We wish to use a 30 kW, 880 r/min, 60 Hz, woundrotor motor as a frequency converter (F) to generate 60 kW at an approximate frequency of 180 Hz (see Fig. 18). If the supply-line frequency is 60 Hz,

calculate the following:

a. The speed of the induction motor (M) that drives the frequency converter
b. The active power delivered to the stator of the frequency converter
c. The power of the induction motor (M)
d. Will the frequency converter overheat under these conditions

Question

We wish to use a 30 kW, 880 r/min, 60 Hz, woundrotor motor as a frequency converter (F) to generate 60 kW at an approximate frequency of 180 Hz (see Fig. 18). If the supply-line frequency is 60 Hz,

calculate the following:

a. The speed of the induction motor (M) that drives the frequency converter
b. The active power delivered to the stator of the frequency converter
c. The power of the induction motor (M)
d. Will the frequency converter overheat under these conditions

Accepted Answer

a. To generate 180 Hz the slip must be

[latex]\begin{aligned}f_{2} &=s f \180 &=s imes 60\end{aligned}[/latex]

from which

[latex]s=3[/latex]

The stator is still fed from the 60 Hz line, consequently, the synchronous speed of the converter is 900 r/min. The converter must be driven at a speed n given by

[latex]\begin{aligned}s &=\left(n_{ s }-n ight) / n_{ s } \3 &=(900-n) / 900\end{aligned}[/latex]

from which

[latex]n=-1800 r / min[/latex]

The converter must therefore be driven at a speed of 1800 r/min. The negative sign indicates that the rotor must run opposite to the revolving field. The induction motor driving the converter must, therefore, have a synchronous speed of 1800 r/min.

b. The rotor delivers an output of 60 kW. This corresponds to [latex]P_{ jr }[/latex], but instead of being dissipated in the rotor, [latex]P_{ jr }[/latex] is useful power delivered to a load (Fig. 20). The power [latex]P_{ r }[/latex] transferred from the stator to the rotor is

[latex]\begin{aligned}P_{ r } &=P_{ jr } / s=60 / 3 \&=20 kW\end{aligned}[/latex]

The power input to the stator of the frequency converter is equal to 20 kW plus the small copper and iron losses in the stator.

c. The remaining power input to the rotor amounting to (60 - 20) = 40 kW, is derived from the mechanical input to the shaft. By referring to Fig. 19 and Fig. 20, we can see how the active power flows into (and out of) the converter.

In summary, the rotor receives 20 kW of electrical power from the stator and 40 kW of mechanical power from the driving motor M. The rotor converts this power into 60 kW of electrical power at a frequency of 180 Hz. Induction motor M must therefore have a rating of 40 kW, 60 Hz, 1800 r/min.

d. The stator of the converter will not overheat because the 20 kW it absorbs is much less than its nominal rating of 30 kW. The rotor will not overheat either, even though it delivers 60 kW. The increased power arises from the fact that the voltage induced in the rotor is three times higher than at standstill. The iron losses in the rotor will be high because the frequency is 180 Hz; however, because the rotor turns at twice normal speed, the cooling is more effective, and the rotor will probably not overheat. The stator frequency is 60 Hz, consequently, the iron losses in the stator are normal.

Question 14-8: We wish to use a 30 kW, 880 r/min, 60 Hz, woundrotor motor a...

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