bittu123

 Profile Answers by bittu123 Questions by bittu123

• Nov 8th, 2011

 

Since you didn't mention which motor you talking about? I am assuming you must be talking about INDUCTION MOTOR. If other, lemme know?

During starting, s=1(or nearly 1 if its just started).
So, R(resistance of rotor) is very less than s.X(equivalent reactance). This made rotor almost inductive circuit. So, during starting, current lags very much. Providing same amount of torque in starting it needs same effective current(i.cos$). So, It has to draw large current(5-7times of running current, since cos$ is 5-7 times smaller in starting).

What happens when it start running????????
- Once it start running, slip(s) becomes very low. So, now R is not that much smaller than s.X(equivalent reactance). circuit is no more inductive and no more that much lagging current is needed. pf(cos$)
improves now so less i(current) is needed.

Hmmmmm, seems right, But why don't we design IM of very high R so that it won't be very lesser even though s=1 during starting and rotor circuit won't become so inductive??????????
-Yes, this may help u getting lesser current during starting, but what if once it start running with full speed. slip became very low so s.X(equivalent reactance) also very less. In such case R>>>> s.X and t will cause unnecessary heat loss. which we don't want. right???

Whats the alternative???
-for getting advantage of both "1. high resistance during starting" & "2. low resistance at running" we apply many methods like......
1. add external resistance to rotor in slip ring IM during starting but once start running we used to cut them.
2. Double cage IM.
3. Deep bar cage rotor motor

bittu123

 Profile Answers by bittu123 Questions by bittu123

• Nov 8th, 2011

 

If You may be talking about DC motor, here is the explanation........

current through rotor is given by i=(V-E)/R.

But during starting no emf is induced, E=0, So, i= (V-0)/R must be greater than if there would have been some value of E.
That's why we use "starter" to start DC motor. since it increases the value of denominator by adding extra resistance during start up and compensate the effect of lowering the numerator values.

By adding starter, i=(V-0)/(r+R)

eng: ahmed gharieb

• Dec 5th, 2011

 

Every thing in the first being hard but with time it get better. The IM is one of this thing, as in the start n=0 so that the equivalent cct of IM
exactly as before

Thanks