# Can you avoid thermodynamic inefficiencies in space propulsion?

In @David Hammen's answer to this question, he highlights the discrepancy between the exhaust velocity that would be achieved by converting all the chemical energy of the fuel into KE of the exhaust (5630 $ms^{-1}$ for $LH_2/LO_2$) and the maximum velocity achievable by converting that energy into heat and then allowing the hot exhaust to expand through a nozzle (roughly 4500 $ms^{-1}$). Is there any way in theory (I'm not really worried about engineering practicality for now) to get around this? Could you use a (hypothetical highly efficient) fuel cell to generate electricity, for instance, and use that to expel the water that comes out of the fuel cell electrically, as charged droplets or ice crystals? Or does the second law of thermodynamics still come back and bite you somewhere?

• There will always be inefficiencies. The trick to building a good engine is to minimise the inefficiency - bearing in mind that the weight of the engine is an inefficiency. – user20636 May 10 '18 at 10:44
• The 2nd law will always bite you. – Organic Marble May 10 '18 at 11:28
• @uhoh I think you'd have to average (weighted by mass) the exhaust velocity of the Xenon ions with the exhaust velocity of the water output from the fuel cell which you presumably just dump over the side. – Steve Linton May 10 '18 at 11:47
• A brief synopsis of the laws of thermodynamics that dictate the rules of an interesting game in physics. Zeroth law: You are forced to play this game. First law: You cannot win this game. Second law: You can't even break even, except on a very cold day. Third law: It never gets that cold. – David Hammen May 10 '18 at 12:15
• – uhoh May 10 '18 at 12:21

I don't know how to answer this as stated except in general terms. So, this:

O2 / H2 fuel cells work on the reaction

2H2 + O2 --> 2H2O + heat

The same reaction that happens when you burn these propellants in an engine.

"the second law of thermodynamics still come back and bite you somewhere" comes into play with this waste heat from the reaction. In a rocket engine it helps you. In a fuel cell it hurts you. You have to have a system to remove the waste heat or your fuel cell explodes. Something has to provide energy to run that system. That energy is not accelerating your reaction mass.

• Except that it's really 2H2 + O2 --> 2H2O + electricity + heat, and the reduction of heat production is a reduction of entropy production. You don't have to have as much "thermodynamic" waste heat to get rid of the entropy created. See this answer to the related question. I agree that the engineering considerations make it really hard to win this way. – Bob Jacobsen May 10 '18 at 14:16
• Yeah. If it didn't produce electrons it would be a poor fuel cell indeed. – Organic Marble May 10 '18 at 15:07
• @BobJacobsen thanks. That is the key point, I guess. Is there any fundamental reason why the fuel cell has to produce a certain proportion of its output as heat? – Steve Linton May 10 '18 at 15:17
• @SteveLinton No fundamental reason. If you can find the right reactants, perhaps the right catalyst, and don’t care about reaction volume or time, you could theoretically create a completely reversible cell. Of course, all of those are hard conditions for a rocket engine... – Bob Jacobsen May 10 '18 at 15:19

The second law always comes back to bite you somewhere.

A chemical rocket that burns hydrogen and oxygen has a specific impulse of about 4500 m/s, or about 80% of the theoretical maximum of 5630 m/s. An efficiency that high is quite incredible. Hydrogen / oxygen fuel cells, for example, in practice operate at about 40 to 60% efficiency, so existing fuel cells used to power any thrust-generating device is a losing proposition.

The theoretical cap on the efficiency of hydrogen / oxygen fuel cells is 83%. Wave that magic wand and yes, you can beat current hydrogen / oxygen chemical rockets. But I'll wave my magic wand right back atcha and somehow improve the efficiency of a hydrogen / oxygen chemical rocket to 84%.

TL;DR: It's really tough to beat a system that already achieves 80% of the theoretical maximum.

• Where does the 83% limit come from? – Steve Linton May 10 '18 at 19:58
• Also 80% of maximum exhaust velocity is only 64% of energy converted into KE – Steve Linton May 10 '18 at 19:59
• @SteveLinton Chemical kinetics in a liquid cell. There’s some irreducible energy lost to molecular vibration and kinetic energy. Different reactants have different values; there are biological reactants with paths over 97% (though they are slow). – Bob Jacobsen May 10 '18 at 21:24

You avoid the 2nd law constraints on efficiency if you avoid creating entropy.

For example, two moles of lH2 and a mole of lO2burned 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 energy. An engine that used only reversible processes, e.g. fuel cells powering an electrostatic accelerator for the exhaust products, need not waste that energy.