ECE 476
Power System Analysis
Review 1
Prof. Tom Overbye
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
overbye@illinois.edu
Announcements
• HW5 is 3.4, 3.10, 3.14, 3.19, 3.23, 3.60, 2.38, 6.9
•
It should be done before the first exam, but does not need
to be turned in
• First exam is Tuesday Oct 6 during class
•
•
Closed book, closed notes, but you may bring one 8.5 by
11 inch note sheet and standard calculators.
Last name starting with A to 0 in 3017; P to Z in 3013
1
Complex Power
S  V I  cos(V   I )  j sin(V   I ) 
 P  jQ
(Note: S is a complex number but not a phasor)
 V I*
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
2
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
3
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.
4
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
5
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
6
Three-Phase - Wye Connection
• There are two ways to connect 3 systems
–
–
Wye (Y)
Delta ()
Wye Connection Voltages
Van
 V  
Vbn
 V   
Vcn
 V   
7
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
8
Per Phase Analysis Procedure
•
1.
2.
3.
4.
To do per phase analysis
Convert all  load/sources to equivalent Y’s
Solve phase “a” independent of the other phases
Total system power S = 3 Va Ia*
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.
9
Line Resistance
Line resistance per unit length is given by
R =

where  is the resistivity
A
Resistivity of Copper = 1.68 10-8 Ω-m
Resistivity of Aluminum = 2.65 10-8 Ω-m
Example: What is the resistance in Ω / mile of a
1" diameter solid aluminum wire (at dc)?
2.65  10-8 Ω-m
m

R 
1609
 0.084
2
mile
mile
  0.0127m
10
Magnetics Review
• Ampere’s circuital law:
F
  H dl  I e
F = mmf = magnetomtive force (amp-turns)
H = magnetic field intensity (amp-turns/meter)
dl = Vector differential path length (meters)

Ie
= Line integral about closed path 
(dl is tangent to path)
= Algebraic sum of current linked by 
11
Magnetics Review
• Ampere’s circuital law:
F
  H dl  I e
F = mmf = magnetomtive force (amp-turns)
H = magnetic field intensity (amp-turns/meter)
dl = Vector differential path length (meters)

Ie
= Line integral about closed path 
(dl is tangent to path)
= Algebraic sum of current linked by 
12
Line Inductance Example
Calculate the reactance for a balanced 3, 60Hz
transmission line with a conductor geometry of an
equilateral triangle with D = 5m, r = 1.24cm (Rook
conductor) and a length of 5 miles.
Since system is assumed
balanced
ia  ib  ic
0 
1
1
1 
a 
ia ln( )  ib ln( )  ic ln( ) 

2 
r'
D
D 
13
Line Inductance Example, cont’d
Substituting
ia  ib  ic
Hence
0 
1
1 


a 
ia ln    ia ln   

2 
 r '
 D 
0
D


ia ln  
2
 r'
0  D  4  107 
5

La 
ln   
ln 
3 
2  r ' 
2
 9.67  10 
 1.25  106 H/m
14
Conductor Bundling
To increase the capacity of high voltage transmission
lines it is very common to use a number of
conductors per phase. This is known as conductor
bundling. Typical values are two conductors for
345 kV lines, three for 500 kV and four for 765 kV.
15
Bundled Conductors, cont’d
Rb
geometric mean radius (GMR) of bundle
 (r ' d12 d13d14 )
 (r ' d12
D1b
d1b
1
4
1
) b
for our example
in general
geometric mean distance (GMD) of
conductor 1 to phase b.
 (d15d16 d17 d18 )
D1c
1
4
 D2b  D3b  D4b  Dab
 (d19 d1,10 d1,11d1,12 )
1
4
 D2c  D3c  D4c  Dac
16
Inductance of Bundle, cont’d
But remember each bundle has b conductors
in parallel (4 in this example). So
0  D 
La  L1 / b 
ln  
2  Rb 
17
Bundle Inductance Example
Consider the previous example of the three phases
symmetrically spaced 5 meters apart using wire
with a radius of r = 1.24 cm. Except now assume
each phase has 4 conductors in a square bundle,
spaced 0.25 meters apart. What is the new inductance
per meter?
r  1.24  102 m r '  9.67  103 m

0.25 M
0.25 M
3
R b  9.67  10  0.25  0.25  2  0.25
0.25 M

1
4
 0.12 m (ten times bigger!)
0
5
La 
ln
 7.46  107 H/m
2 0.12
18
Transposition
• To keep system balanced, over the length of a
transmission line the conductors are rotated so each
phase occupies each position on tower for an equal
distance. This is known as transposition.
Aerial or side view of conductor positions over the length
of the transmission line.
19
Inductance of Transposed Line
Define the geometric mean distance (GMD)
Dm
 d12 d13d 23 
1
3
Then for a balanced 3 system ( I a  - I b - I c )
0 
Dm
1
1  0
a 
I a ln  I a ln

I a ln


2 
r'
Dm  2
r'
Hence
0 Dm
Dm
7
La 
ln
 2  10 ln
H/m
2
r'
r'
20
Inductance with Bundling
If the line is bundled with a geometric mean
radius, R b , then
0
Dm
a 
I a ln
2
Rb
0 Dm
Dm
7
La 
ln
 2  10 ln
H/m
2 Rb
Rb
Line Conductors, cont’d
• Total conductor area is given in circular mils. One
circular mil is the area of a circle with a diameter
of 0.001 =   0.00052 square inches
• Example: what is the area of a solid, 1” diameter
circular wire?
Answer: 1000 kcmil (kilo circular mils)
• Because conductors are stranded, the equivalent
radius must be provided by the manufacturer. In
tables this value is known as the GMR and is
usually expressed in feet.
22
Review of Electric Fields
To develop a model for line capacitance we
first need to review some electric field concepts.
Gauss's law:
A D da
= qe
(integrate over closed surface)
where
D = electric flux density, coulombs/m 2
da = differential area da, with normal to surface
A = total closed surface area, m 2
q e = total charge in coulombs enclosed
23
Gauss’s Law Example
•Similar to Ampere’s Circuital law, Gauss’s Law is
most useful for cases with symmetry.
•Example: Calculate D about an infinitely long wire
that has a charge density of q coulombs/meter.
Since D comes
radially out integrate over the
cylinder bounding
D
d
a

D
2

Rh

q

qh
e
A
the wire
q
D 
ar where ar radially directed unit vector
2 R
24
Line Capacitance, cont’d
To eliminate mutual capacitance we'll again
assume we have a uniformly transposed line.
For the previous three conductor example:
Va  V
Since q a = C Va

qa
2
C 

Va
ln D
r
25
Bundled Conductor Capacitance
Similar to what we did for determining line
inductance when there are n bundled conductors,
we use the original capacitance equation just
substituting an equivalent radius
R cb
 (rd12
1
d1n )
n
Note for the capacitance equation we use r rather
than r' which was used for R b in the inductance
equation
26
Line Capacitance, cont’d
For the case of uniformly transposed lines we
use the same GMR, D m , as before.
2
C 
Dm
ln
Rbc
where
Dm
R cb

 d ab d ac dbc 
 (rd12
1
d1n )
n
1
3
(note r NOT r')
ε in air   o  8.854  10-12 F/m
27
Line Capacitance Example
•Calculate the per phase capacitance and susceptance
of a balanced 3, 60 Hz, transmission line with
horizontal phase spacing of 10m using three conductor
bundling with a spacing between conductors in the
bundle of 0.3m. Assume the line is uniformly
transposed and the conductors have a 1cm radius.
28
Line Capacitance Example, cont’d
Rbc
Dm
C
Xc

1
(0.01  0.3  0.3) 3

1
(10  10  20) 3
 0.0963 m
 12.6 m
2  8.854  1012

 1.141  1011 F/m
12.6
ln
0.0963
1
1


C
2 60  1.141  1011 F/m
 2.33  10 -m (not  / m)
8
29
Transmission Line Equivalent Circuit
•Our current model of a transmission line is shown
below
Units on
z and y are
per unit
length!
For operation at frequency  , let z = r + jL
and y = g +jC (with g usually equal 0)
30
Derivation of V, I Relationships
We can then derive the following relationships:
dV  I z dx
dI
 (V  dV ) y dx  V y dx
dV ( x)
dI ( x)
 zI
 yV
dx
dx
31
Setting up a Second Order Equation
dV ( x)
dI ( x)
 zI
 yV
dx
dx
We can rewrite these two, first order differential
equations as a single second order equation
d 2V ( x)
dI ( x)
z
 zyV
2
dx
dx
2
d V ( x)
 zyV  0
2
dx
32
V, I Relationships, cont’d
Define the propagation constant  as
  yz    j 
where
  the attenuation constant
  the phase constant
Use the Laplace Transform to solve. System
has a characteristic equation
( s   )  ( s   )( s   )  0
2
2
33
Determining Line Voltage, cont’d
V ( x)  K1 cosh( x)  K 2 sinh( x)
V (0)  VR
 K1 cosh(0)  K 2 sinh(0)
Since cosh(0)  1 & sinh(0)  0  K1  VR
dV ( x)
dx
 zI  K1 sinh( x)  K 2 cosh( x)
 K2 
zI R

IR z
z

 IR
y
yz
V ( x)  VR cosh( x)  I R Z c sinh( x)
where Zc 
z
y
 characteristic impedance
34
Lossless Transmission Lines
For a lossless line the characteristic impedance, Zc ,
is known as the surge impedance.
Zc 
jwl
l

 (a real value)
jwc
c
If a lossless line is terminated in impedance
VR
Zc 
IR
Then I R Z c  VR so we get...
35
Lossless Transmission Lines
V ( x)  VR cosh  x  VR sinh  x
I ( x)  I R cosh  x  I R sinh  x
V ( x)
 Zc
I ( x)
2
V(x)
Define
as the surge impedance load (SIL).
Zc
Since the line is lossless this implies
V ( x)  VR
I ( x)  I R
If P > SIL then line consumes
vars; otherwise line generates vars.
36
Transmission Matrix Model, cont’d
 VS   A B  VR 
With    
 I 
I
C
D
 R
 S 
Use voltage/current relationships to solve for A,B,C,D
VS  VR cosh  l  Z c I R sinh  l
VR
I S  I R cosh  l  sinh  l
Zc
 cosh  l
A B
1
T  


 sinh  l
C
D


 Z c
Z c sinh  l 

cosh  l 

37
Equivalent Circuit Model
The common representation is the  equivalent circuit
Next we’ll use the T matrix values to derive the
parameters Z' and Y'.
38
Equivalent Circuit Parameters
VS  VR
Y'
 VR  I R
Z'
2
Z 'Y ' 

VS  1 
 VR  Z ' I R
2 

Y'
Y'
I S  VS  VR  I R
2
2
Z 'Y ' 
Z 'Y ' 


I S  Y ' 1 
 VR  1 
 IR
4 
2 


 1  Z 'Y '

Z
'

 VR 
VS 
2
 I     Z 'Y '   Z 'Y '    I 
 R
 S
Y ' 1 

1

 

 
4  
2  
39
Three Line Models
Long Line Model (longer than 200 miles)
l
tanh
sinh  l Y ' Y
2
use Z '  Z
,

l
2 2 l
2
Medium Line Model (between 50 and 200 miles)
Y
use Z and
2
Short Line Model (less than 50 miles)
use Z (i.e., assume Y is zero)
40
Power Transfer in Lossless Lines
If we assume a line is lossless with impedance jX and
are just interested in real power transfer then:
2
P12  jQ12
V1
V1 V2

90 
90  12
Z
Z
Since - cos(90  12 )  sin 12 , we get
V1 V2
P12 
sin 12
X
Hence the maximum power transfer is
P12Max
V1 V2

X
41
Ideal Transformer
• First we review the voltage/current relationships
for an ideal transformer
–
–
–
no real power losses
magnetic core has infinite permeability
no leakage flux
• We’ll define the “primary” side of the
transformer as the side that usually takes power,
and the secondary as the side that usually delivers
power.
–
primary is usually the side with the higher voltage, but
may be the low voltage side on a generator step-up
transformer.
42
Ideal Transformer Relationships
Assume we have flux m in magnetic material. Then
1  N1m
d 1
2  N 2m
d 2
d m
d m
v1 
 N1
v2

 N2
dt
dt
dt
dt
d m
v1
v2
v1
N1




 a = turns ratio
dt
N1
N2
v2
N2
43
Current/Voltage Relationships
If  is infinite then 0  N1i1  N 2i2' . Hence
i1
N2
 
or
'
N1
i2
i1
N2 1


i2
N1 a
Then
v1 
i 
 1
a 0  v
 2



1  
0
  i2 

a
44
Transformer Equivalent Circuit
Using the previous relationships, we can derive an
equivalent circuit model for the real transformer
This model is further simplified by referring all
impedances to the primary side
r2'  a 2 r2
re  r1  r2'
x2'  a 2 x2
xe  x1  x2'
45
Simplified Equivalent Circuit
46
Calculation of Model Parameters
• The parameters of the model are determined based
upon
–
–
–
nameplate data: gives the rated voltages and power
open circuit test: rated voltage is applied to primary with
secondary open; measure the primary current and losses
(the test may also be done applying the voltage to the
secondary, calculating the values, then referring the
values back to the primary side).
short circuit test: with secondary shorted, apply voltage
to primary to get rated current to flow; measure voltage
and losses.
47
Per Unit Calculations
• A key problem in analyzing power systems is the
large number of transformers.
–
It would be very difficult to continually have to refer
impedances to the different sides of the transformers
• This problem is avoided by a normalization of all
variables.
• This normalization is known as per unit analysis.
actual quantity
quantity in per unit 
base value of quantity
48
Per Unit Conversion Procedure, 1
1. Pick a 1 VA base for the entire system, SB
2. Pick a voltage base for each different voltage level,
VB. Voltage bases are related by transformer turns
ratios. Voltages are line to neutral.
3. Calculate the impedance base, ZB= (VB)2/SB
4. Calculate the current base, IB = VB/ZB
5. Convert actual values to per unit
Note, per unit conversion on affects magnitudes, not
the angles. Also, per unit quantities no longer have
units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)
49
Per Unit Example
Solve for the current, load voltage and load power
in the circuit shown below using per unit analysis
with an SB of 100 MVA, and voltage bases of
8 kV, 80 kV and 16 kV.
Original Circuit
50
Three Phase Per Unit
Procedure is very similar to 1 except we use a 3
VA base, and use line to line voltage bases
1. Pick a 3 VA base for the entire system, S B3
2. Pick a voltage base for each different voltage
level, VB. Voltages are line to line.
3. Calculate the impedance base
ZB 
VB2, LL
S B3

( 3 VB , LN ) 2
3S 1B

VB2, LN
S 1B
Exactly the same impedance bases as with single phase!
51
Three Phase Per Unit, cont'd
4. Calculate the current base, IB
I3B
S B3
3 S 1B
S 1B



 I1B
3 VB , LL
3 3 VB , LN VB , LN
Exactly the same current bases as with single phase!
5. Convert actual values to per unit
52
Three Phase Transformers
• There are 4 different ways to connect 3
transformers
Y-Y
-
Usually 3 transformers are constructed so all windings
share a common core
53
3 Transformer Interconnections
-Y
Y-
54
Load Models
• Ultimate goal is to supply loads with electricity at
constant frequency and voltage
• Electrical characteristics of individual loads matter,
but usually they can only be estimated
–
–
actual loads are constantly changing, consisting of a large
number of individual devices
only limited network observability of load characteristics
• Aggregate models are typically used for analysis
• Two common models
–
–
constant power: Si = Pi + jQi
constant impedance: Si = |V|2 / Zi
55
Bus Admittance Matrix or Ybus
• First step in solving the power flow is to create
what is known as the bus admittance matrix, often
call the Ybus.
• The Ybus gives the relationships between all the bus
current injections, I, and all the bus voltages, V,
I = Ybus V
• The Ybus is developed by applying KCL at each bus
in the system to relate the bus current injections,
the bus voltages, and the branch impedances and
admittances
56
Ybus Example
Determine the bus admittance matrix for the network
shown below, assuming the current injection at each
bus i is Ii = IGi - IDi where IGi is the current injection into the
bus from the generator and IDi is the current flowing into the load
57
Ybus Example, cont’d
We can get similar relationships for buses 3 and 4
The results can then be expressed in matrix form
I  Ybus V
YA
YB
 I1  YA  YB
 I   Y
YA  YC  YD
YC
2
A
  
YC
YB  YC
 I 3   YB
I   0
YD
0
 4 
0  V1 
YD  V2 
 
0  V3 
YD  V4 
For a system with n buses, Ybus is an n by n
symmetric matrix (i.e., one where Aij = Aji)
58
Ybus General Form
• The diagonal terms, Yii, are the self admittance
terms, equal to the sum of the admittances of all
devices incident to bus i.
• The off-diagonal terms, Yij, are equal to the
negative of the sum of the admittances joining the
two buses.
• With large systems Ybus is a sparse matrix (that is,
most entries are zero)
• Shunt terms, such as with the  line model, only
affect the diagonal terms.
59
Modeling Shunts in the Ybus
Ykc
Since I ij  (Vi  V j )Yk  Vi
2
Ykc
Yii 
 Yk 
2
Rk  jX k Rk  jX k
1
1
Note Yk 

 2
Z k Rk  jX k Rk  jX k Rk  X k2
Yiifrom other lines
60
Two Bus System Example
Yc
(V1  V2 )
I1 
 V1
Z
2
1
 12  j16
0.03  j 0.04
 I1 
12  j15.9 12  j16  V1 
 I    12  j16 12  j15.9  V 

 2
 2
61
Power Flow Analysis
• When analyzing power systems we know neither
the complex bus voltages nor the complex current
injections
• Rather, we know the complex power being
consumed by the load, and the power being
injected by the generators plus their voltage
magnitudes
• Therefore we can not directly use the Ybus
equations, but rather must use the power balance
equations
62
Power Balance Equations, cont’d
*
n


Si  Vi I i*  Vi   YikVk   Vi  Yik*Vk*
 k 1

k 1
This is an equation with complex numbers.
Sometimes we would like an equivalent set of real
power equations. These can be derived by defining
n
Yik
Gik  jBik
Vi
Vi e ji  Vi  i
 ik
i   k
Recall e j  cos  j sin 
63
Real Power Balance Equations
n
Si  Pi  jQi  Vi  Yik*Vk* 
k 1

n
 Vi Vk
k 1
n
j ik
V
V
e
(Gik  jBik )
 i k
k 1
(cos ik  j sin  ik )(Gik  jBik )
Resolving into the real and imaginary parts
Pi 
Qi 
n
 Vi Vk (Gik cosik  Bik sinik )  PGi  PDi
k 1
n
 Vi Vk (Gik sinik  Bik cosik )  QGi  QDi
k 1
64
Gauss Iteration
There are a number of different iterative methods
we can use. We'll consider two: Gauss and Newton.
With the Gauss method we need to rewrite our
equation in an implicit form: x = h(x)
To iterate we first make an initial guess of x, x (0) ,
and then iteratively solve x (v +1)  h( x ( v ) ) until we
find a "fixed point", x,
ˆ such that xˆ  h(x).
ˆ
65
Gauss Iteration Example
Example: Solve x - x  1  0
x ( v 1)  1  x ( v )
Let v = 0 and arbitrarily guess x (0)  1 and solve
v
0
1
2
3
4
x(v )
1
2
2.41421
2.55538
2.59805
v
5
6
7
8
9
x(v)
2.61185
2.61612
2.61744
2.61785
2.61798
66
Gauss Power Flow
We first need to put the equation in the correct form
*
n


 Vi   YikVk   Vi  Yik*Vk*
 k 1

k 1
n
Si  Vi I i*
n
n
k 1
k 1
S*i  Vi* I i  Vi*  YikVk  Vi*  YikVk
S*i

*
Vi
n
 YikVk
k 1
 YiiVi 
n

k 1,k i
n

1  S*i
Vi 
 *   YikVk 

Yii  V
k 1,k i

i
YikVk
67
Gauss Two Bus Power Flow Example
•A 100 MW, 50 Mvar load is connected to a generator
•through a line with z = 0.02 + j0.06 p.u. and line
charging of 5 Mvar on each end (100 MVA base).
Also, there is a 25 Mvar capacitor at bus 2. If the
generator voltage is 1.0 p.u., what is V2?
SLoad = 1.0 + j0.5 p.u.
68
Gauss Two Bus Example, cont’d
The unknown is the complex load voltage, V2 .
To determine V2 we need to know the Ybus .
1
 5  j15
0.02  j 0.06
5  j14.95 5  j15 
Hence Ybus  


5

j
15
5

j
14.70


( Note B22  - j15  j 0.05  j 0.25)
69
Gauss Two Bus Example, cont’d
n

1  S*2
V2 
 *   YikVk 
Y22  V2 k 1,k i

 -1  j 0.5

1
V2 
 (5  j15)(1.00) 

*
5  j14.70  V2

Guess V2(0)  1.00 (this is known as a flat start)
v
0
1
2
V2( v )
1.000  j 0.000
0.9671  j 0.0568
0.9624  j 0.0553
v
3
4
V2( v )
0.9622  j 0.0556
0.9622  j 0.0556
70
Gauss Two Bus Example, cont’d
V2  0.9622  j 0.0556  0.9638  3.3
Once the voltages are known all other values can
be determined, such as the generator powers and the
line flows
S1*  V1* (Y11V1  Y12V2 )  1.023  j 0.239
In actual units P1  102.3 MW, Q1  23.9 Mvar
2
The capacitor is supplying V2 25  23.2 Mvar
71
Slack Bus
• In previous example we specified S2 and V1 and
then solved for S1 and V2.
• We can not arbitrarily specify S at all buses
because total generation must equal total load +
total losses
• We also need an angle reference bus.
• To solve these problems we define one bus as the
"slack" bus. This bus has a fixed voltage
magnitude and angle, and a varying real/reactive
power injection.
72