# Why this formula says jet engine is efficient at any speed?

Wikipedia gives the following equation for the efficiency of the jet engine, here:

$$\eta_p = \frac{2}{1 + \frac{v_e}{v}}$$

Where V is the aircraft speed and Ve is the exhaust speed. It is backed by some reference to the book I have no access to so cannot clarify what do they want to say by that.

I am still deeply puzzled, how it could be the efficiency increases all the time with the increase of the aircraft velocity, even after it exceeds the exhaust speed many times. I think I have read somewhere that the efficiency declines and not keeps growing after the aircraft speed approaches exhaust speed.

• If you look at the plot in the Wikipedia article, you will see that the propulsion efficiency for jet engines is only defined up to $v = v_e$. – Bianfable Apr 29 '20 at 11:50

## 2 Answers

This is the result from the simple momentum balance. In order for the propulsor to produce thrust, the exit speed after the propulsing element ($$v_e$$) must be higher than the incoming speed ($$v$$):

$$T=\dot{m}(v_e-v)$$

where $$\dot{m}$$ is the total mass flux through the propulsor.

So the correct way to read the efficiency formula is: the closer the exit speed is to the incoming speed, the more efficient it is! Therefore, the more mass flux you can generate at a smaller speed difference, the more efficiency you can attain without sacrificing the thrust. This is the overarching reason why bypass ratio makes engines more efficient.

This is the so called Propulsive efficiency! It is the main factor for new engines. The smaller the difference between the flight velocity and the jet stream velocity, the higher the Efficiency. Therefore you want a big Fan, which produces a small velocity difference (on big crosssections).

https://en.wikipedia.org/wiki/Propulsive_efficiency