API 2000 4.6.2.1 Equation 4B

[Dosi]

Hello,

i am a german engineer and my actual challenge is to check a design for a storage tank in Dubai.
The designer of the tank uses equation 4B of API 2000 to calculate the required minimum flow area of device.
Nm³/h = 12503 P1 A sqrt{ K / (MTZ*(K-1)) *[(P2/P1)^(2/k) – (P2/P1)^((k+1)/k)]}

=> A = Nm³/h / [12503 P1 sqrt{...}]

My problem is, that if the difference between P2 (pressure at device outlet) and P1 (pressure at deive inlet) ist very small, their ratio becomes close to 1. That means that the term [(P2/P1)^(2/k) – (P2/P1)^((k+1)/k)] is getting very small. Ergo the whole denominator becomes very small which finally means, that the required flow Area A is getting bigger.
That makes absolutely no sense to me at all.
If there is even no difference between the two Pressures P1 and P2, their ratio becomes 1 - the [] term becomes 0 and so the whole equation is insolveable.

So my question is: Am I too stupid to understand the equation correct or has the designer chosen a wrong way to calculate the required venting area?
And if it's the designers fault, what is the right way of calculating the required area?

It's my first time working with any API, so please be patient with me and excuse my non-technikal english.

Thank you for your help.

Regards
Lars

 

[IFRs]

If delta P is small then the area must increase for a given flow. If there is no delta P, by definition there is no flow. This makes sense to me. What am I missing?

 

[Dosi]

Well, propably you're not missing anything and I or better the tank designer uses the wrong equation.

The designer uses that equation in connection with outbreathing and inbreathing requirements. I understood him in that way, that he calculates an area A that is the required area for ventilation.
"how big must my ventilation valve be, if the tank 'breathes in'?" (or 'out')
He first uses clause 4.3.2.1 and 4.3.2.2 to determine the required venting capacity for inbreathing and outbreathing (thermal and liquid movement)and takes that as "actual flow". Than he calculates the maximum design pressure a.p.C. F.4.1 of API 650, and considers that Value P as "delta P" for outbreathing. P2 is athmospheric pressure. So P1 is P2 + delta P (maximum design pressure).
With those pressures and the above equlation he finally calculates the diameter of a required Vent.

I attached this chapter of his calculation.

Thank you so much for helping me. And again - this whole problem is probably very simple for you all, but I had no experience with any API until now and it's a foreign language for me. So please excuse...

 

[IFRs]

What you have done looks correct. But, you have only checked normal in and out breathing. You may also want to consider the following, if any apply.

Fire Exposure (emergency venting)
Circumstances Resulting from Equipment Failures and Operating Errors
Pressure Transfer Blowoff
Inert Pads and Purges
External Heat Transfer Devices
Internal Heat Transfer Devices
Vent Treatment Systems
Utility Failure
Change in Temperature of the Input Stream to a Tank
Chemical Reactions
Control Valve Failure
Steam Out
Uninsulated Tanks with Exceptionally Hot Vapor Spaces

 

[Dosi]

But somehow it cannot be correct.
The Designer determined Inbreathing with P1 only a very little smaller than P2 => The required A gets infenitly big...?

 

[IFRs]

It takes pressure to create flow. If the area of the opening is small it takes more pressure to create the same flow. It's like the difference between yawning and whistling. When your mouth is open wide it takes little effort to expel a lot of air fast. When your lips are closed you have to blow hard to get the same flow.

 

[Dosi]

Thanks for your patience

So what you're telling me is that with equation 4B (or 4A) it's possible to find out how big the flow on a given Area (and Pressure) can be.
=> with my mouth open that wide and a certain pressure that much air will flow, right?

So what I need to know then, is how to determine the correct delta P for inbreathing. Is it also the maximum design pressure?

The designer chosed delta P ~ 0. So the equation doesn't work, correct?

 

[IFRs]

Each tank code includes basic design pressure and vacuum limitations. API 650 tanks are limited to 1 inch of water as an external pressure (vacuum) unless Appendix V is used. They are limited to 2.5 psi internal pressure unless Appendix F is used. Other codes use other values. A look at the tank name plate should clarify what the tank was designed for.

 

[Dosi]

The Design Data says:

Design Pressure : Full liquid
Design Vacuum: ATM

Does that help?

 

[IFRs]

If this is an API 650 tank, designed to a recent version of the standard and the design data does not specify that any Appendix was used for design, then the maximum design vacuum is 1 inch of water and the maximum design pressure is 2 1/2 psi. I would, if I were you, limit the pressure to the weight of the roof plates rather than 2 1/2 psi so you don't get any fluttering of the plates. We are reaching the limits of what the internet can do for you. You should NOT take what we do here and run with it assuming what I have said applies to your tank. Each tank is different, each operation is different. Be careful with taking what I say here as the whole picture. It is not. I do not know enough about your tank and it's operation. But, you have a good start!

 

[Dosi]

Thank you for your support.

You got me a little more into this whole subject.

 

[Duwe6]

I go a little farther than IFR does on my 650 pressure calc's. The object is to not allow an empty tank to 'lift off' by overpressuring the flat floor plates and causing the tank to rise.

I take the weight of the roof plates [roof is sloped so area is a cone, not a disk]
plus the weight of the shell plates
plus the weight of appurtanances and nozzles [reinforced manways are rather heavy]

Divide the total of the above by the circular area of the roof or floor - disk. Being a Yank, my preferred units for working on tanks are psi and square feet. Divide by 144 to convert pounds/ft EXP2 to psig. Use that for your MAWP.

I've *never* seen an allowable of 2.5 psig that would not bulge out the bottom of an API-650 tank without a LOT of liquid in the tank. Also, 2.5 psig should be about, or greater than, the pressure that will blow a Frangible-Seam Roof off of the tank shell. Frangible Roofs are supposed to blow off prior to allowing overpressure to cause a catastrophic failure of the floor, shell, or floor-to-shell joint.

 

[IFRs]

Duwe6 is right. Note that I said I would limit the internal pressure to the weight of the roof plates. Then I don't have to worry about anchoring the tank or uplift or frangibility, etc.

 

[JStephen]

The 2.5 psig is the maximum pressure that an API-650 tank CAN be designed for. Very few API-650 tanks are designed for that value, and it should never be assumed that an API-650 tank was intended for that pressure.

If you have all the tank details (construction drawings), you can evaluate the maximum overpressure that the tank is adequate for. Note that API-650 has been revised several times, and the answer you get will depend on which version you use. See Section 5.8.5 in the current revision.