Power Factor Corrections

LessonAbout 1.5 h

Power Factor Correction

Power Factor Correction is the process of adjusting the power factor of an AC power system towards unity by installing reactive power compensation equipment.

1. Methods to Correct Low Power Factor

  • Arc Lamps: low lagging power factor operation
  • Induction Motors: AC motors operating under partial load
  • Industrial Furnaces: heating equipment drawing reactive power
  • High Line Losses: increased current causes larger conductor \(I^2R\) losses
  • Equipment Overloading: generators, transformers, and lines become thermally overloaded
  • Large Voltage Drop: higher line current increases voltage drop

1.1 Static Condensers

  • Leading Reactive Power: capacitors supply leading kVAR to offset lagging load kVAR
  • Zero Active Power: static condensers absorb virtually zero real power
  • Parallel Connection: connected in parallel across load or busbar

Static Capacitor kVAR Rating

\[ Q_c = P (\tan\theta_1 - \tan\theta_2) \]
\[ \theta_1 = \cos^{-1}(\text{pf}_1) \qquad\qquad \theta_2 = \cos^{-1}(\text{pf}_2) \]

Where: \(Q_c\) = required capacitor reactive power (kVAR), \(P\) = load active power (kW), \(\text{pf}_1\) = original lagging power factor, \(\text{pf}_2\) = corrected target power factor, \(\theta_1, \theta_2\) = initial and final power factor angles (deg)

Capacitor Reactance and Capacitance

\[ X_c = \dfrac{V_p^2}{Q_{\text{phase}}} \qquad\qquad C = \dfrac{1}{2\pi f X_c} \]

Where: \(X_c\) = capacitive reactance per phase (\(\Omega\)), \(V_p\) = phase voltage across capacitor (V), \(Q_{\text{phase}}\) = capacitor reactive power per phase (VAR), \(f\) = supply frequency (Hz), \(C\) = capacitance per phase (F)

EE Board Exam October 1980

Example. A balanced \(500\ \text{kVA}\), 3-phase, \(440\ \text{V}\), \(60\ \text{Hz}\) inductive load operates at a power factor of \(75\%\) lagging. Determine the total capacitor kVAR required to improve the power factor to \(95\%\) lagging.

Solution.

\[ P = 500(0.75) = 375\ \text{kW} \implies Q_c = 375\left[\tan(\cos^{-1} 0.75) - \tan(\cos^{-1} 0.95)\right] = 207.46\ \text{kVAR} \]

Answer: \(207.46\ \text{kVAR}\)

1.2 Synchronous Condensers

  • Over-Excited Motor: synchronous motor operated with over-excitation
  • Variable kVAR: leading kVAR smoothly controlled by varying field current
  • Mechanical Drive: can simultaneously drive shaft load and supply kVAR

Synchronous Condenser Load Relations

\[ P_t = P_L + P_{\text{syn}} \]
\[ Q_t = Q_L - Q_{\text{syn}} \]
\[ Q_{\text{syn}} = P_{\text{syn}} \tan\theta_{\text{syn}} \]

Where: \(P_t\) = total combined active power (kW), \(P_L\) = original load active power (kW), \(P_{\text{syn}}\) = synchronous motor active power input (kW), \(Q_t\) = total net reactive power (kVAR), \(Q_L\) = original load reactive power (kVAR), \(Q_{\text{syn}}\) = synchronous motor leading reactive power (kVAR), \(\theta_{\text{syn}}\) = synchronous motor power factor angle (deg)

Synchronous Condenser Rating (Cosine Law)

\[ S_{\text{syn}} = \sqrt{S_L^2 + S_t^2 - 2 S_L S_t \cos(\theta_L - \theta_t)} \]
\[ \text{pf}_{\text{syn}} = \dfrac{P_{\text{syn}}}{S_{\text{syn}}} \]

Where: \(S_{\text{syn}}\) = apparent power rating of synchronous condenser (kVA), \(S_L\) = original load apparent power (kVA), \(S_t\) = total combined apparent power (kVA), \(\theta_L\) = original load power factor angle (deg), \(\theta_t\) = total corrected power factor angle (deg), \(\text{pf}_{\text{syn}}\) = operating leading power factor of synchronous motor

EE Board Exam April 1988

Example. An existing industrial plant has an average load of \(900\ \text{kW}\) at \(0.60\) power factor lagging. To raise the overall power factor to \(0.92\) lagging, a synchronous motor driving a DC generator is installed. If the input to the synchronous motor is \(250\ \text{kW}\), calculate its kVA rating.

Solution.

\[ S_L = \dfrac{900}{0.60} = 1500\ \text{kVA} \qquad\qquad S_t = \dfrac{900 + 250}{0.92} = 1250\ \text{kVA} \]
\[ S_{\text{syn}} = \sqrt{(1500)^2 + (1250)^2 - 2(1500)(1250)\cos(\cos^{-1} 0.60 - \cos^{-1} 0.92)} = 753\ \text{kVA} \]

Answer: \(753\ \text{kVA}\)

2. Formula Summary

Formula Name Formula Variables
Static Capacitor kVAR Rating \(Q_c = P (\tan\theta_1 - \tan\theta_2)\) \(Q_c\) = capacitor reactive power (kVAR)
\(P\) = active power (kW)
\(\theta_1, \theta_2\) = pf angles (deg)
Capacitive Reactance per Phase \(X_c = \dfrac{V_p^2}{Q_{\text{phase}}}\) \(X_c\) = capacitive reactance (\(\Omega\))
\(V_p\) = phase voltage (V)
\(Q_{\text{phase}}\) = phase reactive power (VAR)
Static Capacitance per Phase \(C = \dfrac{1}{2\pi f X_c}\) \(C\) = capacitance (F)
\(f\) = frequency (Hz)
\(X_c\) = capacitive reactance (\(\Omega\))
Total Combined Active Power \(P_t = P_L + P_{\text{syn}}\) \(P_t\) = total active power (kW)
\(P_L\) = load active power (kW)
\(P_{\text{syn}}\) = synchronous motor power (kW)
Total Combined Reactive Power \(Q_t = Q_L - Q_{\text{syn}}\) \(Q_t\) = total reactive power (kVAR)
\(Q_L\) = load reactive power (kVAR)
\(Q_{\text{syn}}\) = synchronous motor kVAR
Synchronous Motor Leading kVAR \(Q_{\text{syn}} = P_{\text{syn}} \tan\theta_{\text{syn}}\) \(Q_{\text{syn}}\) = motor leading kVAR
\(P_{\text{syn}}\) = motor active power (kW)
\(\theta_{\text{syn}}\) = motor pf angle (deg)
Synchronous Condenser Rating \(S_{\text{syn}} = \sqrt{S_L^2 + S_t^2 - 2 S_L S_t \cos(\theta_L - \theta_t)}\) \(S_{\text{syn}}\) = synchronous motor rating (kVA)
\(S_L\) = load apparent power (kVA)
\(S_t\) = total apparent power (kVA)
\(\theta_L, \theta_t\) = pf angles (deg)
Synchronous Motor Power Factor \(\text{pf}_{\text{syn}} = \dfrac{P_{\text{syn}}}{S_{\text{syn}}}\) \(\text{pf}_{\text{syn}}\) = synchronous motor power factor
\(P_{\text{syn}}\) = active power (kW)
\(S_{\text{syn}}\) = apparent power (kVA)