Electric Motor Power Output and Efficiency

AI Thread Summary
The discussion centers on calculating the output power, power loss, and efficiency of an electric motor operating at 850 RPM with a torque of 5 ft-lbf and an input power of 700 watts. The output power is calculated using the formula that incorporates torque and angular velocity, but an initial calculation yielded an implausibly high output of 5800 watts. Participants emphasized the importance of converting RPM to radians per second to ensure accurate calculations. The confusion arose from the expectation that output power should not exceed input power, highlighting the need for correct unit conversions. Ultimately, the clarification helped align the calculations with the physical principles of motor efficiency.
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Homework Statement


An electric motor operates at a steady state condition with a rotational speed of 850RPM while producing a constant torque of 5ft-lbf. A watt meter reads 700watts supplied to the motor.
What is the output power at the motor shaft?
What is the power lost in the motor which ends up to the surroundings?
What is the motor efficiency?


Homework Equations


(a) Power=torque*angular velocity
(b) Efficiency=output power/input power


The Attempt at a Solution


I converted the torque to 6.78N*m.
Then I plugged in my values or angular velocity (850RPM) and torque into equation (a).
My result was: 5800W
Is this not too large? The input power was only 700W, shouldn't my output power be less than this? Wouldn't this mean that power is not lost, but gained?
 
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You should convert the 850 rev/min to radians/s.
 
Gear300 said:
You should convert the 850 rev/min to radians/s.

Thanks, my answer makes more sense now!
 
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