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I have a flat piece of wood shaped like this: enter image description here And it is turning around axle. What i would like to know is, if it is rotated e.g. at 1000 RPM, how much power would that take?

I know that drag force should be equal to this: $$ F_D=\frac{1}{2}C_D\rho_{air}Av^2 $$ And power:

P = Fd ⋅ v

I am using P air = 1.2 and Cd = 1.1 (i don't know if this is correct) .

First i calculated speed based on outer radius(27cm), but because of V^2 that was a big mistake, now i have split horizontally (or vertically if you are watching the picture) in parts of 0.5cm, and then calculated power for each part individually and then add them together, but i am getting what i think is lower value than what it should be (in range of 150-200W), i am getting ~87W at 660RPM. I am guessing i have a wrong approach to this problem.

EDIT 1: Here is python code how i calculated power with my way and also with Yaniv way.

RPM = 500
RPS = RPM / 60
def estimatedPower(RPS):  # RPS = revolution per second
    finalPower = 0
    forceSmallPiece = 0
    forceBiggerPiece = 0
    for i in range(0, 54):
        radius = i/2
        if radius > 11:
            A = 0.0008
            V = radius/100*2*3.1415 * RPS
            F = (1.1*1.2*A*V*V)/2
            forceBiggerPiece += F
        else:
            A = 0.000375
            V = radius/100*2*3.1415 * RPS
            F = (1.1*1.2*A*V*V)/2
            forceSmallPiece += F
        P = F * V
        finalPower += P*2    # *2 is because there are 2 blades
    print("power: ",finalPower)
    return finalPower, forceSmallPiece, forceBiggerPiece

finalPower, forceSmallPiece, forceBiggerPiece = estimatedPower(RPS)


AngularSpeed = RPM * 2 * 3.1415/60

Fd = forceBiggerPiece
Fd1 = forceSmallPiece


torque1 = Fd * 0.19
torque2 = Fd1 * 0.055

torque = (torque1 + torque2) * 2    # *2 because there are 2 fan blades
p = torque * AngularSpeed

print("Power with torque: ",p)
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First, pay attention to the fact that the total force acting on the wood is zero. Since your device is turning, it would be much easier to calculate the power as torque times the angular velocity: P = tau * w. When Tau is the total torque around the rotation axis (Nm) and w is the item angular speed (rad/sec).

If the drag force is coming from a forced flow having a velocity much greater than your wood linear velocity at the ends - I suggest to calculate the torque as follows:

  1. Deal only with one half of the wood and then multiply the result by two
  2. Divide the taken half into two areas - a 16x16 cm square and a 11x7.5 rectangle
  3. For each area, calculate the drag force according to your formula
  4. Assume the force on each area acts exactly on its center. Figure what are resulted torques are by multiplying the forces by their distance from the rotation axis (19 cm and 5.5 cm respectively)
  5. Sum up both torques and multiply by two as mentioned in step #1 - since there are two symmetric halves of that wood

That's it. Remember that 1000 RPM equals roughly to 105 rad/sec - and are ready to go.

However, since the drag force in your case is due to the rotation itself - we need another approach. lets attack it in two alternative ways.

The numerical approach:

  1. Deal only with one half of the wood and then multiply the result by two
  2. Divide the taken half into two areas - a 16x16 cm square and a 11x7.5 rectangle
  3. divide the radius to small sections (dx), in each calculate the torque as Fd times r. When r is the distance from the rotation axis to specific section. Pay attention - the area of each section is dx times its height (16 cm and 7.5 cm). In your example you took the whole rectangle area..
  4. Sum up both torques and multiply by two as mentioned in step #1 - since there are two symmetric halves of that wood

The analytic approach: You have to integrate the force times the distance over the whole distance. Lets right down the force formula in its differential form: $F=\int 0.5 \cdot\rho\cdot C_d \cdot(w\cdot r)^2\cdot l\cdot dr $ Note that $l\cdot dr $ is the differential representation of A. The same way, the torque formula can be written down as: $Torque=\int 0.5 \cdot\rho\cdot C_d\cdot (w\cdot r)^2\cdot l\cdot r\cdot dr $ Yep, we just multiply the force by r - which is the distance from each infinitesimal section to the rotation axis.

If we integrate the torque formula from 0 to a total radius R we will get: $Torque= 0.125\cdot \rho\cdot C_d \cdot w^2 \cdot l\cdot R^4$ Check the units - we get $\frac{kg*m^2}{sec^2}$ which equals to Nm - the units of torque. You can now use this formula to calculate both torques and go on. Just be careful to deal correctly with the boundaries of the far away area - since it is not integrated from 0.

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  • $\begingroup$ I get very similar result, 79 vs 87W before at 660RPM (I edited OP). I edited OP and added python code how i calculate both powers. I think i have messed up something with speed, Can you check if i made some obvious mistake? Because 80W sound like it is too low. $\endgroup$
    – samo ja
    Apr 9 '20 at 16:44
  • $\begingroup$ I see here two problems. 1. In each integration step you use the whole area. This is not true, you should use only the relevant slice 2. I confused you with the force arm, will update my explanation... $\endgroup$ Apr 9 '20 at 17:26
  • $\begingroup$ I don't know where you saw that i used whole area, I only used Area of 0.5cm x 16cm(0.0008m^2) when radius bigger than 11cm and 0.5cm x 7.5cm (0.000375m^2) if it is smaller. $\endgroup$
    – samo ja
    Apr 9 '20 at 17:43
  • $\begingroup$ yep, but in each step you took one of those areas instead of the differential one. Look at my extended explanation above.. $\endgroup$ Apr 9 '20 at 18:19
  • $\begingroup$ I know this is maybe too much to ask, but can you add for example how to calculate needed power just for bigger piece (16cm x 16cm) at whatever RPM you would like? $\endgroup$
    – samo ja
    Apr 10 '20 at 10:39

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