We have a tank filled with nitrogen gas at 140 PSI. The tank undergoes depressurization time to time round the clock from 140 to atmospheric through a vent line. This vent line is 24/7 ice-covered. To explain the icing, I conclude that from 140 to atmospheric, the temperature plunges below freezing converting moisture content in surrounding air into ice. What I’m not sure 1) how it goes to negative temperature, and 2) how to calculate venting rate to protect icing? Thanks!
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Why vent-line ice-covered at nitrogen tank depressurization? 2
- Thread starter ghensky
- Start date
When you depressurize a gas, it will tend to expand in order to do that it will steal heat from the surrounds the faster you depressurize the faster this occurs the colder the piping will get, hence the umidity in the air is freezing when this occurs. The defrosting depends on your local conditions and on the time based upon each discharge.
You can do a simple hand calculation to get a initial estimative this rate using the thermodynamic tables for nitrogen and simplified heat flow equations. Of course the icing on the piping will cause some influence.
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1
- #5
LittleInch
Petroleum
What I’m not sure
1) how it goes to negative temperature,
Do some research into "Joule Thomson" cooling. E.g.
and 2) how to calculate venting rate to protect icing? Thanks!
~well a simple calculation will show what your temperature drop is.
Then calculate the amount of heat that your pipe can absorb from the surrounding air when it is above zero C, say 5C for a margin - this will be a small number
Then you can work out what mass flow of gas at your low temperature you can flow ( this will also be a small number)
You normally just insulate the pipe and make sure it can handle the low temperature to avoid ice build up.
Once the ice is formed it takes quite a lot of energy to melt it.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
1) how it goes to negative temperature,
Do some research into "Joule Thomson" cooling. E.g.
and 2) how to calculate venting rate to protect icing? Thanks!
~well a simple calculation will show what your temperature drop is.
Then calculate the amount of heat that your pipe can absorb from the surrounding air when it is above zero C, say 5C for a margin - this will be a small number
Then you can work out what mass flow of gas at your low temperature you can flow ( this will also be a small number)
You normally just insulate the pipe and make sure it can handle the low temperature to avoid ice build up.
Once the ice is formed it takes quite a lot of energy to melt it.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
georgeverghese
Chemical
Cause is isentropic expansion of the contents of the tank. See if you can model this on a process simulator, using the batch depressure utility routine. If you must avoid this icing, see what the model tells you; a finned heat exchanger just upstream of the depressure valve might do the trick.
Um...er...how? If gas in cylinder is already @ ambient temperature, it will pick up no to negligible heat; perhaps better installed down stream of PRV so some delta T exists...and even then, only if required; OP has not stated whether chilled N2 @ atmospheric pressure is deleterious to the process being supplied.
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
- Thread starter
- #9
Thank you guys for your very precise answers. It's a shame that there is no reply option to each answer post.
We are not preventing the icing. I speculate the tank wall is double layered as there is no icing on the tank exterior. Thanks LittleInch for pointing out the Joule-Thompson effect.
We are not preventing the icing. I speculate the tank wall is double layered as there is no icing on the tank exterior. Thanks LittleInch for pointing out the Joule-Thompson effect.
LittleInch
Petroleum
The tank might not need insulation as the time period for this de-pressurisation isn't stated. also temperature drop in practice depends on rate of change of pressure. The rate of change at your de-pressurisation valve is very high, hence the temperature becomes very low.
Your tank on the other hand is much slower to change pressure and hence in reality the tank may become colder, but is able to stay above 0 deg C by ambient heat for the surrounding air and has a much bigger surface area than your pipe. It will probably drip with condensation though.
BTW it's "Thomson" without the extra "p"
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Your tank on the other hand is much slower to change pressure and hence in reality the tank may become colder, but is able to stay above 0 deg C by ambient heat for the surrounding air and has a much bigger surface area than your pipe. It will probably drip with condensation though.
BTW it's "Thomson" without the extra "p"
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
- Thread starter
- #11
Well, my background is mechanical so I'm a bit rusty but trying to control the rate to prevent icing could be a challenge. The easy way is to throttle the valve in the field and see if there's an acceptable throttled valve position where icing doesn't occur. One of the two following outcomes will likely occur:
1) The valve will not have the tight control to limit flow where icing will not occur. It'll either be shut, no icing or flow will occur or you crack the valve open and it immediately begins to ice over.
2) The vent rate that prevents icing will not work for you. At some point you'll have to open the valve enough where icing occurs and get back to square one.
You can what-if this and attempt to calculate it but if you have a working model in the field where throttling isn't an issue it'd be easier to just do some field tests.
If you're intent on calculating it, it may not be that simple but it could be possible to get a rough answer. Additional information would be necessary and there's no simple equation to calculate it. Even if a flow rate is calculated and someone says you need a flow rate of 147 lb/hr or kg/hr, how are you going to throttle the valve to that? Do you have a flow meter?
Thanks,
Ehzin
1) The valve will not have the tight control to limit flow where icing will not occur. It'll either be shut, no icing or flow will occur or you crack the valve open and it immediately begins to ice over.
2) The vent rate that prevents icing will not work for you. At some point you'll have to open the valve enough where icing occurs and get back to square one.
You can what-if this and attempt to calculate it but if you have a working model in the field where throttling isn't an issue it'd be easier to just do some field tests.
If you're intent on calculating it, it may not be that simple but it could be possible to get a rough answer. Additional information would be necessary and there's no simple equation to calculate it. Even if a flow rate is calculated and someone says you need a flow rate of 147 lb/hr or kg/hr, how are you going to throttle the valve to that? Do you have a flow meter?
Thanks,
Ehzin
georgeverghese
Chemical
Adiabatic isentropic expansion in the tank; adiabatic isenthalpic for the JT letdown at the valve. If you are using a process simulator, there should be an option to model this minimal heat leak into the tank from the atm.
LittleInch
Petroleum
Try this page and scroll down to pressure drop
Interestingly, the calculation seems to show that for your stated pressure ( about 10 barg) venting down to 0 barg, you should only be dropping 2-3 degrees C
So either your gas is already very close to freezing temp or your pressure is a lot higher.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Interestingly, the calculation seems to show that for your stated pressure ( about 10 barg) venting down to 0 barg, you should only be dropping 2-3 degrees C
So either your gas is already very close to freezing temp or your pressure is a lot higher.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
LittleInch
Petroleum
The OP says gas at 140 psi.
There's clearly something not right here.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
There's clearly something not right here.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
georgeverghese
Chemical
@LI,
Initially, what we have is isenthalpic dp across the JT valve, so the vent nozzle and exit pipe is still warm. In subsequent time steps, there is also isentropic (adiabatic) expansion of the tank content. This then leads to lower and lower vent nozzle temps, since the isentropic dT in the tank is greater than isenthalpic dT at the valve. The tank wall appears to be be warm ( even though the internal content is cold), mainly because of poor heat transfer at the tank inside wall, which is due to zero velocity of the tank content. The thermal inertia of the relatively thicker tank wall also adds to this masking effect. Gas temp at the valve is better reflected at the JT valve due to the much higher velocity / higher htc / thinner pipe wall here.
Gas expansion in the tank is adiabiatic isentropic since there is no corresponding kinetic energy gain for the VdP work done ( there is no KE gain, gas velocity remains zero). In contrast, VdP work done by the gas at the JT valve is almost exactly compensated by the KE gain ( lower pressure ). Hence this expansion is adiabatic isenthalpic at the valve. These effects are a direct observation one can derive from first principles when you look at the Bernoulli equation construct for these 2 locations.
Initially, what we have is isenthalpic dp across the JT valve, so the vent nozzle and exit pipe is still warm. In subsequent time steps, there is also isentropic (adiabatic) expansion of the tank content. This then leads to lower and lower vent nozzle temps, since the isentropic dT in the tank is greater than isenthalpic dT at the valve. The tank wall appears to be be warm ( even though the internal content is cold), mainly because of poor heat transfer at the tank inside wall, which is due to zero velocity of the tank content. The thermal inertia of the relatively thicker tank wall also adds to this masking effect. Gas temp at the valve is better reflected at the JT valve due to the much higher velocity / higher htc / thinner pipe wall here.
Gas expansion in the tank is adiabiatic isentropic since there is no corresponding kinetic energy gain for the VdP work done ( there is no KE gain, gas velocity remains zero). In contrast, VdP work done by the gas at the JT valve is almost exactly compensated by the KE gain ( lower pressure ). Hence this expansion is adiabatic isenthalpic at the valve. These effects are a direct observation one can derive from first principles when you look at the Bernoulli equation construct for these 2 locations.
Ghensky,
You mention speculation regarding the tank being double jacketed. If this is a facility in which you work that has control over this equipment can this not be confirmed? And what is the actual internal temperature of the equipment. Georgeverghese mentions very little temperature drop assuming the tank internal temperature is close to ambient. If the nitrogen volume is ambient then you should have no significant drop in temperature across the valve. Ignore what the outside of the tank is showing you and dig for more information regarding what's going on inside the tank.
Thanks,
Ehzin
You mention speculation regarding the tank being double jacketed. If this is a facility in which you work that has control over this equipment can this not be confirmed? And what is the actual internal temperature of the equipment. Georgeverghese mentions very little temperature drop assuming the tank internal temperature is close to ambient. If the nitrogen volume is ambient then you should have no significant drop in temperature across the valve. Ignore what the outside of the tank is showing you and dig for more information regarding what's going on inside the tank.
Thanks,
Ehzin
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1
- #19
georgeverghese
Chemical
If we take a numerical example, we round off to say initial conditions at 1000kpa abs and 300degK in the tank (near enough to OP's starting conditions).
P. v^k = constant for adiabatic isentropic expansion in the tank. This can be translated to (T^k )/(P^(k-1)) = const.
Hence T2^k = [T1^k/P1^(k-1)].(P2^(k-1)). If gamma = 1.4 for nitrogen, we assume for an irreversible isentropic depressurisation process that k = 1.3 say (k would be less than gamma for irreversible process).
Partway through depressurisation, when tank press P2=500kpa say, we find tank temp drops to 256degK = -17.3degC. The corresponding JT (isenthalpic) drop from 500kpa, 256K to atm at the valve would be much smaller (you can find out how much this is from N2 thermo data if you wish), hence exit temp would be at least -20degC or so.
P. v^k = constant for adiabatic isentropic expansion in the tank. This can be translated to (T^k )/(P^(k-1)) = const.
Hence T2^k = [T1^k/P1^(k-1)].(P2^(k-1)). If gamma = 1.4 for nitrogen, we assume for an irreversible isentropic depressurisation process that k = 1.3 say (k would be less than gamma for irreversible process).
Partway through depressurisation, when tank press P2=500kpa say, we find tank temp drops to 256degK = -17.3degC. The corresponding JT (isenthalpic) drop from 500kpa, 256K to atm at the valve would be much smaller (you can find out how much this is from N2 thermo data if you wish), hence exit temp would be at least -20degC or so.
ghensky has still not stated whether chilled N2 @ atmospheric pressure is deleterious to the process being supplied...so is there a practical reason for the OP beyond simple curiosity?
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
CR
"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]
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