Assuming we have a convergent nozzle I've read that the maximum thrust is achieved just in the moment the nozzle exit -minimum area- is chocked, i.e., the nozzle is adapted in the sense the pressure in the exit area equals the ambient pressure. How can I demostrate this with formulas?

  • $\begingroup$ With a subsonic aircraft and a choked convergent nozzle, the gas pressure in the exhaust will be higher than ambient pressure. What exactly do you want to catch in equations? Thrust, efficiency, exhaust pressure? $\endgroup$
    – Koyovis
    Jun 7, 2017 at 4:56

1 Answer 1


Let's look at two cases, choked exhaust and complete expansion:

1. Choked exhaust

With a convergent exhaust pipe, the jet engine thrust reaches a maximum at the speed of sound of the exhaust gas stream.

The velocity of the gas flow increases if it was subsonic at the entrance of the pipe. In a convergent nozzle the maximum gas exhaust velocity is M = 1, the speed of sound at the temperature of the hot exhaust gas. With a choked exhaust, at M = 1 in the exhaust outlet, the static pressure is higher than ambient pressure.

The exhaust area needs to be reduced until the gas exit velocity is M = 1, which at for instance 800 ºC is 657 m/s. The pressure $p_e$ at the exhaust outlet will then be:

$$ p_e =\frac {\dot{m}\cdot R \cdot T_e}{V_e \cdot A_e }$$

which is greater than the ambient pressure $p_0$.

The net thrust $F$ of a pure jet engine is $$ F = \dot{m} * (V_e - V_0) + A_e * (p_e - p_0) $$

$R$ is the gas constant. Parameters you need to know:

  • outlet mass flow from the turbine $\dot{m}$ in kg/s

  • gas outlet temperature $T_e$ in ºK

  • speed of sound at $T_e$ in m/s, which for a choked exhaust is equal to $V_e$

  • outlet area $A_e$ in $m^2$

  • airspeed $V_0$ in $m/s$ and ambient pressure $p_0$ in $N/m^2$

If we use the following example of an aircraft flying M 0.85 at 30,000 ft, choked converging exhaust, exit area 0.1 $m^2$, mass flow 70 kg/s, exhaust temperature 1073 K, we get:

F = 70 * (657 - 258) + 0.1 * (328,106 - 30,100)

= 27,962 N from kinetic energy + 29,801 N from pressure difference, about the same amount.

2. Complete expansion

$p_e$ is now equal to $p_0$. This condition occurs if total pressure at the turbine exit $p_{Tt} \leq \epsilon_{kr} * p_0$, with $\epsilon_{kr}$ for a hot exhaust gas being around 1.95.

Analogous to the choked exhaust case: $$ A_e =\frac {\dot{m}\cdot R \cdot T_e}{V_e \cdot p_e }$$

For the same conditions at 30,000 ft follows: $A_e$ = 1.09 $m^2$

and F = 70 * (657 - 258) = 27,962 N

The net thrust in this case is a lot lower because the turbine exit pressure is lower than in the choked case, and therefore the propulsive power of the jet engine is lower. Usually, in the case of a turbojet the turbine exhaust pressure $p_{Tt}$ is a lot higher than $\epsilon_{kr} * p_0$, which will lead to the choked exhaust case above.

Turbofans with a high bypass ratio have a low enough $p_{Tt}$ to allow complete expansion, most of the gas generator power being used for the bypass air compression.


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