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Slipring Motor Power Consumption


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Dear All,


I am using slipring motor of 425KW, 6.6 KV. motor for fan application. Motor speed is controlled using grid resistance and it is running at 80% speed. As I know for fan applications power = (speed) 3.


Does this formula applies to slip ring motor also? or


it applies to VFD + Induction motor. As i know if slip ring motor running at 80% speed, remaining slip power dissipates at resistors and there will no energy saving.


Finally my intension is to use VFD+S.C. motor instead of slipring motor.

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Hello santhosh80


Welcome to the forum


Yes, the power required by the fan is proportional to the speed cubed.

When you slow the fan down using any method, the torque required will reduce.

When you use a slip ring motor to slow the fan down, the torque reduces and the speed of the motor reduces increasing the slip.


Because the torque has reduced, the power drawn from the supply will reduce, but the power dissipated in the rotor resistors increases.

The rotor power is still part of the total power into the motor which is reducing with torque, so the power consumed by the fan is much less than at full speed and the power drawn from the supply is reduced by a lesser amount.


At 80 percent speed, 80 percent of the power consumed (less than full power) is drawn by the load and the remaining 20% is dissipated in the rotor resistors.


Using the slip ring motor to reduce the speed will result in energy savings, but eliminating the slip losses will yield higher savings.


Short the rings and us a VFD and you will eliminate the slip losses.


Best regards,


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Thanks a lot! Marke.



I agree as per your reply, that power consumption of the motor reduces when we reduce speed of the fan irrespective of type of motor. But if we see the main draw back of the slipring motors is low p.f at low speeds when compared to induction motor + VFD.


As the power factor reduces in slipring motors when we reduce the speed, does this has no effect in the input power consumption of the motor at lower speeds?








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Hello Santosh


I do not expect the power factor to drop as the motor is slowed down by adding rotor resistance. I would expect that the power factor would be high.

The problem is that the magnetic field is rotating at synchronous speed and the rotor is spinning at less than synchronous speed. Therefore, there are high slip losses. If we take a motor running at half speed, half the power in to the motor is driving the load and the other half is heating the rotor resistors due to slip losses. The efficiency of the motor and rotor resistors reduces as the rotor resistance is increased.


Best regards,


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