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Im[ ]
1 1
Guess 1.05 0
0 0 0.457
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1.045 0.83 1.050 0.83
1 0 0.535 1.049 0.93 1.050 0.93
2 0 0.545 1.050 0.96 1.050 0.96
j
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∠ − ° ∠ − °
+ ∠ − ° ∠ − °
+ ∠ − ° ∠ − ° 17](https://image.slidesharecdn.com/lecture11-160904123658/85/Lecture-11-17-320.jpg)
![Multi-variable Example, cont’d
1 2
1 2 1 2
1
1 1 2 1
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( ) =
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2 2 ( )
Matlab code: x1=x10; x2=x20;
f1=2*x1^2+x2^2-8;
f2=x1^2-x2^2+x1*x2-4;
J = [4*x1 2*x2; 2*x1+x2 x1-2*x2];
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x x x x x f
−
+ −
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x
x
[x1;x2]-inv(J)*[f1;f2].
38](https://image.slidesharecdn.com/lecture11-160904123658/85/Lecture-11-38-320.jpg)
Lecture 11
- 1. EE 369 POWER SYSTEM ANALYSIS Lecture 11 Power Flow Tom Overbye and Ross Baldick 1
- 2. Announcements • Start reading Chapter 6 for lectures 11 and 12. • Homework 8 is 3.1, 3.3, 3.4, 3.7, 3.8, 3.9, 3.10, 3.12, 3.13, 3.14, 3.16, 3.18; due 10/29. • Homework 9 is 3.20, 3.23, 3.25, 3.27, 3.28, 3.29, 3.35, 3.38, 3.39, 3.41, 3.44, 3.47; due 11/5. • Midterm 2, Thursday, November 12, covering up to and including material in HW9. 2
- 3. Wind Blade Failure Photo source: Peoria Journal Star Several years ago, a 140 foot, 6.5 ton blade broke off from a Suzlon Energy wind turbine. The wind turbine is located in Illinois. Suzlon Energy is one of the world’s largest wind turbine manufacturers; its shares fell 39% following the accident. No one was hurt and wind turbines failures are extremely rare events. (Vestas and Siemens turbines have also failed.) 3
- 4. Thermal Plants Can Fail As Well: Another Illinois Failure, Fall 2007 4
- 5. Springfield, Illinois City Water, Light and Power Explosion, Fall 2007 5
- 6. Gauss Two Bus Power Flow Example •A 100 MW, 50 MVAr load is connected to a generator through a line with z = 0.02 + j0.06 p.u. and line charging of 5 MVAr on each end (100 MVA base). •Also, there is a 25 MVAr capacitor at bus 2. •If the generator voltage is 1.0 p.u., what is V2? SLoad = 1.0 + j0.5 p.u. 6 j0.05 j0.05
- 7. Gauss Two Bus Example, cont’d 2 2 bus The unknown is the complex load voltage, . To determine we need to know the , which is a 2 2 matrix. The capacitors have susceptances specified by the reactive power at the rated voltage. Line V V × Y bus 11 22 1 1 series admittance = 5 15. 0.02 0.06 5 14.95 5 15 Hence . 5 15 5 14.70 ( Note: 15 0.05; 15 0.05 0.25). j Z j j j j j B B = = − + − − + = − + − = − + = − + + Y 7
- 8. Gauss Two Bus Example, cont’d 1 1 2 * 2 2 2 2* 22 1, 22 ( 1) 2 2 Note that =1.0 0 is specified, so we do not update . We only consider one entry of ( ), namely ( ). 1 S Equation to solve: ( ). 1 1 0.5 Update: 5 14.70 ( n k k k k V V h V h V V Y V h V Y V j V j V ν = ≠ + ∠ = − = ÷ − + = − ∑ ( ) (0) 2 ( ) ( ) 2 2 ( 5 15)(1.0 0) )* Guess 1.0 0 (this is known as a flat start) 0 1.000 0.000 3 0.9622 0.0556 1 0.9671 0.0568 4 0.9622 0.0556 2 0.9624 0.0553 v v j V v V v V j j j j j ν − − + ∠ ÷ = ∠ + − − − − 8
- 9. Gauss Two Bus Example, cont’d 2 * * 1 1 11 1 12 2 1 1 ˆFixed point: 0.9622 0.0556 0.9638 3.3 Once the voltages are known all other values can be determined, including the generator powers and the line flows. ˆ( ) 1.023 0.239 , V j S V Y V Y V j P jQ = − = ∠ − ° = + = − = − 1 1 2 2 In actual units 102.3 MW, 23.9 MVAr The capacitor is supplying 25 23.2 MVAr P Q V = = = 9
- 10. Slack Bus In previous example we specified S2 and V1 and then solved for S1 and V2. 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 the previous example, this was bus 1. 10
- 11. Gauss for Systems with Many Buses * ( 1) ( ) ( )* 1, ( ) ( ) ( ) 1 2 ( 1) With multiple bus systems we could calculate new values of the voltages as follows: S1 ( , ,..., ) But after we've determined , it is i i n v vi i ik kv ii k k i v v v i n v i V V Y V Y V h V V V V + = ≠ + = − ÷ ÷ = ∑ ( ) a better estimate of the voltage at bus than , so it makes sense to use this new value. Using the latest values is known as the Gauss-Seidel iteration. v ii V 11
- 12. Gauss-Seidel Iteration ( 1) ( ) ( ) ( ) 2 12 2 3 ( 1) ( 1) ( ) ( ) 3 13 2 3 ( 1) ( 1) ( 1) ( ) ( ) 4 14 2 3 4 ( 1) ( 1) ( 1 2 3 Immediately use the new voltage estimates: ( , , , , ) (bus 1 is slack), ( , , , , ) ( , , , , ) ( , , v v v v n v v v v n v v v v v n v v v n n V h V V V V V h V V V V V h V V V V V V h V V V + + + + + + + + = … = … = … = M 1) ( 1) ( ) 4, , ) Gauss-Seidel usually works better than the Gauss, and is actually easier to implement. Gauss-Seidel is used in practice instead of Gauss. v v nV V+ + … 12
- 13. Three Types of Power Flow Buses There are three main types of buses: – Load (PQ), at which P and Q are fixed; goal is to solve for unknown voltage magnitude and angle at the bus. – Slack at which the voltage magnitude and angle are fixed; iteration solves for unknown P and Q injections at the slack bus – Generator (PV) at which P and |V| are fixed; iteration solves for unknown voltage angle and Q injection at bus: special coding is needed to include PV buses in the Gauss-Seidel iteration. 13
- 14. Inclusion of PV Buses in G-S * * 1 ( ) ( )* ( ) 1 To solve for at a PV bus we must first make a guess of using the power flow equation: Hence Im is an estimate of the reactive power injectio k i i n i i ik k i i k n v v v i i ik k V Q S V Y V P jQ Q V Y V = = = = − = − ∑ ∑ ( ) ( ) n. For the Gauss iteration we use the known value of real power and the estimate of the reactive power: v v i i iS P jQ= + 14
- 15. Inclusion of PV Buses, cont'd ( 1) ( )* ( 1) ( ) ( )* 1, ( 1) ( 1) ( 1) ( 1) Tentatively solve for 1 In update, set . But since is specified, replace by . That is, set i v i v n v vi i ik kv ii k k i v i i v i i i i i V S V Y V Y V V V V V V V V ν ν + + = ≠ + + + + = − ÷ ÷ ∠ = ∠ = ∑% % % 15
- 16. Two Bus PV Example Bus 1 (slack bus) Bus 2 V1 = 1.0 V2 = 1.05 P2 = 0 MW z = 0 .0 2 + j 0 .0 6 Consider the same two bus system from the previous example, except the load is replaced by a generator 16 j0.05j0.05
- 17. Two Bus PV Example, cont'd ( ) ( )* ( ) 22 2 1 ( ) ( )* ( ) ( )* 21 221 2 2 2 ( )* ( )* ( 1) ( ) ( )2 2 2 212 1( )* ( )* 22 221, 22 2 (0) 2 ( ) ( 1) ( 1) 2 2 2 Im , Im[ ] 1 1 Guess 1.05 0 0 0 0.457 k n v v k k n k k k k v v v Q V Y V Y V V Y V V S S V Y V Y V Y YV V V v S V V j ν ν ν ν ν ν ν ν ν ν ν ν = + = ≠ + + = − = − + = − = − ÷ ÷ = ∠ ° + ∑ ∑% % 1.045 0.83 1.050 0.83 1 0 0.535 1.049 0.93 1.050 0.93 2 0 0.545 1.050 0.96 1.050 0.96 j j ∠ − ° ∠ − ° + ∠ − ° ∠ − ° + ∠ − ° ∠ − ° 17
- 18. Generator Reactive Power Limits The reactive power output of generators varies to maintain the terminal voltage; on a real generator this is done by the exciter. To maintain higher voltages requires more reactive power. Generators have reactive power limits, which are dependent upon the generator's MW output. These limits must be considered during the power flow solution. 18
- 19. Generator Reactive Limits, cont'd During power flow once a solution is obtained, need to check if the generator reactive power output is within its limits If the reactive power is outside of the limits, then fix Q at the max or min value, and re- solve treating the generator as a PQ bus – this is know as "type-switching" – also need to check if a PQ generator can again regulate Rule of thumb: to raise system voltage we need to supply more VArs. 19
- 20. Accelerated G-S Convergence ( 1) ( ) ( 1) ( ) ( ) ( ) ( Previously in the Gauss-Seidel method we were calculating each value as ( ) To accelerate convergence we can rewrite this as ( ) Now introduce "acceleration parameter" v v v v v v x x h x x x h x x x α + + = = + − 1) ( ) ( ) ( ) ( ( ) ) With = 1 this is identical to standard Gauss-Seidel. Larger values of may result in faster convergence. v v v v x h x xα α α + = + − 20
- 21. Accelerated Convergence, cont’d ( 1) ( ) ( ) ( ) Consider the previous example: 1 0 (1 ) Matlab code: alpha=1.2;x=x0;x=x+alpha*(1+sqrt(x)-x). Comparison of results with different values of 1 1.2 1.5 2 0 1 1 1 1 1 2 2.20 2.5 3 2 v v v v x x x x x xα α ν α α α α + − − = = + + − = = = = 2.4142 2.5399 2.6217 2.464 3 2.5554 2.6045 2.6179 2.675 4 2.5981 2.6157 2.6180 2.596 5 2.6118 2.6176 2.6180 2.626 21
- 22. Gauss-Seidel Advantages Each iteration is relatively fast (computational order is proportional to number of branches + number of buses in the system). Relatively easy to program. 22
- 23. Gauss-Seidel Disadvantages Tends to converge relatively slowly, although this can be improved with acceleration. Has tendency to fail to find solutions, particularly on large systems. Tends to diverge on cases with negative branch reactances (common with compensated lines) Need to program using complex numbers. 23
- 24. Newton-Raphson Algorithm The second major power flow solution method is the Newton-Raphson algorithm Key idea behind Newton-Raphson is to use sequential linearization General form of problem: Find an such that ( ) 0 x f x = 24
- 25. Newton-Raphson Method (scalar) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2( ) ( ) 2 1. Represent by a Taylor series about the current guess . Write for the deviation from : ( ) ( ) ( ) 1 ( ) 2 higher order terms. v v v v v v v f x x x df f x x f x x x dx d f x x dx ν ν ∆ + ∆ = + ∆ + + ∆ + 25
- 26. Newton-Raphson Method, cont’d ( ) ( ) ( ) ( ) ( ) ( ) 1 ( ) ( ) ( ) 2. Approximate by neglecting all terms except the first two ( ) ( ) ( ) 3. Set linear approximation equal to zero and solve for ( ) ( ) 4. Sol v v v v v v v f df f x x f x x x dx x df x x f x dx ν ν − + ∆ ≈ + ∆ ∆ ∆ = − ( 1) ( ) ( ) ve for a new estimate of solution: v v v x x x+ = + ∆ 26
- 27. Newton-Raphson Example 2 1 ( ) ( ) ( ) ( ) ( ) 2 ( ) ( 1) ( ) ( ) ( 1) ( ) ( ) 2 ( ) Use Newton-Raphson to solve ( ) 0, where: ( )= 2. The iterative update is: ( ) ( ) 1 (( ) 2) 2 1 (( ) 2). 2 v v v v v v v v v v v v v f x f x x df x x f x dx x x x x x x x x x x − + + = − ∆ = − ∆ = − − = + ∆ = − − 27
- 28. Newton-Raphson Example, cont’d ( 1) ( ) ( ) 2 ( ) (0) ( ) ( ) ( ) 3 3 6 1 (( ) 2) 2 Matlab code: x=x0; x = x-(1/(2*x))*(x^2-2). Guess 1. Iteratiting, we get: ( ) 0 1 1 0.5 1 1.5 0.25 0.08333 2 1.41667 6.953 10 2.454 10 3 1.41422 6.024 10 v v v v v v v x x x x x x f x xν + − − − = − − = ∆ − − × − × × 28
- 29. Sequential Linear Approximations Function is f(x) = x2 - 2. Solutions to f(x) = 0 are points where f(x) intersects x axis. At each iteration the N-R method uses a linear approximation to determine the next value for x 29
- 30. Newton-Raphson Comments • When close to the solution the error decreases quite quickly -- method has what is known as “quadratic” convergence: – number of correct significant figures roughly doubles at each iteration. • f(x(v) ) is known as the “mismatch,” which we would like to drive to zero. • Stopping criteria is when f(x(v) ) < ε 30
- 31. Newton-Raphson Comments • Results are dependent upon the initial guess. What if we had guessed x(0) = 0, or x(0) = -1? • A solution’s region of attraction (ROA) is the set of initial guesses that converge to the particular solution. • The ROA is often hard to determine. 31
- 32. Multi-Variable Newton-Raphson 1 1 2 2 Next we generalize to the case where is an - dimension vector, and ( ) is an -dimensional vector function: ( ) ( ) ( ) ( ) Again we seek a solution of ( ) 0. n n n n x f x f x f = = = x f x x x x f x x f x M M 32
- 33. Multi-Variable Case, cont’d i 1 1 1 1 1 2 1 2 1 1 2 1 2 The Taylor series expansion is written for each f ( ) ( ) ( ) ( ) ( ) ( ) higher order terms ( ) ( ) ( ) ( ) ( ) higher order terms n n n n n n n n n f f f x x f x x x x x x x f x x x f f f x x f x x x x x x x f x x x ∂ ∂ + ∆ = + ∆ + ∆ +… ∂ ∂ ∂ ∆ + ∂ ∂ ∂ + ∆ = + ∆ + ∆ +… ∂ ∂ ∂ ∆ + ∂ x M 33
- 34. Multi-Variable Case, cont’d 1 1 1 1 2 1 1 2 2 2 2 2 1 2 1 2 This can be written more compactly in matrix form ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) n n n n n n n f f f x x x f x f f f f x x x x f f f f x x x ∂ ∂ ∂ ∂ ∂ ∂ ∆ ∂ ∂ ∂ ∆ ∂ ∂ ∂= + ∂ ∂ ∂ ∂ ∂ ∂ x x x x x x xx f x +Δx x x x x L L M M O O M L higher order terms nx ∆ + M 34
- 35. Jacobian Matrix 1 1 1 1 2 2 2 2 1 2 1 2 The by matrix of partial derivatives is known as the Jacobian matrix, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) n n n n n n n n f f f x x x f f f x x x f f f x x x ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂= ∂ ∂ ∂ ∂ ∂ ∂ J x x x x x x x J x x x x L L M O O M L 35
- 36. Multi-Variable N-R Procedure Derivation of N-R method is similar to the scalar case ( ) ( ) ( ) higher order terms ( ) ( ) ( ) To seek solution to ( ) 0, set linear approximation equal to zero: 0 ( ) ( ) . + ∆ = + ∆ + + ∆ ≈ + ∆ + ∆ = = + ∆ ∆ = − f x x f x J x x f x x f x J x x f x x f x J x x x 1 ( 1) ( ) ( ) ( 1) ( ) ( ) 1 ( ) ( ) ( ) ( ) ( ) ( ) Iterate until ( ) v v v v v v v v ε − + + − = + ∆ = − < J x f x x x x x x J x f x f x 36
- 37. Multi-Variable Example 1 2 2 2 1 1 2 2 2 2 1 2 1 2 1 1 1 2 2 2 1 2 Solve for = such that ( ) 0 where ( ) 2 8 ( ) 4 First symbolically determine the Jacobian ( ) ( ) ( ) = ( ) ( ) x x f x x x f x x x x x f f x x x x f f x x x x = = + − = − + − ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ x f x J x 37
- 38. Multi-variable Example, cont’d 1 2 1 2 1 2 1 1 1 2 1 2 1 2 1 2 2 4 2 ( ) = 2 2 4 2 ( ) Then 2 2 ( ) Matlab code: x1=x10; x2=x20; f1=2*x1^2+x2^2-8; f2=x1^2-x2^2+x1*x2-4; J = [4*x1 2*x2; 2*x1+x2 x1-2*x2]; [x1;x2] = x x x x x x x x x f x x x x x f − + − ∆ = − ∆ + − J x x x [x1;x2]-inv(J)*[f1;f2]. 38
- 39. Multi-variable Example, cont’d (0) 1 (1) 1 (2) ( ) 1 Initial guess 1 1 4 2 5 2.1 1 3 1 3 1.3 2.1 8.40 2.60 2.51 1.8284 1.3 5.50 0.50 1.45 1.2122 At each iteration we check ( ) to see if itν − − = − = − = − − = − = − x x x f x (2) is 0.1556 below our specified tolerance : ( ) 0.0900 If = 0.2 then done. Otherwise continue iterating. ε ε = f x 39