FdEng/HNC – Nuclear Engineering Module SC1124 (L4) – Electric Power

1. FdEng/HNC – Nuclear Engineering Module SC1124 (L4) – Electric Power SINGLE PHASE - POWER & POWER FACTOR

2. A.C. SINGLE PHASE A REVEIW

3. Contained in this presentation are a number of questions on single phase

4. Reproduce waveform to include current V I With voltage applied across the resistor current will flow in proportion as I = V/R Where V and Iare r.m.s values Therefore the current and voltage will be IN PHASE

5. Reproduce waveform to include current V I Current will be at maximum when the voltage across C is at a minimum and at minimum when the voltage across is at maximum. THE CURRENT IS SAID TO BE 90O OUT OF PHASE WITH THE VOLTAGE AND LEADING

6. VSUPPLY In a series circuit the current is the same in any part of the circuit. It is the voltage across each component that is different Reproduce resultant waveform including voltage + = When R and C are connected together then the resultant voltage is between 0 and 90 degree out of phase with the supply current. If R is larger than XC then goes towards 0. If XC larger than R then tends towards 90 NOTE IN EXAMPLE R = XC = 2W I = 0.5A

7. Sketch impedance triangle for above. State formula to obtain impedance Z = Vs/Is = ÖR² - XC² Sketch a labelled including phase angle phasor diagram for above

8. Find (a) Impedance (b) Current (c) Phase angle NOTE -XCis indicating the convention of capacitive reactance in the fourth quadrant (a) Z = ÖR² -XC² = Ö80² - 60² = 100W (b) I = Vs/Z = 230/100 = 2.3 Amps (c) q = cos-1 R/Z = cos-1 80/100 = 36.86o

9. Reproduce waveform to include current V I When voltage is applied the current builds slow so that when voltage is maximum current is minimum and when current maximum voltage is at a minimum. THE CURRENT IS SAID TO BE 90O OUT OF PHASE WITH THE VOLTAGE AND LAGGING

10. VR VL In a series circuit the current is the same in any part of the circuit. It is the voltage across each component that is different - each adding by phasor addition to the total VS I Reproduce resultant waveform including voltage VS I Sketch a phasor of above circuit

11. Find (a) Impedance (b) Current (c) Phase angle (a) Z = Ö R² + XL² = Ö 6² + 18² = 19 (b) I = V/Z = 230/19 = 12.1Amps (c) q = cos-1R/Z = 6/19 = 71.6o

12. Sketch V waveform for the circuit V I Find (a) Impedance (b) Current (c) Phase angle NOTE Since XC and XL are equal then their effect on the circuit is equal and opposite cancelling out. So Z is just the resistance and the circuit in phase and the p.f will be unity. See next slide (a) Z = Ö R² +( XC² - XL²) = Ö 25² + (25² - 25²) = 25 (b) I = V/Z = 200/25 = 8Amps (c ) 0o See Note

13. Circuit currentVoltage across CVoltage across L Voltage across R and supply

14. VL VR I Ref VS VC Draw a phasor diagram labelling VR, VC , VL& VS

15. IRL State a reason for C 1. For power factor correction 2. Could also be used to prevent voltage spikes to the supply IC IS V Sketch phasor diagram for above circuit. Remember V is common on parallel circuits and is used as reference

16. Sketch power waveform P V I Since voltage and current are in phase then as power P = V x I then power becomes all positive because: for positive half cycles (+ V) x (+ I) = +P for negative half cycles (- V) x (- I) = +P Note: we are using r.m,.s values of V and I

17. Sketch power waveform P V I Since V and I are out of phase then at time either V is positive and I negative or V- and I+ leading to negative power. As the amplitude of both - and + are equal then overall power is zero.

18. Sketch power waveform P V I With R and C in series the overall power is that provided by the resistor the effect of C is to add its reactive power thus the circuit has apparently more power supplied than used. This is illustrated next.

19. kVA Apparent power kW True power (resistive) VAR Reactive power

20. kVA kVAR kW Sketch a power triangle State the formula for finding power in a single phase circuit Power P = V x I x cos q or kVA cos q State the meaning of power factor Ratio of true to apparent power or p.f. = kW/kVA or p.f. = True power____ Apparent power

21. kVA (S) Kvar(Q) kW (P) Sketch a power triangle State the formula for finding power in a single phase circuit Power P = V x I x cos φ or kVA cos φ State the meaning of power factor Ratio of true to apparent power or p.f. = kW/kVA or p.f. = True power____ Apparent power

22. An electric motor connected to 230V single phase 50Hz draws 10 amps and supplies 2kW. Find (a) p.f. of the circuit (b) kVAr (a) p.f. = kW/kVA = 2000/230 x 10 = 2000/2300 = 0.87 (b) kVA = Ö kW² + kVAr² so kVAr = Ö kVA - kW = Ö 2.3² - 2² = 1.135

23. State the reason for improving the power factor of a circuit To reduce the circuit current. This leads to smaller cables and switchgear State the reason a large value resistor is placed across capacitor banks Safety (so that capacitors do not remain charged when no circuit voltage is applied)

24. Supplementary - Revision Examples of Series and Parallel connection of Capacitance Three capacitor 4µF, 6µF and 10µF are connected in (a) series (b) parallel. Find the total circuit capacitance in each case. (a) 1/CT = 1/C1 + 1/C2 + 1/C3 = 1/4 + 1/6 + 1/10 (4 x-1+ 6 x-1 + 10 x-1) = 0.516 so CT = 0.516 x-1 = 1.935µF (b) CT = C1 + C2 + C3 = 4 + 6 + 10 = 20µF NOTE x-1 = 1/x on calculator