Pressure relieving devices/overflow slots in an API 650 tank

[mmikola]

Could you give me direction on where to go for sizing the venting and overflow slots on an API 650 process tank used in ethanol production? I used API 2000 to find the inbreathing, outbreathing and emergency venting requirements and I know the pump-in of the tank so I just need to know how to determine the actual net area of opening for overflow and venting in order to satisfy the requirements. It stated in the API 650 that the slots are to be sized to discharge at the tank pump-in rates. I was trying to relate the flowrate to the Bernoulli Equation to find a velocity of the fluid equivalent to the pressure head created by a liquid level to the top of the slot and then find an area using the continuity equation (Q=VA), however, the area I calculated was very small. For the tanks with an internal floating roof I used API 650, Appendix H which was pretty straight foreward.

 

[JStephen]

Vents are sized per Appendix H, which gives the open area based on circumference.

For overflow slots, check orifice flow as it sounds like you are doing; also check flow over a sharp weir if applicable. And yes, the required overflow slot may be fairly small for low pumping rates.

 

[mmikola]

Yes, however, Appendix H can only be used for a tank with a fixed roof and a floating internal roof. These process tanks do not have a floating roof and need to be sized per API 2000 venting requirements. So...I should used the Bernoulli Equation for this then once I determine the in and out-breathing and emergency venting flowrates for the air flow then a separate Bernoulli for the fluid pump-in rates? Also for the Bernoulli, when designing strictly for the air venting, do I just set the design pressure inside the tank at initial state and ignore any air movement in the tank? At the same time, when designing the overflow of fluid, do I ignore the internal design pressure in the tank and just used the pressure head created? If I use the lower pressure, the velocity is smaller and the area of the slot becomes larger (A=Q/V).