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ECE 530 – Analysis Techniques for Large-Scale Electrical Systems Lecture 3: Power Flow Analysis Prof. Hao Zhu Dept. of Electrical and Computer Engineering University of Illinois at Urbana-Champaign [email protected] 1 Announcements • • Homework 1 will be posted today. Due Sep 11 (Thur), in class 2 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, threephase 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 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 p circuit 4 Power System Component Models: Transformers • Transformer equivalent model 5 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: Si = Pi + jQi – constant impedance: Si = |V|2 / Zi 6 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 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 actual quantity quantity in per unit base value of quantity 8 Per Unit Conversion Procedure, 1f 1. Pick a 1f VA base for the entire system, SB 2. Pick a voltage base for each different voltage level, 3. 4. 5. VB. Voltage bases are related by transformer turns ratios. Voltages are line to neutral. Calculate the impedance base, ZB= (VB)2/SB Calculate the current base, IB = VB/ZB 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 Per Unit Solution Procedure 1. 2. 3. Convert to per unit (p.u.) (many problems are already in per unit) Solve Convert back to actual as necessary 10 Per Unit Example Solve for the current, load voltage and load power in the circuit shown below using per unit analysis with an SB of 100 MVA, and voltage bases of 8 kV, 80 kV and 16 kV. Original Circuit 11 Per Unit Example, cont’d Z BLeft 8kV 2 0.64 100 MVA Z BMiddle Z BRight 80kV 2 64 100 MVA 2 16kV 2.56 100 MVA Same circuit, with values expressed in per unit. 12 Per Unit Example, cont’d 1.00 I 0.22 30.8 p.u. (not amps) 3.91 j 2.327 VL 1.00 0.22 30.8 p.u. 2 VL SL 0.189 p.u. Z SG 1.00 0.2230.8 30.8 p.u. VL I L* 13 Per Unit Example, cont’d To convert back to actual values just multiply the per unit values by their per unit base VLActual 0.859 30.8 16 kV 13.7 30.8 kV SLActual 0.1890 100 MVA 18.90 MVA SGActual 0.2230.8 100 MVA 22.030.8 MVA I Middle B 100 MVA 1250 Amps 80 kV I Actual Middle 0.22 30.8 Amps 275 30.8 14 Three Phase Per Unit Procedure is very similar to 1f except we use a 3f VA base, and use line to line voltage bases 1. 2. 3. Pick a 3f VA base for the entire system, S B3f Pick a voltage base for each different voltage level, VB. Voltages are line to line. Calculate the impedance base ZB VB2, LL S B3f ( 3 VB , LN ) 2 3S 1Bf VB2, LN S 1Bf Exactly the same impedance bases as with single phase! 15 Three Phase Per Unit, cont'd 4. Calculate the current base, IB I3Bf 5. S B3f 3 S 1Bf S 1Bf I1Bf 3 VB , LL 3 3 VB , LN VB , LN Exactly the same current bases as with single phase! Convert actual values to per unit 16 Three Phase Per Unit Example Solve for the current, load voltage and load power in the previous circuit, assuming a 3f 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 1f 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 3f Per Unit Example, cont'd 1.00 I 0.22 30.8 p.u. (not amps) 3.91 j 2.327 VL 1.00 0.22 30.8 p.u. 2 VL SL 0.189 p.u. Z SG 1.00 0.2230.8 30.8 p.u. * VL I L Again, analysis is exactly the same! 18 3f Per Unit Example, cont'd Differences appear when we convert back to actual values VL 0.859 30.8 27.6 kV 23.8 30.8 kV Actual SL SGActual 0.1890 300 MVA 56.70 MVA Actual I Middle B 0.2230.8 300 MVA 66.030.8 MVA 300 MVA 1250 Amps (same current!) 3 138 kV I Actual Middle 0.22 30.8 Amps 275 30.8 19 3f Per Unit Example 2 •Assume 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 = 103.3 + j76.0 MVA 20 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 = Ybus V 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 21 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 22 Ybus Example, cont’d By KCL at bus 1 we have I1 = I G1 I D1 I1 I12 I13 V1 V2 V1 V3 ZA ZB I1 (V1 V2 )YA (V1 V3 )YB 1 (with Yj ) Zj (YA YB )V1 YA V2 YB V3 Similarly I 2 I 21 I 23 I 24 YA V1 (YA YC YD )V2 YC V3 YD V4 23 Ybus Example, cont’d We can get similar relationships for buses 3 and 4 The results can then be expressed in matrix form I Ybus V YA YB I1 YA YB I Y YA YC YD YC 2 A YC YB YC I 3 YB I 0 YD 0 4 0 V1 YD V2 0 V3 YD V4 For 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 24 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. 25 Modeling Shunts in the Ybus Ykc Since I ij (Vi V j )Yk Vi 2 Ykc Yii Yk 2 Rk jX k Rk jX k 1 1 Note Yk 2 Z k Rk jX k Rk jX k Rk X k2 Yiifrom other lines 26 Two Bus System Example I1 Yc (V1 V2 ) V1 Z 2 1 12 j16 0.03 j 0.04 I1 12 j15.9 12 j16 V1 I 12 j16 12 j15.9 V 2 2 27 Using the Ybus If the voltages are known then we can solve for the current injections: Ybus V I If the current injections are known then we can solve for the voltages: 1 Ybus I V Zbus I where Z bus is the bus impedance matrix However, this requires that Ybus not be singular; note it will be singular if there are no shunt connections! 28 Solving for Bus Currents For example, in previous case assume 1.0 V 0.8 j 0.2 Then 12 j15.9 12 j16 1.0 5.60 j 0.70 12 j16 12 j15.9 0.8 j 0.2 5.58 j 0.88 Therefore the power injected at bus 1 is S1 V1I1* 1.0 (5.60 j 0.70) 5.60 j 0.70 * S2 V2 I 2 (0.8 j 0.2) (5.58 j 0.88) 4.64 j 0.41 29 Solving for Bus Voltages For example, in previous case assume 5.0 I 4.8 Then 1 12 j15.9 12 j16 5.0 0.0738 j 0.902 12 j16 12 j15.9 4.8 0.0738 j1.098 Therefore the power injected is S1 V1I1* (0.0738 j 0.902) 5 0.37 j 4.51 S2 V2 I 2* (0.0738 j1.098) (4.8) 0.35 j 5.27 30 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 31 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 Power Balance Equations From KCL we know at each bus i in an n bus system the current injection, I i , must be equal to the current that flows into the network I i I Gi I Di n Iik k 1 Since I = Ybus V we also know I i I Gi I Di n YikVk k 1 The network power injection is then Si Vi I i* 33 Power Balance Equations, cont’d * n Si Vi I i* Vi YikVk Vi Yik*Vk* k 1 k 1 This is an equation with complex numbers. n Sometimes we would like an equivalent set of real power equations. These can be derived by defining Yik = Gik jBik Vi = Vi e ji Vi i ik = i k Recall e j cos j sin These equations can also be formulated using rectangular coordinates for the voltages: Vi = ei + jfi 34 Real Power Balance Equations n n k 1 k 1 Si Pi jQi Vi Yik*Vk* Vi Vk e jik (Gik jBik ) n Vi Vk k 1 (cosik j sin ik )(Gik jBik ) Resolving into the real and imaginary parts Pi Qi n Vi Vk (Gik cosik Bik sin ik ) PGi PDi k 1 n Vi Vk (Gik sin ik Bik cosik ) QGi QDi k 1 35 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 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 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 General form of problem: Find an x such that f ( xˆ ) 0 • More tractable for linear systems 38 Newton-Raphson Power Flow In the Newton-Raphson power flow we use Newton's method to determine the voltage magnitude and angle at each bus in the power system. We need to solve the power balance equations Pi Qi n Vi Vk (Gik cosik Bik sin ik ) PGi PDi k 1 n Vi Vk (Gik sin ik Bik cosik ) QGi QDi k 1 39 Power Flow Variables Assume the slack bus is the first bus (with a fixed voltage angle/magnitude). We then need to determine the voltage angle/magnitude at the other buses. 2 n x V 2 Vn P2 (x) PG 2 PD 2 Pn (x) PGn PDn f ( x) Q2 (x) QG 2 QD 2 Qn (x) QGn QDn 40 N-R Power Flow Solution The power flow is solved using the same procedure discussed with the general Newton-Raphson: Set v 0; make an initial guess of x, x( v ) While f (x( v ) ) Do x( v 1) x( v ) J (x( v ) ) 1f (x( v ) ) v v 1 End While 41 Power Flow Jacobian Matrix The most difficult part of the algorithm is determining and inverting the n by n Jacobian matrix, J (x) f1 (x) x 1 f 2 (x) J (x) x1 f (x) n x1 f1 (x) x2 f 2 (x) x2 f n (x) x2 f1 ( x) xn f 2 ( x) xn f n ( x) xn 42 Power Flow Jacobian Matrix, cont’d Jacobian elements are calculated by differentiating each function, fi ( x), with respect to each variable. For example, if fi (x) is the bus i real power equation fi ( x) fi ( x) i n Vi Vk (Gik cosik Bik sin ik ) PGi PDi k 1 n Vi Vk (Gik sin ik Bik cosik ) k 1 k i fi ( x) Vi V j (Gik sin ik Bik cosik ) j ( j i) 43