Motor Power Consumption 1

[oneijack]

I wasn't sure to post this in electrial or mechanical, but here is my situation. I want to reduce power consumption on an extruder production line. I have the following:

Incoming power: 480v 3 phase AC (of course) and assuming power factor is 1.

Motor: 250hp dc motor, 1750rpm, 500v armature voltage direct coupled to gearbox with ratio of 17.5:1 for output of 100rpm.

I do not need 100rpm. If I only need 75 rpm output (max), I can change the ratio to 23:1 ratio, which should drop the dc motor amps.


I have several questions:
Will this save me in electrical consumption? If, yes would a rough estimate of energy savings be 25% (100rpm-75rpm)/100=25%, estimating amps are directly proportional to gearbox ratio change.

Since dc amps are less with 23:1 ratio, therefore ac amps less and less power consumtion.

I am confused with going from ac to dc power consumption formulas, and everyone I ask has a different answer.

Any help would be appreciated.

 

[desertfox]

Hi onejack

The motor will increase the torque available if you reduce
your speed from 100 to 75rpm.
The torque used will be determined by your load and not by altering the rpm of the motor.
We need to know the power consumption of the extruder and what torque it needs to operate.


regards

desertfox

 

[desertfox]

Hi again

Have a look at this site:-

http://lancet.mit.edu/motors/motors4.html

regards

desertfox

 

[Javier72]

You are going to have lower consumption.

Power = speed * torque

then, your needed power, with less speed, is going to be lower. In DC motors, the amps needed are function of the power needed, then, if the power drops, the amps drops too. Be careful because the efficiency changes too.

 

[hydtools]

I see no savings if your power demands are the same.

Ted