# Turbofan: Efficiency wrt. the bypass ratio: “a lot of slow air > a little faster air”?

Reading / viewing up on how jet engines work, this video explains at the 9:02 mark that, for turbofan engines, ".. it is more aerodynamically efficient to have a lot of air moving relatively slowly rather than a little air moving relatively quickly." I.e. a lot of air bypasses the combustion chamber and is only accelerated by the main frontal fan.

This claim has me puzzled, given that kinetic energy is dependent on the square of the velocity, i.e. $E_k = \frac12 m v^2$, I would have guessed the inverse. Does "aerodynamically efficient" mean something very specific here?

I am not an aerodynamics specialist, so the following is almost certainly a huge oversimplification (or maybe downright wrong), but I think it might help with intuition.

Suppose you have an amount of energy $E$ available to spend, and you are trying to accelerate an object of mass $M$. Suppose you can impart the energy in the form of kinetic energy to an air mass of mass $m$. What speed will this accelerate the mass $M$ to?

By energy and momentum conservation, you have $$E=\frac{1}{2}m v_m^2+\frac{1}{2}M v_M^2$$ and $$mv_m+Mv_M=0$$ giving $$v_M=\sqrt{\frac{2 e m}{M (m+M)}}$$ which as a function of air mass $m$ looks like this:

Plot[Sqrt[(2 e m)/(M (m + M))] /. {e -> 1, M -> 100}, {m, 0, 500}, PlotRange -> All]


which means that according to this simple model, it's more energy efficient to use large values of $m$, ie, to eject large volumes of slow air rather than a small volume of fast air.

Aerodynamic efficiency is not necessarily a measure of how much energy you impart to the ejected air per fuel consumption, but rather is how much momentum you impart to the aircraft per fuel consumption.

What you're after is thrust, which is force, which is momentum per second. To get the same force, which is $F = ma$, you can either use a large mass with a small acceleration, or a small mass with a large acceleration. Which requires more energy? (The small mass with the large acceleration, because the imparted velocity is higher. if $v = at$, $2ma = m2a$, but $2m(v)^2/2 < m(2v)^2/2$, because of the squaring of velocity.)

That's why helicopters have long rotor blades, to move a lot of air slowly, instead of little fans, to move a small amount of air quickly.

From an engineering standpoint, the propulsive efficiency (which is relevant for engines, rather than the aerodynamic effiency) is defined as:

$$\eta_{prop}=\frac{\mathrm{Power Available}}{\mathrm{Jet Power}}=\frac{P_a}{P_j}$$

In this, the jet power is the amount of energy added to the flow:

$$P_j=\frac{1}{2}m V_j^2-\frac{1}{2}m V_0^2=\frac{1}{2}m \left(V_j-V_0\right)^2=\frac{1}{2}m \left(V_j-V_0\right)\left(V_j+V_0\right)$$

And since we know that that $T=m \left(V_j-V_0\right)$ We get:

$$P_j=\frac{1}{2}T\left(V_j+V_0\right)$$

The power available follows from $P=Fv$, or for our engine:

$$P_a=T V_0$$

Filling both in our effiency equation gives:

$$\eta_{prop}=\frac{P_a}{P_j}=\frac{T V_0}{\frac{1}{2}T\left(V_j+V_0\right)}=\frac{2 V_0}{V_j+V_0}=\frac{2}{1+\frac{V_j}{V_0}}$$

This directly shows that increasing the jet velocity reduces the value of $\eta_{prop}$