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I don't want a solution; the solution is already given.

A 575 V, 60 Hz, 150 hp, 0.80 power factor lagging, 2-pole, 3-phase induction motor has the following equivalent circuit parameters as seen from the stator side:

enter image description here

Core losses = 600 W, friction/windage losses = 300 W, stray losses = 100 W The line current drawing by this induction motor is 150 A.

The rotor current of the given induction motor is 41 A by using the below formula as solved by the instructor:

enter image description here

However, if I try using the equivalent circuit in the figure below to find the rotor current:

enter image description here

Where is the mistake in the second approach?

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    \$\begingroup\$ You need to show your working. \$\endgroup\$
    – Andy aka
    Jan 6, 2022 at 10:31
  • \$\begingroup\$ @Andy aka Done. \$\endgroup\$
    – OMAR
    Jan 6, 2022 at 11:46
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    \$\begingroup\$ In the statement of the problem, I believe that "150 hp" should be "150 A." The problem contains no value or basis for calculating Pcore, AKA Piron. I can revise the question to show the equivalent circuit if you are unable to do that. More later. \$\endgroup\$
    – user80875
    Jan 6, 2022 at 15:30
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    \$\begingroup\$ I think both the first and the second approach are correct. You have been given the input voltage current and power factor. You have also been given the motor equivalent circuit and iron losses. Those two sets of parameters should agree with each other. I have not done the complete calculation of the first set from the second set, but I think they must not agree with each other. Also an induction motor with 29% stator copper and iron losses is an extraordinarily inefficient motor. An induction motor with only 1.5% FL slip is quite a bit more efficient than average in that respect. \$\endgroup\$
    – user80875
    Jan 6, 2022 at 17:38
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    \$\begingroup\$ @ Andy aka: Here is the equivalent circuit in question with some differences in labeling and including the iron loss branch that is mentioned in my comment. i.stack.imgur.com/fu9Ud.jpg \$\endgroup\$
    – user80875
    Jan 6, 2022 at 17:56

1 Answer 1

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The original statement of the problem is not valid. The question presents an equivalent circuit for an induction motor and apparently states that the line current for a 1.5% slip operating point is 150 amps with a 0.8 power factor. If the equivalent circuit is solved, the solution will give a much higher operating current. If the equivalent circuit is solved for no load and no mechanical losses, slip = 0, neglecting the core loss, the current is simply line current = phase voltage / [R1 + j(X1 + Xm)] = 246.6A, 64 percent higher than the stated line current.

If the statement of the problem provided the correct line current for the equivalent circuit and other parameters given, the two methods given for solving the problem would yield the same answer.

The equivalent circuit is assumed to be essentially the following version of the Steinmetz equivalent circuit. R1 & X1 are the rotor resistance and leakage reactance. X2 is the rotor leakage reactance. R2/s is the rotor resistance divided by the slip, combining the rotor winding resistance with the mechanism for representing mechanical power developed in the rotor. An ideal transformer connected between the stator and rotor circuits has been eliminated by adjusting the X1 and R2 values to "reflect" the turns ratio. Bm and Gm are the reactance of the magnetizing and a resistance representing the iron losses.

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