Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Propeller and thrust

  1. Nov 10, 2006 #1
    :confused:

    I am working on a project where I need to understand flight physics. I am having some trouble getting the whole torque vs power vs thrust thang.

    I'm basing my project off a existing ultralight, the LightSport Ultra from Free Bird Innovations. Here are the specs:

    530lbs gross weight
    121 sq ft wing area
    70" 2-blade Warp Drive propeller
    Rotax 447 dual-carb, 2-stroke, 41.6hp engine
    850fpm climb rate
    55mph cruise speed, 62mph top speed

    these are the specs for the engine:

    41.6hp / 31kW @ 6500rpm
    47Nm / 34.7 lb-ft @ 6000rpm
    6800rpm max

    So the way I get it is that the engine spins the shaft, which turns the propeller (assuming direct drive). The faster the propeller turns the more thrust it generates.

    If you look at the engine above, it generates 34.7 lbft at 6000rpm. Using power = (torque x speed)/5252 you get 39.6hp. How do I know this is enough torque to spin my propeller? How do I know this is enough power to spin the propeller at 6000rpm?

    Now let's say I drop in a different engine, a Rotax 582 with 65hp @ 6500rpm and 55.3 ft-lb @ 6000rpm and a 6800rpm max. Obviously this motor cannot spin the propeller any faster, because it has the same max rpm as the Rotax 447. But it has more torque, so it can spin a bigger prop? Or a prop with more blades?

    Basically, I just want to know for a given propeller how much torque and power I need to turn it at a given speed, and what thrust it will give me at that speed. Any help would make me very happy! :smile:
     
  2. jcsd
  3. Nov 10, 2006 #2
    size and pitch [twist] of the prop control the amount of air moved per rev

    I think like a boat prop the HP is more important then torque for the prop size
    as most motors torque peak is at a much lower RPM then HP peak

    ask a prop maker what they use
     
  4. Nov 13, 2006 #3
    Right. But for a specific prop diameter and pitch how do you determine what torque you need to spin it at a certain speed?
     
  5. Nov 13, 2006 #4
    Again, I would talk to a propellor maker. they can help you out. Or, Google "Propellor design" and I think you'll find what you are looking for.:cool:
     
  6. Nov 16, 2006 #5
    I have been working on something similar...replacing the engine in an ultralight with an AC motor. So I have been looking at flight physics a lot.

    I have found plenty of help on remote control sites, since many RC planes use electric motors.

    There is a commonly used formula by Bob Boucher in his book "The Electric Motor Handbook". It is:

    Power = k * P * D^4 * N^3 where:

    Power is in Watts
    P = prop pitch in feet
    D = prop diameter in feet
    N = RPM in thousands

    The k is a coefficient that depends on the propeller type. There are websites that list tons of numbers for all known RC props. However, I don't know how that would relate to a real prop. I've been looking for that info for a while.

    You can also use a formula by somebody named Abbott that is:

    Power = P * D^4 * N^3 * 5.33 * 10^-15 where

    Power is in Watts
    P = prop pitch in inches
    D = prop diameter in inches
    N = RPM

    I think the 5.33 and 10^-15 are just conversion factors, but I'm not exactly sure.

    Hope this helps.
     
  7. Nov 16, 2006 #6
    Hi Mtagg,
    OK, let me try and give a semi-brief rundown of how you have to approach this. I won't solve all the equations for you, as you need to do that for yourself, but I will get you going in the right direction. You will have to analyze a couple flight conditions, and the first one to look at is always the level-flight (unaccelerated) cruise case where Lift=Weight and Thrust=Drag.

    For propulsion sizing in cruise we focus on the Thrust = Drag. For any given airspeed we can also write the equation for Power Required (Pr) as:

    Pr = Thrust * Velocity = Drag * Cruise Airspeed

    If you have an estimate of the drag coefficient (Cd) for the ultralight you can compute the total Drag at the cruise airspeed as:

    D = Cd* 0.5* air density* Airspeed^2* Wing Reference Area

    The general equation for Power Available (Pa) from a Prop+Reciprocating Engine combination (written in terms of propeller efficiency, engine mechanical shaft efficiency, and engine design parameters) is given as:

    Pa = Nprop* Nmech* RPM* Disp* PressMean/ 120

    Where:

    Nprop = Propeller Efficiency (Prop Manufacturer can provide this)
    Nmech = Engine Mechanical Shaft Efficiency (Engine Manufacturer can provide this)
    RPM = Engine operating RPM
    Disp = Total Engine Displacement (all cylinders)
    PressMean = Mean Effective Cylinder Pressure

    Make sure your units for Disp and PressMean are consistent when using this equation. The efficiency measures are dimensionless (no units). As for the propeller efficiency, this is generally quoted by the manufacturer with respect to a dimensionless parameter called the Advance Ratio (J) which has the equation:

    J = Cruise Velocity/n/D where:

    n = propeller angular speed (in revs/SECOND...convert from RPM)
    D = propeller diameter

    The prop manufacturer will often provide plots of the propeller's efficiency over a range of Advance Ratios for various different propeller pitch angles.

    From all the above, the only other fact you need is that at the level-flight cruise condition, Power Available has to equal Power Required to remain in that steady-state flight condition.

    Climb and descent performance is a bit more involved than cruise. But see if you can solve the cruise case first with what I have given you.

    Rainman
     
    Last edited: Nov 17, 2006
  8. Nov 17, 2006 #7
    Forgive me, but sometimes I can't help myself...

    As a teacher I can't help but to want to provide several different references for someone to read and learn about aircraft and flight design. As others here have said, it is sometimes questionable to refer someone to wikipedia, but the following wiki on propeller design and performance (esp. with respect to variation of prop pitch) does indeed provide a lot of good insights and things to consider if you are faced with a propeller design situation.

    http://en.wikipedia.org/wiki/Propeller#Aircraft_propellers_.28airscrews.29

    Rainman
     
  9. Nov 22, 2006 #8
    Ive changed things around a bit. I am using a Cessna 310R as my model, because I have been able to find lots of info on it. I still don't understand the torque issue, but I'm going to go by your equations.

    For a Cessna 310R flying at max cruise speed of 360kph at 3000ft under ISA conditions:

    Cd = .027 (according to Wikipedia)
    rho = .002176 slug/ft^3
    V = 328 feet/sec
    Area = 179 ft^2 (according to airliners.net)

    So if drag = Cd x .5 x rho x V^2 x Area and Power required = drag x airspeed then you just use V^3 instead of V^2. So putting in the Cessna numbers:

    Pr = .027 x .5 x .002176 x (328)^3 x 179 = 185552Watts = 186kW or about 248hp.

    But I don't understand how this correlates to how fast the propellor spins or how much torque you need to spin it.

    You say that
    If I assume that both efficiencies are 100% (just for kicks) then you are saying that Disp x PressMean/120 is equal to torque. That doesn't make sense to me.
     
  10. Nov 27, 2006 #9
    Hi Mtagg,

    Now that I have recovered from my turkey overdose, let's see if I can help you a bit more. :smile:
    This calculated power is the power needed to overcome aerodynamic drag and keep the airplane flying at the 328 feet/sec airspeed. It correlates to how fast the prop spins and how much torque you need when you realize that the engine+prop combination must provide Pr after accounting for all the efficiency losses. That is where the second equation comes in...

    Let me briefly run through the derivation of how you arrive at this equation...that might help you see the sense of it (and it is just an estimate, since we know that cylinder pressure varys with time during the power stroke of the cycle)...

    Are you familiar with the equation for Indicated Power (IP)? This is the theoretical maximum power available from an internal combustion engine's thermodynamic process. The equation is:

    IP = (n/2)*N*W where:
    n = crankshaft rotation rate (rev/sec)
    N = Total # of cylinders
    W = Total work output per cylinder (i.e. area between the curves of the Otto Cycle's P-V diagram)

    We know that we must also account for both shaft losses and propeller aerodynamic losses so, we can then say that the shaft brake power is:

    Pshaft = Nmech*IP (Nmech is the mechanical efficiency of the engine)

    And now if we account for the propeller losses we get the total power available to move the airplane:

    Pa = Nprop*Nmech*IP = Nprop*Nmech*(n/2)*N*W

    So if you convert "n" into RPM it should be easy to see where the "120" comes from in the denominator of the final equation I gave you. All we need to do now is transform the "N" and "W" into parameters related to the engine itself (such as the bore and the stroke and the mean effective cylinder pressure). If we assume that all the useful work is done during the power stroke, then "W" in the above equation becomes the force on the piston (Mean effective pressure*area of cylinder bore) times the distance through which the piston moves (the stroke). So our equation for "W" will now become:

    W = Pi*bore^2*stroke*PressMean/4

    Now substitute this equation for "W" into the Pa equation above... after that the only other piece of knowledge you need is that the total displacement of the engine is:

    d = Pi*bore^2*stroke*N/4

    Do the math and convince yourself that it reduces to what I gave you. BTW, as for some "gut feel" numbers for prop and shaft efficiences... you would be hard-pressed to find a propeller that has an efficiency of better than about Nprop=0.92 (and that is operating at its ideal pitch for whatever its operating advance ratio is). Furthermore, shaft efficiencies are even lower than prop efficiences due to higher heat/friction losses, so unless someone else here could give you a better state-of-the-art number, I don't think you will see Nmech much higher than about 0.86.

    But now you have an equation that is based upon only a SINGLE state-of-the-art performance parameter (Mean Effective Pressure). A higher performance engine will have a higher PressMean. This is a good parameter to compare engine performance because it is independent of the total engine displacement. Finally, the following link will give you a bunch of good Powerpoint slides that may be more than you need to know... but if you ever want to dig into the real details they may come in handy:

    https://me.queensu.ca/courses/MECH435/notes/Engine_Performance.ppt

    Most airplane engines are turbocharged and you would expect them to have a PressMean of anywhere from 150 psi to as high as 200 psi.

    I hope this helps!
    Rainman
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Propeller and thrust
  1. Thrust from Propellers (Replies: 1)

  2. Propeller problem (Replies: 16)

  3. Propeller Thrust (Replies: 5)

Loading...