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ECE 530 – Analysis Techniques for
Large-Scale Electrical Systems
Lecture 3: Power Flow Analysis
Prof. Hao Zhu
Dept. of Electrical and Computer Engineering
University of Illinois at Urbana-Champaign
[email protected]
1
Announcements
•
•
Homework 1 will be posted today.
Due Sep 11 (Thur), in class
2
Static Power System Analysis
•
•
One of the most common power system analysis tools
is the power flow, which tells how power flows
through a power system in the quasi-steady state time
frame
The power flow can be used to model the full, threephase system, but usually (practically always) for
transmission system analysis the system is assumed to
be balanced. Hence a per phase equivalent model is
used.
3
Power System Component
Models: Transmission Lines
•
•
Power flow timeframe models for common power
system devices, including transmission lines,
transformers, generators and loads, are developed in the
prerequisite course ECE 476.
Transmission lines will be modeled using the p circuit
4
Power System Component
Models: Transformers
•
Transformer equivalent model
5
Power System Component
Models: Loads
•
•
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
6
Power System Component
Models: Generators
•
•
•
Engineering models depend upon application
Generators are usually synchronous machines
For generators we will use two different models:
– a steady-state model, treating the generator as a constant
power source operating at a fixed voltage; this model will be
used for power flow and economic analysis
– a short term model treating the generator as a constant voltage
source behind a possibly time-varying reactance
7
Per Phase 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
8
Per Unit Conversion Procedure, 1f
1. Pick a 1f VA base for the entire system, SB
2. Pick a voltage base for each different voltage level,
3.
4.
5.
VB. Voltage bases are related by transformer turns
ratios. Voltages are line to neutral.
Calculate the impedance base, ZB= (VB)2/SB
Calculate the current base, IB = VB/ZB
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)
9
Per Unit Solution Procedure
1.
2.
3.
Convert to per unit (p.u.) (many problems are already
in per unit)
Solve
Convert back to actual as necessary
10
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
11
Per Unit Example, cont’d
Z BLeft
8kV 2

 0.64
100 MVA
Z BMiddle
Z BRight
80kV 2

 64
100 MVA
2
16kV

 2.56
100 MVA
Same circuit, with
values expressed
in per unit.
12
Per Unit Example, cont’d
1.00
I 
 0.22  30.8 p.u. (not amps)
3.91  j 2.327
VL  1.00  0.22  30.8  
    p.u.
2
VL
SL 

 0.189 p.u.
Z
SG  1.00  0.2230.8  30.8 p.u.
VL I L*
13
Per Unit Example, cont’d
To convert back to actual values just multiply the
per unit values by their per unit base
VLActual  0.859  30.8 16 kV  13.7  30.8 kV
SLActual  0.1890 100 MVA  18.90 MVA
SGActual  0.2230.8 100 MVA  22.030.8 MVA
I Middle
B
100 MVA
 1250 Amps

80 kV
I Actual
Middle  0.22  30.8 Amps  275  30.8 
14
Three Phase Per Unit
Procedure is very similar to 1f except we use a 3f
VA base, and use line to line voltage bases
1.
2.
3.
Pick a 3f VA base for the entire system, S B3f
Pick a voltage base for each different voltage
level, VB. Voltages are line to line.
Calculate the impedance base
ZB 
VB2, LL
S B3f

( 3 VB , LN ) 2
3S 1Bf

VB2, LN
S 1Bf
Exactly the same impedance bases as with single phase!
15
Three Phase Per Unit, cont'd
4.
Calculate the current base, IB
I3Bf
5.
S B3f
3 S 1Bf
S 1Bf



 I1Bf
3 VB , LL
3 3 VB , LN VB , LN
Exactly the same current bases as with single phase!
Convert actual values to per unit
16
Three Phase Per Unit Example
Solve for the current, load voltage and load power
in the previous circuit, assuming a 3f power base of
300 MVA, and line to line voltage bases of 13.8 kV,
138 kV and 27.6 kV (square root of 3 larger than the
1f example voltages). Also assume the generator is
Y-connected so its line to line voltage is 13.8 kV.
Convert to per unit
as before. Note the
system is exactly the
same!
17
3f Per Unit Example, cont'd
1.00
I 
 0.22  30.8 p.u. (not amps)
3.91  j 2.327
VL  1.00  0.22  30.8  
    p.u.
2
VL
SL 

 0.189 p.u.
Z
SG  1.00  0.2230.8  30.8 p.u.
*
VL I L
Again, analysis is exactly the same!
18
3f Per Unit Example, cont'd
Differences appear when we convert back to actual values
VL
 0.859  30.8  27.6 kV  23.8  30.8 kV
Actual
SL
SGActual
 0.1890  300 MVA  56.70 MVA
Actual
I Middle
B
 0.2230.8  300 MVA  66.030.8 MVA
300 MVA

 1250 Amps (same current!)
3 138 kV
I Actual
Middle  0.22  30.8   Amps  275  30.8 
19
3f Per Unit Example 2
•Assume a 3f load of 100+j50 MVA with VLL of 69
kV is connected to a source through the below
network:
What is the supply current and complex power?
Answer: I=467 amps, S = 103.3 + j76.0 MVA
20
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
21
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
22
Ybus Example, cont’d
By KCL at bus 1 we have
I1 = I G1  I D1
I1  I12  I13
V1  V2 V1  V3


ZA
ZB
I1  (V1  V2 )YA  (V1  V3 )YB
1
(with Yj  )
Zj
 (YA  YB )V1  YA V2  YB V3
Similarly
I 2  I 21  I 23  I 24
 YA V1  (YA  YC  YD )V2  YC V3  YD V4
23
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); however this will not be true
in general when we consider phase shifting transformers
24
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 p line model, only
affect the diagonal terms.
25
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
26
Two Bus System Example
I1
Yc
(V1  V2 )

 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
27
Using the Ybus
If the voltages are known then we can solve for
the current injections:
Ybus V  I
If the current injections are known then we can
solve for the voltages:
1
Ybus
I  V  Zbus I
where Z bus is the bus impedance matrix
However, this requires that Ybus not be singular; note it
will be singular if there are no shunt connections!
28
Solving for Bus Currents
For example, in previous case assume
 1.0 
V

0.8

j
0.2


Then
12  j15.9 12  j16   1.0   5.60  j 0.70 
 12  j16 12  j15.9  0.8  j 0.2    5.58  j 0.88


 

Therefore the power injected at bus 1 is
S1  V1I1*  1.0  (5.60  j 0.70)  5.60  j 0.70
*
S2  V2 I 2
 (0.8  j 0.2)  (5.58  j 0.88)  4.64  j 0.41
29
Solving for Bus Voltages
For example, in previous case assume
 5.0 
I


4.8


Then
1
12  j15.9 12  j16   5.0   0.0738  j 0.902 
 12  j16 12  j15.9   4.8   0.0738  j1.098

 
 

Therefore the power injected is
S1  V1I1*  (0.0738  j 0.902)  5  0.37  j 4.51
S2  V2 I 2*  (0.0738  j1.098)  (4.8)  0.35  j 5.27
30
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
31
Power Flow Analysis
•
•
•
Classic paper for this lecture is W.F. Tinney and C.E.
Hart, “Power Flow Solution by Newton’s Method,”
IEEE Power App System, Nov 1967
Basic power flow is also covered in essentially power
system analysis textbooks.
At Illinois we use the term “power flow” not “load
flow” since power flows not load. Also, the power
flow usage is not new (see title of Tinney’s 1967 paper,
and note Tinney references Ward’s 1956 paper)
– A nice history of the power flow is given in an insert by
Alvarado and Thomas in T.J. Overbye, J.D. Weber,
“Visualizing the Electric Grid,” IEEE Spectrum, Feb 2001.
32
Power Balance Equations
From KCL we know at each bus i in an n bus system
the current injection, I i , must be equal to the current
that flows into the network
I i  I Gi  I Di 
n
 Iik
k 1
Since I = Ybus V we also know
I i  I Gi  I Di 
n
 YikVk
k 1
The network power injection is then Si  Vi I i*
33
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.
n
Sometimes we would like an equivalent set of real
power equations. These can be derived by defining
Yik = Gik  jBik
Vi = Vi e ji  Vi i
ik = i   k
Recall e j  cos  j sin 
These equations
can also be formulated
using rectangular
coordinates for
the voltages:
Vi = ei + jfi
34
Real Power Balance Equations
n
n
k 1
k 1
Si  Pi  jQi  Vi  Yik*Vk*   Vi Vk e jik (Gik  jBik )

n
 Vi Vk
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
35
Slack Bus
•
•
•
•
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.
In an actual power system the slack bus does not really
exist; frequency changes locally when the power
supplied does not match the power consumed
36
Three Types of Power Flow Buses
•
There are three main types of power flow buses
– Load (PQ) at which P/Q are fixed; iteration solves for voltage
magnitude and angle.
– Slack at which the voltage magnitude and angle are fixed;
iteration solves for P/Q injections
– Generator (PV) at which P and |V| are fixed; iteration solves
for voltage angle and Q injection
37
Newton-Raphson Algorithm
•
•
Most common technique for solving the power flow
problem is to use the Newton-Raphson algorithm
Key idea behind Newton-Raphson is to use
sequential linearization
General form of problem: Find an x such that
f ( xˆ )  0
•
More tractable for linear systems
38
Newton-Raphson Power Flow
In the Newton-Raphson power flow we use Newton's
method to determine the voltage magnitude and angle
at each bus in the power system.
We need to solve the power balance equations
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
39
Power Flow Variables
Assume the slack bus is the first bus (with a fixed
voltage angle/magnitude). We then need to determine
the voltage angle/magnitude at the other buses.
 2 




 n 
x  

V
 2




 Vn 
 P2 (x)  PG 2  PD 2 




 Pn (x)  PGn  PDn 
f ( x)  
Q2 (x)  QG 2  QD 2 






 Qn (x)  QGn  QDn 
40
N-R Power Flow Solution
The power flow is solved using the same procedure
discussed with the general Newton-Raphson:
Set v  0; make an initial guess of x, x( v )
While f (x( v ) )   Do
x( v 1)  x( v )  J (x( v ) ) 1f (x( v ) )
v
 v 1
End While
41
Power Flow Jacobian Matrix
The most difficult part of the algorithm is determining
and inverting the n by n Jacobian matrix, J (x)
 f1 (x)
 x
1

 f 2 (x)
J (x)   x1


 f (x)
 n
 x1
f1 (x)
x2
f 2 (x)
x2
f n (x)
x2
f1 ( x) 
xn 

f 2 ( x) 
xn 


f n ( x) 

xn 
42
Power Flow Jacobian Matrix,
cont’d
Jacobian elements are calculated by differentiating
each function, fi ( x), with respect to each variable.
For example, if fi (x) is the bus i real power equation
fi ( x) 
fi ( x)

i
n
 Vi Vk (Gik cosik  Bik sin ik )  PGi  PDi
k 1
n
 Vi Vk (Gik sin ik  Bik cosik )
k 1
k i
fi ( x)
 Vi V j (Gik sin ik  Bik cosik )
 j
( j  i)
43