6.061 Introduction to Electric Power Systems, Problem Set 3 Solutions

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.061/6.690 Introduction to Power Systems Problem Set 3 Solutions Problem 1: Voltage across the resistance is given by a voltage divider: VR = VS R R + jωL + 1 jωC The capacitance to maximize voltage across the resistance is the one that balances (cancels) inductor impedance. This is: C= 1 ≈ 265µF ωX The phasor diagram for voltages is, at resonance, shown in Figure 1. The magnitude of the receiving end voltage is plotted in Figure 2 Vx Vr Vc Figure 1: Voltage Phasors at Maximum Output Voltage Finally, to get the sending end real and reactive power, see that current OUT of the source is I s = Vs � 1 R + jX + 1 jωC + jωC � Real and reactive power are simply P + jQ = Vs Is∗ , and these are plotted in Figure 3. 1 Problem Set 3, Problem 1 120 115 Output Voltage 110 105 100 95 90 85 80 100 150 200 250 300 350 400 Capacitance, microfarads 450 500 550 Figure 2: Voltage vs. Compensating Capacitor Value Problem 2: Inductive reactance is X = 2π × 60 × .02 ≈ 7.54Ω, so receiving end voltage is Vr = Vs R R2 − jXR = Vs 2 ≈ 76.6 − j57.7V R + jX R + X2 A phasor diagram of this case is shown in Figure 4. With the capacitor in place, the ratio of input to output voltages is: Vr = Vs 1 R|| jωC 1 R|| jωC + jωL = Vs 1 1 − ω 2 LC + jωL R To make the magnitude of output voltage equal to input voltage, it is necessary that: � 1 − ω 2 LC �2 + � ωL R �2 =1 X R �2 =0 Or noting X = ωL and Y = ωC (XY )2 − 2XY + � This is easily solved by: 1 ± Y = X 2 �� 1 X �2 − 1 R2 Problem Set 3, Problem 1 1500 Watts and Vars 1000 500 0 Real Power Reactive Power −500 −1000 100 150 200 250 300 350 400 Capacitance, microfarads 450 500 550 Figure 3: Real and Reactive Power Vs. Capacitor Value VS = 120 VR = 76.6 −j 57.7 VX = 43.4 + j 57.7 Figure 4: Phasor Diagram: Uncompensated With X = 7.54Ω and R = 10Ω, this evaluates to Y = .0455S, so that C = .0455 377 ≈ 120µF . To construct the phasor diagram, start by assuming the output voltage is real (Vr = 120), Then the capacitance draws current Ic = .0455j × 120 ≈ j × 5.46A. Current through the inductance is Ix = 12 + j5.46, and the voltage across the inductance is Vx = −41 + j90.48. Source voltage is Vs = 78.8 + j90.48, which has magnitude of 120 V (all of this is RMS). The resulting phasor diagram is shown in Figure 5. Maximum voltage at the outupt is clearly achieved when ω 2 LC = 1, when C = 351.8µF . R ≈ 1.33 × 120 ≈ 159V. A plot of relative output vs. Maximum output voltage is Vr = Vs ωL input voltage is shown in Figure 6 Problem 3: Noting that the switch will have the same average voltage as the source since they are separated only by an inductance, and that the switch is shorted by itself a fraction D of the time, then < Vsw >= VC (1 − D) = VS , then the capacitor voltage must be: VC = 10 Vs = = 20V 1−D 1 − .5 3 VS = 78.8 + j 90.4 Vx = −41.2 +j 90.4 VR = 120 Figure 5: Phasor Diagram: compensated to equal voltage Chapter 2, Problem 11, vr vs. C 1.4 Voltage magnitude ratio 1.3 1.2 1.1 1 0.9 0.8 0.7 0 100 200 300 400 500 C, microfarads 600 700 800 Figure 6: Voltage transfer ratio vs. Capacitance di Then, if VC is about constant, so will be L dt . We can write expressions for the change in inductor current for both the ON and OFF times: LΔI = DT VS = (1 − D)T (VC − VS ) Note these expressions are consistent with our initial calculation for VC , Plugging in numbers: ΔI = .5 × 10−4 × 10 ≈ 0.05A .01 Getting capacitor ripple is slightly less obvious, but note that the capacitor is simply dis­ charging during the ON period, and if the capacitor voltage is about constant, the discharge 20 = 12 A So in this case, current will be Id = VRC = 40 ΔVC = VC DT = R C 4 1 2 × 12 × 10−4 1 ≈ V −6 50 × 10 2 A simulation of this is implemented by the MATLAB code appended. Voltage buildup is shown in Figure 7. A few cycles at the end of this are shown in Figure 8. PS3, Problem 3: Boost Converter Current, A 2 1.5 1 0.5 0 0 0.005 0.01 0.015 0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.02 0.025 0.03 0.035 30 Volts 20 10 0 t, sec Figure 7: Buildup Transient for Boost Converter Problem 4: With no compensating capacitor, the receiving end voltage is: Vr = Vs 8000 × 100 R = R + jX 100 + j12 The magnitude of this is just about 7940 volts and the angle is about 6.8 degrees, current lagging. An approximate phasor diagram for this is shown in Figure 9. To get the rest of the problem, see that the impedance of the receiving end capacitance and resistance is: Zr = R|| 1 R = jωCr 1 + jωRC The voltage transfer ratio is Zr R Vr = = Vs Zr + jX R + jωL − ω 2 RLC To make this have a magnitude of unity, we need � 2 1 − ω LC �2 L + ω R � �2 =1 Using shorthand notation Y = ωC, and X = ωL, we have (1 − Y X)2 + ( 5 X 2 ) =1 R PS3, Problem 3: Output Ripple Current, A 1.05 1 0.95 0.0294 0.0295 0.0296 0.0297 0.0298 0.0299 t, sec PS3, Problem 3: Output Ripple 0.0295 0.0296 0.03 0.0301 0.03 0.0301 20.6 Volts 20.4 20.2 20 19.8 19.6 0.0294 0.0297 0.0298 t, sec 0.0299 Figure 8: Ripple Current and Voltage for Boost Converter Vs Vx Vr Figure 9: Phasor Diagram of Voltages and carrying out the squares we find a solution to a quadratic: 1 Y = ± X � ( 1 2 1 ) − ( )2 X R The minus sign yields the smallest capacitance: 1 − Y = 12 � 1 1 − ≈ .000602S 2 12 1002 So Cr = Y /377 ≈ 1.6µF . The rest of the problem is worked in a straightforward way. Current through the line is: Il = Vs jXl + Zr Current through the sending end capacitance is Is = Vs jωCs and then P + jQ = Vs (Il + Is )∗ These are plotted for this case in Figures 9 and 11. 6 Problem Set 4, Problem 4 8060 8050 Receiving Voltage 8040 8030 8020 8010 8000 7990 7980 7970 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 Compensating Capacitance, microfarads 2.8 3 Figure 10: Receiving end voltage magnitude vs. compensating capacitor Problem 5 This thing is most conveniently viewed by using the phasor diagram shown in Fig­ ure 12. The voltage at the terminals of the current source is V = VS + jXI The phase angle betwen voltage V and current I is ψ. We know that and we know the system voltage VS , but we don’t know the terminal voltage V . To find that, we can invoke the ’law of cosines’, something we should have learned in elementary triginometry. This involves only the magnitudes of the various phasors: VS2 = V 2 + (XI)2 = 2V XI sin ψ This can be solved for the magnitude of terminal voltage V , using the straightforward solution for a second order expression and is: V = −XI sin ψ ± � (XI sin ψ)2 + VS2 − (XI)2 Simplifying and noting that only the positive sign gives a reasonable answer: V = � VS2 − (XI cos ψ)2 − XI sin ψ This is evaluated by the script appended, and for power factor of 0.9 (current lagging), unity and 0.9 (current leading), respectively, terminal voltage is: 7 5 Real Power 6.5 Problem Set 4, Problem 4 x 10 6.45 6.4 6.35 1 1.2 1.4 1.2 1.4 1.6 1.8 2 2.2 2.4 2.6 2.8 3 1.6 1.8 2 2.2 2.4 2.6 Compensating Capacitance, microfarads 2.8 3 4 Reactive Power 5 x 10 0 −5 −10 1 Figure 11: Real and reactive power vs. compensating capacitors V1 = 1.0395e+04 V2 = 9.9594e+03 V3 = 9.5235e+03 Then, for varying current, voltages are as shown in Figure!13. To get the same plot vs. real power, note that P = V I cos ψ and do a cross-plot. The result is shown in Figure 14. 8 I ψ V jXI ψ V S Figure 12: Phasor Diagram for Current Injection Problem Set 3, Problem 5 10400 10300 10200 Terminal Voltage 10100 10000 9900 9800 9700 9600 9500 0 50 100 150 200 Current Figure 13: Voltage vs. Current 9 250 4 1.06 Problem Set 3, Problem 5 x 10 1.04 Terminal Voltage 1.02 1 0.98 0.96 0.94 0 0.5 1 1.5 2 Real Power Figure 14: Voltage vs. Real Power 10 2.5 6 x 10 Scripts % 6.061 Problem Set 3, Problem 1 (p2.9 of text) % Parameters R=10; % resistance X=10; % reactance Vs=120; % voltage om = 120*pi; % frequency Cb = 1/(om*X) % capacitance to balance C = .5*Cb:.001*Cb:2*Cb; % capacitance Cd = 1e6 .* C; % in microfarads Z = R + j*X + 1 ./ (j * om .* C); % series impedance % current I = Vs ./ Z; Vr = R .* I; % voltage P = real(Vs .* conj(I)); Q = imag(Vs .* conj(I)); figure(1) clf plot(Cd, abs(Vr)); title(’Problem Set 3, Problem 1’) ylabel(’Output Voltage’) xlabel(’Capacitance, microfarads’) grid on figure(2) clf plot(Cd, P, Cd, Q, ’--’) title(’Problem Set 3, Problem 1’) ylabel(’Watts and Vars’) xlabel(’Capacitance, microfarads’) legend(’Real Power’, ’Reactive Power’) grid on -------------% Chapter 2, Problem 11 Vs = 120; L = .02; om = 2*pi*60; R = 10; Xl = om*L; Yc = 1/Xl - sqrt((1/Xl)^2 - 1/R^2); C = Yc/om; Zc = 1/(j*Yc); Zl = j*Xl; 11 Zo = 1/(1/R + j*Yc); vr = Zo/(Zl + Zo); % Construction of phasor diagram Vr = Vs * abs(vr); Ir = Vr/R; Ic = Vr/Zc; Ix = Ir + Ic; Vx = Zl*Ix; V_s = Vr + Vx; fprintf(’Chapter 2, Problem 11\n’) fprintf(’Value of Capacitance = %g microfarads\n’, 1e6*C) fprintf(’Phasors: Receiving End Voltage = %g\n’, Vr); fprintf(’Inductor Voltage = %g + j %g\n’, real(Vx), imag(Vx)) fprintf(’Sending Voltage = %g + j %g\n’, real(V_s), imag(V_s)) fprintf(’Check: |V_s| = %g\n’, abs(V_s)) fprintf(’Angle of V_s = %g radians = %g degrees\n’, angle(V_s), (180/pi)*angle(V_ s)) % capacitance for maximum voltage: Cm = 1/(om^2 * L); Cc = 0:Cm/500:2*Cm; vr = 1 ./ (1 - om^2 * L .* Cc + j*om*L/R); figure(1) plot(1e6 .* Cc, abs(vr)) title(’Chapter 2, Problem 11, vr vs. C’) ylabel(’Voltage magnitude ratio’) xlabel(’C, microfarads’) --------------% trivial boost converter model global vs L C R vs = 10; f = 10e3; alf = .5; L = 10e-3; C = 50e-6; R = 40; T=1/f; Dton = alf/f; Dtoff = (1-alf)/f; 12 dton = Dton/10; dtoff = Dtoff/10; v0 = 0; i0 = 0; t = []; i = []; v = []; for n = 0:300; [tc, S] = ode45(’upon’,[n*T n*T+Dton] , [i0 v0]’); t = [t tc’]; ic = S(:,1); vc = S(:,2); i = [i ic’]; v = [v vc’]; i0 = ic(length(tc)); v0 = vc(length(tc)); [tc, S] = ode45(’upoff’,[n*T+Dton (n+1)*T] , [i0 v0]); ic = S(:,1); vc = S(:,2); t = [t tc’]; i = [i ic’]; v = [v vc’]; i0 = ic(length(tc)); v0 = vc(length(tc)); end figure(1) clf subplot 211 plot(t, i) title(’PS3, Problem 3: Boost Converter’) ylabel(’Current, A’); grid on subplot 212 plot(t, v) ylabel(’Volts’); xlabel(’t, sec’); N = length(t); grid on figure(2) clf subplot 211 plot(t(N-500:N), i(N-500:N)) title(’PS3, Problem 3: Output Ripple’) ylabel(’Current, A’); 13 xlabel(’t, sec’); grid on subplot 212 plot(t(N-500:N), v(N-500:N)) title(’PS3, Problem 3: Output Ripple’) ylabel(’Volts’); xlabel(’t, sec’); grid on vr = max(v(N-500:N))-min(v(N-500:N)) ir = max(i(N-500:N))-min(i(N-500:N)) -----------function Sdot = up(t, S) global vs L C R il = S(1); vc = S(2); vdot = (1/C) * ( - vc/R); idot = (1/L) * (vs); Sdot = [idot vdot]’; ----------function Sdot = up(t, S) global vs L C R il = S(1); vc = S(2); vdot = (1/C) * (il - vc/R); idot = (1/L) * (vs - vc); Sdot = [idot vdot]’; ----------% 6.061 Problem Set 3, Problem 4 Vs = 8000; om = 120*pi; R = 100; X = 12; % first, with no capacitance at all Vr = Vs*R/(R + j*X); fprintf(’Uncompensated V = %g\n’, abs(Vr)) fprintf(’Angle = %g radians = %g degrees\n’,angle(Vr), (180/pi)*angle(Vr)) Yc = 1/X - sqrt(1/X^2 - 1/R^2); Cc = Yc/om; fprintf(’Cap to Compensate = %g microfarads\n’, 1e6*Cc) 14 Cd = 1:.001:3; C = 1e-6 .* Cd; % in microfarads % in farads Zr = R ./(1 + j*om*R .* C); % receiving end impedance Il = Vs ./ (Zr + j*X); Ic = Vs *j*om .*C; Vr = Zr .* Il; % line current % sending end comp current % receiving end voltage Ps = real(Vs .* conj(Il + Ic)); Qs = imag(Vs .* conj(Il + Ic)); figure(3) plot(Cd, abs(Vr)) title(’Problem Set 4, Problem 4’) ylabel(’Receiving Voltage’) xlabel(’Compensating Capacitance, microfarads’) figure(4) clf subplot 211 plot(Cd, Ps) title(’Problem Set 4, Problem 4’) ylabel(’Real Power’) subplot 212 plot(Cd, Qs) ylabel(’Reactive Power’) xlabel(’Compensating Capacitance, microfarads’) ------------% problem set 3, problem 5 X = 4; I = 250; Vs = 10e3; % part 1: phi = acos(.9); V1 = sqrt(Vs^2-(X*I*cos(phi))^2)+X*I*sin(phi) V2 = sqrt(Vs^2-(X*I)^2) V3 = sqrt(Vs^2-(X*I*cos(phi))^2)-X*I*sin(phi) I = 0:1:250; V1 = sqrt(Vs^2-(X*cos(phi) .* I) .^2)+X*sin(phi).* I; V2 = sqrt(Vs^2-(X .*I) .^2); 15 V3 = sqrt(Vs^2-(X*cos(phi) .* I) .^2)-X*sin(phi) .*I; figure(1) plot(I, V1, I, V2, I, V3) title(’Problem Set 3, Problem 5’) ylabel(’Terminal Voltage’) xlabel(’Current’) I = 0:1:300; V1 = sqrt(Vs^2-(X*cos(phi) .* I) .^2)+X*sin(phi).* I; V2 = sqrt(Vs^2-(X .*I) .^2); V3 = sqrt(Vs^2-(X*cos(phi) .* I) .^2)-X*sin(phi) .*I; P1 = V1 .* cos(phi) .* I; P2 = V2 .* I; P3 = V3 .* cos(phi) .* I; figure(2) plot(P1, V1, P2, V2, P3, V3); title(’Problem Set 3, Problem 5’) ylabel(’Terminal Voltage’) xlabel(’Real Power’) axis([0 2.5e6 9400 10600]) % problem set 3, problem 5 X = 4; I = 250; Vs = 10e3; % part 1: phi = acos(.9); V1 = sqrt(Vs^2+(X*I)^2 + 2*Vs*X*I*sin(phi)) V2 = sqrt(Vs^2+(X*I)^2) V3 = sqrt(Vs^2+(X*I)^2 - 2*Vs*X*I*sin(phi)) I = 0:1:250; V1 = sqrt(Vs^2+(X .*I) .^2 + 2*Vs*X*sin(phi) .* I); V2 = sqrt(Vs^2+(X .*I) .^2); V3 = sqrt(Vs^2+(X .*I) .^2 - 2*Vs*X*sin(phi) .* I); figure(1) plot(I, V1, I, V2, I, V3) title(’Problem Set 3, Problem 5’) ylabel(’Terminal Voltage’) xlabel(’Current’) 16 I = 0:1:300; V1 = sqrt(Vs^2+(X .*I) .^2 + 2*Vs*X*sin(phi) .* I); V2 = sqrt(Vs^2+(X .*I) .^2); V3 = sqrt(Vs^2+(X .*I) .^2 - 2*Vs*X*sin(phi) .* I); P1 = V1 .* cos(phi) .* I; P2 = V2 .* I; P3 = V3 .* cos(phi) .* I; figure(2) plot(P1, V1, P2, V2, P3, V3); title(’Problem Set 3, Problem 5’) ylabel(’Terminal Voltage’) xlabel(’Real Power’) axis([0 2.5e6 9400 10600]) 17