Why vent-line ice-covered at nitrogen tank depressurization? 2

[ghensky]

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!

 

[ghensky]

The internal N pressure is 140 psi for this tank and doing no harm to shell. Neither the frozen vent is of any concern. However, I was just looking for a calculation lately how to determine the vent valve opening rate so Joule Thomson effect is subdued. Thanks for all the good information. I wish each reply had its unique reply button to continue specific subtopic. Thanks George, cr and Ehzin.

 

[georgeverghese]

Would suspect the gas rate at which the JT and isentropic effects would be subdued would be impractically low. You've only got heat transfer from ambient through the short thin walled vent line and the thicker walled tank. Heat transfer calcs for this would moreover be somewhat tedious and far more lengthy than these isentropic / JT chilling effect calcs.

 

[DriveMeNuts]

I hope that ice covered line isn't carbon steel, as it will become brittle at low temperatures.

 

[LittleInch]

George, ~I knew I was missing something but of course, assuming no or very little heat input into the tank the start temperature of the expanding gas inside the tank is the root cause of the low temperatures as it falls as the gas expands.

I'm more used to constant temperatures of gas coming out of pipes, but I've done de-pressurisation runs also where you get low temperatures quite easily.

I agree with George - if there is no heat input into your tank then the flow control through the valve is not worth talking about as the temperature drop is <3 C across the valve. It's the rate of expansion of the gas in your tank which is the issue and lack of heat input into the tank which is causing the icing. It could be that faster is better because there is less time for the ice to form a thick layer before the gas flow ceases.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[georgeverghese]

The P.v^k = constant approach is valid for gases that approach ideal gas behaviour, like N2 well removed from critical conditions. For other non ideal gases like steam at high pressure, CO2 and the like, use the relevant thermo tables / charts for these which also list entropy values and follow the constant entropy path to get T2 for depressurisation processes such as this case.

 

[ghensky]

Thanks again George and Little.

My conclusion for the original problem why nitrogen vent line is getting frozen: The pressurized N is expanding to atmosphere above a critical velocity. The work done by the system is reducing pipe wall temperature, which is freezing the surrounding moisture known as Joule-Thomson effect. If the venting rate could be controlled and kept below the critical velocity, freezing won’t happen. However, this would slow down the process and hence better to live with this as it is not damaging anything.

I’m still to find a way to calculate the critical velocity. But that’s possibly another thread. Thanks everyone once again.

 

[LittleInch]

Ghensky,

I think you misunderstand.

The key issue is the expansion of the gas in the tank with virtually no heat coming in from outside.

There is some JT effect but it's very small compared to the adiabatic depressurisation as outlined by George.

The rate of depressurisation would have to be very slow unless you can put heat into the tank.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[ghensky]

Ok I'm now confused but thanks, Little, for pointing out to the post of George made on Feb 2. I totally missed that.

George, you mentioned two points:

1. Tank is adiabatic isentropic: since there is no corresponding kinetic energy gain for the VdP work done (there is no KE gain, gas velocity remains zero).
2. 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.

For 1, how is the tank adiabatic when energy is added into the tank during pressurization?
For 2, how is the expansion adiabatic isenthalpic? I understand why it is isenthalpic, low pressure is compensated by high volume but because heat is removed, how it is adiabatic?

Thanks for the clarification.

 

[georgeverghese]

@ghensky

1. Adiabatic simply means no external heat input to the system. We say there is a dH loss during depressurisation wrt to the the enthalpy reference basis, where the reference basis (or datum) is the gas which is initially present in the tank at 10bar, 20degC or so.
2. Again the JT letdown is an adiabatic process since there is no external heat input. By approximation we say it is adiabatic - isenthalpic. In reality it is not quite so (strictly speaking), there is actually a small reduction in enthalpy, and this dH is purely due to the VdP term which results in a corresponding KE gain. In most cases, the KE gain is considered negligible and therefore the dH is approximated to be be zero.

 

[ghensky]

Thanks George, now I see what you meant.