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ECE 530 – Analysis Techniques for Large-Scale Electrical Systems Prof. Hao Zhu Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign haozhu@illinois.edu 1 Lecture 3: Power Flow Analysis

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Announcements Homework 1 will be posted today. Due Sep 11 (Thur), in class 2

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Static 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 frame The 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. 3

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Power 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 circuit 4

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Power System Component Models: Transformers Transformer equivalent model 5

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Power System Component Models: Loads Ultimate goal is to supply loads with electricity at constant frequency and voltage Electrical characteristics of individual loads matter, but usually they can only be estimated – actual loads are constantly changing, consisting of a large number of individual devices – only limited network observability of load characteristics Aggregate models are typically used for analysis Two common models – constant power: S i = P i + jQ i – constant impedance: S i = |V| 2 / Z i 6

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Power System Component Models: Generators Engineering models depend upon application Generators are usually synchronous machines For 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 analysis – a short term model treating the generator as a constant voltage source behind a possibly time-varying reactance 7

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Per 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 transformers This problem is avoided by a normalization of all variables. This normalization is known as per unit analysis 8

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Per Unit Conversion Procedure, 1 1. Pick a 1 VA base for the entire system, S B 2. Pick a voltage base for each different voltage level, V B. Voltage bases are related by transformer turns ratios. Voltages are line to neutral. 3. Calculate the impedance base, Z B = (V B ) 2 /S B 4. Calculate the current base, I B = V B /Z B 5. Convert actual values to per unit Note, per unit conversion on affects magnitudes, not the angles. Also, per unit quantities no longer have units (i.e., a voltage is 1.0 p.u., not 1 p.u. volts) 9

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Per Unit Solution Procedure 1. Convert to per unit (p.u.) (many problems are already in per unit) 2. Solve 3. Convert back to actual as necessary 10

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Per Unit Example Solve for the current, load voltage and load power in the circuit shown below using per unit analysis with an S B of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV. Original Circuit 11

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Per Unit Example, cont’d Same circuit, with values expressed in per unit. 12

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Per Unit Example, cont’d 13

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Per Unit Example, cont’d To convert back to actual values just multiply the per unit values by their per unit base 14

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Three Phase Per Unit 1. Pick a 3 VA base for the entire system, 2. Pick a voltage base for each different voltage level, V B. Voltages are line to line. 3. Calculate the impedance base Procedure is very similar to 1 except we use a 3 VA base, and use line to line voltage bases Exactly the same impedance bases as with single phase! 15

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Three Phase Per Unit, cont'd 4. Calculate the current base, I B 5. Convert actual values to per unit Exactly the same current bases as with single phase! 16

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Three Phase Per Unit Example Solve for the current, load voltage and load power in the previous circuit, assuming a 3 power base of 300 MVA, and line to line voltage bases of 13.8 kV, 138 kV and 27.6 kV (square root of 3 larger than the 1 example voltages). Also assume the generator is Y-connected so its line to line voltage is 13.8 kV. Convert to per unit as before. Note the system is exactly the same! 17

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3 Per Unit Example, cont'd Again, analysis is exactly the same! 18

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3 Per Unit Example, cont'd Differences appear when we convert back to actual values 19

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3 Per Unit Example 2 Assume a 3 load of 100+j50 MVA with V LL 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 = 103.3 + j76.0 MVA 20

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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 21

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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 22

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Y bus Example, cont’d 23 =

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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 ); however this will not be true in general when we consider phase shifting transformers 24

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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. 25

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Modeling Shunts in the Y bus 26

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Two Bus System Example 27

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Using the Y bus However, this requires that Y bus not be singular; note it will be singular if there are no shunt connections! 28

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Solving for Bus Currents 29

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Solving for Bus Voltages 30

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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 31

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Power Flow Analysis Classic paper for this lecture is W.F. Tinney and C.E. Hart, “Power Flow Solution by Newton’s Method,” IEEE Power App System, Nov 1967 Basic 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. 32

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Power Balance Equations 33

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Power Balance Equations, cont’d These equations can also be formulated using rectangular coordinates for the voltages: V i = e i + jf i 34 = = =

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Real Power Balance Equations 35

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Slack Bus 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. 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 36

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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 37

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Newton-Raphson Algorithm Most common technique for solving the power flow problem is to use the Newton-Raphson algorithm Key idea behind Newton-Raphson is to use sequential linearization More tractable for linear systems 38

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Newton-Raphson Power Flow 39

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Power Flow Variables 40

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N-R Power Flow Solution 41

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Power Flow Jacobian Matrix 42

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Power Flow Jacobian Matrix, cont’d 43

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