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Peak power of motor

by sharpnova
Tags: motor, peak, power
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sharpnova
#1
May14-13, 03:48 AM
P: 13
How do you work out the peak power of a motor based on its maximum torque and rpm?

I've figured out that you can get the maximum torque via (coil area) * (# turns) * (current) * (magnetic field)

And I get that the peak power should be proportional to the rpm. But it doesn't seem to be as simple as:

(maximum torque) * (rpm) <-- aka, torque * time taken for a revolution

I've looked all over and have found a bunch of instances of people taking the max torque and multiplying it by a sec^-1, but this sec^-1 value seems to always be the max torque divided by 9.549.. which seems completely arbitrary to me and leads me to believe I'm missing something more fundamental here.
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jbriggs444
#2
May14-13, 05:40 AM
P: 907
Quote Quote by sharpnova View Post
(maximum torque) * (rpm) <-- aka, torque * time taken for a revolution
If you assume that torque is constant regardless of rpm then this idea is close. But you want to take torque divided by the time taken to rotate through one radian.

Think about it this way. If you have a torque, that's a force at a radius from the axis of rotation. If you rotate through one radian, then the force has acted through a distance equal to the radius. That effectively converts the torque (force times radial distance) to an energy (force times tangential distance).

To convert from energy to power you divide by the time it took to deliver that much energy.

So you end up with a formula of torque / (seconds per radian ).
Or, equivalently. torque * ( radians per second ).

I've looked all over and have found a bunch of instances of people taking the max torque and multiplying it by a sec^-1, but this sec^-1 value seems to always be the max torque divided by 9.549.. which seems completely arbitrary to me and leads me to believe I'm missing something more fundamental here.
One rotation is 2 pi radians.
One second is 1/60 of a minute.
9.549 is 60 / (2 pi)

So the magic 9.549 number is the conversion factor from rpm to radians per second -- you divide rpm by 9.549 to get radians per second.
rcgldr
#3
May14-13, 09:17 AM
HW Helper
P: 7,054
For an idealized motor, peak torque occurs at zero rpm and torque decreases linearly as rpm increases until it becomes zero at maximum rpm. Peak power would occur at 1/2 of peak rpm, with 1/2 the peak torque.

sharpnova
#4
May14-13, 12:20 PM
P: 13
Peak power of motor

Makes perfect sense. Thanks
sharpnova
#5
May18-13, 09:04 AM
P: 13
Actually I'm being told there is still a problem with the answer. The person I'm helping with homework apparently got this question wrong. The way they worked it out (and the way I explained it to them) (and the problem itself) are as follows:

Question:
An electric motor has 81.8 turns of wire wrapped on a rectangular coil, of dimensions 3.52 cm by 5.83 cm. Assume that the motor uses 15.6 A of current and that a uniform 0.744 T magnetic field exists within the motor. What is the maximum torque delivered by the motor? Answer in units of Nm

My solution (which was correct):
Torque is proportional to the number of turns and the current flowing through the wire. So we have 81.8 turns * .0352 m * .0583 m * 15.6 A * .744 T = 1.948327 Nm

Followup question:
If the motor rotates at 3230 rev/min, what is the peak power produced by the motor? Answer in units of W

My solution (which was apparently incorrect):
The torque is the force at a distance of one radius from the axis of rotation. So you'll need to divide it by the time it takes to move through one radian of tangential distance. This will convert the torque to an energy unit and consequently, the energy/time into a meaningful peak power. So we take our rpm of 3230 and divide it by 60 to convert it to rps. Then divide that by 2*pi to convert it to radians per second. Then multiply by the torque. So we have 1.948327 * 3230 *2 pi / 60 = 649.011494 W

Where was my misstep?
jbriggs444
#6
May18-13, 09:27 AM
P: 907
I believe that the point is this...

The peak torque will presumably occur twice per rotation. It will not be uniform.

Ideally, the force will vary like a sine wave. If the current is reversed for half of the rotation, the force graph will have the bottom of the sine wave reflected so that it is above the x axis.

Look at the area under that curve. Instead of a force of magnitude 1 at a radius of 1 delivering 2 pi units of energy over a complete rotation if you integrate the two halves of the sine function, you'll discover that it delivers 4 units instead.


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