Hello chaterpillar
I assume that at present, the motor operating at 4.5 Amps is operating close to it's rated load.
In this case, it will be delivering close to rated torque.
The mechanical speed reduction will increase the torque to the fan by the speed reduction ratio. Therefore the shaft torque at the fan will be 4.8 times higher than the motor shaft torque.
a) If you use a VSD to reduce the shaft speed, you effectively reduce the power rating of the motor and the shaft torque stays constant.
Therefore, if you use a motor of the same base speed (1720) rpm and reduce the speed down to 360RPM, you will need to increase the size of the motor by a factor of 4.8. You would need to use a 15HP motor.
If you increase the number of poles, you will increase the base torque output of the motor, so you could use a 7.5HP 8 pole motor, or a 4HP 16 pole motor.
You would need to match the VSD to the motor size.
c) If you run at a constant speed, you will save on the belt losses, but you will add the VSD losses and I would not expect apreciable savings. If you change the fan speed depending on the air demand, then there will be savings due to the reduced fan speed. In other words, if the air flow changes, you could save energy. If the airflow is constant, you will not.
Best regards,
