The thrust $F$ of a rocket engine is given by $$F=\dot{m}v$$ where $\dot{m}$ is the mass flow and $v$ is the exhaust speed.

Now the first factor, $\dot{m}$, is almost entirely due to the turbo pumps (or if pressure fed: due to the pressurization of the tanks). A little bit is actually due to the thrust $F$ itself, as the acceleration of the vehicle forces the fluids backwards.

How about the second factor, the exhaust speed $v$? How much of it is due to the pumps, and how much comes from the chemical processes inside the combustion chamber? Alternatively: at what speed are the fuel and the oxidizer injected into the combustion chamber?

I'm interested in the order of $$\frac{v_{\rm injection}}{v_{\rm exhaust}}$$ which I of course expect to be a very small number. Also, my main interest is in rockets with both liquid fuel and liquid oxidizer.

  • $\begingroup$ I guess it comes down to how much does the volume expand, since combustion converts your liquid propellants into a mixture of exhaust gases. Another effect is expansion due to heating and flowing through the convergent-divergent nozzle. $\endgroup$
    – Dohn Joe
    Mar 28, 2019 at 9:46
  • $\begingroup$ If you have the m ass flow rate, you can derive the volume flow rate of the propellants. This, combined with the throat diameter gives a propellant velocidty, although the injector may be the bottleneck, not the throat. $\endgroup$
    – user20636
    Mar 29, 2019 at 16:20
  • $\begingroup$ Note for a solid rocket, there's no injection at all. $\endgroup$ Jun 13, 2019 at 16:35

3 Answers 3


There isn't a particularly meaningful answer to this, but I hope I can provide some insight.

Mostly it boils down to the observation that injection velocity is not particularly meaningful/constant-or-optimised between rocket designs.

Injection mass flux is the interesting engineering quantity ($v \times \rho \times A$), where $v$ is velocity, $\rho$ is density and $A$ is cross sectional area. Hence $\frac{v_i}{v_e} = \frac{\rho_e A_e}{\rho_i A_i}$.

However unlike for the exhaust, where maximizing $v$ is critical, a pintle injector would work almost exactly as well if it had double the area and half the injection velocity or vice-versa. $\rho$ is also a significant source of fluctuation.

The subtleties of the trade-offs are a bit complex. Enough so, that designs vary significantly.

For example:

  • A gas generator cycle feeds the fuel/oxidiser into the injectors pretty much as it comes out of the tanks.

    As do pressure-fed, electric-pump-fed, and tap-off cycle engines.

  • In a staged combustion cycle some or all of the propellant will have already been through a combustion chamber, increasing its temperature and lowering its density.

    In expander cycles the expansion (usually of the fuel) due to heating is directly the source of energy used to pump the propellants.

This change in density would effect injection velocity, for a given injector geometry.

Finally, unlike exhaust velocity which is fairly well defined, injection velocity is a little less clear.

Take a look at: https://upload.wikimedia.org/wikipedia/commons/1/1f/Pintle_3.png

There are a number of constrictions near the outlet. Where you choose to take injector to end, and combustion chamber to begin, will effect your answer you get. It should also be clear you care fairly free to alter the geometry to the same effect.

  • $\begingroup$ Great answer. Could you please clarify the variables that you introduce? Also, I recommend to use lowercase $v$ for the velocity, as in the question and in the other answers. $\endgroup$ Jun 13, 2019 at 21:31
  • $\begingroup$ @EverydayAstronaut fine... $\endgroup$
    – ANone
    Jun 14, 2019 at 9:38

$v_{\rm injection}$ of liquids, as a rule of thumb, doesn't get higher than 10 m/s (dictated by a general limitation of pressure losses). Given that exhaust velocity could be 2500 - 4500 m/s, you can calculate the ratio.

For gaseous fuel/oxidizer it's of course different.

Now, where to get that original estimation...

  • 3
    $\begingroup$ 10 m/s sounds reasonable. If this could be backed by any kind of credible source ... and thanks for pointing out that gaseous/liquid should be clarified. $\endgroup$ Mar 30, 2019 at 15:11

For the most common engines, all of the exit velocity is “due to combustion” because the pumps are driven by fuel-oxidizer combustion.

One of the key differentiators among engine designs is what they do with the exhaust of that pump-driving combustion: how much thrust is generated from those combustion products? More is better, of course, but complexity, weight and cost matter too.

  • $\begingroup$ Good point that combustion driving the turbo pumps should be excluded. Just edited the question accordingly. $\endgroup$ Mar 30, 2019 at 15:20

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