Question 5.9: Power-Factor Correction A50-kW load operates from a 60-Hz 10...

First, we find the load power angle:

\theta_L=\mathrm{arccos}\left(0.6\right) =53.13^\circ

Then, we use the power-triangle concept to find the reactive power of the load. Hence,

Q_L=P_L\tan{(\theta_L)}=66.67\mathrm{~kVAR}

After adding the capacitor, the power will still be 50 kW and the power angle will become

\theta_\mathrm{new}=\mathrm{arccos}\left(0.9\right) =25.84^\circ

The new value of the reactive power will be

Q_\mathrm{new}=P_L\tan{(\theta_\mathrm{new})}=24.22\mathrm{~kVAR}

Thus, the reactive power of the capacitance must be

Q_{C}=Q_\mathrm{new}-Q_{L}=-42.45\mathrm{~kVAR}

Now, we find that the reactance of the capacitor is

X_C=-\frac{V^2_{\mathrm{rms}}}{Q_C} =\frac{\left(10^4\right)^2 }{42,450} =-2356~\Omega

Finally, the angular frequency is

\omega=2\pi 60=377.0

and the required capacitance is

C=\frac{1}{\omega\left|X_C\right| }=\frac{1}{377\times 2356} =1.126~\mu\mathrm{F}