Question: If LOX/LH2 were used in a fuel cell powering an ion engine could it provide a greater delta-v than with a conventional engine?


  • Start in space in a stable heliocentric orbit (no atmospheric drag to consider or minimum acceleration rate)
  • propellant mass is 90% of total mass
  • any reasonable assumptions of thrust to weight based on existing engines, or fuel cell electrical efficiency
  • propellants can include other non-reactive materials but energy source is LOX/LH2
  • ignore energy needed to keep cryogenic material cold
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – called2voyage
    Commented May 10, 2018 at 14:48

5 Answers 5


Fuel cell efficiencies seem to be quoted as up to 60%. Combining 1 mol of O2 with 2 mol of H2 produces 572 kJ, so a 1We, 60% efficient fuel cell consumes about 3 $\mu Mol$ of O2 and 6 $\mu Mol$ of H2 per second. This gives a thrust of about 40 $\mu N$ (based on figures for Dawn). We can probably ignore the mass of Xenon, so we can compute the $I_{sp}$ as thrust divided by fuel consumption rate divided by $g$, giving $4 \times 10^{-5} \over 6 \times 10^{-8}g$ (all in SI units) for a respectable, but hardly startling $I_{sp}$ of about $60s$.

The problem I think is that we are dividing our available energy between a small amount of Xenon moving very fast, and a large amount of water, which is just waste. Because KE is $1/2 m v^2$ whereas momentum is $mv$ we get less momentum this way than if we accelerated everything to the same velocity.

  • $\begingroup$ You have some reaction energy E and some reactant-out (i.e. water) mass M. If you apply that energy to that mass, P = sqrt(2M E). If you add more mass m, e.g. additional reaction mass, and spread that energy optimally, you get more momentum, because now P = sqrt(2 M+m E) i.e. is larger. There are engineering issues with accelerating water and xenon, etc, but this looks like it wins in theory. of course, you have to be carrying the xenon, which doesn't provide reaction energy. So that needs a more careful look. $\endgroup$ Commented May 10, 2018 at 17:39
  • $\begingroup$ As a neat side calculation, using the 1 second numbers from above: Mass is 108x10-9 kg, so the converting the 1J energy gives a water exhaust velocity of about 4500m/s, right about where we started. $\endgroup$ Commented May 10, 2018 at 17:51

TLDR: Yes, but probably no, well maybe.

Fundamentally, the second law of thermodynamics is about entropy: If you create it, part of the energy budget of a running engine has to be spent on getting rid of it. Combustion processes create entropy as a result of creating heat, and the rest follows.

To the extent that fuel cells (and human cells) avoid creating entropy, they can be more efficient than the Carnot limit. The notional ideal fuel cell reaction: "moleculeA + moleculeB = moleculeC + electricity" doesn't generate any entropy because it's fully reversible (for H/O cells, the reverse reaction is electrolysis). That means that, in theory, it can be fully 100% efficient.

Although there are practical difficulties, people have shown ~90% efficiencies in the biological version of this. (I realize that paper is a bit old and almost off topic, but it shows that biological fuel-cell reactions are intrinsically not combustion-limited)

There are some inherent practicalities, however. Chemistry has a little bit of physics in it, however, and for any particular reaction those might mean that the result is not created at rest, is created in an excited state, or entropy (effectively, heat) is intrinsically created as part of the reaction. That will reduce the efficiency, even before engineering considerations are involved.

This makes it a bit hard to define efficiency unambiguously because when you divide "electricity out" by "energy in", there are several possible choices of "energy in". Practical fuel cells can routinely get very large fractions (over 95%) of the possible electrical energy out, but due to the underlying reaction physics, etc, not all the heat energy of the inputs are available.

For $H_2$ and $O_2$, not counting engineering costs and energy costs of getting them to reaction temperature and pressure, a fuel cell can recover 83% of the possible input ("chemical caloric") energy. The rest becomes heat, and it's then an engineering question if that's also useful ("thermal energy recapture" is the term of art). That number varies a bit with temperature of reaction (higher is better, in general) and some other considerations; I've never looked at the kinetics of a cryogenic $lH_2$ and $lO_2$ fuel cell and there may be some intrinsic thermal losses there.

So then it becomes a matter of engineering: Getting the reactants to the right place at the right temperature and pressure, processing the product water, heat and electricity, etc.

It's going to be really hard, though, to beat the numbers in the question. The "normal" thermal engine already has great 2nd-law conditions to operate in the hot side stays really hot, and the cold side is pretty cold.

So in practice, I think the best you could do is a complex, heavy engine that's just a few percent better. That's not going to be a winner.

What you might find, though, is some other set of reactants that has a better energy-mass ratio. If they don't have to burn hot and quick, there might be more possibilities.


I agree with all previous answers. Here's a look with different units, and with real-world engines.

Total impulse is just force times time, or F∆t, and that translates directly to ∆V: F∆t = m∆V => ∆V = F∆t/m. You just have to plug in the vehicle mass for m and voila!: ∆V. So for a given vehicle mass, the greater the total impulse, the greater the ∆V.

If you burn a total mass of 1 kg of hydrogen and oxygen in an SSME (at vacuum environment) you get 4,440 N-s of total impulse.

If you process a total mass of 1 kg of hydrogen and oxygen in a fuel cell whose efficiency is 100%, using the same enthalpy of reaction quoted above by Steve Linton, you get 1.59 x 10^7 Joules of energy, which will run an NSTAR engine (2.3 kW power consumption) for slightly over 6900 seconds. At 91 mN of thrust, this is a total impulse just short of 630 N-s, ~14% of the chemical engine's total impulse, so ~14% of the chemical engine's ∆V.

14% of the SSME specific impulse (453 s) is 63 s, in good agreement with Steve Linton's analysis. But that's a coincidence, because he assumed 60% fuel cell efficiency, and I'm seeing the real-world efficency of the NSTAR engine.

If that fuel cell is only 60% efficient, the total impulse from the NSTAR drops to 377 N-s, 8.5% of the chemical engine value. The results scale linearly with fuel cell efficiency.

Agreed with nearly all: you're way ahead to burn the O2 and H2 in a chemical engine.


Absolutely not.

A rocket carries a fixed amount of energy and propellant mass. Ignoring efficiencies of the various devices involved, the question boils down to

Is it better to eject all propellant at equal velocity or to drop some mass at zero velocity and eject the remainder at higher speed?

Energy is proportional to velocity squared, but impulse is linear. For a given amount of energy, it's always better to eject more mass at lower speed. The highest impulse is gained when all available propellant mass is ejected with uniform velocity.

Consequently, just dumping the water produced by the fuel cell and propelling the rocket with a xenon-based ion engine is definitely out of the game. One question remains: Is the fuel cell and ion engine combination more efficient if the water is used as propellant?

The answer is a straight no. Burning 1kg of LH2/LOX releases a fixed amount of energy, which can be used to accelerate 1kg of combustion products down the nozzle. A conventional rocket engine with a high chamber pressure and a long nozzle could do that with close to 100 percent efficiency, even though real engines usually run around 50 or 60 percent.

In the best case, our fuel cell and ion engine rocket could also do that with a hundred percent efficiency, which isn't any better than what a good rocket engine could do. Needless to say, real fuel cells and ion engines are far worse than the 60 percent real rocket engines achieve. To my knowledge, ion engines that run on water don't even exist.

If you were willing to gain a few percent of efficiency at the cost of a heavier and more complex propulsion plant, you would just lengthen the nozzle of your conventional engine.

  • $\begingroup$ What you say is right, if we keep the amount of energy available constant. But as always in rocketry the actual question is not about the amount of energy available, but the mass of the propellant. We have to compare "burn something and eject the product at 4 km/s" and "burn something and eject something else at 40 km/s" in terms of Delta-v per mass $\endgroup$
    – asdfex
    Commented May 10, 2018 at 16:52
  • 1
    $\begingroup$ @asdfex - Here, the question is different. It is Should we use all the LH2/LOX as propellant or only a small amount of xenon. The answer is: use all the LH2/LOX. $\endgroup$
    – Rainer P.
    Commented May 10, 2018 at 16:59
  • $\begingroup$ "equal velocity" isn't what combustion rockets do, because they emit their products hot: With a range of velocities. "Equal velocity" is better. "Cold" is better than "hot" for the same reason. $\endgroup$ Commented May 10, 2018 at 18:57

The numbers show a possible “yes”.

Two moles of $H_2$ and a mole of $O_2$ burned in a standard engine creates exhaust at up to about 4500m/s.

The 572kJ there, if converted via 100% efficient, reversible processes, could propel those products at about 5600m/s. That’s about a 25% improvement in specific impulse.

How can that be?

Not-quite fundamentally, the issue is the temperature of the exhaust gases. They’re at a high temperature, so they’re a spectrum of speeds, and that’s fundamentally inefficient (@RainerP discussed this): some parts are faster, some slower, and the fast ones need more energy due to the $v^2$ in $1/2 m v^2$ than you gain back from the slow ones. Monochromatic, single-speed exhaust is most energy efficient; having combustion gas exit at e.g. 3500K leaves a lot of energy in that gas (the two moles of product created above carry off about 140kJ of energy at 3500K)

More fundamentally, this is about needing to get rid of the entropy created by combustion through the exhaust. That requires some heat there, which is waste. A fuel-cell to ion accelerator engine, if built to use fully-reversible processes, need not do that.

  • $\begingroup$ The most efficient chemical engines exhaust to a vacuum, so even though the chamber pressures might be 10 MPa or so, the pressure at the nozzle exit is more like 1-10 kPa. That's a lot of expansion! And that cools the gas a lot. 3500 K is a chamber temperature; the exhaust temperature for a vacuum engine will be in the hundreds of K, not thousands. $\endgroup$ Commented May 11, 2018 at 7:41
  • $\begingroup$ @TomSpilker Good point. But does the pV work done by that cooling go to the engine, or to the expansion speed of the gas plume? The nozzle probably captures a lot of it. Will look into that... $\endgroup$ Commented May 11, 2018 at 10:54
  • $\begingroup$ Most of the pV work goes to the kinetic energy of the gas jet. There is some lost to the nozzle surface, but it turns out to be only a couple percent (Sutton, Rocket Propulsion Elements). The part of the jet that loses heat to the nozzle is a relatively thin boundary layer. For a very interesting exploitation of that, check out how they "cooled" the Saturn V F-1 engine nozzle extension! (Actually, they just prevented it from heating as much as it would, if directly exposed to the primary exhaust gases, then cooled it radiatively) $\endgroup$ Commented May 11, 2018 at 18:01

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