I'm trying to understand something about isentropic compression of Hydrogen. If you are assuming constant specific heats, the work done during an isentropic compression is

$$W = \frac{k}{k-1}\cdot RT_0\cdot\left(\left(p_\mathrm f/p_\mathrm i\right)^{0.285} - 1\right).$$

Let's start with hydrogen at $p = 1\ \mathrm{bar}$ at $25\ \mathrm{^\circ C}$ and go to $p = 880\ \mathrm{bar}$. If you plug in these values, and use $k = 1.41$, you come up with $W \approx 26\,000\ \mathrm{kJ/kg}$.

Now my question is, if this is an isentropic process (i.e. it's adiabatic) surely the work done is also directly changing internal energy? If I use the expression for temperature and pressure relations in an isentropic change, then you can find the temperature at the second stage to be $ T_{2} = T_{1}* 880^{0.285}$ and from there, the temperature at the second stage to be $= 2\,057\ \mathrm K$.

Working from an ideal gas table for hydrogen, the change in internal energy is a bit different from the work expression evaluated; the change in internal energy found from the tables based on the change in temperatures is ~ 36000 kJ/kmol which does not equal this when converted to kJ/kg.

I just want to confirm that this is a valid way of arriving at the same answer – finding the isentropic temperature and then determining the change in internal energy, instead of plugging in the expression for the work.

...as an addendum, when I look at the enthalpy differences between these two pressures, it's possible that this does the trick because the enthalpy difference is 54,651 kJ/kmol (~26,000 kJ/kg for H2; there are some triangulation steps and approximations I'm leaving out because they really cannot be that significant) but I'm still curious to see if this is in general a correct approach for an exam.

  • 2
    At 880 bar, I would not rely on ideal gas relationships. – Pieter Oct 14 at 10:57

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