This document provides an overview of complex power in electrical systems. It defines phasor representation using complex exponentials to simplify analysis of constant frequency AC circuits. It describes how real and reactive power can be calculated from voltage and current phasors and discusses power factor. The document also discusses reactive compensation using capacitors to improve power factor by supplying reactive power locally. It provides an example of power factor correction and introduces balanced three-phase power systems with both wye and delta connections.

1. EE369
POWER SYSTEM ANALYSIS
Lecture 2
Complex Power, Reactive Compensation, Three Phase
Tom Overbye and Ross Baldick
1

2. Reading and Homework
Read Chapters 1 and 2 of the text.
HW 1 is Problems 2.2, 2.3, 2.4, 2.5, 2.6, 2.8,
2.11, 2.13, 2.16, 2.18, 2.23, 2.25 and Case
Study Questions A., B., C., D. from the text;
due Thursday 9/3.
HW 2 is Problems 2.26, 2.27, 2.28, 2.29, 2.30,
2.32, 2.33, 2.35, 2.37, 2.39, 2.40 (need to
install PowerWorld); due Thursday 9/10.
2

3. Review of Phasors
Goal of phasor analysis is to simplify the analysis of constant
frequency ac systems:
v(t) = Vmaxcos(ωt + θv),
i(t) = Imaxcos(ωt + θI),
where:
• v(t) and i(t) are the instantaneous voltage and current as a
function of time t,
• ω is the angular frequency (2πf, with f the frequency in Hertz),
• Vmax andImaxare the magnitudes of voltage and current sinusoids,
• θv and θI are angular offsets of the peaks of sinusoids from a
reference waveform.
Root Mean Square (RMS) voltage of sinusoid:
2 max
max
0
1
( ) , so 2 .
2
T
V
V v t dt V V
T
= = =∫ 3

4. Phasor Representationj
( )
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).
Given complex phasor (magnitu
V
V
j t
j
v t V t
v t V e
V
θ
ω θ
θ θ
ω θ
+
= +
= +
 =  
de and angle),
we can determine sinusoidal waveform
(magnitude and phase) and vice versa. 4

5. Phasor Representation, cont’d
The RMS, cosine-referenced voltage phasor is:
,
( ) Re 2 ,
cos sin ,
cos sin .
V
V
j
V
jj t
V V
I I
V V e V
v t V e e
V V j V
I I j I
θ
θω
θ
θ θ
θ θ
= = ∠
=
= +
= +
• (Note: Some texts use “boldface” type for complex numbers, or
“bars on the top”.)
• Also note that the convention in power engineering is that the
magnitude of the phasor is the RMS voltage of the waveform:
• contrasts with circuit analysis. 5

6. Advantages of Phasor Analysis
0
2 2
Resistor ( ) ( )
( )
Inductor ( )
1 1
Capacitor ( ) ( ) (0)
C
= Impedance ,
= Resistance,
= Reactance,
Z = , =arctan .
t
v t Ri t V RI
di t
v t L V j LI
dt
v t i t dt v V I
j C
Z R jX Z
R
X
X
R X
R
ω
ω
φ
φ
= =
= =
= + =
= + = ∠
 +  ÷
 
∫
Device Time Analysis Phasor
(Note: Z is a
complex number but
not a phasor).
6

7. RL Circuit Example
2 2 1
( ) 2 100cos( 30 ), so 100 30 ,
60Hz,
4 , 2 3 ,
4 3 5, tan (3/4) 36.9 ,
100 30
,
5 36.9
20 6.9 Amps,
( ) 20 2 cos( 6.9 ).
v t t V
f
R X L fL
Z
V
I
Z
i t t
ω
ω π
φ
ω
−
= + ° = ∠ °
=
= Ω = = = Ω
= + = = = °
∠ °
= =
∠ °
= ∠ − °
= − °
7

8. Complex Power
max
max
max max
( ) ( ) ( ),
( ) = 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 V t
i I t
p t V I
t
ω θ
ω θ
α β α β α β
θ θ
ω θ θ
=
+
+
= − + +
= − +
+ +
Instantaneous Power :
8

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
θ θ ω θ θ
θ θ
θ θ
φ θ θ
= − + + +
=
= −
= −
−
∫
Instantaneous Power is
Power F
sum
acto
of average
r Angl
and varying te
e
rms :
9

10. Complex Power, cont’d
max max
max max
max max
max max
1
( ) [cos( ) cos(2 )],
2
1
[cos( ) cos(2 2 ( ))],
2
1
[cos( ) cos(2 2 )cos( )]
2
1
sin(2 2 )
2
V I V I
V I V V I
V I V V I
V
p t V I t
V I t
V I t
V I t
θ θ ω θ θ
θ θ ω θ θ θ
θ θ ω θ θ θ
ω θ
= − + + +
= − + + − −
= − + + −
+ +
Re - interpretation of instantaneous Power :
sin( ),V Iθ θ−
10
Instantaneous power into resistive component
Instantaneous power into electric and magnetic fields

11. Complex Power
[ ]
*
cos( ) sin( ) ,
,
= Real Power (W, kW, MW),
= Reactive Power (VAr, kVAr, MVAr),
= magnitude of power into electric and magnetic fields,
= Complex power (VA, kVA, MVA),
Power Factor
,
(pf
V I V IS V I j
P jQ
V
P
Q
I
S
θ θ θ θ= −
= +
=
− +
) = cos ,
If current leads voltage then pf is leading,
If current lags voltage then pf is lagging.
φ
(Note: S is a complex number but not a phasor.)
11

12. Complex Power, cont’d
12
φ
|S|
P
Q
2 2
S P Q= +
S P jQ= +
cos( )
P P
S
pfφ
= =
1
tan
Q
P
φ −  
=  ÷
  2 2
P
pf
P Q
=
+
Power Triangle

13. Complex Power, cont’d
2
1
Relationships between real, reactive, and complex power:
cos ,
sin 1 pf ,
Example: A load draws 100 kW with a leading pf of 0.85.
What are (power factor angle), and ?
cos 0.85 31.8 ,
10
P S
Q S S
Q S
S
φ
φ
φ
φ −
=
= = ± −
= − = − °
=
0kW
117.6 kVA,
0.85
117.6sin( 31.8 ) 62.0 kVAr.Q
=
= − ° =−
13
negative since
leading pf
Load consumes -62 kVAr, i.e. load supplies +62 kVAr ⇒ capacitive load

14. 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 real power and conservation of
reactive power follows since S = VI*.
14

15. Conservation of Power Example
Earlier we found
I = 20∠-6.9° amps
*
* *
2
* *
2
100 30 20 6.9 2000 36.9 VA,
36.9 pf = 0.8 lagging,
( ) 4 20 6.9 20 6.9 ,
1600W ( 0),
( ) 3 20 6.9 20 6.9 ,
1200VAr , ( 0).
R R
R R
L L
L L
S V I
S V I RI I
P I R Q
S V I jXI I j
Q I X P
φ
= = ∠ ° × ∠ ° = ∠ °
= °
= = = × ∠ − °× ∠ °
= = =
= = = × ∠ − °× ∠ °
= = =
= 1600W + j1200VAr
Power flowing from
source to load at bus
15

16. Power Consumption in Devices
2
Resistor Resistor
2
Inductor Inductor L
2
Capacitor Capacitor C
Capaci
Resistors only consume real power:
,
Inductors only "consume" reactive power:
,
Capacitors only "generate" reactive power:
1
.C
P I R
Q I X
Q I X X
C
Q
ω
=
=
= − =
2
Capacitor
tor
C
(Note-some define negative.). CX
V
X
= −
16

17. Example
*
40000 0
400 0 Amps
100 0
40000 0 (5 40) 400 0
42000 16000 44.9 20.8 kV
44.9k 20.8 400 0
17.98 20.8 MVA 16.8 6.4 MVA
V
I
V j
j
S V I
j
∠ °
= = ∠ °
∠ ° Ω
= ∠ ° + + ∠ °
= + = ∠ °
= = ∠ °× ∠ °
= ∠ ° = +
First solve
basic circuit
I
17

18. Example, cont’d
Now add additional
reactive power load
and re-solve, assuming
that load voltage is
maintained at 40 kV.
70.7 0.7 lagging
564 45 Amps
59.7 13.6 kV
33.7 58.6 MVA 17.6 28.8 MVA
LoadZ pf
I
V
S j
= ∠45° =
= ∠ − °
= ∠ °
= ∠ ° = +
18
Need higher source voltage to maintain load voltage magnitude when
reactive power load is added to circuit. Current is higher.

19. 59.7 k V
17.6 M W
28.8 M VR
40.0 k V
16.0 M W
16.0 M VR
17.6 M W 16.0 M W
-16.0 M VR28.8 M VR
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
19

20. Reactive Compensation
44.94 k V
16.8 M W
6.4 M VR
40.0 k V
16.0 M W
16.0 M VR
16.8 M W 16.0 M W
0.0 M VR6.4 M VR
16.0 M VR
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.
20

21. 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 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.
21

22. 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
80 (60 )
60-
ta
80
S j
S j Q
Q
φ
φ
−
−
= + = = °
= = °
= + −
= cap
cap
n18.2 60 26.3 kVAr
33.7 kVAr
Q
Q
° ⇒ − =
=
22

23. Distribution System Capacitors
23

24. Balanced 3 Phase (φ) Systems
A balanced 3 phase (φ) 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 3φ.
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.
24

25. Balanced 3φ -- Zero Neutral Current
* * * *
(1 0 1 1
3
Note: means voltage at point with respect to point .
n a b c
n
an a bn b cn c an a
xy
I I I I
V
I
Z
S V I V I V I V I
V x y
= + +
= ∠ ° + ∠ −120° + ∠120°) = 0
= + + =
25

26. Advantages of 3φ Power
Can transmit more power for same amount of
wire (twice as much as single phase).
Total torque produced by 3φ machines is
constant, so less vibration.
Three phase machines use less material for
same power rating.
Three phase machines start more easily than
single phase machines.
26

27. Three Phase - Wye Connection
There are two ways to connect 3φ systems:
– Wye (Y), and
– Delta (∆).
an
bn
cn
Wye Connection Voltages
V V
V V
V V
α
α
α
= ∠ °
= ∠( ° −120°)
= ∠( ° +120°)
27

28. 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
α α
α
α
α
= − = ∠ − ∠( − °))
= ∠( + °)
= ∠( − °)
= ∠( + °)
Line to line
voltages are
also balanced.
(α = 0 in this case)
28

29. 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.
6
*
3
3 1 30 3
3
j
Line Phase Phase
Line Phase
Phase Phase
V V V e
I I
S V I
π
φ
= ∠ ° =
=
= 29

30. Delta Connection
Ica
Ic
Iab
Ibc
Ia
Ib
*
3
For Delta connection,
voltages across elements
equals line voltages
For currents
3
3
a ab ca
ab
b bc ab
c ca bc
Phase Phase
I I I
I
I I I
I I I
S V Iφ
= −
= ∠ − 30°
= −
= −
=
30

31. Three Phase Example
Assume a ∆-connected load, with each leg Z =
100∠20°Ω, is supplied from a 3φ 13.8 kV (L-L) source
13.8 0 kV
13.8 0 kV
13.8 0 kV
ab
bc
ca
V
V
V
= ∠ °
= ∠ −12 °
= ∠12 °
13.8 0 kV
138 20 amps
138 140 amps 138 0 amps
ab
bc ca
I
I I
∠ °
= = ∠ − °
100∠20° Ω
= ∠ − ° = ∠10 °
31

32. 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
= − = ∠ − ° − ∠10 °
= ∠ − °
= ∠ − ° = ∠7 °
= × = × ∠ ° × ∠20°
= ∠20°
= +
= ° = 0.94
32

33. Delta-Wye Transformation
Y
Line
phase
To simplify analysis of balanced 3 systems:
1)Δ-connected loads can be replaced by
1
Y-connected loads with
3
2)Δ-connected sources can be replaced by
Y-connected sources with
3 30
Z Z
V
V
φ
∆=
=
∠ °
33

34. Delta-Wye Transformation Proof
For the
Suppose
side
the two sides have identical te
we get
rminal behavior
Hence
.
ab ca ab ca
a
ab ca
a
V V V V
I
Z Z Z
V V
Z
I
∆ ∆ ∆
∆
∆
−
= − =
−
=
34
+
-

35. Delta-Wye Transformation, cont’d
For the side we get
( ) ( )
(2 )
Since 0
Hence 3
3
1
Therefore
3
ab Y a b ca Y c a
ab ca Y a b c
a 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 I
V V Z I
V V
Z Z
I
Z Z
∆
∆
= − = −
− = − −
+ + = ⇒ = − −
− =
−
= =
=
35

36. Three Phase Transmission Line
36