0
\$\begingroup\$

I have an induction generator on the same shaft with induction motor. Both machines are same (have same nameplate: \$ U=400\,V\$, \$f=50\,Hz\$, \$n=1435\, rpm\$, \$P=9.3\,kW \$, \$ \cos(\phi) = 0.90 \$, \$I = 17.05\,A\$). The motor is supplied via three phase inverter and the generator is connected to three phase AC/DC converter. The induction motor is supplied by 50 Hz three phase voltage and has four poles (\$p_p=2\$). I expect that the mechanical speed of the motor and also the generator is same \$n_m < 1500\,rpm\$ (but close to 1500 rpm). The corresponding mechanical frequency is \$f_m < 1500/60 = 25\,Hz\$). If I want the generator to produce electrical power I need to achieve negative slip (\$ s = \frac{(f_s-f_m)}{f_s} \$) in the generator. So I have to supply its stator with electrical frequency \$f_s < f_m\$ so for example 20 Hz would be sufficient. Is it correct?

\$\endgroup\$
0
\$\begingroup\$

To operate the motor at nominally 1500 RPM, the stator must be supplied with 400 V, 50 Hz as the nameplate states. If you were to also connect the induction generator to 400 V, 50 Hz, you could return about 9 kW to the 400 V. 50 Hz source by operating the motor at about 1565 RPM. Since the has 65 RPM of slip at full load, the induction generator needs to have about 65 RPM of negative slip. Since the motor must have about 65 RPM of slip at 1565 RPM, the frequency needs to be 1630 X 4 / 120 = 54 Hz.

In addition, the AC/DC converter must be a regenerative inverter capable of supplying 400 V, 50 Hz to the induction generator for magnetizing current. Once the induction generator is operating, it will supply voltage against the auxiliary supply and shut it off. The inverter's DC bus capacitor will then supply the magnetizing current.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.