I mean, for a turbofan: $$ T= \dot m_{air,h} (1+f) v_{e,h}+ \dot m_{air,c} v_{e,c}- (\dot m_{air,h}+ \dot m_{air,c} ) v_0+ (P_{e,h}-P_{amb} ) A_{e,h}+ (P_{e,c}-P_{amb} ) A_{e,c} $$

but $$ \dot m_{air}=\rho Av_0 $$

do I have to impose $\dot m_{air}$? in this case how the thrust expression change (in particular the terms with $v_0$)?

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    $\begingroup$ In your second equation, $v_0$ should be the speed of the air moving through the area $A$, not the speed of the aircraft, right? $\endgroup$
    – Bianfable
    Jul 30, 2019 at 18:14
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    $\begingroup$ It works the same way as it does in the air. The only difference is that there is no ram air at the intake, the compressor has to suck air in. $\endgroup$ Jul 30, 2019 at 18:34
  • $\begingroup$ @Bianfable Yes, I think it should. But in my propulsion course we often approximate $v_{inlet}=v_{aircraft}$. I think that my doubt come from this! $\endgroup$
    – AMasc
    Jul 31, 2019 at 10:07
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    $\begingroup$ The question you probably should ask when someone says something like "but $\dot m_{air}=\rho Av_0$" is why does that hold? You can turn this question around by asking yourself: what happens if the aircraft is stationary on the ground, pointed directly into the wind, with a wind speed of (say) 30 kt? Now consider the opposite case: what happens if there is no wind (perfectly calm air), but the aircraft is somehow being propelled forward at a speed of 30 kt with no friction losses against the ground? From the aircraft's perspective, what is the difference between the two cases? $\endgroup$
    – user
    Jul 31, 2019 at 16:30
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    $\begingroup$ As a practical example, you might note that somewhat modified jet engines are widely used for electric power generation: en.wikipedia.org/wiki/… $\endgroup$
    – jamesqf
    Jul 31, 2019 at 17:47

2 Answers 2


To start the engine, the fan blades have to be spun up. This is typically handled by either blowing air through them from some outside source, and/or by using an APU to generate power to drive the shaft they're connected to.

Once the engine is running, it sucks in enough air to keep going on its own.

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    $\begingroup$ And the airflow has velocity $v_0$ which differs from zero when the aircraft is stationary. $\endgroup$
    – Koyovis
    Jul 31, 2019 at 5:43
  • $\begingroup$ Therefore, should I impose $\dot m_{air}$ and calculate $v_0$ knowing the intake geometry? In this case I have one more question: $\dot m_{air}$ is fixed with altitude variation or not (or at least is it a reasonable approximation)? $\endgroup$
    – AMasc
    Jul 31, 2019 at 10:14

$v_0$ is the airflow speed through the engine, not the airspeed. As per @Bianfable’s comment.

How the engine works at standstill: the compressor sucks air into the inlet, it initiates the flow through the engine. Axial compressor blades similar to a propeller, centrifugal compressors by slinging air out towards the compressor outlet.

Maximum thrust at standstill should be a given for the engine under consideration. At standstill, if the exhaust is choked, the mass flow follows from:

  • Thrust.
  • Speed of sound at the exhaust gas temperature.
  • Exhaust pressure and area.
  • Bypass ratio.

With the mass flow given, you can compute $v_0$


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