# Calculating temperature change of helium due to Joule-Thomson effect

We are trying to fill a rocket helium tank that is 78 liters with helium, final pressure of $$\pu{10500 psi}$$. We plan to pressurize high pressure tanks to $$\pu{20000 psi}$$ using a compressing pump system, so that on launch day we don't have to wait for the 78 liter tank to fill slowly by the compressing system. We plan to fill the rocket tanks from the high pressure tanks in a cascade manner (the number and volume of the tanks may vary, depending on the solution of the JT equation(s).

The problem is that helium has a negative JT coefficient. It will heat up during the transfer from the high pressure tanks to the lower pressure tank if I understand it correctly.

So, the concern is how hot will it get. Will the O-rings/seals melt? I have some equations, but it seems to me there may be too many unknowns.

I've seen others make certain assumptions on this site, but its not clear to me how that works. Can anyone advise how you would proceed?

• Can we please focus on a typical single tank transfer as an example. Tank volumes, pressures before transfer, etc? Jan 21, 2019 at 0:20
• Good question. I'll have to give more details tomorrow after I get to my desk and determine what tank volume is. Pressure before is 20,000,desired final pressure in the 78 liter receiver is 10,500. The supply tank size may vary, I'll provide an update tomorrow. Jan 21, 2019 at 2:26
• So, now that I'm back at my desk it's all coming back to me. The design so far is to use four high pressure storage tanks pressurized to 20,000 psi. Open one to fill the rocket tank as much as it can towards the 10,500 psi GHe. Then close that valve and open the next one, and cascade until the desired pressure is reached. Jan 21, 2019 at 18:08
• The 78 liters converts to 4730 cu. in. for the rocket tank. The pressure vessels are 890 cu. in. P3=(P1*V1+P2*V2)/(V1+V2) Jan 21, 2019 at 18:20
• So initially with P1 @ 20,000psi, V1 at 890 cu. in and the rocket tank at 14.7 psi and 4760 cu. in. the settling pressure is 3162 psi. On the next tank we get to 7,108 psi. The third tank gets us to 10,130 psi. The fourth tank could get us all the way to 12,443 psi, but we'd stop when we reach the desired pressure. The question is, how to calculate the heat gain due to the helium transfer? Jan 21, 2019 at 18:30

The effect of pressure on enthalpy (per mole) of gas is given by $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P\tag{1}$$For a real gas, the equation of state in terms of the compressibility factor Z=Z(P,T) is given by$$PV=ZRT\tag{2}$$If we substitute Eqn. 2 into Eqn. 1, we obtain: $$\left(\frac{\partial H}{\partial P}\right)_T=-\frac{RT^2}{P}\left(\frac{\partial Z}{\partial T}\right)_P\tag{3}$$Integrating Eqn. 3 between P=0 and arbitrary P at constant temperature yields the so-called Residual Enthalpy $$H^R$$:$$H^R(P,T)=-RT^2\int_0^P{\left(\frac{\partial Z}{\partial T}\right)_{P'}\frac{dP'}{P'}}=-RT^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{4}$$where P' is a dummy variable of integration.
If the final pressure coming out of the valve is low (so that the gas exiting the valve is in the ideal gas region), we can write: $$\Delta H=-H^R+C_p\Delta T=0$$where, for a monoatomic gas like Helium, $$C_p=\frac{5}{2}R$$. Therefore, $$\Delta T=-\frac{2}{5}T^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{5}$$ This expression would be evaluated using the data presented in the reference above.