(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 25-kW, 230-V shunt motor has an armature resistance of 0.124 Ω and a field-circuit resistance of 95 Ω. The motor delivers rated output power at rated voltage when its armature current is 73.5 A. When the motor is operating at rated voltage, the speed is observed to be 1150 r/min when the machine is loaded such that the armature current is 41.5 A.

a. Calculate the rated-load speed of this motor. In order to protect both the motor and the dc supply under starting conditions, an external resistance will be connected in series with the armature winding (with the field winding remaining directly across the 230-V supply). The resistance will then be automatically adjusted in steps so that the armature current does not exceed 200 percent of rated current. The step size will be determined such that, until all the extemal resistance is switched out, the armature current will not be permitted to drop below rated value. In other words, the machine is to start with 200 percent of rated armature current and as soon as the current falls to rated value, sufficient series resistance is to be cut out to restore the current to 200 percent. This process will be repeated until all of the series resistance has been eliminated.

b. How much resistance should be cut out at each step in the starting operation and at what speed should each step change occur?

2. Relevant equations

armature voltage is proportional to the speed

Vt= Ea+Ia Ra

3. The attempt at a solution

I have solve for the maximum resistance when the motor is starting Rt= 1.44 ohms

and i have solve for the three resistances to be cut out at each step Ra= 0.78 ohms, Rb= 0.39 ohms and Rc= 0.2 ohms..

MY PROBLEM IS FINDING THE SPEED DURING THE CUT OUT OF RESISTANCE...

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# Homework Help: Shunt motor problem

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