Power System Analysis and Linear/Nonlinear Elements Overview

ECE 476 Power System Analysis Lecture 11: Ybus, Power Flow Prof. Tom Overbye Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign overbye@illinois.edu

Announcements Please read Chapter 2.4; start on Chapter 6 H5 is 3.4, 3.10, 3.14, 3.19, 3.23, 3.60, 2.38, 6.9 It should be done before the first exam, but does not need to be turned in First exam is Tuesday Oct 6 during class Closed book, closed notes, but you may bring one 8.5 by 11 inch note sheet and standard calculators. Covers up to end of today's lecture Last name starting with A to 0 in 3017; P to Z in 3013 Won will give optional review on Thursday; no new material 1

Power Flow Analysis We now have the necessary models to start to develop the power system analysis tools The most common power system analysis tool is the power flow (also known sometimes as the load flow) power flow determines how the power flows in a network also used to determine all bus voltages and all currents because of constant power models, power flow is a nonlinear analysis technique power flow is a steady-state analysis tool 2

Linear versus Nonlinear Systems A function H is linear if H( 1 1 + 2 2) = 1H( 1) + 2H( 2) That is 1) the output is proportional to the input 2) the principle of superposition holds Linear Example:y = H(x) = c x Nonlinear Example:y = H(x) = c x2 y = c(x1+x2) = cx1 + c x2 y = c(x1+x2)2 (cx1)2 + (c x2)2 3

Linear Power System Elements Resistors, inductors, capacitors, independent voltage sources and current sources are linear circuit elements 1 j L I V = R I V = V = I j C Such systems may be analyzed by superposition 4

Nonlinear Power System Elements Constant power loads and generator injections are nonlinear and hence systems with these elements can not be analyzed by superposition Nonlinear problems can be very difficult to solve, and usually require an iterative approach 5

Nonlinear Systems May Have Multiple Solutions or No Solution Example 1: x2 - 2 = 0 has solutions x = 1.414 Example 2: x2 + 2 = 0 has no real solution f(x) = x2 - 2 f(x) = x2 + 2 no solution f(x) = 0 two solutions where f(x) = 0 6

Multiple Solution Example 3 The dc system shown below has two solutions: where the 18 watt load is a resistive load The equation we're solving is 2 = 9 volts 1 +R 2 = = I 18 watts Load R Load R What is the maximum PLoad? Load One solution is R Other solution is R 2 = Load 0.5 Load 7

Bus Admittance Matrix or Ybus First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Ybus. The Ybus gives the relationships between all the bus current injections, I, and all the bus voltages, V, I = YbusV The Ybus is developed by applying KCL at each bus in the system to relate the bus current injections, the bus voltages, and the branch impedances and admittances 8

Ybus Example Determine the bus admittance matrix for the network shown below, assuming the current injection at each bus i is Ii = IGi - IDi where IGi is the current injection into the bus from the generator and IDi is the current flowing into the load 9

YbusExample, contd By KCL at bus 1 we have I I I 1 1 1 G D V V V V 1 3 = + = + 1 2 I 12 I 13 I 1 Z Z A B 1 = + = ( ) ( ) (with Y ) I V V Y V V Y 1 1 2 1 3 j A B Z j = + ( ) Y Y V Y V Y V 1 2 3 A B A B Similarly I = = + + I I I 2 21 Y V 23 + 24 + + ( ) Y Y Y V Y V Y V 1 2 3 4 A A C D C D 10

YbusExample, contd We can get similar relationships for buses 3 and 4 The results can then be expressed in matrix form = + + = I Y V bus Y 0 Y I I I I Y Y Y Y + V V V V 1 1 A B A B + Y Y Y Y Y 2 2 A A C Y Y D C Y D 0 Y 3 3 B C B C 0 0 Y 4 4 D D For a system with n buses, Ybus is an n by n symmetric matrix (i.e., one where Aij = Aji) 11

Ybus General Form The diagonal terms, Yii, are the self admittance terms, equal to the sum of the admittances of all devices incident to bus i. The off-diagonal terms, Yij, are equal to the negative of the sum of the admittances joining the two buses. With large systems Ybus is a sparse matrix (that is, most entries are zero) Shunt terms, such as with the line model, only affect the diagonal terms. 12

Modeling Shunts in the Ybus Y kc = + Since ( ) I V V Y V ij i j k i 2 Y from other lines ii Y kc = + + Y Y ii k 2 + 1 1 R jX jX R R jX X k k k k = = = Note Y k 2 k 2 k + Z R jX R k k k k k 13

Two Bus System Example ( ) 1 c Y V V 1 2 = + = 12 16 j I V 1 1 + 2 0.03 12 0.04 V V Z j + 12 15.9 16 12 j + 16 j I I j 1 1 = 12 15.9 j 2 2 14

Using the Ybus If the voltages are known then we can solve for the current injections: = Y V I bus If the current injections are known then we can solve for the voltages: = = Y I V Z I Z 1 bus bus where is the bus impedance matr ix bus 15

Solving for Bus Currents For example, in previous case assume 1.0 0.8 0.2 Then 12 15.9 12 12 16 12 15.9 Therefore the power injected at bus 1 is = V j + 16 j 1.0 5.60 5.58 0.70 0.88 j j j = + + 0.8 0.2 j j j * = = 1.0 (5.60 + = + S 0.70) 5.60 0.70 1 1 V I V I = j j 1 * = 0.2) ( 5.58 = + (0.8 0.88) 4.64 0.41 S j j j 2 2 2 16

Solving for Bus Voltages For example, in previous case assume 5.0 4.8 Then = I 1 + 12 15.9 16 12 j + 12 16 j 5.0 4.8 0.0738 0.0738 0.902 1.098 j j j = 12 15.9 j Therefore the power injected is * = = = S (0.0738 0.902) 5 0 .37 4.51 1 1 V I V I = j j 1 * ( 0.0738 = 1.098) ( 4.8) j = + 0.35 5.27 S j 2 2 2 17

Power Flow Analysis When analyzing power systems we know neither the complex bus voltages nor the complex current injections Rather, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudes Therefore we can not directly use the Ybus equations, but rather must use the power balance equations 18

Power Balance Equations From KCL we know at each bus i in an n bus system the current injection, , must be equal to the current that flows into the network I i n = = I I I I i Gi Di ik = 1 k I Y V Since = we also know bus n = = I I I ik k Y V i Gi Di = 1 k * = The network power injection is then S V I i i i 19

Power Balance Equations, contd * n n * * * = = = S V I V ik k Y V V ik k Y V i i i i i = = 1 1 k k This is an equation with complex numbers. Sometimes we would like an equivalent set of real power equations. These can be derived by defining Y G jB + ik ik ik j = V V e V i i i i i ik i k j = + Recall e cos sin j 20

Real Power Balance Equations n n j * * = + = = S ( ) P jQ V ik k Y V V V e G jB ik i i i i i k ik ik = = 1 1 k k n = + (cos sin )( ) V V j G jB i k ik ik ik ik = 1 k Resolving into the real and imaginary parts n = + = P ( cos sin ) V V G B P P i i k ik ik ik ik Gi Di = 1 k n = = Q ( sin cos ) V V G B Q Q i i k ik ik ik i k Gi Di = 1 k 21

Power Flow Requires Iterative Solution In the power flow we assume we know S and the . We would like to solve for the V's. The problem is the below equation has no closed form solution: i Y bus * n n * * * = = = S V I V ik k Y V V ik k Y V i i i i i = = 1 1 k k Rath er, we must pursue an iterative approach. 22

Gauss Iteration There are a number of different iterative methods we can use. We'll consider two: Gauss and Newton. With the Gauss method we need to rewrite our equation in an implicit form: x = h(x) (0) To iterate we fir st make an initial guess of x, x , ( +1) v ( ) v = and then iteratively solve x find a "fixed point", x, such that x ( ) until we (x). h x = h 23

Gauss Iteration Example = Example: Solve - 1 0 x x + ( 1) ( ) v v = + 1 x x (0) = Let = 0 and arbitrarily guess x v 1 and solve ( ) v ( ) v v x v x 0 1 2 3 4 1 2 2.41421 2.55538 2.59805 5 6 7 8 9 2.61185 2.61612 2.61744 2.61785 2.61798 24

Stopping Criteria A key problem to address is when to stop the iteration. With the Guass iteration we stop when + ( ) v ( ) v ( 1) ( ) v v with x x x x If x is a scalar this is clear, but if x is a vector we need to generalize t he absolute value by using a norm ( ) v x j Two common norms are the Euclidean & infinity n 2 i = = x x max x x i i 2 = 1 i 25

Gauss Power Flow We first need to put the equation in the correct form * n n * * * = = = S V I V ik k Y V V ik k Y V i i i i i = = 1 1 k n k n * i * * i * i = = = S V I V ik k Y V V ik k Y V i i = = 1 1 k k * i n n S V = = + ik k Y V ii i Y V ik k Y V * i = = k i 1 1, k k * i n S V 1 = V ik k Y V i * i Y = k i ii 1, k 26

Gauss Two Bus Power Flow Example A 100 MW, 50 Mvar load is connected to a generator through a line with z = 0.02 + j0.06 p.u. and line charging of 5 Mvar on each end (100 MVA base). Also, there is a 25 Mvar capacitor at bus 2. If the generator voltage is 1.0 p.u., what is V2? SLoad = 1.0 + j0.5 p.u. 27

Gauss Two Bus Example, contd The unknown is the complex load voltage, V . To determine V we need to know the 1 5 0.02 0.06 5 14.95 Hence 5 15 (Note - 15 0.05 B j j = + 2 Y . 2 bus = 15 j + j + 5 15 j j = Y bus + 5 + 14.70 0.25) j j j 22 28

Gauss Two Bus Example, contd * 2 * 2 n 1 S V = V ik k Y V 2 Y = k i 22 1, k + 1 14.70 j -1 0.5 j = + ( 5 15)(1.0 0) j V 2 * 2 5 V (0) 2 = 1.0 0 (this is known as a flat start) Guess V ( ) 2 ( ) 2 v v v V v V + 0 1 2 1.000 0.9671 0 .9624 0.000 0.0568 j 0.0553 j 3 4 0.9622 0.9622 0.0556 0.0556 j j j 29

Gauss Two Bus Example, contd = = 0.9622 0.0556 0.9638 3.3 V j 2 Once the voltages are known all other values can be determined, such as the generator powers and the line flows * 1 * = + = , Q S In actual units P ( ) 1.023 102.3 MW 0.239 = V Y V 12 2 Y V = j 1 11 1 23.9 Mvar 1 1 25 2 = The capacitor is supplying V 23.2 Mvar 2 30

Slack Bus In previous example we specified S2 and V1 and then solved for S1 and V2. We can not arbitrarily specify S at all buses because total generation must equal total load + total losses We also need an angle reference bus. To solve these problems we define one bus as the "slack" bus. This bus has a fixed voltage magnitude and angle, and a varying real/reactive power injection. 31