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ECE 476 POWER SYSTEM ANALYSIS. Lecture 11 YBus and Power Flow Professor Tom Overbye Department of Electrical and Computer Engineering. Announcements. Homework #4 is due now 5.26, 5.27, 5.28, 5.43, 3.4 Homework 5 is due on Oct 4 3.12, 3.19, 27, 60
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ECE 476POWER SYSTEM ANALYSIS Lecture 11 YBus and Power Flow Professor Tom Overbye Department of Electrical andComputer Engineering
Announcements • Homework #4 is due now • 5.26, 5.27, 5.28, 5.43, 3.4 • Homework 5 is due on Oct 4 • 3.12, 3.19, 27, 60 • Oct 2 class (Tuesday) will be in 50 Everitt Lab • First exam is 10/9 in class; closed book, closed notes, one note sheet and calculators allowed • Power plant, substation field trip, 10/11 during class. • Start reading Chapter 6 for lectures 10 to 13
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
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
Ybus Example, cont’d For a system with n buses, Ybus is an n by n symmetric matrix (i.e., one where Aij = Aji)
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 p line model, only • affect the diagonal terms.
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
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.
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.
Three Types of Power Flow Buses • There are three main types of power flow buses • Load (PQ) at which P/Q are fixed; iteration solves for voltage magnitude and angle. • Slack at which the voltage magnitude and angle are fixed; iteration solves for P/Q injections • Generator (PV) at which P and |V| are fixed; iteration solves for voltage angle and Q injection • special coding is needed to include PV buses in the Gauss-Seidel iteration
Two Bus PV Example Consider the same two bus system from the previous example, except the load is replaced by a generator
Generator Reactive Power Limits • The reactive power output of generators varies to maintain the terminal voltage; on a real generator this is done by the exciter • To maintain higher voltages requires more reactive power • Generators have reactive power limits, which are dependent upon the generator's MW output • These limits must be considered during the power flow solution.
Generator Reactive Limits, cont'd • During power flow once a solution is obtained check to make generator reactive power output is within its limits • If the reactive power is outside of the limits, fix Q at the max or min value, and resolve treating the generator as a PQ bus • this is know as "type-switching" • also need to check if a PQ generator can again regulate • Rule of thumb: to raise system voltage we need to supply more vars
Gauss-Seidel Advantages • Each iteration is relatively fast (computational order is proportional to number of branches + number of buses in the system • Relatively easy to program
Gauss-Seidel Disadvantages • Tends to converge relatively slowly, although this can be improved with acceleration • Has tendency to miss solutions, particularly on large systems • Tends to diverge on cases with negative branch reactances (common with compensated lines) • Need to program using complex numbers