# Induction Motor Problem



## Sharon (Oct 5, 2010)

*Good Morning....I would like to see how you would solve this problem. There were a couple of steps in the solution that disturbed me like dividing the reactance of the rotor by .4(slip) and the appearance of 6.97 in the calculation for I(l). Here is the problem:*

A six pole, three-phase, wye connected, 440 V, 60 Hz induction motor has rotational losses of 700W and 4% slip. The line-neutral impedance in the stator is .7 ohms resistance and .8 ohms reactance. The line-neutral impedance in the rotor is .3 ohms resistance and .8 ohms reactance. The transformation ratio is unity. Core reactance is 40 ohms and core resistance can be neglected. What is most nearly the output power. The solution says 18 kW.

Thanks.....


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## knight1fox3 (Oct 5, 2010)

Sharon said:


> *Good Morning....I would like to see how you would solve this problem. There were a couple of steps in the solution that disturbed me like dividing the reactance of the rotor by .4(slip) and the appearance of 6.97 in the calculation for I(l). Here is the problem:*
> A six pole, three-phase, wye connected, 440 V, 60 Hz induction motor has rotational losses of 700W and 4% slip. The line-neutral impedance in the stator is .7 ohms resistance and .8 ohms reactance. The line-neutral impedance in the rotor is .3 ohms resistance and .8 ohms reactance. The transformation ratio is unity. Core reactance is 40 ohms and core resistance can be neglected. What is most nearly the output power. The solution says 18 kW.
> 
> Thanks.....


Hmm....seems like there is some information missing from the problem. Either the size of the motor or the power being drawn from the line. Unless I'm missing something, Pin - Pgap - Plosses = Pout. From the information provided, I don't see how Pin can be calculated.


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## mull982 (Oct 5, 2010)

knight1fox3 said:


> Sharon said:
> 
> 
> > *Good Morning....I would like to see how you would solve this problem. There were a couple of steps in the solution that disturbed me like dividing the reactance of the rotor by .4(slip) and the appearance of 6.97 in the calculation for I(l). Here is the problem:*
> ...


I believe you'd have to set up the equivelent motor model with the impedences given in the problem including the variable rotor resistance R2/s with the slip that is given in the problem. Once you have the equivelent circuit you can combine all of the impedances to come up with the overall equivelent impedance of the motor. You can then use this equivelent impedance to calculate the input current using V/Z for the motor or possibly even calculate input power using V^2/R.

Let me know if you need help and I'll try to work it out.


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## DK PE (Oct 5, 2010)

Sharon said:


> *Good Morning....I would like to see how you would solve this problem. There were a couple of steps in the solution that disturbed me like dividing the reactance of the rotor by .4(slip) and the appearance of 6.97 in the calculation for I(l). Here is the problem:*
> A six pole, three-phase, wye connected, 440 V, 60 Hz induction motor has rotational losses of 700W and 4% slip. The line-neutral impedance in the stator is .7 ohms resistance and .8 ohms reactance. The line-neutral impedance in the rotor is .3 ohms resistance and .8 ohms reactance. The transformation ratio is unity. Core reactance is 40 ohms and core resistance can be neglected. What is most nearly the output power. The solution says 18 kW.
> 
> Thanks.....


I don't know why they would divide the rotor reactance by the slip unless they were somehow recognizing the frequency in the rotor circuit is slip*60Hz but without more detail I'm guessing. I don't know where 6.97 comes from either.

I punched this out quickly so won't guarantee accuracy but using a basic equivalent circuit I calculated an output power of 19.3kW 

I would get 18kW if the 700W losses they gave is per phase but that would be sneaky and wrong in my opinion.

One thing to remember in these problems is you typically work the problem on a per-phase basis so you have to use applied voltage of 440/sqrt(3) and then when you calculate the developed power that is per-phase also so you need to multiply by 3. Under test pressure these are easy to forget.

Is it possible to post some of the solution in order to critique it?


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## Sharon (Oct 7, 2010)

DK PE said:


> Sharon said:
> 
> 
> > *Good Morning....I would like to see how you would solve this problem. There were a couple of steps in the solution that disturbed me like dividing the reactance of the rotor by .4(slip) and the appearance of 6.97 in the calculation for I(l). Here is the problem:*
> ...




Here is the solution:

Solution:

The impedance as seen by the input line-to-neutral, assuming the core resistance is negligible, is the impedance of the stator windings in series with the parallel combination of the core reactance, jXnl, with the rotor impedance (the traditional model of the induction motor Thevenin equivalent impedance).

Z= R1 + jX1 + {(jXnl)[a^2R₂/s + ja^2X₂] / (jXnl +a^2R₂/s + ja^2X₂]}

From Problem statement:

R₁= 0.7Ω s=.04

R₂= 0.3Ω Xnl = 40Ω

X1 = 0.8Ω a = 1

X2 = 0.8Ω

Z = 0.7Ω + j0.8Ω + {j40Ω(0.3Ω/0.04 + j0.8Ω) / (j40Ω + 0.3/0.04 + j0.8Ω)}

= 0.7Ω + j0.8Ω + (6.97Ω + j2.07Ω)

= 8.19 Ω ﮮ 20.5°

The second step in the calculation above is included because there is a very important parameter in the calculation. The real part in parentheses is the resistance of the real part of the air gap power. The number is needed to calculate the air-gap power. If the core resistance was not negligible, this number would be the parallel combination of the air-gap power and the core loss.

The input line current to the machine is

I(line-neutral) = V(LL) / √3|Z| = 440V/√3(8.19) = 31.02 A

The air gap power is

P(air gap) = (3)(31.02)^2(6.97) = 20.12 kW

The power that is magnetically transferred to the rotor is

P(rotor) = (1 – s)P(air gap) = (1 – 0.04)(20.12 kW)

= 19.32 kW

The output power is the rotor power minus the mechanical losses.

Pout = Protor – Ploss = 19.32 kW – 0.070 kW

= 18.62 kW (18 kW)

Hope you can follow this.


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## DK PE (Oct 7, 2010)

Sharon said:


> *Good Morning....I would like to see how you would solve this problem. There were a couple of steps in the solution that disturbed me like dividing the reactance of the rotor by .4(slip) and the appearance of 6.97 in the calculation for I(l). Here is the problem:*
> A six pole, three-phase, wye connected, 440 V, 60 Hz induction motor has rotational losses of 700W and 4% slip. The line-neutral impedance in the stator is .7 ohms resistance and .8 ohms reactance. The line-neutral impedance in the rotor is .3 ohms resistance and .8 ohms reactance. The transformation ratio is unity. Core reactance is 40 ohms and core resistance can be neglected. What is most nearly the output power. The solution says 18 kW.
> 
> Thanks.....


Wow Sharon, thanks for typing out the full solution, that must have taken some time with the special symbols (gotta learn how to do that)

The difference in answers is they used an exact motor model vs. the one I used is the simplified version with the magnetizing branch moved to the source side of the stator circuit. For the approximate circuit, you just use Zin = Rs +jXs + Rrotor/slip + jXr with a 440/sqrt3 source which gives a line current of 30.4A vs. their 31 A. They are getting the 6.97 as the real part of air gap vs. the Rrotor/slip = 0.3/0.04 = 7.5 ohms in approx. model. Approx model gives a air gap power to the rotor of 20.8kW vs. their 20.1kW. The rest is taking out some of the losses and as I mentioned earlier, my calcs were 19.3kW vs. their ans. of 18.6kW. I have to think more about how they defined "core" losses as I think of those being in magnetizing branch which impacts efficiency and power factor but not output power...need to think more...

When you earlier mentioned dividing the reactance by slip, I think you meant to say resistance = R rotor/slip which is indeed the motor model.

The 6.97 doesn't appear in the calculation for Iline, for that you just divide line-neutral voltage by impedance Z = 8.19 ohms. The 6.97 does come into play in calculating the power generated which is just I^2*R (but remember the 3X as we're in per phase)

If you draw out the circuit, you can see that that:

{j40Ω(0.3Ω/0.04 + j0.8Ω) / (j40Ω + 0.3/0.04 + j0.8Ω)} is simply the product/sum formula for impedance in parallel.

I would hope that a solution using approximate model would be close enough for you to be able to use the "is most nearly" and get the correct answer although you never know.

Since you spent all the time typing this solution, feel free to ask any questions...


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## DK PE (Oct 7, 2010)

To clarify earlier post, I scribbled solution literally on the back of an envelope and attached. A couple of points to remember:

- Solution given here moves magnetizing branch back to input, book solution leaves in parallel with rotor circuit

- Most models I have seen use I 2 * Rrotor/s as power to rotor, not the real part as the given solution provides

- Most of these use the line to neutral voltage, don't forget and use line-line as source in equiv circuit

- multiply X3 to get final power to take care of per-phase

-If given other than 1:1 transformation, need to reflect rotor impedance back using turns ratio2

- In this case, the approx model was within 4% of given answer, you can improve this with an adjustment to applied voltage I won't go into details

-line current can be obtained by phasor addition of magnetizing branch, think I used term line current in earlier post but meant I1


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## DK PE (Oct 9, 2010)

I thought I would look once more for why the solution brought in the "real part of the combination of the rotor and magnetizing branch" to get the 6.97 term in solution posted and can't seem to find it. My source is _Electromagnetic and Electromechanical Machines, Leander W. Matsch_. He uses the magnetizing branch to determine the input impendance and hence the current, but when it comes time to calculate the power, he is back to using just the given rotor resistance. I won't list the edition of this text I used in undergrad but let's just say it is written on papyrus :laugh: For simplicity, I still prefer _Theodore Wildi: Electrical machines, drives and power systems_.

Using the approximate equivalent circuit I provided earlier with the adjusted input voltage enhancement I mentioned, (multiply the input voltage by [1 -X1/Xm ]) (... this adjustment compensates for the voltage drop when you move the magnetizing branch back to input side of stator) and use that only to calculate the current gives 29.8A. Using this new current in the same way as shown on envelope above gives power transferred to rotor - rotor I2R losses of 19.18kW and then you subtract 700W rotational losses and get 18.48kW which is within 1% of the answer they obtained. Not claiming they are wrong, just can't find justification for the method they used.

One other item I forgot to mention under the "almost tricks" category. There is always a chance that you will be given a resistance between two terminals.... and a statement they are connected in wye, remember the two equivalent resistances would then be in series and you would have to get the per phase by dividing by 2. Don't lose your mind and start diving by sqrt(3) or 3 or pi or whatever....


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