPBESSIVE

FORCE = nFc

COMPRESSIVE

FOBCE

E.

II

JI<l

zt<

TENSILE FORCE

CENTROIDDISTRIBUTION

DISTRIBUTION

CENTROID

Figure 4-16. Anchor bolt loading force distribution.

prevents the strength of the bolting in the concrete frombecoming fully developed. It is advisable to set the boltsat least 18 inches apart. To accommodate this minimumspacing a wider base ring with gusset plates can be usedor the skirt can be tapered with a conical skirt. As shownin Figure 4-14, with a tapered skirt the apex angle shouldnot exceed 15 ".

BASE PLATE THICKNESS DESIGN

Base plate design involves the loadings passed on fromthe tower to the foundation. The base plate is a circularring plate used to distribute these load-s around the cir-cumference of the bolt circle. Anchor bolts normallyvary in diameter from one to three inches-bolts smallerthan one inch are more likely to strip or shear off; boltslarger than three inches require large wrenches and cre-ate excessive problems for construction personnel. Forthese reasons it is desirable to attemDt to adhere to theone to three inch size range.

In the case of a concrete foundation, the relativestrength of the concrete to steel becomes a significant

F

F

:o"

os

€s

MAXIMUM TENSILEUPLIFT FORCE- q

I

I+-

factor. The modulus of elasticity of steel is approxi-mately 30 x 106 psi and that of concrete approximately2.O to 4.O x 106 psi. Defining the ratio of the two as n.we wnte

(4-43)

since E.

and E. :

and e, = e. because of the base plate-concrete bond we haveos(induced) = noc(induced) (4-44)

Listed in ?ble 4-7 are values of the moduli ratio n andthe various concrete mixes from Brownell and Young[3]. Figure 4-16 shows a detail ofthe compressive forceof the concrete, F", multiplied by the value of n shownopposite the maximum tensile stress, Fr of the base platesteel.


Table 4-6Bolis, Threaded Parts, and Rivets

Tension [51Allowable loads in kips

187

Bolts and RivetsTension on gross (nominal) area

Nominal Diameter. d. in.11lz13ls11la4s 'l1la3laASTM

DesignationFi

Ksi Area (Based on Nominal Oiameter), in.'?

0.3068 0.4418 0.6013 0.7854 0.9940 1.227 1.485 1.767

A307 bolts 20.0 6.1 8.8 \2.O 15.'7 19.9 29.7 35.3

A325 bolts 44.0 13.5 19.4 26.5 34.6 43.'7 77 .754.0 65.3

A490 bolts 54.0 16.6 23.9 32.5 42.4 53.',1 66.3 80.2 95.4

.\502-l rivets 23.0 '7.1 to.2 13.8 18. I 22.9 ?8.2 34.2 40.6

A502-2,3 rivets 29.0 8.9 12.8 r7 .4 22.8 28.8 35.6 43.1 51.2

The above table lists ASTM specified materials that are generally intended for use as structural fasleners.For dynamic and fatigue loading, only A325 or A490 high-strength bolts should be specified. See AISC Specification. Appendix B. Sect. 83.For allowable combined shear and tension loads. see AISC SDecification Secl. l 6.3.

Threaded Fasteners [51Tension on gross (nominal) area

Nominal Diameter. d. in.ASTM

DesignaiionF, F, FtKsi Ksi Ksi

11la

0.3058 0.4418 0.6013 0.7854 0.9940 1.227 1.485 1.767

1 11ls 13/8 'l1lz

A.r6 19. I l l.558 5.9 8.4 r5.0 19.0 23.4 28.4 33.14572. Cr. 50 65 2t .5 6.6 9.5 t2.9 2t.4 31.9 38.016.9 26.4A588 23. I 10.27.1 r 3.9 18. r 23.0 28.3 34.3 40.8

92 1208l 105

A,149d<lI <d < lr/:

39.631.7

12. i I 7.5 23.8 3l.l3.1.5 12.6

Thc abole lable lists ASTM specified nulc.iul\ !!ailirblc in round blr sr(xk rhat lrc genrr!lly intcnded lirr u\c in rhreaded appljcaoons such rs rie rods. crossbracing and similar uscsThe rensile capacir! ol thc lh.cadcd porlion ol an upsrl r(xl shall bc largrr lhan lh! b( ) lrca rrnrs 0.6F..F, = specified minimunr tensilc strcngth oflhc lasrener nutcrill.t. = 0.llF, = allowable tensile srress in rhrcldcd iasrener.

Table 4-7Design of Supports lor Vertical Vessels [31

Values of Constants q, C", Z,Function of k

C" Cr

andJasa

ZJAverage Values ot Properties ot Three

Concrete Mixes

0.0500. 1000. 1500.2000.2500.3000.3500.4000.4500.5000.5500.600

0.6000.8521.0491.2181.3701.5101.&01.7651.8842.0002.1132.224

3.0082.8812.7722.66r2.551a Ana

2.3332.2242.t t32.0001.884r.765

0.4900.4800.4690.459o.4180.438o.4270.4160.4040.3930.3810.369

0.760o.'766o.7'7 |0.7760.7'790.7810.7830.7840.7850.7860.785o.784

2000250030003750

800100012001400

Water Content oi nU.S. Gallons 28-day Ultimate 30 x 106per 94Jb Sack Compressiveot Cement Streigth, psi Ec

AllowableCompressiveStrength, psi

7tlz63/c65

l5t210

8

'188 Mechanical Design of Process Systems

Equation 4-44 is shown as a linear proportion by thestraight line shown in Figure 4-16. Even though the ten-sile strength of the bolt is, by Equation 4-44, equalto the ratio n times the concrete allowable comDressivestrength. it is not necessarily evenly distributed about theneutral axis as shown in Figure 4-16. This "offset fac-tor," known as the "k Factor," is determined from

os : noc

(d - kd) kd

ork = I

Equation 4-45 is solved by iteration using the followingsteps: Thke values for C", C,, Z, and j in Thble 4-7 for agiven value of Z. Normally, k = 0.333, C" : 1.588,C. = 2.376, Z = 0.431, and j : 0.782 to start the pro-cess. Then the following equations are solved:

/^^\M _ (W,r(z)l!!l" lt)l

rt--/\.. lBclrJrl;l

\'.1(Ah)N'' r(BC)

FIr = / -\

rt,r l!91c,\2/

fc: fi +wE

BPW : base plate width, in.

Bpw:(D.)-(Di)2

tz = BPW - t;

- (c.)(1,000)

= modulus of elasticity of base plate metal, psi: compressive strength of concrete, psi, denoted inThble 4-7 as o"

(4-4s) oq.*, : (sFC)t*X**]

(4-53)f(ER)(SFC)

After computing an initial value of k, this process shouldbe repeated five times in order to converge on a value fork.

Once a value for k is determined, we now solve for themaximum induced stress at the outer periphery in theconcrete,

using L =D" - (skirt OD)

(4-s4)

(4-s5)

tro"

(4-46)

(4-47)

(4-48)

(4-4e)

(4-s0)

(4-51)

(4-s2)

, ln.

we solve for the base plate thickness, BTHK,

BrHK = L [tf,]"' (4-s6)

where ou1 : allowable working stress for base plate metal,psi

By using Equation 4-56 one assumes no gusset plates onthe base plate-skirt connection. To reduce the requiredbase plate thickness in Equation 4-56 the additionalstrength of gusset plates can be used, because with thegusset plate stiffening the base plate at the skirtjuncture.the base plate between the gusset plates can be consid-ered as a rectangular, uniformly loaded plate with twoedges simply supported (at gusset plates), a third edgedfixed (skirt side), and the fourth edge free. The deflec-tions and bending moments are tabulated by Timoshenko[6] and are shown in Thble 4-8. The process of using gus-set plates to stiffen the base plate is begun by making thenumber of gusset plates equal to the number of anchorbolts. Doing this we write

NG : number of gusset plateslet NG: N bolts

F-/IIat5U=

NG + N bolts

M- : 1E,1o"1.""(SG)'?

M,' : lE,1o"1."-(L)2

(4-57)

f,J|L = -- /^l(h - (Enxt )) (]Jrc.r

where SFC : compressive stress on concrete at the boltcircle and

RAT:!SG

(4-58)

(4-59 |

= b/2\=r /

-0.500f"1,-o.428f "t,*0.319tP-o.22'7 f:r2-0.119f"F-o.124f"t2-0.125f "t2-0.125t"t2-0.125f"1,

: = gusset spacing (x direction) inches.: = bearing-plate outside radius minus skirt outside radius (y direction)

lnches.

Where o"1*n,; is determined by Equation 4-55, using thegreater of M, and M, we have

lortalo 'ornN : [_oJ

where M : M, or Mr in.-lb/in.

This iteration can be repeated as many times as desired toreduce the base plate thickness. In normal practice, it isunusual to use more gusset plates than anchor bolts.

The bearing pressure on the base plate must bechecked to prevent exceeding the allowable compressivestress of the concrete. Computing the uplift force oneach anchor bolt we have,

- 96MD" WF ,,-' N1D"z a P:; N'''

N" WF 12MD"o^ -_ ----:: + -----: +

-.

DSI. A" A. 21,

o" ( 1,200 psi for weight and windwhere, M : ftJbs

Table 4-8Maximum Bending Moments in a Bearing


COMPRESSION RING AND GUSSET PLATEDESIGN

Typical designs and dimensions of chair and base plate

designs are shown in Figure 4-15. The compression plate

thickness is determined by

I rr,rc If^^ : t _____________ ::- l

l4('.rr(A - BSI(4-63)

where A and C are dimensions in Figure 4-15

BS : nominal bolt size x 1.25Fi : bolt uplift force, determined by Equation 4-61

o.11 : allowable stress of compression ring metal, psi

The gusset plate thickness is determined by the follow-ing:

a],,2/F \18,000 Gw ta - (F,)ta - l"'i=' = 0 \4-@l

I,J|.ruwhere tc: gusset plate thickness, m.

Gw: gusset plate width, in. (A in Figure 4-15)GH: gusset plate height, in. (see Figure 4-15)

The minimum skirt-to-base plate weld size is determinedby

bt2I

00.0078f"b,o.0293f"bl0.0558f"b,0.w72f.b,0.123fJ20. 131f"b,0.133f"b,0.133f"b,

. [+r'a ] [w. I(4-60) *: ["o*l - [.o"j

(4-61)

(4-62)

w = € : weld size2F*

Fw = (1.33Xo"r)(0.55)

and

(4-65)

where o"1 : smaller of the allowable stress values for thebase plate and skirt metals

M : moment at base plate induced by wind orseismic forces, in.-lb

D.r : outside diameter of skirt, in.

ANCHOR BOLT TORQUE

There have been many recipes proposed for the com-putation of bolt torque over the years. The mystery ofbolt loads is unveiled by such authorities as Bickford [7]and Faires [8]. Their extensive research into bolt loadingproduced the following recommended formulation:

T : CDFi, in.lb (4-66)

where C : 0.20 bare steelC : 0.15 for lubricated boltD = nominal bolt diameter, in.F1 : anchor bolt uplift force, lbs

r[(D")'? - (D)'?]where A" =

4

r[(D")a - (DJa]

Plate with Gussets [61

M, {'\v

t-64

190 Mechanrcal Design ol Procg55 g151snlt

2.3<+<2.6

groutl-i

-f--

In most tower applications, Fel-Pro C5A is a verycommon bolt lubricant. The field of bolt desisn and boltlubricants is almost as involved as tower desien and theinterested reader is relerred to the excelleni work ofBickford [7].

Figure 4-17 shows the two most common types of an-chor bolts, "J" and "L." For large towers where largeloads are anticipated, the bolt in Figure 4-18 is used.

WIND ANALYSIS OF TOWERS

Analyzing wind loading on towers requires combiningloads induced by wind, internal or external pressure, andweight. Such an analysis must be made to ensure that thetower shell thickness is sufficient to withstand the com-bined loads.

Wind and seismic analyses are completed separately,with their respective bending moments being used to de-termine the tower shell thickness values at each section.

Before examining the design criteria, let's consider thefollowing terms:

ow = stress due to wind or earthquakeop = stress due to internal or external pressureo" : stress due to weisht

concrete

L = 17Du

Figure 4-17. "J" and "L" type anchor bolts are used for small vessel..

orout --,L

T_

concrele

Figure 4-18. Straight type bolt-used for large vessels, espe-cially towers.

Boltom of sleeveor top of concrete

PD-4t

Referring to Figure 4-13, we see that the stress ele-:lent in the shell is affected by the combined loads more::r the longitudinal direction than the circumferential di-:ection. However, for the longitudinal axis the internalrr external pressure stress is governed by the relation


such that the leeward side governs, then

loo-o*-o*l < Bfactor

If Equations 4-70 and 4-71 hold,

lo* -f oo - o*,1 ) o"1E

where ou1 : allowable stress in tenslo, for a givenmaterial at a given temperature and pressure

E : weld efficiency

Another form of Equation 4-71 may be determined byrewriting the equation as

op ) o*,

or

P(R, - 0.40 _ WE -;Rl -TI

[: w - t.zo pn,tlInwhrchfi{ r

L fSp" I(4-73)

Equation 4-73 is another form of Equation 4-71, interms of the vessel dimensions, where W is the totalweight of the tower above the section being analyzed.

2. Combination of wind (or earthquake) load, externalpressure and the weight of the tower-On windward side

OWS = O*-Op-O"n

On leeward side

OLs: -O* op - O*,

For most applications with external pressure we have

ols > ows

| - o* -op-o*, ) rw-op-owt

oo*0

Since the value of oo is for external pressure, we mustapply the B factor in Equation 4-67 . After these criteriaare satisfied, we turn our attention to the determinationof wind loads that induce o".

(4-67)

: mean diameter of vessel, in.: internal or external pressure, psi: Iongitudinal stress, psi: shell thickness less corrosion allowance, in.

,rhere DP

op

T

There are two conditions where Equation 4-67 is usedto combine stress values:

L. Combination of wind (or earthquake) Ioad, intemalpressure, and weight of vessel.

For windward side,

\'s:qw+op-o*r

For leeward side

(4-68)

,rLs : op - ow - owr W69)

Comparing Equation 4-68 with Equation 4-69 we have

ow + op- o*,1 ) 1oo-o*-o*,1

:f and only if l oo l > l o*,1

(4-70)

(4-7 t)

which is true for most applications, when the internalpressure stress is greater than that induced by the weightof the tower above the section. However, for a fewcases, the stress induced by the weight is greater thanIhat induced by internal pressure for low-pressure thick-ivalled applications. The thick walls at low pressurecould be for controlling tower deflections due to wind.For these limited cases the allowable stress is that deter-mined by UG-23 (b) of the ASME vessel code, which isknown as the B factor. The B factor is commonly associ-ated with external pressure, because the case of theweight stress exceeding the internal pressure stress israre, but it must be emphasized that the B factor is theallowable value of stress for longitudinal compressiveloading like that encountered in towers. Thus, the B fac-tor is more comprehensive than its external pressure ap-plication would indicate. Therefore, if Equation 4-71 isreversed and

o., | > lool (4-72)


WIND DESIGN SPEEDS

The procedures for determining wind design speedsfor structures, towers, and stacks varv from Counirv tocountry, depending on how well records have been kept.The wind velocity is a function of the temperature gradi-ent and terrain roughness. The first representation of amean wind velocity profile in horizontaily homogeneousterrain is the power law, first proposed in 1916. This lawstates

,t: r.\r:)" (4-74)

where Vo : mean wind speed at a reference height ZZ. = reference height (normally 33 fr orl0 m)Q : a constant dependent upon roughness of

terrainZ = height above ground

Other proposals have been made to determine windspeeds. Simiu [9] states that the logarithmic law is a su-penor representation of strong wind profiles in the loweratmosphere. What is relevant to the reader is to be famil-iar with whatever standard is used. The discussions andexamples presented in this text are slanted toward thosestandards in the United States. However, the techniouesand base principles of engineering are applicable to allcountnes.

In the United States there are four basic codes sovern-ing wind- ANSI A58. | 1982. the Uniform, thi Basic,and the Standard Building Codes [10]. The ANSI-A58.1- 1982 differs from the ANSI-A58. I - 1972 in thatthree optional methods of determining wind design loadson a structure are given [11]. These options are as fol-lows:

l. Choose a design wind speed (50-year mean recur-rence interval) off the U.S. map provided on thedocument. The national map is a graphic display ofisopleths (lines of equal wind speed) of the maxi-mum values of the mean speed for which recordshave been kept. i.e.. basic wind speeds rhat can beexpected to occur within a particular period. This"particular period" is called the return period. Theproblem with a nalional map consisting bf isoplethsis that localized wind speeds can vary as much as30 mph over the speed shown on the isopleth (par-ticularly in mountainous regions).

Hurricanes are fairly well accounted for on thesemaps. Tornadoes are considered to be nonexistent,because it is not economically feasible to design anentire building for tornado wind speeds. The rea-son for this is that the probability of a structure be-

ing hit by one is extremely small; however, nuclearsites are designed to withstand tornado winds.

2. Using site and structure factors calculate the designwind speed. The factors on rhe ANSI l98Z tesrused are as follows:

(a) Importance Coefficient, I, a hospital or nu-clear plant would be designed moie conserva-tively than a barn on a farm.

(b) Variation of wind speed with building heightand surrounding terrain.

(c) Gust response factor.(d) Velocity pressure coefficient, K2.

3. Test a model of the tower and its surroundings in awind runnel. Even though rhe 1972 ANSI stindarddoes not mention this, the 1982 version sDecifiescertain requirements lor wind tunnels.

These three options are new to both the ANSI-ASg. Istandard and to the three building codes-the Uniform.Basic. and the Standard. The larrei three codes do not au-tomatically adopt newly revised ANSI standards, thusmaking for inconsistency in wind code provisions in theUnited States.

The basic wind pressure in the ANSI-A5S .l-19j2 rs

q3a= pv2l2 : (0.5X0.00238)(5,280/3,600fV30: 0.00256 v3o @-'75,

where q.s : basic wind pressure at 30 ft, above gradelb/ft,

V36 : basic wind speed, mph

The effective velocity pressures of winds for buildingsand structures, qF, is

9r : KzGrQ:o (4-76 t

where K2 = velocity pressure coefficient that dependsupon the type of exposure and height Z abovethe ground

Ge : dynamic gust response factor

In the 1982 ANSI-A58.1 Code the effective velocinpressure for wind is partially a combination ol Equarion.4-7 5 and 4-76,

qz : 0.00256 KzGV)2V = basic wind speed, mphI : importance factor : I

\417)

A value of V can be approximated for the United Statesfrom the isopleths shown in Figure 4-19.

One of the major differences between the ANSIA58.1-1972 and 1982 is how the velocitv Dressure coef-ficient, K2, is determined. In the 1972 Cod'e the value ot

-

J


:;;

:

-:

z

OJ

.9TL

o;\- -; e. b\ .E oot;9\ ] b9;.0\ ; *-t!cb\ ; !b69\ 3. iaEg\ o-i=H9

a'*-w* ii .s iird\ * :# "E/\ * *tid

,r\ i ^;it9\i !ii.s- \F 9!.o P'-+!-€ : R :: \8. E i a;

" : ]{ IEE:I .: i-'q

s -i Hf;n*:,'' s 18 ;:I g;ni:-* l(U*963

o-!o$ " !;i;

let o -6 6 r. E'=i" f i:.E4:' ig ?i-6

>iif or No6o5z

)2

f

8

\pxi4'

:

E

.l!L rl' I jll\

194 Vechanical Design of Process Syslem,

Table 4-9Velocity Pressure Exposure Coefficient, Kz [1 1l

Height aboveGround Level, Z

0- l5202530405060708090

100120140160180200250300350400450500

o.t20. 150.1'70.190.23o.270.300.33o.370.400.420.480.530.580.630.670.780.880.98r.071. 16t.24

o.370.42o.460.500.570.630.68o.73o.770.820.860.930.991.051.1 1

1.161.281.391.491.581.671.75

0.800.870.930.981.061.13l. 19| .241.29| .341.381.451.521.58I .631 .681.',791.881.972.O52.122. l8

1.201 .27r.32r.371.461.521.581 .631.67| .'7 |1.7 51.811.871.921.972.012.102.182.252.312.362.4r

K7 is a linear function of the height Z from heights ofthirty to nine hundred feet. This results in a triangularwind distribution on the tower. In the 1982 Code thevalue of K2 is a parabolic function (can be approximatedwith a step function) for wind loading depicted in Table4-9 and for dynamic gust response, K7 is governed bylhe power law, Equation 4-74.

lz\2',r* \r,)

forz > ls feet

Kz= @-78)

forZ < 15 feet

where values of Z" and d are given in Thble 4-10. Theparabolic function is a reflection of the old classical ap-proach used in the ASA 58.1-1955, but is a more refineddistribution. The treatment of K2 in the dynamic gust re-sponse analysis is a new development in U.S. codes.

The force exerted on a tower immersed in a movinsfluid is a function of the properties ol the tower shapiand properties of the fluid. The fluid properties of im-portance are the viscosity, density, and elasticity. Writingthis relationship in functional form we have

F = f(p, Y, I, p, a)

-tpvt,-t;

r _ - /pvi\pv+-'\r/

where a = velocity of sound = 0 in our case, because wincispeeds are extremely low compared to sonic speeds

This equation shows that there is a relationship dic-tated by the dimensions of the parameters involved. Ap-plying dimensional analysis makes the equation

P(Y2!2,}:.

where each of the two components is a dimensionless pa-rameter. The equation can be solved for the first dimen-sionless combination by

(4-19

Equation 4-79 implies that the parameters F/(pVri:and (pYllp) have certain definite values that will be equa.if a geometrically similar body with the same orientatio:is moved through the same fluid or another fluid fo:which pVflp has the same value as the first body. Tsisuch bodies are said to be dynamically similar and dr -

namic similarity is the key to wind tunnel tests. Assuminsthat p has no influence on the force F, we can deduce fror

Equation 4-79 (see any basic fluid mechanics text) and ob-tain

orF: Cp pYz12 (4-80)

where Cp is a dimensionless empirical constant. Equation4-80 states that, for a body of given orientation and shape

that is immersed in a moving fluid, the force experiencedis proportional to the kinetic energy per unit volume of themotion of the fluid (p/2)V2 and a characteristic area f2. Cpis a dimensionless quantity that characterizes the force andis called the /orce coefficient. Two bodies that are im-mersed in moving fluids are said to be similar (geometricsimilarity) if their Reynolds numbers are equal. Then theflows are dynamically similar and have equal force coef-ficients. The Reynolds number pVl y. is called a similairyparameter. Figure 4-20 shows the influence of the Rey-nolds number, corner radius, and surface roughness onthe force coefficient on various bodies. The values of Cpare determined empirically and are shown in the figure.Sometimes this coefficient is referred to as the drag orpressure coefficient.

Kuethe and Schetzer [12], use the Kutta-Joukowskitheorem to show that the force per unit length acting on aright cylinder of any cross section whatever is equal topVf and acts perpendicular to V. The symbol f is circu-Iation flow about the cylinder and | = r'DV. The Kutta-Joukowski principle is exemplified in Figure 4-211131.Here the pressure distribution around the cylinder ismaximum ninety degrees to the air flow. Dependingupon the relative stiffness of the tower sections and massdistribution. this perpendicular lorce vector can cause a

phenomenon known as ovaling, which will be discussedlater.

In computing the wind forces on a tower, Equation 4-80 takes the following form in using ANSI A58. 1- 1982:

F : q2GCpAg (4-81)

whele qz : wind pressure at height Z, EgrJation 4-77,

_ lb/fcG : gust response factor for main wind-force

resisting systems of flexible structuresCp = force coefficientAr = cross-sectional area of tower and other

attachments, ft2

The gust response factor, G, when multiplied by themean wind load, produces an equivalent static wind load


Table 4-10Exposure Category Constanls [111

195

Exposure Category 4 Do

2

3.04.57.0

10.0

---j ''--!1,lz'

(b)

III

,-"-01

BcD

15001200900700

0.0250.0100.00s0.003

1.8

r/h = O.O21

r/h = 0.167

1.2

0-4

o.4

1.2

t/h= 0.333

t/h= O.5

tO. 2 4 8105 2

Ae

-

sanded $rface (d)

---Smooth srrface

4 ato6 2 4 8tO7

Figure 4-20. The curves depict the influence ofthe Reynoldsnumber, corner radius, and surface roughness on the drag co-efficient, square to circular cylinders; r is the corner radiusand K is the sand grain size [9].

Figure 4-21. A sequence ofpressure fields forming around a cylinder at Nq6 = I 12,000 for approximately one third of one cycleof vortex shedding (Flow-Induced Vibration by R. Blevins. @1977 by Van Nostrand Reinhold Company, Inc. Reprinted byoermission.)


e-82)

where p : structural damping coefficienr (percentage ofcritical damping). For normal working stressconditions, 0.01 < P < 0.02 for towers.

that would induce deflections equal to those of a gustywind. MacDonald [14] refers to this approach as a quasi-static loading analysis. Quasi-static means that at any in-stant the stress and deflection induced in the tower arethe same as if the instanlaneous mean wind load were ao-plied as a static load. Thus. the significanl factor is iden-tifying the single highest peak value of instantaneousmean wind speed, or that is, predicting the future worstpeak value. Baker and others found at the end of thenineteenth century that there is a simple relationship be-tween the gust frontal area and gust duration. This rela-tionship provides a means of determining the size of thegust, and is illustrated in Figure 4-22. The figure indi-cates that the worst wind condition for a Darticular toweris not necessarily the maximum value of the wind veloc-ity, but rather the highest wind speed of the particularsize of gust capable of totally enveloping it. To compen-sate for this in a simple quasi-static analysis, ANSIA58.1-1982 gives rhe gust factor as

Table 4-1 1

Probability ol Exceeding Wind Design SpeedP" = 1- (1 - P")N

AnnualProbability Design Lite ot Structure in N years

P, 1510 15 25 500.10 0.100 0.410 0.651 0.'194 0.928 0.9950.05 0.050 0.226 0.401 0.537 0.723 0.9230.01 0.010 0.049 0.096 0.140 0.222 0.3950.005 0.005 0.025 0.049 0.072 0.118 0.222

100o.9990.9940.6340.394

D_ probability of exceeding design wind speeddunng n years, where P : l-(1 - p.)"annual probability of wind speed exceeding agiven magnitude (Table 4-l l)exposure factor evaluated at two-thirds themean height of the structure

_ 2.35(C, )0 5

11 = -(Zl301rt"

structure size factor (Figure 4-23)average horizontal dimension of the buildingor structure in a direction normal to the wind.ft (see Example 4-2)

- lp 11 ?tr.,/s \r2tJ: L,.o.l +t + |

\p 1 + 0.002ciS:

112 p!2

IvMEAN VELOCITY: V

:r[o3ora

^

0

20 30 r0 5060 80 t00 200 300 a005006008001000 2000

hlftl

Figure 4-23. Structure size factor, s [l l].

The Engineering Mechanics of Pressure Vessels r97

GUST DURATION EFFECTIVE GUST DIAMETER

)165 tt

For a tower with many obstructions, such as piping,ladders, platforms, and clips that are comparable in sizeto the vessel, the gust response factor can be determinedby:

^,--ll'--4-l|-JOUnOt'O".?

3

5

15

Figure 4-22. Diagram of relationship between gust duration and gust diameter.

,I

i

I

i

r.: = n r< /t.zsp * r.l.:zr,),s \' '" "--\ 6 1+ o.oolc/(4-83)

The gust response factors given in Equations 4-82 and4-83 are for flexible structures, such as towers, wherethe height exceeds the minimum horizontal dimension at

least by five to one or the structure exhibits a natural fre-quency less than one. The fact that the tower may have anatural frequency less than one is significant.

Simiu and Scanlan [9] point out that for natural fre-quencies greater than one, the response spectra are de-pendent on the structure's height. However, for naturalfrequencies less than one, the spectra distribution has lit-tle influence on structural response, and the magnitudeofturbulent fluctuation components, such as wind gusts,at or near the natural frequency of the tower could signif-icantly affect the structural response. For this reasonEquation 4-82 should be used for towers with particu-larly low natural frequencies.

Figure 4-24 shows a plot of wind gust velocity versusthe structural response of a structure. The cyclic loading

I

s 0.9


Figure 4-24. Quasi-static structural response spectra versuswind velocity [ 14].

induced in the tower can result in fatisue failure of vari-ous vessel components.

Equation 4-81 contains the last parameter that must bedefined, Ar, the total cross-sectional area of the towerand attachments that are perpendicular to the wind. Thisarea is computed by first determining the equivalent di-ameter of the area facing the wind. This can be expressedAS

De : (vessel OD) + 2(vessel insulation thickness)+ (pipe OD) + 2(pipe insulation thickness)+ (platform projection)* (ladder projection) (4-84)

Equation 4-84 does not consider extraneous equipmentattached to a tower, such as reboilers. The engineer mustadd the OD of the reboiler, plus twice the insulationthickness, plus any other equipment diameters to Equa-tion 4-84. Doing this and multiplying by a length overwhich D" is effective determines As. Figure 4-25 showsthe effective or equivalent diameter.

WIND-INDUCED MOMENTS

After the wind pressure distribution is obtained fromEquation 4-77 , the distribution of section force vectors isobtained from Equation 4-81. The force vectors, shownin Figure 4-26, act through the centroids of the pressuredistribution sections. Referring to Figure 4-26, we seethat the wind moment distribution is obtained from thewind force vectors through the following relationships:

Figure 4-25A. Effective diameter can vary with height.

FJ,Ma+F"(2,-Z;+F,"rbMb + (F" + FbXZb - Z") + F,r"M. + (F, + Fb + FcXZc - Zd) + FdtdMd + (F, + Fb + F. + Fi(Zd - Z") + F"t"

or in a general equation,

platform

DE = effective diameterof area resistingwind

-z -r\-p -t c;4n_t'Ll,t,, n

i: lM" = M"-1 * (2"

1 (4-85 r

Figure 4-258. Wind area and force calculations for conical sections.


d = plattorm angle

WIND.INDUCED DEFLECTIONS OFTOWERS

Thll process towers and stack are treated like canti-lever beams in computing deflections induced by wind.Like a cantilever beam, when the tower deflects it trans-lates and rotates at the same time. These translations androtations are most expediently computed by the methodof superposition. The three cases to consider in the su-perposition are a cantilever beam with a uniform load,an end load, and an end couple. These three cases andtheir accompanying equations are shown in Thble 4-12.

The first case of the uniform load reDresents the windload on the side of the tower, the second case o[ the edgeload represents the wind shear at the various shell sec-tions, and the third case of the end couple represents thecase of couples produced at the shell section junctures bythe translation and rotation of the upper sections. Thiscombined loading is shown in Figure 4-26.

Adding the three cases we obtain the following:

/i = section length, ftQi : wind shear at each section junctureMi : moment induced by wind profile, in.-lb

For rotation we have

/n- I \l\-r.l o

^, _\?,'l "{w,r, *q,r,_,,)Er, \6 2 l

Total deflection : y

,s + F ,t."=F

(4-87)

(4-88)

6, = llY{*!{,*M')' Er\8 3 2l(4-86)

= lateral translational deflection of section i, in.: length of section i: concentrated wind load (wi/), lb: wind profile, lb/ft

WIND-INDUCED VIBRATIONS ON TALLTOWERS

Chapter 2 discussed the phenomenon of vortex shed-ding inducing vibrations in piping systems. This chapterfocuses on the nature and techniques of analyzing vortexshedding.

Over the years many researchers have made wind tun-nel tests, proposed various analytical procedures, andconducted field tests of various structures subjected towind loads. Wind-induced vibration was first noticed on

where 61

!1

W1

insulation OD

Mechanical Design of Process Svstems

Table 4-12Cantilever Beam Formulas

Formula1 w--.'

dITTtrMT-1i2

4,-\T-)

EndLoad

UniformLoad

EndCouple

^ w!26EI

: wf'8EI

= Qi,3EI

= lul{2ET

: Q/'2El

:vdEI

^ wl26EI

A:0t,

l iw{ , wrEI\6 -

'

, w/, , Mo{-2EI-Er' +M)

I\- olr\2"1 "tw.t w/ \A = l I'+ I'+M|Er, \6 2 |

tall stacks by Baker at the turn ofthe century. Since then,many advances have been made in the field of aerody-namics allowing designers to adequately design tallstructures. This chapter discusses tall process towers andChapter 5 discusses tall stacks. The differences betweenthe two will become more clear in the following discus-sion. Staley and Graven u5l summarized the state oftheart of wind vibrations. Their studies indicate that eventhough vortex excitation of higher modes has been ob-tained in wind tunnel tests, existing free-standing stackshave always been observed to vibrate during vortex exci-tation at a frequency and with a mode shape associatedwith the fundamental mode. Furthermore. the shaDe ofthe dynamic lorce amplitude or existence of nearly con-stant frequency over the height of the stack (or "lock-in") implies that dynamic response will almost entirelybe induced by the first mode. Staley and Graven con-cluded that all higher modes should be neglected in thedynamic analysis and that the frequency and associatedcritical wind velocity ofthe fundamental mode should beconsidered. For this reason the Rayleigh method is theindustrially accepted method because it is used to deter-mine an approximate value for the lowest natural fre-quency of a conservative system based on an assumedconfisuration of the first mode.

What is clear in wind tunnel tests and field observa-tions is that at low Reynolds numbers the tower is dy-namically stable, vulnerable only to forced vibrationsand at higher Reynolds numbers a possibility of self-ex-cited vibration will be present. From many field obser-vations it can be concluded that the first peak vibrationamplitudes occur at the critical wind velocity Vr, whichcorresponds to a Strouhal number of 0.2 with the forcedvibration as the basic source of excitation. Thus. it is sis-nificant that the peak amplitudes of vibration determinedby forced vibration theory are in very good agreemenrwith field observations. This will be seen later in thischapter in Example 4-4.

Even though the Rayleigh method is the industrialllaccepted method for the present, there are other methodsused to describe the vibration phenomena of tall processtowers and stacks. One such method was devised by N.O. Myklestad, a great pioneer in the theory of vibrations.The Myklestad method used in cantilever beams is es-sentially a Holzer procedure applied to the beam prob-lem. Its strong point is utilizing field and point transfermatrices to obtain relations that govern the flexural mo-tion and vibrations of lumped-mass massless elasticbeam systems. This method is used in such applicationsas aircraft wings where the structural component is sub-

jected to high Reynolds numbers. Since we have alreadydelineated the difference between cylinders subjected tohigh and low Reynolds numbers and the fact that modeshigher than the fundamental mode can be neglected, theMyklestad method has lost favor to the Rayleigh method.We are primarily interested in forced vibration peak am-plitudes of relatively low natural frequencies. Althoughthe Myklestad analysis is excellent for relatively clean

aerodynamic surfaces such as wings and missiles, itspractical use in process towers with attached ladders,platforms, and piping is questionable. Even for stacks.

low Reynolds numbers allow for the fundamental modeto dictate.

Before the Rayleigh method is applied to our analysis,let us summarize some basic precepts. Equation 4-80calculated the pressure force exerted on a cylinder by a

static wind. When dynamic effects settle in maximum ac-tual amplitudes, these amplitudes often exceed those un-der static conditions. The net result is to multiply Equa-tion 4-80 bv a masnification factor. To understand the

The Engineering Mechanics of Pressure Vessels 2O1

magnification factor we must consider some basic prin-ciples.

Consider Figne 4-27 in which a system with a singledegree of freedom is subjected to viscous damping andan externally imposed harmonic force. The spring isdenoted by stiffness k, the friction coefficient by c, mass

by m, displacement by x, impressed force as F sin cJt so,

we have

-X+.x +ki: Fsin<rt (4-89)

From the theory of differential equations we know thatthe solution of Equation 4-89 is the superposition of thegeneral or complementary solution of the homogeneousEquation 4-89 and the particular solution of the same re-lation. Writing this in equation form we have

X=X"*Xpwhere X" is the complementary function and Xo is theparticular solution. This classical differential equation is

,-il]ur=--l[: r-i-

4-.=qlq.,+ r'.+t'-..r.-"1

,=

==

14.=s\"

T"I*

IAL*

i-7- -,

I

r6"_L--r ,r

6!

6.

A

4_.=o"lr,-.+r,.1

4_; qlL,.+r,-,+r, .l

[.-rII_-r.L

l)'.

olFrt

*, = ]+ r|",1 lr-eol

Figure 4-26. Schematic diagram of wind loadings and deflections of a tower.

f fStru"t = forcing function

damper-represents tower'sstiffness

solved in numerous sources and will not be delved intohere. See Vierck [6] for a complete discussion of thesolution. The final solution takes the form of the follow-lng:

X(t) : e t''(A cos (,Dt + B sin ropt)


we have

Figure 4-27. The vibration of a tower is modeled as a sinsledegree of freedom. which i5 exposed to an exrernally impos=edharmonic force and subjected to viscous damping.

x., "T -,,f-1_12r*

The maximum actual amplitude X of forced vibrationis obtained by multiplying the static deflection X,, b1fraction X/X,,.

The fraction or ratio X/X* is called the dynamic mag-nification factor, D. These formulations indicate that thenondimensional amplitude X/X,, and the phase angle, 0.are functions of the frequency ratio r and the dampingfactor f and are plotted in Figure 4-28. These curves in-dicate that the damping factor has a large influence onthe amplitude and phase angle in the frequency regionnear resonance. From Equation 4-91 we see that at reso-nance the dynamic magnification factor, D, is inversellproportional to the damping ratio, or

Freouencv r.tio. = (;/o

Figure 4-28. The dynamic magnification factor versus thefrequency ratio for various amounts of damping. (From Slruc-tural Dynamics by M. Paz. @ 1980 by Van Nostrand ReinholdCompany, Inc. Reprinted by permission.)

(.4-9t)

In-'

t:

Letting

X., sin (c,rt - d)(4-90)

{t -l t+ (r'ttc/c,2(mk)ri2 is the critical damping factor that is thecriteria for critical damping such thatI : nonvibrating motion : overdampingI : harmonic vibration : underdampinga few percent of c. for a tall, slender structure suchas a towerstatic deflection of the spring acted upon by thefbrce F/Kc,,,/o : frequency ratio of forced vibration frequencyto free vibration frequency

X:.(T -l t+ (2rt

U

E

EP

I -r

KM

and tan 0

The damping ratio, €, is not known and extremely dif-iicult to measure at best. A practical method for experi-mentally determining the damping coefficient of a sys-tem is to initiate free vibration, and measure throughdecreasing amplitudes of oscillatory motion, as shown inFigtre 4-29. This decrease or decay is termed the loga-rithmic decrement, 6, and is defined as the natural loga-rithm of the ratio of any two successive peak amplitudes,X1 and X2 in free vibration. Expressing this in equationtbrm we have

^x,x2

The evaluation of damping from the logarithmic dec-rement is given analytically by

X(t) : Ce-fdr cos(@Dt - cr)

It can be shown [17] that the dynamic magnificationfactor, D, and the logarithmic decrement, A, are relatedusing the previous expression as

(4-92)

Most research data available for practical use are pre-sented in terms of the logarithmic decrement, 6. Table 4-13 provides values of 6 versus D for various structures.These values are acceptable for use in actual design ofprocess towers and stacks.

Applying the dynamic magnification factor to Equa-tion 4-80 we have

CeDpV2fz (4-93)


The force coefficient can be readily obtained from Fig-ure 4-29. Equation 4-92 yields the maximum transverseforce per unit area of the projected surface of a cylinderat resonance.

Equation 4-93 may be rewritten with the velocity inmiles oer hour as

F = 0.00086(CrD)(H)V1'?, for air at 50'F

and

F = 0.01I l3pCrDVr'?(dH)

(4-94a)

(4-94b)

These equations apply when the top third of the toweris the controlling length. Often, the top fourth of thestack may be best to use as the controlling length. An ex-ample ofthis would be a section on top ofthe tower thatis one fourth the total tower height and is significantlygreater in diameter than the section below (see Example4-4). Thus, for the top foufih of the tower Equation 4-93becomes

F = 0.00065(CrDXd)(H)Vr'?, for air at 50'F (4-94c)^T

and

0.07728pCeDVr'?(dH) (4-94d)

where d = outside diameter of either upper r/: or r/+ oftower, ft

H = total height of tower, ftVr : first critical wind velocitY,Vr : 3.40d/T, mphT : first period of vibration, Hzp : density of air at any specified temPerature,

lb/ft

Figure 4-29. The Reynolds number versusthe drag coefficient for a circular cylindertet.

Fz=9h

6


Table 4-13Conservative Values for Logarithmic Decrement

and Dynamic Magnilication Factor for TallProcess Towers

LogarithmicDecrement

6

Dynamic MagniticationFactor

DLow damping: rocky-stiff

soil, low-stressed pilesupport, or structural

0.052

frameAverage damping: moderately

stiff soil, normal spread

High damping: soft soil,fbundation on highlystressed friction piles

0.080

o.t26 25

A : Steel frameB : Reinforced or prestressed concreteLow stress levels

(a) 0.005 ( c ( 0.010(b) 0.005 < c ( 0.010

Working stress(a)0.01 <c (0.02(b)0.01s < c < 0.03

Near yield(a)0.M(c(0.06(b)0.0s<c(0.10

c":2(Mk)05=28I).'' \386/ ln.

Coefficients l1

C. = critical damping factor

k : tower stiffness, lb/in.W = total tower weight, lbr

Structural

lbi-sec

C : damping factor = tower stiffness, Ib/in.

" ^ tLl\c./

D.- : r/6

M : tower massc

For tall, slender towers of constant diameter, the firstperiod of vibration is given by the expression

T : (l/0.5l)(WHa/gEI)o 5

where g = 32.2 ftlsecH : total height of tower, ft

(4-95)

The Rayleigh method applies only ro undamped sys-tems, but is found to be sufficientlv accurate for comDut-ing the fundamental frequency of process rowers. e;enthough towers have varying shell thicknesses down theIength that result in unevenly distributed mass and stiff-ness. The Rayleigh method is basically the conservationof energy, i.e., the total kinetic energy of the system iszero at the maximum disDlacement but is a maximum at

the static equilibrium point. For the potential energy o;the system, the reverse is true. Thus,

(K.E.)-,, = (PE.).,- = total energy of the system

The resulting equation will readily yield the natural fre-quency of the system.

To estimate the period of vibration using the Rayleighmethod the tower is considered as a series of lumpecmasses. These lumped masses are determined by considering the weights of

1. Shell and heads2. Trays and internals3. Manways and nozzles4. Insulation and fire proofing

These are summed for each section and the overall:ower is considered as lumped masses at the centroid of:ach section along its entire length. The assumption isnade that the stiffness is constant along the entire lengthJf the tower; an assumption that greatly simplifies the;omputations for the various deflections of the section:entroids. The more sections, the greater the overall ac-

.-uracy achieved. Such a beam with lumped masses isshown in Figure 4-30. For such a simple, one-mass, vi-brating system, Timoshenko et al. [19] have shown thatrhe angular natural undamped frequency (rad/sec) forsuch a system is

(4-96)

Integrating Equation 4-96 numerically across the sectioncentroids of the tower results in

- : [BOMry" + Wzyb + ... + W,y.)/(W1y]+ w2y3 + ... + w"yillo 5

or

(4-e7)

(4-98)


(4-9e)

The moments obtained are used to determine the de-flections induced by vortex shedding. The method of de-flection computation is based on the area-moment (con-jugate beam) method applied to a cantilever beam. In thismethod the slope of the elastic bending curve of the ac-tual beam is equal to the shear at the same point on theconjugate beam, which is an idealized beam correspond-ing to the actual member. The deflection y of the actualbeam (or tower) at any point relative to its original posi-tion is equal to the bending moment at the correspondingpoint on the conjugate beam that has the same M/EI areaof the actual beam. Figure 4-31 shows weights of thevessel sections distributed about the section centroidsalong with beam lengths used in the analysis. The conju-gate beam method of computing deflection is demon-strated in Table 4-14. For an indepth analysis of themethod the reader is referred to Higdon et al. [20]. Theexamples presented at the end of this chapter will clarifythis approach.

OVALING

Ovaling is a resonance phenomenon more common instacks rather than process towers. However, towers ex-hibit this phenomenon mostly during construction, be-fore insulation and appurtenances are added to the ves-sel. To avoid ovaling, the designer should consider thefollowing guidelines.

The cylinder is considered as a ring that has a naturalfreouencv of

- 7.58r. vE'' 6oDt(4-100)

M, : wr () * *, * nJ + w,(?. ")

. *,(?)

- lewv\o'\wv'/

T_

The section weights, Wi, are computed by using cumula-tive weights down the tower. Summing moments aboutthe base in Figure 4-30 we obtain the moment distribu-tion in the tower as follows:

Mr=0/- -\

M, = W, lKr + K2l-\21

u. = w, lI1 +n.\2

* *,lvr-*,\-\ 2 I

," [(i -")/(,8 *,,)]"

, Kil)l

Figure 4-30. A tower modeled as a sectionless beam with distributed lumped masses.


Table 4-14AVibration Deflections Based on the Coniugate Beam Method

,f

w.w"vqq

Xs, = o,

+ +I

+I

+l)t'+ +*t*t +

{t-, I r_, L, l- 15 '

| -1.

M, S, +A, = A,

ZE)l)

P

Mr Pt*l.z: Ittr, t4r

E'Itia, + e,\\?/

Lr=Pr

M2 = W1L1 , RI+R'-' 2

/v. M,\\E,I, Err,/

2

XL2: 52

52*41 :,A2rr 19EzI:

/a, + a.\ Pz * ir: : r:t?,

xL2: P2

M1 : W1(L1 *L)+ w2L2

- R, +R.-2 irur, M, \

\E.I- - E,I,/

I: M:E:I:

53+A.4 = Aj Ptl p+: pt14i14)\21xL3 = Pr

2

XL3: S1

/eo + ellrl

Ptr ps: ptwr(Lr + L2+ L1) +W2(L2 + L3) +W:L,

Lo=l+R"2

-v xl-a = Pa

'. Mr lM.,M.\ Sa A,=AaEoL, \E5I5 E4I4/

2

xl-o = So

M,:Mi r+Li r

i

\-wLll: l

For cylindricalshells,

r _ Ri +Ri+r'2

/ M,*, M, I\E, + rli+ | E,l,/

2

xl-i = S

s' +D

S:=Ar

FT/Ai + Ai*ri Pi * irr+r : IIrl

xL; : P1

I. v. /u.-' M"\ s" A"

E"r. \E"I" E"I"/lil = p

2

xLn: S"

M"+r : M"+L"

xFw,L,l

Co_mputation of lateral y deflections. For formulas of y see Table 4-16C.deflection of real beam : bending moment diagram oi conjusate beamslope of real beam elasric curve -- shear of conjugate beam -moment diagram of real beam : load diagram of conjugate beamy1 : fu)(12), in.

Table 4-148Vibralion Deflections Based on the Coniugate Beam Method-Section Break Method

lttrttlV Y "Y V Vw" w'r w.{

Mi li M/EiliDs, = o,= Mrdx/Eili p = M;d2xlE;11Pi

Mr :0 Ir MrErIt

Mr - ^ sr +Ar:Ar2El1

/e' +l?lxLq:P1

/- ^\Mj W,lr(r + r(?l

= W,Lj- \21Ir M, /vt, M2\ s2 + A, = A/

EtI, \t'313 l'212l

2

_M,' 2r..

xL2 = 52

{rfxL2 = P'

Pt 't pt : ttz

M, = w,lR' , R, r!\-w,fIL$,\ Ir

\2 2l '\ 2 l: Wr(Lr + Lr) + WrL2

Er53+A*:4, Pr*t.,:/:

2

/Mr Mrl\E.I.

- Err',/lA, + Aqllrl

xL3 = P3

u, = w l/! +R, +Ri*w,(R,+ Ri t"' \2 2l '\ 2 I: W1(L1 + L2 + Lr) + W2L2

It, r, \MJ lrvrl f, wlrl ra+45:Ad

EoL \E4I4 E4I4/

Po*/r: = lr

2

xL4: 54

/eo + ,+,\\) I

xL4 = P4

M,:w(q+*,**.*&\\2 2l

+w,l&+n.+!l'\2 2l

+ w.lR. + &l-\21

6

s5 + Dsi15 l&

E:I:/tut, , tuto \\E+ - E.L/

2

xL5 = 55

f,L,+a"\ Ps *ro = rslrl

xL5: P5

, sr --,lvl" = lvl^ r+L" r, W,-l /tut"-, , tnl" \\EJ" - E t/ _jI L" = P^

2

2

xL":S" S"=A"

I, M"Ik

M.+ r

y = (pi)(12) ft

= M"+L"

<Fw,t:l

@ : ebrupt section breakk : n + (number of abrupt section breaks)Infigureabove,k=n*1


Table 4-14CCentroids ol Shell Volumes

Conical Section

- b(4a - 3t)

':'(T i

3H(D": - Di) + 8Hr@., - Di") tano

6[{D"r -Dit - 2H{D., - D.,) randjv

k-t--l

The vortex shedding frequency is given by

f 0.2v,"D

process columns, because these vessels usually haremany external, attached appurtenances. What is morecommonly done with towers is to stiffen up shell sectiontto offset ovaling resonance. See Chapter 5 for more in-formation on ovaling.

CRITERIA FOR VIBRATION ANALVSIS

While there is no absolute parameter available for de-termining whether a vibration analysis is required, thereare certain guidelines for designing towers.

1. If the critical wind velocity, V1, exceeds 60 mpf.then a vibration analysis is not required. Very feucases of severe vortex excitation have been ot -

served for wind velocities in this ranee.

(4-101)

where v : 45 mph or 66 fps

If for any section of the tower fi < 2f,, ovaling vibrationis imminent. The resonance wind velocity that wouldtheoretically induce ovaling is

60 f.D(4-1o2)

where s = Strouhal number : 0.2 for this application

To counter ovaling vibration, ovaling rings or helicalstrakes are added. These normallv are not oractical for


2. If the first critical wind velocity, V|, is greater thanthe wind design speed, a vibration analysis is notrequired.

3. The limiting minimum height-to-diameter ratios H/d are as follows:

z- 4-_}q-+-+q

+

+-i:

.1+_

q-+-+-

'--+---1

r+_

Figure 4-31. The vibration ensemble in which each sectionweisht is located at the section centroid.

4. The Zorilla criterion for vibration analysis is as fol-lows:

vibration analysis mustbe performed

- ". vibration analysis should( r, " -' o_.-'il - (4-lo4)

vibration analysis neednot be performed

5. If the total force on the tower induced by the firstcritical wind velocity V1 does not exceed l/rs of theoperating (corroded) weight W or

H/dH/dH/d

> 13 unlined stacks )> 15 lined stacks )

) 15 process columns I

(4- 103)

LD, \ ZW

20 <,^,

25 < ---LDI

1":oViHd')" 1

w o15 (4- 105)

Further guidelines and procedures for stacks are dis-cussed in Chapter 5.

SEISMIC DESIGN OF TALL TOWERS

There are several ways to analyze earthquake forcesimposed on a structure. The procedures outlined in theUniform Building Code [10] are the simplest and moststraightforward, but do not account for all of the signifi-cant dynamic properties of structures. Large, complexstructures, such as so-story buildings, nuclear powerplants, large dams, and long suspension bridges, requirea more thorough dynamic analysis. Fortunately, theUBC method is accurate enough for most tall, processtower/stack design problems and is presented here.

In seismic analysis the design spectrum is not a speci-fication of a particular earthquake ground motion; it is aspecification of the strengths of structures. For this rea-son the tower must be ductile enough to absorb energywithout ultimate yield. This implies that for the structureto absorb energy that exceeds maximum design condi-tions the overall structure deformation will be ductile


rather than brittle. The result is that while more risorousanalyses are very helpful in determining design ciiteria.practical design procedures are simplifications of thecomplex dynamic phenomenon used as'.quasi" staticcriteria applied with elastic srress limits.

The Uniform Building Code 1982 [10] requires thar allfreestanding structures in seismic zones to be desisnedand constructed to with5tand a total lateral force t-baseshear) given by

I : moment of inertia, ftg = 32.2 ftlsec2

When Equation 4-108 is applied to sreel wirh a value ofE : 30 x 106 psi we have

(4- l09 r

V : ZIKCSW

where ZI

Kcs

: seismic zone factor (see Figure 4-32): occupancy importance factor : I for allprocess towers and stacks: structure type coefficient

: structure period response factor= slte structure interaction factor= total operating weight of tower above ground

The structure type coefficient, K, is as follows:

K = 2.0 for vertical vessels on skirt supportsK = 2.5 for vertical vessels on skirts when

t,h.n > 1.5 tskin

The structure period response factor, C, is determinedby

^lL :

-sec (4_107)

15"rF

where T = structure period of vibration, sec, withc","" : 0.12

For short, stiff structures, such as horizontal vessel sup-ports, in lieu of making a period calculation, the re-sponse factor C may be taken as equal to C."".

For most industrially accepted design methods, the ef-fects of the soil-structure interaction are considered.This is done in the Uniform Building Code by using theratio of the fundamental elastic period of vibration of thetower, T, to the characteristic site period, T,.

Formulations used to determine the fundamental natu-ral period ofvibration for seismic response vary as to thetype of structural cross-section considered. The gener-ally accepted equation for towers of uniform cross-sec-tion is

and for E : 29 x 106 psi,(4- 106)

(4-110)

where D,,, : mean diameter of tower, ft

For a tower with uniform cross section and tapered(conical) skirt the following relationship can be used incomputing the fundamental period:

r = 2" (o qod)"

(4-111 )

where 6 = the calculated deflection at top of tower inducedby 1007" of irs weight applied as a laleral load

With towers of varying cross sections and attachingequipment, a method used to determine the fundamentalfrequency was developed by Warren W. Mitchell in anunpublished work [21]. The solution is based on the Rav-leigh method ofequating porenlial and kinetic energies ina vibrating system. The resulting formulation is readilluseful in computing fundamental periods of cylindrical.tapered-cylindrical, and step-tapered-cylindrical struc-tures common to the petrochemical industry (CpI). Theformulation is as follows:

,: ln)'\100/

(4 Il2r

where THww

'l--

where

t.re lrql"\EIei

(4- 108)

LI .

E_

fundamental period, sectotal heighr, ftweight per unit of height, lb/ftshell thickness, in.modulus of elasticity, psi

/ \, t::--1= 17.65e v t0-") l[)'^1tz*o''

\D",i ! r

\,F-4D-fEo, + a.y

: period, sec

= overall height of tower, ft

E

distributed weight (lb/f0 of each sectionconcentrated loads attached to the towerat any level, that add mass but do notcontribute to the stiffness of the towermodulus of elasticity (106 psi) for eachsectioncoefficients for a given elevationdepending on the ratio of the height ofthe elevation above grade to the overallheight of the tower (h,/H)


3E

xlloollR;lloll0o||o; o''ll,rEll'"ll:!l]f;

tloLJ

nt .€ol 65ol !oLl 6

6l -R5CNI R

ol -

6l


Ao, A.y : differentials in the values of a and .y,

from the top to the bottom of eachsection of uniform weight, diameter,and thickness. 6 is determined fromeach concenttated mass. Values of a. 6.and "r are shown in Table 4-15.

In applying Equation 4-ll2 the following factorsshould be considered:

n Ifa tower's lower section is several times wider in di-ameter and shorter than the upper sections, then thetower's period can be more accurately determined bycomputing the upper section's period, assuming thatthe tower is fixed as to translational and rotational dis-placement. If a tower's shell diameter or thickness issignificantly larger than that of the supporring skirt,the period calculated by Equation 4-112 may beoverly conservative for earthquake design and a moreaccurate method may be desirable.

D For conical tower sections the Mitchell eouation can-not be used because of lack of data for the

'coefficients

a, B, and 7. The Rayleigh equation (Equation 4-97) ismore comprehensive and ubiquitous in application.

Once the fundamental period of vibration is deter-mined, the numerical coefficient for the site structure in-teraction (seismic site-structure resonance coefficient),S, can be determined. As previously stated, the soil-structure interaction is considered in most industriallyaccepted methods. The value of S is determined by thefollowing formulas:

(4-l l3a)

For T/T. > 1.0,

S : 1.2 + 0.6 IT,

s > 1.0(c) (s) < 0.14

(4-l l3b)

The characteristic site period, T,, falls into the followingtlme:

0.5 < T. < 2.5 sec

When T. is not properly established, S is taken as 1.5,except when T exceeds 2.5 seconds, S can be determinedby assuming a value of 2.5 seconds for Ts.

VERTICAL DISTRIBUTION OF SHEARFORCES

For towers having an overall height-to-base-width ra-tio greater than 3.0, a portion of the total earthquakeforce. V. shall be applied ro rhe top of the tower aciord-ing to the following relationships:

For,

h

;<3.0,F,:0h3.0<: < 6.12. F. = 0.07 TV (4-ll4)l)

h

;> 6.12, F, : 0.lsv

where F, : total force applied at top of structureh : overall height of tower, ftD = diameter of tower, ftV = total base shear from Equation 4-106

The remainder of the total seismic force is distributedand applied to the mass distribution in the structure ac-cording to the following equation:

tt/ 1,

F^ : (V F,) """\-w r,LJ "I\

14-1 l5 r

For T/T, ( 1.0,

/ \.T ITI'S = 1.0 +: - 0.5 l:lT, \T,i

where F* : lateral force applied to a mass at level x, lbW* = weight of mass at level x, lbh, = height of level x above the base (normally

measured from bottom of the base plate ofthe tower), ft

Ewh : the sum of the products of w" and h, for allthe masses within the structure, ftlb

The seismic moments are computed from the followingexpression:

- 0.3 E)'\r,/

M : V, L, _, * F*,C; (4-116)

where Lr,-, : length of section below shear force, ftCi - L,lZ for a cylinder

/.\f 2 ^ ^21c - lil lrl-+ zrrr' + rr'l

foracone tsee Figure 4-33r\+/Lri+rlr,+r;I

For an illustration of seismic analysis, see Example4-3.


Table 4-15Coefficients for Determining Period of Vibration of Free-Standing Cylindrical Shells

Having Varying Cross Sections and Mass Distribution'

213

h"H

nx

H

I .000.990.980.970.96

0.950.940.930.920.91

0.900.890.880.870.86

0.850.840.830.820.81

0.800.'790.780.'770.760.750.74o.730.720.7 |0.700.690.680.6'70.660.650.640.630,620.61

0,600.590.58

0.560.550.540.530.520.51

2.1032.02r1.9411.8631.787

1 .'7 t41.6421.5131.5061.4401.3771.3161.2561 .1991.1431.0901.0380.9880.9390.8920.8470.8040.7620.7220.683

0.6460.6100.5760.5430.5120.4810.453o.4250.3990.374o.34970.32690.3052o.2846o.2650o.2464o.2288o.21220. 19650. l8l60.16761.15450.14210.13050.1196

8.3478.12l7.8987 .678'7 .461

7 .248'7 .O3'76.8306.6266.4256.2276.O325.8405.6525.4675.2855. 1064.9304.7584.5894.4244.2614.1o23.9463.7943.6453.4993.3563.2173.081

2.9492.8202.6942.57 |

2.33652.22402.1r482.OO891.90621.80681.71071.61'7'71.52'791.44131 .3579I .217 51.20021.12591.0547

0.500.490.480.4'70.460.450.440.430.420.41

0.400.390.3 8o.370.360.35o.340.330.320.31

0.300.290.280.270.26o.250.24o.230.22o.21

0.200. l90. 180.170. l60. 150. 140. 130.120.11

0.100.090.080.070.060.050.040.030.020.010.

0.98630.92100.85840.7987o.74r80.68760.63610.58720.54090.49710.455'70.416'70.38010.34560.31340.28330.25520.22910.20500.18260.162000.14308o.125160.109970.09564o.0826'70.071010.060560.051260.04303

0.035790.029480.024000.019310.01531

0.011960.009170.006890.005060.00361

o.002490.001650.001040.000620.000340.000160.000070.000020.000000.000000.

0.955730.951430.946830.94r 890.93661

0.9309'7o.924950.91854o.911'730.904480.896790.888&0.880010.870880.86123

0.851050.840320.829010.817100.804s90.79t40.77160.76320.7480o.'7321

0.7 r 550.69810.68000.66100.@13

o.62070.5992o.57 690.55360.52950.50440.4783o.45120.42310.39400.36390.33270.30030.26690.23230. 1966o.159'10.12160.08230.04180.

l .0000001.0000001.000000l .0000001.000000

0.9999990.9999980.9999970.999994o.9999890.9999820.9999't I0.999956o.9999340.9999050.9998670.9998170.999154o.9996140.9995'76

0.999455o.9993090.999t330.9989230.9986760.9983850.9980470.9976560.9972050.996689

0.996101o.9954340.99468 r0.9938340.9928850.991830.990650.989340.987890.986300.98455o.982620.980s20.978230.97 573

0.973010.97W70.966880.963440.959'73

0. 10940.09980.09090.08260.0'749

0.06780.06120.05510.04940.04420.03950.03510.0311o.o2'7 5o.02420.02120.01850.01610.01400.0120

0.0102930.0087690.00-t4260.0062490.0052220.oo43320.0035640.0029070.0023490.001878

0.0014850.0011590.0008930.0006770.0005040.0003680.0002630.0001830.0001240.000081

0.0000510.0000300.0000170.0000090.0000M0.0000020.0000010.0000000.0000000.0000000.

r- ,, E wA. . ,ruiE pB

'vi,cher rormura: t {#l ti--S;;-


CONICAL HEAD

Figure 4-33. The equivalent radius for cones.

TOWER SHELL DISCONTINUITIES ANDCONICAL SECTIONS

Most vessel codes do not discuss the analytical compu-tation of tower shell discontinuity stresses, which areprevented by welding stiffening rings to the outside shellof conical sections. In addition, most codes do not con-sider discontinuity stresses on cylindrical shell sections.The ASME Section VIII Division I uses a safety factor offour to one to compensate for not computing thesestresses.

Conical sections can be tieated quite simply by utiliz-ing the equivalent circle technique. Bednar [22] shows

that the cone-to-cylinder stresses computed by the equiv-alent circle method are very close in magnitude to thosecomputed by more exact methods. Because of its closeapproximate answers and simplicity, the equivalent cir-cle method is normally the method used for treating con-ical sections in towers. The method will only be outlinedhere, as others l22l have already derived it.

Figure 4-33 shows how the sections of a truncatedcone and a conical head are approximated by an equiva-lent circle, which is used to compute the section modulusand moment of inertia. These formulations are used intower design and are demonstrated in the examples thatfollow.

Conical shells used in tower sections have a half anexanglecv ( 30degrees. Whenh. ( 0.10H, rhecon..anbe approximated by considering two cylinders shownwith dotted lines (Figure 4-34). In pracrice, stiffeningrings must be used when required by the vessel code.

Figure 4-34. When h" ( 0.1 H, the cone can be approximatedby considering the two cylinders shown with dotted lines.

r"o= OO SMALL END

r-= OD LARGE END

TRUNCATED CONE

,."=[*&l

Lt\/2T'

EXAMPLE 4-l: WEAR PLATEREOUIREMENT ANALYSIS

A horizontal vessel containing hot oil is to be com-pletely analyzed using the Zick method to determinewear plate requirements (Figure 4-35).


o: : 37.845 psi

o2-r: 02 + oo : 1,715.34t0r'

o7 = circumferential stress at horn of saddle

8R : 8(1.750) = 14.00 + L < 8R

A - l5oo 0.857 = k o.u5R t.750

- 7,828.981

215

Vessel material : 5A-516-70Saddle material : 5.4.-36

Temp:300'FDesign pressure = 671 psi

0 = 120"

r:lrso-91 =tzo'\21t l5e I.r = __: l:: + 301 : 1.396 rad -

180 \12 l

From Equation 4-1 at the saddle,

., = 10*13910) [t# (, _ ,,1] [H#]

or = 50.501 psi

oo : -!D : iilllt'.t : 1,67i.50

orr : or + op : 1,728.00psi

At Midspan

From Equation 4-3

3(7,828.981X 10.0)' ?r(21.0)11.0)

Q : 7,828.981 lbcA:0

80.0'

- o 600]

4(0.94r)[0.375 + 1.s6[(21)(0.94)]0 5l

_ l2(0.05x7,828.98 1)(l.7s0)(10x0.941)r

-284.547 - 928.358

|,212.905 psi < < 1.25 dr,rr : 21,875

ring compressive stress in shell over saddle

7,828.981

o-oFigure 4-35, Horizontal pressure vessel containing hot oil.

(0.941)[0.375 + 1.56[(2 1X0.941)]0rl

. I I - cos(u4) I

lr - I.990 + sin (l t4) cos (l14,

os : 865.678 psi < < 19,000 psi = 0.5 o,

Since the ring compressive stress and the circumferen-tial stress at the saddle are less than one half of yieldstress and the allowable stress, respectively, wear platesare not required.

EXAMPLE 4-2: MEGHANIGALDESIGN OF A PROCESS COLUMN

A detailed mechanical design is required for the pro-cess column shown in Figure 4-36. The design criteriaare as follows:

Design temperature: - 150"FDesign wind speed : 100 mphInternal pressure at top head: 150 psigInternal pressure at boftom head: 162 psigShell material: 304 SSSkirt material: 436External pressure requirement: NonePWHT: YesRadiograph: FullAmbient temperature : lO"F min, 100oF max

Wind distribution is to be computed from ASNI-A58.1-1982:

10'-o'

Ioli-lc\t I*l-l


towER ANo TNTEFATS (Nor ro scAL€) WINO ENSEMBLEVIBBATON ENSEMALE

lSHELL AND TTJBE

IHEAT EXCHANGER

l:'

-------+"oo t$

I'roP aEo

1.,

l-

IIll;,I ltt rI rtt Ir rrtlltttr_!_rt-!-!

'.. -7

,tt -'..

__ _.j--L___ -------lr:

-T___i-L______--L__

csruliiinny -------lli- |,"r"*"" r* I: I

-ti

I

-TlBOTIOMBEO

l*IU.__i1___

F.-i-----+r"-*.-,J:ll"lN.RMAL LrourD .."r.

-f I

FI--l Iri +-

Figure 4-36. Tower analyses ensembles.

9z

Kz

: 0.00256 Kz (IV)'z; V = 100 mph

:2.58


Effective Cross-sectional Area (Figure 4-37)

Top portion of tower,

D" = 66.5 tn. + (12.75 + 12) in. + (12.75 + 1l) in.

12 in. pipeplus

insulation

+ 12 in. : 127.00 in.\-7-

laddersand

platforms

Bottom portion,

D. : 47.O il:'. + (12.'75 + r2) in.

+ (12.75 + 11) in. + 12 in. : 107.50 in.

Gust Response Factor

V = 100 mph, h = 104.292 ft, f : 0.981 Hz

Structural damping coefficient : 0.01

/szzor\c = 1""'-- I { l27.ool

\r04.2921

.(t+*)(ro75o) = 11s3s5

From A.58.1 Table A9, s : 1.00

- f 0.5 Ih (10.5X0.981)1104.2921r:_=#=lu.t4Jsv (1.00X100)

1i'0 PtpE

6" THtcKINSULATION

/r s\'"Kz : 2.58 IZJ for z < rs ft

I 7 \D 286

o, = 0.00256(2.58)1" I'" \900/I7 \0.286: 66.048 (eoo-1 . z > ts rt

12 in. pipeplus

insulation

Tower is in Exposure Category C, for which

a : 7 .0, Ze : 900, D" = 0.005

.At 15 ft,

K,:2.58f l1\''' = o.rot- \900/

ForZ)15ft,

/ 7 \0286K, : 2.58 l:l'.

\900/

For 15 ft and under,

qz : 0.00256(0.801)(Iv)':

From Thbles I and 5 of A58.1-1982,

I = 1.0

az = 0.00256(0.801)( 1001 : 15 fr20.506 lb/fC, z <

(100F

tz'?erre1*

u'r","*-rr*I.$'*"u'o''on

ffJY

Figure 4-37. Effective cross sectional area.


9:h

ir rs.:os\

\ t'? /: o.oe5:c1o4.292

ffi: ooo"

0.0055, y = 6.145

I 18.305

1o4,292ft _

: 9.859

I-

D_ fly : (10.743)(0.0055X0.145) : 0.009

o = oo.*,(fr-o|"'",0r,,.

s : l.l

For a tower with many attachments and connecting pip-ing,

c - 0.6s , [g!q * [(3 32x0 147)l?(l l)105

1 0.01 I { (0.002X9.85s).1(4-83)

G = 0.65 + 1.076 : 1.726

From Simiu[9],

9:9zGFigure 4-38 shows the wind pressure distribution q plot-ted along the tower length.

Centroid of a Spandrel Segment

The centroid ofa parabolic segment is shown in Figure4-39. Applying the general equation, Equation 4-117, toour case we obtain

Figure 4-38. Wind force distribution (q) along tower is parabolic above 15 feet. Section wind force distributions are com-bined into a force vector located at the centroid of the winisection.

Using this equation we compute the wind force distribu-tion. From Figure 4-20, Cr = 0.6. Solving for sectiorproperties we have the following:

(4-117)

From Figure 4-39 we obtain a general expression for the

. t^ \i _ tn + Ql llaqn I ngn.tl ,.t .t .r

2(n + 2(}\ \dqn + nqn r/

composlte area,

t7 -7 .\Af - ::l-----:l--I {dqn + nqn r). lb/ft

(n+q)

Solving for the wind resultant force acting through eachcentroid we obtain,

F : AO"CI

(4-118) I 7 \2t1aa :66.048 1.-l'" \900/

Section A

+- _>..

#----->r

*--->..

q--->E

30'oo *+r-

\+-{)-->

f00tl5.oort

'|

+;

Z : 104.292 tt

The Engineering Mechanics of Pressure Vessels 219-

ql1l--t^

1FN.r<re g^1

i- Llhl rN

)|) l^lNINI

il:il-o

-l;l Ist?l

.l+lNlcl,_l-l

Nt I

-,I^,1:lYINtd

I.tT-l'-lol I"l.elE rlgil(!

$l-

il-

<t)

q)

tt

IN

!L!tN+

^lNrf.f ^ |

r',.'NS

=x":I Lr-' t:-o

Nir++JiiQ,ni+++u,- !L >

tL>

'i! | -;-

N

Rt(G -----i:--q --J9

.9

=No- tNo

:*()+

NI

N

I

(l

@qo

6F

(JO

:l' l

:-^ ldlo* l.

-___t91tl

N

: ^l

,Yl6 .lrr llld! il-isl

dlB N' lltqlo.l' nu' xl

d;d

Y

tl

€ - t:F[?'iYrl

<ft

'l

i l'lNl.rl--l\tIN

ll

lt

IN

'N

220

9z

q"

9n- t


/rr}4 rq?\0 28o

66.048 l:: : --l : 35.659\vwi(l.726X3s.659) : 6r.547

/ qR \0 286( 726t l:;l 166.048) = 60.461

(6.292)l(7)(6t.s47) + (2)(60.461)l

2+7

385.735 lb/ft

r9r [r2rrzrror.5+zr r- r2x60.461)1,. .^..2t2 . A, [ {ix6tJ47) + {2x60r'61) l'" "''3. 151 ft

o 11211t11+r.tsz1nL a)@e8sz)+ {2 x,r4.571)l ,,., ,.,+ t2\44.s1rll"'''l

Section D

q":48.852

e"-r : (r.726)ffiul-

(12.75 )[(7X48.8s2)

(66.048):44.573

+ (2)(44.s73)l

610.739

nL 0@e8sz)

: 6.430 ft

Section E

q" = 60.461

/ qn \o 286

o. , = 1l.726] lîl (66.048) = s9.007\vw/

^ _ (8.00x7)(60.461) + (2X59.007)l

Section B

:48

-s

(66.048):43.097

+ (2X43.097)l

. 103 lb/ft

(2)(7)(60.461)

I 165.919

J2 [ (7x60.461)+ (2 X59.007,,l

{8.00}+ (2)(59.007)l

s l{zxzlt++.stz)nl 0e4.sb+ (2 )(43.097)l (3.7s )+ (2)@3.097 t)

z: tl 0)t44.s7,

: 4.009 ft

Section C

: 1.881 ft

Section F

q" = 59 007

q.-, = (t.i26) (*#)""'

(43.50X7)(s9.007)

(66.048) = 4s.352

+ (2)(48.852)l

(66.048) : 39.943

+ (2)(39.943)l

: 2,468.640 lb/ft .713lblft

(2) (7 )(43.o97) + (2X39. 943)

32 | ('7 )|43.097) + 12)(39.943)

296

eInlz: zl

32 1

(2X7X59.007) + (2)(48.8s2)

32 [ (7)(s9.007) ] ,0, ,0,

]n oo,

= 22.128 ft

+ (2X48.8s2)

Section G

Section I

: 118.786 lb/ft

z : e [r2rtrr32i+rt rle9.:]s{ c.oo,- 32 | e)e9.943) + {2x38.378)l '- --'

= 1.506 ft

Section H

q" = 38.378

- /r? on\o 2s6

9"_r : (1.726) l:::-:::; (66.048) = 36.63s\7wl

^ _ (3.00)t(7x38.378) + (2X36.635)l

: ll).972lblft

i : e l2IlI38.3Z8rj i?I36.61s)l ,r nn,- 321 (7x38.378) + (2)(36.635t'- "-'

: 1.507 ft

q" = 39.943

/ro oo\0 28o

C" | = (1.726) l-:::::l (66.048) : 38.378 9n_r :\vlru/

^ - (3.00)ft?X39.943) + (2X38.378)l At :


Section J

35.347

35.347

q(2" - Z" ,) = (3s.347X15.0)530.205 lb/ft

ll : z.so r,2

Now solving for the section forces we have

F = ArD"Cr

FA : (385.7rs) ('']=oo) ,0." : 2,44e.4r7 tb" \ 12 /

FB = (481.103) {'t]ôol to.ut = 3.055.004 lb\ rl /

Fc : (2.468.&0) {l?lq. ^ 0l (0.6) = t5,675.864 lb\tzl

FD = (6r0.73e) {toLtol ,0.u, = 3,2E2.i22 rb\12 I

Fr = n65.919r l'07 501 ro.ur = Eel.8r5 rb- \12l

FF = es6.ii3) (ue- ) ,0 u, : r ,5e5. r55 rb\ 12 /

F6 = (u8.786) {'ol=to) to.u,t = 638.475 lb\ Lz I

r, = (n3.e7z) (!Z t\ ,o u, : 612.6oo rb" \ i2 /'

Ft = (72.6si) I'ol=tol to.u, = 3e0.746 lb\ r2 I

Fr = (s30.205) Ito],to) ,0.u, = 2,84e.852 lb\12 I

q":36.635

/, < ^^\o

286

q"_r = (1.726) lj':Y:l t66.048) = 35.347\vUU/

^ _ (2.00)[(7)(36.63s) + (2X35.347)]

: 72.697 lblft

- s[Z=-l32 1

(2) (7 ) (36. 63 s) + (2) (3 5.347 )32[ (7X36.635) + (2\(3s.347)]

,r.oo,J

Fc -: 1.003 ft31,,141.650 lb


Solving for section moments we use the following ex-presslon:

n- I

M"=M" t+(z^,-z^)DF,_,

+ F,z. (4-8s)

M^ : (2449.417)(3.151) : 7,718.113 ft-lb

Ms = 7,718, 113 + (2449.4r7)(8.oo)+ (3,055.004X4.009) : 39,560.960 ft-lb

Mc : 39,560.960 + (5504.421X43.5)+ (15,675.864)(22.128) = 625,g7g.rt tt-tt

MD : 625,878.792 + (2r,180.285)(12.75)+ (3,282.722)(6.430) : 917,035.328 ft-lb

ME : 9r7,035.328 + (24,463.007)(3.75)+ (891.815)(1.881) : 1,010,449.109 ft-lb

Mn : 1,010,,149.109 + (25,354.822)(7 .00)+ (1,s9s.155)(3.s2s) : I,193,555.784 ft-lb

Mc : 1,193,555.784 + (26,949.9'77)(3.OO)+ (638.475X1.506) : |,275,367 .258 tt-tb

Mu: 1,275,367.258 + (27,5 88.4s2)(3.00)+ (612.600Xr.491) : 1,359,046.001 ft-lb

Mr = 1,359,046.001 + (28,201.052X2.00)+ (390.746X1.003) : 1,415,840.023 ft-lb

Mr : 1,415,840.023 + (28,591.798)(15.0)+ (2,849 .8s2)(',7 .s0) : 1,866,090.883 ft-lb

Section Moments of Inertia

Section a-rl+-in. t

Figure 4-40. The tube bundle is modeled in banks ofconcen-tric circles used to approximate the section moment of inertia.The tube bundle enhances the section stiffness.

For l-in. OD i6 BWG tube,

I : 0.0210 in.a : metal cross section

A : 0.191 in.2 : metal area

K : number of tubes per circle

n : number of circles

Thus, from the parallel-axis theorem the composite mo-ment of inertia is

I: DKIG

,. " ffs+.sol, _ ile\"] : 0.756 rta" 64 1\t2 I \12/J

Section b-3/a in. t

The shell and tube heat exchanger section moment ofinertia is approximated by a set of concentric circles oftubes. The concentric circle pattern approximates that ofthe exchanger tube sheet. Using the parallel-axis theo-rem, we arrive at the section moment of inertia. Refer-ring to Figure 4-40 we analyze the exchanger as follows:

+ AL)

Values of I are tabulated in Table 4-16.For enclosing shell, :/a-in. t,

I " 1s+.zs)o - (54.00)41 - 23,676.070 in.'64'

)-l = rr4.i35.44r in.a + 23.676.070 in.'Lt'

: 138,411.511 in.a

or for the total cross-sectional area

Ib : 6.675 fC

1 "OD 16 BWG TUBE

af\

-

10

; ;i#.uZr 86 r.765.174 .7s,230.641

The Engineering Mechanics of pressure Vessels

Table 4-16Values of I for Tube Bundle

16.364.299 t6,364.29914.127 .503 30,491.80212.104.53110.284.531 s2,880.9278.574.543 6r,45s.4707. r38.100 68,593.s705.871.903 74,465.473

17.00 2 89.000 0.191 0.021

14.50 210.250 0. l9l o.021 85,937.10688.197.85689,860.69691,063.9539l,901.44592,456.266

Jf, 92,79t.181t718 92.980.407

t9 93.074. 195t6 93.109.836

2ll0 93.119.576

93.120.846t55 2t,6t4.595 l1:1.735.1,+1

Section c- I/z-in. t )ectlon t-rA-tn. rL

, - " [/ss.oo\- /s+ oo\*l64 [\ t2 / \12 /J

: 1.533 ft4

Section d-5/s-in. t

,: "[or4f _/+:.oo\.164 l\ t2 I \t2lj

: 0.917 fta

Section e- t t/ro-in. t

r = L[(r#| (:gl: I .013 ff

,=.[+:1!f _/+z.ool.l64 l\ 12 I \12lJ

: 1.1 10 fta

Section g-r3/r6-in. t

r - ,r ll43.62sl- _ 142.001"1

64 L\ t2 I \ t2 /J: 1.208 fta

Referring to Figures 4-36 and 4_41 we calculate thesectron moments of inertia as follows:

Section h

26.193 + 21.812s,.r:(2 cos 6.934"

\-t- 24. I80 in.

lso

Ia +ydla h.

$ lo a"\


reo=21.8125 in

rho=26.1925 in

tto:29.1109 in

1o= 51.o in

Figure 4-41. The equivalent radii for the skirt sections.

Deflections

! _ tb.292tt ltz.++s.+n xo.zoztl"' (i.157 x lo") [- g I

: 2.416 x 10-s ft

. (8.oot [(e,oss.oo+)(a.oo)- (2.787 10'9 [ 8

- (:2,449.4I7)(S.OO) - {Z,Zrr. r r:l1-3-21: 3.088 x l0 5 ft

. (43.50)'? I' (6.402 10') [

o

+ (s,s04.421)(43.5) + Q2lOo soo)l'32-)

: 0.055 ft

(15,675.864X43.5)

(3 ,282 .722)(t2 .7 5)t /.- .-\+ | ,- ^.\q 1

t. - o- | lr+6 rol - l4o dol l = l.5i3 frr'' 64[\12/ \12ll

Section i

/zs.rrr + zo.rs:\r-- = t___t 27.856 in.* \ 2 cos 6.934' /

"[/ss.zr,z\* /s+.oaz\tl64 L\ 12 / \ 12 /l

Section j

/sr.oo + :s.rrr\'' \ 2 cos 6.934' /

" l/so.uor\t hs.ozo\*lL j' ll"" '" 1 '1 "" "1 | = 7.846 fta' 64 L\ 12 i \ 12 i I

Values of each section's wind force, shear, bendingmoment, and moment of inertia are summarized in Thble4-17 for the entire tower.

(3.'/r, [tal r. s rs)t:.zs)(4230 x r0) t 8

, (24,463.OO7)(3.75) (917,03s.328t-J-rl= 0.002

(7.00), [(r,sqs.rssxz.oo)t-635,10)L 8(4.

8

+ razs,ttt.to4]2l

(21 , 180.285)( l2.75)

0.017 ft

(2s ,354.822)(7 .0O)

('1

+

: 0.006 ft

, (1,0r0,449.109t-2)


Table 4-17Summation of Section Wind Force, Shear, Bending Moment and

Moment of lnertia

Section Fr (lb) Qi (lb) Mi (ft{b) li (fta) El (lb-ft'?)

a 2.449.4t7 2.449.417 7,118.113 0.756 3.157 x 10'q

3,055.004 5.504.421 39,560.960 6.6'7 5 2.78'7 x lOto

15,675.864 21. 180.285 625,878.'192 1.533 6.402 x 10!)

1l

3,282.',722 24.463.00'l 917.035.328 3.829 x 1090.917

891.815 25,354.822 1,010,449. 109 4.230 t t}e1.013

26.949.971 1,193,555.784 1.110 4.635 ^ loa

s 638.475 27,588.452 1,275,367.258 I .208 5-045 x 101r

612.600 28,20 r.052 1,359,046.001 6.402 x 10'1.533

390.746 28,591.798 1.415.840.023 2.54',7 1.064 x 1010

2,849.852 3l,441.650 1,866,090.883 3 .27 6 >< l0t'r7 .846

(6.2e2)(8.00) [(rJ8i ,, 109 t

(3,05s.004)(8.00)

t26.s4s.s71 t3.0t , { 1.19J.555.7S4)l

3'l(2,,149.4 r 7x8.00)

6

I+ ?.718.I r3lI

= 0.001 ft

2

3.899 x 10 5ft

, 13.oo1r [(612.oo](3.0u/-' (6.402 \ lOa) [ 8^ ( 14.2e2X43. s0) [A'r:-l

(6.402 x lOY) [

, (27,s88.4s2X3.0)t- , (r,27s,367.25U]|-)l

( 1s,67s.864)(43. s)

(5,504.421)(43.s)

6

+ 39,560.960]

= 0.001 ft

, 12.001/ [(3e0.7abx2.00)": tr.oo+. ro9[ 8

+ (28,201.052)(2.00) + (r,3s9,046.001t'32ll

: 2.626 x t0 a ft

" (15.00), Iorn:-l (J.276 l0'9 L

= 0.027 ft

^ 15't .792yr2.7s11^' ,,: oJ29 . lo) [

(3 ,282 .122)(12 .'t s)

(21 , 180.2 85X12.75)

o

I+ 625.878 792 |j

(2.849.852)( 15.00)

= 0.148 ft

I

^ r70.542rr3.75,1{8q1.815)r3.?51't- r+.::ot tolL .

_ r 24 .46 3 .001 )\ 3 .15 , _ q tr,u:S.:282

: 0.060 fr

(28,59r.798)(r5.00)

8

, ( 1,415,840.023)l-21: 0.006 ft

226 Mechanical Design ol Process Systems

^ _ r74.292n7.00) [{ t.5q5.155x7.00)-)6 - (4.6iai " ro1 t 6

+ (2s,354.822)(7.00) + 1 I2 -,010,449.t09]

: 0.124 ft

^ (8r.2s2)(3.00)[(638.4Dt 1 (5!45 x rO l-

Tower Section Stress Galculations

Section a-ll+-in. t

For tension on the windward side, using Equation 4-?1

/po\ / ron^na \\4tl - \?r((D" - D,XDa + Di,

lzw\- \"<n. * o, 1

150 psi, E = 1.0, t : 0.25 in., T : -100'F,

0.423 kips

54.0 in., D" = 54.50 in.

[rrsO.Orrs+.orl I r6(54.50,{l2r{7.trA.rr:rlt {4x015) I = [.{015x t08J0)6J86.2t]

[ 2(1.8t4) I- tr(0lsxl08 do = 8,100 ! 161.009 - 42.57

o = 8,218.43 psi tension on windward side

q = 7 ,896.42 psi compression on leeward side

Internal pressure circumferential stress,

PD ( 1s0.0)(54.0)o^=' 2t 2(0.2sX1.0)

: 16,200 psi < 18,800 = o"

The circumferential stress governs in this section and isless than the allowable stress.

Section b- :/a-in. t

[t r:o.orrs+.orl

L(4)(0.37s) I

, [ 16(54.75)(39.s60.960)(12) ]= [r{ 0J7s x 10850X5-9{3J63)l

75X3.00)

(26 ,949 .977)(3 .0O)

6

* , ,oa aaa ,ool"--'---"1: 0.060 ft

P

w

Di

^ _ 184.2s2t(3.00r fr6l2.-r8-(6J02^toq)L-

+ (27,588.452)(3.0)

2

: 0.052 ft

.2s2)(2.oo) l3s0.7)64 x t0ro) |.

6

' , ,ra ao., ,aol-'- '''" -'"1

60x3.00)

(87

(1.(

+l

46)(2.o0)

2

: O.O23 ft

6

+ 1,359,046.001]

^ _ (89.292x 15.00) [(2.849.8s2r 5.00)- '" - rJ.276 < tolor I o

+ (2S,s91.798)(1s.oo) + I

2 1,4r5,840.023.j

: 0.067 ft

064 x t0ro)

(28,591.798X2.00)

": \-,1 * \-,rrt L/" L/ -1

0.088 ft + 0.561 ft = 0.649 ft'7 .787 ir.

_ [ :rrs.zor.oor It.(0 3is,(t08 itl

5,400 + 550.16 - 237.30

5,712.86 psi tension

4,612.5 4 psi compression

(150.0)(54.0) : 10.800 osi2(0.375)(1.0)

< 18,800: o"

Section c-\lz-in. t

16(43.3?5)(1,010,,149. 109)(12)

r(0.688)(85.375X3,645.39r)

q : 2,289.24 + 12,509.56 - 482.98

o = 14,315.82 psi tension < 18,800 psi : o,

o : - 10,703.30 psi compression

o" - (lsq'Pl!4? oo) : 4,578.488 psi' 2(0.688)

Section f-:/+-in. t

I zw,sor.sztt I- t"(0.o88x85.3?tl

krso.olt+z.orl- [ (4)(0.75) I

, Iro(+s.soxr, ts:,555.7s4x12t= I "(0Jsxs550)(3,6562s0)

I

Izr+o.sst.sssl I- t"(oisx8s5o)l

q: 2,100.00 + 13,533.81 - 466.199

o : l5,l67.61psi tension < 18,800 psi

o = -11,900.01 psi compression

*l-t (L 6) (43.25) (9 r7,03 s .328\ (12)

r(0.688)(8s.375)(3,&s.39r)

op _ (1s0.00x42.00) _2(O.75)

4,200.00 psi

I zt+:,+rs.oor I- t-l

lr(0.688t(85.375r1

2,520.00 ! 11,320.361 - 470.584

13,369.777 psi tension < 18.800 psi : o,

- 9,290.944 psi compression

(150'0x42'o) = 5.o4o.oo osi

2(O.62s)

Section g-t:71u-in. g

Ir rso.olrs+.orl0r = t-lt (4x0.s) I

krso.ox+z.orlo:t-l[ (4)(0.62s) I

Section d-s/s-in. t

. I rorss.oolrozs.s 78.isu'l2t]= [-lr(0. sot ( t os. ool (5.9a 1 . oo) I

lzot.tzz.oll- t"rcJ0l t0r)l

o : 4,050.00 ! 6,497 .53 - 440.@6

o : 10,106.884 psi tension < 18,800

r = -2,888.176 psi compression

(150.0x54.0)o^ = :: 6.luuDSl' 2(0.s0x1.0)


Section e-l l/re-in. f,

[rrso.otr+z.or]o=t-l[ {4x0.688) I

*[-t

krso.otr+z.orlo:1-ll(4)(0.813) I

. 116(43.625)( |,27 s.367 .258)(t2= [ "(0-813X85525X3'667140

1. 6(43. 625) (1,27 s,3 67 .258) (r2)

Izt+s.ozl.+ssr I- t"(0-8t3l8s.6rtl


q = 1 ,937 .2'l ! 13,319.972 - 439 .64

o : 14,817.602 psi tension < 18,800 psi : da

q : -ll,822.342 psi compression

All section stresses are less than the allowable stress of18,800 psi. Thus, the tower thicknesses are acceptable.

SKIRT AND BASE PLAIE DESIGN

First we determine the size and number of anchor boltsreouired.

^,: [+-*]/*". (4-42)

Using an A- 193-87 high-strength bolt with an allowa-ble stress of 40,000 psi per AISC and assuming a boltcircle of 107 in., the required bolt area, ,4.6, is

[{4){ l2){ 1.866.090.8S3) _

^. *,. ,rrlI toz "--- -)

(12X40,000)

D": OO of base plate : 106.75 + 2(2.375): 111.50 in.

D; : ID of base plate = 111.50 - 2(6.j 5): 98.00 in.

-rn2 - nllA. '-o------1- = 2221 .302 in '

4

-/na - fr4lI. = " "",' "' : 3.059.323.38064

_ 8MD,, W.' N(Di + Dl) N

" = iryl). iv). /vl)\A.i \A./ \2r.i(4-,11

(4-.+ l

_ (8)(12X1,866,090.883X1 1 1.s0)(.12)(22,036.250)

= 70,219.061 Ib

_ (12)(70,219.061) + 63,8rs.721

63,815.727

n

: l.o I I ln-'

We select a lsls-in. d bolt of 8-thread series with aminimum root area of 1.680 in., Thus, using Figure 4-15, the new bolt circle becomes

BC = 102 in. + 2(.2.375) : 106.75 in.

The new required area becomes

2,22t .302 2,22\.302

( 1.866.090.883X12)(l I L50)2(3,059,323.380)

o" : 379.340 + 28.'129 + 408.069

o. : 816.138 psi < Fb : 1.33(900)= I,197 psi

The concrete bearing strength criteria are met, so we ca.continue to the base plate design.

h = p" - D, _ lll.50 - 98.00

22

= 6.'75 : Base f, width = BP t W

I(4_1j

lo.

20,000 = allowable working stress for steel, psi

I ,000 psi

oJo, = 19

0.333

n

k

[*tgg- -63,81s.727f

( l2)(40,000)

= 1.615 in.'? < 1.680 in.2

. r( 106.75 )lJOrr sDaclng

t2

= 21.947 in. > 18 in. minimum

Maximum bearing pressure on contact area:

with k : 0.333, c" : 1.588, C, = 2.376,2 = 0.431,

J - 0.782, and six iterations of Equation 445 using Th-

ble 4-7, we obtain a value of k of 0. 186.

For k = 0.186,o. : 655.834 psi and o, = 26,850.892 psi

The allowable stress in the base ring = 36,000 psiThe allowable shength in the concrete : 3,000 psi

Using Equation 4-54 and solving for the maximum in-duced stress in the concrete,

^ : r6s{ o,o, [zro.tg.]xtoo.zsr + o.zsl

[ 2(0. r83x 106.75r I

o.1.,"y : 769.139 psi


Equation 4-64 becomes

ti - 0.845 ta - 0.046 = 0

tc = 0.901 in

Make gusset plates 15116 in. thick

Calculating the minimum skirt-to-base-plate weld sizewe have from Equation 4-65

*(* irun'n n.\rDsr/

. kt r2X r ,8oo,090. s83rl 61.815 727- t rll02.00tr I lrt102.00)

: 2,939.611

For wind or earthquake,

o* : 1.33 o.(0.55) : 1.33( 12,700X0.s5): 9,290.05 psi

2.939.611weld size w

-

= 0.158(2)(9290.0s)

Use at least a 3/re-in. weld on each side of skirt.Anchor bolt torque is determined by Equation 4-66.

For lubricated bolts with Fel-Pro C-5A,

C : 0.15

T : (0.15)(1.62s)(64,605.803) : 15,747.664 in.-tb

or

T : 1,313 ftlb with torque wrench

Checking the skirt thickness for reaction of the boltingring against the skirt we have from Equation 4-39,

/ pp \..t 1.76 l--_5- I r' ' (4-19)\m(GH)o"11/

4M-n2

(4-56)

[ (70.2le.06rx2.3zsr l' /rozl' 'r = l. /o t_

@-64) t, ':a'4?

x q "00X

'z0"000)l t'z /

t = Q.672 in. < r3lroin. skirt thickness at chairSkirt thickness meets chair ring reactioncnterla.

A sketch of the skirt and chair design is shown in Figure 4-42.

f. lo5,lJ octmarnlfu\ : L l-lI o,ll I

111.50 - 102.00 _ , ?<n i-2

fr I v?6q r lqrlo tBTHK = t4.75t l - " '""| = 1.613 in.

L 20.000 I

Make base t 15/8 in. thick

Solving for the compression ring thickness using Equa-tion 4-63 we have

[ (b4,605.803x2.37s) I ^-^-[4{20.000)14.75 - r 1.25I 1.625'1].j

Make compression ring :/+ in. thick

Using Equation 4-64 and Figure 4- l6 in calculating thecompression ring thickness, we have

t-! ztE ,

18.000 G*r" - {F,)tl -1.500

Fi = 64,605.803

Grr : 9.00 in.; G* = 4.25 in.


NOTE:TOFOUE BOTTS 1313 tr-|bs WttHFEL-PRO C-sA USING TOROT'E

f 11td ^-$.-87

BoLls

Figure 4-42. The skirt detail.

Section cSection Gentroids

Referring to Figure 4-21 we have the following:

Section a

toio tot r.rt> - 3(0.25)

ul"r"'", - o.rr)

t : s.857 in. = 0.488 ft

50Lr = 0.488 ft + :_: = 2.988 frz

Section b

t,=+=2.5ort

-30b(4a

ROOy.=;=4.00ft

Lz : 2.50 + 4.00 = 6.50 ft

Section d

_ 2.00lo= Z : r.w tt

L: : 4.00 + 1.00 : 5.00 ft

Section e

- - 24oo = l2.ooft'-2

L.+ : 1.00 + 12.00 : 13.00 ft

Section f

v :2'75 : 1.3?5 ft-2

L5 : 12.00 + 1.375 : 13.375 ft

,F-l

Section g

i," = 8oo:4.ooft'"2

Ld = 1.3'75 + 4.00 : 5.375 ft

Section h

v,=17s=0.875ft"2Li : 4.00 + 0.875 : 4.875 ft

Section i

t, :350: l.75oft'2

La : 0.875 + l.'75 = 2.625 ft

Section j

,, _ 3HrDl - Di) r 8{D" - D, ) ran a'molD., - Di, = 43.25 - 42.00 = 1.25

D;. - Di : 1,870.563 - 1,',761.00 = 106.563

t.\cr = arctan lll : Ze.-scs'

\t2l

3(12X106.563) + 8(I\2(1.25) tan (26.565)

6[106.563 + 2(12)(1.25) tan (26.56s)]

: 6.247 rn.

t = 0.521 ft

Lq = 1.750 + (1.0 - 0.521) : 2.229 ft

Section k

I ) ?Sj'^ : '1-' = 6.125 fr2

Lro = 0.521 + 6.125 = 6.646 ft

Section I

3.75= l.X/l tf"2


Ltt = 6.125 + 1.875 : 8.00 ft

Section m

_ 7.00

2

Lr: : 1.875 + 3.50 : 5.375 ft

Section n

- 3.00'"2L13 : 3.50 + 1.50 = 5.00 ft

Section o

/roz.oo - +:. ou s\o = arctan | | 6.934\ 2(240.00) I

= 43.625,D,.: 42.125, D". - D,. = 1.500 in.

: r,903.r41, Di, = 1,774.516, D;. - Di : t28.62s

231

D".

D3,

3(36)(128.62s) + 8(361(1.50) tan (6.93,1)

6[128.625 + 2(36X1.50) tan (6.934)]

: 18.556 in.

y. = 1.546 ft

Lr,r = 1.50 + 1.546 = 3.046 ft

Section p

D., : 52.381 in., D1 : 50.756 in., D". - Di. = 1.625

DZ" = 2,743.'t9s, Dl. : 2,s'76.r97,

D3. - Di = 167.598 in.

3(204X167.598) + 8(2}4)r(t.625) tan (6.934)6[167.s98 + 2(204)(r.62s) tan (6.934)]

: I13.044 in.

yp : 9,420 tt

L,. : (3.00 1.546) + 9.420 : 10.874 ftLro : 17.00 - 9.420 : 7.580 ft


l .' P N :. 9 9 .- 9 | - i p ;i .. 9s H $ R q H E 3 € ! H: q I qS;^x^x^^x^^^;^;xx

2.625' 4.a75 5.375'. r3.375' 13.00'

Figure 4-43, The vibration ensemble of lumped masses.

Vortex.lnduced Vibration

Referring to Figure 4-43 we have the following:

M.:0Mb: (0.423)(2.95 8) : 1.251 *o-tM" : 1.251 + (1.814)(6.50) : 13.042 kip-ftMa : 13.042 + (15.201X5.00) : 89.047 kp-ftM" : 89.047 + (16.192X13.00) : 299.543kip-ftMr : 299.538 + (29.004)(13.375) = 687.472 ktp-ftMc : 687.46'7 + (30.813Xs.375) : 853.091 kip-ftMh : 853.086 + (35.084X4.875) = 1,024.126 kip-ftMi - |,024.r2r + (36.016)(2.625) = 1,118.668 kip-ftIvl = 1,118.663 + (37.s32)(2.229) : r,202.322 ktp-ftM,. = 1 ,2o2 .322 + (37 .913)(6 .646) : | ,454 .292 ktp-ttMt : |,454.292 + (43.171)(800) = 1,799.660 kip-ftM. : 1,799.660 + (4s.028)(5.375) = 2,M1.685 kip-ftM" : 2,041.685 + (47.662)(s.o0) = 2,279.995 Y,tp-ftM. : 2,2'79.995 + (50.684)(3,046) : 2,434.379 kip-ftMF = 2,434.379 + (s 1.937X10. 874) : 2,999. r42 kip-ftMs^, = 2,999.142 + (63.816)('7.58) : 3,482.867 kip-ft

{, = M' , (30 x tO")tb/in.,(t44tin.b/ft,-

E.I,

: 4.320 x lOe \blf€

1,251

* : 6r?ffir,rral : 4.523 x ro-5

687 ,472 : 1.038 x 10 4

(4.32 x 10)(1.533)

8s3,091,4 ?) " lft"t, r Sll' - 1.288 ' 10-4

& = er+;ffi1j3,, = r.345 x 1o 5

t _ 1.O24.t26t' = A Jx) = 1.546 r to '

!" l'118 668 : 1.689 r lo ''- (4.32 x l0e)(1.533)

| .202.322<ro - 14.32 .lo"ltt.gtot

I L<L )A) _14?|v|^4' (4.32 r t0")(0.917)

r,, - f'7q9'660 4.tt2 t ro "(4.32 x 10,X1 .013) -

!,,. 204l'685 -4.258xlo1' \4.32 ^ l0')(1.1l0)

r 1_70 00<' 169 x l0 4

<r4 - (4J2 /' 10\1108) - - '

, 2,434,379r,< -

--r.u78xl0a

(4.32 v l0)( 1.s32)

(4.32 x l0)(0.756)

13,042(4.32 x 10)(6.675)

: 3.830 x 10-7

= 4.523 x l0-1

r,- = 2'99q'l42 = q.902 ,. ro ''(4.12 \ I0'X7.011)

r. . 1482'q67 l.l5o , Io a

(4.12 / 10)(7.01l)

/. ,\^ 18,*t + l,l ,J, = l-l Li

\rl(3.83 x l0 )(2.958) : 5.665 x l0 '


s,, t4.258 + 4.369x10 "r(5.00) _ 2.t57 ,, l0 ,

2

Sr+(,1.369 + 3.678)( r0-4) (3.046) = 1.226 x l0 r

Sr: =

Sro :

(3.678 x 10 1+ 9.902 x l0 )2

2.538 x 10-'

(9.902 x l0 5 + 1.150 x 10 r)

(10.874)

(7.58)

S:=

2

t(3.83 x l0 ) + (4.523 x 10 ?)l

2.?15 x 10 6

[(4.523 x l0 ?) + (1.345 x l0-)]2

3.476 x 10 5

[(1.345 x l0-) + (4.523 x 10 )]

: 8.111 x 10 1

Ar : 8.111 x l0 a

A: = 8.111 xl0-4+2.538xl0r:3.349xl0 l$ : 3.349 x l0 3 + 1.226 x l0 I = 4.575 x l0-rAt : 4.575 x l0-r + 2.15'7 x 10 r : 6.?32 x 10 r

As : 6.'732x l0 3 + 2.249 x l0 3 = 8.981 x 10 lAo = 8.981 x l0-3 + 3.113 x l0-r : 1.209 x l0 r

N : 1.209 x 10 2+ 1.704 x 10 3 = 1.379 x 10-2

As : 1.379 x 10 2+3.506 x 10-1 : 1.414 x 10 lAs = 1.414 x 10-2 + 4.246 x l0-1 : 1.457 x 10 r

No : |.45'7 x 10 2 + 6.908 x 10 a = 1.526 x 10 r

Arr : 1.526 x 10-2 + 6.251 x 10-1 : 1.589 x 10 r

Arz : 1.589 x 10 '? + 9.966 x l0 a - 1.688 x l0 2

Ar: = 1.688 x l0 2 + 3.814 x 10 a = 1.'726 x 10 2

N+: 1.726 x l0-2 + 3.476 x l0-5 : 1.730 x l0-'Ars : 1.730 x 10 2 + 2.115 x l0 6 : 1.730 x 10':Aro = 1.730 x 10 2+5.665 x 10 r = 1.730 x 10'?

/\p = lA'+ A'. lr-,'\21

(6.s0)

a =q+\-q:s+,1j=n(s.00)

(r3.00)

: 3.814 x l0 4

t(4.523 x 10 t + (1.038 x l0-4)lSs:

: 9.966 x 10 a

5" - ,-1 011 - !?88110'-,5325, b.2sr \ l0 -')

(1.288 + 1.546)(10-1 (4.875) = 6.908 x l0 I

s8 : (1.546 + L689)(10 ) e.625) : 4.246 x r0-l

s, : (1 689 + L457Xl0r) e.2zg) = 3.506 x r0 I

Sro =

Sr r

s,, = (4 112 + t258xl0 ) (s.37s) : z.z4g x 1.o l

(13.375)

3.349 x 10+t8.111 x 10 1

x l0-2

x10r 1I

4.575 x l0+,

t,,r.262

\.349

( 10.874)

(3.046)

r,. : (ry, )(7.58) : 3.s74 " ro ,

(1.457 + 3.671)( 10-) (6.646) : 1.704 x 10 r P,. : (

9!\?9(8.00) : 3.113 x ro l

: 1.207 x l0 r

234

Pr.:(

P,r:(


(s.00)4.575 x l0-3

Pr:( L730 x l0-2

+,

730 x 10-,r_d

J

6.732 x 10-+,

1.414 x l0-+,

_n

-)1.730 x 10+

, (6.s0)

: 2.827 x 10-2 : 1.125 x l0-,

6.7 32 x t0-3 +,

8.981 x 1015 175I

F3

Pl

1.730 x l0-

= 4.223 v l0-z

(2.958)

: 5.117 x l0 ']

p16 : 3.074 x l0 3 ft : y(16) : 0.037 in.tt5 : 3.074 x 10-3 + 2.262 x t0-2

: 2.569 x 10-2 ft : Y(15) : 9.39t n.pro : 2.569 x 10'? + 1.207 x I0 2

: 3.776 x l0-2 ft : V(t4y = 9.45, .r.

pB : 3.'176 x l0-2 + 2.827 x l0-2: 6.603 x 16-z 1 : y(13) : 0..792 in.

az = 6.603 x 10'? + 4.223 x 10 2

: 1.083 x l0-rfr : y(12) : 1.399 1r.p11 : 1.083 x l0-r + 8.428 x 10 ': 1.925 x 10 ! ft : y(11) : 2.310 in.p.6 : 1.925 x l0-r + 9.281 x l0-2

P'':( 8.981 .v l0-r +,

1.209 x 10(8.00)

:8 .428 x 1Q-2

1.379 x l0-1P'r=( (6.646)

= 9.281 x lO-'z

P":( 1.457 x 10-+T

1.414 x l0-z

-i)

(2.229)

(2.62s)10,X

.200

.457

3.2(

'r.4

10 2: 2.854 x 10 ' ft : y(10) : 3.425 'n.

: 2.854 x 10-r + 3.200 x l0-2: 3.174 x 10-' ft : y(9) : 3.809 in.= 3.174 x l0-r + 3.915 x l0 2

: 3.565 x 10-'ft: y(8) : 4.278 in.: 3.565 x l0-r + 7.593 x 10 ,: 4.324 x 10 I ft : y(7) : 5.189 in.: 4.324 x 10-r + 8.807 x l0-2: 5.205 x 10-r ft = y(6) : 6.246 tn.: 5.205 x 10-r + 2.283 x 10 ': 7.488 x l0-r ft : Y(5) : 8.986 in.: 7.488 x lO-t + 2.246 x 10 |

: 9.734 x l0 rft: y(4) = 11.681 in.: 9.734 x 10-r + 8.650 x 10-2: 1.060 x l0-'ft : y(3) : 12.720 in.: 1.060 + 1.125 x l0 I

: r.r72 ft : y(2) : 14.064 ir'.: 1.172 + 5.117 x l0 ,: r.224ft: y(1) : 14.683 in.

Pr:( +,

1.526 x l0

: 1.915 x 10-2

xl0,Pr:( 1.526 +,

1.589 x l0

: 7.593 x l0 2

(.4.87 5)

(5.37s)P.:l'\1.589 x l0 ' +

,

-,\

)

1.726 x. 10+,

_n

)

1.688 x l0-

8.807 x 10-'

/r.osa x ro tPr:( (r3.37 s)

( 13.00)

= 2.283 x l0 |

P":( |.726 x. t0 2 +,

1.730 x 10 Section weights and displacements for computing thetower's period of vibration are listed in Table 4-18.

The first period of vibration, T, is determined as fol-lows:: 2.246 x l0-1

P,:( 1.730 x 10-, +t

1.730 ,\ l0-

: 8.650 > l0-2

(s.00) ,=r"\EleDwv

(4-98)

-

--__-4


Table 4-18Tower Vibration Def lections

SectionDellection

y (in.) w (lb) wy (in.lb) Wy' (lb-in.'?)

14.683 423 6.2 r0.909 91.194.771

14.064 1,391 t9 ,563 .024 2'7 5,134.3'70

12.720 13,387 170,282.640 2,165,995. r 81

11.681 9l I I1,575.871 t35,217 .749

8.986 12.812 t15,128.632 I,034,545.8876.246 1,809 fi,299.0r4 70,5'73.64r

e 5. 189 4.271 22,162.219 rr4,999.7544.278 3,98'7 .096 t] .056.797

3.809 1,516 5 ,774.444 2r ,994.85',/

1 3.425 381 | ,304.925 4,469 .368

2.310 s )sR 12,145.980 28,057.214

1 .300 1.857 2.414.100 3,138.330

0.792 2.634 2.086.128 1,652.213

0.453 3.O22 1.368.966 62Q.142

0.308 | )s? 385.924 l 18.865

0.037 20,493 439.523 16.262

D*, = 3l6,t2s:ss D*r,:3.g64,78s.40.

(386.4)(386,129.39s): 1.024 sec/cycle

f : l/T : 0.976 Hz = 0.981 Hz : the initial assumedvalue

Considering the upper third of the tower as being theeffective length for vortex shedding the first criticalwind velocity is as follows:

104.292T_ : 34.764 ft - d

The second critical wind velocity, V2, is

Yz = 6.25 Vr : 65.163 mph

From Equation 4-94a and Table 4-13 we have

F, : 0.00086(0.6x60)(4.583)(104.292)(10.37 r,= 1,592.930 lb

To determine if we have a problem with vortex-inducedvibrations, we must compare the force amplitude of1,592.93 lb against the corresponding maximum windforce amplitude for the same region (either top l/r or r/+

of tower-in this case, the top r/:). Using Figure 4-39 wehave the following:

L : 34.764 ft, from above

1M.292 tt-34.764 ft : 69.528 ft

n. = al 716x15 659) 61 547 lb/trl

/60 srr\o 286

o" , = 1l.726) l"-^^"^-"I (66.048)\900/

9" r : 54.808 lb/ft2

3

54.00 + 1.00

t2

3.40 d

T

fisec

: 4.583 ft

(3.40)(4.s83) ft

1.024 se:

cycle

: 10.375 mph


_ (34.7 64)l(7)(6r. s47) + (2)(54.8o8)l(2+7)

/r rr nn\F = t2087.559r l'-;;""1 (0.6)\ rz I

or

F*,0 : 13,256 lbr > 1,608.56 lbr : F"

Since F*;"6 ) F,iu.",ion, the wind stresses are greaterlhan those at resonance vibrarion, so no further vi6rationanalysis is required.

If the vibration amplitude force had been greater thanthe maximum wind force, further investigation wouldhave been required. Dynamic stresses ar the-crirical windvelocity can be approximated by taking the ratio of thevibration force amplitude, F", to the maximum windforce amplitude, F1,, and multiplying this ratio by thebending stress term in Equation 4-29. The pressurestress, which is a primary stress, and the weight loadstress in tension are unaffected. Shear is not consideredin the equation because it is almost always negligible.Defining the ratio of vibration force amplitude to themaximum wind force amplitude as R. Equirion 4-29 be-comes

oD: t(?J--[**"un#.J.l 2w \= \",rO. + DD7

where oo = dynamic stress, psioD < static allowable stress

Staley and Graven [15] state rhar when dynamicstresses are combined with axial compressive stress. theresult can be compared to the allowabie sutic srress. Thesame is generally true for tension, but one must be cog-nizant of discontinuity stresses at the locations or irregi-lar changes of geometry, such as welds. The latter can beavoided by using stiffening rings. Certainly a more accu-rate and detailed analysis, such as the octahedral shearstress theory of yield, can be used, but such a detailedanalysis can be avoided in most tower designs. A de-tailed fatigue analysis is mandatory in many aipplicarionsand should always be used in case of doubt. Weaver [24]discovered in wind-tunnel tests that vortex sheddins can-not prebently be analyzed as a response spectra beiauseof its random nature and unpredictable motion. Thisgreatly complicates the study of vortex excitation by useof finite element methods, but efforts are being made.

:2,087.559 . Finally, if the Reynolds number is greater than approx-imately 350,000, a vibration analysis is not required, be-cause in these regions the vortices break-up. In our case.

= l? )55 000 rh-_ DcVp

where D. : effective wind diameter at top r/: or r/+ o:tower : 127.0 in.critical wind velocity = 15.292 ft/sec0.071 lb./fc1.285 x l0 5lb-/ft-sec

N"" : : 894,215(1.285 x 1o-1 ;lb.

It-sec

Since N*" : 894,215 > 350,000, a vibration analvsis isnot required since we are outside rhe range of vortex for-matron.

Vortex formation has been observed at NRe > 3.5 x106, but wind velocities encountered would not causeReynolds numbers that high.

Equivalent Diameter Approach versusANS|-A58.t-1982

In determining wind loadings in this example we usedthe formulation to compute wind forces:

p : q.CC,A, (4-81 r

in which Ae is computed using the total width of thetower, insulation, ladders, platforms, and attached pip-ing as an equivalent or effective diameter of a cylinber.called the effective or equivalent cylinder. This iquiva-lent cylinder represents the total wind area. Suih ananalysis is called a quasi-analysis, because it is not exact.The equivalent cylinder concept used for conical sectionsis similar when compared to the exact analysis of a cone.

The ANSI-AS8.1-1982 uses a more refined andequally complex analysis to determine the wind load-ings. The relationship used for wind force is as follows:

F: q7G2CrA61 + qzcciArr

where G2 : gust response factor for cladding and compo,nents calculated at heisht Z

Acr = area of insulation tclaJdingt of tower. and al.external attachments such as platforms. ladders. and piping rhat resist wind

416 : area of the tower shell itself that resists winc

/rzzoo\ /n\ rhlj:i--:, "

"f fi tts.292t l:' lrO.Orr,) l\ Lz I \sec/ 11-

Nn"

The term G7 is given in Table 8 of ANSI-A5S.1-1982rhat is determined by the following expressions:

Gz : 0.65 + 3.65 Tz

- 2.35 (D")osL1 :-

(z/3o1tt"

For category A,

D" = 0.025,o:3.0,2e = 1,500

For category B,

D. : 0.010, q - 4.5.22: 1,200

For category C,

D" = 0.005, a :7.0,2e: 9OQ

For category D,

D. : 0.003,q : 10.0, ze = 700

G2, which is used for cladding and components, varieswith height and is a parabolic distribution. The term G isused with the tower shell and only is constant along theheight of the tower.

Comparing the two methods we set Equation 4-81equal to the comparable expression given by ANSI-A58.1-1982.

F - q/GC,A, - q7G7C,A, r qrCc,e,

We now define the following variables:

Acr . ArrAr Ar

from which

qzcCA:q2C,A1(xQ.+yG)G-xGzayGG1t -y; > xc'

Now, for many, if not almost all cases,

c>G,

This is certainly true as one moves up the tower in com-puting Gz. It can be safely said that

G ) Gau,er


where G21u"ry = average value of G7 across the height of thetower.

Also, rarely is y as great as 0.5, making the inequalityabove more credible.

After applying the real numbers for several cases, it isseen that the equivalent diameter method using theANSI-A58.1-1982 gust factor for flexible structures,G, is more conservative than the ANSI method of usingthe two gust response factors G2 and G. Thus, beingmore conservative than ANSI A58.1 1982, one meetsthe minimum requirements ofthe standard, as it is statedin the title, "minimum design loads for buildings andother structures." Certainly, using the formula for latticestructures, Equation 4-83, is a conservative approach.

For designing a tower without a computer softwarepackage, the equivalent diameter method is recom-mended. In such a design, one is faced with numerouscalculations, which leads to a greater possibility of error.Also, the use of two gust factors with one varying inheight adds considerable complexity to the problem.When using a high-speed electronic computer the use oftwo gust factors would be a very good method to use,although cumbersome to verify. Certainly, some couldargue that with less conservatism a cheaper vessel is pro-duced. Such a consideration must be analyzed in eachseparate circumstance. For some, the additional man-hours may offset the economics of the vessel or time maybe the ruling criterion.

EXAMPLE 4.3: SEISMIC ANALYSIS OF AVERTICAL TOWER

A client has a vertical tower that is to be moved from aDlant in Jackass Flats. Nevada to a location northeast ofLos Angeles, California. The vessel must be analyzedfor seismic zone 4 to determine if it can be moved. Thisresult is to be compared to a wind analysis for an 80-mphwind.

Seismic Analysis

V = ZIKCSW (4- 106)

For zone 4,2 : t,I = 1, K : 2.0, W : 15,571 lbSince the tower is not of uniform thickness, equation

4-108 cannot be used. Either the Rayleigh equation(Equation 4-97) or a modified form of the RayleighEquation, the Mitchell Equation (Equation 4-112), canonly be used. For illustration purposes the MitchellEquation will be applied and then compared to the moreaccurate Rayleigh method.


Using values in Table 4-15 we determine the values tobe used in Equation 4-112. Connecting piping exerts aconcentrated load o12.7 kips at the support point mid-way in Secrion @-@. using the values in Table 4-15 weconstruct Table 4-19,

where A = !twa" + *Btt;

Using Table 4-18 we have

111 \2

u00/: 0.673 sec/cycle

S : 1.2 +0.6(1.468) - 0.3(1.468f

s : 1.434

Solving for V we have

v : (1)(1X2.0X0.078X1.434X1s,571) : 3,484.0 lb

Using Equation 4-114 to find F,, we have the following:

: 36 + Fr = 0.15V

Fr : (0.15)(3,484.0) : s22.60

From Equation 4-115 we obtain

F- : (V - Ft) -YYhY

D*'*"i= I

: 0.0057 wr,hyUsing the more accurate Rayleigh method, Equation4-97 , the value of T is

T : 0.734 sec/cycle

in which the Mitchell Equation is in 8.3 % error (which isquite normal). For application ofthe Rayleigh Equationsee Examples 4-2 and,4-4.

Now, we must solve for the bending moments inducedby the seismic forces. First we find the base shear usingEquation 4-106. To accomplish this we have the follow-rng:

1""'r - l:125 = 1.t25 < t.5... K - 2.0(r.i" l.U

Flexibility facror = C: -]- = O.OZA15(l1tr':

The characteristic site period, T,, is determined by asoils consultant to fall within the following range:

0.5<T.<0.55

To be safe, we will use the lower value of 0.5. Now,

T O ?14_ - ;; = 1.468 > 1.0 in which Equation 4-tl3bl s u.)u

applies. Thus, we have

S : 1.2

h72D2

n: Ee(,f,)'.a,

: 13.484 - 522.60) w)hv

515,380

To solve this equation we must set up the table shownin Thble 4-20. After determining the values for W, h",W)h). F,- and V, we solve for the seismic moments usingEouation 4-116:

+ 0.6 El - 0.. El'\TJ \TJ

M - Vxi Ly_r + Fx Ci DtM, - (0.30)(5.083) + (0.49)Q.s42) : 2.770 2.'770M3 : (0.49)(7.50) + (1.19X3.75) : 8.138 10.908Ma : (1.68)(5.417) + (0.18X2.708) = 9.s88 20.496M5 : (1.86X2.00) + (0.09x1.00) = 3.810 24.306M6 = (1.95X8.00) + (0.28)(4.00) = 16.72o 4r.026M7 : (2.23X4.00) + (0.16)(2.00) : 9.240 50.266Ms : (2.39X10.00) + (0.32)(s.00) : 25.sm 75.766M, : (2.71X10.00) + (0.22X5.00) : 28.200 103.966

M16 : (2.93)(9.083) + (0.23)(4.542) : 27.658 13r.624Mrr : (3.16)(8.00) + (0.10X4.00) : 25.680 rs7.304M12: (3.26)(2.917) + (0.01)(1.458) : 9.524 166.828

The wind moment for an 80-mph wind was calculatedro be 106,716 ft-lb. Since 166,828 ft-lb > 106,716 ft-lbseismic phenomena govern.

The skirt and base plate analysis is identical for seis-mic and wind analyses. Just as in Example 4-2, the seis-mic forces and moments are used instead of the windforces and moments. In the case of this tower a thickerbase plate was welded on, the number of gusset plateswere doubled, and anchor bolts of a high strength alloywere used to meet the seismic criteria. In an earthquakezone other than zero, a comparison of seismic to windshould always be made.

-

t


Table 4-19Numeric Integraiion ol Period

ot Vibralion, T sec/cycle

E l9l'to"u0i

whvkipsfft H

WAa +Aa P WAB/H

0.0911 .00

0.923

2.103

1.506

0.597 0.0541.000 1.000

1.000 1.000

2.'7

0.079

0.878

0.820

o.219

0.045

o.567 5.840

0.939 1.000 1.000

0.100 0.329 0.033 1.234 x 10-60.998o."t45 0.610 0.998

0.t42 0.067 0.010 9.541 x l0 5

0.9970.1r7 0.543 0.997

0.125 o.278 0.035 0.00100.9860.607 0.265 0.986

0.552 0. 1680.161 0.097 0.016 0.0015

0.9730.973

0. 161 0.r24 0.020 0.0080

0.904o.414 0.0,14 0.904

0.155 0.037 0.006 0.021 1

0.7630.276 0.007 o.763

0.285 0.007 0.002 0.0649

0.5040. 151 0,0004 0.504

0.3t2 0.0004 0.0001 0.1175

0.1600.040 0.160

0.3800.000

0.0412

A:0.,140 B:0.261


Table 4-20Wind Load Distribution

x"'o-ri. Lzt ztttp w\-rkips !r w,h, Fx v, *,1-u *,oTi72 ffi

0.49 2.770 2.7703.289 63.2 207 .86 l. l9

1.68 8.138 10.9080.542 56.7 30.73 0. 18

1.86 9.588 20.4960.284 53.0 15.05 0.09

1.95 3.810 24.3061.0 48.0 48.00 0.28

2.23 16.720 41.0260.645 42.0 27 .09 0.16

2.39 9.240 41.0261.613 15.0 56.46 0.32

2.7 1 2s.500 15.7661.550 25.0 38.75 0.22

+\:

.':

-\3

+

+ ->s,* i+2s.k-+

+ 1.0' -l

/:'+G

$

€

$

@

o5l

+ --> rsr ..:

+

+

+

-+

+

2.93 28.200 103.9662.581 15.5 40.01 0.23

3.16 27 .658 131.6242.493 '7 .O 17 .45 0.10

3.26 25.680 157.3041.109 1.5 l .66 0.01 9.524 166.828

15.571 5 15.38 3.270 3.27

EXAMPLE 4-4: VIBRATION ANALYSIS FORTOWER WITH LARGE VORTEX-INDUCED

DISPLACEMENTS

A phone call from a plant manager reveals that an ex-isting tower needs to be analyzed for wind vibrations.The tower was designed, built and installed overseas andis vibrating so badly all the natives drove off the plantsite in fear of the tower falling over.

The tower with the appropriate wind load distributionis shown in Figure 4-44. The tower is divided into windzones at 30 ft,40 ft, and 75 ft and according to shell di-ameter and thickless. The variation of wind zones basedon the shell diameter and thickoess is necessary since thetower's section moment of inertia will vary.

To begin the analysis we start with defining the effec-tive diameter of each section as illustrated in Figure 4-45. Thus we have the following:

Zone 1-Sections 7. 8. and 9

D" : [32 in. + 2 (4) in.] + [6.625 in. + 2(3.5) in.]+ [2.375 in. + 2(3.0) in.] + [4.5 in. + 2(3) in.]

D" = 40.00 + 13.625 + 8.375 + 10.50

D" : 72.5 in. : 6.042 ft

Zone 2-Section 6


D. = 54.25 in. = 4.521 ft

Zone 6-Section 2

D" : [24.50 in. + 2(4.5) in.] + [3.50 in, + 2(4) in.]

D. : 33.50 in. + 11.50 in.

D" = 45.00 in. = 3.75 ft

Zone 7-Section I

: 136.625 in. + 2(4.5) in.l + [6.625 in. + 2(5) in.]+ [3.50 in. + 2(4.5) in.] + [6.625 in. + 2(5) in.]

= 46.625 in. + 16.625 in. + 12.50 in. + 16.625 in.

= 104.875 in. : 8.740 ft

Moments of Inertia

r:#(D".-Di)

\ : hl36.62s)4 - (36.000)41 = 5,876.389 in.a

: 0.283 fta

Transition Piece-Section 2

Referring to Figure 4-45,

,"r:( 18.375 + 12.375

2 cos 26.565'

D.

D"

D.

De:[32in.+8in.]: 4.042 ft

Zone 3-Section 5

D" : 25.25 in. + 2(2.5) in.

D" : 2.521 ft

Zone 5-Section 3

+ 14.5 in. I 2(2) in.l 48.5 in.

5" : [25.25 in. + 2(2.5) in.] + [4.5 in. + 2(2.5) in.]+ [3.5 in. + 2(3.5) in.]

D" = 30.25 in. + 9.50 in. * 10.50 in.

D" : 50.25 in. = 4.l88ft

Zone 4-Section 4

req 17. 190 + D.q = 34.380 in

r, = #(34.380)4 - (33.630)11

Iz : 5791.250 in.a : 0.279 fta

\ : 1, Kz4.sq4 - e4.00)11 : r,400.ri2 in.a

: 0.068 fc

\ = L64Kz4.i5)o - (24.00)11 = 2,133.181 in.a

- 0.103 fta

= 30.25 in.

9" = 125.25 in. + 2(4.5) in.l + [4.5 in. + 2(2.5) in.] +[3.5 in. + 2(3.5) in.]

D" : 34.25 in. + 9.50 in. * 10.50 in.


* "; ".". ' %: *,* '"*

l\'

T(

Figure 4-44. Tower wind ensemble.

- - lrs.as + n.azs\ _ ," ,'* = \-ffi/

:14.174in.

+ D.e = 28.348 in.

y : f,11zt.z+ty4 - (27.72141

b : 2,704.843 in.a : 0.130 fll

I8 = #rc2.00)4 - (30.00F1 = r1,i1r.wzin.a

= 0.565 fll

r, : fftfrz.oof - eo.6zr4l : 8,2e2.684n.a

= 0.2t00 ff

The Engineering Mechanics of Pressure ry'essels

Wind Moment Calculations

Sections I and 2

M2 : es8.4zs) (#. tr.r) * o,uno.rrrr(U)+ 4,450

Mz: lL,99O.762 + 18,158.661 + 4,450.00Mz : 34'599.423 ft-lb

Sections 2 and 3

M3 = (788.425)(15.2W + 17.0O) + Q,690.r72)(6.75

ht\+ 17.00) + 4,450 + (1,453.50) ttM: : 25,394.381 + 63,891.585 + 4,450 + 12,354.75M: = 106,090.716 ft-lb

Section 3 and 4

lvl4 = (788.425)(32.209 + 10.00)+ (2,690.172)Q3.75 + 10.00)

Figure 445. Tfalsition piece of section 2 of Figarc 444.

I. : 1 [(25.ooy - (24.00)41g-

: 0.139 fll

u.= fir<zs.zsf - @4.oof1

= 0,177 tr

Section 7

Referring to Figure 446

: 2,888.744 in.a

: 3,667.316 in.a rroi+ (1,453.s0X10.00) + Q21.5s2\lrl

+ (268.541(+)

Figure 4-46, Section 7 of Figure 4-44.


Mt:33,278.631 + 90,793.305 + 4,450 + 14,535.00+ 3,607 .76 + 402.821

Mq = 147 ,067 .517 ft-lb

Sections 4 and 5

M5 : (788.425X42.2o9 + B.O) + (2,690.172)(43.7 s+ 8.0) + 4,450 + (1,453.50x10 + 8.0)+ (721.552)(5 + 8)

+ (268.547x1.5 - 8) - (J4s.41) l:l\21

+ (3e.328) lll\zl

M: : 39,586.031 + 139,216.401 + 4,450+ 26,163.00 + 9,380.176+ 2,551.197 + 1,397.&4 + 19.664

M5 = 222,7&.113 ft-Ib

Section 5 and 6

M8 : (788.425X67.292 +8) + (2,690.172X68.833 + 8)+ 4,4s0 + (1,453.50X35.083 + 8)+ (721.552)(30.083 + 8) + (268 .547)(26.583 + 8)+ (349.41 l)(21.083 + 8)+ (39.328Xr7.583 + 8) + (522.662)(4.542 + 8)

/n\+ 17s4.042t l:l\zl

Mr = 59,362.095 + 206,693.985 + 4,450 + 62,621.t41+ 27,478.86s + 9,287.16r + 10,161.920+ 1,006.128 + 6,555.227 + 3,016.168

Ma = 390,632.690 ftib

Sections 8 and 9

Sections 7 and 8

Me : Q 88.425)(7 5.292 + 2.9 17) + Q,690.r72)(.1 6.833+ ).:917) + 4,450 + (1,453.50)(43.083 + 2.9t7)+ (721.5s2x38.083 + 2.9t7) + Q68.547)(34.583+ 2.9t7) + (349.41r)(29.083 + 2.917)+ (39.328)(25.583 + 2.917)+ (522.662)(r2.sQ + 2.917)

-, 17s4.042\A + 2.gt.l,) + ,Zt+.OOr(2.717\\z I

Ms : 61,661.931 + 214,541.217 + 4,450 + 66,861.00+ 29,583.632 + 10,070.513 + 11,181.152+ 1,120.848 + 8,079.832 + 5,215.709+ 401.003

Ms = 413,166.837 ft-lb

Wind Deflections

M6 : (788.425)(50.209 + 8.0) + (2,690.172)(5r.75+ 8.0) + 4,450 + (1,453.50)(18.0 + 8.0)+ (72r.5s2)(13 + 8) + (268.547X9.5 + 8)+ (349.411)(4 + 8) + (39.328X0.5 + 8)

/^\+ rs22.662t lll

\21

Mo : 45,893.431 + 160,737.'177 + 4,450+ 37,79r.00 + t5,152.592 + 4,699.573+ 4,192.932 + 334.288 + 2,090.648

M6 = 2'7Q,892.241 ft-lb

Sections 6 and 7

M? : (788.425Xs8.209 + 9.083) + (2,690.r'12)(59.75+ 9.083) + 4,450 + (1,453.50)(26.0 + 9.083)+ (721.552)(2r + 9.083) + (268.547)(17.5+ 9.083) + (349.411X12 + 9.083)

- {3e.328X8.s r 9.083) + rszz.ooz, (9 983)\z I

Mz : 53,054.695 + 185,172.609 + 4,450 + 50,993.141+ 2t,706.449 + 7,138.785 + 7,366.632+ 691.504 + 2,373.669

M7 : 332,94'1 .484 ft-lb

1B), [r:.+ZS.:SZtr r:) , +,+sOl' t4.176 x t0"x0.28J) [ 8 2 )

!z

: 0.00113 ft

_ (17)'(4.r76 \ l0)(0.068)

+ (r,4s3.5_0)(17.0) +34,s2s.42]: 0.04081 ft821(10F l1+,vzz.ost11ro1

Ar?6t-To"xorort :

l{2,+tt.serot .o)

t3

Y::

E

, (99o.o99xro) 106,090.721 _* ff _ _.-l: 001658 fr

(sf ks,szz.rso)(s)(4J?6 x t06x0x9)t 3

+ (388,738)(8) * t+'t.OOt.SZ] = 0.00989 ri8 ' 2 l-(8f ko,:ro.s:+Xs)

Af?6 x ro)(oJ??)-t 3

+ t522.66UG) * Zzz.lo+.tnl = 0.0r I 15 ri8 ' 2 )-

(4.176 x 10)(0.260)


, 622.662t8t ".. ,., . . "lr 'T + 222.764.1131 = 0.t3055 rt

(56.50X9.083) ft6.833.596)(9.081)-'6 - 14.176 x t0\0260)t 2

+ (572 730x9 083) + 2io.8s2.z41f = 0.143 ft6l

(71.583X8) ft2,+oo.rz r'11r;

Lz

+ (572.730)(9.033) +270,8s2.24rf : 0.012 ft82j

(8), [tr.+oo. rzox a rt' - ,aJ6 v 1g\05sr[ .l

+ (7s4.042)(8) + 332,s47.481 : 0.00507 ft821(2.gt667tt [rt. roo.:os x z.r reor I

18 - t+.tzo x ronxo,+oott :

- t274.u2\2.st667, * 39!.632.69] = 0.00201 ft821(13.5)( l7) [tf.+la.sertr t r r-" - (4.176-,. looxol68)t z

+ (1'453 50)(17) * ro.rnn.or]: 0.05519 ft6l(30.50Xr0) ft+.erz.owxror' \4.176 x 10"x0.103)[ 2

, (990'0?9x t0) + 106.090.721 : 0.07709 rr6l(40.50)(8) [rs.lzz. rvolr r

't + - ,4. 17,6, x to\rn t rorl Z

, {388.738x8) , , ,, ".1.s2l = 0.09560 fr- --- 6 + l+/,uo I

(48.s0x8) ft6.310.934)(8)' (4.176 x t0')(0.177)[ 2

(4.176 x 10)(0.565)

+(7s4'y2)(8) + n2,s4i.481: 0-09071 rt6l17 3.58r(2.g t67 ) [rS. rOO.:OStrZ.l rOrr-'* -

1+, tuo x to"1o,+ogt 2

* 121!!!)t2!tfr)+ J90.632.691 = 0.05r74 ft6)

6 : total deflection at top of vessel

\- \-: 2y, + LtA, I

: 0.743 ft

: 8.910 in. at top for static gust wind

Referring to Figure 4-47 , we determine

Mr :0

M, : (4.71)(6.961) : 32.786 kip-ft

Mz : 32.786 + (4.823)(8.789) : 75.u5 kip-ft

M4 : 75.175 + (7.533X13.25) : 174.987 kip-ft

Ms = L'74.987 + (10.013X9.00) : 265.104 kip-ft

M6 : 265.104 + (12.023)(8.00) = 361.288 kip-ft

Mi : 361.288 + (14.253X8.862) : 487.598 kip-ft

Mr : 487.598 + (ri.693)(8.221) : 633.032 kip-ft

Ms : 633.032 + (21.233)(5.458) : 748.922 l<tp-tt

Mrc : 748.922 + (23.143)(1.458) = 782.664 ktp-tt

T:M/I

T, = Mt - 32'786 : rv.5t2.54'' tz 0.279


T^

T4

T6

:M.:I3

:Mo:r4

= M'=I5

:Mu=I6

:Mr:11

:Mt=Is

:M,:Ie

=M'o=Ie

= M dx/I

75,t75 _0.068

17 4,987 _0.103

265,104 _0.139

361,288 _o.t77

487598 _0.130

633,032 _0.565

'748,922 _0.400

782,@

I,105,514.71

1,698,902.91

|,907 ,223.02

2 ,041 ,r7 5 .14

3,7 50,7 53.85

| ,r20 ,4r0 .62

1,872,305.00

-q-T-

---->4.71kT8

-------e

--r.2.71k

------€)----------->2 .48x

Tro

s.

Sro

n /An - 1,956,660.00

_ (1,956,660.00 + 1,872,305.00)(1.458)

:2,791,315.49

(1,872,305.00 + t,r20,410.62)(5.458)

2

8,167 ,120.93->

2.01k

------e->

2,23k

-------€

^ (1.120.410.62 + 3.750.753.85) ^ ^^"\'-_:___:__________________:__________rv,,,r"2

--> 344k

-----e-----l3.54K

:20,022,921.55

S7_ (.3,75O,7s3.8s + 2,041,175.r4)

(8.862)

: ,5 664 017 15

(2,O41,175.14 + 1,90',1 ,223.O2) (8.00)2

15 ,793 ,592 .64

_ (1 ,907 ,223 .02 + | ,698 ,902 .9r)2

: 16,227,566.69

Sr:

(s)' \-"/

----------> 1.91x

(;\ S5

Figwe 4-47. Tower vibration ensemble.

(9.00)

t

Sr:

$=

Po:Sz:

Pro:

(1,698,902.91 + 1,105,514.71)

The Engineering Mechanics of Pressure ry'essels

(56,645,395.32 + 72,438,9U .96)(8.00)

2

516,337 ,533.2

(72,438,987 .96 + 88,666,554.65)

724,974,941.7

. _ (10,958,436.42 + 30,98r,357.9't) $.22t)lE_----_-2-

= 172,393,524.9

^ (30.981.357.97 + s6.&5.39:.32) $.862)rt = --------------T-

:388,274,143.9

(13.25)2

18 ,579 ,266 .73

(1,105,514.71 + lr7,512.54)(8.789)

2

5,374,593.25

(117,512.54\ 6.961\

a

Q9,002.40

P4

Ps:

P::

Pz=

_ (88,666,554.65 + 107.245,821.4) ,,...,z - lr5'zJ)

: 1 ,297 ,919,492

(to7,245,821.4 + 1 t2,620,4t4.7)2

966,202,r7 5.0

(112,620,414.7 + 113,029,417.1)

2

785,374,239.6

ttto : 2,034,868.99

w : 2,034,868.99 + 37,523,072.96= 39,557 ,941 .95

ps : 39,557,941.95 + 172,393,524.9= 211,951,466.9

pt : 211,951,466.9 + 388,274,143.9: @0,225,61O.8

t4 = ffi0,225,610.8 + 516,337,533.2: |,116,563,144

t'.s : |,116,563,144 + 724,974,941.7: 1,841,538,086

lt4 = 1,841,538,086 + 1,297,919,492: 3,139,457,578

14 = 3,139,457,578 + 966,202,175.0= 4,1O5,659,753

(8.789)

(6.961)

o=(*, )u2.79t.315.49.- ,_^.---:---------:- | l -4)x)

2,O34,868.99

Q,79r,3r5.49 + 10,958,436.42)

e, = Ds, = 1M1 dx)/!

2,791,315.49

2,791,315.49 + 8,t67,120.9310,958,436.42

10,958,436.42 + 20,022,921.5530,98r,357 .97

30,981,357 .97 + 25,ffi4,037.3556,&5,395,32

56,&5,395.32 + 15,793,592.&72,438,987.96

72,438,987 .96 + 16,227 ,566.6988,666,554.65

88,666,554.65 + 18,579,266.73rc7,245,82t.4

tu,245,821.4 + 5,37 4,593.25112,620,414.7

lL2,620,414.7 + 409,002.40rt3,o29,4r7.r

37 ,523 ,U2 .96

458)


p2 : 4,105,659,753 + 785,374,239.6: 4,891,033,993

' l44Ei

r,,: t:0]1'868 li :4.it0x r0 aft = 0.006in.(4.32 x l0r)

The tower section weights and displacements are com-bined in Thble 4-21 to determine the period of vibrationof the tower.First critical wind velocity, V,

-. 3.40 d.T

16 q6L= ";-=re.24tt

. = (,+*) $740) +(,uaA.,r', = ,.,,,

From Equation 4-101, at resonance

fy : vortex shedding frequency : natural frequency

V, : fvD - to 91)(7.1221 : 34.540aS U.l Sec

: 23.550 mph

Considering the top portion (Section 1) we have

v - (o eT(lfa) = 423s L = 2E.eo mphu.z sec

Yz:

t5 -

Y8:

J9-

5.1 15 in.

3.102 in.

1.667 in.

0.589 in.

0.1l0 in.

4,891,033,993

144(30 x t05

4,105,659 ,7 53

(43' x tOt

3,1,39,457 ,578(4.32 x l0e)

I,841,538,086

":z1 1bt

|,1t6,563,r44(43' x iort

600,225,610.8

@tt]ott2rr,951,466.9

$8 1639,557 ,94r.95

w2t16

: l.lJ n = lj-)v ln.

: 0.727 ft = 8.721 in.

: 0.950 ft : 11.405 in.

= 0.426 ft :

: 0.258 ft :

: 0.139 ft :

: 0.049 ft :

: 0.009 ft =

Table 4-21Values for Determining Tower's Period of Vibration

Detlectlonw1_!:-!

13.59 4.7 r0 64.008.90 869.880.95

8.725.r23. l0

7102,4802,010

1.289.3323.63t .20

.606.231 .00

14.711.26

064.0665.01 l.7 t19.316.10

1.67 2,230 3.724.10 6.219.250.59 3.444 2,029.60 1.197 .460.11 1 54n 389.400.01 1.9i0 19.10 0.19

Dtr : 114,02s.23 Dwy, = r,182.443.81First Period of Vibration, T

ILwy' i 11. t82.441.8r)r = z,r \/etrwv = zr 1/(386.4X114,020.23, = 1.03 sec/cycle o f = O.9j Hz

tou J t

-


Since the field measurements indicated an air velocityat resonance to be 30 mph and a stack deflection of 13inches, this analysis agrees with empirical results. Fromthe calculations for the first critical wind velocity, it ap-pears that the larger diameter of Section t has a largerinfluence on this deflection. For this reason we use thetop I/+ of the tower rather than the top 1/:. Now,

Y1 : 6.25; Vr : (6.25)(28.90) : 180.63 mph

A tower that has been fabricated and installed in thefield is beyond design changes. Unlike stacks (see Chap-ter 5), vortex strakes are difficult to install on many tow-ers and impossible on others. Shortening the towerheight is impractical, since the tower's internals are nec-essary (unlike a stack). Consequently, the only resolu-tion is to mount guy wires to the tower's upper section(normally 2/3 the height). Except for special applications,guy wires are to be avoided in practice. They use a lot ofspace and plant maintenance people sometimes musttemporarily remove one or two to gain access to an areafor equipment installation or some other reason. Prob-lems then may arise in keeping the tower from fallingover during this temporary time interval, rememberingto reconnect the guy wire(s), and making sure the wiresare properly tensioned once they are reconnected. De-spite these disadvantages, guy wires were essential inthis application.

EXAMPLE 4.5r SADDLE PLATE ANALYSTSOF A HORIZONTAL VESSEL

A proposed horizontal vessel design shown in Figure448 is fully loaded with corn syrup used by a confec-tionery manufacturing plant in Fayetteville, Arkansas.The corn syrup has a specific gravity of .y = 1.4 un6 . ut90"F. The thickness of the head and shell is t/z in. sincethe corn syrup is at 90'F, there is practically no thermalexpansion of the vessel, so only uniform compression isconsidered in evaluating the saddles. Even though a Zickanalysis indicates that the vessel is grossly overstressed,the saddle in Figure 4-48 is to be evaluated.

To analyze the saddle plate, refer to Figure 4-48 c.Each section of the saddle plate, A-B, B-C, C-D, isconsidered separately. Each section supports a portion ofthe vessel weight indicated by the dotted lines. SectionsA-B and C-D support equal weights.

Section A-B and C-D

x : 4.27 ft : 51.24 in., Ri : 6.0 ft = i2rn.

From Appendix A, Equation A-8, the fluid volume invessel above Section A-B is as follows:

2 r\36)\'l 2 )z

= 59.948.76 trl.3 = 259.52 gal in one head

From Equation A-1 in Appendix A the partial volume ofliquid in the cylindrical portion is calculated.

., (72)2(150)(l2t lott+0.+St ^ --l2 L 180 I

: 9,351 ,647 .46 in.3

= 40,483.32 gal

Total fluid volume above Section A-B is

Y : 4O,483.32 gaI + 2(259.52) gal = 4t,002.36 gal: 5,481.22 ft3

The total fluid weight is then

Wres Wrco = t5-48t.22t fr' tOZ.qt Ib, r r .+r

: 478,839.22 tb

Metal Weight Above Each Section, A-B and C-DFor outside surlace on h.ud, thuiur" V, ir,. tt'i.k,

_ o,12r'lst.zq _ 15l.2ar1l2 | l(361 j

_ "<t?.r,lsr.t+ -2r\36 .25 )(7 2 .5 ),

: 62,434.25 in.3

The inside volume in the head was determined in com-puting the fluid volume as being 59,948.76 in3. Themetal volume in one head is then

VM : 62,434.25 in.3 - 59,948.76 in.3 : 2,485.49 in.l

For two heads,

Yu = 2(2,485.49) : 4,970.98 jn.3

The metal volume in the cylinder portion above SectionA-B is determined as follows:For outside surface,

_ (72.5)2050)(12)

2

= 9,512,090.41

l""l;';"' - o'r]

in.l

I50 It TAN/TAN


i-8

wi=3 46 tl

Figure 4-48. Horizontal vessel containing corn syrup.

The inside volume was determined in computing thefluid volume as being 9,351,647.46 in3. The metal vol-ume in the cylinder is then

V : 9,512,090.41 in.3 - 9,351,647.46 in.3: 160,442.95 ir.3

The total metal volume above Section A-B is

vM - 4,970.98 in.r + 160,442.95 in.3 : 165,413.93 in.3

The metal weight is

wy = 1r65.413.93) in.J (0.283) .lb, = 46,8t2.14 lbrn.J

Combining with fluid weight the total weight,

W : 46,812.14 lb + 478,839 .22 lb: 525,651.36 lb

For each saddle,

rr AB - vvcD -

Section B-C

525,651.36 lb : 262,825.68 lb : Q

Similarly to Section A-B, for the head, the liquid vol-ume is determined from Equation A-7 in Appendix A.

r : {#[" r- - t#l : 135,483 43 in3

For total volume,

v = 2(135,483.43) : 270,966.86 in.3 for one head

For both heads,

v : 2(270,966.86) : 54r,933.73 in.3

Liquid volume for cylinder portion is

v : r(72)'?(r50)(r2) in.3 - 2(9,351,647.46) in.3: 10,611,534.46 in.3

tttlABCD

lcl

TlrotIl-I

T

i\A, f.i.,---" i"['i"'

{

The total liquid volume above Section B{ is

Vr : 10,611,534.46 in.3 + 541,933.73 in.3

= 11,153,468.19 in.3

Metal Volume Above Section B-C

For outside surface on a single head, usin€ EquationA-7

u - r(7?.s\2[rt * - tr#J : B7,oe7.7e in.3

The inside volume was determined from calculating theliquid volume as being 135,483.43 in.3 Thus the metalvolume for a single head is

Yu = 137,097.43 in.3 - 135,483.43 in.3 = 1,614'35 in.3

For two heads,

Vr,,r : 2(1,614.36)n.3 : 3,228.72 in.3

The metal volume for the cylindrical portion is deter-mined using Equation A-l and the total volume ofa cyl-inder as follows:

(72)z(r50X12)

V : 19,963,181.93 in.3 for inside volume

For outside volume,

_ (72.5)2(rs0)(r2)

2

I0.551

IV = 20,205,196.86 in.3

The metal volume is

Vr,a = 20,205,196.86 in.3 - 19,963,181.93 in.3: 242,O14.93 in.3

For both sides of centerline in Figure 448 c,

Yu = 2Q42,014.93)in.3: 484,029.86 in.3

Combining both the cylindrlcal and head metal weightswe have

The Engineering Mechanics of Pressure rGssels 251

vru : 484,029.86 in.3 + 3228.72 in.3 : 487,258.58 in.3

The total metal weight above Section B-C is

.^ -^^-. lbWs = (487.258.58)in.i (0.2833) ft3

: 137,894.18 lb

The total liquid volume above Section B-C is

w, _ (11.151,198.I?)in.r (62.4) E (1.4)ftj- | ,723 rn.3

: 563,869.78 lb

The total weight above Section B-C is

Wr = 137,894.18 lb + 563,869.78 lb : 701,763.96 lb

For each saddle,

Wrc:Q: 701,763.96 lb : 350,881.98 lb

Saddle Plate Buckllng Analysls

The critical buckling stress for a plate is determinedfrom Equation 4-17a.

(4-r7a)

where h =ldi ts + 2hG - 1)

h = 0'50 in'

(12x3.46X0.50)(12)(3.46X0.s) + 2(0.sxls - 1)l

= 0.59 in.

AIso,

b. : Kt. (+1s)h = (1.28X0.5) = 0.64 in. (use.0.597 in.)

Adding more length to web plate will net. increase thelocal buckling strength for pure compression. The samealso holds for bending and shear. Substituting the valueof b, above into Equation 4-17a ws have

rQz.0)2(15O't02')

r(72.5)?(150)(12)

*f0a6.'12"\ -


(1.28)r, (29 x 106)vc,_ -,-____ - :4.980.860psi,,1, _ 1l l(3.46x l2 )1,

\ e/\ 0.s I

Substituting this value into Equation 4-18 we deter-mine the buckling load for compressive loading as fol-lows:

Fs : n(A, + 2be tJo. (4-18)

FB : (4)[7.5 + 2(0.597)(0.5) : 161,321.389 lb

Since 161,321.4 lb < 262,825.7 lb < 350,881.98 lb

we must use more stiffening plates if we are to use at/z-in. saddle plate.Now

FB." : 351,000 lb

From Equation 4-18 we have

351,000 : n[7.5 + 2(0.597X0.5)]o".

351,000

n

The effective plate width normal to the web plate axis is

d" : di (0.25 + 0.91\'?)

where \ = (l)[9"\w,/ \o*i

^ = (u*;)(':'-'*f': oo'

w" : (3.46)(12)(0.25 + 0.91 (0.41)'?) : 16.74 1n.

10.392( l2)n = = 7.449 -use E stiffeners16.74

o^, = l5l'ooo : 43.875.0 osi-8

FB = (8)[7.5 + 2(0.597X0.s)](43,875.0): 2,842,W7.0 lb

Since 2,&42,047 lb > > 351,000 lb, eight stiffeners aresufficient.

Horizontal Reactlon Force on Saddle

From Equation 4-19 the horizontal reaction is

F:Q

o = (rso -:) = 'r -f = no"

^ 2(262.825.68) + 350.881.98v: z

: 438,266.67 lb

: 8s,294.56 lb

The effective area resisting this force is

^" = (,u),. = (9 (o 5o) : 12 ss 1n,

This results in a stress of

8s,294.56 tb _ ^-^ ^_o = tzttg irtl : /'uJ6 60 PSr

Referring to Table 4-6, the allowable stress for A-36 is0.60 o, : 22,000 psi. Since 7,058.86 psi < 22,000osi. the saddle is sufficient for the horizontal reaction.

bBP

BPWC

1{OTATION

dimension from saddle centerline to tansent ofhead (Figure 4-2) ft, in.effective area of concrete, ft2plate width (Equation 4-15) in.bearing pressure, psibase plate thickness, in.constant for bolt torque (Equation 4-66), di-mensionless; friction coefficient (Equation 4-89) dimensionless; structure period responsefactor (Equation 4-106) dimensionlesscorrosion allowance, in.critical damping factor (Equation 4-90), di-mensionlesscompressive strength of concrete (Thble 4-7),psi

CA:c.:cs=

z- - B * sin B cos p

D = diameter (Equation 4-27), in.; dynamic magnification factor (Equation 4-9 1), dimensionless

Dr : effective wind diameter (Figure 4-22), in.D. : outside diametet in.D, = inside diameter. in.E : welding joint efficiency (Table 4-2), dimen-

sionless: modulus of elasticity. psiF = wind force (Equation 4-94)F; : bold uplift force (Equation 4-39), lbrf. : natural frequency of a ring (Equation 4-100),

Hzf, : vortex shedding frequency Equation 4-101, Hz

Gr : dynamic gust response factor, dimensionlessGg = gusset plate height (Equations 4-39 and 4-63),

ln.G* : gusset plate width (Equation 4-63), in.H : depth of vessel head (Figure 4-2), in.I : moment of inertia (Equation 4-24), in1 ;occu-

pancy importance factor (Equation 4-106), di-mensionless

I" = moment of inertia of effective area of con-crete, in1

K : coefficient of buckling for shear (Equation 4-15 and Figure 4-3), dimensionless

k : dimensionless parameter for concrete (Thble 4-7)

K' = plate buckling coefficient (Equation 4-15), di-mensionless

Kz : velocity pressure coefficient (Thble.4-9 andEquation 4-78)

L : length of a horizontal vessel from seam toseam (Figure 4-2), ft, in.

L" = effective column length (Equation 4-19), in.M : bending moment, in.lb, ft-lbm : bolt spacing (Equation 4-39), in.

Mc : compressive bending moment in the shell of ahorizontal vessel (Figure 4-2), tt-lb

Mr : tensional bending moment in the shell of a hor-izontal vessel (Figure 4-2), ft-lb

N : number of anchor bolts (Equation 440), di-mensionless

P : buckling load for compressive loading (Equa-tion 4-18), lb6; probability of exceeding winddesign speed during n years (Thble 4-11) andAppendix A), dimensionless

Pu : annual probability of wind speed exceeding agiven magnitude-see (Appendix A), dimen-sionless

R : mean radius of shell (Figure 4-2), ft, in.Ri : inside vessel radius (Equation 4-13), in.& : outside vessel radius (Equation 4-73), in.r : inside radius of vessel (Figure 4-2), ft

Q : reaction at saddle (wl2), lbl


qF : velocity pressure of wind on structures (Equa-

tion q-i6), rcJf(g:o : basic wind pressure at 30 ft, lbrift'?

S : Strouhal number used (Equation 4-102), di-mensionless; structure size factor (Equation 4-82)

T = bolt torque as defined (Equation 4-66), in.-lbTr : exposure facior for wind (Thble 4-11), dimen-

sionlesst6q : compression plate thickness (Equation 4-63),

in.t8 : gusset plate thickness, in.tr, : head thickness (Equation 4-7), in.( : shell thickness (Equation 4-1), in.vo : theoretical ovaling velocity (Equation 4-102),

mph or ft/secvr : first critical wind velocity (Equation 4-94),

mphv30 = basic wind speed at thirty feet used as design

wind speed (Equation 4-75), mphW = vessel weight (Equation 4-40), lbrx.t : static deflection of a spring acted upon by a

force (Equation 4-90). in.xO = displacement as a function of time (Equation 4-

90), in.y = total lateral displacement of tower (Equation

4-88, Figure 4-21), in., ftZ : elevation or height above a reference point,

such as the ground (Equation 4-74), ftZ = reference height in which basic wind speed is

considered (30 ft or 10 m), ft

Greek Symbols

a : ir - (tr 1180)(012 + B/20) (Equation 4-6), de-grees

B = (180 - 012), degreesA = (?./180x5di 12 t 30), degrees6; = lateral translational deflection oftower, (Equa-

tion 4-88 and Figure 4-26), in.d : angle of contact of saddle with shell (Figure 4-

1), degrees, radians; rotational displacementof tower (Figure 4-26), degrees

\ = (t/bxE/ocil used in Equation 4-18, dimension-less

p : radius of gyration : (I/Af .5

6 = general term for stress, psio" : allowable stress values (Table 4-3) psid. : allowable stress induced on concrete (Equation

4-40), psi; general tern for compressive stress(Equation 4-16), psi

ogp = critical stress in a flat plate defined in Equa-tion 4-15, psi


: elastic buckling stress (Equation 4-16), psi;28-day ultimate compressive strength of con-crete (Thble 4-7), psi

: stress due to weight, lbr: pressure stress induced by either internal or

external pressure, psi; longitudinal stress inEquation 4-67 , psi

: tensile stress in steel, psi: stress induced by wind or earthquake response

spectra, psi: minimum yield stress for a ductile material,

psiz = Poisson ratio for a given material, dimension-

less

d : concrete bearing parameter (Equation 4-20),dimensionless

REFERENCES

l. ASME Boiler and Pressure Vessel Code, SectionVIII Division I , American Society of MechanicalEngineers, New York.

2. Zick, L. P., "Stresses in Large Horizontal Cylindri-cal Pressure Vessels on Two Saddle Supports,"Welding Research Journal Stpplement, 1971.

3. Brownell, L. E. and Young, E. H., Process Equip-ment Design, John Wiley and Sons, New York,1959.

4. U.S. Steel, Steel Design Manual, U.S. Steel, Pitts-burgh, Pennsylvania, 1981.

5. American Institute of Steel Construction, Manual ofSteel Construction, Eighth Edition, AISC, Chicago,Illinois. 1980.

6. Timoshenko, 5., Theory of Plates and Shells, Mc-Graw-Hill Book Co., New York, 1959.

7. Bickford, J. H., An Introduction to the Design andBehavior of Bohed Joints, Marcel Dekker, Inc.,New York, 1981.

8. Faires, Y. M., Design of Machine Elements, TheMacmillan Co.. New York. 1962.

9. Simiu, E. and Scanlan, R. H., Wind Effects onStuctures, John Wiley and Sons, New York, 1978.

International Conference of Building Officials, Unl-form Building Code, Whittier, California, 1982.American National Standards Institute, Inc., "ANSIA58.1-Minimum Design Loads for Buildings andOther Structures- 1982," New York.

12. Kuethe, A. M. and Schetzer, J. D., Foundations ofAerodynamics, John Wiley and Sons, New York,1959.

13. Blevins, R. D., Flow-Induced Vibration, Van Nos-trand Rheinhold Co., New York, 1977.

14. Macdonald, A. J., Wind Inading on Buildings, Ap-plied Science Publishers, Ltd., London, England,1980.

15. Staley, C. M. and Graven, G. G., The Static andDynamic Wind Design of Steel Stacks, ASME 72-Pet-30, New York.

16. Vierck, R. K., Vibration Analysis, Harper and Row,New York, 1979.

' 17 . Paz, M., Structural Dynamics, Van NostrandRheinhold Co. New York, 1980.

18. Australian Standard 1170, Part 2-1983 SAA Load-ing Code, Part 2-Wind Forces, p. 55.

19. Timoshenko, S., Young, D. H., Weaver, W., Vibra-tion Problems In Engineering, John Wiley and Sons,New York, 1974.

20. Higdon, A., Olsen, E. H., Stiles, W B., Weese, J.A., and Riley, W. F., Mechanics of Materials, JohnWiley and Sons, New York, 1976.

21. Mitchell, Warren W., "Determination of the Periodof Vibration of MultiDiameter Columns by theMethod Used on Rayleigh's Principle," an unpub-lished work prepared for the Engineering Depart-ment of the Standard Oil Company of California.San Francisco, California, 1962.

22. Bedna\ H. H., Pressure Vessel Design Handbook,Van Nostrand Rheinhold Co.. New York. 1981.

23. Boardman, H. C.. "Stresses at Junction ofCone andCylinder in Thnks With Cone Bottoms or Ends,"Pressure Vessel and Piping Design, coTlected, pa-pers, ASME, N.Y., 1960.

24. Weaver, William, Jr., "Wind-Induced Vibrations inAntenna Members," American Society of Civil En-gineers, Paper No.3336, Yol. 127, Part 1, N.Y..N.Y., 1962.

10.

11.oE

oP

o.ow

oy

PARTIAL VOLUiIE OF A CYLINDER

v" : RiL |to' _ ,inol - panial volume shown (A-l)' 2 \180' rl In snaoeo regron

L : length of cylinder

R : inside radius of cylinder

Examplg lFigure A.tl

For a cylinder with 144-in. ID find the partial volumeof a fluid head of 60 in., if L : 100 ft:

|= w.+r

v, : (721!zoo) ["tlggrsri - sin (160.81")l' 2 [ r80 I

Yp : 7,707 ,650.2 in.3 : 33,366.5 gal

PARTIAL VOLUIIE OF AHEIIISPHERICAL HEAD

,, _ rry':(3Ri -D)rP - ------------- (A-2)J

V. = partial volume shown in shaded region

Example

For vertical volume in Figure A-2a find partial volumefor a head with Ri : 50 in. and y = 35 in.:

Appendix A

Partial Volumes and Pressure Vessel

Calculations

Figure A.l. Sketch for calculating partial volume of a cylin-der.

r(3sft3(50) - l00l : 64,140.85 in.3 : 277.7 ga|

Example

For horizontal volume in Figure A-2b find partial vol-ume for a head with Rr : 50 in. and y : 35 in.'

277 .7 = 138.85 gal

255

J- __ _-/.\I --x^L?-q

' a i t --


Figure A-2. Partial volume of vertical hemispherical head.(8) Partial volume of horizonral hemispherical head.

PARTIAL VOLUMES OF SPHERICALLYDISHED HEADS

Horlzontal Head

The partial volume of a horizontal head (Figure A-3) is

v="lJGt:lT-{p-v-F ryl (A-l)

Vertical Head

The partial volume of a vertical head (Figure A-4) is

-. ?rv(3x2 + v2)v : --:--:----------6

or

,, rry2(3o - v)3

(a) Example-Spherically Dished HorizontalHead

A spherically dished head with a 114-in. @ OD is spunfrom l-in. plate. Determine the partial volume of l0 in.of liquid. From vessel head manufacturer's catalog wedetermine the following:

IDD : 16.786 in. (Figure A-5), p : 193 1n.

^ ll4 - 2(1.0)K, =

-=-:

)O.U ln.z

a : 159.43" : 2.78

L : 108 - 16.786:91.2r in.

-_r--T-lY' IllI ln,lv Itlt?

Figure A-3. Partial volume of spherically dished horizontalneaqs.

(A-4)

Figure A-4. Partial volume of spherically dishedheads.

--.-{-} -- -

(A-5) vertical

n

(91.21)(562 - 6.7862)

V : 38,893.21 in.3 = 168.37 gal

Example- Spherically Dished VerticalHead

For the same head above, determine the partial volumeof a head of liquid of 9 in.

x : 55.456 in.

u_r(9)[3(55.a56f+g'z] : 14,874 in.3 : 64.4 gal

PABTIAL VOLUTES OF ELLIPTICAL HEADS

The exact partial volume of a horizontal elliptical head(Figure 4-6) is as follows:

u = (I93)'-(Rl - n1i (4-6)

6RI

Vertlcal Elliptical Heads

1..,fi082 --61s6-,P - J(lo-s: - 5FFr=\:.,O/l-

Volume of top portion @ of Figure A-7 is

v^: oR't[" - t'I- 2 r 3GDD),1

Volume of bottom portion O is

. . 2r(tDD)R''? rRl I uj Iv^= ' ' - _______: lv2 l' 3(IDD)'?j

Appendix A: Pressure ry'essel Formulations

End View of Horizontal Head

Flgure A-6. Partial volume of horizontal elliptical

Yi = 6.786"

Figure A-5.

(A-7)

(A-8)Figure A.7. Partial volume of vertical elliptical head.


Horizontal Head Exampte

Find the partial volume of a 2:1 (Ri/IDD = 2) ellipti-cal head that is 108-in. OD. The level of the liquid ii 35in.. and the head is spun from l-in. plate.

Avertical head

IDD: 108 - 2(1.0) : 26.50 in.

From Equation ,4-6 and Figure A-8 we have the follow_rng:

., (IDD) a r--t : 6R v{K,' - yi.f

a:138.80":2.42

v _ (19.0)12.42t .,?r s _ / rEtrl

6(53)

V : 17,512.94 in.3 : 75.81 gal

Vertical Head Example

For some head above, determine the partial volume fora vertical head with 19 in. of liquid. Using Equation A-gwe have the following:

v : 2r(lDD)R,2 _ "n, [, _ y,, I6 2I'' 3(rDD4

,, - 2rQ6.50)(53.U2 zrt53.0) [- -z-'lteo-

V = 77,951.81 in.3 - 1310.75 in.3

V : 76,641.06 in.3 = 331.78 gal

IDD

-X

Bhorizontal head

cvertical knuckle region

H=IDO-KR

Dhorizontal knuckle region

Figure A-9. Partial volumes of torispherical heads: (A) verti-cal, (B) horizontal, (C) vertical knuckle region, (D) horizontalknuckle resion.Figure A-8.

Vr : knuckle volumeVo : dish volumeKR : knuckle radius

y : height of liquidIDD : inside depth of dish

p : inside dish radius

PARTIAL VOLUilES OFTORISPHERICAL HEADS

For Figures A-9 and A-10,

Appendix A: Pressure vessel Formulations

Figule 4F10.

end view of dishvolume

Figure A-11. Sketch for example partial volume calculationof horizontal torispherical head.

Figure A-12.

For vertical heads (Figure A-9c) the knuckle-cylinderpartial volume is

v-: ? <t, + 4rM2 + ri2)

The partial volume of the dish region of a vertical head is

., _ ?ry(3x2 + y2)vD - 6-

The total partial volume in a vertical head is

v": oH (r^2 + 4ru2 + 12) + ?rY(3x'z + Y'z)

' 6'" '" " 6

wherey:IDD-KR

Horizontal Todspherical Hcad$

Partial Volume of Dish O (Figure A-l l)

Vo:o {F:1tr - vG, - R-5 _ L(&, - yf ) (A-12)

A-12)

(A- 13)

Volume of Knuck-Cylinder Region @ (Figure

The total partial volume for a horizontal torisphericalhead is as follows:

V1 : V6+ V6

- \GI:TF -

vo = *FI9 + Rr - KR) * (*, - K*)'l

L(&'? - yf)2

(A-e)

(A-10)

(A-11)

(A-14)* "[# + Gi - KR) + (& - KR),]

wherel: p _ IDD


Horlzontal Head Example

A 102-in. @ OD flanged and dished (torispherical)head made to ASME specifications (KR ) 0.60p andKR > 3th, tr, = head thickness) is spun from l-in. plate.The head is horizontal and the liquid level is 35-in. deter-mine the partial volume.

From the vessel head manufacturer's catalog and Fig-ure A-12 we determine the following:

p = 96 in., KR : 6.125 in.. IDD : 17.562 in.

tooR, =

= : 50in., L = 96.0 - 17.562 = 78.4J8 in.

z

From Equation A-14 we have

vr = Q.532)vaq6t--rsry - uOai- tcl

3

The head is vertical and the liquid level is 18-in. Deter-mine the partial volume.

From the vessel head manufacturer's catalog we deter-mine the following:

p : 132 in., KR = 3 in.,IDD = 20.283 in.

llR - trl 5lR, = '-- =-"

-', = 67.50 in.;2.

x = 67.50 - (31 - H2lo5 : 66.446 in.

For kluckle-cylinder region,

r,, = Rr :61.50;ri = Ri - KR : 67.50 - 3.00 : 64.50in.

67.50 + 64.50rm=-=ob.ul

h = 120.283 - (3.0 + 15.0)l :2.283 in.

-() )9,11vv = " -l-'' l(67.50), + 4(66.0)2 + (64.5011

o

* z(17.283)[3(64.500)'? + (17.283)'?]

6

Yv = 31,247.726 in.r + 115,645.832 in.3

Vv = 146,893.558 in.r : 635.903 gal

(78.438)(50' - 15')

+ (5o.oo - 6.12s) + (50.00 - 6.125f1)

Vr = 14.091.,14 in.r = 147.59 ga.

Vertical Head Example

A 138-in. d OD F&D (flanged and dished) head normade to ASME specifications is spun from 1llz-in. plate.

f, /.) < r1, 14(6.125)

T JT

Appendix A: Pressure Vessel Formulations

INTERNAL PBESSURE ASIIE FORIIULATIOI{SwtrH ouTsrDE DlllENslol{s

Cylindrical ShellLongitudinal Joint

i= PRoE + 0.4P

PB"'- 2'E + 1AP

t= PDo

2oE + 1-BP

t= PRo

2dE + O-8P

| _ 0.885P1'-;E+o-sP

. PLM

Circumferential Joint

2:l ElliDsoidal Head

D_ oEt'-R-O3t

^ 2oElRo - 1.4t

^ 2oElD. - 1.8r

o_ 2oEl

^ qEt

0.885L - 0.8t

Sphere and Hemispherical Head

ASME Flanged and Dished Headwhen UR = 16qh

When UB < 161b

2oE+P(M-0.2)^ 2oEt' ML-(M -0.2)

Conical Section

r= PDo- 2 cos o(oE + 0.4P)

^ 2SEl cos aY=-Do - 0.8t cos c


INTERI|IAL PRESSURE ASME FORMULATIONSWITH INSIDE DIMENSIONS

1-\i-lt-----Ti -'------t l'-

t= PRi

oE - 0.6P

t= PRi

2oE + O.4P

Cylindrical ShellLongiludinal Joint

Circumferential Joint

2:1 Ellipsoidal Head

I'ti + u.bt

' -F;- o.4t

^ 2oElOr + 0.2t

^ 2oElR + 0.2t

0.885L + 0.lt

^ 2oEtLM + 0.2t

Sohere and Hemisoherical Head

ASME Flanged and Dished Headwhen UR = 16E3

l-_,

sE - 0.1P

t= '-"'2oE - O.2P

When UR < 16?e

FOR VALUES OF M SEE SUPPLEMENT

Conical

t= PDi

2 cos d(oE - 0.6P)

Section

^ 2oEt cos ao_Di + I.2t cos a

aAppendix A: Pressure Vessel Formulations

Supplement for ASME Formulations

1. For a cylindrical shell, when the wall thickness exceeds onehalf the inside radius or P > 0.385dE, the lormulas in ASMECode ADDendix 1-2 shall be used.For hemispherical hsads without a straight flange, the effi-ciency of ihe head-to-shell joinl is to be used if it is less thanthe efficiency oI lhe seams in the head.For elliDsoidal heads, where the ralio ol lhe major axis isother than 2:1, reler to ASME Code Appendix 1-4(c).To use the lormulations tor a conical section in the table, thehalf apex angle, €r, shall not exceed 30". lf d > 30o, then aspecial analysis is required per ASME Code Appendix1-5(e).For an ASME flanged and dished head (torispherical head)when Ur< 1643 the lollowing values ol M shall be used:

263

4.

M= 1

/ fL\oit.!;/

i

I

I

Values ot Factor M

UTMUrM

1.001 .007.OO1.41

1.251.037.501.44

1.501.068.001.46

1.751 .088.501.48

2.001 .109.001.50

2.251.139.501.52

2.501.1510.01.54

2.751.1710.5t.co

3.001 .1811.0'1.58

3.251.201 1.51.60

3.501.22'12.0

1.62

4.001.2513.01.65

4.501.2a14.01.69

5.001.31

1.72

5.501.3416.01.75

6.00't.36164s1.77

6.501.39

. The maximum allowed ratio: L-t= D. When Ur > 16?3 (non-ASME Code construction), the values of M may be calculated by

xrl -@

A standard is a collection of current practices, past ex-periences, and research knowledge. Standards that aredeveloped by consensus groups (e.g., ASTM, ANSI),trade associations (e.g., AISC, ACI), or governmentgroups (e.g., HUD, CPSC) carry more authority thanother standards because they reflect wider ranges of ma-terials.

The ANSI A58.1-1982 is a collection of informationthat is considered to be the state-of-the-art in the designof buildings and other structures. Local and regionalbuilding codes adopt portions of the ANSI standard fortheir own use. These local and regional codes are devel-oped to represent the needs and interests of their respec-tive areas and are written in legal language to be incor-porated into state and local laws. Because these buildingcodes are regional or local in scope, they often do notinclude everything in the ANSI standard, which is na-tional in perspective. For this reason, one must be cer-tain that a local code written for one area is applicable tothe site being considered.

The ANSI standard does not have as much authority asthe ASME vessel codes, and, unfortunately, does nothave a referral committee or group to officially interpretthe document. Therefore, one must make decisionsbased on past experience and accepted methods of de-sign. The ANSI standard (Paragraph 6.6, p. 16) statesthat in determining the value for the gust response factora rational analysis can be used. A note below the-para-graph states that one such procedure for determining thegust response factor is in the standard's appendix. Thenote at the top ofthe appendix (p. 52) states clearly that itis not a part of the ANSI 458.1 mirninum design stan-dard. What all this implies is that one may follow theguide of the ANSI standard's appendix or use another ra-tional analysis, which includes another wind standard.Thus, one caz use another standard for design purposes.

Appendix B

National Wind Design Standards

One of the most widely accepted international standardsis the Australian Standard 1170. Part 2-1983. SAALoading Code Part 2-Wind Forces.

The Australian Standard I 170 is more applicable to theprocess industries because in it are shape factors forgeometries that are more common in that industry, e.g.,circular shapes. However, before applying the shape fac-tors of the Australian standard to the ANSI or any othernational standard, one must be very careful to correctlyconvert the factors. This is because the codes have dif-ferent basis upon which these factors are deiermined,and a direct application of other parameters is not possible. This is discussed later after we discuss the basis forthe various standards.

CRITERIA FOR DETERMINING WINDSPEED

Wind is caused by differential heating of air masses bythe sun. These masses of air at approximately one mileabove the ground circulate air around their centers ofpressure. At this altitude, the velocity and direction ofthe wind is almost entirely determined by macro-scaleforces caused by large scale weather systems. Below thisgradient height, the wind is modified by surface rough-ness, which reduces its velocity and changes its directionand turbulence. A secondary criterion, except for ex-treme wind conditions, is the temperature gradient,which affects the vertical mobility of turbulent eddiesand therefore influences the surface velocity and the gra-dient height. Therefore, the exact nature of the surfacewind at any point depends, first, on the general weathersituation, which determines the gradient wind and thetemperature gradient, and, second, on the surroundingtopography and ground roughness which, together with

265


the temperature gradient, modify the gradient wind tothe surface wind.

Wind motion is further complicated by the rotation ofthe earth, which induces additional forces that cause theair moving across the earth's surface to be subjected to afbrce at right angles to the wind velocity vector. Theseadditional forces are known as Coriolis forces.

Each country has adopted its own standard for measur-ing wind velocity. The U.S. National Weather Serviceand U.S. codes use the fastest-mile wind sDeed. which isdefined as the average speed ofone mile of air passing ananemometer. Thus, a fastest-mile wind speed of 120 mphmeans that a "mile" of wind passed the anemometer dur-ing a 30-second period. Other nations, namely Australiaand Great Britain, use the two-second gust speed. This isbased on the worst 2-second mean as measured by a cupanemometer. The mean gust speeds are recorded over aperiod of time such that a mean recurrence interval is de-termined. The mean recurrence interval is the reciprocalof the probability of exceeding a wind speed of a givenmagnitude at a particular location in one year. The risk,or probability, R, that the design wind speed will beequaled or surpassed at least once in the life of the toweris given by the expression

R:l-(l-P,)"

where P, : annual probability of exceedance (reciprocalof the mean recurrence interval)

n : life of the tower or stack

The risk that a given wind speed of specified magni-tude will be equaled or exceeded increases with the pe-riod of time that the tower is exposed to the wind. Valuesof risk of exceeding design wind speed for a designatedannual probability and a given design life ofthe structureare shown in Table B-1.

For example, if the design wind speed for a tower isbased on an annual probability of 0.02 (mean recurrenceinterval of 50 years) and the projected tower life is 25years, there is a 0.40 probability that the design wind

Table B-1Probability of Exceeding Wind Design Speed

Pr = 1-(1 - Po)*

AnnualProbability Design Lile of Structure in N Years

PAI0.10 0. 1000.05 0.0500.01 0.0100.005 0.005

Figure B-1. Cup generator anemometer.

speed will be exceeded during the life of the structure.The United States and Australian wind codes use the 50-year recurrence interval.

The instrument for measuring the wind in the UnitedStates, Great Britain, and Australia is the cup-generatoranemometer shown in Figure B-1. This device is oper-ated by the wind striking the cups, which drive a smallpermanent alternator. The indicator, which incorporatesa rectifier, is simply a voltmeter calibrated in miles perhour. In most recent cup-generator models the generatoroutput is used to activate a pen-chart recorder which pro-vides a record of continuous wind soeed.

WIND SPEED RELATIONSHIPS

As stated previously, another method can be substi-tuted for the appendix in ANSI A58. l. What this meansis that another code could be used instead of the appen-dix. To do this one must be careful to utilize the correctconversion factors between standards. To accomplishthis we refer to Figure B-2. For a 100-mph fastest milewind speed in ANSI 458.1 we wish to determine theequivalent fastest mile wind speed for a 2-second gustusing either the Australian or British code. From FigureB-2 we read from the ordinate 1.54 fior 2 sec. Knowinsthat one mile ol wind moving at 100 mph will pass thianemometer in 36 sec, we read 36 sec on the curve andarrive at V,/V3666 = 1.30. Thus, the equivalent fastestmile wind speed is

/r sa\V - t;:^lrl00) mph : tt8.4 mph\1.30i

for a 2-sec gust. For I l0 mph, the values becomes

V : (1.18)(ll0) mph : 129.8 mph

5 l0 15

0.410 0.651 0.7940.226 0.40t 0.5370.049 0.096 0.1400.025 0.049 0.072

25 50 100

0.928 0.995 0.9990.'723 0.923 0.9940.222 0.395 0.634o.tt8 0.222 0.394

a

Appendix B: National Wind Design Standards

Figure B-2, Ratio of probable maximum wind speed averaged over t seconds to hourly mean speed.

Thus, the 1.18 factor would have to be used in the 2-secgust code if that code were to be substituted for Appen-dix A of ANSI A58.1-1982.

Similarly, the Canadian code we must convert to ob-tain an equivalent fastest mile wind speed from the meanhourly. The mean hourly implies that the wind moves an

average of 100 mph across the anemometer in a period of3,600 sec. Reading Figure B-2 we have V'/Vru* = 1.6.

Thus

lj: ozor

which yields an equivalent velocity of 76.9 mph. Withthe Canadian code one must use 0.769 in use of shape

constants and the various other parameters when usingwith ANSI A58.1. A comparison of the major windcodes is given in Thbles B-Z, B-3, B-4, and B-5.

ANS| A58.r-1982 WIND CATEGORIES

In the ANSI A58.1-1982 there are four wind catego-

ries-A, B, C, and D. The categories are described as

follows:

Category A-A very restricted category in which thewind speed is drastically reduced. Most petrochemicaland power facilities do not fall within this category.The wind force is reduced because the structure is con-sidered to be among many tall structures. One exam-ple would be a ten-story building in downtown Man-hattan, New York, where the taller buildings wouldblock the stronger air currents.

Category B-A classification that encompasses some tallstructures, but not enough to block the majority ofwind gusts. An example of this category would be a

tower in the midst of a large petrochemical facilitywhere there were other towers that would block some

of the wind force. A forest surrounding a tower is an-

other example.Category C-The most common classification for petro-

chemical applications. This category is open terralnwhere the tower would receive full impact from thewind with minimum ground resistance to the wind. Anexample of this category would be an open field or an

alrDort.Category D-A classification for wind moving over wa-

ter. A beachhead, in which there is flat beach up to arow of buildings would be in Category D. Miamibeach, from the ocean front up to the facade of hotels,is a good example. Behind the hotel fronts would be

Category C. Another example of this classificationwould be a tall vertical vessel on an offshore structure.


Table B-2Malor U.S. and Foreign Building Codes and Standards Used in Wind Design

Code or Standard Edition Organization AddressAustralian Standard I 170,Part 2-Wind Forces

British Code of BasicData for Design of Buildings(cP3)Wind Loading Handbook(commentary on CP3)National Building Codeofcanada (NRCC No. 17303)The Supplement to theNational Building Code ofCanada (NRCC 17724)

ANSI A58.1,1982

Uniform Building Code

Standard Building Code

Basic Building Code

Standards Associationof Australia

British StandardsInstitution

Building ResearchEstablishment

National ResearchCouncil of Canada

National ResearchCouncil of Canada

American NationalStandards InstituteInternational Conferenceof Building OfficialsSouthern Building CodeCongress International

Building Officials andCode AdministratorsInternational, Inc.

Standards House80 Arthur Street/North Sydney,N.S.W. AustraliaBritish Standards Institution2 Park StreetLondon, WIA 285, EnglandBuilding Research StationGarston, Watford, WD2 7JR, EnglandNational Research Council ofCanada

Ottawa, Ontario K1A OR6Canada

1430 BroadwayNew York, New York 10018

5360 South Workman Mill RoadWhittier, California 90601900 Montclair RoadBirmingham, Alabama 35213

17926 South Halsted StreetHomewood, Illinois 60430

1983

1972

1974

1980

1980

1982

1982

1982with1983 rev.

1984

Table B-3Reference Wind Speed

Australian British Canadian United StatesFeference (SAA, 1983) (BSl, 1982) (NRCC, 1980) (ANS|, 19s2)Averaging time

Equivalent referencewind speed to fastestmile 100 mph

2-3 secondgust speed

118.4

Mean hourly

76.9

Fastest mile

100

2-secondgust speed

I18.4

'l'*"1iil

Appendix B: National Wind Design Standards

Table B-4Parameters Used in the Maior National Standards

Australian British Canadian United Siates1982)Parametel

Wind SpeedTerrain roughnessl,ocal terrainHeight variationRef. speed

Wind PressurePressure coefficients

GustsMagnitudeSpatial correlation

Gust frequency

Analysis procedure

2-sec gusts

Tbbles inappendix includesfigures

Gust speedReduction forlarge areaDynamic considerationfor h/b > 5

This standard is consid-ered by many the bestfor use in the processindustries. Figures andtables are easy to read.The standard actuallyprovides the user withequations to cutves.The analysis procedureis straight-forward.

2-sec gusts

Thbles, includesfigures

Dynamicconsiderationnot included

Overall a very goodcode, its weakest partis the lack of dynamicconsideration .

3NoneYes

Mean hourly

Figures andtables incommentaries

Dynamicconsiderationfor h/b > 4in. or forh > 400 ftAn excellent windstandard. Theanalysis procedureis straight-forwardand the docu-ments-code andsupplement con-tain tables and fig-ures easy to read.

4NoneYes

Fastest mile

Thbles, figuresand notes

Dynamic considerationfor h/b > 5

Although the appendixis technically not con-sidered a part of thestandard, it containsfigures difficult to read,namely Figure 6. Formany structures thedata extend beyond thelimits of the curves inFigures 6 and 7. In themethod in the appendix,one must assume an ini-tial natural frequency,resulting in an iterativeprocess. This method isextremely difficult indesigning petrochemicaltowers without the useof a computer.

,1

4YesYes

4YesYes

Gust speedNone

Gust effect factor Gust response factorGust effect factor Area averaging


Table B-5Limitalions of Codes and Standards

Code or Standard Statement of Limitation LocationAustralian StandardI 170, Part 2- 1983

National BuildingCode of Canada(NRCC, 1980)

British CP3

United StatesANSI A58.I

Uniform BuildingCode

Basic BuildingCode (BOCA, 1984)

Standard BuildingCode, 1982 (SBCCI,t982)

"Minimum Design Loadson Structures""...EssentiallyaSetof Minimum Regulations . . ."

". . . Does Not Apply toBuildings. . . That AreofUnusual Shape or LocationFor Which Special InvestigationsMay Be Necessary . . .""Minimum Design Loads . . .""Specific Guidelines Are GivenFor. . . Wind Tunnel Investigations. .. For Buildings. .. HavingIrregularShapes...""The purpose . . . is to provideminimumstandards...""The Basic Minimum Wind SpeedsAre Shown in Figure 912.1 . . .""The Purpose of This Code is toProvide Minimum Requirements . . .""The Building Official May RequireEvidence to Support the DesignPressures Used in the Designof Structures Not Included inThis Section."

Title

Guide to theUse of the Code

Section 1 (Scope)

TitleParagraph 6. I

Section 102

Section 912.1

PrefaceArticle 1205.2(a)

Appendix C

Properties of Pipe

272 Mechanical Design ol Process Systems

t weighl ot pipe per toor (pouDds)weight ol lPcler p€r foot (pou!ds) =squqre leet ou&id€ ludoco per loot :squorc leet inside surlace F€r loot =inside drea (squdre hches)dred o{ rnetcd (squore irches)moment ol ir6rtia (inches.)

PROPERTIES OF PIPE +

Th6 tollowilg tormulqs dre used in lh€ computorior of th6 volues,bo\|'n in lhe toble:

i Tlr€ lsEilic ste6ls rnay be sbout 5% les!, dDd the crEte.itic stdin_legs sleels qbout 27o greate! thon the values shown in tbiE tqblewhich ore bcsed ort weigbts lor cdrboIt steel.

* achedule numben

Stordord weight pipe qnd schedule rlo qte the scrae in oll sizsslhrough lo-isdr; lrom lz-ilch thtough 24-irch, stqndard {eight pip6has d croll thicloess oI %-ircb.

Extro BtroDg woight pipe ond schedule gO dla the sdEe in oll sireslhrough 8-i[ch, llom 8-inch thlough 24-trch. ertrd strong weightpipe bos a wall thicloess oI ){-incb.

Double €nr(l 3troDg weight pip6 bss no c-orrespodding scheduleauEbe!,

o: ANSI836.10 st€el pipe schedule Dub.b€rs

b: ANSI 836.10 steel pip€ DoDilrol woll thichress dosiglqtio!

c: ANSI 836.19 6tdiDle3s steel pipe schedule uuEbols

10.6802(D-00.340sd,0.2518D0.26r0d0.785d,0,785(Dr-d?)0.049r(D.-d)AÊ o'

saction moduluB (inches3)

lodius oI gyrqtion (i!ches)

= 0.0982(D.-ci.)D

= 0.25t/D,'D,+--

An = oted of Eetql (square i4ches)d = inside didoeter (incb€s)

D = outside diqrn€ter (incheB)

R, = radius ol gFcrion {iiche3)t = pip€ woU thickness (inches)

nordnolpiF .izeou|lide

in

achedule wcllthick.

in.

i!3idedioEr-

in"

inside

rq. rn

metdl

sq. in

Bq lioutaid6surtdce,Per lt

sq ttinlideaurldce,po! tl

stoightF!Il,Ib*

Feiqhto[ wcterpsr It.

moD€ttlOI

inertic,

aoctionEodu-lu&

rardiurgyrc-lior|,iEb

%0.405

4080

srdxs

I0s40s80s

0.01s0.0680,0s5

0.307

0.269

0.2I5

0.07400.05680.036{

0.05480.07200.092s

0.r060.1060.106

0.08040.07050.0563

0.1860.2150.315

0.0321

0.02160.0157

0.0m880.001080.00t22

0.004370.0052s0.00600

0.127t0.12150.llt6

%0.540

40

80srdxs.

l0s40s80s

0.06s0.088

0.119

0.410

0.364

0.302

0.13200.1041

0.0716

0.09700.12500.1574

0.141

0.111

0.111

0.10730.09550.0794

0.3300.4250J35

o,05120.0451

0.0310

0.002790.00331

0.003?8

0.010320.012300.0139s

0.16940.15280.1547

%0.675 40

80

;;;xs

ssl0s40s80s

0.0650.0650.0st0.126

0.7I00.545

0.4930-423

0.3960.23330.19100.1405

0.15820.12{60.16t00.2173

0-2200.t770.r770,t77

0.1859o.t4270.12950.1106

0.5380.4230.5680.739

0.I7160.1011

0.08270.0609

0.011970.005850.007300.00862

0.02850.017370.021600.02s54

0.27500.21690.20900.199r

%0.840

4080

160

;;xs

n(s

r0s40s80s

0.0650.0830.10s0.1470.1870.294

0.710

0.6740.8220.sd60.466

0,252

0.39590.357

0.3040.23400.1706

0.0499

0.15830,19740.25030.3200.3830.504

0.220o.220

0.2200.2200.2200.220

0.18530.17650.16280.14330.t2200.0660

0.5380.571

0,851

r.0€81.304

1.714

0.171

0.15470.13160.10130.07400.0216

0.01200.0I4310.0r7100.020100-022t30.02125

0.02850.m4I0.04070.01780.0s270.0s77

0.27S0

0-28920.2613

0.25050.24020.2rs2

i.05010

s0160

;;;xs

xxs

l0s10s80s

0.06s0.083

0.1130.1540.2180.308

0.9200.8840-821

0-7 42

0.6140,434

0.6550.6140.s330-1320.2961

0.1d79

0.20u0-23210.3330.4350.5700.718

o.2750-275o.2750.2750.2750,215

0.24090.23140.2t570.19130.16070.1137

0,6840.857l.r3l1.4141.937

2.441

o.2a820.2661

0.2301

0.r8750.12840.0641

0.024510.029700.03?00.04480.os270.0s79

0.046?0.0s660.07060.08s30.10040.1104

0.3{90.3430.3340.321

0.3040.28{0

IJ.3t5

40

80

t60xxs

10s40s80s

0.065

0.1090.133

0.179

0.2500.358

1.185

1.097

I.0490.s570.815

0.599

1.t030.9150.86{0.t190.5220.28r8

0,4130.4940.6390.8361.076

0.3{40.3440.344

0.3{40.344o.314

0.3100.2872o,27460.25200.213{0.r570

0.8681.401

1.679

2.t722.4443.659

0.4780.409o,37 4

0.31to.22610.1221

0.0500

0.07570.087{0.10560.t2s20.140s

0,07600.ll5r0.13290.16060.19000.2t37

0.4430.428o.1210.1070.3870.36t

r%I.66'0

40

80

160

l';

*ts

r0s40s80s

0.06s0.109

0.1400.r910.2500.382

1.530

t.1421.380

l27A1.160

0.896

1.839

1.633

1.496

1.283

r,0570.631

0.326u.531

0.6690.88r1.I071,534

0.4340.4340.434

0.{340.43{0.434

0.{010.3780.361

0.3350.3040.2345

1.107

1.805

2.2732.9913.7655.2t1

0.7970.7tl0.618

0.{58o.2r32

0.10380.r60s0.1s48o.24t80.28390.341

0.12500.1934

0.23160.29130.342

0.411

0.55{0.5s00.5400.s240.5060.472

rt41.900 r0s

0.0650.109

t.7701,682

2.161

2.2220-375

0.6130-497

0.4970.4630.440

1.274

2.08s1.067

0.9620.ts800.2469

0.16630.2599

0.6490.634

*Couftesr of ITT Gtinncll.

F

Appendix C: Properties of Pipe

PROPERTIES OI' PIPE (Continued)

noainailprpe !ir(outridediotreteriE"

.chedule!uEber'

trclltbicL-

i|r.

insidedicnr-oler,in-

i!!ide

Bq. i|r.

metclqted,aq. i|l.

6q ltoutridesurtdce,Frft

sq ltinaid€BUttdc€,per It

w6ightper It,lbt

weiEhlol wlterp€! lt,

olinertiq,inJ

modu-lus,in.r

!adiu!9l.rc_UoE

b

rhJ.900

{080

160

xs

xxs

40s80s

0.1450.2000.28r0.4000,5250.650

1.6r01.500

1.338

1.100

0.8s00.600

2,036t-7611.406

0.s500.5670.283

0.7ssr.068

1.4291.885

2-2872.551

0.{970.4970.{970.49?0.1970.497

0.421

0.3930.3500.288o.2230.157

2.7183.63r4.8596.4087.7108.678

0.8820.7650.6080-1120.2180.123

0.3100.39r0.4830.5680.61400.6340

0.3260.{120.5080.5980,64700.6670

0.6230.605

0.581

0.s190.52000.4980

2.3r5;;80

160

''''.

;;xs

xx!;

...'

l0s40s80s

0.0650.1090.1540.2180.3430,4360.5520.587

2.2152-ts72,0871.939

1.689

r.5031.251

1.001

3.963.653:362.9532.240t.774t-2290.187

o.4720.116r.075t.4172.1902.6563.I993.641

o.622o.822o,6220.822o.822o,6220.622o.622

0.5880.5650.541

0.508o.4420.3930.328

0-262

1.604

2.6383.5535.O22

7.1419.029

10.882

12.385

1.7I61.582

1.455

1.280

0.971

0.7690.5330.341

0.3t50.4990.666

0.868

I.1631.3I21.4421.5130

0.26520.1200.s610.73t0.9791.104

r.21401.2740

0.8170,8020.r810.7560-7290.7030.6t100.64d0

2%2.875

;;80

ta:

:..xs

)0(s

l0s40s80s

0.0830.1200,2030.2760,3750.5520.6?50.800

2.709

2.4692.323

2.t251.771t.525t.275

4,754.243.552.4641.825

t,276

0.1281.039

t.7012.2512.9154.034.6635.2t2

0.75s0.7530.7530.7530.7530.7530.7530.753

0.7090.6900.6460.60s

0.4510.3990.334

2.1753,531

5.7937.661

I0.0113.70

15.860

t7.729

2.49S

2.0761.837

1.535

1.087

0.7920.554

0.7100.9881.530

1.925

2.3532.872

3.08903.2250

0.4s40.687L0641.339

1.637

1.s98

2.14902.2430

0.9880.9750.9d,0.9240.8940.84{0.81400.7860

3

3.500

;i80

160

d;xs

xxs

-'

l0s10s80s

0.0830.1200.2160.3000.4370.600

o;1250.850

3.3343.2603.0682.9002.62A2.3002.0501.800

4.738.357.39

5.424.15

3.2992.543

0.891

t.2742.2283.024.21

3.17

7.073

0.9160.9160.9160.9160.9160.9160.9160.916

0.8730.8s30.8030.7590.6870.6020.5370.471

3.034.337.58

10.25

14.32

18.58

2t.48724.Os?

3.783.6r3.202,8642-3481.80t1.431

t.103

1.301

ta223.02

3.905.035.99

6.50106.8530

o,1441.011

t.1242-2282.8763,433.71503.9160

1.2081.195

1.164

r.1361.091

1.o17

1.0140

0.9840

3y2

*Un 4080

;;xs

xt(s

5S

r0s40s80s

0.0830.120o.2260.3180.636

3.8343.t603.5483.3642.728

ll.l09.898.895.845

r.0211,463

2.6803.686.721

t-017t-o47t-041t.0471.047

1.00{0.98{0.9290.8810.716

3.474.579.ll

12.51

22.450

5.0r4.814.283.852.530

1.960

2.7564.796.289.8480

0.9801.378

2.3513.t44.92d0

1.385L.312t.3371.3071.2100

4.5N

;;80

t20

t60

s;;XS

)c;

tGs

40s80s

0.0830.1200.18802370.3370.4370.5000.531

0.6740.8000.925

4.334{.2604.t244.8263.826

3.s003.4383.1522.9002.650

14.75

14.2S

13.35?

t2-73I1,5010.33

9.62r9.247.806.6025.513

1.152

2.5473.-17

{.{l

6.2836.62Lr09.294

t0.384

l.l781.178

1.178

1.178

1.178

t.1781.178

1.178

1.178

t.178].l78

1.135

l.ll51.082

1.054

r.0020.9490.9160.9000.8250.75S

0.694

8.56010.?9

l{.9818.96

2r.36022.51

27.5131.61335,318

6.{06.175.800

4.984.484.1604.O2

2.8642.391

2.8u3.965.85007.239.61

12.71t013.21

r5.2916.6610

17.7130

1.249

1.162

2.600.0

3.21

5.18s.67605.906.797.10507.8720

1.562

1.549

1.5250

1.510

t.4771.445

l-12501.416

t.3711,3380

1.3060

;;80

t20r60

-:.xs

,ots

5S

10s4os80s

-

0.1090.1340.2580.3750.5000.6250.7s00.8751.000

5,3455.2955.0474.8134.5634.3134.0633.8133.563

22.1122,0220.0r18.t916,35

14.61

t2s7I1.413

1.868

2.2454.306.lI7.959.70

t1.3412.880

l{.328

1.t56l.{s61.455

1.156

1.456

1.456

1.456

1.4s6

1.456

1.399

1.386

1.321

t.2601.195

Ll291.064

0.9980.933

?,77t1.8220-7427,0132.9838.55{3.8I0t'|.'134

9,739.53

t.897.09

s-824.9S1

1.232

6.958.43

I5.1720.6825.74

30.0

36.645039.11l0

2.4983.035.{51.439.25

10.80

!2.1013.175014.0610

r.929t.9201.878

1.839

1.799

!.760

1.6860

t.6s20


PROPERTIES OF PIPE (Conti:rued)

pipe size schedulewqllihick-

in,

irgidediam-

in.

inside

sq. rn.

metcl

sq, ia.

Bq ltoutside

pe! lt

sq ltinside

per lt

weightper It.lbt

weight

per lt,lb

ofinertiq,in.'

lu5,in.1

radiu3sYro-tion.in.b

6

6.625

40

80

120

160

sld

xxs

5S

t0s

40s

80s

0.109

0.134

0.2IS

r0.2800.4320.562

0.7t 80.864

1.000

1.125

6.4076.3s76.187

6.06s5.76I5.501

5.1894.897

4.8254.37S

32.231.730.r0028.8926.O7

23.77

21.t 518.83

18.192r5.025

2.2312-7334.4I05.588.40

10.70

r3.3315.64

17.662I9.429

1.734t.734t.7341.734

I.734t.7341.734

1.134t.134t.'t34

t.6771.664

1.620

1.588

I.508L4401.358

r.2s2l.2l I1.145

5.37

9.29

15.020

t8.s724.5',1

36.39

45.30

60.07666.084

r3.9813.74

r3.I00l2.sII1.2910.30

9.168.171.2846.SI7

1I.8514.40

22.660028.t440.549.65S.0

66.3

72.119076.5970

3.584.35

6.8400

8.50t2.2314.s8

17.81

20.0s21.1120

23.1244

2.304

2.2952.27002.2452.195

2.153

2.1042.060

2.0200

L98s0

88.625

20

30

40

60

80

std

xs

l0s

;;80s

0.109

0.148

0.219

0.2s00.211

0.3220.406

0.500

8.407

8.32S

8.187

8.125

8.07t7.981

7.813

7.625

55.554.5

52.63051.851.2

50.047.945.7

2.916

3.94

5.8006.58

7.268.40

I0.48t2.76

2.2582.2582.2582.25a2.2s82.2582.2582.258

2.2012.180

2.1502.1272.t132.0892.0451.996

s.9l13.40

1s.640

22.3824.70

28.5535.64

43.39

24.0723.59

22.90022.4822.t821.69

20.79

19.80

26.4535.4

5t.320057 -7

63.4

72.5

88.8

I0s.7

6.13a.2l

11.9000

13.3S

14.69

16.81

20.58

24.52

3.01

3.002.9700

2.5622,S53

2.938

2.909

2.879

I8.625

100

r20t40r60

0.593

0.7I80.8I20.906

1.000

1.125

7.4397.189

7.00I6.8136.625

6.375

43.540.6

36.5

34.454

31.903

14.96

t1.84ls.s32t.9723.94226.494

2.2582.2582.2582.2582.2582.2s8

1.948

I.882L833t.784I.7341.669

50.87

60.63

74.69

81.437

90.I1{

t8.8417.60

16.69

15.80

14.9{5I3.838

12t.4140.6

I53.8I65.9177.1320190.6210

28.t432.6

35.738.541.0740

44.2020

2.8472.847

2.7772.7482.',1t90

2.6810

l010.750

20

30

40

60

80

I00t20

t40160

stdxs

5S

I0s

4;;80s

0.134

0.16s0.219

0.2500.307

0.36S

0.500

0.5930.718

0.843

0.8751.000

l.I2sL2s01.500

to.48210.420

10.312

10.250

10.136

10.020

9.750s.5649.314

9.064

9.000

8.750

8.500

8.2507.75D

86.385.3

83.5282.580.7

78.97 4.7

7L8

64.5

63.62

60.1

56.753.4547.15

4.52

5.49'1.24

8.26I0.07ll.sl16.10

I8.9222.6326.24

27.!430.6

34.037.31

43.57

2.8152.815

2.815

2.815

2.815

2.8r52.815

2.8152.815

2.8152.815

2.815

2.815

2.815

2.8I5

2.7442.7242.102.6832.654

2.6232.5532.5042.438

2.3732.36

2.1912.2252.182.03

r5. ts18.70

24.6328.04

34.24

40.48

54.74

64.33

76.93

89.20

92.28.04.13

u5.65126.82

t48.I9

37.4

36.9

36.2

35.8

3S.0

34.I32.33l.l29.5

2e.027.626.1

24.623.220.5

63.7

100.46

Ir3.7137.5

160.8

2t2.0244.9288.2324333.46

399

424.t7478.59

11.8S

I4.30I8.692l.r6

29.9039.4

45.653.2

60.3

82.O4

68.4

?4.3

79.65

89.04

3.75

3.743.72

3.71

3.69

3.603.563.523.50

3.433.39

t2)2.750

;i30

40

;;80

100

120

140

;,;

10s

4;;

80s

0.I560.1800.2500.3300.375

0.406

0.5000.562

0.687

0.7s00.843

0.875

t.0001.125

L250r.312

12.438I2.390t2.2s0t2.09012.000

I1.938I1.75011.626

I1.376I1.25011.064

11.000

10.750

10.500

I0.25010.t26

t2t.4120.6

u7.9114.8

I l3.lIII.9I00.4106.2

101.6

99.4096.1

95.0090.886.682.5080.5

6.177.t I9.84

12.88

14.s8

15.74

19.24

2r.5226-O4

28.27

31.532.64

36.94I.l45.1647.1

3.343.34

3.34

3.34

3.34

3.34

3.343.34

3.343.34

3.34

3.34

3.34

3.343.34

3.34

3.24

3.21

3.t13.143.133.08

3.04

2.9782.94

2.897

2.AA

2.4142.7492.682.651

20.9924.20

3s.3843.7'l49.56

53.S3

65.4273.16

88.51

96.207.20

10.3

25.49

39.68

53.660.27

s2.752-2

{9.749.048.S

47.0

46.044.0

43.r41.64t.l3S.3

37.535.8

34.9

t22.2I40.5r91.9248.5219.3300

362401

475510.7

562s78.S

642701

75s.S781

19.20

22.93

30.r39.043.8

47.1

62.8

74.5

80.1

88.r90.7

100.7

109.9

r22.6

4.45t.444.42

4.3S

4.38

4.374.33

4.3r4.214.254.22

1-t74.13

4.091.01

Appendix C: Properties of Pipe 275

PROPERTIES OF PIPE (Continued)

noEit'alpipo rircoutsidedidr!€teri|r

.chedule|tumb€r'

wqlltbicL-

itr.

ilridediqra-

i!-

idside

aq, in

|rretol

aq. in

sq ttoutgidelurlcc€,per lt

rq Itiagideauddc6,perlt

lreight!'er ILtbt

weisht

pe! ll,lb

olinerlid,in.

aeclionmodu-lu&inJ

!adiu!qryr6-Uonia.b

l414.@o

l0

20

;;40

;;80

t00I20140160

t0s0,1560.r880.2100.219

0.250

0.281

0.3120.344

0.375

0,{370.469

0.s000.593

0.6250.750

0.9371.093

1.250

1.406

13.6S8

13.624

13.580

13.562

13.S00

13.438

13.3I213.250

13.r2513,062

13.000

12,814

12.750

12.500

12.t28ll.8r4rr.50011.188

t47.20145.80

144.80

111.50

143.1

141.80

140.5

t39.20137.9

r35.3134.00

t32,7r29.0t27.1t22.7

109.6

103.s

98.3

8.169.109.48

10.80

l2.llt3.4214.16

16.05

18.62

19.94

2t.2124,98

26.25

38.544.350.1

3.67

3.67

3.67

3.673.87

3.673.673.67

3.673.67

3,613.673.573.673.67

3.S8

3.57

3.55

3.553.533.S2

3.503.483.413.443.42

3.40

3.35

3.34

3.27

3.17

3.09

3.01

2.929

23.O

27.7

30.932.2

36.71

4t.245.68

50.254.5763.3767.872.09

84.91

8S.28

108,13

130.73

150.67

t70.22189.12

63.1

62.8

62.1

60.9

50.3

59.7

s8.7s8.057.5

55.3

50.047.545.042.6

194.6

216.2225-l255.4285.2

314

344.3

429156.8

484

589

687

825930

tt27l0l7

2't.830.932.2

36.5

40;I4{.9t9.2s3.361.2

69.1

80.3

84.1

98.2117.8

I32.8l{6.8159.6

{.904.881.574.574.864.854.914.834.824.801.794-744.744;13

4.694.634.584,534,48

l616.0@

io

30

40

80

r00120140

160

0.165

0.1880.250

0.312

0.3750.500

0.843

1.031

r.2181.437

15.670

15.624

r5.500r5.37615.250

15.000

14.688

r4,3t4r3.93813.564

13.126

12.814

192.90

19r.70188.7

185.7

182,6

116.7

169.4

160.9

152.5

144.5

135.3

129.0

a,2L

9.3{12.37

15.38

18.4I24.35

31.6

40.1

48.5s6.665.772.1

4.194.ls4.194.19{.I94.194.194.I94.194,19

4.19{.I9

4.104,094.064.033.993.93

3.853.753.653.55

3.44

3.35

28

32

42.05

52.36

645882.17

107.50

136.46

r64.83192,29

223.64

24S.ll

83.583.081.8

80.5

79.1

73.469.766,1

58.5

257

292384

473562

732933

lt5?r365l5s6I7601894

32.2

48,05S.2

70.39I.S

144.6

170,6

194.5

220.0

236,1

5.18s.13s.375.305.24

5.12

l8

r8.o00

i;20

30

40

6080

100

120

140

150

l0s0.165

0.1880.2s00.312

0.3?5

0.437

0.500

0.5620.750

0.937

l.ls61.37S

1.562

1,781

t7,670t7.62417,500

17.376

17r50t7.12617.00

r6,87615.500

16.126

15.688

15.250

14.876

r4.438

245.20243.90210-S

237.1

233-7

230.4

227.0

223.7

213.8204.2

193.3

182.6

173.8

t53.7

9.2410.52

r3.9411,34

20.76

24.t|27.49

30,8{0.650.261.2

71.8

80.7

90.7

4.7 L

4.71

4.71

1.?l1.?l1.71

4.71

4.7 |4.71

4.7 |

4.63{.614.58

4.554.524.48

4.45

4.424.324.224.tI3.993.89

3.78

31

47.3959.0370.59

82.06

93.45

104.?5

138.r7170.75

207.96244.t4214.23

308.51

I06.2105.7

104.3

102.8

101.2

99.9

s8,4

97.092.788.583.779.275.371.0

368

4t7549

678

807

93I1053

tt72

1834

2180

2499

27sO

3020

40.8

46.461.0

89.6

I03.4117.0

130.2

168.3

203.8242.2

217.6306

336

6.31

6.306.286.2S

6.23

6.2r

6.106.045.975.905.84

5,77

2020.000

i;20

30

40

60

80

100

;;xs

10s0.r880.218

0.250

0.5000.593

0.812

0.875

1.03r1.281

1s.634

I9.56419.500

1s.250

r9.000r8.8I4r8.37618.250

17.s38

17.438

302.40

300.60298.6291.0283.S

278.0265.2

252-7238.8

11.70

13.55

15.5r23.r230.6

36.248.952,861.475.3

s.245.24

5.245.24

5.24

5.24

5.24s.245.24

5.14

5.125.I I5.044,97

4.934.8r1.784.704.57

40

4652.73

78.50t04.I3r22.91r66.40178.73

208,87

13t,0130.2

t29.5126,0

t22.8t20.4lls.01t3.4109.{t03,4

574

757lll41457

1704

22s?24052772

3320

s't.456,3

75.7n t.4I45.7170.4

225.?

240.9277.2332

7.006.99

6.986.94

6.90

6.79



notlrindlpip6 .iz€

rn.

schedule wollthick-

idridsdi(rm-

i!L

idaide

sq in.

met(ll

aq in

6q Itoutsidesutldce,per lt

sq ltir16ide6urlcce,perlt

weightpo!It,tbf

rroight

pe! ll,lb

Ino|'lentolin€rti(r,ia..

aectionnodu-lus,

tcdiutgytq-tiorin.b

2020.000

r20I40160

1.500

1.750

1.968

17.000

16.500

I6.064

227.0213.8202-',l

47.2100.3

Iu.5

5.24s-245,24

4.451-324.21

296.37341.1037S.01

98.3

92.6

87.S

376042204590

376422459

6.S6

6,48

6.41

22

22.000

i;20

30

;;80

100

120I40r60

ioxs

t0s0.188

0.218

0.250

0.375

0.500

0.6250.7s00.875

1.125

I.375I.6251.87s

2.t25

2L.8242I.56421.500

2r.25021.00020.?s020.50020.25019.750

19.2s0

18.750

I8.25017,750

367.3365.2

363.1

354.7

346.4338.2330.1

322.1

306.4231.0276.1

261.6247.4

12.88

t4.9217.t825.48

33.774t.9750.0758.07

73.7889.09

104.02

1t8.55I32.68

5.765.76

5.?65.76

5.76

5.76

5.765.765.765.765.76

5.75

5.655.635.565_50

s.435.375.305.175.044.91

4.78

4.65

44

87

l15143

170

197

25r303

354

403

451

t59.1158.2

157.4

I53.7150.2

146.6

l{3.1r39.6132.8

t26.2u9.6113.3

t07.2

766

88S

l0l01490

1953

2400242932454029

475S

6054

69.7

80.491.8

135.4

t77.52t4.2zs1 -2

295.0366.3

432.6493.S

550.3602,1

1,71

7.107.63

7.56?.52

7.177,33

7.31

7.237.151.07

24.000

l020

30

io

60

80

100

t20t40160

srd

::

0.2500.375

0.500

0.562

0.62s0.687

0.7500.2180.875

0.968

I.2181.531

l.8I?2.0622.343

23.50023.25023.00022.87622.75022.62822.50023.564

22.25022.06421.564

20.93820.31619.876

19.314

4344254154ll406

402398

436.1

388.6

382

365

3443283t0

18.65

27.83

41.4

s0.354.816.29

63.5470.047.2

r08.I126.3

r42.1159.4

6.286.28

6.286.28

5.28

6.286.286.286.28

6.28

6.28

6.286.288.28

6.28

6.09

6.025.995.965.925.89

5.835.78

5.485.335.205.06

63.41

94.62

125.49

140.80

t56.03vt.r?186.2{s5

216

238.11

296.36367.40429.39

483.135{1.94

188.0

183.8

180.1

178.1

t78.2t74.3t72.4r88.9

165.8

158.3

1{9,3141.4

134.5

t27.O

l3l61943

2550

28403140

342037I0lt5242S6

465056706850

783086309460

109.6

161.9

2r2.5237.0281.4

245.2

309

96.03s4.7388

47357I

719

788

8.408.35

8.31

8.294.218.2S

8-228.41

8.18

Lt58.077.96

7.877.r97.70

28

26.000

10

20

srdxs

0.2s00.3I20.37s0.5000.625

0.7500.875

L000l l25

25.500

25.37625.250

25.000

24.75024.500

24.25024.00023.750

s10.7505.8500.7490.9

481.1

471.4

461.9452.4

443.0

I9.S525.1830.I940.0649.S2

59.4969.07

78.548?.91

6.81

6.81

6.81

6.8I6.8t6.81

6.81

6.81

6.686.646.61

6.54

6.48

6.4I6.356.28

6.22

I03r36l6s202235

267

299

221.12t9.2217.1

2t2.8208.6204.4

200.2I96.1192,1

I64620?6

24183259

40I3474454S8

6149

6813

126.6

r59.7I90.6250.7308.7364.94I9.94?3.0

524,1

9.10

9,08

9.06

9.02

8.988.93

8.89

8.85

8.80

2A

28.000

l0

20

30

srdxs

0.2500.3I20.375

0.500

0.6250.750

0.875

1.000

Ll25

27.5002',1.376

27.25027.40026.7S0

26.50026.25026.000

25.750

594.0588.6583.2

572.6

562.0

54I.2530.S

520.8

21.8027.t432.54

13.20

s3.7564.21

74.5684.8294.98

7.33

7.33

7.33

1,207.t7

7.077.006.94

6.8t6.7r

74

92

lll

183

2ta253

288323

257.3255,0252.6244.0243.4238.9234.4230.0225.6

2098260r3105408S

50385964

7740s590

t49.8185.8

221.e291.8359.8426.0{90.3552.8613.6

9.8r9.79

9,7',1

9,72

9.68

9.61

9.609.55

9.51

30

30.000

l0

20

30

srdxs

l0s0.250

0.3I20.375

0.500

0.625

29.500

25.31629.25029.000z8.'ts0

683.4

617.8

672.0

650.5649.2

23.3729.19

34.90

46.34

57.68

7.85

7.857.857.85

7.8s

7.727.69

7.597.53

79

99

9

r58

296.3293.7251.2

286.2281.3

258S

3201

3823

62I3

172.3

213.4

2S4.8

335.5

4t4.2

10.52

10.50

t0.4810.43

10.39



nominclpipe si:eoutsidediamelet,

schedule wolllhick-neat,ilr.

insidedidtn- inside

sq. in,

rrlelal

Bq. in.

sq Itoulsidesultcce.per ft

sq Itinside

per It

weightper Il,lbf

weight

per fttb

olinertid,in.'

modu-Iug,in.3

rddiusgYra-tion,in.b

30

30.000

40 0.750

0.875

1.000

1.t25

28.s0028.2s028.00027 .',t 50

637.9

620.?

6I5.7604.7

68.92

80.06

9t.l lt02.05

7.857.85

7.85

7.44

7.3S

7.33

234272

3t0347

276.627 t.B267.O

262.2

137 r8494

9591

10653

49t.4566.2

63S.4

't t0.2

10.34

10.30

10.25

r0.22

32

32.000

l0

20

30

10

rioXS

0.250

o.3120.375

0.s000.62s0.688

0.750

0.875

1.000

LI25

31.500

3r.3763t.25031.000

30.750

30.62430.s0030.2s030.000

29.750

779.2

7'13,2

766.9

754;1

742.5

736.6730.5

718.3

706.8

694.7

24.9s31.02

31,2s49.48

61.59

73.63

85.5297.38

109.0

8.38

8.38

8.38

8.38

8.388.38

8.388.38

8.38

8.38

8.25

8.21

8.18Lll8.0s8.02

7.98

7.92

7.85

7.ts

85

106

t2'l168

209

230250

291

331

371

337.8

335.2

332.5

327.2

321.9

319.0

3t6.73l1.6306.4

301.3

3l4 t3891

4656

6l{07578

8298

8990

t8372n6801302s

196.3

243.2

291.0

383.8

473.6518.6

561.9648.2

730.0

814.0

tt.22u.20I l.l8I l.l4I LoSI1.0711.05

I l.0I10.95

10.92

34

34.000

10

20

30

40

srdXS

0.2s00.312

0.375

0.5000.625

0.688

0.7s00.875

1.000

I.I25

33.500

33.376

33.2s033.000

32.750

32.624

32.500

32.2s032.000

31.7s0

881.2

s74.9867.8

855.3

841.9

835.9

82S.3

8r6.4804.279r.3

26.5032.99

39.61

52.62

65.5372.O0

78.34

91.0t103.67

116.13

8.S0

8.90

8.S0

LS08.90

8.90

8.90

8.S0

8.S0

8.90

4.77

8.74

8.708.64

8.54

8.51

8.44

8.38

8.31

90

tt2135

l?s223245

310

353

395

382.0

379.3

3',18,2

370.8

365.0

359.5

354.1

348.6

343.2

371s4680

s5977385

9124

9992

10829

l2s0l141t415719

22t.9275.3

329.2

434.4

s36.7587.8637.0735.4

830.2

924.1

11.93

I l.9lII.89I1.85I1.80I1.78r 1.76

tt.12

I I.63

36.000

l0

2D

30

40

;,;xs

0.250

0.312

0.37S

0.500

0.625

0.7500.875

1.000

Ll25

35.500

35.376

35.250

35.000

34.75034.500

34.250

34.000

33.750

98S.7

s82.9975.8962.1

948,3

934.7

920.5907.9894.2

28.1r34.S5

42.01

69.5083.01

96.50109.96

123.I9

9.42

9-429-429.42

9-429.42

9.429.429.42

9.29

s.269.23

9.16

s.l09.03

8.978.908.89

96

ll9143

I90236

282324374419

429.1

426.1

423.1

4t1.14I l.l405.3

399.4

393.6

387.9

4491

6684

8785

t0a72I2898I4903I685Ir8763

249.S

309.1

370.2

48S.I604.0716.5

82',1.9

s36.21042.4

12.64

t2.6212.59

12.55

t2.5112.16

12.42

12.38

t2.34

42

42.000

20

30

40

srdxs

0.250

0.3750.500

0.6?5

0.750I.0001.250

1.500

41.500

41.25041.000

40.7s040.50040.00033.50039.000

1352.6

r336.31320.2

1304.1

1288.2

I256.6r22S-31194.5

32.8249.08

65.I881.2897.23

r28.81I60.03I90.S5

10.99

10.s9

10.99

t0.99r0.9910.99

10.s9

10.99

I0.8610.80

I0.7310.67

10.60

10.47

10.3{t0.21

l12167

222

330

438

544

649

586.4s79.3s72.3565.{558.4544.8

53t.25I7.S

7t26to62714037

1737320589

2708033233

3918I

339.3506.1

668.4827.3985.2

1289.5

1582.5

1865.7

I4.7314.71

14.67

14.62

14.59

14.s0

t4.41t4.33

278 Mechanical Desien of Process Svstems

INSI'LATION WEIGI{T FACTORS

To determine the seight per foot of any pipinginsulation, use the pipe size and nominal insulationthickness to find the insulation l'eight factor F in thechart shorvn belorv. Then multiply F by the densityof the insulation in pounds per cubic foot.

Erample. For 4" pipe rvith 4" nominal thicknessinsulation, F : .77. It the insulation density is12 pounds per cubic foot, then the insulation rveightb .77 x 12 : 9.24lb/tt.

NominalPipe Size

Nominal Insulation Thickness

1%" 2rA" 3%" 4" 4%" 5%" 6"

Ir%lt/i2

.057

.051

.066

.080

.10

.11

.r4

.16

.21

.29,29

.31

.30

.38

.40

.39

.48

.47 .59

2%33%4

.091

.r0.19

.23.30

.36

.34

.41

.39

.46

.54

.cr

.58

.66

.63

.70

.68.83.81

.96.97

1.10

68

10

.24

.34

.43

.34

.38

.59

.45

.66

.58

.64

.80

.93

.88

.971.171.32

t.04r.131.36

1.201.341.56 1.75

1.99

t2l4

18

.50 .68.70.74.87

.88

.901.01\.\2

1.07l.l I1.24|.37

1.3.{1.491.64

1.52

1.7 41.92

1.812.01

1.992.O7

2.292.51

2.242.342.582.82

2.502.622.883.14

2024

.70

.83.96

1.131.23t.44

1.50 1.792.10

2.092.44

2.402.80

2.733.16

3.063.54

3.403.92

LOAD CARRYING CAPACITIES OF THREADED HOT ROLLED STEEL RODCONFORMING TO ASTM A.36

Nominal RodDiameter, in. lz % v4 1 1 1r/e r% 2 2y4 21/2 2y4 3 3r/q 3'h

Root Area ofThread, sq, in. .068 ,126 .202 .302 .419 .693 .889 1.293 1.7 44 2.300 3.023 3.?19 4.619 5.621 6.724 7.918

Max. Safe Load,lbs. at RodTemp. of 650"F

610 1130 1810 21L0 3??0 4960 6230 8000 11630 15700 20700 21200 33500 41580 50580 60480 ?1280

zi.?

z

B

zF

z

z,t

7F

z.(

Au-r'/\w

{l\E-I4/ a^

t_J-----,

\]J


WEIGHTS OF PIPING MATERIALS l " prpe r.sr3' o.D.

Boldface type is ueight inpounds. Lightface t]'pe benerthweight is veight factor forinsulation.

Instrlation thicknesses andweights are based on averageconditions and do not constitutea recommendation for specificthicknesses oI materials. Insula-tion Neights are based on 85/6magnesia and hvdrous calciumsilic&te et 1l lbs,i cubic foot. Thelisted thicknesses and Neights ofcombination covering are thesums of ihe inner layer of dia-tomacecus earth at 2l lbs/cubicfoot and the outer layer at11 lbs,/cubic foot.

Insulotion rveights include al-lorvcnces for wire, cemerrt, can-vas, bands and paint, but notspecial surface finishes.- To find the weight of coveringon flanges, valves or fittings,multiply the \veight frctor by theuoight.pcr foot of covering nsedon slrarght prpe.

Vf,tve \veights 3re rpproxi-mate. When possible, obtainNeights from the nranufacturer.

Cast iron valve $eights are forflangcd end valves; steel $eighLsfor welding end velves.

AII flanged fitting, fl&ngedvalve and fllnge $'eights includethe DroDorlion.l \leieht of boltsor siudi to make up all joiots,

Temperature Range 'F

tr{agnesiaCalcium

Combina-tion

Fiber-Sodium

ffi&Njs{|s.:ssr

'-11T}4lN/9N

@tr\qJ

+€Fsc

* 16 lb cu. ft. density.

280 Mechanical De:ign of Process Systems

l/a" wen r.660, o.D.

Tenrpcraturc Range "F

o uonz

Fiber-Sodium

WEIGHTS OF PIPING MATERIALS

f'^t+,!

HJ-4L.E:::tttl

n_Lt{- i--r\LJ

Ma,gnesia! Calcium! Silicate

zF

z3F

z

7

zI

ffieffifs-is$! T:lii--qF

Boldface type is s'eight inpounds. Lightface type benerthweight is weight factor Jorinsulation.

Insulation thicknesses andweights are based on averageconditions and do not constitutea recommendation for specincthicknesses of ma,terials- Insula-tion weights are based on 85%magnesia and hydrous calciumsilicate &t 11 lbs/cubic foot. Thelisted ihicknesses and i{eights ofcombination covering are the6ums of the inner layer ol dia-tomaceous earth at 21 lbs/cubicfoot and ihe outer laycr at11 lbs/cubic foot.

Insulstion weights include al-Iowances for wire, cement, csn-vas, bands end peint, but notspeeial surface finishes,-

To find the weight of coveringon flanges, valves or fittings,multiply the weightfactor by thewerghl per loot ol coverrng usedon strargnt prpe.

Valve weights are approxi-mate. lVhen possible, obtain$'eights from the manufacturer.

Cast iron velve weiqhts arc forflanged end valves; sGel weightsfor weldins end valves.

.All flanged fitting, flangedvalve and flange weights includethe Drooortionrl weiqht of boltsor si,udi: to make up all joinl,s.

.-al/A4

,N/>

1.<3@

l[' )

+€rc

Nom. Thick.,In.


A

Schedule No. 40 80 160

Wall Designation srd. XS xxsL Lic kness-In. .145 .200 .28r .400

Pipe-Lbs/Ft 3.63 4.86 6.41

lVater-Lbs/Ft .88 ,77 .61 .41

fl.2IJJ

n {"uz^"E {_OE r-i\(, F!-+

e /.,e^

Lt_!(----1--l\IJ

L.R. 90' Elbow.E 1.1 1.{ 1.8

S.R. 90' Elbow.6

.3

L.R. 45' Elbow .2.8.2

1

Tee .6 .63.1.6

3.7.6

Lateral 1.35.41.3

Reducer.6.2

.7t

1.2t

c"p .3.5.3

.7

Temperature Range 'tr' t00-1c9 200-299 300-309 400-499 500-599 600-699 700-799 300-s99 c00-3c9 1000-1009 1100-1200

2\om. Thick., In. 1 1 r% 2 2 2r/6 2% 2% 3 3 3

Lbs/Ft .84 .84 1.35 2-52 3.47 3.47 3.47 4.52 4.52 4.SZ

N Combina-

z

Nom. Thick.,In. 2)i 214 2% 3 3 3

Lbs/Ft 4.20 4.20 4.20 5.6t 5-62 5.62

Fiber-Sodium

\om. Thick., In. I 1 r% lr/4 2 2 2rz 2% 3 3

Lbs/Ft 1.07 1.07 1.07 1.E5 1.85 3.50 3.50 4.76 4,16 6.16 6.16

sffi$:ffiss]sd}.'.=N!

Prc-rsure Raiingpsr

Cast Iron SteelBoldface tlpe is rfeight in

oounds. Lichttace hDc bencathivcight is - rveight' iactor lorinsul:rtion.

Insub.tion ihickncsses and*eights arc based on averageconditioris and do not constituteD recommendation for specificthicknesses of m$terials. lnsula-tion Neights are bcsed oD 8570

3,?il*'11 ll'9"lxli:l:","'.*'"is,3

125 250 150 3C0 400 600 900 1500 2500

Screwed orSlip-On

3.5 7 8 9 9 l91.5

19 3l

\Yelding Neck9 t2 t2 19

1.5l91.5

341.5

Lap JoiniE 9 9 l9 l9 3l

1.5

Blind 1.57

1.59 10 t0

1.5l91.5

19 3l

,a .'11I / ,tJ

e,\z t44\tc lF -ll

S.R. 90" nlbow10 23

3.8263.9

46 listed thiclinesscs cn(l scights ofcombin.tion covering are thesums of t,he inner lrver of dir-tomaceous earth at 2l lbsr'cubicfooi and the outcr lD,l'er 5t11 lbs/cul)ic foot.

Insulltion weights includc al-lo$lnccs for 'iviro, ccment. ctn-vcs, bllncls {Lnd plint, but notstrccial sulf.rce linishcs.-

To find the rvcight of covcringon tlonlles, vxlvcs or fittitrgs,multipll; tlic wciFht fr, tor bv the\\cighi t)cr foot of covcring uscdon stftLisht lliDc.

Vxlvc- \\(iihts arc :rlrptori-matc, Whcn Dossiblc, obtoinrvoights from thi mtnufacturcr.

Cast iron vtlve weights rre forlllnged cnd vrlves; stccl \eightsfor rveldine cnd valvcs.

.\ll flriised fitting, flangedvalve rnd 1|rngc ivcights includcthe l,rorntlion l N(iglrt ol l)oltsor sluds to m.tku up:rll ioints,

L.R. 90' Elbow13

45" Elborv9 l1

3.4 3.523 39

Teet75.6

20 305.8

706

1=<l* k333uuJ<[JFSO

flanged l3onnetGxte 6.8

70.1.5

t25

Flanged BonnebClobe or Angle

40 45 1705

Irlanged BonnetCheck

30 4D ll0

Itressure SeaIRorrret-Crie

421.9

42

Pressurc SealBonnet-Globe

' 16 lb cu. ft. density.


WEIGHTS OF PIPING MATERIALS r.eoo" o.D. l/2" Ywn

rd s cights ofing are thelrver of dir-2l lbs: cubic

:cr lD,J'er 5n

bs includc al-ccmenl. ctn-rint, but nothcs.lt of covcring

or fittiogs,frctor bv the

:ovcring'uscd

rrc apptori-siblc,

-obtlinirnufarcturcr.'eights rre forstccl \eights

;ing, flanged.ights includecight of l)oltsup :rll ioints.


2" ptpn zs. B, o.D. IVEIGHTS OF PIPING MATERIALS

Temperature Range oF

MagnesiaCalcium

Combina-tron

Fiber-Sodium

!r

A,/ /a)

,-61

,N/DIN' '{I

1.<l@rfl

[],._/

+<tFsO

z

z

zF

5z

z

z

z

't

u'N

u,r'Ihd-J.-t

-r--r-\

/>fin

{_L_!

Nr$+fi$N*scr.i-s

Boldface tyDe is weisht inpounds, Lighifbce type bdneathweigit. b weight factor forlnsutailon.

Insulation thicknesses andweights are based on averageconclrtrons and do not constitute& recommendation for soecificthicknesses of materials. I_nsula-iioo veights are based on 85/omagnesia and hydrous calciumsilicate at 11 lbs/cubic foot. Thelisted thicknesses and weishts ofcombination coverinc are thesums of the inner laler of dia-tomsceous earth at 2l lbs/cubicfoot and the outer layer at1l lbs/cubic foot., Insulaiion weights include al-lowances Iol wlre, cemen!, can-vas, bands and oaint. but notspecial surface finishes.

To 6nd the weieht of coverincon flanses. valvds or fittinssimultiply tlie weight factor by tIeweight.per foot of covering usedon-slr&lghl prpe.

valve wergnF are approxl-mate. When possible, obtainweights from th; msnuiacturer.

Cast i.on valve weights are forflanged end valves: sGel weiehtsfor ielding end vaives.

All nsnsed fittios. flancedvalve and flange weigF* inclidethe prcportional weight of boltsor 6tuds to make up all joints,

' 16 lb cu. It. density.

fr

Au-r'

w{T\E'-/.>\F--1

/-Aq-!_,

\]J

7F

z-l

'



2.87s" o.D. 2/2" Ywn

Boldface type is s'eight inpounds. Lightface type beneathweight is weight factor forinsulation.

Insulation Lhicknesses andweights are based on averageconditions and do not constitutea recommendation for specificthicknesses oI materiels- Insula-tion weights ere based on 85/6magnesio and hvdrous calciumsilicate at ll lbs/cubic foot. Thelisted thicknesses end rveights ofcombination covering are thesums of the inner laver of dia-tomaceous eerth at 21 lbs/cubicfoot and the outer b,yer at1l lbs/cubic foot.

Insulation weiqhts include al-lowances for rvird, cement, can-vas, bonds and peint, but notspecial surf&ce linishes,-

To find the weighi of coveringon flanges, valves or fittings,multiply the ileight factor by theweight per foot of covering usedon straight pipe.

Valve 1Aeights are approxi-mate. Whe[ possible, obtainweights fron the manufscturer.

Cast iron velve weiqhts sre forflanged end valves; sGeI *eightsfor welding end valves.

All flanged fitting, flangedvalve ond fiange iveights includethe proportional \ieight of boltso! studs to make up &ll joints,

zo

fz

z

z

z

'l

Temperature Range "F

MagnesiaCalcium

Combina-tron

Fiber-Sodium

ffi$q1$Nl-s$

N.-al

T.AA

a-4,N

L4

.|-{

@flr)

+<it4+ 16 lb cu. lt. density.

284 Mechanical Design of Procesr Systems

3 " "t"" B.boo' o.D. WEIGHTS OF I'IPING N{ATERIALS

Schedule No. 40 EO 1C0

Wall Dcsignation std. xs xxsTlrick ness-In . .216 .300 .438 .600

Pipe-Lbs/Ft 7.54 10.25 14.32 t8.56\1'xter-Lbs/Ft 3.20 2.86 2.35 l.E0

W|4 {I/zr\E {it: {1\3 r.'.'grl F4q

(-r__)

\JJ

L.R. 90' Elbow4.6.8

6.1.8

8.4.8

lo.7.8

S.R.90'Elbow .5 .5

L.R. 45' Elbow .3 .34.4.3

5.4.3

Tce7.4.8 .8

12.2.8

14.8.8

Lsteral 1.8l9

Rcducer .3 .33.7.3

4.7.z

cup .5.1.8.5 .5

3.7.5

'li nrpcrrlur. ncngc'F 100,14r 200-:0c 300-3c9 100-lm 500-599 600-699 700-7s9 800-80s 900-g?9 1000-1099 1100-r200

Magnesia? Calcir.rmY Silicete

\orn. Thllk., In. I 1 1k 2 2 2% 3 3 3 3% 3%

LLs Ft t-25 2.08 3.01 3.01 4.07 5.24 s.24 5.24

z

\ont.'.t'hick., IIL 2\ 3 3 3 3% 3%

II-1i Ft 5.07 6.94 6.94 6.94 9.17 9.17

Fiber-Sodium

\om. TLick., In. 1 I 1 1rz 1tz 2 2 3 3 3% 3%

Ll's, Ft 1.61 1.61 1.61 2.74 3.9E 3.9E 6.99 6.99 8.99 8.99

rffiO r-Fn? s{ ils

N-i.sqF{i.llqn

Pressurc Ratingpsr

t-cst lron SteelBoldface iype is u'eight in

pounds. Lightface type bene3th$eieht is weight fachor forinsul&tion.

Insulation thicknesses andweighis are based on average.onditions and do not constitutea recommendetion for sPecificthicknesses of materials. Insule-tjon Neights are based on 85%macnesia and hvdrous calcium"ili;r. 't l1 lhs/crrhin foot The

125 250 150 300 400 600 900 1500 2500

Screwed otSlip-On

9 17 91.5

t7 201.5

201.5

6l 102

Welding Necktl1.5

191.5 1.5 1.5

381.5

6l1.5

ll31.5

l,ap Joint9 19

1.5l91.5

36 601.5

991.5

Blindl0 l9 l0 20 24

1.524 38

1.56l1.5

r05

2 lâE /'11

B,Nu /9N< .:E BJ ti]

ll---J

S.R- 90' Elbow263.9

464

323.9

53 674.1

984.3

r504.6

listed thicknesses and $eights ocornbination covering are thsums of the inncr layer of dittomrceous earth at 21 lbs/cubiIoot and the outer laJcr a11 lbs,lcubic foot.

Insulation rveights include allorvences for \rire, cenrent, crnves,.blnds and Irrrirrtr buL nostrccrcl surlace nnlsnes.-

To hnd the Neight of coverinlon flanges, valves or fitting:multiply the Neight factor b\ thwelghl per loot ol coverlng useion straight pipe.

Valve weights are spproximate. When possible, obtairweights from the ma,nufacturer.

Cist ilon valve weiehts are foflansed end valves; steet weightfor weldine end valves.

All flrnged 6tting, flange.valve and llanse rleiqhts includthe proportionlel \r eight.of boltor studs to mirl(e uP all Jorntt

L.R. 90' Elbow304.3

504.3

404.3

634.3

45" Elbow413.6

2E 46 603.8

933.9

1354

Tee395.9

676 5.9

E16

1026.2

l5l 23E6.9

{-<t,k€j r\J+<trc

Flanged BonnetGate

667

70I

125 t551.8

2605

4105.5

tr'langed BonnetGlobe or Angle 7.2

t2l 604.3

95 t551.8

495

Flanged BonnetCheck

467.2

100 60 70+.4

r204.8

t504.9

4405.n

Pressure SealBonnet- Cate

2083

235

Pressure SealBonnet-Globc

r35 1803

* 16 tb cu, ft. densitY,

ave.agenstttulespecificInsula-

>n 857acalciumrot. Theights ofrre theof dia-

)s/cubictvcr at

l rL-

not

covefnlgfittings,

)r bv therng usect

approxr-obtain

acturef.Ls are for. weights

flangeds includeof bolts

Il joints.

-

Appendix C: Properties of pipe 28.tt

WEIGHTS OF PIPING MATERIALS 4.ooo'o.D. 3/2" ewr.Schedule No. 40 EO

Wall Dasignation srd. XS xxsThickness-In. .318 .636Pip€-Lbs //Ft 9.tr t2.51 22.8sWater-Lbs / Ft 4.28 3.85 2,53

fr?ut {J-/z^,F [/)El#: {l\3 /)\

/.-NIrt

L.R. 90' Elbow6.4.9

8.7.9

l5.4.9

S.R. 90' Elbow4.3.6 .6

L.R. 45' Elbov4.4

Tee9.9.9

t2.6.9

20.9

Lateral 1.8261.4

Reduce!3.1.a .3

6.9.3

cuP2.t.6

2.a.6 .6

Temperature Range 'F r00-199 200-209 300-399 400-499 500-599 600-699 700-799 800-899 900-999 1000-1099 1100-1200

agnesta Nom. Thick., In. 1 1% 2 2% 2ti 3 3 3% 314 3%

: srlLcate Lbsi/Ft r.E3 1.83 3.71 4.EE 4.88 6.39 6.39 7.80 7.80 7.80

6 titz)mbina- Nom. Thick., In. 2% 3 3 316 3% 3%

Lbs/Ft 6.49 E.7l 8.7 | r0.6 r0.6 10-6

Fiber-Sodium

Nom. Thick., In. 1 1 r% 1X 2 2 3 3 3% 3%Lbs/Ft z.4l 2-41 3.65 5.07 5.07 E.66 8.66 r0.62 10.62

,ffi3S4 itS" NIM

Efsfs$

Pressure Ratingpsr

Cast Iron Ste"l-_-Boldface _tvpe is *eighi in

Pounds. Llghtf:rce t)pe benecthlvelglrt. is \aeight fcctor formsul& on.

Insulation thicknesses and\eights are lssFd on averagccondlt)ons xn.l Jo noI constiiutea recommendation for specificthicknesses of rnetcriu,ls. I_nsrrh-tion lveights are b:rscd on 8b7,m:rgnesir and hvdrous lrlciumsilicsle rt I I Ibs'cul,ic foot. Thelistcd thickncsses :rnrl \.eiqhts ofcombination coveriDg ar:e thesums of the inner hier of dia-tomil(eous e.Lrth at,21 lbs/cubicIoot and thc outer lll,!.er at11 lbs:cubic foot., Insulltion $eights inclutle al-Iowances ioa wlre, cement, can-vas, blncls &nd Dcint, but, notspcciu.l surface fi nishes.

To firrd ihe $eigl,t of coveringon llrnges, vxlves or fiftincs.multit,lj thc $eight f"(bor l,v thcwuight per foot ol cov|jrinlj'usc(lon straiqht DiDe.

Vxlvc \,eigl,ts rrc epprori-ma,te. When possible. obtlinrveights from th_e mtnuiacturer.

( ust iron v{Llvc Neiq}rts arc lorflangtd entl velvesistaci leishtsIor rveldirrg end vdves.

,\ll fluhged fit tins, fllrnsc,lvxlve xnd flxnge rrcigl'rs inclu,lethc proporlional weight of bolts

t25 250 150 300 400 600 900 r600 2500Screwed orSlip-On

13 2l 13 21

Welding Neckl41.5

32

Lap Joint13 2l 26

1.5261.5

Blind14 23 15

1.525 35

O ,'42tdE-qBNO /. 3\<.:E Ptn

8.R.90'Elbow 4.L49 82

4.3

L.R. 90" Elbow404.4

624.4

544.4

45" Elbor3l 5l 39 75

3.9

Tee546

868.2

706

t336.4

1"<3

HKPfqJ+<tl€

Flanged BonnctGste

a27.1

t43 90 155 1804.8

3605

5t0

Flanged BonnetGlobe or Angle

74 1377.7

160

Fhnged Bonnet7.3

1257.7

t2s

Pressure SealBonnet-Gate t40 | 295

2.5 | :.83E03

Pressure SerlBonnet-Globe

* 16 lb cu. ft. density.thc proporlional weight of boltsor studs to mcke up:rll joints.


4" prcn 4.500' o.D. 1YEIGHTS OF PIPING MATERIALS

'l'cmtx,miurr lLrngc'Ir

IlagnesiaCelcium

Coml)inl-iion

Fiber-Sodium

f'2!xtr2

{,\t-i .t{i\HIe-

\IJ

ozF

z

zoFIz

/''ll/A,N/>

F{3@

,lr1

+€rc

zF

7

Boldface type is rveight inpounds. Lightf:lce tl'pe benextlr$'eight is weight lactor lorinsulation.

Insulation thicknesses endweights are based on averageconditions and do not constitutca recommendation fol spccificthicknesses of materials. Insull-tion weights are based on 85t;magnesia and hydrous calciumsilicate &t 11 ibs/cubic foot. Thelisted thicknesses and \\'eights ofcombination covering are thesums of ihe inner layer of dia-tomaceous earth et 2l lbs,/cubicfoot and the oute! la\.cr at11 lbs/cubic foot.

Insulation weighL includc al-lowances Ior uire, cement, can-vas, bands and paint, but notspecial surface finishes.-

To find the weight of coveringon flanges, valves or fittings,multiply the we;ght frctor by thcNeight per foot of covering uscdon straight, pipe.

Valve wcights arc approrii-mcte. When possiblc, obtrinlveights from thc manuf&cturer.

Cast iron valve lvcights &rc forflanged end valves; stecl \cightsfor lelding cnd valves.

All fleriged fittins, flangedvalve rnd flange rvcights inciudethe proportional rveight ol boltsor studs to make up all joiDts.

l Stits

\\ attr-l-bs/I t

Nom.'l'hick., In.

" 16 lb cu. ft. densitv.

C


WEIGHTS OF PIPING MATERIALS 5.563" O.D. 5" PtPe

/11/r4,N/>

ll' 'llIH

t{@0

++3rc

zF

z

B

zF

z

z

zF

zti

J

(-!j

wfl-\4'e.,-'1-l

c_i_)a-1--r

Tcmperature Range 'F

Fiber-Sodium

Combina-tion

MagnesiaCalcium

ffir$s{lrs$sj-N$

Els:i-:5$

BolJfrce type is rreight inpounds. l,ighbf.lce tYpe beneeth$'eight is weight lactor forinsul.rtion.

lnsulation thicknesses andlreights rre besed on everageconditions and do not constitutea recommendotion for specificthicknesses of m&teri3ls. Insuh-tion weights :rre based on 85%magnesia and hvdrous calciumsilicate at 11 lbs/cubic foot. Thelistcd thicknesses and \'eights ofcombination covering are thesums of the inner layer of dia-tomoceous earth at 2l lbs/cubicIoot and thc outcr l&r-er at1l lbs/cubic {oot.

Insulotion l eights include al-lorvances for {ire, cement, can-vas, bands and p&int, but notspecial surfrrcc {inishes.

To find the rveight of coveringon llanges, volvcs or fittings,multitt]'thc wcight f$ctor by thc\reight pcr foot of covoring usedon straight pipe.

Vdve rveights arc opproxi-mate. When possible, obtainweights from the manuflcturer.

Cast iton valve rveights are forflonged end valves; steel rleightsfor welding end valves-

All flangetl Iitting, flrngedvslvc and flange weights includethe proportional weight of boltsor studs to rnake up all joints.

15.6 | r7 .7

Flanged BonnetCheck

* 16 lt cu. ft. density.


6" ,t n 6.625. o.D. WEIGHTS OF PIPING X{ATERIALS

Tcmpcraturc llange 'F

Ma,gnesiaCalcium

Combinl-tion

tr'iber-Sodium

gJ-f

{na-1J

{1\E:cl

E_=_=r

!._!____,

\t/

z

z

'

2oF

D

z

z

,41#4l

,N/9sEq-A

lt' '{t

t{3@

ir)+<iffi

3

4q-x$

sfil$dN-s{Jss;s

Boldface iype is weight inoounds. Liehtface tr.pe beneathq eight is - weight

' iactor forinsulation.

Insulation thicknesses andweights are based on averageconditions and do not constitutea recommendation for specificthicknesses of matedals. Insula-tion weights &re based on 85%magnesia and hvdrous calciums;liAte at 1l lbs/cubic foot. Thelisted thicknesses and weights ofcombina,tion covering are thesums of the inner layer of dia-tomaceous ea,rth at 21 lbs/cubicfoot and the outer layer at11 lbs/cubic foot.

Insulation $eights include al-lowances for rriie, cement, can-vas, bands snd paint, but notsDecial surface finishes.-

To find the $eight of coveringon ffanges, valves or 6ttings,multipit the u eight frctor b-\' the!\eight per foot of covering usedon straight pipe.

Valve $eights are {rppror -mate. When possible, obtainweights from the manufecturer.

Cast iron valve weights are forflenged end valves; steel weightsfor selding end valves.

All ffanged titting, flangedvclve and flenge Neights in.ludethc DroDortronal wcight of boltsor studi to mrke ut rll joints.

z

=z

liom. Thick., In.

* 16 lb cu. ft. densitJ'.

il

Appendix C: Properties of PiP. 249

WEIGHTS OF PIPING MATERIALS 8.625. O.D. 8" "r",

Ae,T,Jr'

uJ{T\r';JlArFr/t\\iJ

ffis{tlts$s is!N

ArAA/>€ela

1-{3t4s^

+<tFsO

zFF

z

,

zF

2

2


MagnesiaCalcium

Combina-tron

Fiber-Sodium

Roldfrce tlpe is $eight introunds. L;qhtfcce tvoe beneathireiglrt. ;s - \rcight ' iacLor lor

Insulation thicknesses andrlcights cre brsed on rverugeconditions cn.I do not corstitutea recommendation for specificthicknesses of materills- Insula-tion \reights cre based on 85lomagnesia and hydrous cslciumsiiicate at 1l lbs/cubic foot. Thelisted thicknesses and leights ofcombination covering are thesums of the inner loyer of dia-tomaceous eadh at 21 lbs/cubicloot snd the outer lol er at1l lbs/cubic foot.

Insulation rveights include al-lowances for uire, cement, can-vas,.b!.nds ond paint, but notsDectal surlace nnrsh€s.-

To find the weisht of coverinson flanges, vrlvis or fittings,multiply the \aeightf&ctor by the\Yeight.per foot of covering usedon strarqn! prDe,

Yalve rvciIhts cre appro\i-mcte. l\rhcn possible, obtxin\leights from th6 manuflcturer.

Cast ilon valve iveights ore forflanged end vrlves;stecl \\'eightsfor \\elding enLl valves.

All flanged fittine, flangedvslve and flongc Neights jncludethe proportionlrl lveight of boltsor studs to make up all joints.

zF

z

1r00-1200

* 16 lb cu. ft, density,

290 Mechanical Design of Process Systcms

10t'prpe ,o.zso" \VIJIGIITS OT PIPING I{ATDRI,\i,S

Trmprrx6url 11''ra. "P

N'IagnesiaCalcium

Combina-uon

Fiber-Sodium

(,zk

.l

zF

z

IA//\w{i\E.-I

4'd',

!-l_,t,t!

z

z

'l

ffiA,/TmAqIS I l\S

N-lsry--rp

Boldfcce type is neight inpounds. Lightfece t) pc benertll$cight. is $eight, Jsctor forlnsut& on.

Insulrt,ion thicknesscs andiveights arc based on avengeconditions and do not constitutea rccommendrtion for specificthicknesses of matcdcls. Insul:r-tion \ieiqhts are bascd oir E59.magnesii and hldrous calciunisilicate at 1l lbs/cubic foot. Thelisted thicknesses and $eights ofcombination covering are thesums of the inner laler of dia-tomeceous earth at 21 lbs/cubicfoot and the outer laver s,t11 lbs/cubic foot.

Insulation \Yeishts include al-lowonces for rdr;, cement, cen-vas, bands and paint, but notspecial surlace frnishes,

To find the wejsht of coverineon flonges, valvds or 6ttings]multipll the neight f:rctor bt theNeight per foot of covering usedon strsight pipc.

|alve \rcishts ore luorori-matc. \\'hcn- possil)le,

-;l)trirr\r0iqhls from thc n)rnufscturcr.

Crrst iron rrlvc $cights rfe for13rngcd t'nrl \.l|lvcsi stccl Neightsfot l-clding end vrlves.

-\11 flrngcd fitting, fllngedvslve and flcngc $eig)r1s inrludethc propottionul \eight of boltsor studs to rn.rke up all joints.

,--ll

/>tP ql

J-<3

@ll.J

++lrc

\\-rtcr-Lbs ' l' i

\om. Thick., In.

\om. TlLn k., I rr.

We)ding Neck

* 16 lb cu. ft. densitl..

Schedule No. 20 30 40 60 80 r00 I20 140 160Wall Designation std. XSThickness-In. .250 .330 .406 .500 .562 .687 I .843 1.000 1.125 1.312Pipe-Lbs/Ft 33.36 43.E 49.6 53.5 65.4 73.2 8E.5 1t07.2 r25.5 r39.7 160.3Wster-Lbs/Ft 5l.t0 49.7 49.0 48.5 47 .0 46.0 4,r.0 I 4r.6 39.3 37 .5 34.9

{?4nuj

7fhF:-

L.R. 90' Elbowlr9 t57

3375

S.R. 90" Elbow80 104

2

L.R. 45" Elbow601.3

7a IEl1.3

7-

'

r32 167 360

J,ls) Lateral1805.4

2735.4

Reducer33 44

.794

Crp30 38 t9

Temperature Range 'tr' r00-199 200-299 300-399 400-499 500-599 600-699 700-7s9 800-899 900-999 1000-1099 1100-1200

{-iryTiuNom. Tbick.,In. 1% 1% 2 2tz 3 3 3% 4 4% 4%

Y Silicate Lbs/Ft 6.04 6.04 E.13 10,5 12,7 15.1 t7.9 17.9 20.4 20.4

{ uomotna-7 tionz

Fiber-Sodiuo

Nom. Thick., In. 3 3% 4 4 4% 4%Lbs/Ft 17.7 2t.9 26.7 26-7 3r.l 31.1

Nom. Thick.,In. 1% 1% lrl 1% r% 2% 2% 4 5

Lbs/Ft 5.22 14.20 14.20 24.@ a.g 32.& 32,&

Bs$!stu" Nls

s\"ssF

Prcssure Ratingpsr

Cast lron rBoJdface t1'pe is rvciqht in

pounds. Lightfaco tlpe bencxth\reigl,f. js reight frctor forldsulallon.

Insulrtion thickncsses andneights are base<l on aver:rteconditions and do not constitule& Iccommendatio4 fot suecifictliicknesses of materirls. Iirsul:r-tion Neighis are l,rsrd on E5%mrgncsiu and hrdrous calciumsilicxte at lL lbs/'cubic loot. Thelisted thickncsscs and Ncishts ofcombinrtion covering aie thesums of thc inner loier of dirr-tomlceous ctrth at 21 lbs/cubicfoot and thc outer la\'er at11 lbsTcubic foot., Insuhtion rvcights includc al-to\l'anccs lor $lrc, ccncnt, can-v:rs, hanrls lrnd Drint, but notspccial surface firishes.

To lin,l bhc ur'rglrt of covelingon flxrgcs, vrlvcs or fittirrae.mult;l'h tl,c \reiglrt i,, tul l,\.the\\(iAht l)cr foot ol coverirrg uscdon strrlalrt DrDc.

Vrtr e rrcights rtc errnlori-m:*c. \1'herr possil'le, obt:rinr'cights from the m:rnuilrcturcr.

Crlst iron vtlye wciqlrts &rc lorflangctl end v0lves: stccl \eichtsJor rrcltiine cnd vclves. -

.\lt flerrgcd fitting, fl:rngedvnlvc rn4 lixfigc \rcigirts includethc proportionrl lcight of boltsor studs to mrkc rrn rll ininlq

250 150 300 400 600 | 900 1500 2500Screwed orSlip-On

7Lt.5

137 l1() 1641.5

26r | 3881.5 | 1.5

E20 l6u

Welding Neck88 163

1.52t21.5

272 | 4341.5 I 1.5

843r.5

19191.5

Lap Joint1641.5

t87 286 | 433 902 1573

96 177 11E 209 261 341 | 4751.5 | 1.5

928 1775

o ,-{t2Ld,. Al| lA

EAz&4dflq

S.R. 90' Elbow265 453

5.2345

5509 669 E15 1474

6.2

L.R. 90' Elbow3756.2 6.2

4856.2

6246.2

l59E6.2

45" Elbow2354.3

3E34.3 4.3

4144.3

4694.5

?0s I rr244.7 | 4.8

Tee403 684 513 754

7.89438.3

136 t8.7 9.3

1.{3*@j rqJ

{={tts0

Flanqed BonDet 6877.a

l29a4

10155

1420 21557

2770 46508

Flaneed BonnetGlobi or Angle

80E9.4

r2009.5.

7r05

1410

Flanged BoDnetCheck

6749.4

I1609.5

560 720 14107.2

26008

33708

Pressure SealBonnet-Gate

19755.5

25606

45t57

Pressure SealBonnet-GIobe


WEIGHTS OF PIPING MATERIALS rz.75o" o.D. 12" prpx

' 16 lb cu. ft. den6rty,

proportionrl Ncight of boltsstuds to make up all joints.


14" ptnE 14" o.D. 1VEIGHTS 0F PIPING IIATERIALS

{?

fh{t}EJJ

-=;t/-\

\t/

z|.

z

t

Temnr.r:1turc Rrngc 'F

2

F

z

z

z|.

I

NlaguesiaCalcium

Conlbina-tlon

Fibe!-Sodium

ffi6{rlsds]sElsisp

Boldlacc tlpe is rrcight inpouncls. Lightface tl'pe bencalh$eight is \eight f.rctor torinsulation.

Insulation lhicknesscs and$eights arc besed on averageconditions and do not constitutea recommendalion for spccilicthicknesses of materials. Insula-tion tieights al.e based on E5fimagnesia and hvdrous ralciumsilicate at 11 lbs/cubic foot. Thelisted thicknesses and rveights ofcombination covering are thcsums of the inner laver of dia-tomaceous e:rrth xt 2l lbs/cubicfoot and the outer lal er et11 lbs,/cubic foot.

lnsulaiion weights include al-lorvances Ior uire, cement, can-vas, bands and paini,, bui notspecial surfece finishcs.

To find the $eight of coveringon flanges, valves or fittings,multipl] the \\'ejght facior by thelYeight pcr foot oI covering usedon straight, pipe.

Valve rveights are approri-mete. When possible, obtainscigbts from tha manufrcturer.

CasL ilon valve Neights are forflanged end vrlvcs; steel *eightstor rYelding end valvcs.

All flanged frttiDg, flangedvrlve end frcnge $rights includethe proportionri \rcighi of holtsor studs to meke up all joints.

/A,-11//,\

t&D' .{

+.{@r)+<trc

Nom. Thick.,In.

* 16 lb cu. ft. density

I


Appendix C: Properries oi Pipe 2gl

ro'o.o. 16t' prpt

GnL!_r'

f>\L4J{l\e4'4!+ir't\

f--.+--l

t4

zF

z

B

Temperature liange 'F

Xlagnesia! Calciumi Silicate

j Cornbina-6 tionz

Ses{-NNiss\sf

Sodium

Bo i.lce t\.pe is rveielrt inpounJs. Lightfi, e ttpe benesthrferqht is \\eislrt frctor foflnsul&lton.

Insulation thicknesses andireights .Ire bascd on avcrrgeconditions and clo not consiir.u[ea recommendrtion for succiticthicknesses of materiais. Iirsul:r-tion \\'eights are brsed on 859%magnesil and hydrous cllciuisilicate rt ll lbs/cuLic [oot. 'IheIisted thicknesses and seights ofcombrlctlon covering ace thesums of the inner laier qI dia-tomsceous eorth at 2l lbs/cubicfoot and the outer layer at11 lbs/cubic foot._ Insuiation weights include al-lovances Ior wirc, cem€nt, can-vas,.bands and-' pxint, but notspecrat surlace hnlshes.

To_ find Lhe weight of coveringon llxngcs, vnlves or fittines.multipl] t|e weight flctor bl thewerght.pcr foot uf covcriDg uscdon strxrqht DlDe,

Valve rri-iel,ts ^re al,rrroxi-

m.Ltc. \\'hcn I'ossil)le, 0l,tcinweights from th_e manui:rciuror-

Cast iroD vrlve $cishts cre forflangcd cud valvesi stiel $eiehtsfor rvelding end v:rlves.

AII flrngcd fitting, frxngedvclve and flxngc ncights i cludethe proportionrl weight of boltsor studs to make up all joints.

dA,N

!!!q

1"<3

@fi1

+<tl4

z

z|.

z1

Et

1100-1200

' 16 lb cu. ft. density.


18" prpo 18" o.D. WEIGI{TS OT' PIPING MATDRIALS

'fonrl)erllturc lhrlac'Ir

MagnesiaCalcium

Combin.r-tron

Fiber-Sodium

zFF

z

B

2oF

fz

z

7F

z

{.!-r'

f>\a-+-!

{T\I-5:I-t\"&\\JJ

ffistfN$NlsqN

Soldface i,r'pc is rrcight inpounds. Ligbtl.rce tlpc benecthtcigl,t is scight factor forinsulation.

Insul&tion thicknesses andFcights arc bascd on averageconditions and do not constitutec lecommcnd:ltion for specificthickncsscs of mstcriols- Insuls-tion $eights &re b:rsed on 85%rnagnesia and h)'drous calciumsilicate at 11 lbs/cubic foot. Thelisted thickncsses and rveights ofcombinction covcling arc thesums oI the inner layer of dia-tom&ceous clLrth at 21 lbs/cubicfoot and the outer laycr at11 ibs/cubic foot,, Insulqtion $'cights include al-]O$an(:os 1or \\-rre, cemcni, can-ves,.1'ends and- pflint, but notSlrcr-li1l sUl I3CC IlnlSnCS.

To find the rvcight of coveringon flrngr-s, valvcs or fittings,multitilj the $ c;glrt fxctor by theNeight t)cr foot of covering usedon.sirlrigLt pipe,

v srvc \{crgn[s crc apl)roxl-mate. \Vhen possiblc, obtain\\cights from the manuf&cturer.

Cast ilon velve \Yeights &re forflanged end valves; stecl weightsfor u clding end valvbs.

All fl:rnged fitting, flangedvalve and fiange \\riHhts includethe proportion:rl \cight of boltsor studs to make up all joints.

r7,N

4!44

D', .SB--rl

@IU

+<trc

.\oro. Thi, k.,ln.

* 16 lb cu. ft. deDsity.


WEIGTITS OF PIPING ]TATURI,\LS zo" o.D. 20" ptpp.

tof\w{l\L=I

F4'1f-l

LJ-!

z

z

Tempcrriurc llrnge "F

,-8./Ai

A,N

le-{

@flr\

+<{rc

z

z

z

Magnesis,Calcium

Combina-!ton

Fiber-Sodium

4dJ$

$fuNjisqlss,rs

Roldfrce tvpe is \\'cight inpounrls. Liglrtfrce t) po bcnerth\\ciglrt. is Neiglrt flctor for

Insulrtion thit knesses rnd$cights rrc b.rscd on avcrrqecurditions rLnd do not roD-qtituiel reconrmt'ntlrtiori for. spccificthickncssrs ol nlrtolirls. Iirsula-tion rveights ruc brscrL on 55.,1rn:rgncsio :rnd hldrous rllriumsilicllte rrt 1l lbs ruhit foot. Thelistc(l tl\i(,lincsscs r!n(l rvcislrts ofcomlrinotidl covoring rio thesums of the irrner l:uer of rlir-tornsceous crrth rlt 21 lbs (ul)icIoot .rnd thc outcr la|er rt11 ll)s'(ul)i0 foot.

hsu|rtion Noights irv.ludc r1-lorvlrnccs for ivir{], (cmont. (1!n-vrs, brnds url prLint, but, notsp(,( ill su frlco {inishos.

To liud tho \\ c;ght of covcring,rrr l1lLrrg' s, vrl!(s or fittirrgs,rnLrltitlt tlrt $eielrt hrrtor l,r thcleig[i lrcl foot JI coverirrA uscdorr stlLlight l)il)c.|itlvu $1 i{lrts rlc rr)r,ro\i-nrxto. \\'lrrn possil,lc, -ol,trinn1'ights from thc nlxnulllcturcr.

OlL\t, irorr vxlvc NoigLts urc forfllugctl cnrl v:rlvcs; stccl Neightsfor lel<lins end vulvrs.

-\ll fitngcrl Iitting, ilrrngcdvlllve &n(l llllngc ryci'alrts il)(ludcthc prol)ortionrl \ycight of l)oltsor studs to mrke up ull joir)ts.

z

z

Pipe-Lbs/ I t$ atcr-l,bs,,lft

1100-r200

Flanged BonnetGlobe or Angle

' 16 Ib cu. ft. deDsity.


24" ptpB 24, o.D. \YItIGIII'S OF PIPING IIATEITL\LS

'I-cnDer:lturc llcngc "F

Magnesia,Calcium

Conrbine-tiolr

Fiber-Sodium

zt-

z

z9

z

zj

z

f,.d1,!J

{Gt\w{i}1_'*,.1

14'1

-t/i\Lr----ti--t\*t"J

ffi+r[1$

N+Sl:N

Boldfrre ,t\pe is weight inpounos, Lrgnlttce tJpe benexthlleight. is $eight factor for

Insulation thicknesses and\reights are based on averageconditions &nd do not constituiea recommendction for specificthicknesscs of msterials. Insule-tion $cights are b.rsed on 857,magncsia and hydrous calciumsilicste rt 1l lbs/cubic foot. Thelisted thicknesses and \eiqhts ofcombination covering are thesums of the inner laver of dia-tomaceous certh at 2l lbs/cubicfoot and the outer laver atll lbs/cubic foot., Insulation *eights include al-loNMces 1ot wlre, cement, can-vas, bands and paint, but notspeciel surface finishes.

To find the geieht of coverinson Banges, vrlvis or 6ttinss]muJtipll the rreight factor by the\aeiglrt.lrrr foot ol covering usedon sLrsrght DlDe.

\'Rlvt $ciehts rre annroxi-mxtc. \\'hen- possiblc,

-obtain

\'eights from thi manuflcturer.Cast iron valve \icights :rre for

frengcd end v:rlvcs, steeJ *eightslor \reldrng end vslves.

A)l flerrged tittins, 63nsedvxlve ar)d {lrngc seights includethc proporiiunxl \reight of boltsor sLuds to mekc up all ioints.

/14

,N/>l,, .{D---S

ffi@

fi1

J-<trc

3

\\'rll Dcsigrr,rtiou

Norn. TLick., In.

* 16 lb cu. ft. deDity.

I

Appendix C: Properties of pipe E7

W!]I(;I]TS OF PIPING MATERIALS za" o.o 26tt prpt

7F

F

fifu-r'hIL4J

{l}E=:l

-4\",TIri\

r-r--ru/llagnesi.r

= Celcirrm

o irrltcateF

/'41t4

,N/>aglg

B,sHt

F<]@

lll')

+<fFqJ


A tion3 ;r:r:::::

Fiber-Sodium

Boldface tvDe is weisht inpounds. Lighiface typ"e be-neath weight is weight factorror lnsulalron.

Insulation thicknesses andweights are based on averageconditions and do not consai-tute a recommendation forspecific thicknesses of mate-rials. Insulation weights alebased on 85% masnesia andhydrous calcium siiicate at 11lbs/cubic foot. The listed thick-nesses and weights of combi-nation covering are the sumsof the inner later of diatoma-ceous ealth it 21 lbs/cubicfoot and the outer laver ati1 lbs/cubic foot.

Insulation weiphts includeallowances for w_ire, cement,canvas, bands and paint. butnot special surface finishes.

To find the weiqht of cover-ing on flanges, valves or fit-tings, multiply the weight fac-tor by the weight per foot ofcovering used on straiqht Dipe.

Valve weishts are aoorbii-mate. When- possible. bbtainweights from manufacturer-

Cast iron valve weights arefor flanged end valves: steelweights forweldinqendvalyes.

All flanged fittlng, flangedvalve and flange welghts in-clude.the prolo.rtiohal weightoI oorls or studs to make uDall joints.

z3

z

ffi$S{''l$N-l-sd\slN|]

* 16 lb cu. ft. densit\-.


28" prpn 28" o.D. WEIGHTS OF PIPING MATERIALS


trIxgnesiaCclcium

Combina-!ron

FiberSodium

ff&?fw{i\E::I-4\.

t-t-!f---Fr\iJ

F

z

F

ffi&NisEN

zFF

z

ll

Boldface type is weight inpounds. Lishtface tvDe be-neath weigii is weighi'factorrot lnsulatron.

Insulation thicknesses andweights are based on averageconolllons and do not constl-tute a recommendation forspecific thicknesses of mate-rials. Insulation weiphts arebased on 8570 masnesia andhydrous calcium silicat4 at 11lbs/cubic foot. The listed thick-nesses and weishts of combi-nation coverind are the sumsof the inner lafer of diatoma-ceous earth at 21 lbs/cubicfoot and the outer laver at11 lbs/cubic foot,

Insulation weiehts includeallowances for wlre, cement,canvas, bands and paint. buthot special surface

-6nishes.

To find the weisht of covei-ing on flanges, v-alves or fit-tings, multiply the weight fac-tor by the weight per foot ofcovering used.on straight pipe.

v arve welghts are approxt-mat€. When possible, obtainweights from manufacturer,

Cast iron valve weishts arefor flanged ehd valves; steelweightsforweldingend valves.

All flansed fittins. flansedvalve and-flanse wiiehts "in-clude.the propo-!tional- wei ghtol Dolts or studs to make ur)all joints,

A#,N/9N

D', 'ilF-Jl

l"<3@

m+<irc

l 16 li cu. ft. density.

if,

Appendix C: Properties of Pipre Ag

\ 'EIC I ITS ()F'PIPIN'; IIIATFIRTALS Bo'o.D. 30" "rpe

45

u-r'

! ii{i\lj:I.4\"

-!----l\tJ

z

z

i


\IagnesiaCalcium

Fiber-Sodirrm

Boldface ti,pe is weight inpounds. Lightface type be-neath w€ight is weight factorIOr lnsulailon.

Insulation thicknesses andweights are based on averageconditions and do not consti-tute a recommendation forspecific thicknesses of mate-lials. Insulation vreights arebased on 85i. maqn-sia andhydrous calcium siticate at 11lbs/cubic foot. The listed thick-nesses and weights of combi-nation covering are the sumsof the inner layer of diatoma-ceous earth at 21 lbs/cubicfoot and the outer layer. at1l lbs/cubic foot-

Insulation weights includeallorvances for w-ire, cement,canvas, bands and paint, butnot special surface ffnishes.

To find the weight of cover-ing on flanges, valves or fit-tings, multiply the weight fac-tor by the weight per foot ofcovering used.on straight pipe.

v alve werEhts are approxr-mate. When possible,;btainweights from manufacturer.

Cast iron valve weights arefor flanged end valves; steelweights for weldingend valves.

All flanged 6tting, flangedvalve anO nanqe werghts rn-clude,the proportionai- wei ghtoI, oolEs or studs !o make upalI Joln!s.

z

zF

7

ffisf,J$Nl${f.,-::r:q}

4l

,\

B,s

i;>tu>/ ltl\ .ll,

@te$-+


{!-r'

I i){l\L-Li

E:-:tf,t\ri\\tJ

zF

2

b


32" prcn 82, o.D. WEIGHTS OF PIPING MATERIALS

Temperature Range .F

MagnesiaCalcium

Z Siliccte

l UOmOrna-5 tion

Fib€r-Sodium

zFtr

z

ffi$fufs],mqJt.rrr.:qs

Boldface type is weight inpounds, Lightface type be-neath weight is weight factoffor insulation.

Insulation thicknesses andweights are based on averageconditions and do not consti-tute a recommendation fo!specifrc thicknesses of hate-rials. Insulation weights arebased on 85% magnesia andhydrous calcium silicat€ at 11lbs/cubic foot. The listed thick-nesses and weights of combi-nation covering are the sumsof the inner lay€r of diatoma-ceous earth at 21 lbs/cubicfoot and the oute! layer at11 lbs/cubic foot.

Insulation weights includeallowances for wire, cement,canvas, bands and paint, butnot special surface finishes,

To find the weight of cover-ing on flanges, valves or fit-tings, multiply the weight fac-tor by the weighi per foot ofcovering used on straight pipe.

Valve weiEhts are approxi-mate. When possible, obtainweights from manufacturer.

Cast iron valve weights arefor flanEed end valves; steelweights forweldingendvalves,

All flanged -fi tting, flangedvarve ano nange werEhts ln-clude the proDortional weishtof bolts o; stjuds to make-upall joints.

AI/.4

dA9.4

D' .fB_{i

Fdl@

D+<irc

' 16 lt cu. ft. density.

!r

Appendix C: Properties of Pipe 3Ol

WEIGHTS OF PIPING MATERIALS 84'o.D. 34" prpt

zFtr

z

F

ztF

z

ATJ-/

b{T\//\"E_=_=iI

"t\\IJTemperature Range "F

MagnesiaCalcium

Fiber-Sodium

ffistitsSql-s$

N

23

Boldface type is weight inpounds. Lightface type be-neath weight is weight factorfor insulation.

Insulation thicknesses andweights are based on averageconditions and do not consti-tute a recommendation forspecific thicknesses of mate-rials. Insulation weights arebased on 85% magnesia andhydrous calcium silicat€ at 11lbs/cubic foot. The listed thick-nesses and weiehts of combi-nation coverine_ are the sumsof the inner laier of diatoma-ceous earth ai 21 lbs/cubicfoot and the outer layer at11 lbs/cubic foot.

Insulation weights includeallowances for u'ire, cement,canvas, bands and paint, butnot special surface finishes.

To find the weight of cover-ing on flanges, valves or frt-tings, multiply the weight fac-tor by the weight per foot ofcovering used on straight pipe.

Valve weights are approxi-mata. When possible, obtainweights from manufacturer.

Cast iron valve weights arefor flanged end valves; steelweights lor weldingend valves.

All flanged fitting, flangedvalve and flange weights in-clude,the proportional weightoI Dolrs or stucts to make uDall joints.

.-al/AJAI

/14

A|i'a

+.{@a

+<irc

r 16 lb cu. ft. deDsity.


36tt prpo s6, o.D. WEIGHTS OF PIPING MATERIALS


Ilrgnesir

Fiber-Sodirm

f.;

{t//\I tt{}dJ44"

L_r-!

\tJ

Crlcilrm.

Ftr

;

Com!ton

zFJ

z

ffi6{-,l$N*S$:T,\1I

F

z

Boldface type is weight inpounds. Lightface type be-neath weight is weight factorIOr lnsulatron_

Insulation thicknesses and.,r,eights are based on averageconditions and do not consti-tute a recommendation forspecific thicknesses of mate-rials. Insulation weights arebased on 85% rnagnisia andhyd.ous calcium silicate at 11ibs/cubic foot.The listed thick-nesses and weights of combi-nation covering are the sumsof the inner layer of diatoma-ceous earth at 21 lbs/cubicfoot and the outer layer at11 lbslcubic foot,

Insulation weights includeallowances for wire, cement,canvas, bands and paint, butnot special surface finishes.

To find the weight of covet-ing on flanges, valves or fit-tings, multiply the weight fac-tor by the weight per foot ofcoveri ng used.on straight pipe.

v arve wergnrs are approxl-mate. When possible, obtainweights from manufacturer.

Cast iron valve weiqhts arefor flanged end valveis; steelweights forweldingend valves.

All flanged fitting, flangedvalve ano nange werghts rn-clude the proportional weiehtof. bolts oi siuds to make'upall lolnls.

.4/.4/.--tl

,\

ll' 'rlF--+l

Fd3F{]

fi^l

+<lFsO

Water-Lbs/Ft

L.R.90' Elbow


TO CONVERT I NTO

Alphabetical Conversion FactorsMULTIPLY 8Y TO COI{VERT ll{T0 MULTIPLY BY

Abcoulomb

Acre

acres

acresacresacre-feetacre'feetamperes/sq cmamperes/5q cmamperes/sq In.ampetes/sq rn.amperes/sq meteramperes/sq mete.ampere,hoursarnpere-hoursampere-turnsampere-turns/cmampere-turns/cmampere-turns/cmampere.turns/in.ampere-turns/ In,ampere-lurns/tn,ampere-turns/metetampere-turns/meterampete-tufns/metelAngstrom unitAngstrom unitAngstrom unit

atesaresAstronomical UnitAtmospheresatmospneresatmospheresatmospheresatmospheresalmospheresalmospheresatmospheres

Barrels (U.S., dry)Earrels (U,S., dry)Barrels (U.S., liquid)barrels (ojl)oarsbarsDarsbarsbarsBatylEolt {US Cloth)BTU8tu8tuBtubtuBtu8tuBtuBtuBtu /hr

A

StatcoulombsSq. chajn (Gunters)RodsSquare links (Gunters)Hectaae or

sq. hectometersq feetsq meterssq mrlessq yardscu feetga onsamps/sq In.amps/sq meteramps/sq cfiamps/sq meteramps/sq cmanps/sq in.coutonbsfaradaysgrlbertsamp-turns/ In.amp{urns/meterSalberts/cmamp-turns/cmamp-turns/metergrlberts/cmamp/Iurns/cmamp{urns/in.gilberts/cmtncnMeterMicron or {Mu)Acre (US)sq. yardsacreSsq metersKilometersTon/sq. inchcms of mercuryft of water (at 4"C)in. of mercury (at 0.C)xgs/ sq cmkgs/sq meterPounds/sq in.tons/sq ft

B

cu, Inchesquarts (dry)ga onsgallons (oil)atmospheresoynes/sq cmkgs/sq meterpounds/sq ftpounds/sq in.Dyne/sq. cm.MetersLiter-Atmosphereergsloot-lbsgram-caioneshofsepower-hrsjouleskrlogram,caloriesxrogram-rheterskrlowatt-hrsfoot-pounds/sec

2.998 x l0'ot0160

lx1O5

.404743,560.0

4,O47.1.562 x l0 '4,840,

43,560.03.259 x 1056.452

l0l0.1550

1,550.010.6.452 x 10 .

3,500.00.037311.2572.540

100.0t.2570.3937

39.370.49500.01o.02440.o1257

3937 x 10-'I x l0-r'1x 10-..0247 |

119,60o.02471

100.01.495 x 10.007348

76.O33.9029.92

1.033310,332.

14.701.058

7056.105.0

42.O0.9869

1061.020 x 10.

2,089,14.50

1.000

10.4091.0550 x 10'o

77a.3

3.931 x 10 |1,054.8

o.2520r07.5

2,928 x lO-.o.2162

Btu/hrBtu/hrBtu/hrBtu/min8tu/minBtu/minBtu/manBtu/sq ft/minBucket (Br. dry)bushelsbushelsbushelsbushelsbushelsbushelsbushels

Calories, gram (mean)Candle/sq. cmCandle/sq. inchcentares (centiares)CentigradecentigramsCentiliterCentiliterCentilitercentiliterscentimeterscentimetelscentrmeterscentimeterscentimeterscentameterscentimeterscentimetelscentimeteFdynescentimeteFdynescentimeter-dynescentimeter-gramscentimetergramscentrmeter-gfamscentjmeters of mercurycentrmeters of mercurycentimeters of mercurycentimeters of rnercurycentimeters of mercurycentameters/s?ccentimeterc/seccentimeters/seccentimeters/seccentimeterc/seccentimeters/seccentimeters/seccentimeterc/sec/seccentimeters/sec/seccentimeters/sec/seccentimeters/sec/secChainChainChains (surveyors'

or Gunter's)circular milscircular milsCircumferencecircular milsCordsCord feetCoulombcoulombs

Sran-cal/sechorsepoweahrswattstoot-lbs/sechoasepowerkilowattswattswatts/sq in.Cubic Cm.cu ftcu in.cu metersliterspeckspints (dry)quarts (dry)

8.T.U. (mean)LambedsLambertssq metersFahrenheitgramsounce ftuid (US)Cubic inchdramslitersfeetincheskilometersmetersmilesmillimetersm ilsyardscm-8ransmeter-kgsPound-feetcfi-dynesmeter-kgspound{eetatmospneresfeet of waterkgs/sq meterpounds/sq ftPounds/sq in.teet/minleet/seckilometels/hrknotsmeters/minm iles/hrfiiles/minfeet/sec/seckms/hrlsecmeters/sec/secmiles/hrlsecInchesmeters

yardssq cmssq milsRadianssq inchescord feetcu. feetStatcoulombslaradays

0.07003.929 x 10-.0.2931

12.960.023560.01757

17.57o.t22l1.818 x 101.2445

2,t50.40.03524

35.244.0

64.032.0

3.9685 x 10-:3.142

.44701.0

(C'x9/5)+320.01

.3382

.61032.7050.0r3.281 x 10-'0.3937

10- 5

0.016,214 x LO-r

10.0

l-094 x 10-I1.020 x l0-'1.020 x 10-.7.376 x 10 |

980.710 -5

7.233 x l0-50.013160.4461

136.0

0.19341.19690.032810.0360.1943

o.022373.728 x 10-.0.032810.0360.01o.02237

792.0O20.12

22.OO5.057 x l0-.0.7854

7.854 x 10-'8

2.998 x 101.036 x 10-!

L

Appendix D: Conversion Factors

(Continued). Alphabetical Conversion Factors

INTO MUI.TIPLY BY

305

TO CONVERT TO CONVERT INTO MULTIPLY BY

coulomb9/sq cmcoulombs/sq cmcoulombs/sq in.coulombs/sq in.coulombs/sq metercoulombs/sq metercubic centimeterscubic centimete6cubac centimeterscubic centimeteascubic centimeterscubic centimeterscubic centimeterscubic cent;meterscubic leetcubic feetcubic feetcubic leetcubic teetcubic feetcubic feetcubic feetcubic feetcubic feet/mincubic teet/mincubic teet/mincubic feet/mincubic feet/seccubic feet/seccubic inchescubic inchescubic inchescubic inchescublc inchescubic inchescubic inchescubic inchescubic inchescubic meterscuDrc meterscubic meterscubic meterscubic meterscubic rneterccubac meterscubac meterscuorc meterscubic yards

cuFic yardscuorc yatoscubic yardscuDrc yards

cubic yards/mincubic yards/mincubic yards/min

coulombs/sq in.coulombs/sq metercoulombs/sq cmcoulombs/sq metercoulofibs/sq crhcoulombs/sq in.cu feetcu Incnescu mete6cu yardsgallons (U. S. liq.)literspints (U.S. tiq.)quats (U.S. liq.)bushels (dry)cu cmscu inchescu meterscu yardsgallons (u.S. liq.)literspints (U.S. liq.)quarts (U.S. liq.)cu cns/secgallons/secliters/secpounds of water/minmillion gals/daySallons/mincu cmscu feetcu metetscu yardsgallonslite.smal-feetpints {U.S. tiq.)quarts tU.S. liq.)bushels (dry)cu cmscu feetcu inchescu yardseallons (U.S. liq.)laterspints (U.S. liq.)quarts (U.S. liq.)cu cmscu feetcu inchescu metersSallons (U.S. ljq.)literspints (U.S. liq.)quarts (U.S. ljq.)cubic ftlsecgallon5/seclrters/sec

0

GramsecondsgramsIrersmeterSqua0ranr5raclransSeconos

1010.1550

10-.6.452 x 10-'3.531 x 10 5

0.0610210-6

1.308 x 10-.2.642 x 10-.0.0012.113 x 10-l1.057 x 10-'0.8036

-24320.O|,728.O0.028320.037047.44052

243259.8429.92

472.0o.12470.4720

62.430.646317

448.831

5.787 x l0-.1.639 x 10-s2.I43 x 10-54.329 x t0-30.016391,061 x 1050.03463o.ot132

28.38106

35.3161,023.0

1.308264.2

r,000.02,113.0r,057.

7.646 x 10527.O

46,656.0o.7646

202.O764.6

1,615.9807.9

0.453.367

12.7 4

olamsdramsdramsDyne/cmoyne/sq. cm.Dyne/sq. cm.Dyne/sq. cm.dynesoynesdynesdynesdynesdynesoynes/sq cfi

EtlErl

Em, Pica

erg5erg5ergsergs

ergs

ergsergs

faradsFaraday/secfaradaysfaradaysFathomfathomsleet{eetfeetfeetfeetfeetteetfeet of waterfeet of waterfeet of water

grafisgrainsouncesErglsq. millimete.

Inch of lVercury at 0'CInch of Water at 4'Cgramsjoules/cmjoules/meter (newtons)kilogramspoun0a t5poundsoars

um.tncnestncnum,Dyne - cm/secBtudyne-centrmetersfoofpounds

gram.cmShorsepower-hrslouleskg-calorieskg-meterskilowatlhrswatt'houtsBtu/minft-lbs/minft-lbs/sechorsepo\derkg-calories/minkilowatts

mrcrofaradsAmpere (absolute)ampere-hourscoulombSNleterfeetcentimeterskrlometersmetersmrles (naut,)miles (stat.)millimetersm ilsarmospneresrn, of mercurykgs/sq cm

0.017450.16672.778 x LO .

10.010.010.0

0.r371429

0.125

1.771827.3437

0.0625.01

9.869 x 10-'2.953 x l0 5

4.015 x l0 .1.020 x 10-r

l0-'l0- 5

1.020 x 10-.7.233 x 10-,2.248 x lO 6

10_6

114,3045

.42331.0009.480 x 10-r'1.0

7.367 x 10-i0.2389 x 10-'1.020 x 10-:3.7250 x 10-la

10 r2.389 x l0 -rr1.020 x 10 |O.2J78x IO tt

0.2778 x 10 -'o

5,688 x 10 '4.427 x lO-67.3756 x l0-l1.341 x l0-ro1.433 x 10 -,

10- r0

1069.6500 x lcr.

26.809.649 x lcr.1.8288046.0

30.483.048 x l0-r0.30481.645 x 10-.1.894 x 10 .

304.81.2 x I Cl.

0.029500.88260.03048

degrees/sec radians/secdegrees/sec revolutions/mindegrees/sec revolutions/secdekagrafis gGmsdekaliters litersdekameters metetsDrams (apothecaries'

or troy) ounces {avoidupois}Drams (apothecaries'

or troy) ounces (troy)Drams (U.S.,

fluid or apoth.) cubic cfi.

Daltondaysdecrgramsdecilitersoecrmetersdegrees (angte)degrees (angte)degrees (angle)

1.650 x 1.0-r.86,400.0

0.10.10.10.01r1t0.01745

3,600.0


TO CONVERT I r'lT0


MULTIPLY BY TO CONVERT INTO MULTIPLY gY

feet of waterfeet of waterfeet of waterleet/ninfeet/minteet/minfeet/ rn infeet/ rninfeet/secfeet/secteet/secfeet/secfeet/secleet/secleet/sec/secfeet/sec/secfeet/sec/secfeet/sec/secfeet/ 100 feetFoot - candle

foofpoundsfoot-pounds

foofpoundsfoot.poundsfoot-poundsfoo!poundsfoot'pounds/ minfoot-pounds/ mintoot-pounds/mintoofpounds/minfoot-pounds/manfoot-pounds/secfoot.pounds/secfoot-pounds/sec{oot-pounds/secfoot-pounds/secFurlongsfurlongsturlongs

gallonsSallonsSalronsgallonsSallonsgallonsgallons (liq. Br. Imp.)gallons (U.S.)gallons of watergallons / m ingallons/mingallons / m ingaussesgaussesgaussesgaussesgilbertsgilberts/cmgilberts/cmgilberts/cmGills (Britash)gillsgillsGradeG€ins

kgs/sq meterpounds/sq ftpoLrnds/sq in.cms/secfeet/seckms/hrmeters/minmiles/hrcms/seckms/hrknot5meters/minmiles/hrmiles/mincms/sec/seckms/hrlsecmeters/sec/secmiles/hrlsecper cent graoeLumen/sq. meterBtuergsgram-caloriesnp-nrsjoulesKg-ca{oneskg-meterskilovr'att-hrsBtu/minfoot-pounds/sec

kg-calories/ minkilowatts8tu/hrBtu/minhorsepowerkg'calories/mankilowattsmiles (U.S.)rodsfeet

cu cmscu feetcu inchescu meterscu yardslitersgallons (U.S. !iq.)gallons (lmp.)pounds oJ watercu ft/secliters/seccu ftlhrlines/sq in.weoers/sq cmwebers/sq in.webers/sq meterampere-turnsamp-turns/cmamp{urns/jnamp{urns/metercubrc cm.literspints (liq.)Radiandrams (avoirdupois)

304.862.43

0.43350.50800.016670.018290.30480.01136

30.481.0970.5921

14.290.68180.01136

30.481.0970.30480.68181.0

10.764r.286 x 10 l1.356 x 1070.32385.050 x l0 '3.24 x 10-.0.13833.766 x 10 ,1.286 x 10 !0.016673.030 x 10 -'3,24 x 10 .2.260 x lO- 5

o.o77171.818 x 10-'0.01945r.356 x 10-'0.125

40.0660.0

3,785.00.1337

231.03.785 x 10-'4.951 x 10-13.7851.200950.832678.34532.228 x lO- I0.063088.02086.452

10 |6.452 x 10-l

10-.0.79580.79582.021

79.58t4?.07

0.11830.25

.0t5710.035s7143

g.ains {troy)grains (troy)grains (troy)grains (troy)grains/l.J.S. galSrains/U.S. galgraans/ lmp. galSrarnsgramsgramsSramsSramsSrarnsgramsgramsSramsSramsgrarns/cmgrams/cu cmgrams/cu cmgrams/cu cmgrams/ literSrams/ litergrams/ litergrarns/litergrams/sq cmgram'calofle5Sram-catoflesgram-caloriesgram-catofle5gram-calonesgram-calo esgmm-calories/secgram-centimetersgram-centimetersgram-centimetersgram-centimetersgram-centimeters

HandnecrareshectareshectogramshectolitershectometershectowattshennesHogsheads {British)Hogsheads (U.S.)Hogsheads (U.S.)horsepowerho15epowerhorsepowerhorsepower (metric)

(542.5 ft lb/sec)horsepower

(550 ft lb/sec)horsepowerhorsepowerhorsepowerhorsepower (boiler)horsepower {boiler)norsepower-nrsnorsepower-ntsnorsepower-nrshorsepower-hrsnorsepower-hrs

grains (avdp)gramsounces {avdp)pennyweight (troy)parts/millionpounds/million galparts/milliondynesgrainsjoules/cmjoules/rneter (newtons)kilogramsmilligramsounces (avdp)ouhces (troy)poundalspoundspounds/inchpounds/cu ftpounds/cu inpounds/mil-footgraans/galpounds/ galpounds/cu ftparts/mallaonpounds/sq tt8tuergSfoot-poundshorsepower-hrskilowatt-hrswatt-hrsBtu/hrBtuergsjouleskg-calkg-meters

H

Cm.

sq feetgramslitersmeterswattsmrllihenriescubic ft.cubic ft.Sallons (U,S.)Bt!/minfoot'lbs/ minfoot-lbs/sechorsepower

(550 ft lb/sec)horsepower (metric)

(542.5 ft lb/sec)kg-calories/minkilowattswattsBt!/hrkilowattsBtuergsloot-lbsgram-caloriesjoules

1.00.064802.0833 x 10-10.04167

17.118142.86

14.286980.7

15.439.807 x 10-!9.807 x 10-r0.001

1,000.o.035270.032150.070932.205 x 10-r5.600 x 10-l

62.430.036133.405 x 10-'

58.4178.3450.062427

1,000.02.04813.9683 x 10-!4.1868 x l0'3.08801.5596 x 10 .1.1630 x l0-.1.1630 x l0-3

t4.2a69.297 x 10 .

980.79.807 x l0-!2.343 x 10-r

l0 -5

10.162.47 |1,076 x 10'

100.0r00.0100.0100.0

1,000.010.1148.42184

4?.4433,000.

550.00.9863

1.014

10.680.7 457

7 45.733.479

9.8032,547.

2.6845 x 10r'1.98 x 1Cl'

641,190.2,684 x 10.

1|

TO CONVERT t1{T0

Appendix D:


i,ULTIPLY BY TO CONVERT INTO

Conversion Factors T7

MULTIPLY 8Y

horsepower-hlshorsepowet-nlshorsepower-hrshourshoulsHund.edvreiShts (long)Hundredweights (long)Hundredwei ghts (short)Hundredweights (short)Hundredweights (short)Hundredweights (short)

inchesinchesinchesinchesinchesinchesanches of mercuryinches ol mercuryinches of mercuryinches of mercuryinches of mercuryinches of mercuryinches of water (at 4'C)inches ot wate. (at 4'C)inches of water (at 4'C)inches of water (at 4'C)inches oI water (at 4'C)inches of water (at 4'C)lnternational AmpereInternational Voltlnternational voltlnternational volt

kg.calorieskg-meterskilowatt-hrsdays

poundstons (long)ounces (avoirdupois)poundstons (metric)tons (long)

I

centimetersmetersmilesmillimetersmtlsyardsatmospneresfeet of waterkgs/sq cmkgs/sq meterpounds/sq ftpounds/sq in.almospneresinches of mercuryKgs/sq cmounces/sq In.pounds/sq ftpounds/sq in.Ampere (absolute)Volts (absolute)Joules (absolute)JOUIeS

)

641.12.737 x LAo.74574.167 x 10-t5.952 x 10-t

1120.05

1600100

0.04535920.0446429

2.5402.540 x 10-r1.578 x 10-,

25.40I,000.0

2.778 x lo-10.033421.1330.03453

345.370.73o.49122.458 x 10 I0.073552.540 x 10-10.57815.2040.03613

.99981.0003

1-593 x 10- "9.554 x 10'

Btu 9.480 x 10-'ergs 107foot-pounds 0.7376kg-calories 2.389 x l0-'kg-meters 0.1020watGhrs 2,778 x lO-'grams 1.020 x 10.dynes 10tjoules/meter(newtons) 100.0poundals 723.3pounds 22,44

K

dynes 980,665.grams 1,000.0joules/cm 0.09807joules/meter(newtons) 9.8Q7poundals 70.93pounds 2,205tons (long) 9.842 x 10-ltons (short) 1.102 x !0 'grams/cu cm 0.001pounds/cu ft 0.06243pounds/cu in. 3.613 x l0-'pounds/mil-foot 3.405 x l0-'opounds/ft 0.6720Dynes 980,665atmospheres 0.9678leet of water 32.81

inches of mercury 24.96pounds/sq ft 2,048.pounds/sq in. 14.22atmospheres 9.678 x l0-'bars 98.07 x 10 'feet ot water 3.281 t 10 1

inches of mercury 2.896 x 10-1pounds/sq ft 0.2048pounds/sq in. 1.422 x 10 tkgs/sq meter lcl'Btu 3.968foot-pounds 3,088.hp-hrs 1.560 x l0 Ijoules 4,185.kg-meters 426.9kiiojoules 4.186kilowatt-hrs 1.163 x 10-'Btu 9.294 x l0 Iergs 9.804 x 10tfoot-pounds 7,233joules 9.804kg-calories 2.342 x 1O-'kilowatt.hrs 2,723 x lO'.maxwells 1,000.0liters 1,000.0centimeters lotfeet 3,281,inches 3.937 x 1Cl.

meters 1,000.0miles 0,62f 4millimeters lOyards 1,094.cms/sec 27.74feet/min 54.68feet/sec 0.9113knots 0.5396merets/ fltn lt.t /miles/hr 0.6214cms/ sec/ sec zl,Iaft/sec/sec 0.9113meters/sec/sec 0.2718mifes/hrlsec 0.6214Btu/min 55.92foot-lbs/min 4,426 x Wfoot-lbs/sec 737.6horsepower 1.341kg-calo.ies/min 14.34watts 1,000,0Btu 3,413.ergs 3.600 x 10rtfootlbs 2.655 x 106gram-calories 859,850.horsepower-hrs 1.341joules 3.6 x lcl.kg-calories 860.5kg-meters 3.671 x lotpounds of waterevaporated from andat212'F. 3.53pounds of water raisedtuon62'to272'F. 22.75teet/hr 6,080.kilometers/hr 1.8532nautical miles/hr 1.0statute miles/hr 1.151

kilogramskilogramskilogramskilogramskilogramskilogramskilogramskilogramskjlograms/cu meterkilograms/cu meterkilog.ams/cu meterkilograms/cu meterkilograms/meterKaloSram/sq. cm,kilograms/sq cmkilograms/sq cm

kilograms/sq cmkalograms/sq cmkilograms/sq cmkilograms/sq heterkilograms/sq meterkilograms/sq meterkilograms/sq meterkilograms/sq meterkilograms/sq rneterkilograms/sq mmkilogram-calorieskilogram-calorieskilogram-calorieskilogram-calorieskilogram.calorieskilogram-calorieskilogram-calorieskalogram meterskiiogram meterskilogram meterskilogram meterskilogram meterskilogram meterskilolineskiloliterskilometerskilometerskilometerskilometerskilometerskilometerskilometerskilometers/hrkilometers/hrkilometers/hrkilometers/hrkilometers/hrkilometers/hrkilometers/hr/seckilometerc/hr/seckilometers/hrlseckilometers/hr/seckilowattskilowattskilowattskilowattskilowattskilowattskilowatt'hrskilowatt-hrskilowatt-hrskilowatt-hrskilowatt-hrskilowatt-hrskilowatt-hrskilowatt-hrskilowatt-hrs

kilowatt-hrs

knotst(hots

xnolsknots

JOulesjoulesjoulesjoulesjoulesioulesiouies/cmjoules/cmjoules/cmjoules/'cmjoules/cm


TO CONVERT INTO


MULTIPLY 8Y TO CONVERT INTO MULTIPLY BY

knotsKNOIS

leagueLight yearLjght yearlines/sq cmlines/sq an.

lines/sq in.lanes/sq in.lines/sq in.links {engineer's)links (surveyor's)literslrterSIitersliterslitersIte15literslrterslitersliters/minliters/rhinlumens/sq ftLumenLumenLumen/sq. ft.lur

megarnesmegohnsmegohmsrnetersmetersmetersmetersmetersmetersmetersmetersmetersmeters/minmbterc/minmeters/minmeters/manmeters/manmeters/minmelers/secfieters/secmeters/secmeters/secme(ers/5ecmelers/secmeters/sec/secmeters/sec/secmeters/sec/secmet€rs/sec/secmeteFkilogramsmeterkilogramsmeterkilogramsmrcrotaradmicrogramsmtcronms

Yards lhlfeet/sec

L

miles (approx.)MilesKilometers8aus5e5Sausses

webers/sq in.webers/sq metertncnesInchesbushels (U.S, dry)cu cmcu feetcu inchescu meterscu yardsgallons (U.S. liq.)pints {U.S. liq.)quarts (U.S. laq.)cu ftlsecgals/secfoot-candlesSpherical candle powerWattLumon/sq. meterfoot'candles

M

kilolines

mrcrohmsohmscentimetersfeettncheskilometersmiles {naut.}miles (stat.)millimetersyardsvarascms/secteet/minteet/seckms/hrknotsmales/hrfeet / minfeet/seckilometers/hrkilometers/minmiles/ hrmiles/mincms/sec/secft/sec /seckms/hrlsecmiles/hr/seccmdynescm-8lamspound-feetfaradsSramsmeEonms

2,027.1.689

3.05.9 x l.0r:9.46091 x 101!1.00.15501.550 x 10-rl0-l

1,550 x 10-'12.O7.920.02838

1,000.00.03531

6r.020.0011.308 x l0-!o.2642

1.0575.886 x 10-'4.403 x r0 '1.0

.079s8

.00149610.760.0929

0.001l0-.1C|'10r'1oi

100.03.281

0.0015.396 x 10-'6.214 x 10-'

r,000.01.0941.179

3.28r0.054580.060.032380.03728

196.83.2813.60.06

0.03728100.0

2.2379.807 x 1CP

105

l0-.

microhmsm icroliteIsMicronsmiles (naut.)miles (naut.)miles (naut,)miles (naut.)miles (naut.)miles (statute)males (statute)miles {statute)miles (statute)miles (statute)miles (statute)miles (statute)miles/hrmiles/hrmiles/hrmiles/hrmiles/hrmiles/hrm iles/hrmiles/hrmiles/hr/secmiles/hr/secrniles/hrlsecmales/hr/secniles/ minmiles/ m inmiles/minmiles/minmiles/minmil-feetmilliersMillim;cronsMilligramsmilligramsnilligrams/litermillihenriesmillilite|smillimetersmillimetersmillimete6millimetercmillimetersmillimetersmillimetersmillirnetersmillion Sals/daym ilslnilsrTr tlsmilsmilsminer's inchesMinims (British)Manims (U.S,, fluid)minutes (angles)minutes (angles)minutes (angles)minutes {angles)myr;agramsmytrametersmynawattS

nepersNewton

metersfeetkilometersmeteasmiles (statute)yaroscentimetersfeetIncheskilometercmetelsmales (naut.)yaroscms/secteet/minfeet/seckms/htkms/minknotsrfieters/minmiles/minc|hs/sec/secfeet/sec/seckms/hrlsecmeters/sec/seccms/5ecteet/seckms/minknots /rn inmiles/hrcu incheskilogramsmetersgra insSramsparts/millionhenraesliterscentimetersleetincheskilometersme(ersm rlesm rlsyardscu ttlseccentimetersfeetIncneSkilometersyaroscu ft/mincuDtc cm.cubtc cm.deSreesquaorantsradianssecondskilogramskalometerskilo,,ratts

N

decibelsDynes

10-.10-.1x 10-.

6,080.27

1,853.l.l.516

2,027.1,609 x l0'

6.336 x 10r.509

r,609.0.8684

1,760.44.7088.

1.467l.609o.o26a20.8584

26.42

44.701.4671.6090.4470

2,642.88.

1.6090.8684

60.09.425 x 10-.

1,000.1x 10-t

0.015432360.0011.00.0010.0010.13.281 x 10-t0.03937l0-.

0.0016.214 x 10-'

1.094 x 10-!1.547232.540 x 10-'8.333 x 10-!0.0012.540 x 10-32.778x 1O-,1.f,0.0591920.0616120.016671.852 x 10-.2.909 x l0-r

60.010.010.010.0

1x105

ohmsliters

TO CONVERT


MULTIPLY BY TO CONVERT INTO

Appendix D: Conlersion Factors 309

MULTIPLY 8Y

OHlvl (lnternational)ohmsohmsouncesouncesouncesouncesouncesouncesounceSounces (fluid)ounces (fluid)ounces (troy)ounces (troy)ounces {troy)ounces (troy)ounces (troy)Ounce/sq. inchounces/sq rn.

ParsecPatsecparts/millionparts/mill,onparts/millionPecks (British)Pecks (Britash)Pecks (U.S.)Pecks (U.S.)Pecks (U.S.)Pecks (U.S.)pennyweights {troy}pennyweights {troy)pennyweights (troy)pennyweights {troy)pints (dry)pints (liq.)pints (liq.)pints (liq.)pints (liq.)pints (liq,)pints (liq.)pints (liq.)pints (liq.)Planck's quanturnrotsePounds (avoirdupois)poundatsp0unoalspounoarspounoatspoundatspounoarspoundspoundspoundspoundspoundspoundspoundspoundspoundspoundspoundspoundspounds (troy)pounds (troy)

0

OHIVI (absolute)megohmsmrcrohmsdramsgrainsgramspoundsounces (troy)tons (long)tons (metricJcu incheslitersgrainsgramsounces (avdp.)pennyweights (troy)pounds (troy)0ynes/sq. cm.pounds/sq in.

P

MilesKilometersgrains/U.S. galgrains/lmp. galpounds/million galcubic incheslitersbushelscubic incheslitersquarts (dry)graansounces (troy)gramspounds (troy)cu rnchescu cms.cu feetcu inchescu mererScu yardsga||onslitersquarts (liq.)Erg - secondGram/cm. sec.ounces (troy)dynesgramsjoules/cmjoules/meter (newtons)kilogramspoundsdramsdynesgrainsgramsjoules/cmjoules/meter (newtons)krlogramsouncesounces {troy)pounoarspounds (t.oy)tons (short)grarnsgrams

1.000510-,10.

16.0

2a3495270.06250.91152.790 x l0-52.835 x 10 5

1.805o.02957

480.031.103481

1.0971420.0

0.083334309

0.0625

19 x 10u3.084 x 10rr0.05840.070168.345

554.69.091901

8.8095828

24.O0.05

4.1667 x 10-r

413.20.01671

24.474.732 x 10-'6.189 x 10-'0.125o.47320.56.624 x 10-111.00

14.5833t3,826.

14.101.383 x 10-10.13830.014100.03108

44.4823 x lcl'7,000.

0.044484.4480.4536

16.014.583332.t7

1.215280.0005

5,760.373.24177

pounds (troy)pounds (troy)pounds (troy)pounds (troy)pounds (troy)pounds (troy)pounds (troy)pounds of waterpounds of waterpounds of waterpounds of water/manpound-feetpound{eet

pounds/cu ftpounds/cu ttpounds/cu ftpounds/cu ftpounds/cu an.pounds/cu in.pounds/cu in.pounds/cu in.pounds/ftpounds/ in.pounds/mil-footpounds/sq ftpounds/sq ftpounds/sq ftpounds/sq ftpounds/sq ftpounds/sq rn.pounds/sq in.pounds/sq in.pounos/sq In.pounds/sq in.

quadrants (angle)quadrants {angle)quadrants (angle)quadrants (angle)quarts (dry)quarts (1,q.)quarts (liq.)quarts (laq.)quarts (liq.)quarts (liq.)quarts (liq.)quarts (liq.)

.ad iansradtansradiansradiansradians/secradians/secradians/secradians/sec/secrao rans / sec/ secradians/sec/secrevolutionsrevoru!onstevolr.rtionsrevolutions/minrevolut,ons/minrevolutions/min

ounces {avdp.)ounces (troy)pennyweights {troy)pounds (avdp.)tons {long)tons (metric)tons (shoft)cu feetcu Inchesga Ionscu ltlseccm-dynescm-gramsmeter-kgsgrams/cu cmkgs/cu meterpoun0s/cu rn,pounds/mil{ootgms/c(1 cmkgs/cu meterpounds/cu ftpounds/mil{ootkgs/metergms/cmgms/cu cmalmospneresfeet of waterinches of mercurykgs/sq rneterpounds/sq in.atmospheres

inches ot mercurykgs/sq meterpounds/sq ft

odeg/eesminutesrad ra nssecondscu Inchescu cmscu feetcu rnchescu rneterscu yards8aIonsliters

R

degreesmrnutesquaoran(5seconds

revol!tions/minrevolutions/secrevs/min/rninrevs/min/secrevs/sec /sec

quadranlsradtansoegrees/ sec€dians/sectevs/sec

12.0240.0

0.a224573.6735 x 10-r3.7324x 1o-'4.1143 x 10 '0.01602

27.640.1 1982.670 x 10-r1.356 x 10'

13,825.0.13830.01602

t6.025.787 x 10-'5.455 x 10-'

2.768 x l0'1,724.

9.425 x 10 6

1.488178.6

2.306 x 1Cr6

4.725 x lO-'0.016020.014144.8826.944 x 10-!0.068042.3072.036

703.1144.0

90.05,400.0

1.57I3.24 x 105

67.20946.4

0.03342

9.464 x l0-.1.238 x 10-!o.250.9463

3,438.0.63662.063 x 105

9.5490.1592

573.09.5490.1592

360.04.06.2a36.00.10470.01667


IO CONVERT INTO

(Continued). Alphabeticel

MULTIPLY BY

Conversion Factors

TO CONVERT INTO MULTIPLY 8Y

revolutions/rhin/minrevolutions/min/mina€volutions/min/minrevolut'ons/secrevolutrons/secrevolutions/secrevolutions/sec/secrevolutions/sec/secrevolutions/sec/secKOO

RodRods (Surveyors' meas-)rcds

Scruplesseconds (angle)seconds (angle)seconds (angle)seconds (angle)SlugSlugSpheresquare centimeterssquare centimeterssquare centimeterssquare centtmeterssquare centimeterssquare cenrmelerssquare centameterssquare feetsquare feetsquare feetsquare feetsquare feet

square feetsquare Incnessquare Inchessquare Inchessquare Inchessquare Inchessquare Incnessquare kilometerssquare kilometers5quare kilometerssquare kilometerssquare kiiometerssquare kilometerssquare kilometerssquare meterssquare meterssquare metetssquare meteassquare meterssquare meters

square miles

square rnilessquare miles

square millimeterssquare millimeterssquare millimeterssquare millimeterssquare mils

radians/sec/secrevs/min/secrevs/sec/secoegrees/ secradians/secrevs/mrnradians/sec /secrevs/min/minrevs/min/secChain (Gunters)I\retersyarosfeet

grains

minutesquadrantsradransKilogramPoundsSteradiansci.cular milssq feetsq inches

sq mrlessq millimeterssq yardsactescircular mils5q crhs

sq merers5q mlessq millimeterssq yardscircular milssq cmssq feetsq millimeterssq mrls

acressq cmssq ftsq Inches

sq milessq yardsacressq cmssq feetsq rncnessq mrlessq millimeterssq yardsacressq feetsq xmssq metercsq yardscircular milssq cmssq {eetsq Inchescircular mils

1.745 x l0 l0.016672.778x lO-.

360.06.283

60.0

3,600.060.0

5,029

16.5

202.778 x 10-.0.016673.087 x 10-.4.848 x 10-6

14.5932.17

1.973 x 10r1.076 x l0-r0.15500.00013.861 x 10 rr

100.01.196 x l0-.2.296 x \o-'1.833 x l0!

929.O144.0

0.092903.587 x 10-r9.290 x rd0.11111.273 x 1066.4526.944 x 10-!

645.2106

7,716 x I0 '247.1

1otr10.76 x 106

1.550 x l0'106

0.38611.196 x 1062.471 x 10 'l0

10.761,550.

3.861 x l0-'10.

1.196640.0

27.88 x 1062.5902.590 x 1063.098 x 106

1,973.0.0I

1.076 x l0-r1.550 x 10-l1.273

square milssquare mrlssquare yardssquare yarossquare yardssquare yardssquare yardssquare yardssquare yards

temperature('c) +213

temperature("c) + 17.78

temperature('F) +460

temperature (" F) -32tons (long)tons (long)tons (long)tons (metric)tons (metric)tons (short)tons (short)tons (short)tons (short)tons (short)tons (short)tons (short)tons (short)/sq tttons (sho.t)/sq fttons of water/24 hrstons ol water/24 hrstons of water/24 hrs

sq inchesactessq cfis

sq inchessq meterssq malessq millimeters

T

absolute temperature ('C)

temperature ('F)

absolute tenperatlre ("F)

temperature ('C)kilogramspoundstons (short)kilogramspoundskilogramsouncesounces (troy)poundspounds {troy)tons (long)tons (metric)kgs/sq meterpounds/sq in.pounds of water/hrgallons/mincu ft/ hr

Statvolts

w

Btu/hr8tu/minergs/secfoot-lbs/minfooflbs/secnorsepowerhorsepower (rnetric)kg-calories/minkilowattsB.T.U. (mean)/min.joules/sec.BtuerSsfoofpoundsEram-caloneSnorsepower-haskilogram-calorieskilogram-meterskalowatt-hrsWatt (absolute)maxwellskilolines

6.452 x 10-'10 -6

2.066 x 10-a8,361.

9.0

0.83613224 x 1O-,8.361x 10'

1,016.2,240.

1.1201,000.2,205.

907.184832,000.29,156.66

2,000.2,430.56

o.a92a70.9078

9,765.2,000.

0.166431.3349

.39370

.003336

1.0

1.8

1.0

Volt/ inchVolt (absolute)

watts

watts

Watts (Abs.)Watts (Abs.)watt-hourswatfhourswatt-hourswatt-hourswatt-hourswatt-hourswatt.hourswatt-hoursWatt (lnternational)weberswebers

3.4r290.05688

107.44.270.73741.341 x l0-'1.360 x 10-!0.01.4330.0010.056884

1

3.4133,60 x l0ro

2,656.859.85

1.341 x l0-10.8605

0.0011.0002lol0'

IAppendix D: Conlersion Factors 311

Synchronous Speeds

Frsoucncy x 120Synd'ronout Spced : N;;|T;G;-

FNEOUENCY TREOUENCY

50 Gyclo 50.y.ls 60 .ycl. 50 cy.lo

2

8

t0

l2

l1

t6

t8

20

2l

30

31

35

38

,t0

3600

lg00

1200

900

720

600

514.3

450

400

360

300

276.9

257 .1

210

225

2 .8

200

189.5

180

3000

t 500

| 000

750

600

500

128.6

375

300

272.7

250

230.8

211.3

200

176.5

166.7

t57.9

t50

1500

750

500

375

300

250

214.3

187.5

166 -7

150

136.4

t25

It5,4

| 07.1

100

93 -7

78.9

12

11

15

18

50

51

56

5S

60

62

61

66

68

70

72

71

76

80

112.9

136.4

t 30.1

t2s

t20

rr5.4

lll.tt07.t

103 .5

100

96.8

93.7

90.9

88.2

85.7

83.3

8l,l78.9

76.9

75

't71.1

r 63.6

t 56.5

t50

141

r 38.5

| 33.3

t 2s.6

121.1

t20

l l6.l

1r2.5

| 09. I

t 05.9

r02.9

100

97 .3

91.7

92.3

90

Courtesy Inge$oll-Rand Co.


Temperature Conversion

NOIA Thc c.ntlr .olsm'| of .'rmbcrt in boldl.ce r.ter3 to the tempe.ot'rre i. degree3, €irher Cenriorodc or fohre.heir, which ii i! dcti..d ro convcrt into th.othe. .col.. lf .o.v.dine from Fohrenh.it ro Ce.ligrode degreei, lhe equivolen. tempe.oture will be found in rhe tefr coiumn, whit€ if Gonve.ri^s trom d.er.erCenligrodc to desrc$ Fohrenhi.t, th. on.wer wi be found in thc cotumn on rhe righr.

C.ntisrode Cenligrode C.nti9.od.Cenligrod.

-2t8-212-207-20 |-196-190

-184-179-173-r69-r68-162-157-l5l-116-t10-r31-129-t23-l l8-l t2-107

-3!0-3t0-360-350-ta0-330-320-3to

-300-290-2ao-2It-2r0-260-250-240

130-220-2to-200-190-t!0-r70-t60

-tol -r50-96 -rao-90 -t30-81 -t20-79 -rr0-73.3 r00-67.8 -90-62.2 30

-273.t7 -159.f-268 -,150-267 -aao-257 -,r30-25l -a20-216 -ato-210 -a00-231 -390

Fohr.6h.it

t30 266f35 275lfo 281ra5 293t50 302t55 3l It50 ?20

459.1-151-136

-100

-361-316

-310

-271-256

-236-220-202-t81-166-ta8.o-t30.0-l 12.0

-t03.0-91 .0-85.0-76.0-67 .0-58.0-19.0-40 .0

-31 .0

-13.0-,(.o

.4.0

-20.6

-16.7-16. 1

-t5.6-15.0-t1.t

-t2.8-12.2

11.7-Ll-10.6-10.0-9.1-8 .9

-8.3

5.7

-5.6

-1.1

-l .7

-{.6o.0

0.6t.l

2.a

3.91.1

5.05.66.I6.77.27.58.38.9

9.410.0t0.6

23.032.0

37 .139.24r.0t2.g11.646.1

14.250 .05r.853.655.157 .259.060 .8

62.661.166.268.069.87t.673.175.2

77 .O78 .a80.682 .181 .286.0a7 .a89.6

9l .,(93.295.096.898.6

| 00 .,(102.2| 0,( .0

105.8147.6| 09.1ln.2113.0111.81 16.61 18.1

120 .2122.0t 23.8

ll.l11.7

12.8r 3.3

l3 .9t1.115.015.6r 6.l16.717 .2t7 .8

18.318.919.420.020.62t .l2t .7

22.823.323.92a.125.025.626.l26.7

27 .227 .a28.328.929 .130.030.63t .l

31 .732 .232.8

33.931.135.035.6

36. I

37 .237 .840.643.346.148.951 .7

3253545555

575859606r52635a

55555f5869707l72

,3

7575

7879EO

EIa2E3l46586a788

89909l92.93949595

979E99

t00105l|0

5t20r25

125.6t27.1t29.2l3l .0132.8

13,(.6r36. a138.2| 40.0t41.8113.6115.4117 .2

149.0r50.8t52.6l s4 .,(155.2158.0159.8161 .6

163.4165.2167.O| 68.8170.6172.1171.2t76 .O

't77 _gt79.6r8l ,1r83.2185.0186.8r 8s.6| 90.1

192.2194.0195.8197.6| 99.4201 .2203.0201.8

206.6208.1210.2212.0

230239215257

5a.t

60.062 .865.668.371.1

73.9

79.1s2.285.0a7 .g90 .693.396.r98 .9

t00.0r02r04107I t0lt3116

r l8t2l121127t29

r35r38r4Ir43l,a6t,a9t5{r60166171177

t82t88193r992012t0216221

727

213219251260

t65t70175tt0tt5t90t952002052t02122152202252302332ao

2tl52502557.502652fo2f52t02452902953003t0320330340350

3603703t0390400at0420430

-50

I23a

67!

9t0tlt2t3t4t5l5

tft8l9202l

?324

7525

2aa9

3l

343536373l3940

{t12a34445154748

a9505t

-59,a

-53 .9-51 .l

18.3-15.6-12.8-10.0

3,t.a-31 .7

-26.1-23.3

-lt-70-65-60-5t

50-{5-{0

-35-30

25-20-I5-10

329338317

371

40t410all11912et37116

161

t73192,a9l5005095t8527536

563572590608626611

680698716731

770

806

:lao 821,150 812460 8601fo 878aEo 896a90 914500 932

inlo the orhcr i.ole.,

forhulot ol thc right moy olro be urcdconyerling Centi!.odc or tohrcnh€ir Des,ee3 c€.r.. .c = j er + rot -ro

= | et-r'rDcaree! Kelvin, "K:oC + 223,2

Dcsr€e3 Fohr., .F : ! r. + rot -.0

c +32

Desre.. non&lne, 't :of +159.7

9

Courtesy Ingersoll-Rand Co.

-5000,{500

-1000-3500-3000

-25002000

-t500-t000-500

0500

r000t5002000

2500300035004000,1500

50006000700080009000

10,00015,00020,00025,00030,000

35,00040,00045,00050,00055,000

60,00070,00080,00090,000

t00,000120,0001,{0,000160,000180,000200,000

720,O00240.000260,0002S0.000300,000

,{00,000500.000600,000800,000

1,000,000

1,200,000t,400,0001,600.000r.800.0002.000,000

0.95l I1.31.5'| .7

I _92.a3.81.75.7

6.67.68.5

10 .4

.4| 3.315.217.1t8.922.826.630.431.237 .9

1t .745.519.353.156.9

75.991.8

I l,{r52189

35.0031.1233.8433.27

32.7032.113l .583t .0230.t7

29.92

28.8628.3327 .82

27.3226.8226.3325.8425.37

21.9023.9923.t022.232l .39

20.58r6.8913.76

.128.903

7 .060

1.3753.1112.7t2

2.1351.325

18.273 |

5.200-'3.290-'I .358-r5 .917-72./16-71.284-'5.816-r

2.523 '9.955-.3.513-rI 1.{3-r3.737 '6.3 7

t.1-,5.9-.| .6-r5.1-'

2.0-,8.2-r03.8 10

| .8-roL 2-r'

903 .7889.0871.3859.5845. t

830.6816 .1802.1

773.9

760.O716.3733 .O719.6706.5

693.9681 ,2668.86s6.3611.1

632.s609.3586.7561.6513.3

522.7A29 .0319.52A2 .l226.1

't79 .3111 .2

t.l87 .568.9

51.2

2t.0

8.363.,(5t.5l

ft gt-t3.26-I| .48-I

6.A1-'2.53 '8.92-!3.67-'9.19.

l .60-'3.56-61.50-6,r.06-'1.30-t

s.08 r2.08-l9.65-'1 .57-'2.31-e

17.1817.19I6.90r6.62| 6.3,r

t6.0615.79r5.51l5 .2311.96

r4.69611.4314.1613 .91| 3.66

t 3.41r 3.17| 2.93l2.69t2.t6

12.23.78

rL3,(l0 ,9110.50

t0.108.296.765.461.37

3.172.73?.t5r .69

| .05.651.,(06.255

:!:

.0238

.0458

.0285

.o179ot t,

Altitude and Atmospheric Pressures

Hs Abr. Hg Ab3.

Appendix D: Conrersion Facrors 313

Ke/'qPSIA

-1526-t373-1220-1068-9r5

-7636t0

-,f58-305

0t53

1586t0

763915

106812201373

1526l83l213621112716

3050

610?76289153

1O,67i12,2O1t3,73015,255t6/81

18,3062t,3572A,10827,15930,51036,61212/1118,8t 651,91861,o20

67,12273,22179,32685,12891,530

66

-70-70

62-57

_26I

28t9-3

1186

t29

-5-t4

-11

-57-57-57

-57-55

52-59-16

-t6

-t9

7775737l70

586661

6l

5957

52

50a8t71513

{t3S343l

23

-3018

252123

21

20t9l8l7t6

.229

.209

.188

.169

. r19

.129

. t09

.091

.071

.o52

.0333

.0t5

.956

.975

.960

.943

.926

.909

.892

.876

.860

.828

.797

.767

.738

.7to

.A7 5

.38,(

.307

.2A1

.r5l

. 9

.0935

t1

12t1

l09

75

3II

1266

-90

-88

122,010 | -- |

152,550 | -l--r 83,060 - | -211,080 -- I -305,100 -- -

22426630,{312

366,t 20127,110,{88.t605!9,r80610,200

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ACI bearing strengths, 180American Institute of Steel Construction. See AISC.Anchor bolts

analysis, preloaded bolt, 184, 186bolt area, required, 184bolt loads, allowable, 187bolt load, minimum required, 184bolt spacing, 186common types of, 190large bolts, undesirability of, 184loading force, distribution of, 186loadings induced on, 184lubricant, 190philosophy, design, 184size and number, 228stress in, 184, 186tension on gross area, 187torque, anchor bolt, 189-190, 229

ASME Piping CodesASME 831.1, 48ASME 831.3, 48ASME B3I.4, 48ASME B3I.5, 48

.ASME 831.8,48ASME Section IlI, 48. Also see Pressure vessels.

for piping, 48for pressure vessels, 48

ASME Section VIII, Division IIfor piping, 48

Aspect ratio, 85

Baseplate design, 186-189anchor bolt size range, 186bearing pressure on, 189concrete foundation for, 186concrete mixes, 186, 187

Index

concrete modulus of elasticity of, 186concrete and steel, relative strength of, 186gusset plates, 188* 189k-factor, offset, 188steel, modulus of elasticity, 187steel-concrete moduli ratio, 186tension on gross area, 187torque, anchor bolt, 189-190, 229

Bernoulli equation, 2Bingham, 6-7Boundary conditions for saddle plate design, 178Buckling coefficients for saddle plate design, 175-178

Centroid, section,212Circumferential stress, moment, 170Codes, vessel

differences in, 159foreign, 159

Cold-spring,49Colebrook equation, 4. Also see Friction factors.Compressible flow

adiabatic flow, 2compressibility effects, 24introduction to, l-2, 24isothermal flow 1

modulus, bulk compressibility, 24non-steady flow, 24sound, velocity of, 24steady flow, 24

Concrete mixes for baseplate design, 186-187Concrete modulus of elasticity, 186Conical sections, 199, 224Cost-plus contractor, 183Creep,49Critical damping factor, 202, 2O4Critical pressure, 83

315


Critical temperature, 83Critical wind velocity, 236

Damping coefficient, 2OZ 2MDeflections, windt 199-2Ol , 242Degree of freedom, 201Discontinuity, 236Drag, 195,203Ductile materials, 50, 52Dynamic magnification factor, 201-204Dynamic response, 200

EJMA. Sze Expansion joints, bellows.Electrical tracing, 103Equivalent length, 2Expansion joints

bellows, corrugated, 77gimbal joint, 79hinged joint, 78-79inJine pressure balanced, 79multi-ply, 80pipe span, allowable, 78pressure thrust, 78-79single ply, 80standards of the Expansion Joint Manufacturers

Association (EJMA), 80stiffness, rotational, 78stiffness, translational, 78tie rods, 78-79

reasons for, 78universal joint, pressure-balanced, 78

Fanning equation, 3Fluid Mechanics, piping. See Hydraulics.Fourier number, l5lFriction factors, 4

Colebrook equation, 4laminar flow, 4Moody friction factors, 4Prandtl solution, 5turbulent flow, 4von Karman solution, 5

Gimbal joint, 79Grashof number, 132, 134, 153Gusset plates, 188-189Gust (wind) effects, 194-196, 236-237Guy wires, 249

Head -*T co'\J 'rv 5oo r{ 'll"i, '

foot of, 2pressure, Istatic, Ivelocity. See Velocity head.

Headsmanufacture of, 160thickness of, 160

Heat transfercontrol mass, 115, 131control volume, 115, 13lelectrical tracing, 103Fourier number, 151Grashof number, 132, 134, 153in jacketed pipe, I 12- I l5LMTD (log mean)

chart for, 114definition of, I 14

Nusselt number, 132, 134, 153in pipe shoes, 135- 136

application of, 156heat balance for, 136temperature distribution in, 136

in pipe supports, 133in piping

temperature distribution in, 134typical applications of, 133- 134

Prandtl number, 112, 139-140in process systems, 103in residual systems

applications of, 132deflections, thermal, 134-135overall heat transfer coefficient, 134

tubular tracers. See Tracing.in vessel skirts

application of, 152- 154coefficients of, 132convection, significance of, 133free convection, 133rate of, 133temperature, distribution oI, 132- 133

Heat transfer design example, 148-150static analysis, i48- 150transient analysis, 150- 152

Heisler's chart, l5lHesse formula, 82Horizontal pressure vessels

saddle bearing plate design, 180ACI bearing strengths, 180bearing plate thickness, 180factor of safety for, 180

saddle plate buckling analysis, 251 252saddle plates

application of , 249 -252boundary conditions for, 178buckling coefficients for, 175- 178design of, 174- 179effective area, 174, 178

effective width, 113, 178, l'79horizontal reaction, 119, 252stiffener plates, I74, 179STTESS

criterion for residual, 178elastic buckling, 179inelastic buckling, 179

U.S. Steel design method, 174-179web plates, 174

wear plate requirements, 215Zick analysis, 166, 215

bending moment diagram, 167constant, circumferential bending moment,introduction to, 166saddle supports, location, criteria for, 172shear stress, 171

shellstiffened by head, 171

unstiffened, saddles away from head, 17lstiffening rings, 172, 174STTESS

allowable compressive, 166circumferential compressive, l7 Icircumferential at horn of saddle, 17lhead used as a stiffener, 171

"Hot-spring," 49Hydraulic radius,

definition of, 2itabulated values, 24

Hydraulicsbasic equations, IBernoulli equation, 2

modified form of, 3

compressible flowadiabatic flow, 2compressibility effects, 24introduction to, l-2, 24isothermal flow, Imodulus, bulk compressibility, 24non-steady flow, 24sound, velocity of, 24steady flow, 24

incompressible flow, 1

non-Newtonian fluidsBingham,6-7introduction to, 5-7Metzer and Reed, 7pseudoplastic, 6-7rheological constants, 8

rheopectic,6-7thixotropic, 6 7time-dependent, 6-7time-independent, 6-7viscoelastic, 6-7

l:;:..

yield-pseudoplastic, 6 7

piping, reasonable velocities in, 25problem formulation, 24two-K method, 8,21viscosity,24-26

Incompressible flow. See Hydraulics.Internal pressure, 159- 160

Jacketed pipeannulus, hydraulic radius for, 112applications of, l12-115, 139 140details of, 104-106, I 12-l l3expansion joints for, 105- 106heat transfer, I 12- I l5

coefficient, film, I l2coefficient, overall, 112rates of, I 12- 115

pressure drop in, I l5- I 17

rules of thumb for, 103versus traced pipe, 103- 106

Joints. expansion. See Expansion joints.Laminar flow, 4. Also see Friction factors.Lumped-mass approach, 204-205Lump-sum contractor, 183

Maximum allowable working pressure, 160Mitchell equation , 210, 212Moments

equations for, 198of inertia, for tube bundle, 222-223wind-induced, 198

Moody friction factors. See Friction factors.Myklestad method, 200-201

Non-Newtonian fluids. See Hydraulics.Nusselt number, 132, 134,153

Ovaling, 205, 208

Pipe loops, 59-68Pipe lug supports , 70-12, 98-99Pipe materials

ductile materials, 50, 52non-ductile materials, 50plastic deformation, 50 52stress-strain curves, 50-51

Pipe shoes, heat transfer in, 135-136Pipe supports, heat transfer in, 133Piping codes. See ASME.Piping expansion joints. See Expansion joints.Piping mechanics

anchor, pipe, definition, 58API,47

170


equipment nozzle loads, 94extraneous piping loads

"cold spring" for, 80vibration

applications for, 100- 101natural frequency of beam elements, 86vortex shedding, 83,87

resonance,83Reynolds number, 195, 200, 2Ol, 236Strouhal number, 84-85vortex force, 83vortex streets, 83

flexibility (compliance) matrix, 53flexibility method, 59-68, 8l

advantages of, 53, 68application of, 95-98

"hot-spring," 49nozzle flexibility factors,

angle of twist, 70circumferential, 70longitudinal, T0Oak Ridge Phase 3 Report, 70rotation deformation of, 70rotational spring rate, 70

pipe loops, 59-68pipe lug supports , 70-72, 98-99pipe restraints

moment restraints (MRS), 5'7 -59, 77 , 88-94rotational 58, 68translational,58,68

pipe roughness, 5prpe stress

circumferential bending/membrane, 7l"cold-spring," 49creep,49"hot-spring," 49internal pressure, circumferential stress, 49longitudinal stress, 49pipe weight, bending stress, 49pressure, 72prestressed piping, 80primary stress, 49-50, 72range, allowable, 42residual stress, 5lsecondary stress, 49-52, 72self-spring,49"shakedown," 52thermal expansion, 49torsional or shear stress, 49

self-spring,49shear flow, 58-59spring supports, 72, 75, 76

guided load column, 72jamming of, 77

stiffnessbeam element, 54concrete,69matrix,53-54method,8l

advantages,53,68applications of, 88-94

piping elements, 55-56, 69translational, 54

Pipe Stress. See Piping mechanrcs.

Piping systemsadiabatic process, 83API 520 Pafi 2, 82ASME 31.I, 82critical pressure, 83critical pressure ratio, 83critical temperature, 83Hesse formula, 82impulse-momentum principle, as applied to a pipe

elbow, 8lnozzle correction factor, 82nozzle discharge coefficient, 82nozzles,83

Prandtl number, ll2, 139-140Pressure vessels

ASME Section VIII Division I, 160components, 159- 160design, philosophy of, 159external pressure, 160heads, 160horizontal

saddle bearing plate design, 180saddle plate buckling analysis, 251-252saddle plate design, 174- 179

application of , 249-252boundary conditions for, 178buckling coefficients for, 175- 178effective area, 174, 178effective width, 173, 178, 179horizontal rcaction, 179, 252stiffener plates, 174, 179stress, criterion for residual, 178stress, elastic buckling, 179stress, inelastic buckling, 179U.S. Steel design method, 174-179wear plate requirements, 215web plates, 174

Zick analysis, 166, Zl5bending moment diagram, 167compressive B-factor, 174constant, circumferential bending moment, 170head used as stiffener, 171saddle support location, 172

shear stress in head/shell, 171

shellstiffened by head, l7lunstiffened, saddles away from head, 171

stiffening rings, 172, 174stress, allowable compressive, 166stress, circumferential con.rpressive, 171

stress, location of, 168- 169

tangential shear, 167- 171

wear plates, 171- 172internal pressure

component thickness, 159maximum allowable working pressure, 160quality of welds, 159

upset conditions, 160vertical

anchor boltsanalysis, preloaded bolt, 184, 186bolt area, required, 184bolt loads, allowable, 187

bolt load, minimum required, 184bolt spacing, 186common types of, 190large bolts, undesirability of, 184loading force, distribution of, 186loadings induced on, 184lubricant, 190philosophy, design, 184size and number, 228stress in, 184, 186tension on gross area, 187torque, anchor bolt, 189-190, 229

ANSr-1982,215baseplate design, 186- 189

anchor bolt size range. 186bearing pressure on, 189concrete foundation for, 186concrete mixes, 186, 187concrete modulus of elasticity of, 186concrete and steel, relative strength of, 186

. gusset plates, 188- 189k-factor, offset, 188steel, modulus of elasticity, 187steel-concrete moduli ratio, 186stress, compressive, on concrete, 188thickness, baseplate, 188

centroid, section,212combined loads on, 181

compression plate, 189cone, truncated, equivalent radius for, 214conical head, equivalent radius for,214conical sections, equivalent radii for,224earthquake, See Seismic design.loads, wind and seismic, 190-191

b"l- !

momentsequations for, 198of inertia, for tube bundle, 222-t3pressure sections, centroids of, 198vectors, section force, 198

wind-induced, 198wind pressure, distribution of, 198

section properties of, 181

seismic analysis of, loads, combined, 190-l9lseismic design

baseplate design, 238coefficients, Mitchell, 210, 213coefficients, structure type, 210criteria, quasi-static, 210criteria,238Mitchell equation, 2lO, 2lZ

compared to Rayleigh equation, 237 -238occupancy importance factor, 210period

characteristic site, 238numeric integration of vibration, 238-239of tower, 210, 2lZ

Rayleigh equation, 212compared to Mitchell equation, 237 238

seismic zone factor/map, 210-211site structure interaction factor, 210, 212

equation for, 212shear forces

earthquake force, total, 212lateral force, equation for, 212vertical distribution of, 212

seismic moments, equation for, 212skirt design, 238structural period response factor, 210Uniform Building Code, 209 210

self-supporting, 180skirts

controlling criteria for, 184design of, 183, 185

cost-plus contractor, 183Iump-sum contractor, 183

stress equation, 183supports, 183, 185thichess, 183- 184

stress, bending, 181combined loading, 181

compressive B factor, l9lcompressive, leeward side, 181

discontinuity, 236elements in, 182tensile, windward side, l8lvacuum, 183

towerscentroids, section, 230-231

319

32O Mechanical Design of Process Systems

definition of, 181

equivalent circle method, 214section moment of inertia, 241-243skirt and baseplate destgn, 228-229

anchor bolts, 228anchor bolt torque, 229compression ring thickness, 229skirt thickness, 229weld size, minimum for skirt-to-base plate,

229skirt detail, 230stress, discontinuity

criteria foq 2 14for conical sections, 214

stresses, wind section, 226-228transition piece, 241, 243-244vibration ensemble, 216

of lumped masses, 232, 246wind deflections

modes of, 199schematic diagram of, 201superposition, method of, 199wind ensemble, 242

vibration, wind-inducedangular natural undamped frequency, 205applications of, 232-236, 241-249area-moment method, 205-207conjugate beam. See Area moment.controlling length, 203critical damping factor, 202, ZO4critical wind velocity, 208-209 , 236, 248,249

total wind force, 209Zorilla criteria, 209

damping coefficient, 203damping ratio, 202-203degree of freedom, single, 201differential equations for, 201-2OZdynamic magnification factor, 201-202, 2O3,

2Mdynamic response, 200example of, 232-236first period of, 204force amplitude, 235force amplitude, dynamic, 200forced vibration theory, 200frequency

natural,248ratio,202vortex shedding, 208, 248

guy wires, disadvantages of, 249Holzer procedure, 200lock-in effect, 200logarithmic decrement, ZO3-204lumped mass approach, 204-205

mode shapes, 200Myklestad method, 200, 201ovaling,205natural frequency of, 205vibration due to, 208wind velocity, resonance, 208

period of vibration, 234-235, 248phase angle, 202Rayleigh equation, ZOO, 201, 204, 205resonance,236Reynolds number, 195, 20O,201,236soil types, 204stresses, dynamic, 236tower

fluid forces on, 203model for, 201-202moment disrribution in, 205stiffness, 205

vibration ensemble, 209of lumped masses, 232

vibration, first peak amplitude, 200vortex shedding, 199vortex strakes, 249wind tunnel tests, 236

wind analysis of, loads, combined, 190-191wind design speed

ASA 58.1-1955, 194ANSI-A58.1-1972, 192

basic wind pressure, 192effective velocity pressure, 192gust response factor, dynamic, 192

ANSI A58. 1- 1982, 196, 236-237effective velocity pressure, 192gust response factor, 192importance coefficient, 192velocity pressure coefficient, 192wind speed, variation of, 192wind tunnel tests, 192

centroid of spandrel segment, for wind section,218

coefficient, drag, 195structural damping, 217

conical sections, 199constant exposure category, 195cross-sectional area, effective, 217cylinder, pressure fields around, 196equivalent diameter method, 236-237

vs. ANSI-A58. 1- 1982, 236-237exposure lactor. 196fatigue failure, 198flexible structures, defined, 197gust duration, 196

vs. gust diameter, 197gust frontal area, 196

ii

gust response, dynamic, 194

gust response factor, 195, 196,217,236-231gust size, 196isopleths, 192- 193

Kutta-Joukowski theorem, 195

loading analysis, quasi-static, 196logarithmic law, 192parabolic area, centroid of, 219parabolic function, 194peak values, types of, 196power law, 192probability of exceeding. 196

response spectra, 198

return period, 192

similarity parameters, 195

structure size factor, 196, 197

surface roughness, 195

towercross-sectional area of, 198

fluid force exerted on, 194-195gust velocity vs. structural response, 197natural frequency of, 197

wind area section properties, 219wind force distribution, 218wind distribution

parabolic, 194, 218-219triangular, 194

wind loadapplications of, 215-231, 241-245equivalent static, 195mean, 195

weld size, skirt-to-base plate, 189welding, joint efficiencies for, 161-165,172Zick analysis, 166, 215

bending moment diagram, 167

compressive B-factor, 174constant, circumferential bending moment, 170

head used as stiffener, l7lsaddle support location, 172

shear stress in head/shell, 171

. shellstiffened by head, 171

unstiffened, saddles away from head, 171

stiffening rings, 172, 174stress, allowable compressive, 166

stress, circumferential compressive, 171

stress, location of, 168- 169tangential shear, 167- 171

wear plates, 171- 172

Residual systems, heat transfer in, 132-135in piping, 154- 155

Reynolds number, 195, 2OO, 2Ol, 236drag coefficient vs., 203

l:r.= r

Newtonian fluids, 21, 30,32 41. 1,19-l{l"t. l-!:.145,147

non-Newtonian fluids. See Hydraulics.Non-Newtonian fluids.

Strouhal coefficient vs., 85vortex shedding, for, 83-85

Saddle plate design, 174- 179application of , 249 -252boundary conditions for, 178

buckling coefficients for, 175- 178

effective area, 174, 178effective width, 173, 178, 179

horizontal react\on, 179, 252stiffener plates, 174, 119stress, criterion for residual, 178

stress, (in-) elastic buckling, 179

U.S. Steel design method, 174-179wear plate requirements, 215web plates, 174

Seismic designbaseplate design, 238coefficients, Mitchell, 210, 213coefficients, structure tYPe, 210criteria, quasi-static, 210compared to wind, 238Mitchell equation , 210, 212

compared to Rayleigh equation, 231-238moments, equation for, 212occupancy importance factor, 210period, characteristic site, 238period, vibration

numeric integration of, 238 239tower,210,212

Rayleigh equation, 212compared to Mitchell equation, 231-238

seismic zone factor/map, 210, 2llshear forces

earthquake force, total, 212lateral force, equation for, 212vertical distribution of, 212

site structure interaction factor, 210, 212equation for, 212

skirt design, 238structural period response factor, 210Uniform Building Code, 209-210

Skirts, 185controlling criteria for, 184

cost-plus contractor, 183design of, 183lump-sum contractor, 183stress equation, 183supports, 185thickness, 183- 184


Strouhal number, 84Reynolds number vs., 85vibration, vortex shedding, 84-85, 200, 20g

Supports, 72,75,76. Also see p\ping mechanics.

Thermal design. See Heat transfer tie rods, 78-79Towers

centroids, section, 230-231definition of, l8lequivalent circle method, 214section moment of inertia, 241-243skirt and baseplate design, 228-229

anchor bolts, 228anchor bolt torqte, 229compression ring thickness, 229skirt thickness, 229weld size, minimum for skirt{o-base plate, 229

skirt detail, 230stress, discontinuity

criteria for, 214for conical sections, 214

stresses, wind section, 226-228transition piece, 241, 243t244vibration ensemble, 216

of lumped masses, 232, 246wind deflections of

modes of, 199schematic diagram of, 201superposition, method of, 199wind ensemble, 242

Tracingof pipes

applications of, 136- 139condensate return for, I l0

condensate load, determining, 1l Iguidelines for, 110-l llspargers, 1l Iseparation keys, I l1typical layout, 111water hammer, 11 I

hot oil, application of, 137-139steam, application of, 136-137versus jacketed pipe, 103- 106with heat transfer cement, 106, 109- I 10

advantages, 106procedure for, 109

film coefficient, natural convection, 108 109heat balance for, I l0heat transfer rates of, I l0

without heat transfer cement, 106-109advantages of, 106disadvantages of, 106equivalent insulation thickness, 107heat balance fog 107

heat transfer, rules of, 107modes of heat transfer, 107outside film coefficient, 107overall heat transfer coefficient, 107procedure for design, 107

of vessels and equipmentagrtators

film coefficients for, 143use of, 115

applications of, 130, 140- 148film coefficient, vessel-side, 147heat duty of, jacketed heads, 146heat transfer coefficients, reasonable values of,

130transient, I l5

criteria for, 115importance of, 130

internal baffle plates, heat duty of, 144jacketed walls, heat transfer film coefficient, 145jackets, types of, 115, l28,13lnon-Newtonians, use of, 146plate channels, equivalent velocity of, 147reasons for, 115

Turbulent f|ow, 4 - Also see Friction factors.

Velocity headintroduction,3,8method,3two-K method, 8, 21values of, 9-20, 21, 22-23, 30-32

Vessels. See Pressure vessels.Vibration, wind-induced

angular natural umdamped frequency, 205applications of , 232-236, 241 -249area-moment method, 205-207conjugate beam. See Area moment.controlling length, 203critical damping factor, 202, 204critical wind velocity, 208-209 , 236, Z4g-249

total wind force, 209Zorilla criteria, 209

damping coefficient, 203damping ratio, ZO2-203degree of freedom. single. 201differential equations for, 201,202dynamic magnification factor, 201 -202, 203, ZO4dynamic response, 200example of, 232-236first period of, 204force amplitude, 235force amplitude, dynamic, 200forced vibration theory, 200frequency

natural,248

,!i

lri:r

ftIio, 202vortex shedding, 2O8' 248

suy wires, disadvantages of' 249- i{olzer procedure, 200

lock-in effect, 200losarithmic decrement, 203 -204

lumfed mass aPProach, 204-205mode shapes, 200Myklestad method, 200, 201

ovaling,205natuial frequencY of. 205

vibration due to, 208wind velocitY, resonance, 208

period of vibration, 234-235, 248

ohase angle, 202

ilayleigh-equarion. 200. 201. 204 ' 205

resonance,236Reynolds number, 195, 200, 2O1' 236

soil types, 204stresses, dYnamic, 236

towerfluid forces on, 203

model for, 201-202moment distribution in, 205

equations for, 205

stiffness,205vibration ensemble, 209

of lumped masses, 232

vibration, first peak amplitude' 200

vortex shedding, 83-87' 199

vortex strakes, 249wind tunnel tests, 236

Viscosity, 24-25von Karman solution, 5Vortex shedding,83-87

aspect ratio, 85

cylinders,83damping vs. amPlitude, 87

guidelines for, 85

mode shaPes, 85

reduced damPing, 85

Weld sizesrecommended values, for Plates, 71

skirt to baseplate, 189

Welding, joint efficiencies for, 161-165,Wind design sPeed

ASA 58.1-1955, 194

ANSI A58.1-1972basic wind Pressure, 192

effective velocitY Pressure, 192

qust response iactor. dynamic. 192

ANsl A58. l-1982, t96, 236-231effective velocitY Pressure, 192

sust response factor. 192

irpottun." coefficient. 192

velocitv pressure coefficient, 192

wind speid, variation of' 192

wind tunnel tests, 192

centroid of spandrel segment, for wind section' : i -r

coefficient, drag, 195

structural damPing, 217

conical sections, 199

constant exposure category, 195

cross-sectional area, effective, 217

cvlinder, pressure fields around, 196

equivaleni diameter method, 236-237vs. ANSI-A58.1- 1982, 236-231

exposure factor, 196

fatigue failure, 198

fle;ble structures, defined, 197

gust duration, 196vs. gust diameter, 197

gust frontal area, 196

iurt t.rpon.., dYnamic. 194

iurt ,.tpont" factor. 195. 1c0.217.236-237gust size, 196isopleths, 192- 193

Kuna-Joukowski Theorem. 195

loading analysis, quasi-static, 196

losarithmic law, 192paiabolic area, centroid of, 219parabolic function, 194

peak values, tYPes of, 196power law, 192probability of exceeding, 196

iesponse sPectra, 198

return period, 192

similarity parameters, 195

structure size factor, 196' 197

surface roughness, 195

towercross-sectional area of, 198

fluid force exerted on, 194-195gust velocity vs. structural response' 197

iatural frequencY of, 197

wind area section Properties, 219

wind force distribution, 218

wind distributionparabolic, 194, 2t8-219triangular, 194

wind loadapplications of, 215-231, 241-245equivalent static, 195

mean, 195

Yield, 159octahedral shear stress theory, 236

172


Zick analysis, 166, 215bending moment diagram, 167compressive B-factot 174constant, circumferential bending moment, 170head used as stiffener, l7lsaddle support location, 172shear stress in head/shell, 171shell

stiffened bv head. l7l

unstiffened, saddles away from head, 171stiffening ings, 172, 174stress, allowable compressive, 166stress, circumferential compressive, 171stress, location of, 168- 169tangential shear, 167- 171wear plates, l7l-172

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