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ECE 476 POWER SYSTEM ANALYSIS Lecture 10 Transformers, Load & Generator Models, YBus Professor Tom Overbye Department of Electrical and Computer Engineering Announcements Homework #4 4.34, 4.35, 5.14, 5.26 is due now Homework #5 is 3.12, 3.14, 3.19, 60 due Oct 2nd (Thursday) First exam is 10/9 in class; closed book, closed notes, one note sheet and calculators allowed Start reading Chapter 6 for lectures 11 and 12 1 In the News: Solar Energy On 9/23/08 US Senate passed a bill that would greatly expand the tax credits available for solar energy installations. The professor Chapman experience is available at http://www.patrickchapman.com/solar.htm. But there are environmental objections to largescale solar projects Picture source: NYTimes, 9/23/08 2 ECE PowerLunches ECE Department (through ECE Student Advisory Committee) sponsors “PowerLunches” to encourage undergraduates to go out to lunch with a professor of their choice – – The lunch, which takes place in the Illini Ballroom, is free for the students (no more than three) and the professor Details can be found at http://sac.ece.uiuc.edu/newpage/pwrlunch.php 3 Load Tap Changing Transformers LTC transformers have tap ratios that can be varied to regulate bus voltages The typical range of variation is 10% from the nominal values, usually in 33 discrete steps (0.0625% per step). Because tap changing is a mechanical process, LTC transformers usually have a 30 second deadband to avoid repeated changes. Unbalanced tap positions can cause "circulating vars" 4 Phase Shifting Transformers Phase shifting transformers are used to control the phase angle across the transformer Since power flow through the transformer depends upon phase angle, this allows the transformer to regulate the power flow through the transformer Phase shifters can be used to prevent inadvertent "loop flow" and to prevent line overloads. 5 Phase Shifting Transformer Picture Costs about $7 million, weighs about 1.2 million pounds 230 kV 800 MVA Phase Shifting Transformer During factory testing Source: Tom Ernst, Minnesota Power 6 ComED Control Center 7 ComED Phase Shifter Display 8 Autotransformers Autotransformers are transformers in which the primary and secondary windings are coupled magnetically and electrically. This results in lower cost, and smaller size and weight. The key disadvantage is loss of electrical isolation between the voltage levels. This can be an important safety consideration when a is large. For example in stepping down 7160/240 V we do not ever want 7160 on the low side! 9 Load Models 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 10 Generator Models 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 11 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 12 Linear versus Nonlinear Systems A function H is linear if H(a1m1 + a2m2) = a1H(m1) + a2H(m2) 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(x1+x2) = cx1 + c x2 Nonlinear Example: y = H(x) = c x2 y = c(x1+x2)2 ≠ (cx1)2 + (c x2)2 13 Linear Power System Elements Resistors, inductors, capacitors, independent voltage sources and current sources are linear circuit elements 1 V = R I V = j L I V = I j C Such systems may be analyzed by superposition 14 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 15 Nonlinear Systems May Have Multiple Solutions or No Solution Example 1: x2 - 2 = 0 has solutions x = 1.414… Example 2: x2 + 2 = 0 has no real solution f(x) = x2 - 2 two solutions where f(x) = 0 f(x) = x2 + 2 no solution f(x) = 0 16 Multiple Solution Example 3 The dc system shown below has two solutions: where the 18 watt load is a resistive load The equation we're solving is 2 9 volts I RLoad RLoad 18 watts 1 +R Load What is the One solution is R Load 2 maximum Other solution is R Load 0.5 PLoad? 2 17 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 18 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 19 Ybus Example, cont’d By KCL at bus 1 we have I1 IG1 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 20 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) 21 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. 22 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 23 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 24 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 25 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 26 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 27 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 28 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* 29 Power Balance Equations, cont’d * n * * Si Vi YikVk Vi YikVk k 1 k 1 This is an equation with complex numbers. n * Vi I i Sometimes we would like an equivalent set of real power equations. These can be derived by defining Yik Gik jBik j i Vi Vi e ik i k Vi i Recall e j cos j sin 30 Real Power Balance Equations n Si Pi jQi Vi Yik*Vk* k 1 n Vi Vk k 1 n j ik V V e (Gik jBik ) i k k 1 (cos ik j sin ik )(Gik jBik ) Resolving into the real and imaginary parts Pi Qi n Vi Vk (Gik cos ik Bik sinik ) PGi PDi k 1 n Vi Vk (Gik sin ik Bik cosik ) QGi QDi k 1 31 Power Flow Requires Iterative Solution In the power flow we assume we know Si and the Ybus . We would like to solve for the V's. The problem is the below equation has no closed form solution: * n Si Vi I i* Vi YikVk Vi Yik*Vk* k 1 k 1 Rather, we must pursue an iterative approach. n 32 Gauss Iteration There are a number of different iterative methods we can use. We'll consider two: Gauss and Newton. With the Gauss method we need to rewrite our equation in an implicit form: x = h(x) To iterate we first make an initial guess of x, x (0) , and then iteratively solve x (v +1) h( x ( v ) ) until we find a "fixed point", x, ˆ such that xˆ h(x). ˆ 33 Gauss Iteration Example Example: Solve x - x 1 0 x ( v 1) 1 x ( v ) Let k = 0 and arbitrarily guess x (0) 1 and solve k x(v) k x(v) 0 1 2 3 4 1 2 2.41421 2.55538 2.59805 5 6 7 8 9 2.61185 2.61612 2.61744 2.61785 2.61798 34 Stopping Criteria A key problem to address is when to stop the iteration. With the Guass iteration we stop when x ( v ) with x ( v ) x ( v 1) x ( v ) If x is a scalar this is clear, but if x is a vector we need to generalize the absolute value by using a norm x ( v ) j Two common norms are the Euclidean & infinity x 2 n 2 x i i 1 x max i x i 35 Gauss Power Flow We first need to put the equation in the correct form n Vi YikVk Vi Yik*Vk* k 1 k 1 Si Vi I i* * Si S*i * Vi * n Vi * Vi I i * n YikVk k 1 n YikVk k 1 YiiVi Vi * n YikVk k 1 n k 1,k i YikVk n 1 S*i Vi * YikVk Yii V k 1,k i i 36