EE/Econ 458
PF Equations
J. McCalley
1
Power system representation
NODE or BUS
(substation)
BRANCHES
(lines or
transformers)
NETWORK
(but unloaded
and unsupplied)
2
Power system representation
LOAD: Extracts MW out
of the node (injects negative
MW into the node)
GENERATOR: Injects
MW into the node
NETWORK
(loaded and
supplied)
3
Power system representation
Approximate
branch model
Best branch
model
NETWORK
(loaded and
supplied)
4
Power system representation
Approximate
branch model
Branch resistance
Branch inductive reactance
Branch capacitive susceptance
Ignore resistance, OK because it is much less than reactance.
Ignore susceptance, OK because its affect on MW flows very small.
Only model reactance, OK for getting branch flows.
5
Power system representation
Here is what we will model as a network (reactance only)
NETWORK
(loaded and
supplied)
6
Power system representation
The impedance is a complex number zij=rij+jxij.
We ignore the resistance: zij=jxij
z12
2
1
z13
z14
4
z23
z34
3
7
Power system representation
Impedance relates voltage drop and current via Ohm’s law:
I ij 
Current(amps)
z ij
(V i  V j )
Voltage drop (volts)
Vi
i
1
Iij
Vj
zij
j
8
Power system representation
I ij 
1
z ij
(V i  V j )
Admittance, yij, is the inverse of impedance, zij:
I ij  y ij (V i  V j )
Vi
i
Iij
Vj
yij
j
9
Power system representation
Label the admittances yij
y12
2
1
y13
y14
4
y23
y34
3
10
Power system representation
Current injections: Ii flowing into bus i from generator or load.
Positive if generator; negative if load.
I1, I4 will be positive.
I3 will be negative.
I2 will be positive if
gen exceeds load,
I1
y12
I2 otherwise negative.
2
1
y13
y14
4
y23
y34
I4
3
I3
11
Power system representation
Voltages: Vi is voltage at bus i.
I1
y12
I2
2
1
V2
V1
y13
y14
y23
V4
4
y34
I4
3
I3
V3
12
Power system representation
Kirchoff’s current law: sum of the currents
at any node must be zero.
I 1  I 12  I 13  I 14
I1
y12
1
2
I12
V1
I14
y14
I2
V2
y13
I13
y23
V4
4
y34
I4
3
Note:
We assume there are no
bus shunts in this system.
Bus shunts are capacitive
or inductive connections
between the bus and the
ground. Although most
systems have them, they
inject only reactive power
(no MW) and therefore
affect MW flows in the
network only very little.
V3
I3
13
Power system representation
Now express each current using Ohm’s law: I ij  y ij (V i  V j )
I 1  I 12  I 13  I 14
I 1  y12 (V1  V 2 )  y13 (V1  V 3 )  y14 (V1  V 4 )
I1
y12
1
2
I12
V1
I14
y14
I2
V2
y13
I13
y23
V4
4
y34
I4
3
I3
V3
14
Power system representation
Now collect like terms in the voltages:
I 1  y12 (V1  V 2 )  y13 (V1  V 3 )  y14 (V1  V 4 )
I 1  V1 ( y12  y13  y14 )  V 2 (  y12 )  V 3 (  y13 )  V 4 (  y14 )
I1
y12
1
2
I12
V1
I14
y14
I2
V2
y13
I13
y23
V4
4
y34
I4
3
I3
V3
15
Power system representation
Repeat for the other four buses:
I 1  V1 ( y12  y13  y14 )  V 2 (  y12 )  V 3 (  y13 )  V 4 (  y14 )
I 2  V1 (  y 21 )  V 2 ( y 21  y 23  y 24 )  V 3 (  y 23 )  V 4 (  y 24 )
I 3  V1 (  y 31 )  V 2 (  y 32 )  V 3 ( y 31  y 32  y 34 )  V 4 (  y 34 )
I 4  V1 (  y 41 )  V 2 (  y 42 )  V 3 ( y 41  y 42  y 43 )  V 4 (  y 43 )
I1
y12
1
2
I12
V1
I14
y14
I2
V2
y13
I13
y23
V4
4
y34
I4
3
I3
V3
16
Power system representation
Repeat for the other four buses:
I 1  V1 ( y12  y13  y14 )  V 2 (  y12 )  V 3 (  y13 )  V 4 (  y14 )
I 2  V1 (  y 21 )  V 2 ( y 21  y 23  y 24 )  V 3 (  y 23 )  V 4 (  y 24 )
I 3  V1 (  y 31 )  V 2 (  y 32 )  V 3 ( y 31  y 32  y 34 )  V 4 (  y 34 )
I 4  V1 (  y 41 )  V 2 (  y 42 )  V 3 ( y 41  y 42  y 43 )  V 4 (  y 43 )
I1
y12
1
2
I12
V1
I14
y14
I2
V2
y13
I13
y23
V4
4
y34
I4
Notes:
1. yij=yji
2. If branch ij does
not exist, then yij=0.
3
I3
V3
17
Power system representation
Write in matrix form:
 I 1   y 12
  
I
 2  
I3  
  
I4  
 y 13  y 14
 y 12
 y13
 y 14
 y 21
y 21  y 23  y 24
 y 23
 y 31
 y 32
y 31  y 32  y 34
 y 41
 y 42
 y 43
y 41
  V1 
 
 y 24
V
 2 
 V 3 
 y 34
 
 y 42  y 43  V 4 
Define the Y-bus:
 y 12

Y  



 y 13  y 14
 y 12
 y 13
 y 21
y 21  y 23  y 24
 y 23
 y 31
 y 32
y 31  y 32  y 34
 y 41
 y 42
 y 43
 y 14
y 41


 y 24


 y 34

 y 42  y 43 
Define elements of the Y-bus:
 Y11

Y 21

Y 
 Y 31

Y 41
Y12
Y13
Y 22
Y 23
Y 32
Y 33
Y 42
Y 43
Y14 

Y 24

Y 34 

Y 44 
 I 1   Y 11
  
Y
I
 2    21
 I 3   Y 31
  
 I 4  Y 41
Y 12
Y 13
Y 22
Y 23
Y 32
Y 33
Y 42
Y 43
Y 14   V 1 
 
Y 24 V 2
 
Y 34  V 3 
 
Y 44  V 4 
18
Power system representation
Forming the Y-Bus:
1. The matrix is symmetric, i.e., Yij=Yji.
2. A diagonal element Yii is obtained as the sum of admittances for
all branches connected to bus i (yik is non-zero only when there
exists a physical connection between buses i and k).
3. The off-diagonal elements are the negative of the admittances
connecting buses i and j, i.e., Yij=-yji.
19
Power system representation
From the previous work, you can derive the power flow
equations.
These are equations expressing the real and reactive power
injections at each bus. If we had modeled branch resistance, we
would obtain:
N
Pk 

V k V j G kj cos(  k   j )  B kj sin(  k   j ) 
j 1
N
Qk 

V k V j G kj sin(  k   j )  B kj cos(  k   j ) 
j 1
where Yij=Gij+jBij.
This requires too much EE, so forget about them. Let’s make
some assumptions instead.
But first, what is θk and θj?
20
Power system representation
N
Pk 

V k V j G kj cos(  k   j )  B kj sin(  k   j ) 
j 1
N
Qk 

V k V j G kj sin(  k   j )  B kj cos(  k   j ) 
θk and θj are the angles of the
voltage phasors at each bus.
j 1
The angle captures the time
difference when voltage
phasors cross the zerovoltage axis.
In the time domain
simulation, the red curve
crosses before the blue one
by an amount of time Δt
and so has an angle of θ=ωΔt
where ω=2πf and f is
frequency of oscillation, 60
Hz for power systems.
21
Power system representation
Simplifying assumptions:
1. No resistance: Yij=jBij
2. Angle differences across branches, are small: θi-θj:
• Sin(θi-θj)= θi-θj
• Cos(θi-θj)=1.0
3. All voltage magnitudes are 1.0 in the pu system.
N
Pk 
 B
kj
( k   j ) 
j 1 ,
jk
This is the basis for the “DC power flow.”
Per-unit system:
A system where all quantities
are normalized to a
consistent set of bases. It will
result in powers being
expressed as a particular
number of “100 MVA”
quantities.
Admittance is also perunitized.
22
Example
Pg2=2pu
Pg1=2pu
1
2
y12 =-j10
y14 =-j10
y13 =-j10
Pd2=1pu
y23 =-j10
N
y34 =-j10
4
Pg4=1pu
3
Pd3=4pu
Pk 
 B
kj
( k   j ) 
j 1 ,
jk
P1  B12 ( 1   2 )  B13 ( 1   3 )  B14 ( 1   4 )
 B12  1  B12 2  B13 1  B13 3  B14 1  B14  4
Collect terms in the same variables
P1   B12  B13  B14  1  B12 2  B13 3  B14 4
Repeat procedure for buses 2, 3, 4:
P2   B 21 1   B 21  B 23  B 24  2  B 23 3  B 24  4
P3   B 31 1  B 32 2   B 31  B 32  B 34  3  B 34 4
P4   B 41 1  B 42 2  B 43 3   B 41  B 42  B 43  4
23
Example
Pg2=2pu
Pg1=2pu
1
2
y12 =-j10
y14 =-j10
y13 =-j10
Pd2=1pu
N
y34 =-j10
4
y23 =-j10
3
Pg4=1pu
Pk 
 B
kj
( k   j ) 
j 1 ,
jk
Pd3=4pu
Now write in matrix form:
 P1   B12
  
P
 2  
 P3  
  
 P4  
 B13  B14
 B12
 B13
 B 21
B 21  B 23  B 24
 B 23
 B 31
 B 32
B 31  B 32  B 34
 B 41
 B 42
 B 43
  1 
 
 B 24

 2 
  3 
B 34
 
 B 42  B 43   4 
 B14
B 41
24
Example
Compare:
 P1   B12
  
P
 2  
 P3  
  
 P4  
 B11

B 21
Y  j
 B 31

 B 41
  30

10
Y  j
 10

 10
 B13  B14
 B12
 B13
 B 21
B 21  B 23  B 24
 B 23
 B 31
 B 32
B 31  B 32  B 34
 B 41
 B 42
 B 43
B12
B13
B 22
B 23
B 32
B 33
B 42
B 43
10
10
 20
10
10
 30
0
10
B14 

B 24

B 34 

B 44 
 b12

 j



10 

0

10 

 20 
 b13  b14
B 41
 b12
 b13
 b 21
b 21  b 23  b 24
 b 23
 b 31
 b 32
b 31  b 32  b 34
 b 41
 b 42
 b 43
 2   30

 
1
 10


  4    10

 
 1    10
 10
 10
20
 10
 10
30
0
 10
  1 
 
 B 24

 2 
  3 
B 34
 
 B 42  B 43   4 
 B14
 b14
b 41


 b 24


b34

 b 42  b 43 
 10    1 
 

0
 2 
 10   3 
 
20   4 
25
 2   30

 
1
 10


  4    10

 
 1    10
 10
 10
20
 10
 10
30
0
 10
Example
 10    1 
 

0
 2 
 10   3 
 
20   4 
  1   30
  

 10
 2  
 3    10
  
 4    10
 10
 10
20
 10
 10
30
0
 10
 10 

0

 10 

20 
1
 2 


1


 4


 1 
But matlab indicates above matrix is singular which means it
does not have an inverse.
There is a dependency among the four equations, i.e., we can
add the bottom three rows and multiply by -1 to get the top
row.
This dependency occurs because all four angles are not
independent; we have to choose one of them as a reference with a
26
fixed value of 0 degrees.
Example
Eliminate one of the equations and one of the variables by setting
the variable to zero. We choose to eliminate the first equation and
set the first variable θ1=0 degrees.
  1   30
  

 10
 2  
 3    10
  
 4    10
 10
 10
20
 10
 10
30
0
 10
 10 

0

 10 

20 
1
 2 


1



4




 1 
 2   20
  
   10
 3 
 4   0
 10
30
 10
0 

 10

20 
1
 1    0 . 025 

 

 4   0 . 15

 

 1    0 . 025 
But we want power flows:
Pkj  B kj ( k   j )
P12  B12 ( 1   2 )  10 ( 0   0 . 025 )  0 . 25
P13  B13 ( 1   3 )  10 ( 0   0 . 15 )  1 . 5
P14  B14 ( 1   4 )  10 ( 0   0 . 025 )  0 . 25
P23  B 23 ( 2   3 )  10 (  0 . 025   0 . 15 )  1 . 25
P34  B 34 ( 3   4 )  10 (  0 . 15   0 . 025 )   1 . 25
27
Example
Resulting solution:
Pg2=2pu
Pg1=2pu
1
P14 =0.25
2
P12=0.25

P13=1.5
Pd2=1pu
P43 =1.25
4
Pg4=1pu
P23 =1.25
3
Pd3=4pu
28
Example
Resulting solution:
Pg2=2pu
Pg1=2pu
1
P14 =0.25
2
P12=0.25

P13=1.5
Pd2=1pu
P43 =1.25
4
Pg4=1pu
P23 =1.25
3
Pd3=4pu
29
How to solve power flow problems
Develop B’ matrix:
 P1   B12
  
P
 2  
 P3  
  
 P4  
 B13  B14
 B11

B 21
Y  j
 B 31

 B 41
B12
B13
B 22
B 23
B 32
B 33
B 42
B 43
1.
2.
3.
4.
 B12
 B13
 B 21
B 21  B 23  B 24
 B 23
 B 31
 B 32
B 31  B 32  B 34
 B 41
 B 42
 B 43
B14 

B 24

B 34 

B 44 
 b12

 j



Get the Y-bus
Remove the “j” from the Y-bus.
Multiply Y-bus by -1.
Remove row 1 and column 1.
  1 
 
 B 24

 2 
  3 
B 34
 
 B 42  B 43   4 
 B14
B 41
 b13  b14
 b12
 b13
 b 21
b 21  b 23  b 24
 b 23
 b 31
 b 32
b 31  b 32  b 34
 b 41
 b 42
 b 43
  30

10
Y  j
 10

 10
10
10
 20
10
10
 30
0
10
 b14
b 41


 b 24


b34

 b 42  b 43 
10 

0

10 

 20 
 20

B '   10

 0
 10
30
 10
0 

 10

20 
30
How to solve power flow problems
Develop equations to compute branch flows:
P B  ( D  A)  
where:
• PB is the vector of branch flows. It has dimension of M x 1.
Branches are ordered arbitrarily, but whatever order is chosen
must also be used in D and A.
• θ is (as before) the vector of nodal phase angles for buses 2,…N
• D is an M x M matrix having non-diagonal elements of zeros; the
diagonal element in position row k, column k contains the
negative of the susceptance of the kth branch.
• A is the M x N-1 node-arc incidence matrix. It is also called the
adjacency matrix, or the connection matrix. Its development
requires a few comments.
31
How to solve power flow problems
How to develop node-arc incidence matrix:
•
•
•
•
•
P B  ( D  A)  
number of rows equal to the number of branches (arcs) and a
number of columns equal to the number of nodes.
Element (k,j) of A is 1 if the kth branch begins at node j, -1 if the kth
branch terminates at node j, and 0 otherwise.
A branch is said to “begin” at node j if the power flowing across
branch k is defined positive for a direction from node j to the other
node.
A branch is said to “terminate” at node j if the power flowing across
branch k is defined positive for a direction to node j from the other
node.
Note that matrix A is of dimension M x N-1, i.e., it has only N-1
columns. This is because we do not form a column with the reference
bus, in order to conform to the vector θ, which is of dimension (N-1) x
32
1. This works because the angle being excluded, θ1, is zero.
How to solve power flow problems
Pg2=2pu
Pg1=2pu
1
2
2
1
5
Pd2=1pu
3
4
4
3
Pg4=1pu
Pd3=4pu
P B  ( D  A)   
10

0

0

0
0

number
node


2
3
4
0
0
0
10
0
0
0
10
0
0
0
10
0
0
0
0  0

0 -1

0  1

0  0
10   0
0
0
-1
-1
-1
- 1

0

0 

1
0 
0

-1

A 1

0
0

10

0

D  0

0
0

0
0
-1
-1
-1
- 1

0

0 

1
0 
0
0
0
10
0
0
0
10
0
0
0
10
0
0
0
1

2


3  branch
4

5 
num ber
0

0

0

0
10 
 2 
 

 3
 4 
33
How to solve power flow problems
 PB 1  10

 
PB 2
0

 
 PB 3    0

 
P
 B4   0
P   0
 B5  
0
0
0
10
0
0
0
10
0
0
0
10
0
0
0
0     4    10  4

 


2
 10  2
0
 


0    2   3    10 ( 2   3 ) 
 


0     3   4  10 (  3   4 ) 

10     3    10  3

Pg2=2pu
Pg1=2pu
1
 PB 1   0 . 25 

 

PB 2
0 . 25

 

 PB 3   1 . 25 

 

P
1
.
25
 B4  

 P   1 .5 

 B5  
 2    0 . 025 
  

 3   0 . 15
  

 4    0 . 025 
2
P12=0.25
P14 =0.25
P13=1.5
Pd2=1pu
P43 =1.25
4
Pg4=1pu
P23 =1.25
3
Pd3=4pu
34