Power in Alternating-Current Circuits — Question 4

Question 4

Partial power-factor correction requires the difference of reactive powers

A load has real power P and initial lagging power factor $p_1$. It is supplied at fixed rms voltage V and angular frequency omega. A parallel capacitor is to improve source power factor to lagging $p_2>p_1$.

1. Derive the capacitance in terms of both phase angles.

2. For $p_1=.600$ and $p_2=.900$, find the fraction of the unity-correction capacitance required.

Question 4 - Solution

Partial power-factor correction requires the difference of reactive powers

(a) Method and calculation

$$\phi_1=\cos^{-1}p_1,\quad\phi_2=\cos^{-1}p_2,\quad Q_{\mathrm{before}}=P\tan\phi_1.$$$$\boxed{C=\frac{P(\tan\phi_1-\tan\phi_2)}{\omega V^2}}.$$

(b) Interpretation and further calculation

$$C_{\mathrm{unity}}=P\tan\phi_1/(\omega V^2).$$$$\boxed{C/C_{\mathrm{unity}}=1-\frac{\tan(\cos^{-1}.900)}{\tan(\cos^{-1}.600)}=.636758}.$$

Physical check

Subtract reactive powers, not power factors; the latter are nonlinear functions of phase.