6.061 / Introduction to Electric Power Systems

Transcription

1 MIT OpenCourseWare / 6.69 Introduction to Electric Power Systems Spring 27 For information about citing these materials or our Terms of Use, visit:

2 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.61 Introduction to Power Systems Problem Set 5 Solutions March 1, 27 Problem 1: The current source will launch two pulses along the line, one going to the right as v + and one to the left as v. These will have associated currents and as the point of injection, v(t) = v + + v i(t) = I (t) = i + i = v + V + From symmetry we can see that the two pulses that get launched must be of the same magnitude, so that: 1 1 v + = v = I = 2,A 25Ω = 2,5,V = 2.5MV 2 2 Time of flight is T = m = 1 ms = 5µS. 1 8 m/s 2 At the right-hand end, when the forward going pulse reaches the short, a second, reverse direction pulse is generated, and there because of the short, v + + v =. The current in the pulse is doubled, so that at peak it has value of 2, A. It takes 2 T = µs for this pulse to get to the left hand end. Since the left-hand end is terminated in impedance =, both pulses appear across the resistor but no reflections are generated. The results are shown in graphical form in Figure 1 Voltage at Termination esistor 2.5 MV 2 µs 5 µ S µ S Current at right hand short 2 A t t Figure 1: Transmission Line Example 1

3 Problem 2: Once corrected, the notes give us the pertinent formula: the ratio of receiving end to sending end voltage is: V r V s = cos(kl) + j sin(kl) 1 Since the wavelength is λ = 8 6 = m = 5km, Here, kl = 2π 5 To find out what happens at the sending end, note that, at x =, voltage and current are: V s = V + + V 1 I s = (V + V ) Since the ratio of reflected to incident wave voltage is: V V + = e 2jkl then, with a little algebraic pain we may conclude that: ( ) V s V = ( ) ( ) e j2ll ( ) V s 1 e j2kl V = ( ) ( ) e j2ll Then, with a little more algebra the sending end current can be found to be: I s = V s j sin kl + cos kl cos kl + j sinkl The calculations from here are carried out in MATLAB. Here is the session. More results show up later >> p2 / Vr Is Open

4 For the final two parts of this problem, note that the derivation done in the notes can be generalized to a termination with a complex impedance. In this case the complex amplitudes of receiving end and sending end voltages are related by: V = V r s cos(kl) + j sin(kl) For the first part, we note that the load is drawing surge impedance load when =. As a practical matter we can use load admittance G = 1/ as a parameter, calculating both load power and voltage as a function of that parameter and doing a cross-plot. The result is shown in Figure 2 x Problem 5.2, eal Loading Volts, MS eal Load, Watts x 1 8 Figure 2: Voltage vs. power ow for the second part, we may approximate things by assuming a real part of the terminating admittance suitable for the power levels involved and then note that reactive power supplied by the termination is: Q s = V 2 Im {Y s } A cross-plot of receiving end voltage against Q s is shown in Figure. The script that implements these calculations is in the appendix. Problem : Voltage across the resistor is 48V

5 x Problem 5.2, eactive Compensation Volts, MS % SIL eal Power % SIL eal Power 12 % SIL eal Power eactive Comp, VAS x 1 8 Figure : Voltage vs. reactive injection So that power is P = 48 = 48kW. The phase voltage on the secondary (wye) side is V, so that the turns ratio of the transformer is = frac Voltage across the load resistor is (see Figure 4) V = 48e j 2 π bc Current on the secondary side of the transformer is in phase with this voltage and flows in phases B and C.Power in each phase is: P B = P C = 277 cos( ) = 24kW The phase current is reflected across to the primary winding through the turns ratio: π = I ba e j 2 I e j 2 cb I ac = π = 4

6 Terminal current is then: I a = I ac I ba = I b = I ba I cb = 2 I c = I cb I ac = e j 2 π e j 2 π e j 2 π ote that = 6.598A Primary side power is, noting that line-neutral voltage is V, P a = cos(6 ) 8,W P b = ,W P c = cos(6 ) 8,W Total primary power is 8, + 2, + 8, = 48,W. eactive power is Q a = sin( 6 ) 1857V A Q b = Q c = sin(6 ) 1857V A Problem 4: For 6.69: This one is actually even easier, at least to start. Since the voltage is 48 V and current is A, power is P = 48 = 48kW. The turns ratio is = Voltage across the resistor is in phase with primary side phase C: V cb = 48e j 2 π ote that since the primary is ungrounded, i A +i B +i C =. And since there is no connection to the Phase A corner, i A = i B, leaving us with i A = i B = i C 2. Then: 2 1 i C = 1.196A 5.52 Understanding that this current is in phase with voltage, we can now write down the three phase currents: 5

7 V c V BC V C I C V a V b V B I B V A I a = Ic I b Figure 4: Currents for Problem I 6.598e j 2 a I b 6.598e j 2 I 1.196e j 2 π c π π Then: P A Q A P B Q B P C = cos( 6 ) 8kW = sin( 6 ) 1.856kV A = cos(6 ) 8kW = sin(6 ) 1.856kV A = kW Q C = Scripts Finally, nothing happens when the ground is restored. To see this, view the equivalent delta side of the transformer as is shown in Figure 5. Since the combination of Phase A and Phase B voltage is equal to the Phase C voltage, the thevenin equivalent, including leakage impedance, has the same back voltage but twice the series impedance. That combination will source half the current of Phase C, so there is really no difference between the grounded and ungrounded situations here. 6

8 Vc + X X X 2X V V V V a + b c c X Figure 5: Delta Equivalent % Problem Set 5, Problem 2 C=e8; % speed of light L=e; % line length = 25; % characteristic impedance Vs = 5e; % line voltage f = 6; lambda = C/f; k = 2*pi/lambda; % part : easy part Vro = Vs /cos(k*l); Iso = (Vs/) * tan(k*l); rr = [1/.8 1 1/1.2]; Vrr= Vs.* rr./ (rr.* cos(k*l) + j*sin(k*l)); IAs = (Vs/).* (-rr.* j*sin(k*l) + cos(k*l))./ (rr.* cos(k*l) -j*sin(k*l)); fprintf( / Vr Is \n ); fprintf( Open %1.1f %1.1f \n,abs(vro),iso); for i = 1:length(rr) fprintf( %1.4f %1.1f %1.1f \n,rr(i),abs(vrr(i)),abs(ias(i))) end % part 1: eal loading only rf = 1.25; % fudge factor Gr = :.1:rf; Gi = ; r = 1./ (Gr + j *Gi); Vr = r./ (r.* cos(k*l) + j*sin(k*l)); V_r = Vs.* abs(vr); P_r = V_r.^2.* (Gr/); figure(1) plot(p_r, V_r) title( Problem 5.2, eal Loading ); ylabel( Volts, MS ); xlabel( eal Load, Watts ) grid on 7

9 %part 2: eactive Loading Gr =.8; Gi = -.:.1:.2; r = 1./ (Gr + j *Gi); Vr = r./ (r.* cos(k*l) + j*sin(k*l)); V_r1 = Vs.* abs(vr); Q_r1 = V_r1.^2.* (Gi/); Gr = 1.; Gi = -.2:.1:.2; r = 1./ (Gr + j *Gi); Vr = r./ (r.* cos(k*l) + j*sin(k*l)); V_r2 = Vs.* abs(vr); Q_r2 = V_r2.^2.* (Gi/); Gr = 1.2; Gi = -.2:.1:.; r = 1./ (Gr + j *Gi); Vr = r./ (r.* cos(k*l) + j*sin(k*l)); V_r = Vs.* abs(vr); Q_r = V_r.^2.* (Gi/); figure(2) plot(q_r1, V_r1, Q_r2, V_r2, Q_r, V_r) title( Problem 5.2, eactive Compensation ); ylabel( Volts, MS ); xlabel( eactive Comp, VAS ) legend( 8 % SIL eal Power, % SIL eal Power, 12 % SIL eal Power ) grid on 8