EE/Econ 458 PF Equations

EE/Econ 458PF Equations

J. McCalley

1

Power system representation

NODE or BUS(substation) BRANCHES

(lines or transformers)

NETWORK(but unloaded and unsupplied)

2

Power system representationLOAD: Extracts MW out of the node (injects negative MW into the node)

GENERATOR: Injects MW into the node

NETWORK(loaded and supplied)

3



Best branch model

Approximate branch model

4


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



Here is what we will model as a network (reactance only)

6


The impedance is a complex number zij=rij+jxij.We ignore the resistance: zij=jxij

7

1 2

34

z12

z14

z34

z23

z13


Impedance relates voltage drop and current via Ohm’s law:

)(1

jiij

ij VVz

I

i j

zij

Vi Vj

Iij

Voltage drop (volts)Current(amps)

8


Admittance, yij, is the inverse of impedance, zij:

)(1

jiij

ij VVz

I

i j

yij

Vi Vj

Iij

)( jiijij VVyI

9


Label the admittances yij

10

y12

y14

y34

y23

y13

1 2

34


I1

Current injections: Ii flowing into bus i from generator or load.Positive if generator; negative if load.

11

y12

y14

y34

y23

y13

1 2

34

I2

I4

I3

I1, I4 will be positive.I3 will be negative.I2 will be positive if gen exceeds load, otherwise negative.


I1

Voltages: Vi is voltage at bus i.

12

y12

y14

y34

y23

y13

1 2

34

I2

I4

I3

V1V2

V3V4


I1

Kirchoff’s current law: sum of the currents at any node must be zero.

13

y12

y14

y34

y23

y13

1 2

34

I2

I4

I3

V1V2

V3V4

1413121 IIII

I14

I13

I12

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.


I1

Now express each current using Ohm’s law:

14

y12

y14

y34

y23

y13

1 2

34

I2

I4

I3

V1V2

V3V4

1413121 IIII

I14

I13

I12

)( jiijij VVyI

)()()( 4114311321121 VVyVVyVVyI


I1

Now collect like terms in the voltages:

15

y12

y14

y34

y23

y13

1 2

34

I2

I4

I3

V1V2

V3V4

I14

I13

I12

)()()( 4114311321121 VVyVVyVVyI )()()()( 14413312214131211 yVyVyVyyyVI


I1

Repeat for the other four buses:

16

y12

y14

y34

y23

y13

1 2

34

I2

I4

I3

V1V2

V3V4

I14

I13

I12

)()()()( 14413312214131211 yVyVyVyyyVI )()()()( 24423324232122112 yVyVyyyVyVI )()()()( 34434323133223113 yVyyyVyVyVI )()()()( 43443424134224114 yVyyyVyVyVI

Power system representationRepeat for the other four buses:

17

)()()()( 14413312214131211 yVyVyVyyyVI )()()()( 24423324232122112 yVyVyyyVyVI )()()()( 34434323133223113 yVyyyVyVyVI )()()()( 43443424134224114 yVyyyVyVyVI

Notes:1. yij=yji

2. If branch ij does not exist, then yij=0.

I1 y12

y14

y34

y23

y13

1 2

34

I2

I4

I3

V1V2

V3V4

I14

I13

I12

Power system representationWrite in matrix form:

18

Define the Y-bus:

4

3

2

1

434241434241

343432313231

242324232121

141312141312

4

3

2

1

V

V

V

V

yyyyyy

yyyyyy

yyyyyy

yyyyyy

I

I

I

I

434241434241

343432313231

242324232121

141312141312

yyyyyy

yyyyyy

yyyyyy

yyyyyy

Y

Define elements of the Y-bus:

44434241

34333231

24232221

14131211

YYYY

YYYY

YYYY

YYYY

Y

4

3

2

1

44434241

34333231

24232221

14131211

4

3

2

1

V

V

V

V

YYYY

YYYY

YYYY

YYYY

I

I

I

I

Power system representationForming 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 representationFrom 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:

20

N

jjkkjjkkjjkk

N

jjkkjjkkjjkk

BGVVQ

BGVVP

1

1

)cos()sin(

)sin()cos(

This requires too much EE, so forget about them. Let’s make some assumptions instead. But first, what is θk and θj?

where Yij=Gij+jBij.


21

N

jjkkjjkkjjkk

N

jjkkjjkkjjkk

BGVVQ

BGVVP

1

1

)cos()sin(

)sin()cos(

θk and θj are the angles of the voltage phasors at each bus.

The angle captures the time difference when voltage phasors cross the zero-voltage axis.In the time domain simulation, the red curve crosses before the blue one by an amount of time Δtand so has an angle of θ=ωΔt where ω=2πf and f is frequency of oscillation, 60 Hz for power systems.

Power system representationSimplifying assumptions:1.No resistance: Yij=jBij

2.Angle differences across branches, are small: θi-θj:• Sin(θi-θj)= θi-θj

• Cos(θi-θj)=1.03.All voltage magnitudes are 1.0 in the pu system.

22

This is the basis for the “DC power flow.”

N

kjj

jkkjk BP,1

)( 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 per-unitized.

Example

23

N

kjj

jkkjk BP,1

)(

y13 =-j10 y14 =-j10

y34 =-j10

y23 =-j10

y12 =-j10

Pg1=2pu

Pd3=4pu

Pd2=1pu

1 2

3 4

Pg2=2pu

Pg4=1pu

414114313113212112

4114311321121 )()()(

BBBBBB

BBBP

41431321211413121 BBBBBBP

Collect terms in the same variables

Repeat procedure for buses 2, 3, 4: 42432322423211212 BBBBBBP

43433432312321313 BBBBBBP

44342413432421414 BBBBBBP

Example

24

N

kjj

jkkjk BP,1

)(

y13 =-j10 y14 =-j10

y34 =-j10

y23 =-j10

y12 =-j10

Pg1=2pu

Pd3=4pu

Pd2=1pu

1 2

3 4

Pg2=2pu

Pg4=1pu

Now write in matrix form:

4

3

2

1

434241434241

343432313231

242324232121

141312141312

4

3

2

1

BBBBBB

BBBBBB

BBBBBB

BBBBBB

P

P

P

P

Example

25

Compare:

4

3

2

1

434241434241

343432313231

242324232121

141312141312

4

3

2

1

BBBBBB

BBBBBB

BBBBBB

BBBBBB

P

P

P

P

44434241

34333231

24232221

14131211

BBBB

BBBB

BBBB

BBBB

jY

434241434241

343432313231

242324232121

141312141312

bbbbbb

bbbbbb

bbbbbb

bbbbbb

j

2010010

10301010

0102010

10101030

jY

4

3

2

1

2010010

10301010

0102010

10101030

1

4

1

2

Example

26

4

3

2

1

2010010

10301010

0102010

10101030

1

4

1

2

1

4

1

2

2010010

10301010

0102010

101010301

4

3

2

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 fixed value of 0 degrees.

Example

27

1

4

1

2

2010010

10301010

0102010

101010301

4

3

2

1

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.

025.0

15.0

025.0

1

4

1

20100

103010

010201

4

3

2

But we want power flows:

)( jkkjkj BP 25.0)025.00(10)( 211212 BP

5.1)15.00(10)( 311313 BP

25.0)025.00(10)( 411414 BP

25.1)15.0025.0(10)( 322323 BP

25.1)025.015.0(10)( 433434 BP

Example

28

Resulting solution:

P13=1.5 P14 =0.25

P43 =1.25

P23 =1.25

P12=0.25

Pg1=2pu

Pd3=4pu

Pd2=1pu

1 2

3 4

Pg2=2pu

Pg4=1pu

Example

29

Resulting solution:

P13=1.5 P14 =0.25

P43 =1.25

P23 =1.25

P12=0.25

Pg1=2pu

Pd3=4pu

Pd2=1pu

1 2

3 4

Pg2=2pu

Pg4=1pu

How to solve power flow problems

30

Develop B’ matrix:

4

3

2

1

434241434241

343432313231

242324232121

141312141312

4

3

2

1

BBBBBB

BBBBBB

BBBBBB

BBBBBB

P

P

P

P

44434241

34333231

24232221

14131211

BBBB

BBBB

BBBB

BBBB

jY

434241434241

343432313231

242324232121

141312141312

bbbbbb

bbbbbb

bbbbbb

bbbbbb

j

1. Get the Y-bus2. Remove the “j” from the Y-bus.3. Multiply Y-bus by -1.4. Remove row 1 and column 1.

2010010

10301010

0102010

10101030

jY

20100

103010

01020

'B


31

Develop equations to compute branch flows:

)( ADPB

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.


32

How to develop node-arc incidence matrix:

)( ADPB

• 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 1. This works because the angle being excluded, θ1, is zero.


33

5 1

4

3

2

Pg1=2pu

Pd3=4pu

Pd2=1pu

1 2

3 4

Pg2=2pu

Pg4=1pu

number branch

5

4

3

2

1

01-0

11-0

01-1

001-

1-00

A

4 3 2

number node

100000

010000

001000

000100

000010

D

4

3

2

100000

010000

001000

000100

000010

01-0

11-0

01-1

001-

1-00

)( ADPB


34

3

43

32

2

4

3

43

32

2

4

5

4

3

2

1

10

)(10

)(10

10

10

100000

010000

001000

000100

000010

B

B

B

B

B

P

P

P

P

P

025.0

15.0

025.0

4

3

2

5.1

25.1

25.1

25.0

25.0

5

4

3

2

1

B

B

B

B

B

P

P

P

P

P

P13=1.5 P14 =0.25

P43 =1.25

P23 =1.25

P12=0.25

Pg1=2pu

Pd3=4pu

Pd2=1pu

1 2

3 4

Pg2=2pu

Pg4=1pu

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EE/Econ 458 PF Equations