API 650, EXTERNAL PRESSURE (Pe)

Because the internal pressure is within 0.25 (or 6.9) kPa of the external pressure.

But internal pressure max limit is 18 kpa as per API650 Cl. F.1.1 and external pressure limit is 0.25 to 6.9 kpa.

The internal pressure differential is relative to the external pressure and vice versa...when the internal and external pressures are the same, it's a wash....

You should ask all of your questions about API 650 tank design on one thread, there's no need to start a new thread with every question...

Thanks sir,

For atm tank

You mean internal pressure is = tank internal atm pressure - 0.25 kpa (vacuum)

= 101.325 - 0.25 kpa

= 101.075 kpa (vacuum inside tank)

External pressure = External atm pressure - tank internal pressure

= 101.325 - 101.075

= 0.25 kpa

when 6.9 kpa external pressure

tank internal pressure = 101.325 - 6.9 (Vacuum)

= 94.425 kpa

External pressure then = external atm pressure - tank internal pressure

= 101.325 - 94.425

= 6.9 kpa

Is it right sir ?

It remains impossible to see what's going on without access to a P&ID.

Do bear in mind that "second person check" as a design procedure is not embraced by a review in an international, anonymous, Engineering forum.

IMO that's a complicated way of looking at it.

It's just saying if A - B = C, then A - C = B.

The relevant point is, the roof sees minus 0.25 to minus 6.9 kPag or 0.25 to 6.9 kPa vacuum. See #6.

Sir please note the following cases,

Case-I Internal maximum design pressure as per code is 18 Kpag, but for vacuum design it is 0 kpag. so case-I is excluded.

Case-II API 650 tank are atmospheric tanks so internal pressure is 101.325 kpa.

Now, the external design pressure is 0.25 Kpa to 6.9 Kpa

i.e, internal pressure (vacuum) = 101.325 - 0.25= 101.075 kpa hence external pressure is 0.25 kpa.

internal pressure (vacuum) = 101.325 - 6.9 = 94.425 kpa hence external pressure is 6.9 kpa.

Conclusion= for external pressure 0.25 to 6.9 kpa internal pressure (vacuum) limited to 94.425 to 101.075 kpa.

please correct me if i wrong.

Without looking up API 650, I assume Case I is for positive pressure (in the roof space) and Case II is for negative.

You say Case-I Internal maximum design pressure as per code is 18 Kpag, but for vacuum design it is 0 kpag. so case-I is excluded. I don't know what you mean by for vacuum design it is 0 kpag. so case-I is excluded. Case I is not for vacuum design, and 0 kPag is not a vacuum or a pressure.

The roof should be designed for 18 kPag, unless the code allows the actual pressure to be used, as I mentioned in the other thread. The overall plant design must ensure that pressure cannot be exceeded.

Case-II API 650 tank are atmospheric tanks so internal pressure is 101.325 kpa.

It's a nominally atmospheric tank but that doesn't mean the internal pressure is exactly 101.325 kPa. The whole point is that you may need to design for a negative pressure condition. As it's atmosphere outside the roof that means 0.25 to 6.9 kPa vacuum inside. The differential is all you need to know to design the roof.

The 18 kPag limit is the max figure for tank roof design when it is under positive internal pressure.

External pressure limit of 0.25 to 6.9 kPa is for negative pressure conditions, could say internal pressure minus 0.25 to minus 6.9 kPag. Limits are lower than for positive pressure because negative is a more onerous design condition.

I don't know the significance of the 0.25 kPa, I'd have thought maximum design vacuum 6.9 kPa (with no minimum) would cover it. If the negative pressure were 0.1 kPa, does the code mean different criteria apply? Perhaps somebody can explain.

I think this probably deals with the size difference of the tank in question...

Not quite with you. Can you elaborate?

The larger the tank the more sq footage, hence the less psi loading allowed....but I'm just guessing...

OK thanks