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

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 ) =  1 H(  1 ) +  2 H(  2 ) That is 1)the output is proportional to the input 2)the principle of superposition holds Linear Example: y = H(x) = c x y = c(x 1 +x 2 ) = cx 1 + c x 2 Nonlinear Example: y = H(x) = c x 2 y = c(x 1 +x 2 ) 2 ≠ (cx 1 ) 2 + (c x 2 ) 2 3

Linear Power System Elements 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:x = 0 has solutions x =  1.414… Example 2: x = 0 has no real solution f (x) = x f (x) = x two solutions where f(x) = 0 no solution 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 What is the maximum P Load ? 7

Bus Admittance Matrix or Y bus First step in solving the power flow is to create what is known as the bus admittance matrix, often call the Y bus. The Y bus gives the relationships between all the bus current injections, I, and all the bus voltages, V, I = Y bus V The Y bus 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

Y bus Example Determine the bus admittance matrix for the network shown below, assuming the current injection at each bus i is I i = I Gi - I Di where I Gi is the current injection into the bus from the generator and I Di is the current flowing into the load 9

Y bus Example, cont’d 10

Y bus Example, cont’d For a system with n buses, Y bus is an n by n symmetric matrix (i.e., one where A ij = A ji ) 11

Y bus General Form The diagonal terms, Y ii, are the self admittance terms, equal to the sum of the admittances of all devices incident to bus i. The off-diagonal terms, Y ij, are equal to the negative of the sum of the admittances joining the two buses. With large systems Y bus 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 Y bus 13

Two Bus System Example 14

Using the Y bus 15

Solving for Bus Currents 16

Solving for Bus Voltages 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 Y bus equations, but rather must use the power balance equations 18

Power Balance Equations 19

Power Balance Equations, cont’d 20

Real Power Balance Equations 21

Power Flow Requires Iterative Solution 22

Gauss Iteration 23

Gauss Iteration Example 24

Stopping Criteria 25

Gauss Power Flow 26

Gauss Two Bus Power Flow Example A 100 MW, 50 Mvar load is connected to a generator through a line with z = 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 V 2 ? S Load = j0.5 p.u. 27

Gauss Two Bus Example, cont’d 28

Gauss Two Bus Example, cont’d 29

Gauss Two Bus Example, cont’d 30

Slack Bus In previous example we specified S 2 and V 1 and then solved for S 1 and V 2. 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