Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Introduction to Power Systems

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1 Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science Introduction to Power Systems Problem Set 5 Solutions February 27, 2011 Problem 1: The inductance and phase velocity are: L = CZ0 2 = µHy/m 1 1 c = = m/S LC Before the load is connected, voltage is uniform and equal to the source and current is zero. So: + = = I + = I = 2 2 2Z 0 2Z0 When the switch is thrown, at the load end, becomes zero, since the load end is matched to the line. For the first interval, the situation is as shown in Figure 1. The voltage at the matched end is +. When the reverse going pulse gets to the sending end, + becomes equal to the source, since is equal to zero. That takes another period of time to propagate to the receiving end, at which point the voltabe is equal to sending end voltage. This is shown in Figure 2. Transit time is: And it takes 2T t for the transient to finish. L 100,000 T t = = 600µS u

2 I + I + + I I + I x=l + + I Figure 1: Pulses along the line I r Z 0 2 Z 0 2L u t Figure 2: Termination oltage 2

3 Problem 2: oltage at the receiving end and impedance at the sending end are: r Z s Z t = s Zt cos kl + jz 0 sin kl Z t cos kl + jz 0 sin kl = Z 0 jzt sin kl + Z 0 cos kl If the line is open at the receiving end, S r = cos kl = = jz t coth kl Z s The rest of the calculations are carried out by Matlab (source appended) and the results are: Line Open at Receiving End Receiving end voltage = Sending End Current = A Line loaded with 60 Ohms Receiving End oltage = Sending end Current = A Receiving end Power = W Sending End Real Power = W Sending End Reactive Power = AR 3

4 Problem 3: (Chapter 6, Problem 6 from text) 13,800 Turns ratio is N = If the primary voltages are: j 2π B = e 3 j 2 π C = e 3 Then, on the secondary side,with respect to Figure 3: bc = 2π j e 3 N ca = j e 2π 3 N ab = ca bc = N I b b i C c C i B B a I c I a Figure 3: Equivalent Circuit with Currents The actual phase voltages are shown in Figure 4 c bc ca b a ab Figure 4: Secondary Side oltages The phase voltages are: 4

5 a = e j π 6 3 N b = j e 5π 6 3N c = e j π 2 3N If the loads are unity power factor and of magnitude I 0, On the wye side of the transformer: This is shown in Figure 5 Ia = 0 6 Ib = 5π j I 0 e 6 j π I e j π I c = I + 0e 2 I A = 0 I b I 0 B = = e N N 5π j I 6 I a I 0 j 5π IC = = e 6 N N C I C I B A B Figure 5: Primary Side Currents As a test, we can see if the real and reactive powers are the same on the primary and secondary sides. On the primary side: P A = 0 I 0 j P e π B = 6 N 5

6 I 0 j P π C = e 6 N 3 I0 P A + P B + P C = N On the secondary side, power is: I 0 P = 3 = 3N 3 I0 N A few numbers are: Problem 6.6 Turns Ratio = Primary Power = j 0 Secondary Powr = j e-28 Finally, if the ground is not connected on the primary side, the thing becomes a singla phase circuit, with B B C = j 3 Then: BC = CA = j sqrt3 2N And this drives a 3 resistance of R eq = 2R. 6

7 Problem 4: Essentially the whole story is contained in the attached script. The first part is straightforward. The second part is done by first doing the delta-wye equivalent, finding the currents in the delta, re-assembling them into the wye and then transforming them across the transformer. The third part is a little tricky, but note that the voltages can be found by: a R a + R n R n R n I a b = R n R b + R n R n I b c R n R n R c + R n I c This is easily inverted in Matlab to get the currents. A summary of what is calculated is: Turns Ratio = Line-Neutral oltage = Phase C current = Phase C real power = Phase A real power = 8000 Low side power = High side power = High side reactive = >> open p4.m >> p6_6 Problem 6.6 Turns Ratio = Primary Power = j 0 Secondary Powr = j e-28 >> p6_7 Part A: Secondary Side Grounded Secondary Currents Ia = j 0 = angle 0 Ib = j -8 = angle Ic = j 8 = angle Primary Currents IA = j = angle IB = j = angle IC = e-16 + j = angle Check: Primary P = 6656 Q = 0 Secondary P = 6656 Q = 0 Part B: Secondary Side Ungrounded Rab = 130 Rbc = 78 Rca = 130 Secondary Currents Ia = j e-16 = angle e-16 Ib = j -8 = angle Ic = j 8 = angle Primary Currents IA = j = angle IB = j = angle

8 IC = e-16 + j = angle Check: Primary P = 6498 Q = 0 Secondary P = 6498 Q = 0 Part C: secondary grounded through a resistor Secondary Currents Ia = j e-16 = angle e-17 Ib = j -8 = angle Ic = j 8 = angle Primary Currents IA = j = angle IB = j = angle IC = e-16 + j = angle Check: Primary P = 6583 Q = 0 Secondary P = 6583 Q = 0 Note that Matlab sometimes gives an odd rendition of zero (as in something times 10 17, Problem 5 Problem 7-10 from the text R 1 = R 2 = 1 R 3 = R 4 = 1 So 4 GMD = m 8

9 Scripts % Problem Set 5 Problem 2 2/26/11 L = 1e5; % line length 100 km Z0 = 30; % characteristic impedance Cap = 2e-10; % capacitance om = 120*pi; % frequency s = 45000; % sending end voltage Ind = Cap * Z0^2; % line inductance u = 1/sqrt(Ind*Cap); % phase velocity k = om/u; % wavenumber %part 1: line open r = s / cos(k*l); % receiving end voltage Is = j*tan(k*l)*s/z0; % sending end voltage fprintf( Line Open at Receiving End\n ) fprintf( Receiving end voltage = %10.1f \n, r) fprintf( Sending End Current = %10.1f A\n, abs(is)) % part 2: resistive loading at 60 ohms Zr = 60; r = s * Zr/(Zr * cos(k*l) + j*z0 * sin(k*l)); Zs = Z0 * (Zr*cos(k*L) + j*z0*sin(k*l))/(j*zr*sin(k*l) + Z0 * cos(k*l)); Is = s/zs; Ss = s*conj(is); Ps = real(ss); Qs = imag(ss); Pr = abs(r)^2/zr; fprintf( Line loaded with 60 Ohms\n ) fprintf( Receiving End oltage = %10.1f \n, abs(r)) fprintf( Sending end Current = %10.1f A\n, abs(is)) fprintf( Receiving end Power = %10.1f W\n, Pr) fprintf( Sending End Real Power = %10.1f W\n, Ps) fprintf( Sending End Reactive Power = %10.1f AR\n, Qs) % chapter 6, problem 6 _0 = 13800/sqrt(3); N = _0/480; _A = _0; _B = _0 * exp(-j*2*pi/3); _C = _0 * exp(j*2*pi/3); 9

10 _s = 480/sqrt(3); _a = _s * exp(-j*pi/6); _b = _s * exp(-j*5*pi/6); _c = _s * exp(j*pi/2); I_a = 100 * exp(-j*pi/6); I_b = 100 * exp(-j*5*pi/6); I_c = 100 * exp(j*pi/2); S_s = _a * conj(i_a) + _b * conj(i_b) + _c * conj(i_c); I_P = 100/N; I_B = I_P * exp(-j*5*pi/6); I_C = I_P * exp(j*5*pi/6); S_P = _B * conj(i_b) + _C * conj(i_c); fprintf( Problem 6.6\n ) fprintf( Turns Ratio = %g\n, N) fprintf( Primary Power = %g + j %g\n, real(s_p), imag(s_p)) fprintf( Secondary Powr = %g + j %g\n, real(s_s), imag(s_s)) % Problem Set 5, Problem 4 (6.7 from Text) % This is mostly book-keeping p = 4160; s = 480; sn = s/sqrt(3); Ra = 50; Rb = 30; Rc = 30; N = p/sn; a = exp(j*2*pi/3); % primary side, line-line % secondary side, line-line % secondary side, line-neutral % three phase resistors % turns ratio: delta-wye % 120 degree rotation % Preliminary: Primary Side pn = p/sqrt(3); A = pn * exp(-j*pi/6); % connection rotates +30deg B = A*a^2; C = A*a; % part a: secondary side grounded a = sn; b = sn*a^2; c = sn*a; % set real to the secondary side 10

11 Ia = a/ra; Ib = b/rb; Ic = c/rc; SS = a*conj(ia)+b*conj(ib)+c*conj(ic); % translate to primary side: IA = (Ia-Ic)/N; IB = (Ib-Ia)/N; IC = (Ic-Ib)/N; SA = A*conj(IA); SB = B*conj(IB); SC = C*conj(IC); SP = SA + SB + SC; % output fprintf( Part A: Secondary Side Grounded\n ) fprintf( Secondary Currents\n ) fprintf( Ia = %g + j %g = %g angle %g\n, real(ia), imag(ia), abs(ia), angle(ia)) fprintf( Ib = %g + j %g = %g angle %g\n, real(ib), imag(ib), abs(ib), angle(ib)) fprintf( Ic = %g + j %g = %g angle %g\n, real(ic), imag(ic), abs(ic), angle(ic)) fprintf( Primary Currents\n ) fprintf( IA = %g + j %g = %g angle %g\n, real(ia), imag(ia), abs(ia), angle(ia)) fprintf( IB = %g + j %g = %g angle %g\n, real(ib), imag(ib), abs(ib), angle(ib)) fprintf( IC = %g + j %g = %g angle %g\n, real(ic), imag(ic), abs(ic), angle(ic)) fprintf( Check: Primary P = %10.0f Q = %10.0f Secondary P = %10.0f Q = %10.0f\n, real(sp), ima % Part B: secondary side ungrounded Rab = (Ra*Rb + Ra*Rc + Rb*Rc)/Rc; Rbc = (Ra*Rb + Ra*Rc + Rb*Rc)/Ra; Rca = (Ra*Rb + Ra*Rc + Rb*Rc)/Rb; % find equivalent delta fprintf( Part B: Secondary Side Ungrounded\n ) fprintf( Rab = %g Rbc = %g Rca = %g\n, Rab, Rbc, Rca) Iab = (a - b)/rab; Ibc = (b - c)/rbc; Ica = (c - a)/rca; Ia = Iab - Ica; Ib = Ibc - Iab; % currents in the delta % now get currents in the wye 11

12 Ic = Ica - Ibc; % repeat the rest of part a: it is now the same SS = a*conj(ia)+b*conj(ib)+c*conj(ic); % translate to primary side: IA = (Ia-Ic)/N; IB = (Ib-Ia)/N; IC = (Ic-Ib)/N; SA = A*conj(IA); SB = B*conj(IB); SC = C*conj(IC); SP = SA + SB + SC; fprintf( Secondary Currents\n ) fprintf( Ia = %g + j %g = %g angle %g\n, real(ia), imag(ia), abs(ia), angle(ia)) fprintf( Ib = %g + j %g = %g angle %g\n, real(ib), imag(ib), abs(ib), angle(ib)) fprintf( Ic = %g + j %g = %g angle %g\n, real(ic), imag(ic), abs(ic), angle(ic)) fprintf( Primary Currents\n ) fprintf( IA = %g + j %g = %g angle %g\n, real(ia), imag(ia), abs(ia), angle(ia)) fprintf( IB = %g + j %g = %g angle %g\n, real(ib), imag(ib), abs(ib), angle(ib)) fprintf( IC = %g + j %g = %g angle %g\n, real(ic), imag(ic), abs(ic), angle(ic)) fprintf( Check: Primary P = %10.0f Q = %10.0f Secondary P = %10.0f Q = %10.0f\n, real(sp), ima % Part C: Secondary grounded through a resistor Rn = 10; RM = [Ra+Rn Rn Rn;Rn Rb+Rn Rn;Rn Rn Rc+Rn]; v = [a; b; c;]; Is = RM\v; % matrix of resistances % vector of voltages % these should be the currents Ia = Is(1); Ib = Is(2); Ic = Is(3); % repeat the rest of part a: it is now the same SS = a*conj(ia)+b*conj(ib)+c*conj(ic); % translate to primary side: IA = (Ia-Ic)/N; 12

13 IB = (Ib-Ia)/N; IC = (Ic-Ib)/N; SA = A*conj(IA); SB = B*conj(IB); SC = C*conj(IC); SP = SA + SB + SC; fprintf( Part C: secondary grounded through a resistor\n ) fprintf( Secondary Currents\n ) fprintf( Ia = %g + j %g = %g angle %g\n, real(ia), imag(ia), abs(ia), angle(ia)) fprintf( Ib = %g + j %g = %g angle %g\n, real(ib), imag(ib), abs(ib), angle(ib)) fprintf( Ic = %g + j %g = %g angle %g\n, real(ic), imag(ic), abs(ic), angle(ic)) fprintf( Primary Currents\n ) fprintf( IA = %g + j %g = %g angle %g\n, real(ia), imag(ia), abs(ia), angle(ia)) fprintf( IB = %g + j %g = %g angle %g\n, real(ib), imag(ib), abs(ib), angle(ib)) fprintf( IC = %g + j %g = %g angle %g\n, real(ic), imag(ic), abs(ic), angle(ic)) fprintf( Check: Primary P = %10.0f Q = %10.0f Secondary P = %10.0f Q = %10.0f\n, real(sp), ima 13

14 MIT OpenCourseWare / Introduction to Electric Power Systems Spring 2011 For information about citing these materials or our Terms of Use, visit:

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.061/6.690 Introduction to Power Systems

Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science.01/.90 Introduction to Power Systems Problem Set 4 Solutions February 28, 2007 Problem 1: Part 1: Chapter,