Author Topic: Q&A Basic Thrust Equation  (Read 528 times)

Offline mead

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Q&A Basic Thrust Equation
« on: 11/01/2017 02:34 AM »
I hope this isn't one that leaves me feeling stupid but....

The basic thrust equation is given as:

F = qVe + (pe - pa)*ae

Where F is thrust, q is mass flow rate, Ve is exhaust velocity, pe is exhaust pressure, pa is ambient pressure, and ae is the exit area.

I have read, and it also makes intuitive sense that an optimized nozzle where pe=pa would produce the most thrust. However, the basic equation seems to contradict that, but I'm sure I'm missing something obvious. Everything else being equal, thrust seems like it would go up as pe goes up. Set me straight!

Online meberbs

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Re: Q&A Basic Thrust Equation
« Reply #1 on: 11/01/2017 03:43 AM »
What you are missing is the relationship between Ve and Pe. These are not independent variables. Working under the assumption that you can only change the nozzle (which is where the advice on nozzle size is relevant) increasing the exit area increases Ve and decreases Pe. The increased Ve is more useful, but if Pe goes below Pa, this breaks down due to other effects.

For an intuitive feel, this section of wikipedia has a picture that shows the effect of over or under expansion. In the underexpansion case (Pe > Pa) at the top of the diagram, you can see that the air expands more after leaving the nozzle to balance the pressure. If the nozzle was bigger, it could have made use of that extra pressure more effectively. You can also see in the overexpanded case that it basically acts like a smaller nozzle anyway, plus it can cause turbulence that can damage the engine in the extreme case.

Offline mead

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Re: Q&A Basic Thrust Equation
« Reply #2 on: 11/01/2017 06:26 PM »
Ah that makes sense...

From

Ve = sqrt((2k/k-1)(R*Tc/M)(1-(Pe/Pc)^(k-1)/k))

As exhaust pressure increases, it reduces the exhaust velocity faster than it contributes to increased thrust in the basic thrust equation. Thanks for pointing me in the right direction!

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