Project Presentation

7/21/2019 Project Presentation

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4-Nov-08 3



v As a result of all these accidents:-

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In yearIn year 19141914

4-Nov-08 5



Power Boiler I

Material SpecificationsII

Nuclear ComponentsIII

Heating Boiler IV

Nondestructive ExaminationVRecommended Rules for Care and Operation of Heating BoilersVI

Recommended Guidelines for the Care of Power BoilersVII

Pressure Vessels

Division 1

Division 2 Alternative Rules

Division 3 High Pressure Vessels Components

VIII

IX Welding and Brazing qualifications

X Fiber Reinforced Plastic Pressure

Vessels

XI Rules For In service Inspection OfNuclear Power Plant components

4-Nov-08 6



ØPressure Vessels are used in many industries (e.g.,

hydrocarbon processing, chemical, power, pharmaceutical, food

and beverage).

qSection VIII is divided into three divisions.

3.1 Division 1

3.2 Division 2

3.3 Division 3

4-Nov-08 7



3.1 Division 1

ØThe ASME Code Section VIII, Division 1 applies

for pressures that exceed 15 psig and through

3000 psig.

ØAt pressures below 15 psig, the ASME Code

is not applicable.

3.2 Division 2

ØThe ASME Code Section VIII, Division 2 applies

for pressures that exceed 3000 psig and through

10000 psig.ØA division 2 design is more attractive for

vessels that require greater wall thickness,

Typically over approximately 2 in thickness .

ØThe choice between using Division 1 andDivision 2 is based on economics.

3.3 Division 3

ØDivision 3 applies to pressure vessels operating

at internal or external pressures generally above

10000 psi.

4-Nov-08 8



4. SCOPE OF ASME PROCESS PIPING (B31.3)

Piping System: Conveys fluid between locations

qPiping System includes:

1) Pipes

2) Fittings (e.g. elbows, reducers, branchconnections, .etc.)

3) Flanges, gaskets, bolting

4) Valves

5) Pipe supports

qAPPLICATIONS- Petroleum refineries - Paper plants

- Chemical plants - Semiconductor plants

- Pharmaceutical plants - Cryogenic plants

4-Nov-08 9



5.GENERAL METHODOLGY

The execution of the project is planned in several stages.

Process Specification

Mechanical Data

Sheet Specification

Project Design

Procedures

Project Assembly Drawing

Fabrication

Testing And

Inspection4-Nov-08 10



6.PROJECT OBJECTIVES

1) TO design a vertical pressure vessel

according to ASME section VIII division

1(Rules of Construction of PressureVessel) for separating crude oil.

2) And design a piping system according to

ASME processes piping B31.3.

4-Nov-08 11



4-Nov-08 12

Pressure

Vessel

Gas Outlet

Crude Oil Inlet

Normalized Oil Outlet

PROCESS SPECIFICATION



VERTICAL PRESSURE VESSEL

DESIGN

Internal PressureInternal PressureREINFORCEREINFORCE--MENT OFMENT OF

OPENINGSOPENINGSVIBRATIONVIBRATION

Wind LoadWind LoadCOMBINATION OFCOMBINATION OF

STRESSESSTRESSES

STRENGTH OF ALLSTRENGTH OF ALL

ATTACHMENTSATTACHMENTS

WEIGHTWEIGHT

OFOF

VESSELVESSEL

DEFLECTIONDEFLECTION

NOZZLE EXTERNALNOZZLE EXTERNAL

FORCES ANDFORCES AND

MOMENTSMOMENTS

SKIRT SUPPORTSKIRT SUPPORT

DESIGNOFDESIGNOF

BASE RINGBASE RING

OPTIMUM VESSELOPTIMUM VESSEL

SIZESIZE

SEISMIC LOADSEISMIC LOAD ANCHOR BOLTANCHOR BOLTLOW TEMPERATURELOW TEMPERATURE

OPERATIONOPERATION

4-Nov-08 13



4-Nov-08 14

ü As it states in Par. U-2 (G) the code dose not

contain rules to cover all design details , where

complete details not given , the manufacturer shall provide details.

ü Design and construction details have been

selected from generally accepted sources ,

utilized the most practical and economical

methods.

NOTE:-



4-Nov-08 15



4-Nov-08 16

According To ASME Code UG-27



1.1 Shell

P SE

PR

6.0-t =

t = 0.6296 Inches

= 15.99 mm

Thickness with corrosion allowance.

t = 15.991 + 3 mm = 18.991 mm.

= 0.747 in.4-Nov-08 17



1.2 Ellipsoidal Head.

P SE

PR

2.02 -t =

t= 0.6253 Inches

= 15.887 mm

Thick with corrosionAllowance.

t = 15.887 + 3 = 18.887 mm

4-Nov-08 18



4-Nov-08 19

Based On Standard ANSI 7-93, Approved 1994.



4-Nov-08 20



Af

F

4-Nov-08 21



4-Nov-08 22

V = PW X D X H



4-Nov-08 23



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4-Nov-08 27



4-Nov-08 28

ACCORDING TO:-



tDSE

W

tsE

Mr +

2

12t =

4-Nov-08 29



I. For wind

tSE

T M

2

12

t1 = t = 0.0614 Inches

= 1.5621 mm

II. For weight

Total thickness

tT

= t1

+ t2

= 0.0614 + 0.01976

= 0.08116 Inches

= 2.0621 mm

t2 =tDSE

W t = 0.01976 Inches

= 0.5 mm

4-Nov-08 30



4-Nov-08 31

Based On Uniform Building Code , 1991 (UBC).



4-Nov-08 32



T

- Seismic zone : 4 Z = 0.4

- s=1 as soil depth is less than 200 ft

= 0.398 sec.

4-Nov-08 33



32

25.1

T

S ´

w

w R

ZIC ´

4-Nov-08 34



4-Nov-08 35

Based on Freese , C.E Vibration Of Vertical Pressure

Vessel ASME, Paper 1959



t

wD

D

H 2

51065.2 ÷

ø

öçè

æ ´ -

g v

wH

´80.0

4-Nov-08 36



4-Nov-08 37



qStress due to internal pressure

S =

t

PD

4

S = 5,740.41 psi

qStress due to wind:

S =

t Mr

´´t 2

12S=1,135.61 psi

qStress due to weight:In erection condition

S =

cmt

w S= 98.211 psi

In operation condition

S =cmt

wS = 365.02 psi

4-Nov-08 38



- STRESS DUE TO

WIND

+ 1135.6 - STRESS DUE

TO WIND

- 1135.6

- STRESS DUE TO

WEIGHT

- 98.2 - STRESS DUE

TO WEIGHT

- 98.2

(NO. INTERNAL

PRESSIRE IN

ERECTION)

1038.4

- 1038.4(PSI)

Wind ward side Lee ward side

4-Nov-08 39



- Stress Due To Int.

Press

+ 5740.0 Int.

pressure

+ 5740.0

- Stress Due To Wind + 1135.6 wind - 1135.6

- Stress Due To Weight - 98.2 weight - 98.2

Sum. + 6777.8 Sum. - 4505.6

(psi) (psi)

The tensile stress + 6,777.8 (psi) in operating condition and

the in ward side governs.

Q the tensile stress = 6,777.8 psi < Allowable stress

= 17,100 psi

\ then the thickness shell t = 0.786 in is satisfactory

Wind ward side Lee ward side

4-Nov-08 40



4-Nov-08 41

ACCORDING TO:-

S.S. Tang : Short Cut Method For Cal culating Tower

Deflection Hydrocarbon Processing November 1968



( )

I

H H Dw

P

8

312

1m =

- I = R3 t I = 52,512.4 in4

= 0.19471 in

The max. allowable deflection. 6 inches per 100 ft. of height.

\ the thickness of skirt is satisfactory Form page 188 (H.B)

Modules of elasticity E = 28 x 106 psi

at T = 170°F at c < 0.30 %

4-Nov-08 42



4-Nov-08 43

Approximate Method Acc. To PV Hand Book 10th Edition.

ACCORDING TO :-



9.1 Maximum compressi on ratio:

Pc = sC

w

A

M

+2

12

Pc = 1,180.9 Ib/lin-in

9.2 Approximate width of base ring

L =

b f

cL= 2.361 in

4-Nov-08 44



No. bolt = 18

bolt size = 2.5 in

using plate

SA 325

Ma St. 15000 psi

use 1 .75 in thickness base ring

4-Nov-08 45



4-Nov-08 46



K = 0.2857

f f cbcb == 400400 psps

CCcc == 11..4343

CCtt == 22..6565

J =J = 00..777777

Z =Z = 00..4444

cbnf

Sa+1

1K =

4-Nov-08 47



10.1 Required are of ancber bolt

B =

jd q

S t

C wzd M -122t B = 7.32 sq.in

From Table (A) Get

d = 5/8 in

10.2 Tensile stresson the ancher bolts.

65.22

11.118´´÷÷÷

ø

ö

ççç

è

æ

st

t f

Sa =

Using 18 anchor bolts.

With corrosion allowance

0.4 + 0.125 = 0.52 in

= 7.32/18 = 0.388 in @ 0.4 in

= 14808.7 psi

14808.7 psiSa =

4-Nov-08 48



10.3 Compressive Load On The Concrete.

Fc = Ft + w

Fc = 152939.32 Ib

boltandercompressive stress in the4.10Sa = nFcb Sa = 4955.03 psi

10.5 The required thickness of base ring\

tB = tB = 0.47 in sc

f /34

4-Nov-08 49



4-Nov-08 50

ASME code UG-26 to UG-44.ACCORDING TO :-



4-Nov-08 51



4-Nov-08 52



11.2 FOR 24 MANHOLE SCH 40#

11.2.1 Wall thinkness required for Manhole

tmh=SE

PR

6.0-tmh = 0.082 in

11.2.2 Area of reinforcement req.

AR = dtrs

= 14.448 dq.in

11.2.3 Area of reinforcement available

A1 = ( Excess in shell ) larger of

=( t tr ) d

A1= 0.84 sq.in

= ( t tr ) (tn + t) x 2A1 = 0.118 sq.in

Or

So that the Larger is 0.84 sq.in4-Nov-08 53



A2 = [ Excess in nozzle neck ] smaller of follow

=(tn trn) 5t = 2.384 sq.in

=(tn trn) 5tn = 2.084 sq.in

A3 = ( inside projection ) tn x 2hh = 2.5 x tn

= 2.366 sq.in

A4 = ( Area of fillet weld inside )

A5 = ( Area of fillet weld outside)

A4 , A5 = 0.1253 sq.in

4-Nov-08 54



11.2.4 Total Area available

AR = A1 + A2 + A3 + A4 + A5

= 5.54 sq.in

QAR > AV

The area required for reinforcement pad

11.2.5 The required area of the pad

= AR AV= 8.484 sq.in

Using plate with thickness equal to the fillet weld 9 mm = 0.354\\

Width of the pad =16.96 in

The out side diameter of the reinforcement pad

= dn + w= 40.949 in

W = 16.949 in

The circum firance

= 41.637 in4-Nov-08 55



4-Nov-08 56



4-Nov-08 57



MONHOLE3FOR1.12 Check the attachment of nozzle1.1.13

load to carried by welds load to be carried by welds.

( A A1 ) S = 25565.21 Ib

load to carried by welds load to be carried by welds a.c.e

(A2+2tn*t)S= 47425.34 ib

Stress values of welds2.1.12

Fillet weld shear = 8374 psiGroove weld tension = 12654 psi

Stress value of nozzle wall shear = 14000 psi

.necleStrength of welds and nozzle3.1.12

Fillet weld shear =16302.69 psi

Groove weld tension = 29483.9 psiStress value of nozzle wall shear = 24592.5 psi

Possible path of failures4.1.13Through a and b = 45786.5 psi

Through a and c = 40895.19 psi

Then Both paths are strongest than the required strength 25566.21 Ib

4-Nov-08 58



12.2 FOR 24 MONHOLE

12.2.1 Calculation the strength of attachment of nozzle

load to be carried by welds

(a a1) s = 232696.8 ib

load to be carried by welds a, c, e

(a2 + 2 tn t) s

= 47425.345 ib12.2.2 Stress values of welds

Fillet weld shear = 8374 psi

Groove weld tension = 12654 psi

Stress value of nozzle wall shear = 14000 psi

12.2.3 Strength of welds and nozzle neck Fillet weld shear =101049.6 Ib

Nozzle wall = 328137.5 Ib

Fillet weld shear =95395.66 Ib

Groove weld tension =177783 Ib

4-Nov-08 59



Through b and d = 373532.5 psi

Through c and d = 293394.9 psi

Through a, c and e= 374442.95 psi

Path 1 and 2 are stronger than the total strength al 232696.8 Ib

Path 3 is stronger than strength of = 47425.34 Ib

The enter fillet weld d strength 95395.66 Ib is greater thanthe reinforcing pad strength (dp do) tc x 17100 = 116594.537 Ib

Possible Path Of Failure4.2.12

4-Nov-08 60



4-Nov-08 61

Local Stresses in Spherical and Cylindrical Shells due to External Loadings, K. R.

Wichman, A. G. Hopper and J. L. Mershon Welding Research Council. Bulletin

107/August 1965 Revised Printing December 1968.

Standards for Closed Feed water Heaters, Heat Exchange Institute, Inc., 1969.

ACCORDING TO:-



4-Nov-08 62



b = ÷÷ ø

ö

ççè

æ

m

o

R

r

0.875 b = 0.154

Rm inch44.59==

T

Rm52.75=

13.2 For 24 Manhole

From fig (1) page 155 (H.B) we get

From fig (2) page. 156 (H.B) we get

µ = 2800

S = 3000

From fig (3) page 157 (H.B) we get

= 1600

Calculates pressure stress

=-÷÷÷

ø

ö

ççç

è

æ

2

2 T m

RT

P =

= 23088.93 psi

4-Nov-08 63



Allowable forces :

= - 18840.67 Ib( )s -µ

Sy Rm

2

FRRF =

MRCM =

å

Syo

r m

R2

= 211456.8 In.Ib

MRLM = ÷ ø öç

è æ -

D s Syo

r m

R2

= - 138441.6 In.Ib

MRM

= 15451 Ib

FRF = - 3111 Ib

4-Nov-08 64



4-Nov-08 65

ACCORDING TO:-



F = CSE

P

= 0.08 in-1 Dop

= 7 ft

The length of vessel =2

4

tD

V

= 45.88 ft

4-Nov-08 66



4-Nov-08 67

ASME UGS-66 and UGS-66.1

ACCORDING TO:-



Sa

= P

t

PRúû

ùêë

é+ 6.0

= 13681.59 psi Actual Stress

The max allowable stress of the materi al 17100 psi

Ratio =

all S

a= 0.8

the Impact test curves at66 From fig UCS

thickness t = 0.787 inch is - 5°F

Reduction of minimum1.66 From fig UCS

metal temperature.

= - 5 20 = - 25°F

4-Nov-08 68



4-Nov-08 69

PIPING SYSTEM

DESIGN

Pipe Thickness

Reacting force

Piping RoutingPiping Routing

Piping supportsMaximum Bending

stressPiping Sketch



4-Nov-08 70

ASME Process Piping B31.3

ITT Grinnell Industrial Piping, Inc. Piping Design And Engineering.

ACCORDING TO:-



1. Pipe Thickness:-

t = PD / 2 ( SE + PY ) t= 0.0135 = 0.5 mm

From properties of pipe table (PDH) Page 12

Ip (moment of inertia) = 3.017 inch 4

Sm (section modules) = 1.724 inch 3

And from expansion factor table (PDH) P11

Y Coefficients (table 304.1) B 31.3

S Stress value material table A.1

Total thickness = t + corrosion allowance= 0.5 + 3 = 3.5 mm

Actual thickness = 5.49 mm

SCH 40

Y = 0.4

S = 17000 psi

At D= 3 and SCH 40

T = 0.216

Caf 750 = 160

4-Nov-08 71



2.Piping Routing : -a

h

L

FX FX

Design data From U shape with equal. Tangent table (PSH)a= 4 m = 13.1 ft h = 4 m =13.1 ft L = 8 m = 26.2 ft

L / a = 2 L / k = 2

So

Kx = 12 Ky = 184-Nov-08 72



3. Reacting force

Fx = K x.c.Ip/L2

= 12*160*3.017/(26.2)2 = 8.4381 b

FX < F allowable

Sb = Kb.c.D/L

= 18*160* 118.11/26.2 = 8655.3 blt

. Maximum Bending stress4

From max distance of span table

(PDH) P 150

Maximum distance between two span

At d= 3 Span = 12 ft.

-supports:. Piping5

4-Nov-08 73



6. Piping Sketch: -

1) Wilding fitting S.R 40o elbow

SCH 40# t = 4.6

2) Flanged fittings welding neck

Steel flange rating 150 # t = 11

4-Nov-08 74

13.1 ft

13.1 ft

10 ft8.1 ft 8.1ft

PRESSURE

VESSEL

To Process

Flange

Elbows

Support



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