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API Preparatory Class

Lesson 7

External Pressure Calculations

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(a) Rules for the design of shells and tubes under external

pressure given in this Division are limited to cylindrical

shells, with or without stiffening rings, tubes, and

spherical shells..

(b) The symbols defined below are used in the procedures

of this paragraph:

A =factor determined from Fig. G in Subpart 3 of Section

II, Part D and used to enter the applicable material chart

in Subpart 3 of Section II, Part D

B = factor determined from the applicable material chart

in Subpart 3 of Section II, Part D for maximum design

metal temperature, psi .

UG28 Thickness of Shells and TubesUnder External Pressure

Page 24 Section VIII

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Do= outside diameter of cylindrical shell course or tube,

in.

E = Not on exam.

L = total length, in. (mm), of a tube between tubesheets,or design length of a vessel section between lines of

support (see Fig. UG-28.1). A line of support is:

(1)a circumferential line on a head. Not on exam!

(2)a stiffening ring.Not on exam!(3)a jacket closure .Not on exam!

(4)a cone-to-cylinderNot on exam!

P = external design pressure, psi

UG28

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Pa= calculated value of maximum allowable external

working pressure for the assumed value of t, psi

Ro= outside radius of spherical shell, in.t = minimum required thickness of cylindrical shell or

tube, or spherical shell, in.

ts= nominal thickness of cylindrical shell or tube, in.

Beginning with UG-28(c) there are step by step

instructions for working these problems. We will go

through these steps one at a time.

UG28

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1. Cyl inders having Do /t values > or = 10:Step 1Assume a value for t and determine the ratios

L/Do and Do /t.

Example: The cylinder has corroded to a wall thickness of

0.530, its length is 120 and the outside diameter is 10. It

operates at 500 oF

So then;

Temperature = 500 oF

t = 0.530

L = 120

Do= 10

Calculate Do/t = 10/.530 = 18.8 call it 19 (no need to be exact)

Now we do L/Do = 120/10 = 12

UG28

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Step 2Enter Fig. G in Subpart 3 of Section II, Part at thevalue of L/Do determined in Step 1. For values of L/Do

greater than 50, enter the chart at a value of L/Do =50.

For values of L/Do less than 0.05, enter the chart at a

value of L/Do =0.05.

In our example problem we must go up the left side of

the Fig. G until we reach the value of L/Do of 12.

Using the chart we have the following;

UG28

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Step 3Move horizontally to the line for the value Do /tdetermined in Step 1.... Which in our case was 19, but

we will round this to 20 since these problems are not

meant to be extremely precise. So now we have.

UG28

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Step 3. Continued.

From this point of intersection move vertically downward to

determine the value of factor A.

Which gives us the following;

UG28

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Step 4

Using the value ofA calculated in Step 3, enter the

applicable material chart in Subpart 3 of Section II, Part D

for the material under consideration. Move vertically to an

intersection with the material/temperature line for the design

temperature see UG-20). Interpolation may be madebetween lines for intermediate temperatures. In cases

where the value ofA falls to the right of the end of the

material /temperature line, assume an intersection with the

horizontal projection of the upper end of thematerial/temperature line.

To use the next figure we enter at the bottom at the value

Factor A = .0028 and then up to our temperature of 500 oF.

UG28

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Factor A = .0028 and then up to our temperature of 500 oF.

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Factor B is 12,000. Plug it into the formula below and wehave our External Pressure allowable, Pa Which will be;

UG28

)tD3(

4B

=Pa o

psi80060

000,48

20x3

12,000x4

=Pa

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All that work just to find Factor B is and plug it into thesimple formula below.

UG28

)

t

D3(

4B=Pao

As regards the final answers to these problems, because of

the difficulty of being precise with the Fig. G there will always

be some difference from one person to the next in the

determination of Factor A. This is allowed for on the exam by

listing choices of answers that are in a range of +/- 5%. In our

previous problem the answer was 800 psi, on the exam the

correct choice would have been offered as 760 to 840 psi, i.e.

Answer Range: 760 840 psi

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To Summarize UG-28

External calculations depart significantly from internal

calculations simply because under external pressure the

vessel is being crushed. Internal pressure wants to tear the

vessel apart.

Because of the crushing or buckling load, the Lengththe

Outs ide Diameterand theThicknessof the vessel are

important. External pressure problems are based on the

thickness of the shell to the outside diameter ratios. There

are two types of external pressure calculations, the type we

will use is when the O.D to (Do) thickness ratio (t) isgreaterthan 10and the other type,not on the test, is when it isless

than 10.

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To Summarize UG-28

In order to solve these types of problems two charts will berequired. The first chart Fig. G is used to find a value called

Facto r A and then Facto r A is used to find a Facto r Bin the

second material specific chart. The value of Facto r Bfound

is the number needed to solve the problem using theformula given in paragraph UG-28 (c)(1) step 6. As stated in

the API 510 Body of Knowledge, these charts will be

provided in the exam body, IF an external calculation is

given on the examination.

One more problem. Find the allowed external pressure on

an existing vessel of a known thickness with a Do/t ratio >

10.

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UG-28

Problem: A vessel is operating under an external pressure,

the operating temperature is 500 F. The outside diameter ofthe vessel is 40 inches. Its length is 70 inches. The vessels

wall is 1.25 inches thick and is of SA-515-70 plate. Its

specified min. yield is 38,000 psi. What is the maximum

external pressure allowed?

Givens:

Temp = 500 F

t = 1.25

L = 70 inches

D0= 40 inches

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UG28

From UG-28 (c)Cyl ind r ical Shel ls and Tubes. Therequired minimum thickness of a shell or a tube under

external pressure, either seamless or with longitudinal

butt joints, shall be determined by the following

procedure.

(1) Cylinders having a

Testing to see if this paragraph applies:10values

t

Do

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UG28

Step 1Our value of Do is 40 inches and L is 70 inches. We

will use these to determine the ratio of:

32=1.25

40=

t

Do

1.75=40

70=

oD

L

Step 2 Enter theFacto r A chart at the valueof 1.75determined above.

Step 3 Then move across horizontally to the curve Do/t =

32. Then down from this point to find the value ofFacto r A

which is.0045

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UG28

Step 4. Using our value of Factor Acalculated in Step 3,

enter the Factor B(CS-2) chart on the bottom. Then

vertically to the material temperature line given in the stated

problem (in our case 500 o F).

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UG28

Step 5 Then across to find the value of Facto r B. We find

that Facto r Bis approximately 13000.

Step 6 Using this value of Facto r B, calculate the value of

the maximum allowable external pressure Pausing the

following formula:

)t

D3(

4B=Pa

o

psi541.66=9652,000=

3(32)4x13,000=P a

Answer Range: 514 568 psi

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Class Quiz

UG-28

A vessel under external pressure has been found to a

thickness of 1.123 ". The vessel is 8'-2" long and operates at

a temperature of 300

o

F. The vessels outside diameter is 54inches. It is made of a material with a minimum yield of

30,000 psi. Presently the external working pressure is 350

psi. May this vessel continue to operate in accordance with

the Code? Show all work and quote code paragraphs used.

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(1) Cylinders having:

UG28

10valuest

Do

Testing to see if this paragraph applies:

48.08=1.123

54=

t

Do

Step 1 Our value of Do is 54 inches and L is 98 inches.

We will use these to determine the ratio of:

1.81=54

98=

oD

L

Step 2 Enter theFacto r A chart at the valueof 1.8

determined above.

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UG28

Step 3. Then move across horizontally to the curve at

approximatelyDo/t = 48. Then down from this point to

find the value ofFactor A which isapproximately .0022

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Step 4 Using our value of Facto r A calculated in Step 3,enter the Facto r B(CS-2) chart on the bottom, then

vertically up to the material temperature line given in the

stated problem (in our case 300 o F).

Step 5 Next step is across to find the value of Facto r B.We

find that Facto r Bis approximately 15000. Note due to

the variance in the reading of the charts answers and

values may vary, but should be within a 5% range of the

solution.

UG28

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Step 6 Using this value of Facto r B, calculate the value ofthe maximum allowable external pressure Pausing the

following formula:

UG28

)tD3(

4B=Pao

psi416.66=144

60,000=

3(48)

4x15,000=P a

Answer Range: 395 - 437 psi

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Later.