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A CIM motor is attached to a small dynamometer which exerts a fixed constant load torque of 100 oz-in on the motor shaft. The CIM is directly attached to a 12VDC source (assume no internal resistance). Assume the CIM's specs are exactly as specified here.
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1) What is the CIM's speed?
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Solution: Assume that the speed vs torque curve is linear, and use the given load torque (100 oz-in) to interpolate the rpm:
rpm = 5310*(1-100/343.4) = 3763.7 rpm
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2) What is the CIM's current?
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Solution: Assume that the current vs torque curve is linear, and use the given load torque (100 oz-in) to interpolate the current:
amps = 2.7 + (100/343.4)*(133-2.7) = 40.64 amps
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3) How much power (watts) is being input to the CIM?
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Solution: watts = volts*amps = 12*40.64 = 487.7 watts
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4) How much power (watts) is the CIM delivering into the load (dynamometer)?
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Solution: watts = oz_in*rpm/1352.3 = 100*3763.7/1352.3 = 278.3 watts
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5) How do you account for the difference between #4 and #3 above?
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Solution: The total power loss is 487.7-278.3 = 209 watts. Most of this loss is due to I^2R heat loss in the motor windings. The motor winding resistance is 12/133 = 0.09023 ohms, so the I^2R loss is (40.64)^2*(0.09023) = 149 watts. The remaining loss (60 watts) is due to a combination of motor bearing friction, windage, and motor core losses (hysteresis losses and induced eddy currents).
Now, suppose I open the circuit and insert a 0.1 ohm resistor in series with the motor. Assume that the load torque remains the same, at 100 oz-in, and that the resistor is large enough that it does not overheat.
Questions:
1) What is the CIM's speed?
2) What is the CIM's current?
3) How much power (watts) is being input to the CIM?
4) How much power (watts) is the CIM delivering into the load (dynamometer)?
5) Explain any assumptions you had to make to get the answers.