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Chapter 02 Fundamentals PS [Autosaved].ppt
1.
CHAPTER 2: Fundamentals 0 © 2012
Cengage Learning Engineering. All Rights Reserved. Chapter 2: Fundamentals
2.
Review of Phasors Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 1
3.
Review of Phasors Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 2
4.
Review of Phasors Goal
of phasor analysis is to simplify the analysis of constant frequency ac systems v(t) = Vmax cos(wt + qv) i(t) = Imax cos(wt + qI) Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 3 Root Mean Square (RMS) voltage of sinusoid 2 max 0 1 ( ) 2 T V v t dt T
5.
Phasor Representation Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 4
6.
Phasor Representation Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 5
7.
Phasor Representation, cont’d Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 6 (Note: Some texts use “boldface” type for complex numbers, or “bars on the top”)
8.
Phasor Representation, cont’d Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 7
9.
Advantages of Phasor
Analysis Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 8 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 w w Device Time Analysis Phasor (Note: Z is a complex number but not a phasor)
10.
RL Circuit Example Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 9 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 w w w
11.
Complex Power Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 10
12.
Complex Power Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 11
13.
Complex Power, cont’d Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 12
14.
Complex Power Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 13
15.
Complex Power Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 14
16.
Complex Power, cont’d Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 15 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
17.
Complex Power, cont’d Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 16
18.
Complex Power, cont’d Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 17
19.
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* Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 18
20.
Conservation of Power
: Example Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 19 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
21.
Power Consumption in
Devices Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 20 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 w 2 citor C C (Note-some define X negative) X
22.
Example Chapter 2: Fundamentals ©
2012 Cengage Learning Engineering. All Rights Reserved. 21 * 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
23.
Example, cont’d Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 22 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
24.
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 Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 23 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
25.
Reactive Compensation Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 24 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
26.
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. Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 25
27.
Power Factor Correction
Example Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 26 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
28.
Distribution System Capacitors Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 27 Provide reactive power to offset inductive loading from devices like motors, lighting loads, etc.
29.
A single-phase source
delivers 100 kW to a load operating at a power factor of 0.8 lagging. Calculate the reactive power to be delivered by a capacitor connected in parallel with the load in order to raise the source power factor to 0.95 lagging. Sketch the power triangle for the source and load. Assume that the source voltage is constant, and neglect the line impedance between the source and the load. [Qc=42.13 kVAR] Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 28
30.
A three-phase motor
draws 20 kVA at 0.707 power factor lagging from a 220 V source. Determine the kVAR rating of a capacitor bank needed to improve the power factor to 0.95 lagging, and the line current before and after the capacitors are added. Comment on the effect of the change in the line current to the system. Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 29
31.
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 Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 30
32.
Balanced 3 –
No Neutral Current Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 31 * * * * (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
33.
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 constant Three phase machines use less material for same power rating Three phase machines start more easily than single phase machines Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 32
34.
Three Phase –
Wye Connection There are two ways to connect 3 systems – Wye (Y) – Delta () Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 33 an bn cn Wye Connection Voltages V V V V V V
35.
Wye Connection Line
Voltages Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 34 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)
36.
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 Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 35 6 * 3 3 1 30 3 3 j Line Phase Phase Line Phase Phase Phase V V V e I I S V I
37.
Delta Connection Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 36 Ica Ic Iab Ibc Ia Ib VPhase = VLine
38.
Three Phase Example Assume
a -connected load is supplied from a 3 13.8 kV (L-L) source with Z = 10020 Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 37 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
39.
Three Phase Example,
cont’d Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 38 * 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
40.
Delta-Wye Transformation Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 39 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
41.
Delta-Wye Transformation Proof Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 40 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
42.
Delta-Wye Transformation, cont’d Chapter
2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 41 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
43.
Three Phase Transmission
Line Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 42
44.
Per Phase Analysis
Per phase analysis allows analysis of balanced 3 systems with the same effort as for a single phase system Balanced 3 Theorem: For a balanced 3 system with – All loads and sources Y connected – No mutual Inductance between phases Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 43
45.
Per Phase Analysis,
cont’d Then – All neutrals are at the same potential – All phases are COMPLETELY decoupled – All system values are the same sequence as sources. The sequence order we’ve been using (phase b lags phase a and phase c lags phase a) is known as “positive” sequence; later in the course we’ll discuss negative and zero sequence systems. Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 44
46.
Per Phase Analysis
Procedure To do per phase analysis 1. Convert all load/sources to equivalent Y’s 2. Solve phase “a” independent of the other phases 3. Total system power S = 3 Va Ia * 4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts) 5. If necessary, go back to original circuit to determine line-line values or internal values. Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 45
47.
Per Phase Example Assume
a 3, Y-connected generator with Van = 10 volts supplies a -connected load with Z = -j through a transmission line with impedance of j0.1 per phase. The load is also connected to a -connected generator with Va″b″ = 10 through a second transmission line which also has an impedance of j0.1 per phase. Find 1. The load voltage Va′b′ 2. The total power supplied by each generator, SY and S Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 46
48.
Per Phase Example,
cont’d Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 47 First convert the delta load and source to equivalent Y values and draw just the "a" phase circuit
49.
Per Phase Example,
cont’d Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 48 ' ' ' a a a To solve the circuit, write the KCL equation at a' 1 (V 1 0)( 10 ) V (3 ) (V j 3 j j
50.
Per Phase Example,
cont’d Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 49 ' ' ' a a a ' a ' ' a b ' ' c ab To solve the circuit, write the KCL equation at a' 1 (V 1 0)( 10 ) V (3 ) (V j 3 10 (10 60 ) V (10 3 10 ) 3 V 0.9 volts V 0.9 volts V 0.9 volts V 1.56 j j j j j j volts
51.
Per Phase Example,
cont’d Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 50 * ' * ygen * " ' " S 3 5.1 3.5 VA 0.1 3 5.1 4.7 VA 0.1 a a a a a a a gen a V V V I V j j V V S V j j
52.
Wye Connection Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 51
53.
Delta Connection Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 52 VPhase = VLine
54.
Example Chapter 2: Fundamentals ©
2012 Cengage Learning Engineering. All Rights Reserved. 53 Assume a -connected load is supplied from a 3 13.8 kVLL source with Z = 10020 i) Find the total power at the load. ii) Find the line currents.
55.
Chapter 2: Fundamentals ©
2012 Cengage Learning Engineering. All Rights Reserved. 54 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
56.
Chapter 2: Fundamentals ©
2012 Cengage Learning Engineering. All Rights Reserved. 55
57.
Delta-Wye Transformation Chapter 2:
Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 56 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
58.
Per Phase Analysis
Per phase analysis allows analysis of balanced 3 systems with the same effort as for a single phase system Balanced 3 Theorem: For a balanced 3 system with – All loads and sources Y connected – No mutual Inductance between phases Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 57
59.
Per Phase Analysis
Procedure To do per phase analysis 1. Convert all load/sources to equivalent Y’s 2. Solve phase “a” independent of the other phases 3. Total system power S = 3 Va Ia * 4. If desired, phase “b” and “c” values can be determined by inspection (i.e., ±120° degree phase shifts) 5. If necessary, go back to original circuit to determine line-line values or internal values. Chapter 2: Fundamentals © 2012 Cengage Learning Engineering. All Rights Reserved. 58
60.
Example Chapter 2: Fundamentals ©
2012 Cengage Learning Engineering. All Rights Reserved. 59
61.
Chapter 2: Fundamentals ©
2012 Cengage Learning Engineering. All Rights Reserved. 60
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