Has anyone done any calculations to show what happens when an atmospheric storage tank is drained without venting? We have had several issues with this lately. One involved steam and we are sure that was related to the vacuum created when the steam condensed. The other was when a tank was drained without proper venting. The tank is about 86 ft tall, 12 ft diameter and was filled about 90% full with water. For some reason the vent was covered then a 2" valve was opened at the bottom to drain the tank. I am sure that the tank will drain some amount because of the head pressure, but it would seem it would stop draining when the pressure drops to some amount and an equilibrium us reached. Any ideas as to how to analyze this event? The tank did deform at the top as it was drained so I am sure it did reach some level of vacuum. It is made of 1/4" CS plate. Thanks for any information you can provide.
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Storage tank imploded while draining
- Thread starter krb
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Both scenarios you describe are very common. Think about it. Twelve feet in diameter is 16,286 square inches. If you apply 1 lbf/in^2 to that many square inches it is 16,286 lbf. Atmospheric pressure is generally 12-15 psia, so if you tried to pull a 20 inch vacuum (-10 psig) you would apply 162,000 lbf to the shell, it is probably going to distort a bit.
The calculation is really very simple. PV=nRT. For a constant mass in the head space at a constant temperature then P2=P1V1/V2. So if you increase the head space 10% (i.e., V2=1.1V1) then if your starting pressure was 14 psia, the ending pressure is 12.7 psia if the vessel stayed intact with 21,000 lbf dP.
The condensing steam calc is the same thing, just way bigger numbers way faster.
David Simpson, PE
MuleShoe Engineering
"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
The calculation is really very simple. PV=nRT. For a constant mass in the head space at a constant temperature then P2=P1V1/V2. So if you increase the head space 10% (i.e., V2=1.1V1) then if your starting pressure was 14 psia, the ending pressure is 12.7 psia if the vessel stayed intact with 21,000 lbf dP.
The condensing steam calc is the same thing, just way bigger numbers way faster.
David Simpson, PE
MuleShoe Engineering
"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
Compositepro
Chemical
It is astounding how similar this tank is to a manometer with one end closed. ![[wink]](/data/assets/smilies/wink.gif "[wink] [wink]")
A tank as large as this one would probably collapse under a vacuum of less than on psi. Given your circumstances it is quite possible to create close to a full vacuum (greater than 14 psi).
A tank as large as this one would probably collapse under a vacuum of less than on psi. Given your circumstances it is quite possible to create close to a full vacuum (greater than 14 psi).
- Thread starter
- #4
Thanks for the comments. Zdas04, I considered this approach but again, I am not sure how much the vessel would drain. I have been looking at how Bernoulli's equation applies, but not sure how to deal with the tank not being vented. In other words, how can I calculate V2? Also, how will temperature (from day to night) impact V2?
Muleshoe, good comment. We know the tank will implode, we have the evidence. I want to understand the mathmatical model behind it.
Thanks again.
Muleshoe, good comment. We know the tank will implode, we have the evidence. I want to understand the mathmatical model behind it.
Thanks again.
Why does everyone want to use Bernoulli for every single fluid mechanics problem in the universe? There is nothing about this that satisfies any of the conditions for that equation.
Do SOME arithmetic yourself. You have an 86 ft tall tank that is 90% full so your hydrostatic gradient is 77 ft of water or 33.2 psi at the outlet nozzle. If you lower the level to 68 ft, you have a vacuum above the fluid level and the pressure at the outlet is still 29 psig (less whatever the vacuum above the fluid is, but it is less than 29 psi) and water is leaving if the valve is open.
David Simpson, PE
MuleShoe Engineering
"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
Do SOME arithmetic yourself. You have an 86 ft tall tank that is 90% full so your hydrostatic gradient is 77 ft of water or 33.2 psi at the outlet nozzle. If you lower the level to 68 ft, you have a vacuum above the fluid level and the pressure at the outlet is still 29 psig (less whatever the vacuum above the fluid is, but it is less than 29 psi) and water is leaving if the valve is open.
David Simpson, PE
MuleShoe Engineering
"Belief" is the acceptance of an hypotheses in the absence of data.
"Prejudice" is having an opinion not supported by the preponderance of the data.
"Knowledge" is only found through the accumulation and analysis of data.
KRB, you said that initially the tank is 86 feet tall and was about 90% full. Let's assume for now that it had a flat top. The calculations are similar for a cone roof tank but you include the volume of the cone roof tank above the top of the shell wall.
If the level of water is drained from 90% to 80%, you have doubled the vapor volume and therefore the pressure in the tank will be 1/2 of the initial pressure or about 1/2 atmospheric (P1*V1 = P2*V2). You won't even drain 10% of the level in this case before the tank collapses under vacuum. The starting level is important. If the tank level is 99%, the amount of water you can drain is much less than say if it's 10%.
You can use the relationship P1/T1 = P2/T2 to see how the internal pressure will change with temperature changes from day to night. A 50F temperature change would equal about a 10% change in pressure.
If the level of water is drained from 90% to 80%, you have doubled the vapor volume and therefore the pressure in the tank will be 1/2 of the initial pressure or about 1/2 atmospheric (P1*V1 = P2*V2). You won't even drain 10% of the level in this case before the tank collapses under vacuum. The starting level is important. If the tank level is 99%, the amount of water you can drain is much less than say if it's 10%.
You can use the relationship P1/T1 = P2/T2 to see how the internal pressure will change with temperature changes from day to night. A 50F temperature change would equal about a 10% change in pressure.
Sadly this sort of tank failure is all too common. I (and I am sure many others) have seen it happen under both the scenarios you describe (draining a tall vessel and condensing steam).
A similar, but less frequent scenario in my experience, is absorption of organic vapors in water. In the storage of volatile organics the vessels are often kept under pressure to hold the contents in the liquid phase. In the case I am aware of, the contents had been drained but of course the vessel was still full of organic vapor at atmospheric pressure. Water was then pumped in to purge the vessel before entry for maintenance. This particular substance is very soluble in water and it behaved just like condensing steam. Although the vessel was designed for about 6 bar internal pressure it was not designed for vacuum and it collapsed like a squashed beer can in seconds.
Katmar Software - Uconeer 3.0
"An undefined problem has an infinite number of solutions"
A similar, but less frequent scenario in my experience, is absorption of organic vapors in water. In the storage of volatile organics the vessels are often kept under pressure to hold the contents in the liquid phase. In the case I am aware of, the contents had been drained but of course the vessel was still full of organic vapor at atmospheric pressure. Water was then pumped in to purge the vessel before entry for maintenance. This particular substance is very soluble in water and it behaved just like condensing steam. Although the vessel was designed for about 6 bar internal pressure it was not designed for vacuum and it collapsed like a squashed beer can in seconds.
Katmar Software - Uconeer 3.0
"An undefined problem has an infinite number of solutions"
what is the minimum design pressure of the tank? perhaps -0.25 psig or -10" water column?
install a protectoseal or other reputable conversation breather vent on it. or even just an open goose-neck vent pipe (with no valve).
all you need to know is how fast will the water drain (do the math with the number of equivalent length of piping out the bottom and the head pressure when the tank is full). then you need to look at the vapor flow in the vent piping or CBV and the dP there.
install a protectoseal or other reputable conversation breather vent on it. or even just an open goose-neck vent pipe (with no valve).
all you need to know is how fast will the water drain (do the math with the number of equivalent length of piping out the bottom and the head pressure when the tank is full). then you need to look at the vapor flow in the vent piping or CBV and the dP there.
moltenmetal
Chemical
Another common one is where the idiots install the protection vent/vacuum relief down low in the structure rather than at the top of the tank to make it easier to maintain...the line fills with 32' of water (or condensate during steam-out) and voila- the tank crumples inward like a Coke can. I had a good paper related to vacuum failures (with great pictures of catastrophic vacuum failures) but lost it years ago and have never found it since. Perhaps it evaporated!
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