The definition of isentropic flow is that the flow is both adiabatic (no heat is transferred into or out of it) and reversible (energy is conserved in it). I am wondering why the flow through a rocket nozzle is isentropic. Shouldn't we take into account the heat transfer between the flow and nozzle and the resistive force of the nozzle experienced by the flow as it passes through?

Besides, a book I am reading says that mass flow rate is constant because the flow is isentropic, which I don't quite understand: As the flow passes through the nozzle, expands, and accelerates, the energy goes from pressure to velocity. But how can we derive from this that mass flow rate is constant?

Last but not least, it makes sense that the mass flow rate of the exhaust when it is passing through the nozzle is constant. However, if we are talking about the mass of exhaust leaving the rocket in a unit of time, is it still mostly constant with small variations? If it is, are (1) the velocity at which the exhaust leaves the rocket and (2) the thrust also constant over time?

Thanks for your patience!


2 Answers 2


Isentropic Flow

First off: the flow in a real rocket nozzle is not isentropic. It is a simplifying assumption that comes reasonably close to real world observations. As is all of ideal rocket theory, which you are most likely studying.

So you are absolutely right, there are losses due to heat flowing to the surroundings and friction losses at the nozzle wall. However, these are relatively small enough that pretending they aren't there still gives good real world approximations.

Do note that this is only the case if the flow is expanded and compressed gradually. Once shock-waves are present in the flow due to over- or under-expansion the assumptions that the flow is isentropic doesn't work anymore.

Constant Mass Flow

For constant mass flow, see this link on choked flow by NASA. Just to quickly summarize that: mass flow is maximum at Mach equal to 1, which happens at the throat of a conventional rocket nozzle. The equation that then gives this mass flow is: $$\dot{m}=\frac{A_t p_c}{\sqrt{R\cdot T_c}} \sqrt{\gamma}\bigg(\frac{\gamma+1}{2}\bigg)^{-\frac{\gamma+1}{2(\gamma-1)}}$$

With $A_t$ the throat area, $p_c$ the chamber pressure, $R$ the gas constant, $T_c$ the chamber temperature and $\gamma$ the ratio of specific heats.

Now I don't like the look of all those $\gamma$'s, so let's replace them with the vandenkerckhove function $\Gamma$, just to make it look nicer (it's exactly the same otherwise)

$$\dot{m}=\frac{\Gamma A_t p_c}{\sqrt{R\cdot T_c}}$$

Now $\Gamma$ and $R$ are constants, $A_t$ should be constant (unless your throat is melting :P) and $p_c$ and $T_c$ should be constant once your engine is running in a steady state, thus your mass flow will be constant. And the mass flow equation given at the start of this paragraph can only be derived if you assume isentropic flow, as can be seen on the linked NASA website. So that is why for isentropic flow the mass flow is constant.

Exhaust conditions

By simple conservation of mass, once the mass flow at the throat is constant, so should the mass flow at the nozzle exit. Otherwise mass would be building up between the nozzle throat and exit. In real life of course there would be fluctuations and the the mass flow might be changed by design due to throttling, but under Ideal Rocket Theory it is constant.

Assuming the previously mentioned chamber pressure and temperature are not purposefully changed the equivalent velocity is also constant* and thus the thrust. Since the thrust is equal to the mass flow times the characteristic velocity, $F=\dot{m}\cdot v_{eq}$.

*(as long as the ambient pressure doesn't change, so for a rocket slowly going up in the atmosphere the equivalent velocity and the thrust will also change)

Comment Questions:

  1. Mass flow is directly dependent on molecular weight through the individual gas constant, which is the universal gas constant divided by its molecular weight $$ R = \frac{R_u}{M_{gas}}$$ Indirectly: changing your propellants/combustion to get a different molecular weight, will also change all of the other parameters except $A_t$ most likely.
  2. Yes and no. You will see $F=\dot{m}v_e$ quite often, but this is only true if the nozzle is expanded optimally, thus the exit pressure being equal to the ambient pressure. So it is better to use the equivalent velocity given by the following formula: $$ v_{eq} = v_e + \frac{p_e - p_a}{\dot{m}}\cdot A_e$$ With $v_{eq}$ and $v_e$ the equivalent and exit velocity respectively, $p_e$ the nozzle exit pressure, $p_a$ the ambient pressure, $\dot{m}$ the mass flow rate and $A_e$ the nozzle exit area. As you can see $v_e$ and $v_{eq}$ are equal if $p_e=p_a$, if you want to know more see this NASA link
  3. Sometimes you want constant acceleration for your rocket, which means you need to throttle down continuously, because your rocket is getting lighter and lighter as it expels all the propellant. Or maybe you want to go slower through the lower atmosphere so you do not have as much drag, but more gravity losses. But how and why to throttle mostly belong to the topic of Rocket Motion which is a book in itself.
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    $\begingroup$ Thanks for your thorough answer! Still have the following questions: 1. Intuitively speaking, shouldn't mass flow also depend on the molecular weight of the exhaust? 2. Is it correct to use the velocity of the exhaust at nozzle exit to calculate thrust? My understanding is that the reason for having a CD nozzle is to achieve the maximum exit velocity, since dv is negatively related to dA when Mach number < 1 and positively related to dA otherwise. 3. Could you give an example when varying mass flow and thus thrust through throttling is more desired? Thanks! $\endgroup$
    – Xi Liu
    Oct 12, 2020 at 18:11
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    $\begingroup$ Is this Vanderkerckhove function a standard thing? Does it appear in other relations? I didn't have much luck Googling it. $\endgroup$ Oct 12, 2020 at 19:45
  • $\begingroup$ @XiLiu I have added answers to your questions to the end of the original answer. You have many questions, but I am happy to keep explaining. $\endgroup$
    – Ruben
    Oct 12, 2020 at 20:52
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    $\begingroup$ @WaterMolecule might be a Dutch thing, but to help you out: $$\Gamma =\sqrt{\gamma}\bigg(\frac{2}{\gamma+1}\bigg)^{\frac{\gamma+1}{2(\gamma-1)}}$$ $\endgroup$
    – Ruben
    Oct 12, 2020 at 20:58
  • $\begingroup$ Hi Ruben thank you for the added answers! I checked that the equation for exit velocity contains 1/sqrt(M(gas)), so when calculating thrust the effect of molecular weight vanishes. However, speaking of energy/mass efficiency, propellants with low molecular weights are preferred. Is my understanding correct? Also, I am wondering whether the effect of molecular weight effect still vanishes in the case of electrostatic engines. Otherwise, what accounts for the low thrust of an ion thruster despite its high exit velocity and specific impulse? What does the equations of mass flow rate look like? $\endgroup$
    – Xi Liu
    Oct 14, 2020 at 13:55

I can't tell you whether or not the flow is isentropic because this is beyond my expertise. But I can answer some of your specific questions:

  1. There is a lot of heat flowing from the exhaust to the nozzle. This is easily seen on any of the numerous launches by SpaceX after the second stage is ignited: the nozzle glows red because it radiates heat. However, I don't know if this heat flow can be regarded as negligible when compared to the total heat involved in the process, i.e. maybe the flow is approximately adiabatic.
  2. The overall mass flow rate depends almost only on throttling. Ambient pressure, e.g. at sea level, might decrease it a little bit. But that shouldn't matter much for powerful engines with hundreds of bars pressure inside the combustion chamber. Or are you concerned about the mass flow rate depending on the position within the flow? It must be slightly lower near the nozzle due to friction.
  3. Thrust $F$, mass flow rate $\dot{m}$ and exit velocity $v$ relate by $F=\dot{m}v$. Again, this is mostly dependent on the throttle by which $\dot{m}$ is directly controlled. The other two quantities adjust accordingly.

I hope to have answered all of your questions. If not, please let me know in the comments.

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    $\begingroup$ " Or are you concerned about the mass flow rate depending on the position within the flow? It must be slightly lower near the nozzle due to friction." Where did the missing mass go? $\endgroup$ Oct 12, 2020 at 14:18
  • $\begingroup$ @OrganicMarble The rate is higher in the center of the flow than near the nozzle wall. $\endgroup$ Oct 13, 2020 at 9:20
  • $\begingroup$ The mass flow rate? I think you're getting it confused with the flow velocity. Unless mass is added or subtracted, the mass flow rate is constant. $\endgroup$ Oct 13, 2020 at 11:43

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