Line pf of 35 kW induction motor with capacitors

The capacitor kVAR is fixed by the half-load condition and then stays connected at full load and at no load.

Input power and reactive power of the motor

Full load: output 35 kW, $\eta = 0.9$, pf 0.9

$$\begin{aligned} P_{FL} &= \frac{35}{0.9} = 38.89\ \text{kW} \\ Q_{FL} &= P_{FL}\tan\phi = 38.89 \times \tan(\cos^{-1}0.9) = 38.89 \times 0.4843 = 18.83\ \text{kVAR} \end{aligned}$$

Half load: output 17.5 kW, $\eta = 0.7$, pf 0.6

$$\begin{aligned} P_{HL} &= \frac{17.5}{0.7} = 25\ \text{kW} \\ Q_{HL} &= 25 \times \tan(\cos^{-1}0.6) = 25 \times 1.3333 = 33.33\ \text{kVAR} \end{aligned}$$

No load: current is 25% of full-load current, pf 0.1

$$\begin{aligned} S_{FL} &= \frac{38.89}{0.9} = 43.21\ \text{kVA}, \quad S_{NL} = 0.25 \times 43.21 = 10.80\ \text{kVA} \\ P_{NL} &= 10.80 \times 0.1 = 1.080\ \text{kW}, \quad Q_{NL} = 10.80 \times \sin(\cos^{-1}0.1) = 10.75\ \text{kVAR} \end{aligned}$$

Capacitor rating (line pf 0.8 at half load)

$$\begin{aligned} Q_C &= P_{HL}(\tan\phi_1 - \tan\phi_2) = 25(1.3333 - 0.75) = 14.58\ \text{kVAR} \end{aligned}$$

(i) Line pf at full load

$$\begin{aligned} Q' &= 18.83 - 14.58 = 4.25\ \text{kVAR (lagging)} \\ \cos\phi &= \frac{38.89}{\sqrt{38.89^2 + 4.25^2}} = 0.994\ \text{lagging} \end{aligned}$$

(ii) Line pf at no load

$$\begin{aligned} Q' &= 10.75 - 14.58 = -3.84\ \text{kVAR (leading)} \\ \cos\phi &= \frac{1.080}{\sqrt{1.080^2 + 3.84^2}} = 0.271\ \text{leading} \end{aligned}$$

Answer: Capacitor $= 14.58$ kVAR; line pf (i) full load $\approx 0.994$ lagging, (ii) no load $\approx 0.271$ leading.