API 650 Tanks Maximum Pressure as per Appendix F 2

[richissa]

Hello everyone,

I am trying to increase to Maximum allowable working pressure provided in Appendix F.4.1.

F.4.1 says that the maximum design pressure for a tank that has been constructed can be calculated using the following formula but subject to the limitations of Pmax in F.4.2

F.4.2 says the maximum design pressure, limited by uplift at the base of the shell, shall not exceed the value calculated from the following equation unless further limited by F.4.3:


F.4.3 says that In order to provide a safe margin between the maximum operation pressure and the calculated failure pressure, a suggested further limitation on the maximum design pressure for tanks is Pmax less or equal to 0.8* Pf (failure pressure)

My question is that can I use Pmax as per F.4.1 instead of Pmax provided in F.4.2 if Pmax less or equal to 0.8* Pf?

Thanks in advance!

 

[JStephen]

I'm not sure I understand the question, but I think the intent is to use the lower of the three values calculated there.

If you're re-rating existing tanks, be aware that the pressre/wind/overturning criteria have been revised considerably in the last several issues, so make sure whether the re-rate is per current code, code of construction, etc.

 

[richissa]

Stephen,

Thanks for the quick reply and I hope this post makes sense. I am trying to increase MAWP to use the higher value.

Per my calcs:

Pmax per F.4.1 is 5.58 psi
Pmax per F.4.2 is 0.539 psi

In this case Pmax F.4.1 (which is higher) is subjected to the limitation of Pmax F.4.2.

However F.4.3 allows to go for a higher pressure if Pmax is less or equal to (0.8* Pf) (where Pf is failure pressure).

My question is can I discard the limitation of Pmax per F.4.2 and go with Pmax per F.4.1 if Pmax is less or equal to (0.8* Pf) as mentioned in F.4.3

Thanks again!

 

[JStephen]

I don't see that F.4.3 "allows to go for a higher pressure", it is another limitation.
I'd say, you figure:
Pmax is 5.58 psi per F.4.1
Pmax is 0.539 psi per F.4.2.
You calculate your failure pressure, just guessing, say it's 9 psi.
Then your maximum design pressure is 0.539 psi, but also is recommended not to exceed 0.8*9 psi = 7.2 psi, which it does not. So it's 0.539 psi.

Note that there is a clause there somewhere that states for certain venting purposes, you take M = 0 in evaluating pressure per F.4.2.

 

[mariog123]

The failure pressure Pf is related to roof failure only and may be expressed in SI units as:
Pf=8*Fy*A*tanθ/D^2+4/PI*DLR/D^2 - units in [Pa]
(where A is the area subject to stress Fy)
or
Pf=Fy*A*tanθ/(125*D^2)+0.0012732*DLR/D^2 in [kPa]

F.4.1 defines "the maximum design pressure" considering a safety coefficient of 1.6 applied to the first term of Pf:
P=Fy*A*tanθ/(1.6*125*D^2)+0.0012732*DLR/D^2=
=Fy*A*tanθ/(200*D^2)+0.00127*DLR/D^2

F.4.2 is a math derivation equivalent with 5.11.2.1 point 1, i.e.
1) 0.6Mw + MPmax = MDLS/1.5 + MDLR
and has nothing related to the failure of roof

F.4.3 is a limit of Pmax calculated according to F.4.2, for frangible tanks, and is Pmax<0.8Pf
I would interpret is as "I don't want to see my roof destroyed by a combination of Pmax and wind, I prefer to have roof failure only at Pf pressure, in case of fire!"

Now about your question:
"My question is that can I use Pmax as per F.4.1 instead of Pmax provided in F.4.2 if Pmax less or equal to 0.8* Pf?"

You say that Pmax calculated as per F.4.1 is less than 0.8* Pf- this is OK for roof and this result is expected, because by definition, F.4.1 takes care of the roof.
However adopting as Pmax only the value given by F.4.1 (as a figure greater than those calculated by F.4.2), it means you are not complying with basic API, 5.11.2.1 point 1, which is an overall overturning conditions.

 

[Samm17]

In short if your tank is anchored (or you add anchors) and your anchors can hold your new design pressure you can ignore F4.2 as it is a check for anchoring as mariog was alluding to.

 

[michek]

Hi,
I have a question about calculating the maximum design pressure for unanchored tankse a question on Annex F. The roof tank is supported by column.
In F.4.1 Dlr equals to the amount of the nominal weight of roof plate and rafters supported by the roof-to-shell junction. The other one being supported by columns.
Am i right?

And i also have to check roof-to-shell junction with forces from roof struture.

Best regards

 

[Duwe6]

Just finished an Appendix F evaluation - phew!

michek, if your roof rafters are 'loose' and not welded to the roof, D[sub]LR[/sub] will be the weight of your roof and appurtanences. But not the unattached rafters and centerpole. Be sure to take credit for any platforms, valves, piping, and purge equipment supported by your roof. What the formulae is 'looking for' is the total weight on the roof available to keep the roof from being lifted up by the internal pressure.

D[sub]LS[/sub] is the total shell weight that is holding the roof down, so it won't rise. [and that will not include the centerpole or loose rafters either]. Finally, the wind moment is to include the possible overturning force that will cause the tank to rotate [tip over] and thus cause the tank to rise and remove hold-down force from the flat floor.

 

[michek]

Thanks Duwe6 for your prompt reply.

Best regards.

 

[JStephen]

I think there are still some discrepancies in API-650. If the internal pressure is greater than the weight of the roof plate, then theoretically, the roof plate would lift off the rafters slightly, and all the roof plate could be considered in resisting overturn. If the internal pressure is less than the weight of the roof plates, that is not the case. But, that distinction is not really made in API.

 

[JStephen]

I think there are still some discrepancies in API-650. If the internal pressure is greater than the weight of the roof plate, then theoretically, the roof plate would lift off the rafters slightly, and all the roof plate could be considered in resisting overturn. If the internal pressure is less than the weight of the roof plates, that is not the case. But, that distinction is not really made in API.

 

[mariog123]

JStephen, I do not second the idea API would implement a difference in approach.
Of course you are right saying a small internal pressure will appear as diminishing the weight of the roof, without producing uplift. However in this case it will change also the force transmitted to the shell (by junction shell-roof).
A bigger internal pressure shall lift-off the plates, but in the end the effect will be also transmitted to shell.
I think that for the discussion is it more important the force balance applied to the shell than the fact plates lift-off or not.

 

[JStephen]

Consider in the formulas in Table 5.21b, for example. For seismic alone, the dead weight term is W2 which includes "any dead load including roof plate acting on the shell". So in that case, only dead weight supported by the shell is considered to resist uplift.
Then in combined pressure + seismic, the roof plate weight is incorporated into the -8th term, which implies all of the roof plate is resisting uplift.
You'd think as pressure was reduced to zero, these two equations would become the same, with possibly a change in allowable stress. But they don't.

 

[mariog123]

I think I don't understand.
For me, "any dead load including roof plate acting on the shell" means I already have there the weight of plates roof i.e. the plates weight is included in W2.

As you says, when in Table 5.21b they considered terms with pressure, they substract "8th" in pressure and use -W1 or -W3 which are "...any dead load other than roof plate acting on the shell...". I understand they preferred to consider the roof plates weight as an equivalent pressure, instead to include the roof plates weight as included in W term. In the end it is rather a math trick.

In the particular case of "seismic" this math trick would be applied, but nobody will understand why there is a term as "– 8th × D^2 × 4.08". I understand they accepted here to include plates of thickness of th in W2 calculation.

In the end, I understand the intention was to consider "– 8th × D^2 × 4.08- W1" as an approximation for "-W2". It may be not exact, but I guess API people thought the two written are equivalent.

 

[JStephen]

The difference is when the roof is structurally supported. The roof plate acting on the shell is then only a fraction of the total roof plate- roughly 2/3 for a single center column, less for a multi-column roof.

Note that the Fp*P - 8th term can be negative, which is including roof weight supported by the columns in the deadweight calculation for holding down the shell.

 

[mariog123]

It is evident that API 650 does not cover all cases; I think the best engineering approach is to perform a "free-body analyses" made at the level of shell-roof joint.
We can separate the tank model in two parts and introduce a pair of vertical internal forces keeping them together.
For the upper part, we can write-down an equilibrium equation involving the connecting force, weighs (plates roof, rafters, etc)- corroded or nominal, as API considers, and the pressure force. From calculation it results the internal (connecting) force. In case there are multiple supports, that subsystem is not statically determinate, can be solved by various methods.

For the shell part, we just need to consider in top that internal force and to continue the analysis.