smart grid topic on power systems

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
• Power Production:
 Different Types:
 Traditional
 Renewable
 Capacity, Cost, Carbon Emission
 Step‐up Transformers

Power Systems
• Power Transmission:
 High Voltage (HV) Transmission Lines
 Several Hundred Miles
 Switching Stations
 Transformers
 Circuit Breakers

Power Systems
• The Power Transmission Grid in the United States:
www.geni.org

Power Systems
• Major Inter‐connections in the United States:
www.geni.org

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

Power Systems
• Power Consumption:
 Industrial
 Commercial
 Residential
 Demand Response
 Controllable Load
 Non‐Controllable

Power Systems
Generation Transmission Distribution Load

Power Systems
• Power System Control:
 Data Collection: Sensors, PMUs, etc.
 Decision Making: Controllers
 Actuators: Circuit Breakers, etc.

Power Grid Graph Representation
Nodes: Buses
Links: Transmission Lines
Generator
Load

Nodes: Buses
Generator
Load
Buses (Voltage)

Nodes: Buses
Generator
Load
Transmission Lines (Power Flow, Loss)

Nodes: Buses
Generator
Load
Consumers

Nodes: Buses
Generator
Load
10 MW 3 MW
7 MW

Transmission Line Admittance
• 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
• 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
• 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
ikk
ikiii yyY
,1
ijij yY 

Y-Bus Matrix
• Example: For a grid with 4‐buses, we have:
• After separating the real and imaginary parts:

















4342414434241
3434323133231
2423242321221
1413121413121
yyyyyyy
yyyyyyy
yyyyyyy
yyyyyyy
Ybus
BjGYbus 

Bus Voltage
• 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
• 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

• 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
j
jkkjjkkjjkk
N
j
jkkjjkkjjkk
BGVVQ
BGVVP
1
1
)cos()sin(
)sin()cos(



• 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
j
jkkjjkk
N
j
jkkjjkk
BVVQ
BVVP
1
1
)cos(
)sin(



• Step 2: For most neighboring buses: .
• As a result, we have:

15to10 ji 





1)(
)(
jk
jkjk
Cos
Sin


 
 





N
j
kjjkk
N
j
jkkjjkk
BVVQ
BVVP
1
1
)( 

• 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.
1ji VV
  k
N
kj
j
kjkk
N
j
kjk
N
j
jkkjk
bBBBQ
BP







,11
1
)( 

• 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.
1ji VV
  k
N
kj
j
kjkk
N
j
kjk
N
j
jkkjk
bBBBQ
BP







,11
1
)( 
DC Power Flow Equations

• 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
j
jkkjk BP
1
)( 
)( jiijij BP  

• Example: Obtain power flow values in the following grid:
puPg
21  puPg
22 
puPg
14  puPl
43 
puPl
12 
1014 jy  1013 jy 
1034 jy 
1023 jy 
1012 jy 

• First, we obtain the Y‐bus matrix:











































44434241
34333231
24232221
14131211
4342414434241
3434323133231
2423242321221
1413121413121
BBBB
BBBB
BBBB
BBBB
jBjj
bbbbbbb
bbbbbbb
bbbbbbb
bbbbbbb
jYbus

• Next, we write the (active) power flow equations:
• This can be written as:
 
 
 
  44342413432421414
43433432312321313
42432322423211212
41431321211413121




BBBBBBP
BBBBBBP
BBBBBBP
BBBBBBP













































4
3
2
1
434241434241
343432313231
242324232121
141312141312
4
3
2
1




BBBBBB
BBBBBB
BBBBBB
BBBBBB
P
P
P
P

• From the last two slides, we finally obtain:
• Therefore, the voltage angles are obtained as:





































4
3
2
1









































1
4
3
2
1





• 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 

• We should also remove the corresponding rows/columns:
• The angular differences (with respect to ):










































4
3
2
1
2010010
10301010
0102010
10101030
1
4
1
2







































4
3
2
20100
103010
01020
1
4
1



1


















































025.0
15.0
025.0
1
4
1
20100
103010
01020
1
4
3
2



025.0
15.0
025.0
14
13
12







• Finally, the power flow values are calculated as:
puPg
21  puPg
22 
puPg
14  puPl
43 
puPl
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










BP
BP
BP
BP
BP
0.25
1.25
1.250.25 1.5

• 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
• 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 ,,

• 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

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

• We should select P1, P2, and P3 to:
• Meet total load Pload = 850 MW
• Minimize the total cost of generation
• Economic Dispatch Problem:
     
850
20050
400100
600150subject to
CCCminimize
321
3
2
1
321
,, 321





PPP
P
P
P
PPP
PPP

• 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?

• 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

• What if we have to satisfy topology constraints?
1P
3P MW400
2P
MW45020013 P





































3
2
1
4
3
2
400
450
20100
103010
01020
P
P



202010)( 33311313  BP

• The same optimal solutions are still valid:
MW393.21 P
MW122.23 P MW400
MW334.62 P
MW450
























805.3
830.19
685.15
4
3
2



203  20013 P
156.8
160.3
41.438.1 198.3

• The same optimal solutions are still valid:
MW393.21 P
MW122.23 P MW400
MW334.62 P
MW450
























805.3
830.19
685.15
4
3
2



203  20013 P
156.8
160.3
41.438.1 198.3
What if 17013 P

• Then the economic dispatch problem becomes:
     
17
400
450
20100
103010
01020
850
20050
400100
600150subject to
CCCminimize
3
3
2
1
4
3
2
321
3
2
1
321
,,,,, 321321
















































P
P
PPP
P
P
P
PPP
PPP

• Then the economic dispatch problem becomes:
     
17
400
450
20100
103010
01020
850
20050
400100
600150subject to
CCCminimize
3
3
2
1
4
3
2
321
3
2
1
321
,,,,, 321321
















































P
P
PPP
P
P
P
PPP
PPP Still a Convex
Program?

• 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”.
MW2801 P
MW1703 P MW400
MW4002 P
MW450
110
170
600 170

• 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”.
MW2801 P
MW1703 P MW400
MW4002 P
MW450
110
170
600 170
What if 15013 P

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.

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.

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