Question

A factory has a load of \(400\ \text{kW}\) at a power factor of \(0.8\) lagging. If the power factor is to be improved to unity, the required KVAR of the capacitor bank is

• A. \(400\ \text{KVAR}\)
• B. \(300\ \text{KVAR}\)
• C. \(500\ \text{KVAR}\)
• D. \(100\ \text{KVAR}\)

Hint:

For power factor correction, capacitor KVAR is \(P(\tan\phi_1-\tan\phi_2)\).

Answer 1

The Correct Option is B

Concept: Reactive power compensation required for power factor correction is: \[ Q_c=P(\tan\phi_1-\tan\phi_2) \] where \(\phi_1\) is initial power factor angle and \(\phi_2\) is final power factor angle.

Step 1: Given active power: \[ P=400\ \text{kW} \] Initial power factor: \[ \cos\phi_1=0.8 \]

Step 2: Find \(\sin\phi_1\). \[ \sin\phi_1=\sqrt{1-\cos^2\phi_1} \] \[ \sin\phi_1=\sqrt{1-0.8^2} \] \[ \sin\phi_1=\sqrt{1-0.64} \] \[ \sin\phi_1=0.6 \]

Step 3: Find \(\tan\phi_1\). \[ \tan\phi_1=\frac{\sin\phi_1}{\cos\phi_1} \] \[ \tan\phi_1=\frac{0.6}{0.8}=0.75 \]

Step 4: Final power factor is unity: \[ \cos\phi_2=1 \] So: \[ \phi_2=0^\circ \] \[ \tan\phi_2=0 \]

Step 5: Calculate capacitor KVAR. \[ Q_c=P(\tan\phi_1-\tan\phi_2) \] \[ Q_c=400(0.75-0) \] \[ Q_c=300\ \text{KVAR} \] Therefore: \[ \boxed{300\ \text{KVAR}} \]