Capacitor bank for 340 kW induction motor

The supply voltage is not stated; since the capacitors are rated 3300 V, the motor is taken as a 3300 V, 3-phase motor and the 6 capacitors in each delta limb are taken as connected in parallel (each sees the full 3300 V line voltage).

Motor input

$$P = \frac{\text{output}}{\eta} = \frac{340}{0.85} = 400\ \text{kW}$$

i) kVAR rating of the capacitor bank

$$\begin{aligned} \tan\phi_1 &= \tan(\cos^{-1}0.8) = 0.75, \quad \tan\phi_2 = \tan(\cos^{-1}0.96) = 0.2917 \\ Q_C &= P(\tan\phi_1 - \tan\phi_2) = 400(0.75 - 0.2917) = 183.33\ \text{kVAR} \end{aligned}$$

ii) Capacitance of each limb (delta connected)

kVAR per limb $= 183.33/3 = 61.11$ kVAR; voltage across each limb $= V_L = 3300$ V.

$$\begin{aligned} C_{limb} &= \frac{Q_{limb}}{\omega V^2} = \frac{61\,111}{2\pi \times 50 \times 3300^2} \\ &= 17.86 \times 10^{-6}\ \text{F} = 17.86\ \mu\text{F} \end{aligned}$$

iii) Capacitance of each capacitor

Each limb has 6 equal 3300 V capacitors in parallel:

$$C = \frac{17.86}{6} = 2.98\ \mu\text{F}$$

        L1
        /\
   C=17.86 uF   each limb = 6 x 2.98 uF
      /    \     in parallel, 3300 V
   L2 ------ L3

Answer: (i) 183.33 kVAR, (ii) 17.86 μF per limb, (iii) 2.98 μF per capacitor.