topic 1: basics of power systems
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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.
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