1. 1 1. Power and RMS Values

2. 2 Instantaneous power p(t) flowing into the box Circuit in a box, two wires +−+− Circuit in a box, three wires +−+− +−+− Any wire can be the voltage reference Works for any circuit, as long as all N wires are accounted for. There must be (N – 1) voltage measurements, and (N – 1) current measurements.

3. 3 Average value of periodic instantaneous power p(t)

4. 4 Two-wire sinusoidal case Power factor Average power zero average

5. 5 Root-mean squared value of a periodic waveform with period T Apply v(t) to a resistor Compare to the average power expression rms is based on a power concept, describing the equivalent voltage that will produce a given average power to a resistor The average value of the squared voltage compare

6. 6 Root-mean squared value of a periodic waveform with period T For the sinusoidal case

7. 7 Given single-phase v(t) and i(t) waveforms for a load Determine their magnitudes and phase angles Determine the average power Determine the impedance of the load Using a series RL or RC equivalent, determine the R and L or C

8. 8 Determine voltage and current magnitudes and phase angles Voltage cosine has peak = 100V, phase angle = -90º Current cosine has peak = 50A, phase angle = -135º Using a cosine reference, Phasors

9. 9 The average power is

10. 10 Voltage – Current Relationships

11. 11 Thanks to Charles Steinmetz, Steady-State AC Problems are Greatly Simplified with Phasor Analysis (no differential equations are needed) Resistor Inductor Capacitor Time DomainFrequency Domain voltage leads current current leads voltage

12. 12 Problem 10.17

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14. 14

15. 15 Active and Reactive Power Form a Power Triangle P Q Projection of S on the real axis Projection of S on the imaginary axis Complex power S is the power factor

16. 16 Question: Why is there conservation of P and Q in a circuit? Answer: Because of KCL, power cannot simply vanish but must be accounted for Consider a node, with voltage (to any reference), and three currents IAIA IBIB ICIC

17. 17 Voltage and Current Phasors for R’s, L’s, C’s Resistor Inductor Capacitor Voltage and Current in phase Q = 0 Voltage leads Current by 90° Q > 0 Current leads Voltage by 90° Q < 0

18. 18 P Q Projection of S on the real axis Projection of S on the imaginary axis Complex power S

19. 19 also so Resistor, Use rms V, I

20. 20 also so Inductor, Use rms V, I

21. 21 also so Capacitor, Use rms V, I

22. 22 Active and Reactive Power for R’s, L’s, C’s (a positive value is consumed, a negative value is produced) Resistor Inductor Capacitor Active Power PReactive Power Q source of reactive power

23. 23 Now, demonstrate Excel spreadsheet EE411_Voltage_Current_Power.xls to show the relationship between v(t), i(t), p(t), P, and Q

24. 24 A Single-Phase Power Example

25. 25 A Transmission Line Example Calculate the P and Q flows (in per unit) for the loadflow situation shown below, and also check conservation of P and Q. 0.05 + j0.15 pu ohms j0.20 pu mhos P L + jQ L V L = 1.020 /0° V R = 1.010/-10° P R + jQ R ISIS I capL I capR j0.20 pu mhos

26. 26

27. 27

28. 28 0.05 + j0.15 pu ohms j0.20 pu mhos P L + jQ L V L = 1.020 /0° V R = 1.010/-10° P R + jQ R ISIS I capL I capR j0.20 pu mhos

29. 29 RMS of some common periodic waveforms Duty cycle controller DT T V0V0 0 < D < 1 By inspection, this is the average value of the squared waveform

30. 30 RMS of common periodic waveforms, cont. T V0V0 Sawtooth

31. 31 RMS of common periodic waveforms, cont. Using the power concept, it is easy to reason that the following waveforms would all produce the same average power to a resistor, and thus their rms values are identical and equal to the previous example V0V0 V0V0 V0V0 0 -V V0V0 V0V0 V0V0

32. 32 2. Three-Phase Circuits

33. 33 Three Important Properties of Three-Phase Balanced Systems Because they form a balanced set, the a-b-c currents sum to zero. Thus, there is no return current through the neutral or ground, which reduces wiring losses. A N-wire system needs (N – 1) meters. A three- phase, four-wire system needs three meters. A three-phase, three-wire system needs only two meters. The instantaneous power is constant Three-phase, four wire system abcnabcn Reference

34. 34 Observe Constant Three-Phase P and Q in Excel spreadsheet 1_Single_Phase_Three_Phase_Instantaneous_Power.xls

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40. 40 Balanced three-phase systems, no matter if they are delta connected, wye connected, or a mix, are easy to solve if you follow these steps: 1. Convert the entire circuit to an equivalent wye with a grounded neutral. 2. Draw the one-line diagram for phase a, recognizing that phase a has one third of the P and Q. 3. Solve the one-line diagramforline-to-neutral voltages and line currents. 4. If needed, compute line-to-neutral voltages and line currents for phases b and cusing the ±120° relationships. 5. If needed, compute line-to-line voltagesanddelta currents using the 3 and±30° relationships. a n a n Z load + Van – Z line I a a c b – V ab + 3Z load a c b I b I a I c Z line Z l Z l 3Z load 3Z load The “One-Line” Diagram

41. 41 Now Work a Three-Phase Motor Power Factor Correction Example A three-phase, 460V motor draws 5kW with a power factor of 0.80 lagging. Assuming that phasor voltage V an has phase angle zero, Find phasor currents I a and I ab and (note – I ab is inside the motor delta windings) Find the three phase motor Q and S How much capacitive kVAr (three-phase) should be connected in parallel with the motor to improve the net power factor to 0.95? Assuming no change in motor voltage magnitude, what will be the new phasor current I a after the kVArs are added?

42. 42 Now Work a Delta-Wye Conversion Example Part c. Draw a phasor diagram that shows line currents Ia, Ib, and Ic, and load currents Iab, Ibc, and Ica.

43. 43 3. Transformers

44. 44 Single-Phase Transformer Rs jXs Ideal Transformer 7200:240V Rm jXm 7200V 240V Turns ratio 7200:240 (30 : 1) (but approx. same amount of copper in each winding) Φ

45. 45 Short Circuit Test Rs jXs Ideal Transformer 7200:240V Rm jXm 7200V 240V Turns ratio 7200:240 (but approx. same amount of copper in each winding) Φ Short circuit test: Short circuit the 240V-side, and raise the 7200V-side voltage to a few percent of 7200, until rated current flows. There is almost no core flux so the magnetizing terms are negligible. + Vsc - Isc

46. 46 Open Circuit Test Rs jXs Ideal Transformer 7200:240V Rm jXm 7200V 240V Turns ratio 7200:240 (but approx. same amount of copper in each winding) Φ + Voc - Open circuit test: Open circuit the 7200V-side, and apply 240V to the 240V-side. The winding currents are small, so the series terms are negligible. Ioc

47. 47 Single Phase Transformer. Percent values are given on transformer base. Winding 1 kv = 7.2, kVA = 125 Winding 2 kv = 0.24, kVA = 125 %imag = 0.5 %loadloss = 0.9 %noloadloss = 0.2 %Xs = 2.2 Rs jXs Ideal Transformer 7200:240V Rm jXm 7200V 240V Magnetizing current No load loss Xs Load loss 3. If standard open circuit and short circuit tests are performed on this transformer, what will be the P’s and Q’s (Watts and VArs) measured in those tests? 1. Given the standard percentage values below for a 125kVA transformer, determine the R’s and X’s in the diagram, in Ω. 2. If the R’s and X’s are moved to the 240V side, compute the new Ω values.

48. Distribution Feeder Loss Example Annual energy loss = 2.40% Largest component is transformer no- load loss (45% of the 2.40%) Modern Distribution Transformer: Load loss at rated load (I 2 R in conductors) = 0.75% of rated transformer kW. No load loss at rated voltage (magnetizing, core steel) = 0.2% of rated transformer kW. Magnetizing current = 0.5% of rated transformer amperes

49. 49 Single-Phase Transformer Impedance Reflection by the Square of the Turns Ratio Rs jXs Ideal Transformer 7200:240V Rm jXm 7200V 240V Ideal Transformer 7200:240V 7200V 240V

50. 50 Now Work a Single-Phase Transformer Example Open circuit and short circuit tests are performed on a single-phase, 7200:240V, 25kVA, 60Hz distribution transformer. The results are:  Short circuit test (short circuit the low-voltage side, energize the high-voltage sideso that rated current flows, and measure P sc and Q sc ). Measured P sc = 400W, Q sc = 200VAr.  Open circuit test (open circuit the high-voltage side, apply rated voltage to the low-voltage side, and measure P oc and Q oc ). Measured P oc = 100W, Q oc = 250VAr. Determine the four impedance values (in ohms) for the transformer model shown. Rs jXs Ideal Transformer 7200:240V Rm jXm 7200V 240V Turns ratio 7200:240 (30 : 1) (but approx. same amount of copper in each winding) Φ

51. 51 Y - Y A three-phase transformer can be three separate single-phase transformers, or one large transformer with three sets of windings N1:N2 Rs jXs Ideal Transformer N1 : N2 Rm jXm Wye-Equivalent One-Line Model ANAN Reflect to side 2 using individual transformer turns ratio N1:N2

52. 52 Δ - Δ For Delta-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye N1:N2 Ideal Transformer ANAN Wye-Equivalent One-Line Model Convert side 1 impedances from delta to equivalent wye Then reflect to side 2 using individual transformer turns ratio N1:N2

53. 53 Δ - Y For Delta-Wye Connection Model, Convert the Transformer to Equivalent Wye-Wye N1:N2 Ideal Transformer ANAN Wye-Equivalent One-Line Model Convert side 1 impedances from delta to wye Then reflect to side 2 using three-phase bank line-to-line turns ratio

54. 54 Y - Δ For Wye-Delta Connection Model, Convert the Transformer to Equivalent Wye-Wye N1:N2 Ideal Transformer ANAN Wye-Equivalent One-Line Model So, for all configurations, the equivalent wye-wye transformer ohms can be reflected from one side to the other using the three- phase bank line-to-line turns ratio Reflect to side 2 using three-phase bank line-to- line turns ratio

55. 55 For wye-delta and delta-wye configurations, there is a phase shift in line-to-line voltages because the individual transformer windings on one side are connected line-to-neutral, and on the other side are connected line-to-line But there is no phase shift in any of the individual transformers This means that line-to-line voltages on the delta side are in phase with line-to-neutral voltages on the wye side Thus, phase shift in line-to-line voltages from one side to the other is unavoidable, but it can be managed by standard labeling to avoid problems caused by paralleling transformers

56. 56 Linear ScaleLog10 Scale Saturation – relative permeability decreases rapidly after 1.7 Tesla Relative permeability drops from about 2000 to about 1 (becomes air core) Magnetizing inductance of the core decreases, yielding a highly peaked magnetizing current

57. 57 Residual magnetism Hysteresis Loss is ½ the Area of the Parallelogram per AC Cycle per Cubic Meter of Core Steel