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

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".

• I believe the <200 mph thing is more to do with the idea that the temperature rise from compression is low enough to not require temperature corrections between surface and freestream until you are above that speed. Aug 17, 2019 at 22:01
• The “it's better to move a lot of air a little than a little air a lot” is because you need to give the air certain momentum to produce the desired amount of lift and if you accelerate less air to higher velocity, you must give it more kinetic energy, which manifests as induced drag. However, the second part of the explanation is form drag and I realized that I am not sure whether it fits your description. Aug 17, 2019 at 22:07
• I'm not sure what you're asking...
– JZYL
Aug 17, 2019 at 22:49
• @Jimmy Let's carry on Hudec's line "you must give it more kinetic energy" - it begins to form a compression wave - won't this happen even a lower speeds? Aug 17, 2019 at 22:55
• Related (maybe a duplicate? I'm having a hard time following your question): aviation.stackexchange.com/q/53709/4108 and aviation.stackexchange.com/q/54808/4108 Aug 18, 2019 at 6:40

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.

• Daniel (and Jan) thanks so much, this perfectly clarifies why a larger, slower airfoil or prop is more efficient +10! Aug 18, 2019 at 5:49
• @RobertDiGiovanni, note that this explans the first part, why longer span airfoil has lower induced drag. However, it has lower induced drag at all speeds, so to see why it's optimum point is at lower speed you also need to know why the form and skin drag increase with speed, and I don't actually know any derivation of that. Aug 19, 2019 at 18:31

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.

• So unless you "trap" the higher pressure over a meaningful area (as in a heavily undercambered wing or parachute) or you go pretty fast, air flow CAN be modeled as incompressible, and pressure/drag contributions are negligible. Fine work! Aug 18, 2019 at 15:56
• @RobertDiGiovanni I believe even for parachute, incompressible is a good assumption unless we are talking about high speed spin chutes. However, you can no longer assume irrotational and inviscid.
– JZYL
Aug 18, 2019 at 19:03