Topic 1: Basics of Power Systems
ECE 5332: Communications and Control for Smart
Spring 2012
A.H. Mohsenian‐Rad (U of T)
Networking and Distributed Systems
1
Power Systems
• 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.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
2
Power Systems
• Power Production:
 Different Types:
 Traditional
 Renewable
 Capacity, Cost, Carbon Emission
 Step‐up Transformers
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
3
Power Systems
• Power Transmission:
 High Voltage (HV) Transmission Lines
 Several Hundred Miles
 Switching Stations
 Transformers
 Circuit Breakers
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
4
Power Systems
• The Power Transmission Grid in the United States:
www.geni.org
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
5
Power Systems
• Major Inter‐connections in the United States:
www.geni.org
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
6
Power Systems
• 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
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
7
Power Systems
• Power Consumption:
 Industrial
 Commercial
 Residential
 Demand Response
 Controllable Load
 Non‐Controllable
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
8
Power Systems
Generation
Dr. Hamed Mohsenian-Rad
Transmission
Distribution
Communications and Control in Smart Grid
Load
Texas Tech University
9
Power Systems
• Power System Control:
 Data Collection: Sensors, PMUs, etc.
 Decision Making: Controllers
 Actuators: Circuit Breakers, etc.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
10
Power Grid Graph Representation
Nodes: Buses
Links: Transmission Lines
Generator
Load
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
11
Power Grid Graph Representation
Nodes: Buses
Links: Transmission Lines
Buses (Voltage)
Generator
Load
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
12
Power Grid Graph Representation
Nodes: Buses
Links: Transmission Lines
Generator
Load
Transmission Lines (Power Flow, Loss)
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
13
Power Grid Graph Representation
Nodes: Buses
Links: Transmission Lines
Load
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Consumers
Generator
Texas Tech University
14
Power Grid Graph Representation
Nodes: Buses
Links: Transmission Lines
Generator
10 MW
3 MW
Load 7 MW
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
15
Transmission Line Admittance
• Admittance y is defined as the inverse of impedance z:
z=r+jx
(r: Resistance, x: Reactance)
y=g+jb
(g: Conductance, b: Susceptance)
y=1/z
 Parameter g is usually positive
 Parameter b:
 Positive: Capacitor
 Negative: Inductor
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
16
Transmission Line Admittance
• 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.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
17
Y-Bus Matrix
• We define:
 Ybus = [ Yij ] where
 Diagonal Elements:
 Off‐diagonal Elements:
Yii  yi 
N
y
k 1, k  i
ik
Yij   yij
 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.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
18
Y-Bus Matrix
• Example: For a grid with 4‐buses, we have:
Ybus
 y1  y12  y13  y14

 y21



 y31

 y41

 y12
y2  y21  y23  y24
 y13
 y23
 y32
 y42
y3  y31  y32  y34
 y43
 y14
 y24




 y34

y4  y41  y42  y43 
• After separating the real and imaginary parts:
Ybus  G  j B
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
19
Bus Voltage
• Let Vi denote the voltage at bus i:
• Note that, Vi is a phasor, with magnitude and angle.
Vi  Vi  i
• In most operating scenarios we have:
Vi  V j
Dr. Hamed Mohsenian-Rad
i   j
Communications and Control in Smart Grid
Texas Tech University
20
Power Flow Equations
• Let Si denote the power injection at bus i:
Si = Pi + j Qi
Active Power
• Generation Bus:
Pi > 0
• Load Bus:
Pi < 0
Dr. Hamed Mohsenian-Rad
Reactive Power
(negative power injection)
Communications and Control in Smart Grid
Texas Tech University
21
Power Flow Equations
• Using Kirchhoff laws, AC Power Flow Equations become:
Pk   V k V j G kj cos( k   j )  B kj sin( k   j ) 
N
j 1
Q k   V k V j G kj sin( k   j )  B kj cos( k   j ) 
N
j 1
• 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.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
22
Power Flow Equations
• 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:
Pk   Vk V j Bkj sin( k   j ) 
N
j 1
Qk   Vk V j  Bkj cos( k   j ) 
N
j 1
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
23
Power Flow Equations
• Step 2: For most neighboring buses:  i   j  10 to 15 .
• As a result, we have:
Sin ( k   j )   k   j

Cos ( k   j )  1
Pk   Vk V j Bkj ( k   j ) 
N
j 1
Qk   Vk V j  Bkj 
N
j 1
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
24
Power Flow Equations
• Step 3: In per‐unit, |Vi| is very close to 1.0 (0.95 to 1.05).
• As a result, we have: Vi V j  1 .
N
Pk   Bkj ( k   j )
j 1
Qk    Bkj    Bkk   Bkj  bk
N
N
j 1
j 1,
jk
• Pk has a linear model and Qk is almost fixed.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
25
Power Flow Equations
• Step 3: In per‐unit, |Vi| is very close to 1.0 (0.95 to 1.05).
• As a result, we have: Vi V j  1 .
DC Power Flow Equations
N
Pk   Bkj ( k   j )
j 1
Qk    Bkj    Bkk   Bkj  bk
N
N
j 1
j 1,
jk
• Pk has a linear model and Qk is almost fixed.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
26
Power Flow Equations
• Given the power injection values at all buses, we can use
N
Pk   Bkj ( k   j )
j 1
to obtain the voltage angles at all buses.
• Let Pij denote the power flow from bus i to bus j, we have:
Pij  Bij ( i   j )
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
27
Power Flow Equations
• Example: Obtain power flow values in the following grid:
P2g  2 pu
P1g  2 pu
y12   j10
y14   j10
y13   j10
P2l  1 pu
y23   j10
y34   j10
P4g  1 pu
Dr. Hamed Mohsenian-Rad
P3l  4 pu
Communications and Control in Smart Grid
Texas Tech University
28
Power Flow Equations
• First, we obtain the Y‐bus matrix:
Ybus
b1  b12  b13  b14

 b21
 j

 b31

 b41



 j



Dr. Hamed Mohsenian-Rad
 b12
b2  b21  b23  b24
 b32
 b42


 jB



 b13
 b23
b3  b31  b32  b34
 b43
 B11
B
j  21
 B31

 B41
B12
B22
B32
B42
Communications and Control in Smart Grid
B13
B23
B33
B43
 b14


 b24


 b34

b4  b41  b42  b43 
B14 
B24 
B34 

B44 
Texas Tech University
29
Power Flow Equations
• Next, we write the (active) power flow equations:
P1  B12  B13  B14  1  B12 2  B13 3  B14 4
P2   B 21 1  B 21  B 23  B 24  2  B 23 3  B 24 4
P3   B31 1  B32 2  B31  B32  B34  3  B34 4
P4   B 41 1  B 42 2  B 43 3  B 41  B 42  B 43  4
• This can be written as:
 P1   B12  B13  B14
P  
 B21
 2  
 P3  
 B31
  
 B41
 P4  
Dr. Hamed Mohsenian-Rad
 B12
B21  B23  B24
 B13
 B23
 B32
B31  B32  B34
 B42
 B43
Communications and Control in Smart Grid
 1 
  
 2 
  3 
B34
 
B41  B42  B43   4 
 B14
 B24
Texas Tech University
30
Power Flow Equations
• From the last two slides, we finally obtain:






 1 
  
 2 
  3 
 
  4 
 
 

 
 
 
• Therefore, the voltage angles are obtained as:
1  
  
 2  
 3  
  
 4  
Dr. Hamed Mohsenian-Rad






1






Communications and Control in Smart Grid






Texas Tech University
31
Power Flow Equations
• 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: 1  0 .
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
32
Power Flow Equations
• We should also remove the corresponding rows/columns:
 2   30  10  10  10 1 
 1   10 20  10 0   
 2 
 
 4  10  10 30  10  3 
 
  
1
10
0
10
20


  4 
  
 1   20  10 0   2 
 4   10 30  10  
  
 3 
 1   0  10 20   4 
• The angular differences (with respect to 1 ):
 2   20  10 0 
    10 30  10
 3 

 4   0  10 20 
Dr. Hamed Mohsenian-Rad
1
 1   0.025
 4    0.15 
  

 1   0.025
Communications and Control in Smart Grid
 2  1  0.025
 3  1  0.15
 4  1  0.025
Texas Tech University
33
Power Flow Equations
• Finally, the power flow values are calculated as:
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  B23 ( 2   3 )  10(0.025  0.15)  1.25
P34  B34 ( 3   4 )  10(0.15  0.025)  1.25
P2g  2 pu
P1g  2 pu
0.25
0.25
P2l  1 pu
1.5
1.25
1.25
P4g  1 pu
Dr. Hamed Mohsenian-Rad
P3l  4 pu
Communications and Control in Smart Grid
Texas Tech University
34
Power Flow Equations
• 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?
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
35
Economic Dispatch Problem
• In the example we discussed earlier, we had:
Power Supply = Power Load
• In particular, we had:
P1g  P2g  P4g  P2l  P3l
• However, generation levels
P1g , P2g , and P4g
assumed given.
• Q: What if the generators have different generation costs?
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
36
Economic Dispatch Problem
• For thermal power plants, generation cost is quadratic:
Generation Cost = C(P) = a1 + a2 x P + a3 x P2
• Example: a grid with three power plants:
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
• Each power plant has some min and max generation levels.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
37
Economic Dispatch Problem
• For thermal power plants, generation cost is quadratic:
Generation Cost = C(P) = a1 + a2 x P + a3 x P2
• Example: a grid with three power plants:
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
• Each power plant has some min and max generation levels.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
38
Economic Dispatch Problem
• For thermal power plants, generation cost is quadratic:
Generation Cost = C(P) = a1 + a2 x P + a3 x P2
• Example: a grid with three power plants:
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
• Each power plant has some min and max generation levels.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
39
Economic Dispatch Problem
• For thermal power plants, generation cost is quadratic:
Generation Cost = C(P) = a1 + a2 x P + a3 x P2
• Example: a grid with three power plants:
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
• Each power plant has some min and max generation levels.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
40
Economic Dispatch Problem
• For thermal power plants, generation cost is quadratic:
Generation Cost = C(P) = a1 + a2 x P + a3 x P2
• Example: a grid with three power plants:
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
• Each power plant has some min and max generation levels.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
41
Economic Dispatch Problem
• We should select P1, P2, and P3 to:
• Meet total load Pload = 850 MW
• Minimize the total cost of generation
• Economic Dispatch Problem:
minimize CP1   CP2   CP3 
P1 , P2 , P3
subject to 150  P1  600
100  P2  400
50  P3  200
P1  P2  P3  850
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
42
Economic Dispatch Problem
• 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
Q: Do they satisfy all constraints?
P3 = 122.2 MW
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
43
Economic Dispatch Problem
• 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
Minimum Cost = 3916.6 + 3153.8 + 1123.9
= 8194.3
P3 = 122.2 MW
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
44
Economic Dispatch Problem
• What if we have to satisfy topology constraints?
P2
P1
P13  200
P3
 2   20  10 0 
    10 30  10
 3 

 4   0  10 20 
Dr. Hamed Mohsenian-Rad
1
 P2  450
  400 


 P3 
450 MW
400 MW
P13  B13 (1   3 )  10  3  20   3  20
Communications and Control in Smart Grid
Texas Tech University
45
Economic Dispatch Problem
• The same optimal solutions are still valid:
P2  334.6 MW
P1  393.2 MW
156.8
38.1
450 MW 41.4
198.3
160.3
P3  122.2 MW
 2   15.685
    19.830
 3 

 4    3.805 
Dr. Hamed Mohsenian-Rad
400 MW
 3  20
Communications and Control in Smart Grid
P13  200
Texas Tech University
46
Economic Dispatch Problem
• The same optimal solutions are still valid:
P2  334.6 MW
P1  393.2 MW
156.8
38.1
What if P13  170
160.3
P3  122.2 MW
 2   15.685
    19.830
 3 

 4    3.805 
Dr. Hamed Mohsenian-Rad
450 MW 41.4
198.3
400 MW
 3  20
Communications and Control in Smart Grid
P13  200
Texas Tech University
47
Economic Dispatch Problem
• Then the economic dispatch problem becomes:
minimize CP1   CP2   CP3 
P1 , P2 , P3 ,1 ,  2 ,  3
subject to 150  P1  600
100  P2  400
50  P3  200
P1  P2  P3  850
 2   20  10 0 
    10 30  10
 3 

 4   0
 10 20 
 3  17
Dr. Hamed Mohsenian-Rad
1
Communications and Control in Smart Grid
 P2  450
  400 


 P3 
Texas Tech University
48
Economic Dispatch Problem
• Then the economic dispatch problem becomes:
minimize CP1   CP2   CP3 
Still a Convex
Program?
P1 , P2 , P3 ,1 ,  2 ,  3
subject to 150  P1  600
100  P2  400
50  P3  200
P1  P2  P3  850
 2   20  10 0 
    10 30  10
 3 

 4   0
 10 20 
 3  17
Dr. Hamed Mohsenian-Rad
1
Communications and Control in Smart Grid
 P2  450
  400 


 P3 
Texas Tech University
49
Economic Dispatch Problem
• The new optimal solutions are obtained as:
P2  400 MW
P1  280 MW
110
0
450 MW 60
170
170
P3  170 MW
400 MW
• The total generation cost becomes: $8,233.66 > $8,194.3
• Here, we had to sacrifice “cost” for “implementation”.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
50
Economic Dispatch Problem
• The new optimal solutions are obtained as:
P2  400 MW
P1  280 MW
110
0
What if P13  150
450 MW 60
170
170
P3  170 MW
400 MW
• The total generation cost becomes: $8,233.66 > $8,194.3
• Here, we had to sacrifice “cost” for “implementation”.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
51
Unit Commitment
• 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.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
52
References
• 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", Lecture
Notes, EE 458, Department of Electrical and Computer
Engineering, Iowa State University, Spring 2010.
Dr. Hamed Mohsenian-Rad
Communications and Control in Smart Grid
Texas Tech University
53