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Please refer to this MIT page which I used for my calculations: https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node86.html

I calculated my aircraft propeller's prop, thrust, and torque curves, and subsequently the efficiency at a certain operating point (given the input thrust requirements and airspeed), by dividing the useful power (=Tu_0) by the input power (torqueRPS), where u_0 is the flight speed.

Now, in the propeller plane, the airspeed is u_disk according to Actuator Disk Theory, and can be calculated using the formulas and the known info. When I calculate the input propeller power according to another formula on that page [Power in = Thrust * u_disk], I get an answer not equal to the one using the torque*RPS formula. Where did this extra power go? I was expecting them to be the same, and the aerodynamic losses to be factored into the useful power being lower.

In my example, the flight speed is 3 m/s. Due to the conditions of the aircraft and environment, this requires 42N of thrust, which requires 4.5N-m of torque at 1503 RPM. This is all due to the propeller characteristics. Now, the useful work is T * airspeed = 42 x 3 = 126W.

Using the calculations on that page, we can get the accelerated speed at the disk is 11 m/s. So the input power (According to the page) is 42 x 11 = 464 W. This is different from the conventionally calculated input power of Torque * RPS = 4.5 * (1500/60) = 700 W.

dumbpropnerd
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Defining "useful" as moving the plane forwards, the rate at which useful work can be done is equal to whatever power which has been imparted to the fluid, i.e thrust x velocity at disc. This is propeller power. The rate at which useful work is done which is equal to whatever power which has gone into actually making the plane move forward, i.e thrust x freestream velocity (since freestream velocity and airspeed are interchangeable depending on the frame of reference). This is propulsive power

Consequently there are two different measures of efficiency, the propulsive efficiency which is a measure of how efficiently the plane uses the propeller power and the propeller efficiency which is a measure of how efficiently the propeller uses shaft power. The equation on the MIT webpage is for propulsive efficiency but you want to know how efficiently your shaft power (2π x torque x RPS) was used, therefore you have to calculate propeller efficiency so you don't use that equation.

Aviator S
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  • Thanks. I meant "2*pi" which is why in my numerical calculation, the numbers work out. Sorry for the mistype, I couldn't get it to edit my comment. – dumbpropnerd Oct 05 '18 at 00:00
  • However, in the MIT link I posted, it defines useful power out as thrust x upstream velocity u_0. That's my confusion. I think input power should be the same using the torque calculation and the velocity at disk calculation... – dumbpropnerd Oct 05 '18 at 00:02
  • @dumbpropnerd sorry I misunderstood the definitons you were using, I amended my answer – Aviator S Oct 05 '18 at 16:33
  • Looked at your amended answer. That's actually the conclusion I initially arrived at, hence I asked "where does the extra power go" between the shaft, and the available power to be used on air (some of which goes to accelerate the air and some goes to generate thrust). But before it becomes "available power" why is it already lower than the shaft power? That's my confusion. I think that they should be the same...I am wondering if the discrepancy is from me using a 3D analysis to find the torque required, and the Disk Theory not accounting for 3D effects. – dumbpropnerd Oct 05 '18 at 19:24
  • The extra power goes into creating tip vortices and swirling the flow, neither of which provide useful thrust therefore, that power is a loss. – Aviator S Oct 05 '18 at 22:25
  • I believe that is part of the loss, but the accelerated air is also part of the loss. According to this link I just found: https://s2.smu.edu/propulsion/Pages/efficiency.htm it shows the two methods of calculating efficiency I used, and equated them as being equal to each other. Thus, in my example, I should have 700 W for shaft power, 700W for thrust X disk velocity (assuming no shaft-losses) and then directly 126 W for useful work with no step down in the middle. From within that 700W, the losses towards accelerating air as well as swirl and vortices occur, I think. – dumbpropnerd Oct 05 '18 at 22:33