Vacuum rating of Tanks

[Mech85]

I have recently received a “short cut” vacuum rating calculation for tanks and vessels that states that the maximum vacuum that would exist in a gravity drain tank before air would enter is given by

Vacuum (inches Hg) = 2.036*H*rho*(g/gc)/Po

Where
H = total tank vertical height in ft
rho = liquid density in lbm/cubic ft
Po = atmospheric pressure in psi abs or lbf/ sq.in
g = gravity acceleration in ft/sec squared
gc = conversion factoring ft lbm/ft lbf ie g/gc =1

Does anyone know the derivation OR source of this calculation? Is it correct as written?

 

[JStephen]

I can't see that it answers the problem you've stated.

Suppose you have a tank 100' high, completely filled with water to the top, with no vent. You lower the level by 1'. That leaves 1' of void space, which would actually be occupied by low pressure water vapor from boiloff of the contents. So the absolute pressure would be the saturated vapor pressure from the steam tables at that temperature, and the vacuum would be close to the atmospheric pressure. It would be unrelated to variations in height of the tank, the density, or gravitational acceleration in that case.

In practical terms, tanks that have heights measured in feet also invariably have vents or P/V valves to prevent this vacuum. On an "atmospheric" tank, vacuum is typically limited to 5 PSF or so.

 

[rmw]

JCStephen,

What happens to the pressure in the example you describe if you keep lowering the level by evacuating at the bottom until the level is lowered by the amount of 34 ft?

Assume no heat transfer for the purposes of discussing this example.

rmw

 

[EGT01]

Perth1,

As your equation attempts to show, the potential to create a vacuum by gravity drain of a liquid is a function of the the difference in elevation of the top of your liquid level and the point of discharge to atmosphere and the liquid density.

For example, if you have a tank located at an elevation of 34 ft above grade and you have a gravity drain from the tank down to grade, then for the liquid being water you could impose full vacuum on the tank if you don't let air in when draining.

See if this helps as an explanation....

http://www.fluidedesign.com/download-free/unusual_aspects-pumps-syst.pdf

As for your equation, it almost looks like someone was trying to convert feet of liquid to psi then to inches of Hg. But I get a constant that is a factor of 10 different so I'm not sure what their equation is trying to do.

Recognizing that g/gc = 1

inHg vac. = ft liqiuid * lb/ft3 liquid * 1 ft2/144 in2 * 29.92 inHg/14.7 psi

Rearranging the constants 29.92 inHg/(144 in2/1 ft2) = 0.2078

inHg vac. = 0.2078 * ft liquid * lb/ft3 liquid / 14.7 psi

 

[JStephen]

I don't see how the Po enters into it the equation, unless it is part of the conversion from psi to mm mercury.

Anyway, in the example I quoted, if you drop the liquid level by another 34', you just get more low-pressure water vapor, at essentially the same pressure/ vacuum as before.

Note that in real life, there is normally air above the water in a tank. So if you take a real water tank, block off the vent(s) and drop the water level, it will create a partial vacuum, but not as dramatic as in the example. The actual pressure or vacuum would depend on the volume of the air space.

Anyway, rather than discuss that equation, it might help if you just indicated exactly what you're trying to check. If this relates to a real tank, what kind of tank is it? How high? What kind of circumstance is developing a vacuum?

Most real-world tanks of any size aren't built to withstand anything near a full vacuum. So if you are wondering what happens to your water tower if you block off the vent- well, it's a goner in that case.

 

[robsalv]

JStephen said: "Most real-world tanks of any size aren't built to withstand anything near a full vacuum. So if you are wondering what happens to your water tower if you block off the vent- well, it's a goner in that case."

There are many examples around the world of tanks having their roof sucked in when draining the test water... we have one here where a plastic shopping bag got sucked into the vent when draining the tank. That was enough of a blockage to suck the roof down. Very embarrasing.

If your vent is smaller than your drain, then there's a more than fair chance you'll suck the roof of your tank down too.

Tanks can't be designed to sustain vaccuum. When you consider there are some tanks you can blow the roof off with lung pressure... it's not hard to see why. Pressure Vessels can resist partial or full vaccuum though, usually by extra thickness or stiffening rings.

I personally don't understand what the formula was trying to get at. If the liquid level dropped and the tank/vessel was open to atmosphere via a vent, atmosphere would immediately rush in to replace the missing volume. Its the rate of replacement that would control the ultimate vaccuum pressure.

My suggestion would be DO NOT rig up a tank to be exposed to any kind of vaccuum.

Cheers

Rob