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Chapter 4 (power factor correction)
- 1. 1 AC POWER CALCULATIONAC POWER CALCULATION Power Factor CorrectionPower Factor Correction
- 2. 2 Power FactorPower Factor As we have seen before (for sinusoidal voltage and current), p.f. = cos (θv−θi) The REAL power (or average power) is transformed into useful energy e.g. heat, mechanical, light, sound, etc For a given Vrms and P, loads with high power factor draw LESS current compared with loads with low power factor P = Vrms Irms cos (θv−θi) Load + Vrms − Irms Vrms = Vrms∠θv Irms = Irms∠θi
- 3. 3 Power FactorPower Factor For a given Vrms and P, loads with high power factor draw LESS current compared with loads with low power factor Pave p.f. = cos θ1 Vrms I1, rms Pave p.f. = cos θ2 Vrms I2, rms cos θ1 > cos θ2 θ1 < θ2 Pave Pave θ1 θ2 rms rms,1 V I 1S = rms rms,2 V I 2S = 1S Q1 Q2 rms,2rms,1 II < 2S
- 4. 4 Power FactorPower Factor For a given Vrms and P, loads with high power factor draw LESS current compared with loads with low power factor Less current results in LESS losses during transmission Utility company (TNB) charge more to loads with LOW power factor Therefore, it is desirable to increase the power factor LoadSource + − I2 R
- 5. 5 Power Factor CorrectionPower Factor Correction Process of increasing the power factor without altering the voltage or current to the original load PL QL θ1 θ2 QC QT (voltage and current to original load retained) Before C added, S = PL + jQL p.f. = cos θ1 After C added, S = PL + j(QL – QC) p.f. = cos θ2 i.e. increased + VL − IL + Vs (rms) −
- 6. 6 Power Factor CorrectionPower Factor Correction PL θ1 QL QC θ2 QT How do we calculate C?How do we calculate C? Voltage across C = Vs (rms) Impedance of C = C 1 j ω − 2 21 2 2 )tan(tan 1 s L s cs c PQ C C Q VV V ω θθ ω ω − ==⇒=
- 7. 7 LV Static CapacitorLV Static Capacitor 3 phase connection cct3 phase connection cct
- 8. 8 LV Capacitor Bank c/w P.F RegulatorLV Capacitor Bank c/w P.F Regulator
- 9. 9 LV Capacitor Bank c/w P.F RegulatorLV Capacitor Bank c/w P.F Regulator
- 10. 10
- 11. 11
- 12. 12
- 13. 13 Prob 1# A mud pump has an output power of 2 kW, an efficiency of 70% and a power factor of 0.5 lagging when operated from a 230V, 50Hz supply. It is required to improve the power factor to 0.95 lagging by connecting a capacitor in parallel with the motor. Determine:
- 14. 14 a)The current taken by the motor and its phase angle. 01 3 3 6050.0cos 45.25 50.0110 104.1 cos 4.1)102)(70.0( == === === − θ θ η A x x V P I kWxPP i M oi
- 15. 15 b) The supply current after power factor correction AI Ioa A Ioa M 40.13 19.18cos73.12 73.1260cos45.25 60cos 19.1895.0cos'6050.0cos 0 0 0 0101 =∴ == == = ==⇒== −− θθ
- 16. 16 c) The current taken by the capacitor. AAacabI A Iac A Iab C M 86.17)18.404.22( 18.419.18sin)40.13( 19.18sin 04.2260sin45.25 60sin 19.1895.0cos'6050.0cos 0 0 0 0 0101 =−=−= == = == = ==⇒== −− θθ
- 17. 17 d) Draw phasor diagram by showing phase angle and magnitude. IICC = 17.86A= 17.86A O IImm =25.45=25.45 a c b 18.1918.1900 606000 I =13.4AI =13.4A IICC = 17.86A= 17.86A
- 18. 18 e) The capacitance of the capacitor. F fV I C fCV fC V X V I c c C µ ππ π π 17.247 )230)(50(2 86.17 2 2 2 1 === ===