# 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 at 11:50

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.