T/flow Rate - Joule Thomson Cooler

You could use energy balance

The Work done by the expansion of Nitrogen should be = equal to the heat being removed on the surrounding.

something like this W= ∫Pdv = mass rate N2 Cp (ΔT),

or you could get the data from different mass rate of N2 versus ΔT, regress it, then you should find your equation for the correlation of the two. This is the most practical approach of your problem.

Helium & Nitrogen, don't have the same Cp(specific heat), don't use that Helium data, Nitrogen has its very own Cp. I suggest you go, try to experiment.

But, for me this is already been nailed by equations alone- relationships between the variables PV=mRT and Cp & Cv has already been known, so this is something that can be calculated given the conditions & constants.

Well, good luck then

http://webbook.nist.gov/chemistry/fluid/

I think you will find the lowest temperature you can achieve is the boiling temperature of the gas at ambient temperature/pressure. See: http://www.aero.org/publications/donabedian/donabedian-1.html The benefit of greater gas flow is an increase in cooling capacity, not lower temperature. Also you will find it critical to make sure your nitrogen supply is very dry or you will clog your J-T orifice with ice.

Thank for your help guys! But I don't get it…

Look at this diagram, I would like to be able to calculate the same: Temperature change as a function of Flow Rate (pressure drop per second)

What formula do i need to use?

Guess what? - you answered your own question. The boiling temperature of LN2 is 77 K which is what your graph shows. The gas flow required to get to that temperature depends on the mass you are cooling, its specific heat and how well it is insulated from ambient conditions. So the answer to your question is 77 K. If you want the flow required to get to that temperature you must supply additional information. Most J-T cooler designers make measurements and construct that curve

These diagram you provided above, is just the plot of the variables, from time o opening to stable cooling state state at time 2.2 min.

Take time from 2.2 to 10 min as the real effect of N2 cooling there. Time 0 to 2.2 is just the transition

I think the transition time is what OP is looking to calculate. The problem is that the J-T cooler is a refrigerator. When you put something in a refrigerator, the time it takes to reach low temperature depends on what it is. A tonic can takes less time to cool than 10 pounds of meat. If the item you are cooling goes through a phase change, you must add the latent heat for that change. Also, specific heat (Cp) for most materials is a function of temperature. These complexities illustrate why performance is usually estimated and then measured.

Hey Noudge79, welderman thank you for your input!

welderman you are right the flow will probably change if the Mass that needs to be cooled is bigger/smaller.

Let's do it for this case:

This system has constant ambient temperature of 180 degC. The nitrogen in the tank has 4000psi and 180 gedC. I need to cool this aluminum mass from 180 degC to 170 degC! (Let's say in 20mins)

If my flow rate is very small I will probably not notice any change in the temp. If its to big the mass will cool down to lower temp. then needed.

So how to find the right flow rate?

This is just like what I analysis stated above,

Energy balance of the two system should be considered,

Considering that the right side vacuum vessel, is something that can be considered as a reservoir, meaning, exhaust pressure of Nitrogen wont accumulate on the vessel, it stays vacuum regardless of how much discharge it has to take.

Again, the analysis is, the Heat(out) due to the expansion of Nitrogen would be equal to the heat input to 200 kg aluminum maintaining at 170 deg. C

∫Pdv(Nitrogen) = m C(ΔT)(Aluminum, C @ (170+180/2)) (First phase)

When this condition is now achieved, the equation will now change to

heat convected by Aluminum to surrounding = Heat (out) by Nitrogen expansion (Second Phase) --> this is minimal compared to the first phase obviously due to the type of heat transfer.

Most likely the same transition patterns of the diagram you provided @ #4.