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EE 369 POWER SYSTEM ANALYSIS Lecture 13 Newton-Raphson Power Flow Tom Overbye and Ross Baldick 1 Announcements Read Chapter 6 and (fourth edition) Chapter 11, concentrating on sections 11.4 and 11.5; (fifth edition) Chapter 12, concentrating on sections 12.4 and 12.5. Homework 10 is fourth edition: 6.27, 6.35, 6.36, 6.44, 6.46, 6.57, 11.24, 11.29; fifth edition 6.28, 6.36, 6.37, 6.45, 6.51, 6.57, 12.24, 12.28; due November 3. Turn in to in box next to my office door. 2 Using the Power Flow: Example 1 A SLA C K3 4 5 MVA A MVA 2 1 8 MW 1 .0 2 pu RA Y 3 4 5 sla ck 5 4 M var A A A 1 .0 2 pu MVA MVA SLA C K1 3 8 MVA T IM 3 4 5 1 .0 1 pu RA Y 1 3 8 1 .0 0 pu A MVA T IM 1 3 8 3 3 MW A MVA A A MVA 1 .0 3 pu Using 1 .0 2 pu 1 3 M var case MVA A A 1 6 .0 M var 1 8 MW 1 .0 2 pu RA Y 6 9 MVA MVA 5 M var 3 7 MW A 1 7 MW A 1 .0 2 pu T IM 6 9 P A I6 9 1 3 M var 1 .0 1 pu 3 M var from MVA MVA A 2 3 MW 1 .0 1 pu GRO SS6 9 A A MVA 7 M var MVA FERNA 6 9 MVA A 1 .0 1 pu WO LEN6 9 2 1 MW M O RO 1 3 8 Example A MVA MVA H ISKY 6 9 7 M var A A 4 .8 M var 1 2 MW MVA A MVA 5 M var 2 0 MW 1 .0 0 pu 1 .0 0 pu H A NNA H 6 9 8 M var A MVA P ET E6 9 5 8 MW DEM A R6 9 A 1 .0 0 pu A BO B1 3 8 MVA A MVA 6.13 MVA MVA 5 1 MW 4 0 M var 4 5 MW 1 5 M var A 1 .0 2 pu BO B6 9 1 2 M var 2 9 .0 M var MVA UIUC 6 9 0 .9 9 pu 1 4 .3 M var 1 .0 0 pu 1 4 0 MW 5 6 MW 1 2 .8 M var A 4 5 M var A MVA 1 3 M var LY NN1 3 8 A 0 MW MVA A 0 M var MVA A MVA 5 8 MW A 1 4 MW 0 .9 9 7 pu BLT 1 3 8 3 6 M var MVA 1 .0 0 pu MVA 4 M var 0 .9 9 pu A M A NDA 6 9 A A A SH IM KO 6 9 1 .0 2 pu H O M ER6 9 3 3 MW MVA MVA 7 .4 M var A MVA 1 0 M var 1 .0 1 pu A BLT 6 9 MVA A 1 .0 1 pu MVA 1 5 MW A MVA 1 5 MW 3 M var H A LE6 9 A 1 0 6 MW 5 M var MVA 1 .0 0 pu 8 M var A MVA 3 6 MW MVA A A A 1 .0 1 pu 6 0 MW MVA 1 0 M var 7 .2 M var MVA A A MVA 1 2 M var 1 .0 0 pu 1 .0 0 pu P A T T EN6 9 MVA 0 .0 M var A MVA 4 5 MW 1 4 MW RO GER6 9 MVA 1 .0 0 pu WEBER6 9 0 M var LA UF6 9 2 M var 1 .0 2 pu 2 3 MW 2 2 MW 0 MW A A 6 M var 1 4 MW A 2 0 MW 1 5 M var 0 M var MVA MVA 3 M var MVA 3 0 M var 1 .0 2 pu JO 1 3 8 JO 3 4 5 LA UF1 3 8 1 .0 2 pu SA V O Y 6 9 4 2 MW 1 .0 0 pu 2 M var 1 .0 1 pu BUC KY 1 3 8 A A MVA A 1 5 0 MW MVA 1 .0 1 pu SA V O Y 1 3 8 MVA A A 0 M var MVA MVA 1 5 0 MW A 0 M var MVA 1 .0 3 pu 1 .0 2 pu A MVA 3 Dishonest Newton-Raphson Since most of the time in the Newton-Raphson iteration is spent dealing with the Jacobian, one way to speed up the iterations is to only calculate (and factorize) the Jacobian occasionally: – known as the “Dishonest” Newton-Raphson or Shamanskii method, – an extreme example is to only calculate the Jacobian for the first iteration, which is called the chord method. Honest: x(v1) x(v ) - J (x(v ) )-1f (x( v ) ) Dishonest: x(v1) x(v ) - J (x(0) )-1f (x( v ) ) Both require f (x(v ) ) for a solution 4 Dishonest Newton-Raphson Example Use the Dishonest Newton-Raphson (chord method) to solve f ( x ) 0, where: f ( x) x - 2 2 1 x (v) df ( x (0) ) f ( x ( v ) ) dx x ( v ) 1 (( x ( v ) )2 - 2) 2 x (0) (v) 1 x ( v 1) x (( x ( v ) )2 - 2) 2 x (0) 5 Dishonest N-R Example, cont’d x ( v 1) x (v) 1 (( x ( v ) )2 - 2) (0) 2x Guess x (0) 1. Iteratively solving we get x ( v ) (honest) x ( v ) (dishonest) 0 1 1 We pay a price 1 1.5 1.5 in increased 2 1.41667 1.375 iterations, but with decreased 3 1.41422 1.429 computation 4 1.41422 1.408 per iteration 6 Two Bus Dishonest ROC Slide shows the region of convergence for different initial guesses for the 2 bus case using the Dishonest N-R Red region converges to the high voltage solution, while the yellow region converges to the low voltage solution 7 Honest N-R Region of Convergence Maximum of 15 iterations 8 Decoupled Power Flow The “completely” Dishonest Newton- Raphson (chord) is not usually used for power flow analysis. However several approximations of the Jacobian matrix are used that result in a similar approximation. One common method is the decoupled power flow. In this approach approximations are used to decouple the real and reactive power equations. 9 Decoupled Power Flow Formulation General form of the power flow problem P ( v ) P (v) θ V θ( v ) P (x( v ) ) f (x( v ) ) Q ( v ) Q (v) (v) V Q(x ) (v) θ V where P2 (x( v ) ) PD 2 PG 2 P(x ) (v) P (x(v ) ) P P n Dn Gn 10 Decoupling Approximation P ( v ) Q ( v ) Usually the off-diagonal matrices, and V θ are small. Therefore we approximate them as zero: P ( v ) 0 θ ( v ) P( x ( v ) ) θ f (x( v ) ) Q V (v) (v) Q( x ) (v) 0 V Then the problem can be decoupled ( v ) 1 ( v ) 1 P (v) Q θ (v) P( x ( v ) ), V Q( x ( v ) ) θ V 11 Off-diagonal Jacobian Terms Justification for Jacobian approximations: 1. Usually r x, therefore Gij Bij 2. Usually ij is small so sin ij 0 Therefore Pi Vi Gij cos ij Bij sin ij 0 Vj Qi Vi V j Gij cos ij Bij sin ij 0 θ j 12 Decoupled N-R Region of Convergence 13 Fast Decoupled Power Flow By further approximating the Jacobian we obtain a typically reasonable approximation that is independent of the voltage magnitudes/angles. This means the Jacobian need only be built and factorized once. This approach is known as the fast decoupled power flow (FDPF) FDPF uses the same mismatch equations as standard power flow so it should have same solution if it converges The FDPF is widely used, particularly when we only need an approximate solution. 14 FDPF Approximations The FDPF makes the following approximations: 1. Gij 0 2. Vi 1 (for some occurrences), 3. sin ij 0 cos ij 1 Then: θ( v ) B 1diag{| V |( v ) }1 P( x ( v ) ), B 1diag{| V |( v ) }1 Q( x ( v ) ) (v) V Where B is just the imaginary part of the Ybus G jB, except the slack bus row/column are omitted. Sometimes approximate diag{| V |( v ) } by identity. 15 FDPF Three Bus Example Use the FDPF to solve the following three bus system Line Z = j0.07 One Two 200 MW 100 MVR Line Z = j0.05 Line Z = j0.1 Three 1.000 pu 200 MW 100 MVR 34.3 14.3 20 Ybus j 14.3 24.3 10 20 10 30 16 FDPF Three Bus Example, cont’d 34.3 14.3 20 Ybus j 14.3 24.3 10 B 24.3 10 10 30 20 10 30 1 0.0477 0.0159 B 0.0159 0.0389 Iteratively solve, starting with an initial voltage guess (0) 2 (0) 0 V 2 1 V 1 3 0 3 2 (1) 0 0.0477 0.0159 2 0.1272 2 0.1091 3 0 0.0159 0.0389 17 FDPF Three Bus Example, cont’d (1) V 2 1 0.0477 0.0159 1 0.9364 V 1 0.0159 0.0389 1 0.9455 3 Pi ( x ) n P P Vk (Gik cosik Bik sin ik ) Di Gi Vi k 1 Vi 2 (2) 0.1272 0.0477 0.0159 0.151 0.1361 0.1091 0.0159 0.0389 0.107 0.1156 3 (2) V 2 0.924 V 3 0.936 0.1384 0.9224 Actual solution: θ V 0.1171 0.9338 18 FDPF Region of Convergence 19 “DC” Power Flow The “DC” power flow makes the most severe approximations: – completely ignore reactive power, assume all the voltages are always 1.0 per unit, ignore line conductance This makes the power flow a linear set of equations, which can be solved directly: θ B 1 P 20 DC Power Flow Example P 21 DC Power Flow 5 Bus Example One Five Four Three A A 360 MW 520 MW MVA MVA A 0 Mvar MVA slack 0 Mvar 1.000 pu 1.000 pu A A 1.000 pu 80 MW 0.000 Deg -4.125 Deg MVA MVA -1.997 Deg 0 Mvar 1.000 pu 0.524 Deg 1.000 pu Two -18.695 Deg 800 MW 0 Mvar Notice with the dc power flow all of the voltage magnitudes are 1 per unit. 22 Power System Control A major problem with power system operation is the limited capacity of the transmission system – lines/transformers have limits (usually thermal) – no direct way of controlling flow down a transmission line (e.g., there are no low cost valves to close to limit flow, except “on” and “off”) – open transmission system access associated with industry restructuring is stressing the system in new ways We need to indirectly control transmission line flow by changing the generator outputs. 23 Indirect Transmission Line Control What we would like to determine is how a change in generation at bus k affects the power flow on a line from bus i to bus j. The assumption is that the change in generation is absorbed by the slack bus 24 Power Flow Simulation - Before •One way to determine the impact of a generator change is to compare a before/after power flow. •For example below is a three bus case with an overload. 131.9 MW 124% One Two 200.0 MW 200 MW 68.1 MW 68.1 MW 71.0 MVR 100 MVR Z for all lines = j0.1 Three 1.000 pu 0 MW 64 MVR 25 Power Flow Simulation - After •Increasing the generation at bus 3 by 95 MW (and hence decreasing generation at the slack bus 1 by a corresponding amount), results in a 31.3 MW drop in the MW flow on the line from bus 1 to 2. 101.6 MW 100% One Two 105.0 MW 200 MW 3.4 MW 98.4 MW 64.3 MVR 100 MVR 92% Z for all lines = j0.1 Limit for all lines = 150 MVA 1.000 pu Three 95 MW 64 MVR 26 Analytic Calculation of Sensitivities Calculating control sensitivities by repeated power flow solutions is tedious and would require many power flow solutions. An alternative approach is to analytically calculate these values The power flow from bus i to bus j is Vi V j i j Pij sin( i j ) X ij X ij i j ij So Pij We just need to get X ij PGk 27 Analytic Sensitivities From the fast decoupled power flow we know θ B 1P (x) So to get the change in θ due to a change of generation at bus k , just set P( x) equal to all zeros except a minus one at position k . 0 P 1 Bus k 0 28 Three Bus Sensitivity Example For the previous three bus case with Zline j 0.1 20 10 10 Ybus j 10 20 10 B 20 10 10 20 10 10 20 Hence for a change of generation at bus 3 1 2 20 10 0 0.0333 1 0.0667 3 10 20 0.0667 0 Then P3 to 1 0.667 pu 0.1 P3 to 2 0.333 pu P 2 to 1 0.333 pu 29

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