Wall Thickness Calculation: ASME B31.3

Paragraph 304.1.1 of B31.3 has described the required thickness of straight section of pipe shall be determined in accordance with:

For t ≥ D/6, the internal pressure design thickness for straight pipe shall not be less than calculated accordance to either of two equations below:

or

Where:

• c  =  sum of the mechanical allowances plus corrosion

and erosion allowances.

• D = Outside Diameter of pipe as listed in tables of

standards or specifications or as measured.

• d = Inside Diameter of pipe.
• E = quality factor from Table A-1A or A-1B

ASME B31.3

• P = Internal Design Gage Pressure
• S = Stress value for material from Table A-1 or

Table A-1M.

• t_m = minimum wall thickness as including mechanical,

corrosion and erosion allowance.

• t = pressure design thickness
• W = weld joint strength reduction factor in accordance to

Paragraph 302.3.5. ASME B31.3

• Y = coefficient from Table 304.1.1 ASME B31.3, valid

for t < D/6, and may be interpolation for t ≥ D/6,

However, for t  ≥ D/6 or P/SE > 0.385, then there will be a special consideration of factors such as theory of failure, effects of fatigue, and thermal stress will be required.

Sample of Calculation:

Design Pressure: 500 psig

Design Temperature: 260^oC (500^oF)

C.A = 1.6 mm (0.063 inches)

Pipe Size: 10 inches

Material : A53 Gr. B, EFW

Mill Tolerance = 1.0 mm (0.0394 inches)

There will be five steps to find the wall thickness of pipe:

1. Calculate the pressure design thickness, t
   1. Add the mechanical corrosion and erosion corrosion to obtain the t_m = t + c
   1. Check if  tm < D/6
   1. Select the next commercially available schedule wall thickness, T.
   1. Check against t_m after deducting mill tolerance.

If t, after deducting mill tolerance  is more than tm, then the selection of wall thickness is acceptable.

Answer:

P = 500 psig

T = 500 F

D = 10.75 inches

S = 19,000 psi from Table A-1, Sh

E = 0.85 (Table A-1B)

Y = 0.4 (Table 304.1.1)

W = 1 (Table 302.3.4)

Mill Tolerance = 12.5%

0.164 inches

t_m = t + c = 0.164 + 0.063 = 0.227 inches < D/6 .

As stated above and in accordance with paragraph 304.1.2(a) that t must be less than D/6, then we need to calculate D/6, which in this case is: 10.75/6 = 1.792 inches.

Since the t = 0.164 inches < D/6 (which is 1.792 inches) then equation selected to calculate wall thickness is applicable without special consideration of factors listed in paragraph 304.1.2 (b).

Now, next step is to find the next commercially available wall thickness after adding mill tolerance, which is 12.5%.

First, we need to find the wall thickness closer to calculated wall thickness, t = 0.164 inches, which is T = 0.250 inches. We need to check against 12.5% Mill tolerance: T = 0.250 (1.00 – 0.125) = 0.21875, and then we need to check whether >  t_m ; which is 0.21875 < 0.227 inches. It found that our selection on Wall Thickness was incorrect since T < t_m. which force us to re-select the appropriate wall thickness, then we will do a second attempt of calculation.

Second attempt:

Select: T = 0.279 inches from ASME B36.10, and calculate against mill tolerance 12.5%, will get the result of T = 0.279 (1.00 – 0.125) = 0.244 inches.

Next step is to check whether  T after deducting with mill tolerance > t_m , which is 0.244 > 0.227 inches. The result calculation shows that our selected wall thickness, 0.279 inches is acceptable.