ECE 476
POWER SYSTEM ANALYSIS
Lecture 2
Complex Power, Reactive Compensation, Three Phase
Professor Tom Overbye
Department of Electrical and
Computer Engineering
Announcements

For lectures 2 through 3 please be reading Chapters
1 and 2
1
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)
Root Mean Square (RMS) voltage of sinusoid
1T
Vmax
2
v(t ) dt 

T0
2
2
Phasor Representation
Euler's Identity: e jq  cosq  j sin q
Phasor notation is developed by rewriting
using Euler's identity
v(t )  2 V cos(w t  qV )
v(t )  2 V Re e j (w t qV ) 
(Note: V is the RMS voltage)
3
Phasor Representation, cont’d
The RMS, cosine-referenced voltage phasor is:
V
 V e jqV  V qV
v(t )
 Re 2 Ve jw t e jqV
V
 V cosqV  j V sin qV
I
 I cosq I  j I sin q I
(Note: Some texts use “boldface” type for
complex numbers, or “bars on the top”)
4
Advantages of Phasor Analysis
Device
Resistor
Inductor
Capacitor
Time Analysis
v(t )  Ri (t )
di (t )
v(t )  L
dt
Phasor
V  RI
1t
i (t ) dt  v(0)

C0
1
V 
I
jw C
V  jw LI
Z = Impedance  R  jX  Z 
R = Resistance
X = Reactance
Z =
R2  X 2
(Note: Z is a
complex number but
X
 =arctan( ) not a phasor)
R
5
RL Circuit Example
V (t )  2 100cos(w t  30)
f
 60Hz
R
 4
X  wL  3
Z
I
i(t)

42  32  5   36.9
V
10030


Z
536.9
 20  6.9 Amps
 20 2 cos(w t  6.9)
6
Complex Power
Power
p (t )  v(t ) i (t )
v(t) = Vmax cos(w t  qV )
i (t)
= I max cos(w t  q I )
1
cos cos   [cos(   )  cos(   )]
2
1
p (t )  Vmax I max [cos(qV  q I ) 
2
cos(2w t  qV  q I )]
7
Complex Power, cont’d
Average Power
1
p (t )  Vmax I max [cos(qV  q I )  cos(2wt  qV  q I )]
2
T
Pavg
1

p (t )dt

T0
1
 Vmax I max cos(qV  q I )
2
 V I cos(qV  q I )
Power Factor Angle =  =qV  q I
8
Complex Power
S  V I  cos(qV  q I )  j sin(qV  q I ) 
 P  jQ
 V I
*
(Note: S is a complex number but not a phasor)
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 lags voltage then pf is lagging
9
Complex Power, cont’d
Relationships between real, reactive and complex power
P  S cos 
Q  S sin 
  S 1  pf
2
Example: A load draws 100 kW with a leading pf of 0.85.
What are  (power factor angle), Q and S ?
  -cos 1 0.85  31.8
100kW
S 
 117.6 kVA
0.85
Q  117.6sin(31.8)  62.0 kVar
10
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*
11
Conversation of Power Example
Earlier we found
I = 20-6.9 amps
S  V I  10030  206.9  200036.9 VA
  36.9 pf = 0.8 lagging
*
SR  VR I  4  20  6.9 206.9
*
PR  1600W
2
 I R
(Q R  0)
SL  VL I *  3 j  20  6.9 206.9
2
QL  1200 var  I X
(PL  0)
12
Power Consumption in Devices
Resistors only consume real power
2
PResistor  I Resistor R
Inductors only consume reactive power
2
Q Inductor  I Inductor X L
Capacitors only generate reactive power
2
QCapacitor   I Capacitor X C
QCapacitor  
VCapacitor
XC
1
XC 
wC
2
(Note-some define X C negative)
13
Example
First solve
basic circuit
400000 V
I 
 4000 Amps
1000 
V  400000  (5  j 40) 4000
 42000  j16000  44.920.8 kV
S  V I *  44.9k20.8 4000
 17.9820.8 MVA  16.8  j 6.4 MVA
14
Example, cont’d
Now add additional
reactive power load
and resolve
Z Load  70.7
pf  0.7 lagging
I  564  45 Amps
V  59.713.6 kV
S  33.758.6 MVA  17.6  j 28.8 MVA
15
Power System Notation
Power system components are usually shown as
“one-line diagrams.” Previous circuit redrawn
17.6 MW
16.0 MW
28.8 MVR
-16.0 MVR
59.7 kV
17.6 MW
28.8 MVR
40.0 kV
16.0 MW
16.0 MVR
Generators are Transmission lines
shown as circles are shown as a
single line
Arrows are
used to
show loads
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
16.8 MW
16.0 MW
6.4 MVR
0.0 MVR
44.94 kV
16.8 MW
6.4 MVR
40.0 kV
16.0 MW
16.0 MVR
16.0 MVR
Compensated circuit is identical to first example with
just real power load
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 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.
18
Power Factor Correction Example
Assume we have 100 kVA load with pf=0.8 lagging,
and would like to correct the pf to 0.95 lagging
S  80  j 60 kVA
  cos 1 0.8  36.9
PF of 0.95 requires desired
 cos 1 0.95  18.2
Snew  80  j (60  Qcap )
60 - Qcap
80
 tan18.2  60  Qcap  26.3 kvar
Qcap  33.7 kvar
19
Distribution System Capacitors
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
21
Balanced 3 -- No Neutral Current
I n  I a  Ib  I c
V
In 
(10  1   1  
Z
*
*
*
*
S  Van I an
 Vbn I bn
 Vcn I cn
 3 Van I an
22
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
23
Three Phase - Wye Connection

There are two ways to connect 3 systems
–
–
Wye (Y)
Delta ()
Wye Connection Voltages
Van
 V  
Vbn
 V   
Vcn
 V   
24
Wye Connection Line Voltages
Vca
Vcn
Vab
-Vbn
Van
Vbn
Vbc
Vab
(α = 0 in this case)
 Van  Vbn  V (1  1  120

3 V   30
Vbc

3 V   90
Vca

3 V   150
Line to line
voltages are
also balanced
25
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
VLine  3 VPhase 130  3 VPhase e
j
6
I Line  I Phase
S3
*
 3 VPhase I Phase
26
Delta Connection
For the Delta
phase voltages equal
line voltages
Ica
For currents
Ia  I ab  I ca
Ic

Ib
Ibc
Iab
Ia
3 I ab   
I b  I bc  I ab
Ic  I ca  I bc
S3 
*
3 VPhase I Phase
27
Three Phase Example
Assume a -connected load is supplied from a 3
13.8 kV (L-L) source with Z = 10020
Vab  13.80 kV
Vbc  13.8 0 kV
Vca  13.80 kV
13.80 kV
I ab 
 138  20 amps
 
I bc  138  140 amps
I ca  1380 amps
28
Three Phase Example, cont’d
I a  I ab  I ca  138  20  1380
 239  50 amps
I b  239  170 amps I c  2390 amps
*
S  3  Vab I ab
 3  13.80kV  138 amps
 5.7 MVA
 5.37  j1.95 MVA
pf  cos 20   lagging
29
Delta-Wye Transformation
To simplify analysis of balanced 3 systems:
1) Δ-connected loads can be replaced by
1
Y-connected loads with ZY  Z 
3
2) Δ-connected sources can be replaced by
VLine
Y-connected sources with Vphase 
330
30
Delta-Wye Transformation Proof
From the  side we get
Vab Vca
Vab  Vca
Ia 


Z Z
Z
Hence
Vab  Vca
Z 
Ia
31
Delta-Wye Transformation, cont’d
From the Y side we get
Vab
 ZY ( I a  I b )
Vca  ZY ( I c  I a )
Vab  Vca  ZY (2 I a  I b  I c )
Since
Ia  I b  I c  0  I a   I b  I c
Hence
Vab  Vca  3 ZY I a
3 ZY
Vab  Vca

 Z
Ia
Therefore
ZY
1
 Z
3
32
Three Phase Transmission Line
33