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A 0.56 kW induction motor is operating from a 240 V 50 Hz supply. Its synchronous speed is 1,050 rpm and its rated speed is 880 rpm. It produces no torque at 1,050 rpm and its rated torque is 6.08 Nm.

Assume a linear τ - ω relationship near synchronous speed, and hence calculate the speed (in rpm) corresponding to 2/3s of the rated torque.

I have used the equation so far that: 0.56 kW/6.08*2/3 = Speed

That answer is wrong, so is anyone able to help me with this one?

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    \$\begingroup\$ How could the synchronous speed at 50 Hz be 1050 rpm? Anyway I think it is just a simple proportion. A shift of 170 rpm produces 6.08 N-m of torque. Then 0.66*170 = ? \$\endgroup\$
    – user57037
    Apr 27, 2018 at 4:22
  • \$\begingroup\$ @mkeith That is stated in the question - the answer of 2/3 * 170 is not correct. \$\endgroup\$
    – J-Dorman
    Apr 27, 2018 at 4:25
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    \$\begingroup\$ It seems from your comment that you know what the correct answer is. Would you be so kind as to edit your question and add in this important fact? For a 50 Hz induction motor, the possible synchronous speeds are 3000 rpm / N where N is an integer. This is why it is impossible for the synchronous speed to be 1050. I didn't claim that 2/3 * 170 is the correct answer. It is one step in calculating the correct answer. The second step is subtracting the result from synchronous speed. \$\endgroup\$
    – user57037
    Apr 27, 2018 at 4:39
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    \$\begingroup\$ What the question is telling you is that torque increases linearly from 0 to 6.08 N-m as speed decreases from 1050 to 880 rpm. So the torque will be 2/3 of rated torque when speed is 2/3 of the way from 1050 to 880. Do you understand? This appears to be school work, so I am trying to just explain the idea and let you figure it out. \$\endgroup\$
    – user57037
    Apr 27, 2018 at 6:01
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    \$\begingroup\$ The best way to thank me would be if you write a good answer to your own question so that the next person can find a good answer. \$\endgroup\$
    – user57037
    Apr 27, 2018 at 18:55

1 Answer 1

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I shall use the following notations : rated speed is \$N_r\$, synchronous speed \$N_s\$, torque \$T\$, rated torque \$T_r\$, operating torque \$T_{op} = \frac{2}{3} \times T_r\$.

When you write "Assume a linear τ - ω relationship near synchronous speed", this means: $$T = K (N_s - N)$$ for any speed \$N\$ near synchronous speed and corresponding torque \$T\$. This is also true for rated torque and speed: $$T_r = K (N_s - N_r)$$ and for operating torque and associated speed: $$T_{op} = K (N_s - N_{op})$$ The two last equations yield: $$N_s - N_{op} = \frac{T_{op}}{K} = \frac{T_{op}}{T_r}(N_s - N_r)$$

which should give you the answer.

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