9
$\begingroup$

I am studying the thrust distribution of jet engines. But I'm now a bit confused.

In Rolls-Royce's "The Jet Engine" book, enter image description here http://aeromodelbasic.blogspot.com/2012/05/thrust-distribution-distribution-of.html

At the start of the cycle, air is induced into the engine and is compressed. The rearward accelera- tions through the compressor stages and the resultant pressure rise produces a large reactive force in a forward direction. On the next stage of its journey the air passes through the diffuser where it exerts a small reactive force, also in a forward direction

I understand the first part of the paragraph that the compressor is providing forward thrust, as it is pushing (so compressing) air rearward. But why is the diffuser also providing forward thrust? And also why the nozzle is providing rearward thrust?

The similar conculsion is also shown here: http://www.pulse-jets.com/phpbb3/viewtopic.php?t=2183, that the diffuser is providing positive thrust by calculating the pressure force.

From my understanding of basic fluid mechanics, shouldn't a nozzle be providing forward thrust, like the sprinkler in the garden or a fire hose? And shouldn't a diffuser be providing rearward thrust, as the outlet speed is slower than the inlet speed, and so m dot X (v - u) is negative?

What's wrong with my understanding?

It seems like I don't know when I need to add a pressure force term P_out A_out - P_in A_in, in addition to the change of momentum term (m dot x (v-u)), while calculating the force of a control volume.

Improve this question
$\endgroup$
5
  • 2
    $\begingroup$ One point where you are wrong is " like the sprinkler in the garden or a fire hose" - jet engine is not a rocket engine. The air exerts force on the engine (and the entire airplane) at points of contact with the engine at an angle where its decompression pushes the engine forward - like the rears of the blades of the compressor. It can't accelerate the jet just by passing through. But I'm stumped on the diffuser too. $\endgroup$
    – SF.
    Jun 12, 2018 at 15:30
  • $\begingroup$ Eminently related aviation.stackexchange.com/q/33068 although not a full explanation of what you are asking. $\endgroup$
    – Salomon Turgman
    Jun 13, 2018 at 0:38
  • 4
    $\begingroup$ I suspect the numbers on the diagram in the Rolls-Royce book were from an analysis of a real engine design and combined with a simplified picture of a typical engine. One of the main lessons from the picture (at least for mechanical engineers) is that the resultant engine thrust is the difference of two large numbers and the structure has to withstand those large forces, not just their small resultant. It's possible that some air is being extracted from the diffuser for cooling, powering auxiliary equipment, etc, which are not shown in the picture but were included in the calculations. $\endgroup$
    – alephzero
    Jun 13, 2018 at 14:41
  • $\begingroup$ It looks like maybe the exit nozzle has a reduced diameter, which would explain the sign of the net thrust there. $\endgroup$
    – Carl Witthoft
    Jun 13, 2018 at 15:18
  • $\begingroup$ You might get answers faster over at aviation.SE . $\endgroup$
    – Carl Witthoft
    Jun 13, 2018 at 15:19

1 Answer 1

Reset to default
1
$\begingroup$

In your force/momentum balance equations you always can include the pressure of the gas. The only time you can ignore it, is if the difference in $PA$ is small compared to other forces.

For example in a diffuser at the entrance you have a lower pressure and lower area than at the exit. This results in a significantly lower $PA$ at the entrance, more than making up for the reward momentum loss, resulting in net forward thrust.

You could calculate the same thrust value by integrating the pressure across the wall of the diffuser. This might result in a more intuitive sense of why the diffuser gets pushed forward, while a converging nozzle will be pushed rearward (and the diverging portion of a supersonic nozzle will be pushed forward again)

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.