ECE 476 – Power System Analysis Fall 2014 Homework 8 Reading: Chapter 6. Problem 1. Problem 6.38 of GS&O, 5th Edition Solution. First, convert all values to per unit. P 150 MW = −1.5 p.u. Sbase 100 MVA Q 50 Mvar = = −0.5 p.u. =− Sbase 100 MVA ∆ymax 0.1 MVA = = 0.001 p.u. = Sbase 100 MVA Ppu = Qpu ∆ymax,pu =− The Y -bus for this system is 1 j0.1 1 − j0.1 Y = 1 − j0.1 1 j0.1 −j10 j10 = j10 −j10 The state vector of unknowns is x = [θ2 , V2 ]T . Next, we construct the power flow equations at bus 2. Since it is a PQ bus, we have two equations as follows: f P2 (x) = 0 = V2 V1 [G21 cos (θ2 − θ1 ) + B21 sin (θ2 − θ1 )] + V22 G22 − P2 = 10V2 sin θ2 + 1.5, f Q2 (x) = 0 = V2 V1 [G21 sin (θ2 − θ1 ) − B21 cos (θ2 − θ1 )] − V22 B22 − Q2 = −10V2 cos θ2 + 10V22 + 0.5. The Jacobian in this problem is " J= ∂f P2 ∂θ2 ∂f Q2 ∂θ2 ∂f P2 ∂V2 ∂f Q2 ∂V2 # = 10V2 cos θ2 10V2 sin θ2 10 sin θ2 . −10 cos θ2 + 20V2 With initial guess of V¯2 = 1∠0◦ , after 4 iterations, the final converged solution is V¯2 = 0.9334∠ − 9.2473◦ . % ECE 476 Fall 2013 % Homework 8, Problem 1, 6.38 of GS&O clear all close all % x = [t2 V2]’ P2 = -1.5; Q2 = -0.5; Xl = 1i*0.1; Y = [1/Xl -1/Xl; 1 -1/Xl 1/Xl]; B = imag(Y); x = [0 1]’; f = [B(2,1)*x(2)*sin(x(1)) - P2; -B(2,1)*x(2)*cos(x(1)) - B(2,2)*x(2)^2 - Q2]; eps = 1e-4; cnt = 0; while abs(f(1))>eps || abs(f(2))>eps J = [B(2,1)*x(2)*cos(x(1)) B(2,1)*sin(x(1)); B(2,1)*x(2)*sin(x(1)) -B(2,1)*cos(x(1))-2*B(2,2)*x(2)]; f = [B(2,1)*x(2)*sin(x(1)) - P2; -B(2,1)*x(2)*cos(x(1)) - B(2,2)*x(2)^2 - Q2]; x = x - J\f; cnt = cnt + 1; end Problem 2. Problem 6.39 of GS&O, 5th Edition Solution. Assume the system power base is 100 MVA. Then, P2 = 0.6 p.u., P3 = −0.8 p.u., and Q3 = −0.6 p.u. The Y -bus is given as −j7 Y = j2 j5 j2 −j6 j4 j5 j4 . −j9 In this problem, the state vector of unknowns is x = [θ2 , θ3 , V3 ]T . The power flow equations are f = [f P2 , f P3 , f Q3 ]T , where f P2 (x) = 0 = V2 V1 [G21 cos (θ2 − θ1 ) + B21 sin (θ2 − θ1 )] + V22 G22 + V2 V3 [G23 cos (θ2 − θ3 ) + B23 sin (θ2 − θ3 )] − P2 = 2 sin θ2 + 4V3 sin (θ2 − θ3 ) − 0.6, f P3 (x) = 0 = V3 V1 [G31 cos (θ3 − θ1 ) + B31 sin (θ3 − θ1 )] + V3 V2 [G32 cos (θ3 − θ2 ) + B32 sin (θ3 − θ2 )] + V32 G33 − P3 = 5V3 sin θ3 + 4V3 sin (θ3 − θ2 ) + 0.8, f Q3 (x) = 0 = V3 V1 [G31 sin (θ3 − θ1 ) − B31 cos (θ3 − θ1 )] + V3 V2 [G32 sin (θ3 − θ2 ) − B32 cos (θ3 − θ2 )] − V32 B33 − Q3 = −5V3 cos θ3 − 4V3 cos (θ3 − θ2 ) + 9V32 + 0.6. The corresponding Jacobian is 2 cos θ2 + 4V3 cos(θ2 − θ3 ) −4V3 cos(θ2 − θ3 ) 4 sin(θ2 − θ3 ) −4V3 cos (θ3 − θ2 ) 5V3 cos θ3 + 4V3 cos (θ3 − θ2 ) 5 sin θ3 + 4 sin (θ3 − θ2 ) J = −4V3 sin (θ3 − θ2 ) 5V3 sin θ3 + 4V3 sin (θ3 − θ2 ) −5 cos θ3 − 4 cos (θ3 − θ2 ) + 18V3 With a maximum power flow mismatch of 0.1 MVA and initial guess of V¯2 = 1∠30◦ and V¯3 = 1∠30◦ , after 5 iterations, the final converged solution is V¯2 = 1∠3.4680◦ and V¯3 = 0.9226∠ − 3.9898◦ . % ECE 476 Fall 2013 % Homework 8, Problem 2, 6.39 of GS&O clear all close all % x = [t2 t3 V3]’ P2 = 0.6; P3 = -0.8; Q3 = -0.6; Y = -1i * [7 -2 -5; -2 6 -4; -5 -4 9]; B = imag(Y); x = [pi/6 pi/6 1]’; f = [B(2,1)*sin(x(1)) + B(2,3)*x(3)*sin(x(1)-x(2)) - P2; B(3,1)*x(3)*sin(x(2)) + B(3,2)*x(3)*sin(x(2)-x(1)) - P3; -B(3,1)*x(3)*cos(x(2)) - B(3,2)*x(3)*cos(x(2)-x(1)) - B(3,3)*x(3)^2 - Q3]; eps = 1e-4; cnt = 0; while abs(f(1))>eps || abs(f(2))>eps || abs(f(3))>eps J = [B(2,1)*cos(x(1)) + B(2,3)*x(3)*cos(x(1)-x(2)) ... -B(2,3)*x(3)*cos(x(1)-x(2)) ... B(2,3)*sin(x(1)-x(2)); -B(3,2)*x(3)*cos(x(2)-x(1)) ... B(3,1)*x(3)*cos(x(2)) + B(3,2)*x(3)*cos(x(2)-x(1)) ... B(3,1)*sin(x(2)) + B(3,2)*sin(x(2)-x(1)); -B(3,2)*x(3)*sin(x(2)-x(1)) ... B(3,1)*x(3)*sin(x(2)) + B(3,2)*x(3)*sin(x(2)-x(1)) ... -B(3,1)*cos(x(2)) - B(3,2)*cos(x(2)-x(1)) - 2*B(3,3)*x(3)]; f = [B(2,1)*sin(x(1)) + B(2,3)*x(3)*sin(x(1)-x(2)) - P2; B(3,1)*x(3)*sin(x(2)) + B(3,2)*x(3)*sin(x(2)-x(1)) - P3; -B(3,1)*x(3)*cos(x(2)) - B(3,2)*x(3)*cos(x(2)-x(1)) - B(3,3)*x(3)^2 - Q3]; x = x - J\f; cnt = cnt + 1; end Problem 3. Problem 6.40 of GS&O, 5th Edition Solution. Using the same code as the previous problem with initial guess of V¯2 = 1∠0◦ and V¯3 = 0.25∠0◦ , after 6 iterations, the final converged solution is V¯2 = 1∠6.4388◦ and V¯3 = 0.1206∠ − 404.7242◦ . Problem 4. Problem 6.42 of GS&O, 5th Edition Solution. The Y -bus is given as −j10 j5 j5 −j10 j5 . Y = j5 j5 j2 −j7 In this problem, the state vector of unknowns is x = [θ2 , θ3 , V2 , V3 ]T . The power flow equations are f = [f P2 , f P3 , f Q2 , f Q3 ]T , where f P2 (x) = 0 = V2 V1 [G21 cos (θ2 − θ1 ) + B21 sin (θ2 − θ1 )] + V22 G22 + V2 V3 [G23 cos (θ2 − θ3 ) + B23 sin (θ2 − θ3 )] − P2 = 5V2 sin θ2 + 5V2 V3 sin (θ2 − θ3 ) + 1, f P3 (x) = 0 = V3 V1 [G31 cos (θ3 − θ1 ) + B31 sin (θ3 − θ1 )] + V3 V2 [G32 cos (θ3 − θ2 ) + B32 sin (θ3 − θ2 )] + V32 G33 − P3 = 5V3 sin θ3 + 2V3 V2 sin (θ3 − θ2 ) + 1.5, f Q2 (x) = 0 = V2 V1 [G21 sin (θ2 − θ1 ) − B21 cos (θ2 − θ1 )] − V22 B22 + V2 V3 [G23 sin (θ2 − θ3 ) − B23 cos (θ2 − θ3 )] − Q2 = −5V2 cos θ2 + 10V22 − 5V2 V3 cos (θ2 − θ3 ) + 0.5, f Q3 (x) = 0 = V3 V1 [G31 sin (θ3 − θ1 ) − B31 cos (θ3 − θ1 )] + V3 V2 [G32 sin (θ3 − θ2 ) − B32 cos (θ3 − θ2 )] − V32 B33 − Q3 = −5V3 cos θ3 − 2V3 V2 cos (θ3 − θ2 ) + 7V32 + 0.75. The corresponding Jacobian is J J = 11 J21 where J12 , J22 5V2 cos θ2 + 5V2 V3 cos (θ2 − θ3 ) −5V2 V3 cos (θ2 − θ3 ) , −2V3 V2 cos (θ3 − θ2 ) 5V3 cos θ3 + 2V3 V2 cos (θ3 − θ2 ) 5 sin θ2 + 5V3 sin (θ2 − θ3 ) 5V2 sin (θ2 − θ3 ) J12 = , 2V3 sin (θ3 − θ2 ) 5 sin θ3 + 2V2 sin (θ3 − θ2 ) 5V2 sin θ2 + 5V2 V3 sin (θ2 − θ3 ) −5V2 V3 sin (θ2 − θ3 ) = , −2V3 V2 sin (θ3 − θ2 ) 5V3 sin θ3 + 2V3 V2 sin (θ3 − θ2 ) J11 = J21 and J22 = −5 cos θ2 + 20V2 − 5V3 cos (θ2 − θ3 ) −5V2 cos (θ2 − θ3 ) . −2V3 cos (θ3 − θ2 ) −5 cos θ3 − 2V2 cos (θ3 − θ2 ) + 14V3 With a maximum power flow mismatch of 0.1 MVA and initial guess of V¯2 = 1∠0◦ and V¯3 = 1∠0◦ , after 5 iterations, the final converged solution is V¯2 = 0.7768∠ − 18.2603◦ and V¯3 = 0.7348∠ − 22.6208◦ . % ECE 476 Fall 2013 % Homework 8, Problem 4, 6.42 of GS&O clear all close all % x = [t2 t3 V2 V3]’ syms t2 t3 V2 V3 x = [t2; t3; V2; V3]; P2 Q2 P3 Q3 = = = = -1; -0.5; -1.5; -0.75; Y = [-1i*10 1i*5 1i*5; 1i*5 -1i*10 1i*5; 1i*5 1i*2 -1i*7]; B = imag(Y); f = [B(2,1)*V2*sin(t2) B(3,1)*V3*sin(t3) -B(2,1)*V2*cos(t2) -B(3,1)*V3*cos(t3) + + - B(2,3)*V2*V3*sin(t2-t3) - P2; B(3,2)*V3*V2*sin(t3-t2) - P3; B(2,2)*V2^2 - B(2,3)*V2*V3*cos(t2-t3) - Q2; B(3,2)*V3*V2*cos(t3-t2) - B(3,3)*V3^2 - Q3]; J = jacobian(f,x); xval = [0 0 1 1]’; fval = subs(f, x, xval); eps = 1e-4; cnt = 0; while abs(fval(1))>eps || abs(fval(2))>eps || abs(fval(3))>eps || abs(fval(4))>eps Jval = subs(J, x, xval); fval = subs(f, x, xval); xval = xval - Jval\fval; cnt = cnt + 1; end
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