# Torque and engine torque after aplying prandtl tip loss factor

I have a question regards to the following formula related to Blade Element Momentum theory for the application to calculate on airplane propellers:

$$dQ=4\pi r^3 \rho V_\infty(1+a) a' \Omega F dr$$

where

• $$Q$$ = angular momentum
• $$r$$ = radius
• $$V_\infty$$ = aircraft's forward velocity
• $$a$$ = axial inflow factor
• $$a'$$= angular inflow factor (swirl factor)
• $$\Omega$$ = angular velocity $$\left(\frac{2\pi n}{60}\right)$$
• $$n$$ = rotations per minute
• $$F$$ = prandtl tip loss factor

As a result of using the tip loss factor which are numbers between 0 and 1 the end result of the calculated torque Q is lower. I have sincere trouble understanding what this means. Either the propeller runs with less resistance meaning that the propeller will run faster (higher rpm's) while keeping the engine power constant. (Shaft torque is equal) Or There is a loss in torque which means the engine has to deliver a higher torque and so the propeller runs slower (rpm's go down) while keeping the engine power constant. (Shaft torque is then not equal)

Which is the case? If it is the latter, then how is the torque calculated that the engine needs to be able to deliver?

In case of a wind turbine I somehow find this much easier to understand. There is a certain amount of power delivered by the wind. After the tip loss correction has been applied the resulting Q can directly be used to calculate the expected power output.

• Do you have a source for that equation? Commented Mar 27, 2023 at 15:10
• Commented Mar 27, 2023 at 16:50

You answered your own question by describing the effect of tip losses on the power output of the wind turbine.

But, like so many, the "tender trap" is getting stuck on the notion that somehow, engine "power" is constant.

Using pure logic, tip losses work both ways: affecting power from the airflow to drive a generator, or from the engine to move air.

So, when calculating torque required from the engine without tip loss factor, and comparing to calculations with tip factor, it becomes apparent that more torque must be applied, resulting in a higher RPM, to generate the desired thrust$$^1$$.

How to increase RPM?

Step on the gas (increase throttle)!

Torque is Newton-meters. The pistons give Newtons, the radius of the crankshaft is meters. Increased RPM is the result of the higher torque of the motor, and the drag of the propeller, finding a new equilibrium angular velocity.

What does this all mean?

Just as, theoretically, stronger wind is needed to generate the same power with tip losses, more fuel is needed to generate the same power with tip losses. (one reason why modern windmills are very high aspect ratio).

So when calculating Thrust Specific Fuel Consumption, expect a loss of efficiency, $$\eta$$, due to tip losses.

$$^1$$ or ... prop AoA can be increased.

• Thanks this is great info. Keeping the engine power constant is what I do in my calculation model. I've done what you suggested, instead keeping the thrust constant. Indeed model shows increase in needed power and decrease in efficiency. So I guess the model is correct after all. I did expect a higher impact from correction factor though. With this prop approx 2% increase in needed power. It's just so confusing that Q increases when I keep the engine power constant, although thrust and efficiency are going down as well indeed. Commented Mar 27, 2023 at 16:57

As a result of using the tip loss factor... torque Q is lower. I have sincere trouble understanding what this means.

The effect of the tip loss factor is to model the more or less sudden drop in lift which happens close to the tip of the blade (picture taken from this answer):

So, being all the other parameters the same, including the tip loss factor in the calculations will result in less torque and less thrust too, exactly because the tip loss factor models this reduced lift at the blade's tip.

To compensate for this reduction and to get the same total thrust, pitch needs to be incremented with the relevant even higher increment of drag and therefore of needed torque. Obviously higher pitch also implies being closer to stall conditions. In general a higher number of blades (at constant solidity and disk loading) smooths out this effect.