1. AC POWER ANALYSIS

2. 2 Content Average Power Maximum Average Power Transfer Complex Power Power Factor Correction

3. 3 Average Power Average Power, in watts (W), is the average of instantaneous power over one period

4. 4 Average Power Resistive load (R) absorbs power all the time. For a purely resistive circuit, the voltage and the current are in phase (  v =  i ).

5. 5 Average Power Reactive load (L or C) absorbs zero average power. For a purely reactive circuit, the voltage and the current are out of phase by 90 o (  v -  i = ±90).

6. 6 Exercise 11.3 Find the average power supplied by the source and the average power absorb by the resistor

7. 7 Solution The current I is given by The average power supplied by the voltage source is

8. 8 Solution Notice that the average power supplied by the voltage source is same as the power absorbed by the resistor. This result shows the capacitor absorbed zero average power. The average power absorbed by the resistor is The current through the resistor is The voltage across resistor is

9. 9 Practice Problem 11.3 Calculate the average power absorbed by the resistor and the inductor. Then find the average power supplied by the voltage source

10. 10 Solution The current I is given by For the resistor

11. 11 Solution The average power supplied by the voltage source is For the inductor Notice that the average absorbed by the resistor is same as the power supplied by the voltage source. This result shows the inductor also absorbed zero average power.

12. 12 MAXIMUM AVERAGE POWER TRANSFER

13. 13 Maximum Power Transfer For maximum power transfer, the load impedance Z L must equal to the complex conjugate of the Thevenin impedance Z th

14. 14 Maximum Average Power The Maximum Average Power delivered to the load is The current through the load is

15. 15 Maximum Average Power In a situation in which the load is purely real, the load resistance must equal to the magnitude of the Thevenin impedance. By setting R L = R th and X L = -X th, the maximum average power is

16. 16 Exercise 11.5 Determine the load impedance Z L that maximize the power drawn and the maximum average power.

17. 17 Solution First we obtain the Thevenin equivalent To find Z th, consider circuit (a) To find V th, consider circuit (b)

18. 18 Solution From the result obtained, the load impedance draws the maximum power from the circuit when The maximum average power is

19. 19 Practice Problem 11.5 Determine the load impedance Z L that absorbs the maximum average power. Calculate the maximum average power.

20. 20 Solution First we obtain the Thevenin equivalent To find Z th, consider circuit (a) To find V th, consider circuit (b) By using current divider

21. 21 Solution From the result obtained, the load impedance draws the maximum power from the circuit when The maximum average power is

22. 22 Example 11.6 Find the value of R L that will absorbs maximum average power. Then calculate that power.

23. 23 Solution First we obtain the Thevenin equivalent Find Z th Find V th By using voltage divider

24. 24 Solution The value of R L that will absorb the maximum average power is The maximum average power is The current through the load is

25. 25 Practice Problem 11.6 Find the value of R L that will absorbs maximize average power, Then calculate the power.

26. 26 Solution First we obtain the Thevenin equivalent To find Z th let To find V th By using voltage divider and Then

27. 27 Solution The value of R L that will absorb the maximum average power is The maximum average power is The current through the load is

28. 28 Complex Power Apparent Power, S (VA) Real Power, P (Watts) Reactive Power, Q (VAR) Power Factor, cos 

29. 29 Complex Power Complex power is the product of the rms voltage phasor and the complex conjugate of the rms current phasor. Measured in volt-amperes or VA As a complex quantity Its real part is real power, P Its imaginary part is reactive power, Q

30. 30 Complex Power (Derivation)

31. 31 Complex Power (Derivation)

32. 32 Complex Power (Derivation) From derivation, we notice that the real power is and also the reactive power or

33. 33 Real or Average Power The real power is the average power delivered to a load. Measured in watts (W) The only useful power The actual power dissipated by the load

34. 34 Reactive Power The reactive power, Q is the imaginary parts of complex power. The unit of Q is volt-ampere reactive (VAR). It represents a lossless interchange between the load and the source Q = 0 for resistive load (unity pf) Q < 0 for capacitive load (leading pf) Q > 0 for inductive load (lagging pf)

35. 35 Apparent Power The apparent power is the product of rms values of voltage and current Measured in volt-amperes or VA Magnitude of the complex power

36. 36 Power Factor Power factor is the cosine of the phase difference between voltage and current. It is also cosine of the angle of the load impedance.

37. 37 Power Factor The range of pf is between zero and unity. For a purely resistive load, the voltage and current are in phase so that  v -  i = 0 and pf = 1, the apparent power is equal to average power. For a purely reactive load,  v -  i =  90 and pf = 0, the average power is zero.

38. 38 Power Triangular Comparison between the power triangular (a) and the impedance triangular (b).

39. 39 Problem 11.46 For the following voltage and current phasors, calculate the complex power, apparent power, real power and reactive power. Specify whether the pf is leading or lagging. a) V = 220  30 o V rms, I = 0.5  60 o A rms. b) V = 250  -10 o V rms, I = 6.2  -25 o A rms. c) V = 120  0 o V rms, I = 2.4  -15 o A rms. d) V = 160  45 o V rms, I = 8.5  90 o A rms.

40. 40 Solution a)S = VI* = (220  30 o )( 0.5  -60 o ) = 110  -30 o VA = 95.26 – j55 VA Apparent power = 110 VA Real Power = 95.26 W Reactive Power = -55 VAR pf is leading because current leads voltage b)S = VI* = (250  -10 o )(6.2  25 o ) = 1550  15 o VA = 1497.2 + j401.2 VA Apparent power = 1550 VA Real Power = 1497.2 W Reactive Power = 401.2 VAR pf is lagging because current lags voltage c)S = VI* = (120  0 o )( 2.4  15 o ) = 288  15 o VA = 278.2 + j74.54 VA Apparent power = 288 VA Real Power = 278.2 W Reactive Power = 74.54 VAR pf is lagging because current lags voltage d)S = VI* = (160  45 o )(8.5  -90 o ) = 1360  -45 o VA = 961.7 – j961.7 VA Apparent power = 1360 VA Real Power = 961.7 W Reactive Power = -961.7 VAR pf is leading because current leads voltage

41. 41 Problem 11.48 Determine the complex power for the following cases: a) P = 269 W, Q = 150 VAR (capacitive) b) Q = 2000 VAR, pf = 0.9 (leading) c) S = 600 VA, Q = 450 VAR (inductive) d) V rms = 220 V, P = 1 kW, |Z| = 40  (inductive)

42. 42 Solution a)Given P = 269W, Q = 150VAR (capacitive) Complex power, b) Given Q = 2000VAR, pf = 0.9 (leading) Complex power,

43. 43 Solution c) Given S = 600VA, Q = 450VAR (inductive) Complex power,

44. 44 Solution d) Given V rms = 220V, P = 1kW, |Z| = 40  (inductive) Complex power,

45. 45 Problem 11.42 A 110V rms, 60Hz source is applied to a load impedance Z. The apparent power entering the load is 120VA at a power factor of 0.707 lagging. Calculate a) The complex power b) The rms current supplied to the load. c) Determine Z d) Assuming that Z = R + j  L, find the value of R and L.

46. 46 Solution Given S = 120VA, pf = 0.707 = cos    = 45 o a) the complex power b) the rms current supplied to the load

47. 47 Solution d) value of R and L If Z = R + j  L thenZ = 71.278 + j 71.278 c) the impedance Z

48. 48 Problem 11.83 Oscilloscope measurement indicate that the voltage across a load and the current through is are 210  60 o V and 8  25 o A respectively. Determine a) The real power b) The apparent power c) The reactive power d) The power factor

49. 49 Solution a) the real power d) the power factor c) the reactive power b) the apparent power

50. 50 Power Factor Correction The process of increasing the power factor without altering the voltage or current to the original load. It may be viewed as the addition of a reactive element (usually capacitor) in parallel with the load in order to make the power factor closer to unity.

51. 51 Power Factor Correction Normally, most loads are inductive. Thus power factor is improved or corrected by installing a capacitor in parallel with the load. In circuit analysis, an inductive load is modeled as a series combination of an inductor and a resistor.

52. 52 Implementation of Power Factor Correction

53. 53 Calculation If the original inductive load has apparent power S 1, then P = S 1 cos  1 and Q 1 = S 1 sin  1 = P tan  1 If we desired to increased the power factor from cos  1 to cos  2 without altering the real power, then the new reactive power is Q 2 = P tan  2 The reduction in the reactive power is caused by the shunt capacitor is given by Q C = Q 1 – Q 2 = P (tan  1 - tan  2 )

54. 54 Calculation The value of the required shunt capacitance is determined by the formula Notice that the real power, P dissipated by the load is not affected by the power factor correction because the average power due to the capacitor is zero

55. 55 Example 11.15 When connected to a 120V (rms), 60Hz power line, a load absorbs 4 kW at a lagging power factor of 0.8. Find the value of capacitance necessary to raise the pf to 0.95.

56. 56 Solution If the pf = 0.8 then, cos  1 = 0.8   1 = 36.87 o where  1 is the phase difference between the voltage and current. We obtained the apparent power from the real power and the pf as shown below. The reactive power is

57. 57 Solution When the pf raised to 0.95, cos  2 = 0.95   2 = 18.19 o The real power P has not changed. But the apparent power has changed. The new value is The new reactive power is

58. 58 Solution The difference between the new and the old reactive power is due to the parallel addition of the capacitor to the load. The reactive power due to the capacitor is The value of capacitance added is

59. 59 Practice Problem 11.15 Find the value of parallel capacitance needed to correct a load of 140 kVAR at 0.85 lagging pf to unity pf. Assume the load is supplied by a 110V (rms) 60Hz power line.

60. 60 Solution If the pf = 0.85 then, cos  1 = 0.85   1 = 31.79 o where  1 is the phase difference between the voltage and current. We obtained the apparent power from the reactive power and the pf as shown below. The real power is

61. 61 Solution When the pf raised to 1 (unity), cos  2 = 1   2 = 0 o The real power P has not changed. But the apparent power has changed. The new value is The new reactive power is

62. 62 Solution The difference between the new and the old reactive power due to the parallel addition of the capacitor to the load. The reactive power due to the capacitor is The value of capacitance is

63. 63 Problem 11.82 A 240V rms, 60Hz source supplies a parallel combination of a 5 kW heater and a 30 kVA induction motor whose power factor is 0.82. Determine a) The system apparent power b) The system reactive power c) The kVA rating of a capacitor required to adjust the system power factor to 0.9 lagging d) The value of capacitance required

64. 64 Solution For the heater P1 = 5000Q1 = 0 For the 30kVA induction motor, the pf = 0.82 then, cos  1 = 0.82   1 = 34.92 o The real and the reactive power for the induction motor

65. 65 Solution The total system complex power S total = S 1 + S 2 = (P 1 + P 2 ) + j (Q 1 + Q 2 ) = 29600 + j17171 The system apparent power S = |S total | = 34.33kVA The system reactive power Q = 17171 kVAR The system power factor

66. 66 Solution The system pf = 0.865 then, cos  1 = 0.865   1 = 30.12 o The new system pf = 0.9 then, cos  2 = 0.9   2 = 25.84 o The rating for the capacitance required to adjust the power factor to 0.9 Q C = P (tan  1 + tan  2 ) = 29600 (tan 30.12 + tan 25.84) = 2833kVAR

67. 67 Solution The value of capacitance is