In order to understand how Jet-Engines works I came across the "famous" formula for net-thrust:


$F_{thrust} = Q_{out} \cdot v_{out} - Q_{in}\cdot v_{in}$

From well known physical law about momentum this sound quite reasonable for me.

But now assume an empty cylindrical chamber with nozzle at the end, mounted statically in a wind channel:

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Regardless of any details, just by continuity equation we must have $Q_{out} = Q_{in}$ and $v_{out} \gt v_{in}$

Nevertheless, this construction would not produce a net propulsion in left direction. It it were true, we just had to put this "empty" construction on a plane...

I was thinking over that quite a time and become already a little crazy. Where is my mistake? Or is the formula above not the whole truth?

Anyway, I'm not sure if

$v_{in} = v_{air}$

although it sounds reasonably for me.

But even when I take the formula

$F_{thrust} = Q_{out} \cdot v_{out} - Q_{in}\cdot v_{air}$

a left thrust would be produced, whenever

$v_{out} \ge v_{air}$

Is this the key-factor where I'm wrong?

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    $\begingroup$ You forgot to add energy inside the duct. The way you draw it, the entrance is blocked by the smaller exit and very little Q will happen. V$_{in}$ is much lower than flight speed, so the picture needs to include what is happening ahead of the duct. $\endgroup$ Oct 2, 2020 at 8:50
  • $\begingroup$ This was intentionally, to make it absurd. I always assumed, that Vin is ~ flight speed and maybe this is part of my problem. In fact it cannot be true, because vin is not zero when the plane is on ground hold. I wonder, however, how the thrust equation can be in form F = Q(out)*v(out) - Q(in)*v(air). Shouldn't it be F = Q(out)*v(out) - Q(in)*v(in) ? $\endgroup$
    – MichaelW
    Oct 2, 2020 at 14:54
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    $\begingroup$ If your v$_{out}$ is higher than v$_{in}$ there must be some pressure drop along the duct so the air can accelerate. A positive v$_{out}$ must have enough pressure to overcome ambient pressure. So where is the much higher entry pressure coming from? This can only be something close to stagnation pressure. Now you have the large frontal area with stagnation pressure working on it – a lot of drag that needs to be included in the model. And stagnation pressure implies v$_{in}$ = 0, so for any Q to happen your frontal pressure is a bit below stagnation pressure. v$_{out}$ is below flight speed $\endgroup$ Oct 2, 2020 at 15:52

2 Answers 2


You forgot to take pressure into account. This formula works assuming your nozzle is well adapted and the exit flow is at atmospheric pressure. In your case the velocity is increased but the pressure drops as stated in Bernoulli's equation. $$P_0+ 0.5\rho V_0^2 = P_1+ 0.5\rho V_1^2$$ This means that the pressure force on your nozzle will counteract the thrust provided by the fluid acceleration.

See this explanation your general thrust equation if the pressure are different is $$\text{Thrust} = (Q * V)_e - (Q * V)_0 + (P_e - P_0) * A_e $$ Given $A_e$ is the exit aera of the nozzle.

  • $\begingroup$ I understand, that the pressure on the right side is not necessarily the same as the pressure on the left side. Bernoulli's law would give the pressure after some calculation. But then I do not understand, why we can assume that the exit flow pressure should be atmospheric pressure when we have a real jet-engine? What is the difference in terms of aerodynamics as compared to the empty case? $\endgroup$
    – MichaelW
    Oct 2, 2020 at 8:32
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    $\begingroup$ The engine is adding some work to the airflow, increasing the pressure internally, thus when applying Bernoulli you can reach atmospheric pressure at a higher velocity on the exit plane. $\endgroup$
    – MaximEck
    Oct 2, 2020 at 8:36
  • $\begingroup$ With this additional term it gives sense. In my simple case I get an under-pressure on the right side, pulling against the thrust from the fluid. However, is atmospheric pressure necessarily reached at the exit plane? Why is the pressure term not given in the formula I have linked to? $\endgroup$
    – MichaelW
    Oct 2, 2020 at 8:46
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    $\begingroup$ Well it's wikipedia... There are often assumption not written in the simplified formula given. Atmospheric is not always reached but the goal is to come as close as possible to it tanks to the nozzle as this would be the most efficient. Supersonic jet engine are the perfect example with the mach diamond in the exhaust proving that the exit pressure is not similar to the surroundings. $\endgroup$
    – MaximEck
    Oct 2, 2020 at 8:52
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    $\begingroup$ For basic understanding the link to the NASA web site in the answer should be sufficient ? You've got links to the begginer propulsion handbook in the text grc.nasa.gov/WWW/k-12/VirtualAero/BottleRocket/airplane/… $\endgroup$
    – MaximEck
    Oct 2, 2020 at 9:10

Ah, the joy of simplicity. The diagram is merely drawn backwards. A good jet has equal pressure at intake and exit. The exit has more volume and higher velocity than the inlet after compression and addition of heat. With jets, the fuel also contributes a slight increase in mass.

As always, start with the simplest case, the ram jet, and try to make it work. A rocket is even easier, there is no air inlet requirement at all, fuel and oxidizer are added to the combustion chamber.

Using the "rocket" analogy, the compressor is holding a "plug" of more dense air over what is essentially a combustion chamber. Perhaps easier to understand it if one pictures the inlet as an even smaller diameter of liquid oxygen.

We note in this system: the inlet has the higher pressure (equal to the pressure on the walls of the container), and the outlet has the higher velocity. The larger outlet stream drives the turbine, which drives the compressor. It is interesting to note the early axial flow Jumo 004 had only has one set of turbine blades driving its compressor. A two stage turbine design, the Jumo 012, with more than double the thrust, was to be the follow on.

  • 4
    $\begingroup$ I did not vote on your answer but I can see why some people downwoted it, your not really answering the question ^^... Furthermore if the jet exhaust is subsonic, you will have a convergent nozzle, so "the exit has more volume and higher velocity" is wrong in that case. $\endgroup$
    – MaximEck
    Oct 2, 2020 at 13:04
  • $\begingroup$ @MaximEck good point about the nozzle. Comparing the diameter of the narrowest point of the nozzle to the air inlet (narrowest point of the compressor), the nozzle must be wider because T is higher. The wider diameter of the exit also allows greater torque (on the longer turbine blades), enabling fewer to drive the compressor blades. $\endgroup$ Oct 2, 2020 at 13:30

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