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Topic 1: Basics of Power Systems A.H. MohsenianRad (U of T) 1 Networking and Distributed Systems ECE 5332: Communications and Control for Smart Spring 2012

Topic 1: Basics of Power Systems - intra.ece.ucr.eduintra.ece.ucr.edu/~hamed/Smart_Grid_Topic_1_Power_Systems.pdf · Topic 1: Basics of Power Systems A.H. Mohsenian‐Rad (U of T)

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Topic 1: Basics of Power Systems

A.H. Mohsenian‐Rad (U of T) 1Networking and Distributed Systems

ECE 5332: Communications and Control for Smart

Spring 2012

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 2

• The Four Main Elements in Power Systems:

Power Production / Generation

Power Transmission

Power Distribution

Power Consumption / Load 

• Of course, we also need monitoring and control systems.

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 3

• Power Production:

Different Types:

Traditional

Renewable

Capacity, Cost, Carbon Emission

Step‐up Transformers

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 4

• Power Transmission:

High Voltage (HV) Transmission Lines

Several Hundred Miles

Switching Stations

Transformers

Circuit Breakers

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 5

• The Power Transmission Grid in the United States:

www.geni.org

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 6

• Major Inter‐connections in the United States:

www.geni.org

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 7

• Power Distribution:

Medium Voltage (MV) Transmission Lines (< 50 kV)

Power Deliver to Load Locations

Interface with Consumers / Metering

Distribution Sub‐stations

Step‐Down Transformers

Distribution Transformers

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 8

• Power Consumption:

Industrial

Commercial

Residential

Demand Response

Controllable Load

Non‐Controllable

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 9

Generation Transmission Distribution Load

Power Systems

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 10

• Power System Control:

Data Collection: Sensors, PMUs, etc.

Decision Making: Controllers

Actuators: Circuit Breakers, etc.

Power Grid Graph Representation

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 11

Nodes: Buses

Links: Transmission Lines

Generator

Load

Power Grid Graph Representation

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 12

Nodes: Buses

Links: Transmission Lines

Generator

Load

Buses (Voltage)

Power Grid Graph Representation

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 13

Nodes: Buses

Links: Transmission Lines

Generator

Load

Transmission Lines (Power Flow, Loss)

Power Grid Graph Representation

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 14

Nodes: Buses

Links: Transmission Lines

Generator

Load

Consum

ers

Power Grid Graph Representation

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 15

Nodes: Buses

Links: Transmission Lines

Generator

Load

10 MW 3 MW

7 MW

Transmission Line Admittance

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 16

• Admittance y is defined as the inverse of impedance z:

z = r + j x                   (r: Resistance, x: Reactance)

y = g + j b (g: Conductance, b: Susceptance)

y = 1 / z

Parameter g is usually positive

Parameter b: 

Positive: Capacitor

Negative: Inductor

Transmission Line Admittance

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 17

• For the transmission line connecting bus i to bus k:

Addmitance: yik

Example:

yik = 1 – j 4   (per unit)

Note that, yii is denoted by yi and indicates:

Susceptance for any shunt element (capacitor) to ground at bus i. 

Y-Bus Matrix

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 18

• We define:

Ybus = [ Yij ] where

Diagonal Elements:

Off‐diagonal Elements: 

Note that Ybas matrix depends on the power grid topology and the admittance of all transmission lines.

N is the number of busses in the grid.

N

ikkikiii yyY

,1

ijij yY

Y-Bus Matrix

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 19

• Example: For a grid with 4‐buses, we have:

• After separating the real and imaginary parts:

4342414434241

3434323133231

2423242321221

1413121413121

yyyyyyyyyyyyyyyyyyyyyyyyyyyy

Ybus

BjGYbus

Bus Voltage

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 20

• Let Vi denote the voltage at bus i:

• Note that, Vi is a phasor, with magnitude and angle. 

• In most operating scenarios we have:

iii VV

jiji VV

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 21

• Let Si denote the power injection at bus i:

Si = Pi + j Qi

• Generation Bus:  Pi > 0

• Load Bus:  Pi < 0 (negative power injection)

Active Power        Reactive Power

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 22

• Using Kirchhoff laws, AC Power Flow Equations become:

• Do we know all notations here? 

• If we know enough variables, we can obtain the rest of variables by solving a system of nonlinear equations. 

N

jjkkjjkkjjkk

N

jjkkjjkkjjkk

BGVVQ

BGVVP

1

1

)cos()sin(

)sin()cos(

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 23

• The AC Power Flow Equations are complicated to solve.

• Next, we try to simplify the equations in three steps. 

• Step 1: For most networks, G << B. Thus, we set G = 0:

N

jjkkjjkk

N

jjkkjjkk

BVVQ

BVVP

1

1

)cos(

)sin(

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 24

• Step 2: For most neighboring buses:                                  .

• As a result, we have: 

15 to10 ji

1)()(

jk

jkjk

CosSin

N

jkjjkk

N

jjkkjjkk

BVVQ

BVVP

1

1)(

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 25

• Step 3: In per‐unit, |Vi| is very close to 1.0 (0.95 to 1.05). 

• As a result, we have:                    . 

• Pk has a linear model and Qk is almost fixed.

1ji VV

k

N

kjj

kjkk

N

jkjk

N

jjkkjk

bBBBQ

BP

,11

1)(

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 26

• Step 3: In per‐unit, |Vi| is very close to 1.0 (0.95 to 1.05). 

• As a result, we have:                    . 

• Pk has a linear model and Qk is almost fixed.

1ji VV

k

N

kjj

kjkk

N

jkjk

N

jjkkjk

bBBBQ

BP

,11

1)(

DC Power Flow Equations

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 27

• Given the power injection values at all buses, we can use

to obtain the voltage angles at all buses.  

• Let Pij denote the power flow from bus i to bus j, we have:

N

jjkkjk BP

1)(

)( jiijij BP

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 28

• Example: Obtain power flow values in the following grid:

puPg 21 puPg 22

puPg 14 puP l 43

puP l 12 1014 jy 1013 jy

1034 jy

1023 jy

1012 jy

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 29

• First, we obtain the Y‐bus matrix:

44434241

34333231

24232221

14131211

4342414434241

3434323133231

2423242321221

1413121413121

BBBBBBBBBBBBBBBB

jBjj

bbbbbbbbbbbbbbbbbbbbbbbbbbbb

jYbus

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 30

• Next, we write the (active) power flow equations:

• This can be written as: 

44342413432421414

43433432312321313

42432322423211212

41431321211413121

BBBBBBPBBBBBBPBBBBBBP

BBBBBBP

4

3

2

1

434241434241

343432313231

242324232121

141312141312

4

3

2

1

BBBBBBBBBBBBBBBBBBBBBBBB

PPPP

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 31

• From the last two slides, we finally obtain:

• Therefore, the voltage angles are obtained as: 

4

3

2

1

1

4

3

2

1

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 32

• However, the last matrix in the previous slide is singular!

• Therefore, we cannot take the inverse. 

• The system of equations would have infinite solutions.

• The problem is that the four angles are not independent.

• What matters is the angular/phase difference. 

• We choose one bus (e.g., bus 1) as reference bus:             .01

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 33

• We should also remove the corresponding rows/columns:

• The angular differences (with respect to     ): 

4

3

2

1

201001010301010010201010101030

14

12

4

3

2

2010010301001020

14

1

1

025.015.0

025.0

14

1

2010010301001020 1

4

3

2

025.015.0025.0

14

13

12

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 34

• Finally, the power flow values are calculated as:

puPg 21 puPg 22

puPg 14 puP l 43

puP l 12

25.1)025.015.0(10)(25.1)15.0025.0(10)(

25.0)025.00(10)(5.1)15.00(10)(

25.0)025.00(10)(

433434

322323

411414

311313

211212

BPBPBPBPBP

0.25

1.25

1.250.25 1.5

Power Flow Equations

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 35

• What if the generator connected to bus 1 is renewable?

• What if the capacity of transmission link (1,3) is 1 pu?

• What if we can apply demand response to load bus 3?   

• What if one of the transmission lines fails? 

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 36

• In the example we discussed earlier, we had:

• In particular, we had:  

• However, generation levels             and      assumed given.

• Q: What if the generators have different generation costs?

Power Supply = Power Load

llggg PPPPP 32421

ggg PPP 421 ,,

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 37

• For thermal power plants, generation cost is quadratic:

• Example: a grid with three power plants:

• Each power plant has some min and max generation levels.

Generation Cost = C(P) = a1 + a2 x P + a3 x P2

C1(P1) = 561 + 7.92 x P1 + 0.001562 x (P1)2 150 MW ≤ P1 ≤ 600 MW

C2(P2) = 310 + 7.85 x P2 + 0.001940 x (P2)2 100 MW ≤ P2 ≤ 400 MW

C3(P3) =   78 + 7.97 x P3 + 0.004820 x (P3)2 50 MW ≤ P3 ≤ 200 MW

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 38

• For thermal power plants, generation cost is quadratic:

• Example: a grid with three power plants:

• Each power plant has some min and max generation levels.

Generation Cost = C(P) = a1 + a2 x P + a3 x P2

C1(P1) = 561 + 7.92 x P1 + 0.001562 x (P1)2 150 MW ≤ P1 ≤ 600 MW

C2(P2) = 310 + 7.85 x P2 + 0.001940 x (P2)2 100 MW ≤ P2 ≤ 400 MW

C3(P3) =   78 + 7.97 x P3 + 0.004820 x (P3)2 50 MW ≤ P3 ≤ 200 MW

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 39

• For thermal power plants, generation cost is quadratic:

• Example: a grid with three power plants:

• Each power plant has some min and max generation levels.

Generation Cost = C(P) = a1 + a2 x P + a3 x P2

C1(P1) = 561 + 7.92 x P1 + 0.001562 x (P1)2 150 MW ≤ P1 ≤ 600 MW

C2(P2) = 310 + 7.85 x P2 + 0.001940 x (P2)2 100 MW ≤ P2 ≤ 400 MW

C3(P3) =   78 + 7.97 x P3 + 0.004820 x (P3)2 50 MW ≤ P3 ≤ 200 MW

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 40

• For thermal power plants, generation cost is quadratic:

• Example: a grid with three power plants:

• Each power plant has some min and max generation levels.

Generation Cost = C(P) = a1 + a2 x P + a3 x P2

C1(P1) = 561 + 7.92 x P1 + 0.001562 x (P1)2 150 MW ≤ P1 ≤ 600 MW

C2(P2) = 310 + 7.85 x P2 + 0.001940 x (P2)2 100 MW ≤ P2 ≤ 400 MW

C3(P3) =   78 + 7.97 x P3 + 0.004820 x (P3)2 50 MW ≤ P3 ≤ 200 MW

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 41

• For thermal power plants, generation cost is quadratic:

• Example: a grid with three power plants:

• Each power plant has some min and max generation levels.

Generation Cost = C(P) = a1 + a2 x P + a3 x P2

C1(P1) = 561 + 7.92 x P1 + 0.001562 x (P1)2 150 MW ≤ P1 ≤ 600 MW

C2(P2) = 310 + 7.85 x P2 + 0.001940 x (P2)2 100 MW ≤ P2 ≤ 400 MW

C3(P3) =   78 + 7.97 x P3 + 0.004820 x (P3)2 50 MW ≤ P3 ≤ 200 MW

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 42

• We should select P1, P2, and P3 to: 

• Meet total load Pload = 850 MW

• Minimize the total cost of generation 

• Economic Dispatch Problem:

85020050400100600150 subject to

CCC minimize

321

3

2

1

321 , , 321

PPPPPP

PPPPPP

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 43

• Is the formulated problem a convex program? Why?

• Convex programs can be solved efficiently.

• An useful software is CVX for Matlab (http://cvxr.com/cvx).

• The optimal economic dispatch solution: 

P1 = 393.2 MW

P2 = 334.6 MW

P3 = 122.2 MW

Q: Do they satisfy all constraints? 

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 44

• Is the formulated problem a convex program? Why?

• Convex programs can be solved efficiently.

• An useful software is CVX for Matlab (http://cvxr.com/cvx).

• The optimal economic dispatch solution: 

P1 = 393.2 MW

P2 = 334.6 MW

P3 = 122.2 MW

Minimum Cost = 3916.6 + 3153.8 + 1123.9= 8194.3 

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 45

• What if we have to satisfy topology constraints? 

1P

3P MW 400

2P

MW 45020013 P

3

21

4

3

2

400450

2010010301001020

P

P

202010)( 33311313 BP

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 46

• The same optimal solutions are still valid:

MW 393.21 P

MW 122.2 3 P MW 400

MW 334.6 2 P

MW 450

805.3830.19685.15

4

3

2

203 20013 P

156.8

160.3

41.438.1 198.3

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 47

• The same optimal solutions are still valid:

MW 393.21 P

MW 122.2 3 P MW 400

MW 334.6 2 P

MW 450

805.3830.19685.15

4

3

2

203 20013 P

156.8

160.3

41.438.1 198.3

What if 17013 P

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 48

• Then the economic dispatch problem becomes:

17

400450

2010010301001020

850 20050 400100 600150 subject to

CCC minimize

3

3

21

4

3

2

321

3

2

1

321 , ,, , , 321321

P

P

PPPPPP

PPPPPP

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 49

• Then the economic dispatch problem becomes:

17

400450

2010010301001020

850 20050 400100 600150 subject to

CCC minimize

3

3

21

4

3

2

321

3

2

1

321 , ,, , , 321321

P

P

PPPPPP

PPPPPP Still a Convex 

Program?

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 50

• The new optimal solutions are obtained as:

• The total generation cost becomes: $8,233.66 > $8,194.3

• Here, we had to sacrifice “cost” for “implementation”.

MW 2801 P

MW 170 3 P MW 400

MW 400 2 P

MW 450

110

170

600 170

Economic Dispatch Problem

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 51

• The new optimal solutions are obtained as:

• The total generation cost becomes: $8,233.66 > $8,194.3

• Here, we had to sacrifice “cost” for “implementation”.

MW 2801 P

MW 170 3 P MW 400

MW 400 2 P

MW 450

110

170

600 170

What if 15013 P

Unit Commitment

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 52

• Economic Dispatch is solved a few hours ahead of operation.

• On the other hand, we need to decide about the choice of power plants that we want to turn on for the next day.

• This is done by solving the Unit Commitment problem.

• We particularly decide on which slow‐starting power plants we should turn on during the next day given various constraints. 

• The mathematical concepts are similar to the E‐D problem. 

References

Dr. Hamed Mohsenian-Rad Texas Tech UniversityCommunications and Control in Smart Grid 53

• W. J. Wood and B. F. Wollenberg, Power Generation,Operation, and Control, John Wiley & Sons, 2nd Ed., 1996.

• J. McCalley and L. Tesfatsion, "Power Flow Equations", LectureNotes, EE 458, Department of Electrical and ComputerEngineering, Iowa State University, Spring 2010.