Maximum Design Pressure per API 650

[richissa]

Hello everyone,

I am getting an extremely high number when trying to calculate the internal design pressure (in of water) as per API 650 Appendix F.4.1.

Per Appendix F.4.1 the internal design pressure shall be calculated using the following equation:
P=(0.962*A*Fy*tanO)/D2

where:
[ul]
[li]A = area resisting the compressive force, as illustrated in Figure F-2 (in.2),[/li]
[li]Fy = lowest minimum specified yield strength(modified for design temperature) of the materials in the roof-to-shell[/li]
[li]junction (lb/in.2).[/li]
[/ul]

In this case I am using steel material A238 Gr. C with Fy= 30000 psi

The problem is (Fy = 30000 psi), when plugin in this value (Fy = 30000 psi) to the formula above, my answer for P (internal design pressure) is around 5400 (in water) which is way above the allowable pressure.

Can someone please advise what I am doing wrong here?

Thanks in advance!

 

[JStephen]

You don't give any other values- but using typical numbers, say:
A = 4 square inches
Fy = 30,000 psi
tan(theta) = 1/12
D = 40 feet
Then P = .962 * 4 * 30,000 * 1/12 /(40^2) = 6.01" of water column.
Maybe you're multiplying by D-squared instead of dividing by it, or using theta instead of tan(theta) or leaving the decimal out of 0.962?

 

[richissa]

Thanks JStephen !