# Why the exhaust pressure should be equal to ambient pressure?

I've came across various sources on rocket operation basics (e.g. this one), and they all pretty much give the same equation for thrust that looks something like this: $F = \dot{m}v_e + (p_e - p_a)A$, where $\dot{m}$ is the mass flow rate, $v_e$ is the exhaust gases velocity, $p_e$ and $p_a$ are the exhaust and ambient pressures respectively, and $A$ is the nozzle area.

Then all of those sources agree that in order to maximize the thrust efficiency (exhaust velocity etc.) we should equalize $p_e$ and $p_a$, so that the second term would vanish. My question is: why?

From the naive point of view the thrust would be maximized whenever both of the terms are maximized - i.e. when the $p_e$ is actually as big as possible rather than equal to $p_a$, so that the second term would return a positive increment in the expression, not a zero.

So what am I missing here?

So what am I missing here?

You are implicitly assuming that effective exhaust velocity $v_e$ is constant. This is not the case. What you are missing is that $v_e$ in a conventional nozzle (a de Laval nozzle) tends to increase as exit pressure decreases. An underexpanded nozzle (one in which $p_e > p_a$) will have a significantly reduced exhaust velocity compared to a perfectly expanded nozzle. The slight gain in thrust due to the positive $(p_e-p_a)A$ term does not offset the significant loss in thrust due to the suboptimal value of the $\dot m v_e$ term.

• Thanks. The only thing left to clarify is why are both these factors at optimum when the pressures are exactly equalized? Jan 19 '17 at 14:07
• I can’t judge on your answer (but don‘t doubt it‘s correctness) but why is the equation then useful in the first place? Don’t you want an equation in which all essential variables that govern the problem are independent, at least in a first approximation? Jun 6 '20 at 6:20

The optimum occurs because of the following mathematical arguments. I am going to give the physical argument with the nozzle flow equations, however, the method of Lagrangian multipliers could be used to mathematically prove the optimum occurs when $(p_e-p_a) = 0$.

Anyways, as you mentioned above, the thrust for a conventional rocket is given as,

$$F = \dot{m} v_e + (p_e-p_a)A_e$$

From the derivation using control volume analysis, it follows that the first term is the momentum flux across the nozzle exit plane, and the second term is the pressure area force acting on the nozzle exit plane.

Now, we want to look at how each parameter above is dependent on the properties of the rocket thrust chamber and nozzle configuration. Rewriting the mass flow rate ($\dot{m}$) using the mass flow parameter for a nozzle with the choked condition at the nozzle throat we have,

$$\dot{m} = \frac{A_t p_1}{\sqrt[]{T_1}} \ \sqrt[]{\frac{\gamma}{R}} \left(\frac{\gamma+1}{2}\right)^{-(\gamma+1)/[2(\gamma-1)]}$$

where $p_1$ is the chamber pressure, $T_1$ is the chamber temperature, $A_t$ is the throat area, $R$ is the specific gas constant for the burned products in the thrust chamber, and $\gamma$ is the ratio of specific heats of the burned products in the thrust chamber. Notice, this parameter has no dependence as to what is taking place at the nozzle exit, or the nozzle exit area for that matter. This is simply a result of the mass flow parameter being applied to the throat choking condition, which alleviates the dependency at the nozzle exit plane. Thus, our first parameter of the first term is not influenced by the nozzle exit condition.

Now for the velocity at the nozzle exit plane ($v_e$) we have,

$$v_e = \sqrt[]{\frac{2\gamma}{\gamma-1} R T_1 \left[1-\left(\frac{p_e}{p_1}\right)^{(\gamma-1)/\gamma}\right]}$$

As observed in this expression, the lower $p_e$ we can achieve the higher the nozzle exit flow velocity we can achieve. So in the context of the momentum flux from the nozzle exit, it is desired to expand the gas to the lowest $p_e$ as possible in order to achieve the highest $v_e$ as possible.

Now for the pressure area force term, it is obvious that if $p_e$ drops below the ambient pressure $p_a$, then we will have a negative pressure area force acting on the nozzle exit plane. Obviously, this is not desirable, so there is a limit to which we wish to expand the chamber pressure $p_1$ through the nozzle to $p_e$ at the nozzle exit plane. Hence, we have two scenarios left to consider, the case where $p_e = p_a$ (perfect expansion and no pressure area force) and the case where $p_e > p_a$ (reduced momentum flux, but pressure area force). The latter condition will result in a reduced momentum flux than the case $p_e = p_a$, but we will have a positive pressure area force, instead of a zero pressure area force. The question then becomes, what is the ratio of the gain in pressure area force vs. the loss in momentum flux. The easiest way to demonstrate the optimum thrust argument is by looking at how each thrust component term grows in relation to $p_e/p_a$. Below is a plot demonstrating this relation. Clearly, the optimum thrust is achieved when $p_e = p_a$, hence, the gains from the pressure area force are not comparable to the loss in momentum flux for the case when $p_a < p_e$. • Is $A_e$ independent of $v_e$ and $p_e$? I'm guessing not since otherwise $F=\dot{m}v_e+p_eA_e-p_aA_e=\dot{m}v_e+p_eA_e+constant$ and maximising $F$ would just involve maximising $\dot{m}v_e+p_eA_e$, which is independent of $p_a$? Also are we able to say $\dot{m}$ is independent of $v_e$ even if both are dependent on $p_1$? Just asking since I've got the same question, thanks. Apr 5 '18 at 12:22

Consider it this way. Expanding the fluid to a pressure above ambient is equivalent to removing a portion of some length from the correct nozzle. The amount of potential energy the fluid possesses after the turbine stage is not entirely converted to kinetic energy if the nozzle is too short. Thus, not attaining the exhaust velocity it will attain if the nozzle was of correct length. The effect of exhaust velocity on thrust is greater than the pressure thrust.