ASME VIII Div. 1- 1 Internal Pressure

it is not intended that the Code be used as a design handbook; rather, engineering judgment must be employed in the selection of those sets of Code rules suitable to any specific service or need.

The designer is responsible for complying with Code rules and demonstrating compliance with Code equations when such equations are mandatory.

The Code neither requires nor prohibits the use of computers for the design or analysis of components constructed to the requirements of the Code. However, designers and engineers using computer programs for design or analysis are cautioned that they are responsible for all technical assumptions inherent in the programs they use and they are responsible for the application of these programs to their design.

Shall include the requirements of: Introduction Subsection A, General Requirements

Subsection B, Methods of Fabrication Subsection C, Classes of Materials Mandatory Appendices Section II, V, and IX when reference May include: Code Cases Nonmandatory Appendices Other sources which do not conflict with

mandatory requirements

(b) Minimum thickness permissible of shells and heads is 1/16 in. (1.5 mm) excluding corrosion allowance (C.A.), after forming & regardless of product form and material. except: Heat transfer plates of plate type heat exchangers. Inner pipe of double pipe heat exchangers and tubes

in shell and tube heat exchangers (where the pipe or tube is NPS 6 [DIN

150] or less. Unfired steam boilers shall be 1/4 in. (6mm) exclusive

of corrosion allowance Compressed air, steam or water service vessels of

UCS-23 materials shall be 3/32 in. (2.5mm) excluding corrosion allowance.

(c) Mill Under tolerance Plate material shall not be ordered thinner than the

design thickness. Vessels made of plate furnished with an under

tolerance of not more than the smaller of 0.01 in. (0.25mm) or 6% of the ordered thickness may be used at the full design pressure for the ordered thickness. If the specification to which the plate is ordered allows a greater under tolerance, the ordered thickness of the materials shall be sufficiently greater than the design thickness so that the thickness of the material furnished is not more than the smaller of 0.01 in. (0.25 mm) or 6% under the design thickness.

(d) Pipe Under tolerance If pipe or tube is ordered by its nominal wall thickness, the material specification manufacturing under tolerance on wall thickness shall be taken into account (except for nozzle wall reinforcement area requirements per UG-37).

After determining the minimum t, the thickness shall be increased by an amount sufficient to provide the manufacturing under tolerance allowed in the material specification.

(e) Corrosion Allowance In Design Formulas All dimensional symbols used in all design

formulas throughout this Division represent dimensions in the corroded condition.

Therefore, material designated for corrosion

purposes cannot contribute to the strength of the vessel. Such material must be deducted when determining design pressure, and added to the required thickness to determine design thickness. (Definitions, Appendix 3)

(b) Special Shapes Vessels other than cylindrical or spherical and those for which

no design rules are provided in this Division may be designed under the conditions set forth in U-2.

U-2(g) This Division of Section VIII does not contain rules to cover all details of design and construction. Where complete details are not given, it is intended that the manufacturer, subject to the acceptance of the Inspector, shall provide details of design and construction which will be as safe as those provided by the rules of this Division.

(c) No Design Rules When no design rules are given and the strength of a vessel

(or part) cannot be calculated with a satisfactory assurance of accuracy, Maximum Allowable Working Pressure shall be established in accordance with the provisions of UG-101. (Proof testing)

UG-20 Design Temperature (a) Maximum Shall be not less than the mean metal temperature under operating conditions for the part under consideration. May be determined by computation or measurement from equipment "in service" under equivalent operating conditions. (b) Minimum Is the lowest expected "in service", Determined same as for maximum temperature, Consideration includes:

Lowest operating temperatures, Operational upsets, Auto refrigeration, Atmospheric conditions, Any other source of cooling.

(c) Maximum Design temperatures that exceed the temperature

limit in the applicability column shown in Section II, Part D, Subpart 1, Tables 1A, 1B, and 3 are not permitted.

In addition, design temperatures for vessels under external pressure shall not exceed the maximum temperatures given on the external pressure charts.

(d) The design of zones with different metal temperatures may be based on their determined temperatures.

(e) Suggested methods for obtaining the operating temperatures of vessel walls in service are given in Appendix C.

The MDMT marked on the nameplate shall correspond to a coincident pressure equal to the MAWP. When there are multiple MAWP’s, the largest value shall be used to establish the MDMT marked on the nameplate. Additional MDMT’s corresponding with other MAWP’s may also be marked on the nameplate (see footnote 37).

When determining design pressure, consideration must include the following. At least the most severe condition if coincident pressure and temperature expected in normal operation. The operating and test conditions including the maximum difference in pressure between The inside and outside of the vessel, Any two chambers of a combination unit.

(b) The MAWP for a vessel part is the maximum internal or external pressure, including static head thereon as determined by the rules and formulas of this Division, together with the effect of any combination of loadings listed in UG-22 which are likely to occur for the designated coincident temperature, excluding any metal specified as corrosion allowance (UG-25).

UG-98 Maximum Allowable Working Pressure

(a) It is the maximum permissible pressure at the top of the vessel in its normal operating position, at the designated coincident temperature specified for that pressure.

The loadings to be considered in designing a vessel shall include those from: Internal or external design pressure (UG-21). Weight of the vessel & normal contents under operating or test conditions (this includes additional pressure due to the static head of liquids).

- 1 cubic foot of water at 62°F = 62.35 lb. - 62.35 lb. /1728 cubic inches = 0.0361 lb. / cu. inch - 0.0361 cu. inches x 12 inches = 0.433 lb./foot (for water)

(d) the attachment of: (1) internals (see Appendix D); (2) vessel supports, such as lugs, rings, skirts, saddles, and legs (see Appendix G); (e) cyclic and dynamic reactions due to pressure or

thermal variations, or from equipment mounted on a vessel, and mechanical loadings; (f) wind, snow, and seismic reactions, where

required; (g) impact reactions such as those due to fluid shock; (h) temperature gradients and differential thermal

expansion; (i) abnormal pressures, such as those caused by

deflagration.

The maximum allowable tensile stress values permitted for different materials are given in Subpart 1 of Section II, Part D.

A listing of these materials are included in Subsection C.

Materials meeting more than one specification and/or grade may be used provided all requirements for the selected specification and/or grade are met.

(a) the user or his designated agent (see U-2) shall specify corrosion allowances other than those required by this Division.

Where corrosion allowances are not provided, this fact shall be indicated in the Data Report.

(b) vessels or parts subject to thinning by Corrosion Erosion Mechanical abrasion shall have provision made for the desired life of the

vessel by a suitable increase in the thickness over that determined by the design formulas, or by using some other suitable method of protection. (See Appendix E).

(c) Material for these purposes need not be of the same thickness for all parts of the vessel if different rates of attack are expected for the various parts.

(d) No additional thickness need be provided when previous experience in like service has shown that corrosion does not occur or is of only a superficial nature.

(e) Telltale Holes May be used to provide some positive indication when

the thickness has been reduced to a dangerous degree. Shall not be used in vessels which are to contain lethal

substances [seeUW-2(a)], except as permitted by ULW-76 for vent holes in layered construction.

(e) Telltale Holes

(f) Openings for Drains Vessels subject to corrosion shall be supplied with a suitable drain opening at the lowest point practicable in the vessel; or a pipe may be used extending inward from any other location to within 1/4 in. (6mm) of the lowest point.

(a) The minimum required thickness of shells under internal pressure shall not be less than that computed by the following formulas except as permitted in Appendix 32 (local thin areas). In addition, provision shall be made for any of the loadings listed in UG22, when such loadings are expected.

The provided thickness of the shells shall also meet the requirements of UG-16, except as permitted in Appendix

32. UG-16(b) requires that minimum thickness be determined

after forming and exclusive of any c.a. UG-16(c) Mill under tolerance UG-16(d) Pipe Under tolerance UG-16(e) Corrosion Allowance in Design Formulas

UG-27 Thickness of Shells Under Internal Pressure

(b) The symbols defined below are used in the formulas of this paragraph. t = minimum required thickness P = internal design pressure R = inside radius of the shell course under consideration S = maximum allowable stress value (see UG-23 and the stress limitations specified in UG-24 E = joint efficiency for, or the efficiency of, appropriate joint in cylindrical or spherical shells, or the efficiency of ligaments between openings, whichever is less For welded vessels: UW-12, for ligaments: UG-53

(c) Cylindrical Shells. The minimum thickness or Maximum allowable Working Pressure of cylindrical shells shall be the greater thickness or lesser pressure as given in (1) or (2) below.

(c)(l) Circumferential Stress (Longitudinal Joints)

(c)(2) Longitudinal Stress (Circumferential Joints15 )

15 These formulas will govern only when the circumferential joint efficiency is less than one-half the longitudinal joint efficiency, or when the effect of supplemental loadings (UG-22) are being investigated.

(c)(l) Circumferential Stress (Longitudinal Joints)

(c)(2) Longitudinal Stress (Circumferential Joints)

Example: Assume P = 150 PSI S = 17, 500 psi R = 24 inches E=1.0

(c)(l) Circumferential stress (Longitudinal Joints)

(c)(2) Longitudinal Stress (Circumferential Joints)

Comparing Internal and External Formulae

Example: Given a cylindrical shell with the following variables, solve for the MAWP of the cylinder using both formulas.

P = ? * The question mark defines what is being solved for. t = 0.500" S = 15,000 psi E = 1.0 R = 18.0“ and Routside = 18.5"

psi 8.4093.18

7500

0.500)x (0.4 - 5.18

0.500x 1.0x 000,15

0.4t-R

SEt=P 1)-(1 1 App

o

Example 1:

Internal Formula

A vessel shell has corroded to an inside radius of 24.18” its working pressure is 500 psi and its stress allowed is 17,500 psi. What is the required thickness?

Other terms sometimes used: Corroded internally, found to have an inside diameter/radius, etc.

In these cases we must use the inside formula of UG-27

UG-27 Internal Pressure Cylindrical Shells

Example 2:

External Formula

A vessel shell has corroded to an outside diameter of 24.38” its working pressure is 500 psi and stress allowed is…..what is the required thickness? Other terms sometimes used: found externally corroded, attacked by corrosion under insulation (CUI) etc.

Here we would have to use the formula of Appendix 1.


Example: Internal Formula or External Formula

3. A vessel shell has corroded to an inside radius of 24.18 ” its working pressure is 500 psi and its stress allowed is ….its original thickness was .500” and the original inside radius was 24.0 ” ** for inside calculations use R = 24.18 (actual)

To use the outside formula we can add the original thickness to the original inside radius.

24” + .500 = 24.5 ” = Ro Original radius outside

Now we can use the formula of Appendix 1-1 if we chose to.


Example: External Formula for Thickness

4. A vessel shell has an outside radius of of 24.0 ” its working pressure is 500 psi and its stress allowed is 15,000 psi. The joint efficiency, E = 1.0. The shell has corroded internally to a thickness of 0.343”. What is its present Maximum Allowable Working Pressure?

Here you must use the O.D. Formula since you cannot determine the present internal corroded radius, not having the original thickness you cannot determine the original I.D.!


Here is an example of working a problem using both inside and outside dimensions having all the information needed.

A cylindrical shell has been found to have a minimum thickness of .353". Its original thickness was .375“ with an original inside radius of 12.0”. What is its present MAWP ?

P = 300 psi t = .353" S = 13,800 psi E = .85 R = 12.0" + (.375-.353) = 12.022 corroded inside radius Ro= 12.0" + 0.375 (orig. t) =12.375” original outside radius


Here is a graphical representation of the problem:


Routside = 12.0" + 0.375 (orig. t) = 12.375”

R corroded = 12 + (0.375 – 0.353) = 12.022”

t Corroded = 0.353" t Corroded = 0.353"

t Orig. = 0.375"

Radius Inside for MAWP using UG-27(c)(1).


psitR

SEt 46.338

12.2338

4140.69=

.353)x (0.6+ 12.022

.353x .85x 13,800=

6.0P 27(c)(1)-UG

Radius Outside for MAWP using App: 1 (1-1).

psi 46.3382338.12

69.4140

.353)x (0.4 - 375.12

.353x .85x 800,13

0.4t-R

SEt=P 1)-(1 1 App

o

Answer: Yes

338.46 psi > 300 psi

Please use this approach for all calculations. Givens: Sketch t = P = R = S = E =

Code Paragraph UG-27 (c) (1)


t 0.6 +R

SEt = P

I.D. ?

t = ?

Problem # 1 Find the Maximum Allowable Working Pressure (MAWP) of a 12 inch inside diameter shell. This shell is seamless and is stamped RT 2. It has an allowable stress value of 16,600 psi and the wall thickness is .406”. No corrosion is expected.


Givens: Plugged in from the question! P = ? t = .406 R = D/2 = 12/2 = 6.0 ” this formula uses the Radius. S = 16,600 psi E = 1.0 per UW-12(d) Seamless shells and heads From UG-27 (c) (1) Circumferential Stress


t 0.6 +R

SEt = P

2436.)0.6(

6.6739

.406)x 0.6 (+) (6.0

x.406 1.0x 16,600 = P


psi 44.10792436.6

6.6739 = P

As can be seen the calculations are simple, it is more a matter of deciding on the correct formula to use, inside or outside, and transferring the givens accurately to the formula. Once again use the approach;

Givens: SKETCH

P =

t =

ect.

I.D. ?

t = ?

About rounding answers. In the ASME Code and for the exam you must round DOWN for pressure allowed so in our solution below we would round down to 1079 psi. Even if our solution had been 1079.999 we cannot round to 1080 , we still round down to 1079 psi.

This is the conservative approach taken by the Codes in general and of course is different for the normal rules of rounding.


t 0.6 +R

SEt = P

psi 44.10792436.6

6.6739

.406)x 0.6 (+) (6.0

x.406 1.0x 16,600 = P

Problem # 2 Find the minimum required thickness of a cylindrical shell designed for a working pressure of 100 psi. The shell's inside radius is 2'-0". The longitudinal joint is type 1 (table UW-12) and no radiography was performed. The shell is made of carbon steel rolled plate with an allowable stress of 15,000 psi.


SKETCH

t = ?Type 1 Category

A

No RT

Givens: t = ? P = 100 psi R = 24" S = 15000 E = .70 ( From Table UW-12 column C) From UG-27 (c) (1) Circumferential Stress


P 0.6 - SE

PR = t

" .2298= 10440

2400=

100)x (0.6 - ) .70x (15,000

24x 100 = t

When rounding thickness required we must round up. The most conservative thing to do. So our example below would round to .230”. Even it had been .2291 we would still round up to .230”.


P 0.6 - SE

PR = t

" .2298= 10440

2400=

100)x (0.6 - ) .70x (15,000

24x 100 = t

Problem # 3

Determine the minimum required thickness of a cylindrical shell designed for an internal pressure of 50 psi, no corrosion is expected.

The shell’s Category A and B, Type 1 welds have been fully radiographed. The material’s stress allowable is 17,500 psi. The vessel will be stamped RT 1.


Long Joint (Circumferential Stress) SKETCH:


I.D. 10'- 0"

t = ?Type 1 Category

A & B

Full RT

Givens: t = ? P = 50 R = 10’ x 12” = 120”/2 = 60" S = 17500 E = 1.0 (RT 1) From UG-27 (c) (1)


.172" to rounds " .1717=17470

3000=

50)x (0.6 - ) 1.0x (17,500

60x 50 = t

P 0.6 - SE

PR = t

1. Calculate the thickness required for a vessel’s seamless shell made of SA-106 gr. B

pipe. The O.D. is 12.75 inches. UW-11(a)(5)(b) has been applied. The shell will operate at a maximum of 500 psi. The stress allowed on the shell material is 15,000 psi. The vessel will be stamped RT 2.

From the question above fill in the givens below.

Givens: SKETCH: t = P = S = E = R or Ro = State Code Paragraph and Formula:

Class Quiz UG-27 Internal Pressure Cylindrical Shells

1) From: Appendix 1-1

Solution

PSE

PRot

4.0

Givens: t = ? P = 500 psi S = 15,000 psi E = 1.0 per UW-12(d)

Ro =

"375.6

2

75.12 "2097.

)5004.0()0.1000,15(

375.6500

xx

xt

Answer: Required t = .2097” Rounds to .210”

Class Quiz

2. What is the maximum allowed working pressure on the following shell? The shell’s inside radius is 52 inches, and the shell’s thickness is .850 inches. The allowable stress for the shell's material is 15,000 psi. The joint efficiency of the shell's Category A joint is 1.0

From the question above fill in the givens below.

Givens: SKETCH: t = P = S = E = R or Ro = State Code Paragraph and Formula:

2) From: UG – 27 (c) (1)

Solution

Givens: t = .850” P = ? S = 15,000 psi E = 1.0 R = 52”

tR

SEtP

6.0

510.52

750,12P

Answer: 242.81 psi (rounds to 242 psi)

)850.6.0(52

850.0.1000,15

x

xxP

psiP 81.242510.52

750,12


You are now familiar with the basic cylindrical shell formula from UG-27. However that formula in its published form is only useful for the calculation of vessel shells that are designed without a corrosion allowance. Most of the time a corrosion allowance must be given to vessel part.

Example:

A vessel is being designed for a specific volume of water. The designer determines the optimum inside diameter and length of the vessel to obtain that volume.

The engineer set the inside diameter at 48” so it must be constructed with that inside diameter, resulting in an inside radius of 24” to be used in the calculation.


In the design calculation the engineer adds the corrosion allowance to the radius. The basic formula of UG-27 would be modified to be;

P 0.6 - SE

.).(R P =

act

t 0.6 + c.a)(R

SEt = P

or

t 0.6 + )125.0(24

SEt = P

Inside diameter = 48.0” Inside radius = 24.0” c.a. = 1/8” = 0.125” Inside radius used in calculations = 24.0 + 0.125 = 24.125” resulting in the following;


The vessel shell would be constructed of the required calculated thickness and then rolled to an inside radius of 24”, it retirement radius would be 24.125” This is no different from what occurs during the evaluation of an in service vessel that has corroded. However here we use actual measurements. Suppose the vessel shell above was built with a thickness of 0.500” and rolled to the 24.0” inside radius. Corrosion has occurred and the new minimum wall thickness is 0.450”. To calculate we would use a radius of 24.0 + (0.500 – 0.450) or 24.050”. This would leave a remaining corrosion allowance of .125 -.050 = .075”

Class Quiz

1. A vessel shell has a design pressure of 200 psi, and an allowable stress of 14,800 psi. The inside radius is 84”. The nameplate is stamped RT1 . The shell has corroded down to 1.28 inches. Its original t was 1.375". What is its current calculated MAWP in accordance with rules of Section VIII Division 1? Givens: t = P = ? S = E = R =

Solution

From: UG-27 ( c )(1) Givens: t = 1.28” P = ? S = 14,800 psi E = 1.0 RT 1 R = 84” = 84’ + (1.375-1.28) = 84.095”

t 0.6 + c.a)(R

SEt = P

)28.16.0(095.84

28.10.1800,14

x

xxP

psiP 23.223)863.84

944,18

)768.0(095.84

944,18P

(rounds to 223 psi)

(a) The required thickness at the thinnest point after forming of ellipsoidal, torispherical, hemispherical, conical and toriconical heads under pressure on the concave side (plus heads) shall be computed by the appropriate formulas in this paragraph, except as permitted in Appendix 32 (Localized Thin Areas). (a) In addition, provision shall be made for any of the loadings listed in UG-22. The provided thickness of the heads shall also meet the requirements of UG-16, except as permitted in Appendix 32. (b) The thickness of an unstayed ellipsoidal or torispherical head shall in no case be less than the required thickness of a seamless hemispherical head divided by the efficiency of the head to shell joint.

(c) The symbols defined below are used in the formulas of this paragraph. t = minimum required thickness of the head after forming P = internal design pressure D = inside diameter of the head skirt; or inside length of a major axis of an ellipsoidal head; or inside diameter of a conical head at the point under consideration, measured perpendicular to the longitudinal axis. Dt = inside diameter of the conical portion of a toriconical head at its point of tangency to the knuckle, measured perpendicular to the axis of the cone. [ = D- 2r(l-cos a)]

r = inside knuckle radius S = maximum allowable stress value in tension as

given in the tables referenced in UG-23, except as limited in UG-24 and (e) below E = lowest efficiency of any joint in the head; for

hemispherical heads this includes head-to-shell joint; for welded vessels, use the efficiency specified in UW-12 L = inside spherical or crown radius. The value of

L for elliptical heads shall be obtained from Table UG-37 a = one-half of the included (apex) angle of the

cone at the center line of the head (see fig. 1-4)

(d) Ellipsoidal Heads with t/L >0.002.

The required thickness of a dished head of semi-ellipsoidal form, in which half the minor axis (inside depth of head minus head skirt) equals one-fourth of the inside diameter of the head skirt, shall be determined by:

NOTE: For ellipsoidal heads with t/L< 0.002, the rules of l-4(f) shall also be met. An acceptable approximation of a 2:1 ellipsoidal head is one with a knuckle radius of 0.17D and a spherical radius of 0.90D.

(e)Torispherical Heads with t/L >0.002. The required thickness of a torispherical head for the case in which the knuckle radius is 6% of the inside crown radius and the inside crown radius equals the outside diameter of the skirt [see (j) below], shall be determined by:

Note: For torispherical heads with t/l <0.002, the rules of 1 -4(f) shall also be met.

(f)Hemispherical: Heads. When the thickness of a hemispherical head does not exceed 0.356L, or P does not exceed 0.665SE, the following formula shall apply:

(i) When an ellipsoidal, torispherical, hemispherical is of a lesser thickness than required by the rules of this paragraph, it shall be stayed as a flat surface according to the rules of UG-47 for braced and stayed flat plates. (o) If a torispherical, ellipsoidal, or hemispherical head is formed with a flattened spot or surface, the diameter of the flat spot shall not exceed that permitted for flat heads as given in Formula (1) in UG-34, using C = 0.25. (p) Openings in formed heads under internal pressure shall comply with the requirements of UG-36 through UG-47.

(a) The required thickness at the thinnest point after forming of ellipsoidal, torispherical, hemispherical, conical, and toriconical (not on exam) heads under pressure on the concave side (plus heads) shall be computed……. (b) The thickness of an unstayed ellipsoidal or torispherical head shall in no case be less than….this is a test to see if you should use this formula or the ones given in Appendix 1. Not on Exam!

(c) The symbols defined below are used in the formulas of this paragraph:

UG-32 Formed Heads Pressure on the Concave Side

t = minimum required thickness of head after forming, in. (mm)

P = internal design pressure (see UG-21), psi (kPa)

D = inside diameter of the head skirt; or inside length of the major axis of an ellipsoidal head; in. (mm)

S = maximum allowable stress value in tension.

E = lowest efficiency of any joint in the head; for hemispherical heads this includes head-to-shell joint; for welded vessels, use the efficiency specified in UW-12

L = inside spherical or crown radius, in. (mm)


There are 5 formed heads listed in UG-32. You will be responsible for the calculations of these 3 only;

Hemispherical, Ellipsoidal and Torispherical.

The next series of slides are example calculations of all three types for thickness required. These calculations will use the exact same conditions for service, stress allowed, Joint E, dimensions, and pressure.

• With all things being equal, which do you suspect will be the thinnest allowed?

• Which do you think will be the thickest required?

• Which is in the middle?


Givens: The same pressure, stress and, dimension values will be used for all heads. Let’s determine which type of head will be the thickest required and which will be the thinnest allowed. P = 100 psi S = 17500 PSI E = .85 for spot RT of Hemi-head joint to shell E = 1.0 for seamless heads ( Ellipsoidal and Torispherical ) L = 48" for the inside spherical radius for the hemi-head L = 96" for the inside crown radius of the torispherical head D = 96" inside diameter of the ellipsoidal t = ? Required wall thickness, inches


Problem # 1

Given the above data find the required thickness of a seamless ellipsoidal head.

UG-32 Formed Heads Pressure On The Concave Side

P 0.2 - 2SE

PD = t

" .2744 =34980

9600

100)x (0.2 - 1.0)x 17,500x (2

96x 100 = t

From UG-32 (d)

Problem # 2 Using the same data, calculate the required thickness of a hemispherical head.


0.2P - 2SE

PL =t

" 1614.029730

4800

)100 2.0()85.0 500,17 2(

48 100

xxx

xt

From UG-32 (f)

Problem # 3 Determine the required t of this torispherical head. (These are also called ASME flanged and dished heads, by the way).


0.1P - SE

0.885PL =t

" .4857 =17490

8496 =

100)x (0.1 - 1.0)x (17,500

96x 100x 0.885 =t

From UG-32(e)

So we have from thickest to thinnest, all things equal: Torispherical = .4857” (Rounds to .486”) Ellipsoidal = .2744 (Rounds up to .275”) Hemispherical = .1614 (Rounds to .162”) There have been several exams where the question was asked, “ Which is required to be thickest” or “ Which can be the thinnest” Remember this.


One last important comment: Hemispherical heads while they can be formed seamless are not considered seamless heads by Section VIII. As mentioned previously they essentially form a Category A seam between the head and the other part. The spot RT of UW-12(d) does not apply to the Joint E used to calculate a Hemispherical head. They are never seamless, their Joint E comes from Table UW-12 based on the Type of weld and the extent of Radiography applied. * Remember This.*


Class Quiz

1. Calculate the required thickness of a 2 to 1 Ellipsoidal head with an inside diameter of 48 inches. The vessels will have a MAWP of 350 psi and will be in lethal service and stamped RT1. The joint used to join the head to shell will be a Type No. 2 from Table UW-12. The stress allowed on the head's material will be 15,000 psi.

Givens: SKETCH: t = P = S = E = D = State Code Paragraph and Formula:

# 1 From: UG-32 (d) Givens: t = ? P = 350 psi S = 15,000 psi E = 1.0 Full RT 1 D = 48” inside diameter

Solution

PSE

PDt

2.02

"5613.)3502.0()0.1000,152(

48350

xxx

xt

Class Quiz

2. A Torispherical head has corroded to a thickness of .353 " ; its inside crown radius is 56 inches. The head's material has a stress allowable of 13,800 psi. The shell is seamless and the spot radiography of UW-11(a)(5)(b) has been applied to the vessel. Can this head remain in service at 100 psi per Code? Givens: SKETCH: t = P = S = E = L = State Code Paragraph and Formula:

2) From: UG-32 (e) Givens: t = .353 P = 100 psi S = 13,800 psi E = 1.0 L = 56” crown radius

Solution

Answer: No the head may not remain in service. "3593.)1001.0()0.1800,13(

56100885.0

xx

xxt

0.1P - SE

0.885PL =t

Class Quiz

3. A Hemispherical head is being considered as a replacement on a vessel with a MAWP of 200 psi. The head's Inside diameter will be 64 inches. What would be its required thickness if the head's material has a maximum allowable stress of 17,500 psi? The Category A type 1 joint that attaches the head will be spot radiographed. Givens: SKETCH: t = P = S = E = L = State Code Paragraph and Formula:

3) From: UG-32 (f) Givens: t = ? P = 200 psi S = 17,500 psi E = .85 Spot RT per UW-12(b) and Table UW-12 Column B L = 32.0” inside spherical radius (D/2)

Solution

Answer: the required thickness = .2154”

PSE

PLt

2.02

"2154.)2002.0()85.500,172(

0.32200

xxx

xt

Class Quiz

4. What would the required thickness for a seamless Ellipsoidal head be given the following variables? The Category B weld that will attach this head would not have UW-11(a)(5)(b) applied. Givens: t = ? P = 200 psi S = 17,500 psi E = .85 No spot RT per UW-12(d) D = 64.0”

4) From: UG-32 (d) Givens: t = ? P = 200 psi S = 17,500 psi E = .85 No spot RT per UW-12(d) D = 64.0”

Solution

Answer: thickness required = .4308”

PSE

PDt

2.02

"4308.)2002.0()85.500,172(

64200

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(a) The minimum thickness of unstayed flat heads, cover plates and blind flanges shall conform to the requirements of this paragraph. Some acceptable types of flat heads and covers are shown in Fig. UG-34. In this figure, the dimensions of the component parts and the dimensions of the welds are exclusive of extra metal required for corrosion allowance.

(c) The thickness of flat unstayed heads, covers and blind flanges shall conform to one of the following three requirements. (c)(l) Circular blind flanges conforming to the standards referenced in Table U-3, when the blind flange is of the types shown in Fig. UG34 sketches (j) and (k).

(c) Flat unstayed heads, covers and blind flanges shall be calculated by the following formula: (c)(2)Circular

(c)(3) square, rectangular, elliptical, obround, segmented or otherwise noncircular

(c)(2)(a) Openings in Cylindrical or conical shells, or formed heads shall be reinforced to satisfy the requirements of UG-37 except as given in (3) below. (c)(2)(b) Openings in flat heads shall be reinforced as required by UG-39. (c)(3) Openings in vessels not subject to rapid fluctuation in pressure do not require reinforcement other than that inherent in the construction under the following conditions: (a) welded, brazed or flued connections with a finished opening not larger than: 3 1/2 in. (89mm) dia. in vessel shells or heads with a required minimum thickness of 3/8 in. (10mm) or less 2 3/8 in. (60mm) dia. in vessel shells and heads over a minimum required thickness of 3/8 in (10mm).

UG-36(c) exemptions to reinforcement consideration, continued... UG-36(c)(3)(b) threaded, studded or expanded connections in which the hole cut in the shell or head is not greater than 2 3/8 in. (60mm) dia. UG-36(c)(3)(c) no two unreinforced openings, in accordance with (a) or (b) above, shall have their centers closer to each other than the sum of their diameters example

(d) Openings Through Welded Joints Additional provisions governing openings through welded joints are given in UW-14. UW-14 (a) Any type of opening that meets the requirements for reinforcement given in UG-37 or UG-39 may be located in a welded joint. UW-14(b) Single openings meeting the requirements given in UG36(c)(3) may be located in head-to-shell or Category B or C butt welded joints, provided the weld meets the RT requirements of UW-51 for a length equal to three times the diameter of the opening with the center of the hole at midlength. Defects that are completely removed in cutting the hole shall not be considered in judging the acceptability of the weld.

UW-14(c) In addition to meeting the radiographic requirements of (b) above, when multiple openings meeting UG-36(c)(3) are in line in head-to-shell or Category B or C butt welded joints, the requirements of UG-53 (ligaments) shall be met or the openings shall be reinforced in accordance with UG-37 through UG-42. UW-14(d) Except when the adjacent butt weld satisfies the requirements for radiography in (b) above, the edge of openings in solid plate meeting the requirements of UG-36(c)(3) shall not be placed closer than 1/2 in. (13mm) to the edge of a Category A, B, or C weld for material 1 1/2 in. (38mm) thick or less.

(b) General. The rules of this paragraph apply to all openings other than: Small openings, UG- 36(c)(3) Openings in flat heads, UG-39 Openings designed as reducer sections, UG-36(e) Large head openings, UG-36(b)(2) Tube holes with ligaments, UG-53 Reinforcement shall be provided in amount and distribution such that the area requirements for reinforcement are satisfied for all planes through the center of the opening and normal to the vessel surface.

Design parameters can be reviewed, such as Code Edition and Addenda Pressure and Temperature Materials Size Loadings and Service Conditions Based upon the design parameters, fabrication requirements can be reviewed such as: Welding Nondestructive Examination Heat Treatment Impact Testing Forming and Bending Limitations Pressure Testing Data Reports and Stamping

Based upon a review of design parameters and fabrication requirements, you are in a better position to determine meaningful inspection involvements using documents such as: Calculations Drawings & Bills of Material Purchase & Receiving Documents Process Control Documents Special Process Procedures (WPS, NDE, HT, etc.)

Design Review Tips

Has the correct construction Code Section been selected and used? Is the Code Edition/Addenda correct? Have the required calculations been prepared? Do the design documents include criteria as supplemented by the QC system? Are design documents reviewed/approved by appropriate personnel? Are you reviewing the latest revision to the design/fabrication document? If the design work is computer generated, have the computer program capabilities and results been verified as accurate?

Date post:	01-Jan-2016
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