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The exhaust velocity of a rocket is given by: $v_e = \sqrt{\frac{2k}{k-1} \cdot R \cdot T_c \cdot (1-(P_e/P_c))^{\frac{k-1}{k}}}$

The velocity tends to a finite asymptote as $P_c$ increases. Therefore, the specific kinetic energy of the exhaust jet, $\frac{1}{2}v_e^2$ (kinetic energy per unit mass of exhaust jet) cannot be increased beyond a finite value.

But, it takes energy to pressurize the propellants before they are injected into the combustion chamber. If we had a nuclear reactor driving the turbopumps, we could pressurize our propellants to arbitrarily high values, using vast amounts of energy to do so. However, as the above equation shows, we are not gaining anything by doing so beyond a certain point.

Where does that pressurization energy go?

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    $\begingroup$ The limit is being imposed by the gas dynamics in the nozzle. Not really a turbopump question since it doesn't matter how you achieve your arbitrarily high chamber pressure. If you fix the nozzle, excess chamber pressure means the flow is underexpanded and thus less efficient, so the losses are presumably mostly going to heat and acoustics, see e.g. braeunig.us/space/propuls.htm. I'll be curious to see if someone can put together a really clear answer though. $\endgroup$
    – Erin Anne
    Apr 3, 2023 at 6:29
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    $\begingroup$ Something else to remember is that $T_c$ and $P_c$ are related. In normal operation that has to do with the combustion reaction itself, but in your scenario you're pumping pressure into the chamber and therefore raising its temperature as well. The energy flow might get a lot clearer if you put together an equation that encodes your assumptions about how the system is working, rather than sticking with the standard one. $\endgroup$
    – Erin Anne
    Apr 3, 2023 at 7:27
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    $\begingroup$ Are you allowing Pe to change in your limit? I suspect not. For a fixed nozzle, increasing Pc does not change ve at all. It does, however directly increase the thrust. The reason is because there is more mass flow. That is where the extra energy goes. The thrust will proportionally increase with turbopump outlet pressure, not asymptote. I will be writing up an answer in a few hours. $\endgroup$
    – A McKelvy
    Apr 3, 2023 at 11:27

1 Answer 1

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Where does that pressurization energy go?

Into the flow and then into the vehicle, exactly as you would expect!

I am assuming (like you) that the most immediate effect of a nuclear turbo-pump like you describe would be an increased chamber pressure.

Let us first consider the jet flow from the rocket engine from a thermodynamic perspective. The flow through a rocket engine basically follows a Brayton cycle, with isentropic compression and expansion, and constant pressure heat addition. I draw a possible P-v diagram below:

Rocket P-V Diagram

If you remember your thermo, then you know that the power output of the engine can be computed by integrating the cycle. Now it becomes obvious why engine designers want high chamber pressures (to increase the height of this cycle boundary!) and large nozzles (to move point four as low and right as possible).

But we can also obviously see that an infinitely increasing chamber pressure should result in an infinitely increasing power output. How might this manifest? If we assume that the losses are negligible and that the majority of the power output is transferred into the vehicle, then we can write an energy balance: $$P_{engine} = \frac{d}{dt}\left(\frac{1}{2}mV^2 + mgh\right)_{vehicle}$$

Here the engine power output will equal the time rate of change of the vehicle's energy. Assuming level flight and a vehicle mass much larger than the mass flow rate gives the following simple expression: $$P_{engine} = FV$$ where F is the engine thrust and V is the vehicle velocity.

Ok, so we consider an arbitrary flight condition, then increasing the power output of the engine has the direct effect of increasing the engine thrust. And if you take the limit as chamber pressure goes to infinity, the thrust also tends to infinity.

Now I will address your argument.

You are correct that the specific energy of the gas asymptotes in that equation in the limit as chamber pressure goes to infinity. But this does not simulate a turbo-pump acting as you describe!

That equation solves for the exhaust velocity in a nozzle, but the variable $P_e/P_c$ is fixed for a given nozzle expansion ratio. What you are doing by taking this limit is investigating the exhaust velocity if expansion ratio is taken to infinity, because that is the only condition that will take $P_e/P_c$ to zero.

So if you have an infinitely long and large nozzle, your gas velocity will asymptote, but this has no impact on the thrust, just the engine massflow efficiency.

To summarize: If you had a nuclear powered turbo pump, you might raise the chamber pressure arbitrarily high, this will increase the thrust arbitrarily high. The temperatures remain unchanged (ideally), Isp remains unchanged, the exhaust velocity is unchanged, etc. The primary effect of your proposed change is an increase in thrust.

Also, regarding Erin Anne's comment, indeed, for a fixed nozzle, popping in your proposed turbo pump would create a horribly underexpanded nozzle. However, this does not destroy the performance of the engine. Any FIXED rocket engine produces more thrust while underexpanded than while perfectly expanded, it is just that it would produce even more thrust if you slapped a nozzle extension on.

I go into more detail on the specific roles of pressure and temperature in a rocket engine in this answer.

EDIT: Following discussion in the comments, I want to address the energy contributions of two engines, one with a gas-generator turbo pump and another with a nuclear-thermal turbo pump.

Assuming the same propellant and chamber pressure, the power requirement of both pumps is identical. The only difference is that the gas-generator engine has to burn propellant to generate the required pump energy, while the nuclear engine practically (comparatively) consumes no mass. This gives the gas-generator engine a lower Isp because it gives the same thrust but at higher mass flows.

My answer above shows that there is no problem (in a neighborhood near non-exotic pressures) increasing the pump power/chamber pressure arbitrarily high. That energy is pumped into the flow and then blown into the vehicle, despite the limit on exhaust velocity.

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  • $\begingroup$ Engine power increases, yes, but increasing P_c manifests itself as an increase in mass flow rate. This is inversely proportional to burn time, if I increase my chamber pressure 10-fold I will just burn through my fuel 10 times faster. Total energy = Power*time, which would be identical in the two cases. $\endgroup$ Apr 3, 2023 at 13:49
  • $\begingroup$ Yes, of course the total energy will be the same. Its all stored in the propellant. $\endgroup$
    – A McKelvy
    Apr 3, 2023 at 13:50
  • $\begingroup$ If I hook up a nuclear reactor to my pump to pressurize the propellant, I've added energy to my system. $\endgroup$ Apr 3, 2023 at 13:51
  • $\begingroup$ @NikhilMurali, ok in that case you will definitely produce more energy and burn for a longer time. Because now you can use the propellant that would otherwise be consumed (energy content wise) in the preburner. I thought your question was specifically about raising pressure to extreme values and the effect on power output. Obviously if you add nukes, batteries, or any other power source for the pumps, then thats energy you dont need to take from the propellant. $\endgroup$
    – A McKelvy
    Apr 3, 2023 at 13:54
  • $\begingroup$ @NikhilMurali, I edited my answer to hopefully clarify this issue $\endgroup$
    – A McKelvy
    Apr 3, 2023 at 14:18

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