1. Review 1

2. Review of Phasors Goal of phasor analysis is to simplify the analysis of constant frequency ac systems: v(t) = V max cos(  t +  v ), i(t) = I max cos(  t +  I ), where:  v(t) and i(t) are the instantaneous voltage and current as a function of time t,  is the angular frequency (2πf, with f the frequency in Hertz), V max and I max are the magnitudes of voltage and current sinusoids,  v and  I are angular offsets of the peaks of sinusoids from a reference waveform. Root Mean Square (RMS) voltage of sinusoid: 2

3. Phasor Representation 3

4. Phasor Analysis (Note: Z is a complex number but not a phasor). 4

5. Complex Power 5

6. Complex Power, cont’d 6

7. Complex Power (Note: S is a complex number but not a phasor.) 7

8. Complex Power, cont’d 8

9. example 9 Z L =jwL=j*1000*1*10^-3 =j

10. 10

11. Example Earlier we found I = 20  -6.9  amps = 1600W + j1200VAr Power flowing from source to load at bus 11

12. Power Consumption in Devices 12

13. Example First solve basic circuit I 13

14. Example, cont’d Now add additional reactive power load and re-solve, assuming that load voltage is maintained at 40 kV. 14

15. Power System Notation Power system components are usually shown as “one-line diagrams.” Previous circuit redrawn. Arrows are used to show loads Generators are shown as circles Transmission lines are shown as a single line 15

16. Reactive Compensation Key idea of reactive compensation is to supply reactive power locally. In the previous example this can be done by adding a 16 MVAr capacitor at the load. Compensated circuit is identical to first example with just real power load. Supply voltage magnitude and line current is lower with compensation. 16

17. Reactive Compensation, cont’d Reactive compensation decreased the line flow from 564 Amps to 400 Amps. This has advantages: – Lines losses, which are equal to I 2 R, decrease, – Lower current allows use of smaller wires, or alternatively, supply more load over the same wires, – Voltage drop on the line is less. Reactive compensation is used extensively throughout transmission and distribution systems. Capacitors can be used to “correct” a load’s power factor to an arbitrary value. 17

18. Power Factor Correction Example 18

19. Distribution System Capacitors 19

20. Balanced 3 Phase (  ) Systems A balanced 3 phase (  ) system has: – three voltage sources with equal magnitude, but with an angle shift of 120 , – equal loads on each phase, – equal impedance on the lines connecting the generators to the loads. Bulk power systems are almost exclusively 3 . Single phase is used primarily only in low voltage, low power settings, such as residential and some commercial. Single phase transmission used for electric trains in Europe. 20

21. Balanced 3  -- Zero Neutral Current 21

22. Advantages of 3  Power Can transmit more power for same amount of wire (twice as much as single phase). Total torque produced by 3  machines is constant, so less vibration. Three phase machines start more easily than single phase machines. 22

23. Three Phase - Wye Connection There are two ways to connect 3  systems: – Wye (Y), and – Delta (  ). 23

24. Wye Connection Line Voltages V an V cn V bn V ab V ca V bc -V bn Line to line voltages are also balanced. (α = 0 in this case) 24

25. Wye Connection, cont’d We call the voltage across each element of a wye connected device the “phase” voltage. We call the current through each element of a wye connected device the “phase” current. Call the voltage across lines the “line-to-line” or just the “line” voltage. Call the current through lines the “line” current. 25

26. Delta Connection I ca IcIc I ab I bc IaIa IbIb 26

27. Three Phase Example Assume a  -connected load, with each leg Z = 100  20  is supplied from a 3  13.8 kV (L-L) source 27

28. Three Phase Example, cont’d 28

29. Delta-Wye Transformation 29

30. Delta-Wye Transformation Proof 30 + -

31. Delta-Wye Transformation, cont’d 31

32. 3 phase power calculation 32