Capacitor Start motors

Thread Starter

keelezibel

Joined Jan 14, 2009
2
Hi everyone,

I have this practical based independent learning project to do. It is on capacitor-start induction motors.

Background

A company uses 200 single phase induction motors for it production operations. The motors deliver a constant load torque of 0.8N-m while in operation. In order to increase production, the company wants to add in another 80 motors. However, there will be a current overload.

Assume that the current drawn by each motor initially is 2A each, if the current drawn by each motor is less than 1.5A, the current rating of the cables will not be compromised.

Power supply = 230V, 50 Hz.

I was thinking of adding capacitors in parallel to the main winding of the motor. However, how do i determine the value of the capacitor to be added?

S = 456VA
Pin = 273.378W
Pout = 120W
p.f. = 0.6

TIA
 

Thread Starter

keelezibel

Joined Jan 14, 2009
2
There is a value 86microfarad. I suppose it refers to the electrolytic capacitor that is connected in series to the auxiliary winding.
 

nagaloo

Joined Jan 27, 2009
29
I know this is a project but
There is such a thing as a Cap start Cap run induction motor. The start winding is used to reduce the running current with an oil filled 370v capacitor (minimum) connected in series and across the line this makes the motor run more efficiently. As an example I had a 10 hp in my shop for repair and the run caps were bad. With out them the no load current was over 25 amp and once replaced it was just over 7 amp. But the motor has to be designed to have this feature.
Another point to consider with this problem is induction motors cause problems with power factor which I think affect total current draw. I would have to dig up some of my text books to be sure how it all goes. But 2 ways to correct PF problems are synchronous motors and properly rated oil filled capacitors connected across the line. Voltage and current get out of phase and either of these methods help bring it back close to 1 again.
Stew
 
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