This is probably a very simple question, but I have been really struggling and need your help.

I have simulated a propeller with these parameters on Ansys: Thrust: 2000 N Angular Velocity: 314 rad/s Torque: 200 N.m Airspeed: 100 m/s Propeller Diameter: 1.2 m

I got the following results (rounded for simplicity): Thrust: 2000 N. Torque: 200 N.m

Now I need the to find the efficiency, and I have used the formula Thrust*Airspeed / ShaftPower For the shaft power, I have used Torque * angular velocity.

Using my numbers, I get an efficiency of 3.18, so there is a mistake somewhere but I cannot understand where. Are my parameters relevant? Could there be an error in the simulation ? Is it possible for 200 Nm torque, 1.2 meter diameter propeller to produce a 2000N thrust? Is my method of calculating shaft power wrong ?


  • $\begingroup$ Given the basic information provided, a momentum assessment of efficiency of this propeller shows it to be about 4 or 5 percent too high, compared with a typically expected maximum of about 88 or 89%. The error noted herein, and that by OP, may require accounting for induced flow. And shaft power, of what: is this prop 2 blades, or perhaps 3 blades? Power input is thrust times air-flow velocity through the propeller disk; power output is thrust times translation velocity of the propeller. That's easily calculated. More information about this propeller, & its application, would be helpful. $\endgroup$ Mar 12 at 17:44

3 Answers 3


The equation Thrust * Airspeed / (Torque * AngularVelocity) is indeed the correct one for propeller efficiency, implying that we get an efficiency of zero at take-off.

enter image description here

For aeroplane propellers there are a few diagrams published, such as the one above from Torenbeek section 6.3 on characteristics of propellers: dimensionless power coefficient $C_P$ against J = airspeed/propTipSpeed. Listing efficiency, which is of course always less than 1.

enter image description here

For your question, a graph plotting thrust against power would be more interesting, and these graphs are mostly found in the helicopter literature, such as referred to in this answer. We can use this as an order-of-magnitude check for your derived parameters. The three equations valid in the hover would equate to propeller data at takeoff:

$$T = C_T \cdot \rho A (\Omega R)^2$$

$$Q = C_Q \cdot \rho A (\Omega R)^2 R$$

$$P = C_P \cdot \rho A (\Omega R)^3$$

Thrust: 2000 N Angular Velocity: 314 rad/s Torque: 200 N.m Airspeed: 100 m/s Propeller Diameter: 1.2 m

$C_T = \frac{T}{\rho A (\Omega R)^2}$ = 2000/(1.225 * $\pi$ * 0.6$^2$ * (314 * 0.6)$^2$) = 0.0407. Which is off the scale! We could be an order of magnitude off here.

What I would do is check all parameters and outcomes using graphs like above as a cross reference.

  • $\begingroup$ +1 for calcs. We can see why James Watt analyzed the first meter of motion, before drag from air became significant. The same can be done here, a measurement of acceleration (same mass, same fuel consumption). Or one can essentially put different props on the exact same airplane and do a full scale velocity vs prop, holding aircraft coefficient of drag and fuel consumption constant. $\endgroup$ Mar 12 at 9:33

Simply put, yes you are wrong somewhere, most likely in how you put together your Ansys simulation.

Thrust times airspeed gives power, and so does angular velocity times torque. You somehow get more power out than you put in, which doesn't make sense.

Ask yourself:

  • Is the thrust vector and the torque vector in the direction you expect? Perhaps you're simulating a windmilling prop?
  • Is the 100m/s freestream velocity or induced velocity?
  • How do my numbers compare to analytical equations for this problem?

All numerical packages suffer from garbage in/garbage out: if you give it nonsensical input, it will happily give you nonsensical output without any complaints. More details on your simulation would be required to exactly pinpoint where you're going wrong.

  • $\begingroup$ @Robert No. The OP clearly has an impossible situation where output power is greater than input power. There's most definitely a direct relation between shaft power and power "state" as you call it. I'm not going to remove my downvote since it shows a clearly lacking understanding of basic physics. Thanks. $\endgroup$
    – Sanchises
    Mar 11 at 15:37
  • 2
    $\begingroup$ @RobertDiGiovanni Power is conserved across mechanical advantage (unlike force, torque and displacement, rotation respectively). So 2kN acting on 100m/s velocity requires 200kW of power, regardless of any mechanical advantage. A rotation of 314rad/s and 200Nm produces ~36kW, regardless of any mechanical advantage. Power input is less than power output; clearly impossible. Please stop being so stubborn and learn the basics first. $\endgroup$
    – Sanchises
    Mar 11 at 16:18

First, we can convert 2000 N thrust to pounds thrust:

2000 N × .225 lbs/N = 500 lbf thrust

Next we convert rad/sec to RPM:

314 rad/sec × 57 degrees/rad × 60 sec/min × 1 revolution/360 degrees = 2983 RPM

Looking good so far.

Now we convert 2983 RPM and 200 Nm torque to Horsepower:

200 Nm × 2983 RPM/7127 = 84 Horsepower

These results are certainly near the range of reality, especially for a static bench test of a 1.2 meter propeller. Thrust will drop as airspeed increases, which is one of the reasons why prop engines are rated in shaft Horsepower. Thrust will vary dependent on pitch, RPM, and airspeed.

Engine Nm torque for a given RPM is the same as the angular drag torque of the prop at the same RPM.

Airspeed: 100 m/s. Converting to miles per hour:

100 m/s × 1 mile/1600 m × 3600 s/hour = 225 mph

This is a bit optimistic for 84 horses, but with reduction of weight and drag, and a variable pitch prop, one could get a bit closer to 225 mph than a Cessna 152, perhaps with something like this.

Regarding the prop efficiency calculation, it would seem far more meaningful to compare fuel consumption vs thrust against an agreed upon "standard propeller" at a given airspeed and RPM. Fuel consumption is directly proportional to prop drag. Efficiency is (obviously) thrust vs drag of the prop, not the aircraft.

  • 1
    $\begingroup$ The OP posted everything in SI units (as one always should), and the first thing you do is convert to "pounds thrust" (for which the correct symbol is lbf, not lbs, by the way). Why? $\endgroup$
    – Bianfable
    Mar 10 at 18:50
  • $\begingroup$ @Bianfable Because it can then be compared with known values to see if OP is "in the ballpark" with the numbers. I recollect around 3 lbs of thrust per HP at speed for props. Please correct me if I'm wrong. $\endgroup$ Mar 10 at 18:53

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