I've been looking for this for quite some time now, and I can't find anything other than calculations where they already assume a chamber pressure. So my questions are two:

  • What are the equations for calculating the chamber pressure for a bipropellant engine? In case it's the opposite, how do you find out the needed propellant flow rate so that you achieve the design chamber pressure?

  • When you have a pressurized gas thruster, that is, with no combustion, what do you have to take into account in order to calculate the parameters of the injector?

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    $\begingroup$ Your second question is already answered at space.stackexchange.com/questions/21577/… $\endgroup$ Commented Jun 2, 2017 at 15:10
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    $\begingroup$ No, it's not. That's why I am asking again. Yes, you have a regulator that you can control, but you should be able to calculate before even running the engine how much feed pressure you need. And I doubt you would feed it with the chamber pressure you want, cause the chamber is open, therefore you won't achieve as much pressure. $\endgroup$ Commented Jun 2, 2017 at 19:16
  • $\begingroup$ If you're not satisfied with the answer(s) because you think they're wrong, asking a new question isn't the way to handle it. With more rep you can add a bounty for better answers, or you can edit the question to be clearer. Otherwise, we're just going to give the same answers again, or nothing. $\endgroup$ Commented Jun 2, 2017 at 19:19
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    $\begingroup$ What else can I do? It's not like I can add a bounty, and I don't really understand what's not clear about the question. $\endgroup$ Commented Jun 2, 2017 at 19:22
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    $\begingroup$ With no combustion there's no point to having a combustion chamber and as result no combustion chamber pressure. The pressure fed from the regulator IS the pressure reaching the nozzle, directly. $\endgroup$
    – SF.
    Commented Mar 18, 2020 at 13:49

3 Answers 3


The existing answers to this question are quite poor.

Chamber pressure is a design choice. While the equations for rocket engine performance can make it seem like chamber pressure is a function of mass flow or throat area, it is really the opposite: these are a consequence of your selection for chamber pressure. A typical liquid rocket engine features deflagrative combustion, that is to say that there is negligible pressure gain produced by the combustion. Instead, all the pressure in the chamber is provided by the injection. So to increase your chamber pressure, you run your pumps faster, or you pump more helium into your propellant tanks. Both of these options increase the pressure upstream of the injectors. There is some pressure loss (~10%) through the injector, and then that is your chamber pressure.

Engine designers will typically start with a thrust value that they want their new engine to meet. The thrust is directly proportional to chamber pressure via the thrust coefficient ($C_f$) with the following equation:

$$C_f \equiv \frac{F}{A_t P_c}$$

The thrust coefficient has a nasty formulation, but it is constant for a given nozzle at a given altitude. Here $F$ is the engine thrust, $A_t$ is the throat area, and $P_c$ is the chamber pressure. So a designer, after choosing their desired thrust, can calculate or make an educated guess on the thrust coefficient their nozzle will provide (typical ranges are 1.2 to 1.8), then compute the required product $P_c A_t$. If they want a small engine with a small throat area, they need a larger chamber pressure, or they can get away with a lower chamber pressure if they use a larger throat area, but you can see that to maximize thrust, you would want high chamber pressure and a large throat area. As I mentioned before, they then configure the propellant supply to provide whichever chamber pressure they choose.

Regarding mass flow, this is the critical factor for engine efficiency (to provide the same thrust with less mass flow is the bee's knees). Just like thrust, it is directly proportional to the chamber pressure, but inversely proportional to the square root of temperature. And, just like thrust, there is a parameter relating the mass flow to the chamber pressure, called the characteristic velocity (or cstar):

$$C^*\equiv \frac{A_tP_c}{\dot{m}}$$

The characteristic velocity ($C^*$) is constant for a particular propellant combination and combustor and tells the engine designers how much mass flow ($\dot{m}$) a particular chamber pressure and throat area combination will produce. To reduce the mass flow, you desire a higher characteristic velocity. The formulation is, again, very nasty, but higher values are achieved with increased flame temperature and lighter molecular weights of the combustion products. This is why Hydrogen is a popular choice for propellant, because it offers both of these advantages.

These two parameters I have mentioned, the thrust coefficient ($C_f$) and the characteristic velocity ($C^*$), represent the performance of the nozzle and combustion chamber respectively. Together, they give a measure of the entire rocket engine. You can formulate the specific impulse (a key engine performance parameter) by combining these two values:

$$I_{sp}= \frac{F}{\dot{m}g} = \frac{C_fC^*}{g}$$

Here $g$ is the acceleration due to gravity at sea level. This discussion highlights how and why designers choose particular chamber pressures (though I didn't include many nuances in the tradeoffs of this choice).

A final note on the injection system. As I mentioned, the two primary methods for providing the injector head (upstream pressure) are pumps or tank pressure. To yield the benefits of high chamber pressure, you need beefy pumps and/or beefy tanks and all that implies.

A final final note on the other answers: The answer of Declan Nnandozie suggests that the isentropic flow relationship between pressure ratio and exit Mach number would be used to compute chamber pressure. This is total nonsense. While you can compute your exit Mach for a given nozzle, and you can thus compute the nozzle pressure ratio, you can't use this to compute your chamber pressure. The exit pressure (he calls environmental pressure) is independent of the environmental pressure. His suggestion only works if you are dealing exclusively with perfectly expanded nozzles (with exit pressure equal to ambient pressure) which is a very narrow operating condition.

The answer of Kaan Guven is more on track, but misses the key design process, which is to say that thrust is the starting point for any rocket design.

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    $\begingroup$ Agreed! Nice answer. $\endgroup$ Commented Feb 16 at 18:27

By chamber Pressure, you mean stagnation/Steady state Pressure.

Here's the Equation

$$ P_0 = p\left[1 + \frac{1}{2} \left(k - 1\right)M^2\right]^{\frac{k}{k-1}} $$

$P_0$ = Stagnation Pressure or Steady-state Pressure your chamber Pressure

$p$ = let's call it environmental pressure (Atmospheric pressure, but it changes with altitude)

$k$ = ratio of specific heat at constant pressure to specific heat at constant volume

$M$ = is the Exit velocity in Mach number

The equation above assumes that the Mach of the chamber/before the nozzle throat is negligible

Ask me more if you still confuse

You might ask me what is Stagnation Pressure and Steady-state Pressure?

And you might question me why I used the word Environmental pressure?


I can only answer your first question. You can use the following equation:

$$ \dot{m} = \frac{P_{c}A_{t}}{C^*} $$ where $$C^* = \frac{\sqrt{RT_{c}}}{\sqrt{\gamma}(\frac{2}{\gamma+1})^\frac{\gamma+1}{2(\gamma-1)}}$$

Edit: Okay, since a lot of people ask I will try to explain this a bit better.

1_ By the assumptions of isentropic flow and choked throat conditions, you get a relationship between mass flow rate, and pressure/temperature/throat area/ratio of specific heats. Chamber temperature is the adiabatic flame temperature of the gas mixture. This also changes with pressure. You can use Gaseq or NASA CEA package, or Canterra. There, you will also see the ratio of specific heats change.

When you make a design, you should define your criterias. For example you may want to start making a safe, and relatively cheap rocket. In that case the clever thing to do is to pick a pressure, and throat area. Then, you can easily calculate the required mass flow rate for the conditions that you first assumed. Or, maybe you are a student and want to make some measurements on a small engine. Then your professor might say, our laboratory can tolerate a mass flow rate of 1 kg/s. Then you fix mass flow rate and iteratively find the other parameters.

2_ It is the same principle. You are trying to use the pressure as much as you can. So, you need a throat section first. There you accelerate your gas up to sonic speed. If the pressure is still enough you need a nozzle and further expand your gas, and decrease the pressure. But be careful. Your gas was cold in the first place. So when you decrease its pressure it gets cold. You don't want your gas to condensate. Then expand it until it reaches the condensation temperature. There you will take all you can take from a cold gas thruster. N.B. cold here, in the jargon, referes to non combustive propellant flow.

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    $\begingroup$ Could you give a source and/or define the variables? $\endgroup$ Commented Jun 1, 2018 at 13:11
  • $\begingroup$ link. Here you can find the notation I use. Also the main source for introduction to rocket propulsion is Sutton's Rocket Propulsion Elements. You can find the same formula if you check equation 3-32. $\endgroup$ Commented Jun 1, 2018 at 13:37
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    $\begingroup$ Temporary, reversible -1 until the source cited is added back into the answer and the variables explained within the answer post. In Stack Exchange answers need to be fairly stand-alone so that if links break the answer does not loose value, and not everybody has a copy of Sutton in their back pocket. Thanks! $\endgroup$
    – uhoh
    Commented Jun 1, 2021 at 23:51
  • $\begingroup$ -1 (Also temp/rev) This answer doesn't say what any of that stuff means, as @uhoh says. Without that, it's a bit useless, sorry. $\endgroup$
    – SusanW
    Commented Aug 16, 2021 at 11:56

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