For a 3-phase slip ring induction motor, the maximum torque is 2.5 times the full load torque and the starting torque is 1.5 times the full load torque. The percentage reduction in rotor circuit resistance to get a full load slip of 3% is _______Neglect stator impedance.a)56.9 %b)62.3 %c)39.4%d)43.2%Correct answer is option 'A'. Can you explain this answer?

Question

Answer

Concept:

The ratio of maximum torque to full load torque is

Where sm is the slip at maximum torque.

sfl is the slip at full load torque.

Slip at maximum torque,

R2 is the rotor resistance

X2 is the rotor reactance

Calculation:

Given that, Tm = 2.5 Tfl

Tst = 1.5 Tfl

When full load slip is 0.03

Answer

To find the percentage reduction in rotor circuit resistance in order to achieve a full load slip of 3% in a 3-phase slip ring induction motor, we can use the torque-slip characteristic of the motor.

Given:
Maximum torque (Tmax) = 2.5 times the full load torque
Starting torque (Tst) = 1.5 times the full load torque

The torque-slip characteristic of an induction motor can be represented by the equation:

T = k(1 - s)

Where T is the torque, k is a constant, and s is the slip.

We know that at full load, the slip (s) is given by:

s = (Ns - N) / Ns

Where Ns is the synchronous speed and N is the actual speed of the motor.

To determine the slip at full load, we need to know the synchronous speed. However, the synchronous speed is not given in the question. So, we cannot directly calculate the slip at full load.

However, we can use the fact that the maximum torque occurs at a slip of 1, and the starting torque occurs at a slip of 0. Therefore, we can write the following equations:

Tmax = k(1 - 1) = 0
Tst = k(1 - 0) = k

From these equations, we can conclude that Tmax is equal to zero, and Tst is equal to k.

Now, we can find the slip at full load (sfl) using the given information that Tmax is 2.5 times the full load torque:

Tmax = k(1 - sfl)
2.5Tfl = k(1 - sfl)

Similarly, we can find the slip at starting (sst) using the given information that Tst is 1.5 times the full load torque:

Tst = k(1 - sst)
1.5Tfl = k(1 - sst)

Dividing these two equations, we get:

2.5Tfl / (1.5Tfl) = (1 - sfl) / (1 - sst)
1.67 = (1 - sfl) / (1 - sst)

Now, we need to find the slip at full load (sfl) when the slip is 3%:

sfl = 0.03

Substituting these values into the equation, we can solve for sst:

1.67 = (1 - 0.03) / (1 - sst)
1.67 = 0.97 / (1 - sst)

1.67 - 1.67sst = 0.97
0.67 = 1.67sst
sst = 0.67 / 1.67
sst = 0.40

Now we have the slip at full load (sfl = 0.03) and the slip at starting (sst = 0.40).

To find the percentage reduction in rotor circuit resistance, we need to use the formula for slip:

s = (Ns - N) / Ns

At full load, the slip is 0.03. We can assume that Ns is 1 (normalized value).

0.03 = (1 - N) / 1
N = 0.97

So, the normalized speed at full load is 0.97.

To achieve a

Similar Electrical Engineering (EE) Doubts

The rotor resistance and standstill reactance of a 3-phase induction motor are respectively 0.018Ω and 0.08 Ω per phase, normal slip at full load is 4%. The percentage reduction in stator voltage to develop full-load torque at half of full-load speed is.

The rotor of a 3-phase induction motor has a resistance per phase of 0.04 ohm. This motor has a 0.2-ohm standstill reactance per phase. Neglect stator resistance. If the external resistance 0.02Ω is added to the circuit then what is the percentage improvement in p.f. during starting?

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