1
$\begingroup$

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?

Improve this question
$\endgroup$
14
  • 1
    $\begingroup$ Can we please focus on a typical single tank transfer as an example. Tank volumes, pressures before transfer, etc? $\endgroup$
    – Chet Miller
    Jan 21, 2019 at 0:20
  • $\begingroup$ 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. $\endgroup$
    – markk1pe
    Jan 21, 2019 at 2:26
  • $\begingroup$ 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. $\endgroup$
    – markk1pe
    Jan 21, 2019 at 18:08
  • $\begingroup$ 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) $\endgroup$
    – markk1pe
    Jan 21, 2019 at 18:20
  • $\begingroup$ 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? $\endgroup$
    – markk1pe
    Jan 21, 2019 at 18:30

1 Answer 1

Reset to default
3
$\begingroup$

To do this analysis, one needs data up to 20000 psi (1400 bars) on either enthalpy vs temperature and pressure (a graph) or compressibility factor Z vs temperature and pressure. The only graphs I have found of the former type go up to only 100 bars, which is a factor of 14 too low. However, I have found data on the compressibility factor Z at temperatures up to 300 K and pressures up to 1000 bars: https://cds.cern.ch/record/1444601/files/978-1-4419-9979-5_BookBackMatter.pdf Although this is still a factor of 1.4 too low, it might provide some idea of the temperature rise that might be expected in the valve.

So here is how the data would be used.

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.

If a reference can be found with data going out to 1400 bar, that would be even better.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thank you @Chester Miller for your effort in putting all this together. If I understand this correctly, I would have to integrate the equation inside the parenthesis, take the derivative of that over ∂T, then multiply that by -2/5T squared, is that what you are telling me? $\endgroup$
    – markk1pe
    Jan 24, 2019 at 17:28
  • $\begingroup$ Yes. You would use the data in the reference, and integrate numerically with respect to lnP, and differentiate numerically with respect to T. You would also be extrapolating to fill in the blank between 1000 bars and 1400 bars. $\endgroup$
    – Chet Miller
    Jan 24, 2019 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.