# Nuclear thermal rocket working fluid performance

After reading an announcement about NASA restarting NTR (nuclear thermal rocket) research I was wondering about how a working fluid would be chosen.

For clarity, the working fluid is the gas that is heated by the nuclear reactor, and the radioactive material in the reactor is the fuel.

Hydrogen is the fluid considered in most articles on the topic. This is because of the (often discussed on this site) advantage of having lightweight high speed combustion products accelerated out of the rocket nozzle.

What would the efficiencies be for other gasses given the same total energy input?

I am specifically curious about comparing H2, argon, N2, and O2.

• Isn't hydrogen usually in the form of H2 (i.e., not monatomic)? Argon (and He, Xe, etc.) would be monatomic. – Mike Harris Aug 11 '17 at 20:23
• @mikeharris correct. I'll fix that – OrangePeel52 Aug 12 '17 at 12:32
• True. Hygrogen in the tanks would be H<sub>2</sub>. But if heated enough (e.g. >3000 Kelvin) it will disassociate. Which might not be all that desirable in an engine. – Hennes Aug 12 '17 at 14:12
• Diatomic hydrogen particle has mass of about two protons. Monoatomic helium is two protons plus two neutrons, so about twice as heavy. – SF. Aug 12 '17 at 20:32
• Using O2 is not a good choice, it will oxidize a lot of parts of the nuclear reactor that should not be oxidized. Noble gases, hydrogen or nitrogen may be used, but not very reactive gases like oxygen or chlorine. – Uwe Aug 12 '17 at 20:58

The relation is quite straightforward.

The energy injected into a unit of propellant before it evaporates and spent on evaporating it is minuscule comparing to energy applied to the propellant as gas; and for gas of similar pressure and temperature the number of particles per unit of volume varies very little. As result, approximation that assumes each particle of propellant, no matter what propellant it is, receives about the same amount of energy from the reactor core, is quite close to accurate. So let's take the equations for kinetic energy and specific impulse as function of exhaust velocity:

$$E = {1 \over 2} m v^2 \\ I_{sp} = {v \over g_0} \\ v = \sqrt{ 2E \over m } \\ I_{sp} = \sqrt{ 2E \over m {g_0}^2}$$

$g_0$ is a constant. We assume each particle receives about the same amount of energy. Therefore,

$$I_{sp} ∝ m^{-1/2}$$

where $m$ is the mass of a particle of the propellant.

If $I_{sp}$ for monoatomic hydrogen (not really doable, but just to establish baseline) was 1 (of some unit, it's not quite important what), then:

• One $H_2$ particle weighs about 2u, $\sqrt{1/2} = 0.7$
• A helium atom is about 4u, $\sqrt{1/4} = 0.5$
• A lithium atom is about 7u, $\sqrt{1/4} = 0.37$
• One $H_2O$ particle is about 18u, $\sqrt{1/18} = 0.23$
• Zinc was proposed in one of questions on this site. One $Zn$ atom is about 65u, $\sqrt{1/65} = 0.12$
• If for some perverse reason you wanted to use xenon, 131u, $\sqrt{1/131} = 0.08$

As you can see, hydrogen leaves the competition in the dirt. Helium is somewhat comparable, but there are no big advantages using it over hydrogen, and for any hydrogen-rich compounds (say, methane), the savings are mostly lost in the mass of the binding atom ruining the total performance.

edit: adding other gases you're asking about: $Ar$: 0.16; $N_2$: 0.19; $O_2$: 0.18. Hydrogen is more than three times better than the best of them (Nitrogen).

• doesn't some of the energy get tied up in the bonds between atoms of the diatomic (and higher) gasses, reducing the total velocity? – OrangePeel52 Aug 12 '17 at 21:21
• @OrangePeel52: Yes, assuming you increase the temperature enough to break them up, which is risky (extremely corrosive). But the bond energies are way lower than energy of the NTR; they are on par with chemical engines, and a small enough fraction the approximation I used holds. Never mind the resulting monoatomic gas will have particles of half the mass of diatomic, and accelerated further will quickly recuperate losses due to separation. – SF. Aug 12 '17 at 23:10