We are operating a DC Motor of 69.8kW and 2080 RPM. Now we are replacing the motor with a DC motor of 74.5kW and 2550 RPM. What will be the kW of new motor at 2080RPM? What are the torque of three cases?
Torque is determined by the speed/torque characterisitcs of the load, so at identical speed, the torque on the new motor will be the same as that on the old one.
Power is torque multiplied by angular velocity divided by efficiency. The last is not stated, so the first is indeterminate. However, provided the circuit protection device(s), the cabling and the motor overload protection device(s) setting(s) are unaltered, the new arrangement will be safe.
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Assuming both motors have same nominal voltage, and same fixed field voltage.
and assuming losses are minimal, the speed/ voltage curve for a DC motor is linear, for the new motor to have 2080 RPM , it should be supplied with 2080/2550 * V nominal =0.81 Vnom.
Nominal current of Motor 2 ( 74.5KW) is 1.067 x current of motor 1.
Motor 2 KW at 2080RPM = 0.81 Vnom x I1 nom x1.067
Vnom x I1nom = 69.8
KW2 at 2080 RPM = 69.8 x 1.067 x 0.81 = 60.3 KW
torque will be 6.7 % higher
Motor Speed and Torque
Re: Motor Speed and Torque
