Calculating possible vacuum pressure
Calculating possible vacuum pressure
(OP)
Hello,
I am curious about determining how much vacuum a vessel can see during draining.
My case would be a vessel 90% full of liquid being drained with nothing else open. Is it as simple as P1*V1=P2*V2, where I would solve for P2 when I have 100% empty vessel?
Does the rate of draining impact the amount of vacuum developed? It seems that if it took 5 days to drain this vessel, developing a vacuum would be more difficult than if it was drained in 5 minutes, but perhaps this logic is incorrect?
Thanks again,
I am curious about determining how much vacuum a vessel can see during draining.
My case would be a vessel 90% full of liquid being drained with nothing else open. Is it as simple as P1*V1=P2*V2, where I would solve for P2 when I have 100% empty vessel?
Does the rate of draining impact the amount of vacuum developed? It seems that if it took 5 days to drain this vessel, developing a vacuum would be more difficult than if it was drained in 5 minutes, but perhaps this logic is incorrect?
Thanks again,





RE: Calculating possible vacuum pressure
Mark Hutton
RE: Calculating possible vacuum pressure
When you say "minus the static pressure" are you referring to the surrounding static pressure? i.e. -14.7 psia if I am at atmosphere? Or static pressure in the vessel from liquid head?
RE: Calculating possible vacuum pressure
Regards
StoneCold
RE: Calculating possible vacuum pressure
I apoligize for my ignorance. What exactly would this energy balance look like? Since I am draining, it seems that my mass will go to zero, and I will not be able to close the enthalpy balance on water.
Any help is appreciated.
RE: Calculating possible vacuum pressure
Mark Hutton
RE: Calculating possible vacuum pressure
Start like this. What is the fluid? What is the operating temperature? Based on that you can find the vapor pressure.
Assuming the liquid volume is fairly large the flashing vapor will have little effect on the bulk temperature so for this first calculation we will ignore that problem. So the pressure in the tank will fall to the vapor pressure of the liquid as the liquid is pumped out of the bottom.
If the liquid drains out in 5 minutes or 5 days, if there is no leaks the answer is very close to the same pressure.
If it drained out instantaneously that would be a non equilibrium problem that would be more difficult to solve but the answer would be between the vapor pressure of the liquid and 0 psia.
Hope this answers your question.
Regards
StoneCold
RE: Calculating possible vacuum pressure
Perhaps I can offer a few comments on your question and the responses you have received. Consider a vessel:
Tank volume= x cuft; 90% full (outage = 10% filled with air)
Ambient conditions: P = 14.7 psia; T = 530 R.
Tank drained (gravity or pump); no air ingress
From the gas law: P(2) = P(1)*V(1)/V(2). (volume is headspace)
From volume x: V(1) = 0.1x ; V(2) = x (empty)
Temperature at empty same as initial temperature.
P(2) = 14.7 * 0.1x/x = 14.7 * 0.1 = 1.47 psia
Depending on tank temperature and liquid stored there may be some amount of flashing which will make the final pressure higher than the 1.47 psia.
Tank pressure drops as headspace increases (level drops). Rate of drain only affects the rate at which the pressure drops. If drain is by gravity, then drain rate is given by the efflux equation. Drain rate by gravity is a function of the head of liquid in the tank. If the tank is an atmospheric tank, its mechanical integrity will be affected by the external pressure as the tank drains. Such a tank will buckle, collapse, and perhaps rupture if it is not rated for external pressure.
RE: Calculating possible vacuum pressure
Perhaps I can offer a few comments on your question and the responses you have received. Consider a vessel:
Tank volume= x cuft; 90% full (outage = 10% filled with air) Ambient conditions: P = 14.7 psia; T = 530 R.
Tank drained (gravity or pump); no air ingress From the gas law: P(2) = P(1)*V(1)/V(2). (volume is headspace)
From volume x: V(1) = 0.1x ; V(2) = x (empty) Temperature at empty same as initial temperature.
P(2) = 14.7 * 0.1x/x = 14.7 * 0.1 = 1.47 psia
Depending on tank temperature and liquid stored there may be some amount of flashing which will make the final pressure higher than the 1.47 psia.
Tank pressure drops as headspace increases (level drops). Rate of drain only affects the rate at which the pressure drops. If drain is by gravity, then drain rate is given by the efflux equation. Drain rate by gravity is a function of the head of liquid in the tank. If the tank is an atmospheric tank, its mechanical integrity will be affected by the external pressure as the tank drains. Such a tank may buckle, collapse, and perhaps rupture if it is not constructed for some external pressure (vacuum).
RE: Calculating possible vacuum pressure
like the above posts say, what is left is the vapor pressure inside the vessel and the pressure of the air at the new expanded volume (assuming you don't consider the air coming into the outlet). At a max with water at 20 deg C youll see about 20 mmHg from the vapor pressure + the final pressure of the 10% air if you have a check valve at the tank outlet.
if this is a tank, id expect this to have PVSV on the top to make sure you didnt pull too much vacuum on the vessel, but assumptions above, you can pull pretty close to complete vacuum.
-Mike