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Basics of Power Systems

Basics of Power Systems 22
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 1Power 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 MohsenianRad Communications and Control in Smart Grid Texas Tech University 2Power Systems •Power Production: Different Types: Traditional Renewable Capacity, Cost, Carbon Emission Step‐up Transformers Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 3Power Systems •Power Transmission: High Voltage (HV) Transmission Lines Several Hundred Miles Switching Stations  Transformers Circuit Breakers Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 4Power Systems •The Power Transmission Grid in the United States: www.geni.org Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 5Power Systems •Major Inter‐connections in the United States: www.geni.org Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 6Power 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 MohsenianRad Communications and Control in Smart Grid Texas Tech University 7Power Systems •Power Consumption: Industrial Commercial Residential Demand Response Controllable Load Non‐Controllable Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 8Power Systems Generation Load Transmission Distribution Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 9Power Systems •Power System Control: Data Collection: Sensors, PMUs, etc. Decision Making: Controllers Actuators: Circuit Breakers, etc. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 10Power Grid Graph Representation Nodes: Buses Links: Transmission Lines Generator Load Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 11Power Grid Graph Representation Nodes: Buses Links: Transmission Lines Buses (Voltage) Generator Load Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 12Power Grid Graph Representation Nodes: Buses Links: Transmission Lines Generator Load Transmission Lines (Power Flow, Loss) Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 13Power Grid Graph Representation Nodes: Buses Links: Transmission Lines Generator Load Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 14 ConsumersPower Grid Graph Representation Nodes: Buses Links: Transmission Lines Generator 10 MW 3 MW Load 7 MW Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 15Transmission 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 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 16Transmission Line Admittance •For the transmission line connecting bus ito bus k: Addmitance: y ik Example: y = 1 –j 4   (per unit) ik Note that, y is denoted by y and indicates: ii i  Susceptance for any shunt element (capacitor) to ground at bus i.  Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 17YBus Matrix • We define: Y =  Y  where bus ij N Diagonal Elements: Y y y ii i ik k1,ki Off‐diagonal Elements:  Yy ij ij Note that Y matrix depends on the power grid topology  bas and the admittance of all transmission lines. N is the number of busses in the grid. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 18YBus Matrix •Example: For a grid with 4‐buses, we have: y y y y y y y  1 12 13 14 12 13 14   y y y y y y y 21 2 21 23 24 23 24  Y bus   y y y y y y y 31 32 3 31 32 34 34   y y y y y y y 41 42 43 4 41 42 43  •After separating the real and imaginary parts: YG jB bus Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 19Bus Voltage •Let V denote the voltage at bus i: i •Note that, V is a phasor, with magnitudeand angle.  i V V i i i •In most operating scenarios we have: V V i j i j Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 20Power Flow Equations •Let S denote the power injectionat bus i: i S = P + j Q i i i Active Power        Reactive Power •Generation Bus:  P  0 i •Load Bus:  P  0(negative power injection) i Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 21Power Flow Equations •Using Kirchhoff laws, AC Power Flow Equationsbecome: N P V V G cos( )B sin( ) k k j kj k j kj k j j1 N Q V V G sin( )B cos( )  k k j kj k j kj k j 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 MohsenianRad Communications and Control in Smart Grid Texas Tech University 22Power 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: N P V V B sin( )  k k j kj k j j1 N  Q V VB cos( ) k k j kj k j j1 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 23Power Flow Equations  •Step 2: For most neighboring buses:                                  10 to 15 . i j Sin ( )  k j k j •As a result, we have:   Cos ( ) 1 k j  N  P V V B ( ) k k j kj k j j1 N  Q V VB k k j kj j1 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 24Power Flow Equations •Step 3: In per‐unit, V is very close to 1.0 (0.95 to 1.05).  i V V1 •As a result, we have:                    .  i j N P B ( ) k kj k j j1 N N Q BB Bb  k kj kk kj k j1 j1, jk •P has a linear model and Q is almost fixed. k k Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 25Power Flow Equations •Step 3: In per‐unit, V is very close to 1.0 (0.95 to 1.05).  i V V1 •As a result, we have:                    .  i j DC Power Flow Equations N P B ( ) k kj k j j1 N N Q BB Bb  k kj kk kj k j1 j1, jk •P has a linear model and Q is almost fixed. k k Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 26Power Flow Equations •Given the power injection valuesat all buses, we can use N P B ( )  k kj k j j1 to obtain the voltage anglesat all buses.   •Let P denote the power flowfrom bus ito bus j, we have: ij P B ( ) ij ij i j Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 27Power Flow Equations •Example: Obtain power flow values in the following grid: g g P 2 pu P 2 pu 2 1 yj10 12 l P1 pu 2 yj10 yj10 yj10 23 14 13 yj10 34 g l P1 pu P 4 pu 4 3 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 28Power Flow Equations •First, we obtain the Y‐bus matrix: bbbbbbb  1 12 13 14 12 13 14  b bbbbbb 21 2 21 23 24 23 24  Y j bus  bb bbbbb 31 32 3 31 32 34 34  bbb bbbb  41 42 43 4 41 42 43 B B B B  11 12 13 14   B B B B 21 22 23 24    j jB j  B B B B 31 32 33 34  B B B B  41 42 43 44 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 29Power Flow Equations •Next, we write the (active) power flow equations: P B B B B B B 1 12 13 14 1 12 2 13 3 14 4 PB B B B B B 2 21 1 21 23 24 2 23 3 24 4 PB B B B B B 3 31 1 32 2 31 32 34 3 34 4 PB B B B B B 4 41 1 42 2 43 3 41 42 43 4 •This can be written as:  P BBBBBB  1 12 13 14 12 13 14 1  PB BBBBB 2 21 21 23 24 23 24 2    PBB BBB B 3 31 32 31 32 34 34 3  PBBB BBB  4 41 42 43 41 42 43 4 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 30Power Flow Equations •From the last two slides, we finally obtain:   1   2    3    4 •Therefore, the voltage angles are obtained as:  1   1   2    3     4 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 31Power 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.   0 • We choose one bus (e.g., bus 1) as reference bus:             . 1 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 32Power Flow Equations • We should also remove the corresponding rows/columns:  2 30101010  1 1 2010 0   2  110 2010 0 2    410 3010 3   4 10 10 30 10  3  1 010 20   4 110 010 20  4 •The angular differences (with respect to     ):   1 1  2010 0 1 0.025 0.025 2 2 1  10 3010 4 0.15 0.15 3 3 1    010 20 1 0.025 0.025  4 4 1 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 33Power Flow Equations •Finally, the power flow values are calculated as: P B ( )10(00.025) 0.25 12 12 1 2 P B ( )10(00.15)1.5 13 13 1 3 P B ( )10(00.025) 0.25 14 14 1 4 P B ( )10(0.0250.15)1.25 23 23 2 3 P B ( )10(0.150.025)1.25 34 34 3 4 g g P 2 pu P 2 pu 2 1 0.25 l P1 pu 2 0.25 1.25 1.5 1.25 l g P 4 pu P1 pu 3 4 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 34Power Flow Equations •What if the generator connected to bus 1 is renewable •What if the capacityof transmission link (1,3) is 1 pu •What if we can apply demand responseto load bus 3    •What if one of the transmission lines fails  Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 35Economic Dispatch Problem •In the example we discussed earlier, we had: Power Supply = Power Load •In particular, we had:   g g g l l PPP PP 1 2 4 2 3 g g g •However, generation levels             and      assumed given. P ,P , P 1 2 4 •Q: What if the generators have different generation costs Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 36Economic Dispatch Problem •For thermal power plants, generation cost is quadratic: 2 Generation Cost = C(P) = a + a x P + a x P 1 2 3 •Example: a grid with three power plants: 2 C (P ) = 561 + 7.92 x P + 0.001562 x (P ) 150 MW ≤ P ≤ 600 MW 1 1 1 1 1 2 C (P ) = 310 + 7.85 x P + 0.001940 x (P ) 100 MW ≤ P ≤ 400 MW 2 2 2 2 2 2 C (P ) =   78 + 7.97 x P + 0.004820 x (P ) 50 MW ≤ P ≤ 200 MW 3 3 3 3 3 •Each power plant has some min and max generation levels. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 37Economic Dispatch Problem •For thermal power plants, generation cost is quadratic: 2 Generation Cost = C(P) = a + a x P + a x P 1 2 3 •Example: a grid with three power plants: 2 C (P ) = 561 + 7.92 x P + 0.001562 x (P ) 150 MW ≤ P ≤ 600 MW 1 1 1 1 1 2 C (P ) = 310 + 7.85 x P + 0.001940 x (P ) 100 MW ≤ P ≤ 400 MW 2 2 2 2 2 2 C (P ) =   78 + 7.97 x P + 0.004820 x (P ) 50 MW ≤ P ≤ 200 MW 3 3 3 3 3 •Each power plant has some min and max generation levels. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 38Economic Dispatch Problem •For thermal power plants, generation cost is quadratic: 2 Generation Cost = C(P) = a + a x P + a x P 1 2 3 •Example: a grid with three power plants: 2 C (P ) = 561 + 7.92 x P + 0.001562 x (P ) 150 MW ≤ P ≤ 600 MW 1 1 1 1 1 2 C (P ) = 310 + 7.85 x P + 0.001940 x (P ) 100 MW ≤ P ≤ 400 MW 2 2 2 2 2 2 C (P ) =   78 + 7.97 x P + 0.004820 x (P ) 50 MW ≤ P ≤ 200 MW 3 3 3 3 3 •Each power plant has some min and max generation levels. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 39Economic Dispatch Problem •For thermal power plants, generation cost is quadratic: 2 Generation Cost = C(P) = a + a x P + a x P 1 2 3 •Example: a grid with three power plants: 2 C (P ) = 561 + 7.92 x P + 0.001562 x (P ) 150 MW ≤ P ≤ 600 MW 1 1 1 1 1 2 C (P ) = 310 + 7.85 x P + 0.001940 x (P ) 100 MW ≤ P ≤ 400 MW 2 2 2 2 2 2 C (P ) =   78 + 7.97 x P + 0.004820 x (P ) 50 MW ≤ P ≤ 200 MW 3 3 3 3 3 •Each power plant has some min and max generation levels. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 40Economic Dispatch Problem •For thermal power plants, generation cost is quadratic: 2 Generation Cost = C(P) = a + a x P + a x P 1 2 3 •Example: a grid with three power plants: 2 C (P ) = 561 + 7.92 x P + 0.001562 x (P ) 150 MW ≤ P ≤ 600 MW 1 1 1 1 1 2 C (P ) = 310 + 7.85 x P + 0.001940 x (P ) 100 MW ≤ P ≤ 400 MW 2 2 2 2 2 2 C (P ) =   78 + 7.97 x P + 0.004820 x (P ) 50 MW ≤ P ≤ 200 MW 3 3 3 3 3 •Each power plant has some min and max generation levels. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 41Economic Dispatch Problem • We should select P , P , and P to:  1 2 3 •Meet total loadP = 850 MW load •Minimize the total costof generation  •Economic Dispatch Problem: minimize CP CP C P 1 2 3 P , P , P 1 2 3 subject to 150P 600 1 100P 400 2 50P 200 3 PPP 850 1 2 3 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 42Economic 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 optimaleconomic dispatch solution:  P = 393.2 MW 1 P = 334.6 MW Q: Do they satisfy all constraints  2 P = 122.2 MW 3 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 43Economic 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 optimaleconomic dispatch solution:  P = 393.2 MW 1 Minimum Cost= 3916.6 + 3153.8 + 1123.9 P = 334.6 MW 2 = 8194.3  P = 122.2 MW 3 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 44Economic Dispatch Problem •What if we have to satisfy topology constraints  P P 1 2 450 MW P 200 13 P 400 MW 3 1  2010 0 P 450  2 2  10 3010 400 3 P B ( )10 2020  13 13 1 3 3 3   010 20 P  4 3 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 45Economic Dispatch Problem •The sameoptimal solutions are still valid: P 334.6 MW P 393.2 MW 1 2 156.8 450 MW 41.4 38.1 198.3 160.3 P 122.2 MW 400 MW 3 15.685  2  19.830 3 P 200 20 3 13   3.805  4 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 46Economic Dispatch Problem •The sameoptimal solutions are still valid: P 334.6 MW P 393.2 MW 1 2 156.8 450 MW 41.4 38.1 198.3 160.3 P 170 What if 13 P 122.2 MW 400 MW 3 15.685  2  19.830 3 P 200 20 3 13   3.805  4 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 47Economic Dispatch Problem •Then the economic dispatch problem becomes: minimize CP CP CP 1 2 3 P , P , P , ,  ,  1 2 3 1 2 3 subject to 150P 600 1 100P 400 2 50P 200 3 PPP 850 1 2 3 1  2010 0 P 450  2 2  10 3010 400 3    010 20 P 4 3  17 3 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 48Economic Dispatch Problem •Then the economic dispatch problem becomes: minimize CP CP CP 1 2 3 Still a Convex  P , P , P , ,  ,  1 2 3 1 2 3 Program subject to 150P 600 1 100P 400 2 50P 200 3 PPP 850 1 2 3 1  2010 0 P 450  2 2  10 3010 400 3    010 20 P 4 3  17 3 Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 49Economic Dispatch Problem •The new optimal solutions are obtained as: P 400 MW P 280 MW 1 2 110 450 MW 60 0 170 170 P 170 MW 400 MW 3 •The total generation cost becomes: 8,233.66  8,194.3 •Here, we had to sacrifice “cost” for “implementation”. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 50Economic Dispatch Problem •The new optimal solutions are obtained as: P 400 MW P 280 MW 1 2 110 450 MW 60 0 170 170 P 150 What if 13 P 170 MW 400 MW 3 •The total generation cost becomes: 8,233.66  8,194.3 •Here, we had to sacrifice “cost” for “implementation”. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 51Unit Commitment •Economic Dispatch is solved a few hours aheadof operation. •On the other hand, we need to decide about the choice of  power plants that we want to turn onfor the next day. •This is done by solving the Unit Commitmentproblem. • We particularly decide on which slow‐startingpower plants we  should turn on during the next day given various constraints.  •The mathematical concepts are similar to the E‐D problem.  Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 52References •W. J. Wood and B. F. Wollenberg, Power Generation, nd Operation,andControl,JohnWileySons,2 Ed.,1996. •J. McCalley and L. Tesfatsion, "Power Flow Equations", Lecture Notes, EE 458, Department of Electrical and Computer Engineering,IowaStateUniversity,Spring2010. Dr. Hamed MohsenianRad Communications and Control in Smart Grid Texas Tech University 53
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