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