4 replies to this topic

[YohanJ]

Hello everyone.

I have a question regarding P/T change across a valve.

Specifically, an pur argon line at room temp and 2700torr is connected to a vacuum chamber.
The chamber is continually pumped to retain about 70torr of pressure.
That is nearly 40times the pressure diff.

I'm curious as to what happens when that valve opens. The expansion must make the gas very cold, but how would you calculate how cold that gets? My initial estimates using isentropic expansion and adiabatic expansion seems to make the gas WAY too cold (near critical point).

I've done some reading and 'choke flow' seems to matter here. Restricted to mach 1, the pressure right after the valve is about 0.48 times of inlet. But how does it actually reach the final chamber pressure... and does the gas get even colder as it keeps going down pressure?


I'd appreciate any thoughts.
Thanks.

[Art Montemayor]

Yohan:

Please refer to my post at: http://www.cheresour...amp;#entry25709

What you are describing has little or nothing to do with the Mach Number or choked flow. It also has nothing to do with Isentropic expansion. You are describing the free expansion of pure Argon from a high pressure (1,326 psig = 2,700 Torr) into a vacuum. Therefore, what will happen is that the Joule-Thomson effect will take place – the adiabatic cooling of a real gas upon free expansion.

Instead of calculating the Joule-Thomson Coefficient, you can obtain it directly by going to: http://webbook.nist....hemistry/fluid/

Some people – especially students who don’t understand the theory or the process – are using simulation programs to do their calculations without having first-hand knowledge or experience of what is happening when you expand the gas blanket that exists on top of a condensed hydrocarbon liquid as opposed to what happens when you simply expand the contents of a pure, real gas from a compressed cylinder. Both are batch-type of processes (not steady state) but one (the vessel with the liquid) affects the ultimate temperature of a typical adiabatic expansion. That is why some simulation programs (such as Hysys) employ an empirical factor – the isentropic factor – to obtain a more realistic answer of the expansion temperature downstream. Your case does not involve this. You should first carefully analyse and understand what the conditions are and how the fluid(s) is(are) affected. With pure, real gases, the Joule-Thomson effect is what results upon free adiabatic expansion. This was clearly and thoroughly proven empirically by James Prescott Joule many years ago and documented in collaboration with Lord Kelvin (Mr. Thomson).

[Zauberberg]

 J_T_Data_for_Argon.JPG   94.37KB   65 downloads

(Extract from Perry's ChE Handbook)

[YohanJ]

Thank you both for a very helpful reply.

According to the chart the mu factor is around 0.35(C/atm), and simplified,
mu = delta T / delta P

So "3atm --> vacuum" , 3atm of pressure change would only bring about 1 degree of temperature change.

Does this seem right to you? This seems intuitively too small to me... But then I've never touched the gas expanding..

Another question is why would 'choked flow' not apply to this case? Is this because you assume valve opening is big enough so that choked flow conditions do not apply?


thanks again.

[Art Montemayor]

Yohan:

You've mis-interpreted my remarks. What I meant was that choke flow has nothing to do with the lowering of the temperature of a gas. YOU HAVE CHOKE FLOW. I think that is pretty obvious. Anytime your gas expansion pressure ratio is more than 2:1, you have choke flow.

Choke flow is sonic flow. It means that the mass flow rate of the gas in question is constant across the orifice or adiabatic expansion device that you use. This physical phenomenom has nothing to do with the resulting temperature - that is, as I said, due to the Joule-Thomson Effect.

I hope I've cleared up these two subjects.