The rotor resistance and standard reactance of a 3 phase induction mot...
Let N
s = 1000 r.p.m
Full load speed = [1 – 0.03] × 1000 = 970 rpm.
Let the normal voltage be V
1 volts.
Speed in second case = 970/2 = 485 r. p.m
We know that
Since Torque is the same in both case
Hence % reduction of stator voltage
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The rotor resistance and standard reactance of a 3 phase induction mot...
Ohms and 0.05 ohms per phase. The stator voltage is 415 V and the rated power factor is 0.8 lagging. The rated power output of the motor is 50 kW.
(a) Calculate the rated rotor current and rotor power factor.
(b) Calculate the rated rotor copper losses and rotor core losses.
(c) Calculate the rated efficiency and the rated slip of the motor.
Solution:
(a) The rated stator current is given by:
I = P / (sqrt(3) x V x PF)
where P is the rated power output, V is the stator voltage, and PF is the rated power factor.
Substituting the given values, we get:
I = 50,000 / (sqrt(3) x 415 x 0.8) = 69.8 A
The rated rotor current can be calculated as:
I2 = s x I1
where s is the slip and I1 is the rated stator current. At rated conditions, s = 0.04 (assuming a typical value for a 3 phase induction motor).
Therefore, I2 = 0.04 x 69.8 = 2.79 A
The rated rotor power factor can be calculated as:
cos(phi2) = R2 / sqrt(R2^2 + X2^2)
where R2 is the rotor resistance and X2 is the rotor reactance.
Substituting the given values, we get:
cos(phi2) = 0.015 / sqrt(0.015^2 + 0.05^2) = 0.292
Therefore, the rated rotor power factor is 0.292 lagging.
(b) The rated rotor copper losses can be calculated as:
Pc = 3 x I2^2 x R2
Substituting the given values, we get:
Pc = 3 x 2.79^2 x 0.015 = 0.586 kW
The rated rotor core losses can be estimated as a fraction of the rated power output. A typical value for a 3 phase induction motor is around 0.5-1% of the rated power output. Assuming a value of 0.75%, we get:
Pc0 = 0.75% x 50,000 = 375 W
Therefore, the rated rotor core losses are 375 W.
(c) The rated efficiency of the motor can be calculated as:
Efficiency = Pout / Pin
where Pout is the rated power output and Pin is the rated input power.
The rated input power can be calculated as:
Pin = 3 x V x I1 x PF
Substituting the given values, we get:
Pin = 3 x 415 x 69.8 x 0.8 = 87.5 kW
Therefore, the rated efficiency is:
Efficiency = 50,000 / 87,500 = 0.5714 or 57.14%
The rated slip of the motor can be calculated as:
s = (Ns - Nr) / Ns
where Ns is the synchronous speed and Nr is the rotor speed at rated conditions.
The synchronous speed can be calculated as:
Ns = 120 x f / P
where f is the supply frequency (assumed to be 50 Hz) and P is the number of poles (assumed to be
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