Presentation on theme: "ECE 530 – Analysis Techniques for Large-Scale Electrical Systems"— Presentation transcript:
1ECE 530 – Analysis Techniques for Large-Scale Electrical Systems Lecture 3: Power Flow AnalysisProf. Hao ZhuDept. of Electrical and Computer EngineeringUniversity of Illinois at Urbana-Champaign
2Announcements Homework 1 will be posted today. Due Sep 11 (Thur), in class
3Static Power System Analysis One of the most common power system analysis tools is the power flow, which tells how power flows through a power system in the quasi-steady state time frameThe power flow can be used to model the full, three-phase system, but usually (practically always) for transmission system analysis the system is assumed to be balanced. Hence a per phase equivalent model is used.
4Power System Component Models: Transmission Lines Power flow timeframe models for common power system devices, including transmission lines, transformers, generators and loads, are developed in the prerequisite course ECE 476.Transmission lines will be modeled using the p circuit
5Power System Component Models: Transformers Transformer equivalent model
6Power System Component Models: Loads Ultimate goal is to supply loads with electricity at constant frequency and voltageElectrical characteristics of individual loads matter, but usually they can only be estimatedactual loads are constantly changing, consisting of a large number of individual devicesonly limited network observability of load characteristicsAggregate models are typically used for analysisTwo common modelsconstant power: Si = Pi + jQiconstant impedance: Si = |V|2 / Zi
7Power System Component Models: Generators Engineering models depend upon applicationGenerators are usually synchronous machinesFor generators we will use two different models:a steady-state model, treating the generator as a constant power source operating at a fixed voltage; this model will be used for power flow and economic analysisa short term model treating the generator as a constant voltage source behind a possibly time-varying reactance
8Per Phase Calculations A key problem in analyzing power systems is the large number of transformers.It would be very difficult to continually have to refer impedances to the different sides of the transformersThis problem is avoided by a normalization of all variables.This normalization is known as per unit analysis
9Per Unit Conversion Procedure, 1f Pick a 1f VA base for the entire system, SBPick a voltage base for each different voltage level, VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral.Calculate the impedance base, ZB= (VB)2/SBCalculate the current base, IB = VB/ZBConvert actual values to per unitNote, per unit conversion on affects magnitudes, notthe angles. Also, per unit quantities no longer haveunits (i.e., a voltage is 1.0 p.u., not 1 p.u. volts)
10Per Unit Solution Procedure Convert to per unit (p.u.) (many problems are already in per unit)SolveConvert back to actual as necessary
11Per Unit Example Solve for the current, load voltage and load power in the circuit shown below using per unit analysiswith an SB of 100 MVA, and voltage bases of8 kV, 80 kV and 16 kV.Original Circuit
12Per Unit Example, cont’d Same circuit, withvalues expressedin per unit.
14Per Unit Example, cont’d To convert back to actual values just multiply theper unit values by their per unit base
15Three Phase Per UnitProcedure is very similar to 1f except we use a 3fVA base, and use line to line voltage basesPick a 3f VA base for the entire system,Pick a voltage base for each different voltage level, VB. Voltages are line to line.Calculate the impedance baseExactly the same impedance bases as with single phase!
16Three Phase Per Unit, cont'd Calculate the current base, IBConvert actual values to per unitExactly the same current bases as with single phase!
17Three Phase Per Unit Example Solve for the current, load voltage and load powerin the previous circuit, assuming a 3f power base of300 MVA, and line to line voltage bases of 13.8 kV,138 kV and 27.6 kV (square root of 3 larger than the 1f example voltages). Also assume the generator is Y-connected so its line to line voltage is 13.8 kV.Convert to per unitas before. Note thesystem is exactly thesame!
183f Per Unit Example, cont'd Again, analysis is exactly the same!
193f Per Unit Example, cont'd Differences appear when we convert back to actual values
203f Per Unit Example 2Assume a 3f load of 100+j50 MVA with VLL of 69 kV is connected to a source through the below network:What is the supply current and complex power?Answer: I=467 amps, S = j76.0 MVA
21Bus 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 = Ybus VThe 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
22Ybus Example Determine the bus admittance matrix for the network shown below, assuming the current injection at eachbus i is Ii = IGi - IDi where IGi is the current injection into thebus from the generator and IDi is the current flowing into theload
24Ybus Example, cont’dFor a system with n buses, Ybus is an n by n symmetric matrix (i.e., one where Aij = Aji); however this will not be true in general when we consider phase shifting transformers
25Ybus General FormThe 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.
31Power Flow AnalysisWhen analyzing power systems we know neither the complex bus voltages nor the complex current injectionsRather, we know the complex power being consumed by the load, and the power being injected by the generators plus their voltage magnitudesTherefore we can not directly use the Ybus equations, but rather must use the power balance equations
32Power Flow AnalysisClassic paper for this lecture is W.F. Tinney and C.E. Hart, “Power Flow Solution by Newton’s Method,” IEEE Power App System, Nov 1967Basic power flow is also covered in essentially power system analysis textbooks.At Illinois we use the term “power flow” not “load flow” since power flows not load. Also, the power flow usage is not new (see title of Tinney’s 1967 paper, and note Tinney references Ward’s 1956 paper)A nice history of the power flow is given in an insert by Alvarado and Thomas in T.J. Overbye, J.D. Weber, “Visualizing the Electric Grid,” IEEE Spectrum, Feb 2001.
36Slack BusWe can not arbitrarily specify S at all buses because total generation must equal total load + total lossesWe 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.In an actual power system the slack bus does not really exist; frequency changes locally when the power supplied does not match the power consumed
37Three Types of Power Flow Buses There are three main types of power flow busesLoad (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 injectionsGenerator (PV) at which P and |V| are fixed; iteration solves for voltage angle and Q injection
38Newton-Raphson Algorithm Most common technique for solving the power flow problem is to use the Newton-Raphson algorithmKey idea behind Newton-Raphson is to use sequential linearizationMore tractable for linear systems