Does the explanation of why slower airfoils are more efficient involve less energy loss from heating or compression of the working fluid?

Question

We read time and time again, how more efficient slower airfoils such as helicopter rotors, props, and sailplane wings are compared with rapidly rotating fan blades or faster (still subsonic wings). We also read where some people consider compression a negligible factor below a certain speed, even to the point of considering airflow < 200 mph as "incompressible"!

Yet looking at this commonplace and ubiquitous phenomena (indeed, while your prop is putting out 600 lbs of thrust, your wing is creating 4x more lifting force), is there a mathematical explanation for this involving the ideal gas law: PV=nRT?

Can we visualize creating and maintaining a compressed "bow wave" ahead of a moving surface involves substantial amounts of energy even at lower speeds?

I have yet to fully understand why "it is better to move a lot of air a little than a little air a lot".

Answer 1

So you asked a whole bunch of things, and I'm not sure I really understand most of what you are asking, but I can maybe clarify this one part for you

I have yet to fully understand why "it is better to move a lot of air a little than a little air a lot".

The equation for thrust is just Newton's second law F=$\dot m \Delta$V, where $\dot m$ is mass flow rate (kg/s) and $\Delta$V is the change in velocity of the air (exit velocity minus inlet velocity). So, if you move 1 kg/s of air at 10 m/s, or you move 10 kg/s of air at 1 m/s, the thrust is the same. Doesn't matter.

But, the equation of kinetic energy (well power really in this case) is $P=(1/2) \dot m (\Delta V)^2$. So if you move 1 kg/s of air at 10 m/s, it takes $0.5*1*10^2$ = 50 W of power to do that. But if you move 10 kg/s of air at 1 m/s it takes $0.5*10*1^2$= 5W of power to do that. Exact same amount of thrust, but with literally 10X less energy required. i.e. you need to burn way less fuel if you can move a lot of air at a low velocity than if you move a little bit of air a high velocity.

Answer 2

On your question of why "moving a lot of air a little than a little air a lot", you can use the Momentum Disc Theory, applicable to wing or propulsor. Assuming incompressible flow, the main result is:

• Thrust: $T=\dot{m}(u_e-u_0)$
• Input power: $P_{in}=\frac{1}{2}\dot{m}(u_e^2-u_0^2)$

where $u_0$ is the free-stream speed, $u_e$ is the airspeed after being accelerated by the wing, propeller or compressor (exhaust speed), $\dot{m}$ is the mass flow rate through the device.

As you can see, thrust is linear to both mass flow rate and exhaust speed, but input power is square of exhaust speed. That's why it's more efficient to move a lot of air a little than a little air a lot.

On your question of compressibility, for airspeed less than Mach 0.3 (around 200mph at sea level), there isn't much to be gained by assuming the flow is compressible. Yes, air is compressible, but so is everything else in the world. The difference is, if an air molecule is moving slow enough, the air molecules ahead of it can get out of its way without being squished. And the mechanism has to do with speed of sound. The faster it is, the less "time" they have to do that. Super dumbed down, but that's the intuition.

For a mathematical proof, you can follow this NASA page, whose result I will cite here. Assuming a compressible, isentropic, non-viscous fluid, we have:

$$-M^2\frac{du}{u}=\frac{d\rho}{\rho}$$

So for M=0.3, 1% change in airspeed is only 0.1% change in density. For most engineering purposes, negligible.