My confusion stems from the relationship between thrust and the pressure distribution in the thrust chamber. For a given chamber pressure the wall static pressure along the nozzle is a function of area ratio and (importantly) not of temperature. Additionally, rocket combustion is typically a near constant pressure process, so unless I am mistaken, the pressure distribution should be the same for a cold and hot gas thruster given the same chamber pressure.

Thrust is directly a function of the pressure distribution in the thrust chamber, so it seems like there should be no difference in thrust for the cold vs hot gas case.

From a momentum perspective there is a clear mechanism for increased thrust. Similar to pressure, the exit Mach number would be the same for cold vs hot, but the gas velocity would be significantly higher for the hot case, giving more thrust.

Is there some change to the pressure distribution for a given geometry with cold vs hot gas? If not, then where does the extra thrust come from?

EDIT: I understand that chamber pressure will not approach nominal levels with unignited liquid propellant. This question only applies for a theoretical situation with gaseous injection. I should also note that choking pressure ratio is also independent of temperature.

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    $\begingroup$ Never heard that a hot gas increases its volume when held at constant pressure and its pressure when held at constant volume? $\endgroup$
    – Uwe
    Jun 17, 2022 at 20:37
  • $\begingroup$ @Uwe Sure, and in this case where the total pressure is the same there should be some "expansion" relative to the cold case which accounts for the increased gas velocity. But there MUST be an increase to the integrated axial wall pressure equal to the increase in thrust. I cannot understand how this jives with the compressible flow equations given fixed total pressure and geometry. $\endgroup$
    – A McKelvy
    Jun 17, 2022 at 21:07
  • $\begingroup$ "Additionally, rocket combustion is typically a near constant pressure process" No, the constant pressure was very hard to achieve by adjusting the propellant flow to the throat diameter and keeping the flow constant. If pressure would increase steadily, the chamber would be destroyed soon. If pressure would decrease, the rocket would loose thrust and crash back to the launch pad. Pressure oscillations (called pogo)were very hard to avoid when developing the Saturn V F-1 engine. If you want a reliable and efficient rocket engine, chamber pressure should be held constant. $\endgroup$
    – Uwe
    Jun 17, 2022 at 21:38
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    $\begingroup$ @AMcKelvy the pumps likely wouldn't even achieve choked flow through the nozzle, certainly not in engines where the propellants are injected as liquids. The volume of the propellants being injected into the chamber is not the same as the volume of the combustion products exiting it through the throat. $\endgroup$ Jun 17, 2022 at 21:58
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    $\begingroup$ No, if you pump the liquid propellants into the chamber without igniting them, you would never reach the nominal chamber pressure. $\endgroup$
    – Uwe
    Jun 17, 2022 at 21:58

1 Answer 1


“Why does igniting propellant provide more thrust?”

It doesn't!

At least with the specified parameters. Consider the following control volume: Control Volume Sketch

For a perfectly expanded case with choked flow, the momentum thrust $\dot{m}V_e$ will be equal to the pressure thrust $F$. Where pressure thrust is defined as:$$F = \int P_xdA$$ where $P_x$ is the axial component of static pressure acting on a given piece of thrust chamber wall $dA$.

Combusting propellants increases the total temperature of a fluid, so you are asking about two cases: one with a very high $T_o$ and one with a low $T_o$ and both with the same total pressure $P_o$. Lets investigate the pressure thrust first.

To calculate pressure thrust, we need to know the static pressure at every axial location in the thrust chamber. For isentropic flow, the static pressure is related to the Mach number by $$P = P_o\left(1+\frac{\gamma - 1}{2}M^2\right)^{\frac{-\gamma}{\gamma-1}}$$ and the Mach number is related to the area ratio at a given axial location by $$\frac{A}{A^*} = \left(\frac{\gamma+1}{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}\frac{\left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{\gamma+1}{2(\gamma-1)}}}{M}$$

A nasty equation. Here $\gamma$ is the ratio of specific heats and $\frac{A}{A^*}$ is the ratio of channel area to throat area. What can we glean from these equations? Mach number is purely a function of the area ratio and of gamma, so for the hot and cold case, the Mach distribution along the thrust chamber will be the same. The static pressure is a function of total pressure, gamma, and Mach, none of which change between the two cases, so the static pressure distribution will also be the same! (neglecting the weak dependence of gamma on temperature).

If the pressure distribution is the same, then the thrust must also be the same between the hot and cold cases.

Now, to the source of your confusion: what about the momentum thrust!? Surely this increases! Lets check.

As you mentioned, the exit velocity for the hot case will increase according to $$V_e = M_e\sqrt{\gamma RT_e}$$ where the subscript $e$ denotes conditions at the exit plane and $T_e$ is the exit static temperature. This follows a similar relationship as the static pressure with $$T = T_o\left(1+\frac{\gamma-1}{2}M^2\right)^{-1}$$ Clearly the exit static temperature is directly proportional to the total temperature, so the hot case will have a higher gas exit velocity. So you say "Yes, that is why the thrust must be higher!", but calm your pants; what about the mass flow?

The mass flow for a choked nozzle is defined by $$\dot{m} = \frac{A^*P_o}{\sqrt{T_o}}\sqrt{\frac{\gamma}{R}}\left(\frac{\gamma+1}{2}\right)^{-\frac{\gamma+1}{2(\gamma-1)}}$$ where $R$ is the ideal gas constant. Aha! Perhaps things become clearer: the mass flow is inversely related to the total temperature. Could the decrease in mass flow exactly counter the increase in velocity? Solving for the momentum thrust, $\dot{m}V_e$ by crudely combining these equations gives $$\dot{m}V_e = A^*P_oM_e\sqrt{\frac{\gamma^2RT_o}{RT_o\left(1+\frac{\gamma-1}{2}M_e^2\right)}}$$ Well well well, would you look at that: the term $RT_o$ cancels from the momentum thrust. The momentum thrust will also be the same for the two cases!

Whew, crisis of aerothermal physics averted. But wait, if thrust is the same, why does anybody bother with those extremely high temperatures? The answer is mass flow efficiency. As we discovered, the higher temperature case used lower mass flow for the same thrust. This gives a higher specific impulse and makes a rocket using this engine much lighter and cheaper than for the same journey using cold gas.

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    $\begingroup$ Regarding the bolded It doesn't! Being pedantic, to what are you referring? What, in particular, is it, what does It not do? $\endgroup$ Jun 18, 2022 at 3:11
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    $\begingroup$ @DavidHammen It was meant as a direct answer to the question title: “why does igniting propellant provide more thrust?” Answer: “it doesn’t”. The it being “igniting propellant” and “providing more thrust” the thing not being done. Now that youve said this though, and me also being pedantic, Ive updated my answer to better reflect the relationship. $\endgroup$
    – A McKelvy
    Jun 18, 2022 at 3:43

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