PSV calculation as per API 520,
F=W /366 * (kT/(k+1)M)^1/2 + AP

One calculation sheet i have defines calculation method as followings;
W /366 x (kT/(k+1)M)^1/2 = F1 applied for Horizontal Dire.
A x P = F2 applied for Vertical Dire.
(where P = 0.00245 W/(d2)^2 x (T/kM )^1/2 )

and another sheet defines for F2 :
F2 = ( w^2 x R x T ) / ( g x Ps x A )
F1 is same.

For DLF, F1 is only applied with 2.

My question is.
1. What is difference of F2 mentioned in above.
2. Why does it apply DLF only for F1.

Have you had a look at API RP 520 to see what the Code actually says ?

There are endless 'Excel' type spreadsheet calcs out there, many have fatal errors. Be careful and ensure that any that you use are properly validated. Remember that you will end up in court, not the writer of that dodgy Excel or similar file.

If the code is clear and easy to understand, there is only one excel sheet.

Is there nobody able to reply?

sgcho,

I have not read API RP 520, however admitting the fact that the DLF of any single application load can not be larger than 2, one should always consider the pop reaction force of a PSV twice as much in a static analysis.

1.Just looking to the formula
F2 = ( w^2 x R x T ) / ( g x Ps x A )
do you think it makes sense a formula in which the PSV set pressure is in denominator of fraction?
Less pressure in vessel means a good chance to have a strong vertical up force in PSV?

2.In my understanding, API 520 does not mention any horizontal or vertical force in PSV body.

Mariog,

there are different influencing factors in the vertical PSV reaction; you can not conclude that less pressure would lead to larger force since less pressure would surely affect the orifice area and discharge rate.
on a open system you are right, there is no need to consider any force on the body of the valve when the flow is "established". in closed systems and also in the pop condition, the reaction forces apply on the valve itself.

IMO, I can conclude that less pressure would lead to larger force when W is the same.
Anyway, that formula is a non-sense (again IMO!) and I explained the reason in an old post. There wouldn't be the set pressure.

When you say "in closed systems and also in the pop condition, the reaction forces apply on the valve itself" is this API 520 opinion/requirement?