Power factor is crucial in industrial applications:
Example: A factory with 1000kW load at 0.7 PF requires 1429kVA. Improving to 0.95 PF reduces requirement to 1053kVA.
Electric motors are major contributors to poor power factor:
Correction Methods: Capacitor banks, synchronous condensers, or power electronic devices.
Power factor in residential applications:
Standards: Many countries require residential equipment to maintain PF > 0.9.
Financial advantages of power factor correction:
Payback Period: Typical payback for PF correction is 1-3 years.
Power Factor (PF) is a measure of how effectively electrical power is being used. It represents the ratio of real power (doing useful work) to apparent power (total power drawn from the source).
A 50HP motor draws 100A at 480V with a power factor of 0.85. Calculate the real and reactive power.
Solution:
Apparent Power: S = √3 × V × I = 1.732 × 480 × 100 = 83.1 kVA
Real Power: P = S × PF = 83.1 × 0.85 = 70.6 kW
Reactive Power: Q = S × sin(φ) = 83.1 × 0.527 = 43.8 kVAR
A 500kW load operates at 0.7 PF. Calculate the capacitor size needed to improve PF to 0.95.
Solution:
φ₁ = arccos(0.7) = 45.57°, φ₂ = arccos(0.95) = 18.19°
Qc = 500 × (tan(45.57°) - tan(18.19°)) = 500 × (1.02 - 0.33) = 345 kVAR
A factory pays $0.10/kVAR penalty for PF below 0.9. Current load: 1000kW at 0.75 PF.
Before Correction: S = 1000/0.75 = 1333 kVA, Q = 882 kVAR
Monthly Penalty: 882 × $0.10 × 730 hours = $64,386
After Correction to 0.95 PF: Q = 329 kVAR, No penalty