Induction Generator (Cage Rotor)

If the rotor of an induction machine rotates above synchronous speed, slip is negative, as are torque, mechanical output power and air gap power. i.e. the machine is operating as a generator. When considering synchronous machines it is normal to re-define the direction of positive current when switching between generators and motors (keeping a positive power flow for both cases). With induction machines we will not do this. If slip is negative, the "input power" to the electrical terminals will be negative, implying that power is flowing out of the electrical terminals.

Consider the machine used in the following examples:

A 480V, 60 Hz, 6-pole, three-phase, delta-connected induction motor with the following parameters:

R₁=0.461 Ω, R₂=0.258 Ω, X₁=0.507 Ω, X₂=0.309 Ω, X_m=30.74 Ω

Rotational losses are 2450W.

Now, suppose that the machine is being driven by a mechanical system such that it is rotating at 1224 rpm. Calculate the following information:

1. slip
2. Line Current
3. Power and Reactive Power at the terminals
4. Airgap Power
5. Torque Developed
6. Mechanical Power
7. Efficiency

Much of the theory for this problem is identical to a motor problem:

1. Slip is given by
   \[ s = \frac{n_s - n_m}{n_s} \]
   and
   \[ n_s = \frac{120 f}{p} = \frac{120 \times 60}{6} = 1200\text{ rpm} \]
   Therefore
   \[ s = \frac{1200 -1224}{1200} = \frac{-24}{1200} = \color{red}{-0.02} \]
2. Now, phase current is given by
   \[ I_1 = \frac{V_1}{Z_{in}} \]
   where phase impedance is given by
   \[ \begin{aligned} Z_{in}& = Z_1+ Z_m \mathbin{\!/\mkern-1mu/\!} Z_2 \\ Z_m & = R_C \mathbin{\!/\mkern-1mu/\!} jX_m \\ \end{aligned} \]
   In this case, \(R_C\) can be neglected, giving \(Z_m=jX_m\) and
   \[ \begin{aligned} Z_{in}& = Z_1+ \frac{jX_m \left( \frac{R_2}{s} +j X_2\right) } {\frac{R_2}{s}+j\left(X_2+X_m \right)} \end{aligned} \]
   Using the above equation
   \[ Z_{in}=-10.3=j5.29\Omega \]
   And noting that the machine is delta connected:
   \[ \begin{aligned} V_1 & =V_{LL} = 480V \\ \vec{I_1} & = -36.7-j18.8A \quad I_1=41.4A \\ \text{Therefore: } \color{red}{I_L} & \color{red}{= 71.7A} \end{aligned} \]
   Note that the real part of phase current is negative, indicating that real power flows out of the terminals. Also, the imaginary part of the phase current is negative, which indicates that reactive power must flow into the terminals. An induction generator cannot operate without a reactive power supply. This reactive power can be supplied from the grid or, in a standalone appliucation, a capacitor bank.
3. Using complex notation,
   \[ S=3VI^* \]
   or
   \[ \begin{aligned} P_{elec}&=3V_1\operatorname{Re}\{I_1\} \quad \color{red}{P_{elec}=-53.0kW} \\ Q_{elec}&=-3V_1\operatorname{Im}\{I_1\} \quad \color{red}{Q_{elec}= +27.2kVAR} \\ \end{aligned} \]
4. Airgap Power is given by
   \[ P_{gap}=\frac{3I_2^2R_2}{s} \]
   This approach requires rotor current to be found. With no core loss resistance:
   \[ \begin{aligned} \vec{I_2} & = \frac{jX_m}{\frac{R_2}{s}+j\left(X_2+X_m \right)} \vec{I_1}\\ I_2 & = \left| \frac{jX_m}{\frac{R_2}{s}+j\left(X_2+X_m \right)} \right| I_1 \end{aligned} \]
   Giving \(I_2=37.8A\). Substituting into the power equation \(\color{red}{P_{gap}=-55.4kW}\)
5. Torque developed can be found from
   \[ \tau=\frac{P_{gap}}{\omega_s} \]
   where synchronous speed in radians per second is given by
   \[ \omega_s = \frac{4 \pi f }{p} = 40\pi \]
   giving \(\color{red}{\tau=-441Nm}\)
6. Mechanical power can be found using
   \[ \begin{aligned} P_{shaft} & =P_{conv}-P_{rotational} \\ P_{conv} & = P_{gap}(1-s) \end{aligned} \]
   Therefore
   \[ \begin{aligned} P_{conv} = -55.4 \times \left(1-\left(-0.02\right)\right) & \quad P_{conv}= -56.5kW \\ P_{shaft} = -56.5 - 2.45 & \quad \color{red}{P_{shaft}= -58.9kW} \end{aligned} \]
7. Efficiency is given by
   \[ \eta=\frac{P_{elec}}{P_{shaft}} = \frac{53.0}{58.9} \quad \color{red}{\eta= 89.9\%} \]

Summary

It can be seen from the above analysis that the equations for an induction motor can all be applied to an induction generator. (As long as output and input power are correctly described as either electrical or mechanical). Induction generation used to be relatively rare. However, it is becoming increasingly common as induction generators are the generator of choice for large wind turbines. These generators are usually wound rotor machines, using the rotor voltage to to supply reactive power via a variable speed drive.