Induction generator excitation capacitor value

An induction motor works as an induction generator when its rotor is driven by a prime mover above synchronous speed ($N > N_s$), so that slip is negative and mechanical power is converted into electrical power.

Motor to generator action

• With $N < N_s$ ($s > 0$) the machine motors.
• Driven at $N > N_s$ ($s < 0$), the rotor cuts the field in the opposite sense; rotor emf, current and torque reverse. Air-gap power $3I_2^2R_2/s$ becomes negative: power flows from shaft to stator terminals.
• The machine still needs lagging reactive power to magnetize its core. On a grid it takes this from the grid. In isolated mode a 3-phase capacitor bank across the stator supplies it, and voltage builds up by self-excitation from residual magnetism.

 Prime mover  +------+
 (N > Ns) ====|  IG  |===+======> Load
              +------+   |
                        === C per phase

Voltage settles where the magnetization curve meets the capacitor line $V = I_c X_c = I_c/(\omega C)$.

Procedure to find the excitation capacitance

1. Obtain the magnetization curve. Run the machine as a motor on no load (speed $\approx N_s$) from a variable-voltage supply, or drive it at synchronous speed. Vary the applied phase voltage $V$ from about 120% of rated downwards and record the no-load (magnetizing) current $I_m$. Plot $V$ against $I_m$.

 V |        _.---- magnetization
   |     .-'       curve
 Vr|---.*   <- required point
   |  /|   capacitor line
   | / |   slope = Xc
   |/__|_____________ Im
      Im,r

2. Choose the operating voltage $V_r$ (rated phase voltage) and read the corresponding magnetizing current $I_{m,r}$ from the curve.
3. Capacitive reactance per phase — the capacitor line must pass through this point:

$$X_c = \frac{V_r}{I_{m,r}}$$

4. Capacitance per phase (star-connected bank):

$$C_Y = \frac{1}{2\pi f X_c} = \frac{I_{m,r}}{2\pi f V_r}$$

For a delta-connected bank (each capacitor at line voltage): $C_\Delta = C_Y/3$. 5. Minimum capacitance: the capacitor line must be less steep than the linear (air-gap) part of the curve; otherwise voltage will not build up. The slope of the air-gap line gives $X_{c,max}$ and so $C_{min}$. 6. Allowance for load: for a lagging load, extra capacitance is added to supply the load VAr, $Q_L = 3V^2\omega\, \Delta C$, so that voltage does not collapse. Also, because the slip is negative, the generated frequency is slightly below the speed-equivalent value $PN/120$, so the calculation should use this expected (slightly lower) operating frequency, not $PN/120$ itself.

Example: 400 V, 50 Hz machine; at rated phase voltage 230.94 V the curve gives $I_m = 4$ A.

$$\begin{aligned} X_c &= \frac{230.94}{4} = 57.74\ \Omega \\ C_Y &= \frac{1}{2\pi \times 50 \times 57.74} = 55.1\ \mu\text{F per phase} \\ C_\Delta &= \frac{55.1}{3} = 18.4\ \mu\text{F per phase} \end{aligned}$$