![8
Complex Power
max
max
max max
( ) ( ) ( )
v(t) = cos( )
(t) = cos( )
1
cos cos [cos( ) cos( )]
2
1
( ) [cos( )
2
cos(2 )]
V
I
V I
V I
p t v t i t
V t
i I t
p t V I
t
Power](https://image.slidesharecdn.com/ece4762011lect2-241124114439-0454e190/85/Power-system-analysis-complex-power-reactive-compensation-8-320.jpg)
![9
Complex Power, cont’d
max max
0
max max
1
( ) [cos( ) cos(2 )]
2
1
( )
1
cos( )
2
cos( )
= =
V I V I
T
avg
V I
V I
V I
p t V I t
P p t dt
T
V I
V I
Power Factor
Average P
Angle
ower](https://image.slidesharecdn.com/ece4762011lect2-241124114439-0454e190/85/Power-system-analysis-complex-power-reactive-compensation-9-320.jpg)
Power system analysis complex power reactive compensation
- 1. Lecture 2 Complex Power, Reactive Compensation, Three Phase Professor Tom Overbye Department of Electrical and Computer Engineering ECE 476 POWER SYSTEM ANALYSIS
- 2. 2 Announcements For lectures 2 through 3 please be reading Chapters 1 and 2
- 3. 3 Review of Phasors Goal of phasor analysis is to simplify the analysis of constant frequency ac systems v(t) = Vmax cos(t + v) i(t) = Imax cos(t + I) Root Mean Square (RMS) voltage of sinusoid 2 max 0 1 ( ) 2 T V v t dt T
- 4. 4 Phasor Representation j ( ) Euler's Identity: e cos sin Phasor notation is developed by rewriting using Euler's identity ( ) 2 cos( ) ( ) 2 Re (Note: is the RMS voltage) V V j t j v t V t v t V e V
- 5. 5 Phasor Representation, cont’d The RMS, cosine-referenced voltage phasor is: ( ) Re 2 cos sin cos sin V V j V j j t V V I I V V e V v t Ve e V V j V I I j I (Note: Some texts use “boldface” type for complex numbers, or “bars on the top”)
- 6. 6 Advantages of Phasor Analysis 0 2 2 Resistor ( ) ( ) ( ) Inductor ( ) 1 1 Capacitor ( ) (0) C Z = Impedance R = Resistance X = Reactance X Z = =arctan( ) t v t Ri t V RI di t v t L V j LI dt i t dt v V I j C R jX Z R X R Device Time Analysis Phasor (Note: Z is a complex number but not a phasor)
- 7. 7 RL Circuit Example 2 2 ( ) 2 100cos( 30 ) 60Hz R 4 3 4 3 5 36.9 100 30 5 36.9 20 6.9 Amps i(t) 20 2 cos( 6.9 ) V t t f X L Z V I Z t
- 8. 8 Complex Power max max max max ( ) ( ) ( ) v(t) = cos( ) (t) = cos( ) 1 cos cos [cos( ) cos( )] 2 1 ( ) [cos( ) 2 cos(2 )] V I V I V I p t v t i t V t i I t p t V I t Power
- 9. 9 Complex Power, cont’d max max 0 max max 1 ( ) [cos( ) cos(2 )] 2 1 ( ) 1 cos( ) 2 cos( ) = = V I V I T avg V I V I V I p t V I t P p t dt T V I V I Power Factor Average P Angle ower
- 10. 10 Complex Power * cos( ) sin( ) P = Real Power (W, kW, MW) Q = Reactive Power (var, kvar, Mvar) S = Complex power (VA, kVA, MVA) Power Factor (pf) = cos If current leads voltage then pf is leading If current V I V I V I S V I j P jQ lags voltage then pf is lagging (Note: S is a complex number but not a phasor)
- 11. 11 Complex Power, cont’d 2 1 Relationships between real, reactive and complex power cos sin 1 Example: A load draws 100 kW with a leading pf of 0.85. What are (power factor angle), Q and ? -cos 0.85 31.8 100 0. P S Q S S pf S kW S 117.6 kVA 85 117.6sin( 31.8 ) 62.0 kVar Q
- 12. 12 Conservation of Power At every node (bus) in the system – Sum of real power into node must equal zero – Sum of reactive power into node must equal zero This is a direct consequence of Kirchhoff’s current law, which states that the total current into each node must equal zero. – Conservation of power follows since S = VI*
- 13. 13 Conversation of Power Example Earlier we found I = 20-6.9 amps * * R 2 R R * L 2 L L 100 30 20 6.9 2000 36.9 VA 36.9 pf = 0.8 lagging S 4 20 6.9 20 6.9 P 1600 (Q 0) S 3 20 6.9 20 6.9 Q 1200var (P 0) R L S V I V I W I R V I j I X
- 14. 14 Power Consumption in Devices 2 Resistor Resistor 2 Inductor Inductor L 2 Capacitor Capacitor C Capa Capacitor Resistors only consume real power P Inductors only consume reactive power Q Capacitors only generate reactive power 1 Q Q C I R I X I X X C V 2 citor C C (Note-some define X negative) X
- 15. 15 Example * 40000 0 400 0 Amps 100 0 40000 0 (5 40) 400 0 42000 16000 44.9 20.8 kV S 44.9k 20.8 400 0 17.98 20.8 MVA 16.8 6.4 MVA V I V j j V I j First solve basic circuit
- 16. 16 Example, cont’d Now add additional reactive power load and resolve 70.7 0.7 lagging 564 45 Amps 59.7 13.6 kV S 33.7 58.6 MVA 17.6 28.8 MVA Load Z pf I V j
- 17. 17 59.7 kV 17.6 MW 28.8 MVR 40.0 kV 16.0 MW 16.0 MVR 17.6 MW 16.0 MW - 16.0 MVR 28.8 MVR 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
- 18. 18 Reactive Compensation 44.94 kV 16.8 MW 6.4 MVR 40.0 kV 16.0 MW 16.0 MVR 16.8 MW 16.0 MW 0.0 MVR 6.4 MVR 16.0 MVR 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
- 19. 19 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 I2 R decrease – Lower current allows utility to use small wires, or alternatively, supply more load over the same wires – Voltage drop on the line is less Reactive compensation is used extensively by utilities Capacitors can be used to “correct” a load’s power factor to an arbitrary value.
- 20. 20 Power Factor Correction Example 1 1 desired new cap cap Assume we have 100 kVA load with pf=0.8 lagging, and would like to correct the pf to 0.95 lagging 80 60 kVA cos 0.8 36.9 PF of 0.95 requires cos 0.95 18.2 S 80 (60 Q ) 60 - Q ta 80 S j j cap cap n18.2 60 Q 26.3 kvar Q 33.7 kvar
- 22. 22 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
- 23. 23 Balanced 3 -- No Neutral Current * * * * (1 0 1 1 3 n a b c n an an bn bn cn cn an an I I I I V I Z S V I V I V I V I
- 24. 24 Advantages of 3 Power Can transmit more power for same amount of wire (twice as much as single phase) Torque produced by 3 machines is constrant Three phase machines use less material for same power rating Three phase machines start more easily than single phase machines
- 25. 25 Three Phase - Wye Connection There are two ways to connect 3 systems – Wye (Y) – Delta () an bn cn Wye Connection Voltages V V V V V V
- 26. 26 Wye Connection Line Voltages Van Vcn Vbn Vab Vca Vbc -Vbn (1 1 120 3 30 3 90 3 150 ab an bn bc ca V V V V V V V V V Line to line voltages are also balanced (α = 0 in this case)
- 27. 27 Wye Connection, cont’d Define voltage/current across/through device to be phase voltage/current Define voltage/current across/through lines to be line voltage/current 6 * 3 3 1 30 3 3 j Line Phase Phase Line Phase Phase Phase V V V e I I S V I
- 28. 28 Delta Connection Ica Ic Iab Ibc Ia Ib a b * 3 For the Delta phase voltages equal line voltages For currents I 3 I I 3 ab ca ab bc ab c ca bc Phase Phase I I I I I I I S V I
- 29. 29 Three Phase Example Assume a -connected load is supplied from a 3 13.8 kV (L-L) source with Z = 10020 13.8 0 13.8 0 13.8 0 ab bc ca V kV V kV V kV 13.8 0 138 20 138 140 138 0 ab bc ca kV I amps I amps I amps
- 30. 30 Three Phase Example, cont’d * 138 20 138 0 239 50 amps 239 170 amps 239 0 amps 3 3 13.8 0 kV 138 amps 5.7 MVA 5.37 1.95 MVA pf cos20 lagging a ab ca b c ab ab I I I I I S V I j
- 31. 31 Delta-Wye Transformation Y phase To simplify analysis of balanced 3 systems: 1) Δ-connected loads can be replaced by 1 Y-connected loads with Z 3 2) Δ-connected sources can be replaced by Y-connected sources with V 3 30 Line Z V
- 32. 32 Delta-Wye Transformation Proof From the side we get Hence ab ca ab ca a ab ca a V V V V I Z Z Z V V Z I
- 33. 33 Delta-Wye Transformation, cont’d a From the side we get ( ) ( ) (2 ) Since I 0 Hence 3 3 1 Therefore 3 ab Y a b ca Y c a ab ca Y a b c b c a b c ab ca Y a ab ca Y a Y Y V Z I I V Z I I V V Z I I I I I I I I V V Z I V V Z Z I Z Z
- 34. 34 Three Phase Transmission Line