CONTENTS 1 THE POWER SYSTEM: AN OVERVIEW 1 2 BASIC PRINCIPLES 5 3 GENERATOR AND TRANSFORMER MODELS; THE PER-UNIT SYSTEM 25

Solutions Manual Hadi Saadat Professor of Electrical Engineering Milwaukee School of Engineering Milwaukee, Wisconsin McGraw-Hill, Inc

CONTENTS 1 THE POWER SYSTEM: AN OVERVIEW 1 2 BASIC PRINCIPLES 5 3 GENERATOR AND TRANSFORMER MODELS; THE PER-UNIT SYSTEM 25 4 TRANSMISSION LINE PARAMETERS 52 5 LINE MODEL AND PERFORMANCE 68 6 POWER FLOW ANALYSIS 107 7 OPTIMAL DISPATCH OF GENERATION 147 8 SYNCHRONOUS MACHINE TRANSIENT ANALYSIS 170 9 BALANCED FAULT 181 10 SYMMETRICAL COMPONENTS AND UNBALANCED FAULT 208 11 STABILITY 244 12 POWER SYSTEM CONTROL 263 i

CHAPTER 1 PROBLEMS 11 The demand estimation is the starting point for planning the future electric power supply The consistency of demand growth over the years has led to numerous attempts to fit mathematical curves to this trend One of the simplest curves is P = P 0 e a(t t 0) where a is the average per unit growth rate, P is the demand in year t, and P 0 is the given demand at year t 0 Assume the peak power demand in the United States in 1984 is 480 GW with an average growth rate of 34 percent Using MATLAB, plot the predicated peak demand in GW from 1984 to 1999 Estimate the peak power demand for the year 1999 We use the following commands to plot the demand growth t0 = 84; P0 = 480; a =034; t =(84:1:99) ; P =P0*exp(a*(t-t0)); disp( Predicted Peak Demand - GW ) disp([t, P]) plot(t, P), grid xlabel( Year ), ylabel( Peak power demand GW ) P99 =P0*exp(a*(99 - t0)) The result is 1

2 CONTENTS Predicted Peak Demand - GW 840000 4800000 850000 4966006 860000 5137753 870000 5315441 880000 5499273 890000 5689463 900000 5886231 910000 6089804 920000 6300418 930000 6518315 940000 6743740 950000 6976978 960000 7218274 970000 7467916 980000 7726190 990000 7993398 P99 = 7993398 The plot of the predicated demand is shown n Figure 1 Peak Power Demand GW 800 750 700 650 600 550 500 450 84 86 88 90 92 94 96 98 100 Year FIGURE 1 Peak Power Demand for Problem 11 12 In a certain country, the energy consumption is expected to double in 10 years

CONTENTS 3 Assuming a simple exponential growth given by calculate the growth rate a P = P 0 e at Solving for a, we have a = 0693 10 2P 0 = P 0 e 10a ln 2 = 10a = 00693 = 693% 13 The annual load of a substation is given in the following table During each month, the power is assumed constant at an average value Using MATLAB and the barcycle function, obtain a plot of the annual load curve Write the necessary statements to find the average load and the annual load factor Annual System Load Interval Month Load MW January 8 February 6 March 4 April 2 May 6 June 12 July 16 August 14 September 10 October 4 November 6 December 8 The following commands data = [ 0 1 8 1 2 6 2 3 4 3 4 2 4 5 6 5 6 12

4 CONTENTS 6 7 16 7 8 14 8 9 10 9 10 4 10 11 6 11 12 8]; P = data(:,3); % Column array of load Dt = data(:, 2) - data(:,1); % Column array of demand interval W = P *Dt; % Total energy, area under the curve Pavg = W/sum(Dt) % Average load Peak = max(p) % Peak load LF = Pavg/Peak*100 % Percent load factor barcycle(data) % Plots the load cycle xlabel( time, month ), ylabel( P, MW ), grid result in Pavg = Peak = LF = 8 16 50 P MW 16 14 12 10 8 6 4 2 0 0 2 4 6 8 10 12 time, month FIGURE 2 Monthly load cycle for Problem 13

CHAPTER 2 PROBLEMS 21 Modify the program in Example 21 such that the following quantities can be entered by the user: The peak amplitude V m, and the phase angle θ v of the sinusoidal supply v(t) = V m cos(ωt + θ v ) The impedance magnitude Z, and its phase angle γ of the load The program should produce plots for i(t), v(t), p(t), p r (t) and p x (t), similar to Example 21 Run the program for V m = 100 V, θ v = 0 and the following loads: An inductive load, Z = 125 60 Ω A capacitive load, Z = 20 30 Ω A resistive load, Z = 25 0 Ω (a) From p r (t) and p x (t) plots, estimate the real and reactive power for each load Draw a conclusion regarding the sign of reactive power for inductive and capacitive loads (b) Using phasor values of current and voltage, calculate the real and reactive power for each load and compare with the results obtained from the curves (c) If the above loads are all connected across the same power supply, determine the total real and reactive power taken from the supply The following statements are used to plot the instantaneous voltage, current, and the instantaneous terms given by(2-6) and (2-8) Vm = input( Enter voltage peak amplitude Vm = ); thetav =input( Enter voltage phase angle in degree thetav = ); Vm = 100; thetav = 0; % Voltage amplitude and phase angle Z = input( Enter magnitude of the load impedance Z = ); gama = input( Enter load phase angle in degree gama = ); thetai = thetav - gama; % Current phase angle in degree 5

6 CONTENTS theta = (thetav - thetai)*pi/180; % Degree to radian Im = Vm/Z; % Current amplitude wt=0:05:2*pi; % wt from 0 to 2*pi v=vm*cos(wt); % Instantaneous voltage i=im*cos(wt + thetai*pi/180); % Instantaneous current p=v*i; % Instantaneous power V=Vm/sqrt(2); I=Im/sqrt(2); % RMS voltage and current pr = V*I*cos(theta)*(1 + cos(2*wt)); % Eq (26) px = V*I*sin(theta)*sin(2*wt); % Eq (28) disp( (a) Estimate from the plots ) P = max(pr)/2, Q = V*I*sin(theta)*sin(2*pi/4) P = P*ones(1, length(wt)); % Average power for plot xline = zeros(1, length(wt)); % generates a zero vector wt=180/pi*wt; % converting radian to degree subplot(221), plot(wt, v, wt, i,wt, xline), grid title([ v(t)=vm coswt, i(t)=im cos(wt +,num2str(thetai), ) ]) xlabel( wt, degrees ) subplot(222), plot(wt, p, wt, xline), grid title( p(t)=v(t) i(t) ), xlabel( wt, degrees ) subplot(223), plot(wt, pr, wt, P, wt,xline), grid title( pr(t) Eq 26 ), xlabel( wt, degrees ) subplot(224), plot(wt, px, wt, xline), grid title( px(t) Eq 28 ), xlabel( wt, degrees ) subplot(111) disp( (b) From P and Q formulas using phasor values ) P=V*I*cos(theta) % Average power Q = V*I*sin(theta) % Reactive power The result for the inductive load Z = 125 60 Ω is Enter voltage peak amplitude Vm = 100 Enter voltage phase angle in degree thatav = 0 Enter magnitude of the load impedance Z = 125 Enter load phase angle in degree gama = 60 (a) Estimate from the plots P = Q = 2000 3464 (b) For the inductive load Z = 125 are 0 60 Ω, the rms values of voltage and current V = 100 1414 = 7071 0 V

CONTENTS 7 v(t) = V m cos ωt, i(t) = I m cos(ωt 60) 100 6000 p(t) = v(t)i(t) 50 4000 0 2000 50 0 100 0 100 200 300 2000 400 0 100 200 300 400 ωt, degrees ωt, degrees 4000 p r (t), Eq 26 4000 p x (t), Eq 28 3000 2000 2000 0 1000 2000 0 4000 0 100 200 300 400 0 100 200 300 400 ωt - degrees FIGURE 3 Instantaneous current, voltage, power, Eqs 26 and 28 Using (27) and (29), we have 0 I = 7071 125 60 = 5657 60 A P = (7071)(5657) cos(60) = 2000 W Q = (7071)(5657) sin(60) = 3464 Var ωt, degrees Running the above program for the capacitive load Z = 20 30 Ω will result in (a) Estimate from the plots P = 2165 Q = -1250

8 CONTENTS Similarly, for Z = 25 0 Ω, we get Q = P = 2000 0 (c) With the above three loads connected in parallel across the supply, the total real and reactive powers are P = 2000 + 2165 + 2000 = 6165 W Q = 3464 1250 + 0 = 2214 Var 22 A single-phase load is supplied with a sinusoidal voltage The resulting instantaneous power is v(t) = 200 cos(377t) p(t) = 800 + 1000 cos(754t 3687 ) (a) Find the complex power supplied to the load (b) Find the instantaneous current i(t) and the rms value of the current supplied to the load (c) Find the load impedance (d) Use MATLAB to plot v(t), p(t), and i(t) = p(t)/v(t) over a range of 0 to 1667 ms in steps of 01 ms From the current plot, estimate the peak amplitude, phase angle and the angular frequency of the current, and verify the results obtained in part (b) Note in MATLAB the command for array or element-by-element division is / p(t) = 800 + 1000 cos(754t 3687 ) = 800 + 1000 cos 3687 cos 754t + sin 3687 sin 754t = 800 + 800 cos 754t + 600 sin 754t = 800[1 + cos 2(377)t] + 600 sin 2(377)t p(t) is in the same form as (25), thus P = 600 W, and Q = 600, Var, or (b) Using S = 1 2 V mi m, we have S = 800 + j600 = 1000 3687 VA 1000 3687 = 1 2 200 0 I m

CONTENTS 9 or I m = 10 3687 A Therefore, the instantaneous current is i(t) = 10cos(377t 3687 ) A (c) Z L = V I = 200 0 10 3687 = 20 3687 Ω (d) We use the following command v(t) 200 100 0 100 200 0 100 200 300 400 ωt, degrees p(t) 2000 1500 1000 500 0 500 0 100 200 300 400 ωt, degrees 10 i(t) 5 0 5 10 0 100 200 300 400 ωt, degrees FIGURE 4 Instantaneous voltage, power, and current for Problem 22

10 CONTENTS Vm = 200; t=0:0001:001667; % wt from 0 to 2*pi v=vm*cos(377*t); % Instantaneous voltage p = 800 + 1000*cos(754*t - 3687*pi/180);% Instantaneous power i=p/v; % Instantaneous current wt=180/pi*377*t; % converting radian to degree xline = zeros(1, length(wt)); % generates a zero vector subplot(221), plot(wt, v, wt, xline), grid xlabel( wt, degrees ), title( v(t) ) subplot(222), plot(wt, p, wt, xline), grid xlabel( wt, degrees ), title( p(t) ) subplot(223), plot(wt, i, wt, xline), grid xlabel( wt, degrees ), title( i(t) ), subplot(111) The result is shown in Figure 4 The inspection of current plot shows that the peak amplitude of the current is 10 A, lagging voltage by 3687, with an angular frequency of 377 Rad/sec 23 An inductive load consisting of R and X in series feeding from a 2400-V rms supply absorbs 288 kw at a lagging power factor of 08 Determine R and X R X + I V FIGURE 5 An inductive load, with R and X in series The complex power is The current given from S = V I, is θ = cos 1 08 = 3687 S = 288 08 3687 = 360 3687 kva 360 103 3687 I = 2400 0 = 150 3687 A Therefore, the series impedance is Z = R + jx = V I = 2400 0 150 Therefore, R = 128 Ω and X = 96 Ω = 128 + j96 Ω 3687

CONTENTS 11 24 An inductive load consisting of R and X in parallel feeding from a 2400-V rms supply absorbs 288 kw at a lagging power factor of 08 Determine R and X R I + V X FIGURE 6 An inductive load, with R and X in parallel The complex power is S = 288 08 3687 = 360 3687 kva = 288 kw + j216 kvar V 2 R = P = (2400)2 288 10 3 = 20 Ω V 2 X = Q = (2400)2 216 10 3 = 26667 Ω 25 Two loads connected in parallel are supplied from a single-phase 240-V rms source The two loads draw a total real power of 400 kw at a power factor of 08 lagging One of the loads draws 120 kw at a power factor of 096 leading Find the complex power of the other load The total complex load is The 120 kw load complex power is θ = cos 1 08 = 3687 S = 400 08 3687 = 500 3687 kva = 400 kw + j300 kvar S = 120 096 1626 = 125 1626 kva = 120 kw j35 kvar

12 CONTENTS Therefore, the second load complex power is S 2 = 400 + j300 (120 j35) = 280 kw + j335 kvar 26 The load shown in Figure 7 consists of a resistance R in parallel with a capacitor of reactance X The load is fed from a single-phase supply through a line of impedance 84 + j112 Ω The rms voltage at the load terminal is 1200 0 V rms, and the load is taking 30 kva at 08 power factor leading (a) Find the values of R and X (b) Determine the supply voltage V V I + 84 + j112 Ω 0 V R 1200 jx FIGURE 7 Circuit for Problem 26 θ = cos 1 08 = 3687 The complex power is S = 30 3687 = 24 kw j18 kvar (a) From S = V I, the current is V 2 R = P = (1200)2 24000 = 60 Ω V 2 X = Q = (1200)2 18000 = 80 Ω I = 30000 Thus, the supply voltage is 3687 1200 0 = 25 3687 A V = 1200 0 + 25 3687 (84 + j112) = 1200 + j350 = 1250 1626 V

CONTENTS 13 27 Two impedances, Z 1 = 08 + j56 Ω and Z 2 = 8 j16 Ω, and a singlephase motor are connected in parallel across a 200-V rms, 60-Hz supply as shown in Figure 8 The motor draws 5 kva at 08 power factor lagging + I I 1 I 2 I 3 08 8 200 0 S V 3 = 5 kva M at 08 PF lag j56 j16 FIGURE 8 Circuit for Problem 27 (a) Find the complex powers S 1, S 2 for the two impedances, and S 3 for the motor (b) Determine the total power taken from the supply, the supply current, and the overall power factor (c) A capacitor is connected in parallel with the loads Find the kvar and the capacitance in µf to improve the overall power factor to unity What is the new line current? (a) The load complex power are S 1 = V 2 Z 1 V 2 Z2 = (200)2 = 1000 + j7000 VA 08 j56 S 2 = = (200)2 = 1000 j2000 VA 8 + j16 S 3 = 5000 3687 = 4000 + j3000 VA Therefore, the total complex power is S t = 6 + j8 = 10 5313 kva (b) From S = V I, the current is I = 10000 5313 200 0 = 50 5313 A and the power factor is cos 5313 = 06 lagging

14 CONTENTS (c) For overall unity power factor, Q C = 8000 Var, and the capacitive impedance is and the capacitance is Z C = V 2 S C = (200)2 j8000 = j5 Ω The new current is C = 10 6 = 5305 µf (2π)(60)(5) 0 I = 6000 200 0 = 30 0 A 28 Two single-phase ideal voltage sources are connected by a line of impedance of 07 + j24 Ω as shown in Figure 9 V 1 = 500 1626 V and V 2 = 585 0 V Find the complex power for each machine and determine whether they are delivering or receiving real and reactive power Also, find the real and the reactive power loss in the line 500 I 12 1626 + V 07 + j24 Ω + 585 0 V FIGURE 9 Circuit for Problem 28 I 12 = 500 1626 585 0 07 + j24 = 42 + j56 = 70 5313 A S 12 = V 1 I 12 = (500 1626 )(70 5313 ) = 35000 3687 = 28000 j21000 VA S 21 = V 2 I 21 = (585 0 )( 70 5313 ) = 40950 5313 = 24570 + j32760 VA

CONTENTS 15 From the above results, since P 1 is positive and P 2 is negative, source 1 generates 28 kw, and source 2 receives 2457 kw, and the real power loss is 343 kw Similarly, since Q 1 is negative, source 1 receives 21 kvar and source 2 delivers 3276 kvar The reactive power loss in the line is 1176 kvar 29 Write a MATLAB program for the system of Example 25 such that the voltage magnitude of source 1 is changed from 75 percent to 100 percent of the given value in steps of 1 volt The voltage magnitude of source 2 and the phase angles of the two sources is to be kept constant Compute the complex power for each source and the line loss Tabulate the reactive powers and plot Q 1, Q 2, and Q L versus voltage magnitude V 1 From the results, show that the flow of reactive power along the interconnection is determined by the magnitude difference of the terminal voltages We use the following commands E1 = input( Source # 1 Voltage Mag = ); a1 = input( Source # 1 Phase Angle = ); E2 = input( Source # 2 Voltage Mag = ); a2 = input( Source # 2 Phase Angle = ); R = input( Line Resistance = ); X = input( Line Reactance = ); Z = R + j*x; % Line impedance E1 = (075*E1:1:E1) ; % Change E1 form 75% to 100% E1 a1r = a1*pi/180; % Convert degree to radian k = length(e1); E2 = ones(k,1)*e2;%create col Array of same length for E2 a2r = a2*pi/180; % Convert degree to radian V1=E1*cos(a1r) + j*e1*sin(a1r); V2=E2*cos(a2r) + j*e2*sin(a2r); I12 = (V1 - V2)/Z; I21=-I12; S1= V1*conj(I12); P1 = real(s1); Q1 = imag(s1); S2= V2*conj(I21); P2 = real(s2); Q2 = imag(s2); SL= S1+S2; PL = real(sl); QL = imag(sl); Result1=[E1, Q1, Q2, QL]; disp( E1 Q-1 Q-2 Q-L ) disp(result1) plot(e1, Q1, E1, Q2, E1, QL), grid xlabel( Source #1 Voltage Magnitude ) ylabel( Q, var ) text(1125, -180, Q2 ) text(1125, 5, QL ), text(1125, 197, Q1 ) The result is

16 CONTENTS Source # 1 Voltage Mag = 120 Source # 1 Phase Angle = -5 Source # 2 Voltage Mag = 100 Source # 2 Phase Angle = 0 Line Resistance = 1 Line Reactance = 7 E1 Q-1 Q-2 Q-L 900000-1055173 1291066 235894 910000-939497 1149856 210359 920000-821021 1008646 187625 930000-699745 867435 167690 940000-575669 726225 150556 950000-448794 585015 136221 960000-319118 443804 124687 970000-186642 302594 115952 980000-51366 161383 110017 990000 86710 20173 106883 1000000 227586-121037 106548 1010000 371262-262248 109014 1020000 517737-403458 114279 1030000 667013-544668 122345 1040000 819089-685879 133210 1050000 973965-827089 146876 1060000 1131641-968299 163341 1070000 1292117-1109510 182607 1080000 1455393-1250720 204672 1090000 1621468-1391931 229538 1100000 1790344-1533141 257203 1110000 1962020-1674351 287669 1120000 2136496-1815562 320934 1130000 2313772-1956772 357000 1140000 2493848-2097982 395865 1150000 2676724-2239193 437531 1160000 2862399-2380403 481996 1170000 3050875-2521614 529262 1180000 3242151-2662824 579327 1190000 3436227-2804034 632193 1200000 3633103-2945245 687858 Examination of Figure 10 shows that the flow of reactive power along the interconnection is determined by the voltage magnitude difference of terminal voltages

CONTENTS 17 Q var 400 300 200 100 0 100 200 300 90 95 100 105 110 115 120 Source # 1 Voltage magnitude Q 1 Q 2 Q L FIGURE 10 Reactive power versus voltage magnitude 210 A balanced three-phase source with the following instantaneous phase voltages v an = 2500 cos(ωt) v bn = 2500 cos(ωt 120 ) v cn = 2500 cos(ωt 240 ) supplies a balanced Y-connected load of impedance Z = 250 3687 Ω per phase (a) Using MATLAB, plot the instantaneous powers p a, p b, p c and their sum versus ωt over a range of 0 : 005 : 2π on the same graph Comment on the nature of the instantaneous power in each phase and the total three-phase real power (b) Use (244) to verify the total power obtained in part (a) We use the following commands wt=0:02:2*pi; pa=25000*cos(wt)*cos(wt-3687*pi/180); pb=25000*cos(wt-120*pi/180)*cos(wt-120*pi/180-3687*pi/180); pc=25000*cos(wt-240*pi/180)*cos(wt-240*pi/180-3687*pi/180); p = pa+pb+pc;

18 CONTENTS plot(wt, pa, wt, pb, wt, pc, wt, p), grid xlabel( Radian ) disp( (b) ) V = 2500/sqrt(2); gama = acos(08); Z = 250*(cos(gama)+j*sin(gama)); I = V/Z; P = 3*V*abs(I)*08 10 4 3 25 2 15 1 05 0 05 0 1 2 3 4 5 6 7 8 Radian FIGURE 11 Instantaneous powers and their sum for Problem 210 (b) V = 2500 2 = 176777 0 V 0 I = 176777 250 3687 = 7071 3687 A P = (3)(176777)(7071)(08) = 30000 W 211 A 4157-V rms three-phase supply is applied to a balanced Y-connected threephase load consisting of three identical impedances of 48 3687 Ω Taking the phase to neutral voltage V an as reference, calculate (a) The phasor currents in each line (b) The total active and reactive power supplied to the load V an = 4157 3 = 2400 V

CONTENTS 19 With V an as reference, the phase voltages are: V an = 2400 0 V V bn = 2400 120 V V an = 2400 240 V (a) The phasor currents are: (b) The total complex power is I a = V an Z = 2400 0 48 3687 = 50 3687 A I b = V bn Z = 2400 120 48 3687 = 50 15687 A I c = V cn Z = 2400 240 48 3687 = 50 27687 A S = 3V an I a = (3)(2400 0 )(50 3687 ) = 360 3687 kva = 288 kw + j216 KVAR 212 Repeat Problem 211 with the same three-phase impedances arranged in a connection Take V ab as reference V an = 4157 3 = 2400 V With V ab as reference, the phase voltages are: I ab = V ab Z = 4157 0 48 3687 = 866 3687 A I a = 3 30 I ab = ( 3 30 )(866 3687 = 150 6687 A For positive phase sequence, current in other lines are I b = 150 18687 A, and I c = 150 5313 A (b) The total complex power is S = 3V ab I ab = (3)(4157 0 )(866 3687 ) = 1080 3687 kva = 864 kw + j648 kvar 213 A balanced delta connected load of 15 + j18 Ω per phase is connected at the end of a three-phase line as shown in Figure 12 The line impedance is 1+j2 Ω per phase The line is supplied from a three-phase source with a line-to-line voltage of 20785 V rms Taking V an as reference, determine the following:

20 CONTENTS a V L = 20785 V b 1 + j2 Ω b a 15 + j18 Ω c c FIGURE 12 Circuit for Problem 213 (a) Current in phase a (b) Total complex power supplied from the source (c) Magnitude of the line-to-line voltage at the load terminal V an = 20785 3 = 120 V Transforming the delta connected load to an equivalent Y-connected load, result in the phase a equivalent circuit, shown in Figure 13 a I a 1 + j2 Ω + V 1 = 120 0 V V 2 n 5 Ω j6 Ω FIGURE 13 The per phase equivalent circuit for Problem 213 (a) (b) The total complex power is (c) 0 I a = 120 6 + j8 = 12 5313 A S = 3V an I a = (3)(120 0 )(12 5313 = 4320 5313 VA = 2592 W + j3456 Var V 2 = 120 0 (1 + j2)(12 5313 = 9372 293 A

CONTENTS 21 Thus, the magnitude of the line-to-line voltage at the load terminal is V L = 3(9372) = 1623 V 214 Three parallel three-phase loads are supplied from a 20785-V rms, 60-Hz three-phase supply The loads are as follows: Load 1: A 15 HP motor operating at full-load, 9325 percent efficiency, and 06 lagging power factor Load 2: A balanced resistive load that draws a total of 6 kw Load 3: A Y-connected capacitor bank with a total rating of 16 kvar (a) What is the total system kw, kvar, power factor, and the supply current per phase? (b) What is the system power factor and the supply current per phase when the resistive load and induction motor are operating but the capacitor bank is switched off? The real power input to the motor is P 1 = (15)(746) = 12 kw 09325 S 1 = 12 06 5313 kva = 12 kw + j16 kvar S 2 = 6 kw + j0 kvar S 3 = 0 kw j16 kvar (a) The total complex power is S = 18 0 kva = 18 kw + j0 kvar The supply current is I = 18000 (3)(120) = 50 0 A, at unity power factor (b) With the capacitor switched off, the total power is S = 18 + j16 = 2408 4163 kva I = 24083 4163 (3)(120 0 ) = 6689 4163 A The power factor is cos 4163 = 0747 lagging

22 CONTENTS 215 Three loads are connected in parallel across a 1247 kv three-phase supply Load 1: Inductive load, 60 kw and 660 kvar Load 2: Capacitive load, 240 kw at 08 power factor Load 3: Resistive load of 60 kw (a) Find the total complex power, power factor, and the supply current (b) A Y-connected capacitor bank is connected in parallel with the loads Find the total kvar and the capacitance per phase in µf to improve the overall power factor to 08 lagging What is the new line current? S 1 = (a) The total complex power is 60 kw + j660 kvar S 2 = 240 kw j180 kvar S 3 = 60 kw + j0 kvar S = 360 kw + j480 kvar = 600 5313 kva The phase voltage is V = 1247 3 = 72 0 kv The supply current is I = 600 5313 (3)(72) = 2777 5313 A The power factor is cos 5313 = 06 lagging (b) The net reactive power for 08 power factor lagging is Q = 360 tan 3687 = 270 kvar Therefore, the capacitor kvar is Q c = 480 270 = 210 kvar, or S c = j210 kva X c = V L 2 S c = (1247 1000)2 j210000 = j74048 Ω C = 10 6 (2π)(60)(74048) = 358µF

CONTENTS 23 Q θ P Q Q c FIGURE 14 The power diagram for Problem 215 I = S 360 j270 = = 20835 3687 A V (3)(72) 216 A balanced -connected load consisting of a pure resistances of 18 Ω per phase is in parallel with a purely resistive balanced Y-connected load of 12 Ω per phase as shown in Figure 15 The combination is connected to a three-phase balanced supply of 34641-V rms (line-to-line) via a three-phase line having an inductive reactance of j3 Ω per phase Taking the phase voltage V an as reference, determine (a) The current, real power, and reactive power drawn from the supply (b) The line-to-neutral and the line-to-line voltage of phase a at the combined load terminals a j3 Ω a V L = 34641 V b b 18 Ω c c 12 Ω n FIGURE 15 Circuit for Problem 216

24 CONTENTS Transforming the delta connected load to an equivalent Y-connected load, result in the phase a equivalent circuit, shown in Figure 16 a j3 Ω 12 Ω V 1 = 200 0 V V 2 + n I I 1 I 2 6 Ω FIGURE 16 The per phase equivalent circuit for Problem 216 (a) V an = 34641 3 = 200 V The input impedance is Z = (12)(6) 12 + 6 + j3 = 4 + j3 Ω The total complex power is (b) 0 I a = 200 4 + j3 = 40 3687 A S = 3V an I a = (3)(200 0 )(40 3687 ) = 24000 3687 VA = 19200 W + j14400 Var V 2 = 200 0 (j3)(40 3687 ) = 160 3687 A Thus, the magnitude of the line-to-line voltage at the load terminal is V L = 3(160) = 2771 V

CHAPTER 3 PROBLEMS 31 A three-phase, 31875-kVA, 2300-V alternator has an armature resistance of 035 Ω/phase and a synchronous reactance of 12 Ω/phase Determine the no-load line-to-line generated voltage and the voltage regulation at (a) Full-load kva, 08 power factor lagging, and rated voltage (b) Full-load kva, 06 power factor leading, and rated voltage V φ = 2300 3 = 13279 V (a) For 31875 kva, 08 power factor lagging, S = 318750 3687 VA I a = S 3V φ = 318750 3687 (3)(13279) = 80 3687 A E φ = 13279 + (035 + j12)(80 3687 ) = 14092 244 V The magnitude of the no-load generated voltage is E LL = 3 14092 = 24408 V, and the voltage regulation is VR = 24408 2300 2300 100 = 612% (b) For 31875 kva, 06 power factor leading, S = 318750 5313 VA I a = S 3V φ = 318750 5313 (3)(13279 = 80 5313 A 25

26 CONTENTS E φ = 13279 + (035 + j12)(80 5313 ) = 12704 361 V The magnitude of the no-load generated voltage is E LL = 3 12704 = 22204 V, and the voltage regulation is VR = 22004 2300 2300 100 = 433% 32 A 60-MVA, 693-kV, three-phase synchronous generator has a synchronous reactance of 15 Ω/phase and negligible armature resistance (a) The generator is delivering rated power at 08 power factor lagging at the rated terminal voltage to an infinite bus bar Determine the magnitude of the generated emf per phase and the power angle δ (b) If the generated emf is 36 kv per phase, what is the maximum three-phase power that the generator can deliver before losing its synchronism? (c) The generator is delivering 48 MW to the bus bar at the rated voltage with its field current adjusted for a generated emf of 46 kv per phase Determine the armature current and the power factor State whether power factor is lagging or leading? V φ = 693 3 = 40 kv (a) For 60 kva, 08 power factor lagging, S = 60000 3687 kva I a = S 3V φ = 60000 3687 (3)(40) = 500 3687 A E φ = 40 + (j15)(500 3687 ) 10 3 = 449 7675 kv (b) P max = 3 E V = (3)(36)(40) = 288 MW X s 15 (c) For P = 48 MW, and E = 46 KV/phase, the power angle is given by 48 = (3)(46)(40) 15 sin δ

CONTENTS 27 or δ = 74947 and solving for the armature current from E = V + jx s I a, we have I a = 46000 74947 40000 0 j15 = 54747 4306 A The power factor is cos 1 4306 = 07306 lagging 33 A 24,000-kVA, 1732-kV, Y-connected synchronous generator has a synchronous reactance of 5 Ω/phase and negligible armature resistance (a) At a certain excitation, the generator delivers rated load, 08 power factor lagging to an infinite bus bar at a line-to-line voltage of 1732 kv Determine the excitation voltage per phase (b) The excitation voltage is maintained at 134 KV/phase and the terminal voltage at 10 KV/phase What is the maximum three-phase real power that the generator can develop before pulling out of synchronism? (c) Determine the armature current for the condition of part (b) V φ = 1732 3 = 10 kv (a) For 24000 kva, 08 power factor lagging, S = 24000 3687 kva I a = S 3V φ = 24000 3687 (3)(10) = 800 3687 A E φ = 10 + (j5)(800 3687 ) 10 3 = 12806 1447 kv (b) P max = 3 E V = (3)(134)(10) = 804 MW X s 5 (c) At maximum power transfer δ = 90, and solving for the armature current from E = V + jx s I a, we have I a = 13400 90 10000 0 j5 = 3344 3673 A

28 CONTENTS The power factor is cos 1 3673 = 07306 leading 34 A 3464-kV, 60-MVA, three-phase salient-pole synchronous generator has a direct axis reactance of 135 Ω and a quadrature-axis reactance of 9333 Ω The armature resistance is negligible (a) Referring to the phasor diagram of a salient-pole generator shown in Figure 17, show that the power angle δ is given by ( ) δ = tan 1 Xq I a cos θ V + X q I a sin θ (b) Compute the load angle δ and the per phase excitation voltage E when the generator delivers rated MVA, 08 power factor lagging to an infinite bus bar of 3464-kV line-to-line voltage (c) The generator excitation voltage is kept constant at the value found in part (b) Use MATLAB to obtain a plot of the power angle curve, ie, equation (326) over a range of d = 0 : 05 : 180 Use the command [P max, k] = max(p); d max = d(k), to obtain the steady-state maximum power P max and the corresponding power angle d max a I d δ θ I q b d δ I a c e V E X d I d jx q I q FIGURE 17 Phasor diagram of a salient-pole generator for Problem 34 (a) Form the phasor diagram shown in Figure 17, we have or V sin δ = X q I q = X q I a cos(θ + δ) = X q I a (cos θ cos δ sin θ sin δ) sin δ cos δ = X q I a cos θ V + X q I a sin θ

CONTENTS 29 or δ = tan 1 X q I a cos θ V + X q I a sin θ (b) V φ = 3464 3 = 20 kv (a) For 60 MVA, 08 power factor lagging, S = 60 3687 MVA I a = S 3V φ = 60000 3687 (3)(20) = 1000 3687 A δ = tan 1 (9333)(1000)(08) 20000 + (9333)(1000)(06) = 1626 The magnitude of the no-load generated emf per phase is given by E = V cos δ + X d I a sin(θ + δ) = 20 cos 1626 + (135)(1000)(10 3 ) sin 5313 = 30 kv (c) We use the following commands V = 20000; Xd = 135; Xq = 9333; theta=acos(08); Ia = 20E06/20000; delta = atan(xq*ia*cos(theta)/(v + Xq*Ia*sin(theta))); deltadg=delta*180/pi; E = V*cos(delta)+Xd*Ia*sin(theta+delta); E_KV = E/1000; % Excitaiton voltage in kv fprintf( Power angle = %g Degree \n, deltadg) fprintf( E = %g kv \n\n, E_KV) deltadg = (0:25:180) ; delta=deltadg*pi/180; P=3*E*V/Xd*sin(delta)+3*V^2*(Xd-Xq)/(2*Xd*Xq)*sin(2*delta); P = P/1000000; % Power in MW plot(deltadg, P), grid xlabel( Delta - Degree ), ylabel( P - MW ) [Pmax, k]=max(p); delmax=deltadg(k); fprintf( Max power = %g MW,Pmax) fprintf( at power angle %g degree \n, delmax)

30 CONTENTS P MW 140 120 100 80 60 40 20 0 0 20 40 60 80 100 120 140 160 180 Delta, degree FIGURE 18 Power angle curve for Problem 34 The result is Power angle = 162598 Degree E = 30 kv Max power = 138712 MW at power angle 75 degree 35 A 150-kVA, 2400/240-V single-phase transformer has the parameters as shown in Figure 19 02 + j045 Ω 0002 + j00045 Ω I 1 I + I 0 2 I 2 I c I m V 1 1000 Ω j1500 Ω E 1 E 2 V 150 kva 2 08 lag FIGURE 19 Transformer circuit for Problem 35 (a) Determine the equivalent circuit referred to the high-voltage side (b) Find the primary voltage and voltage regulation when transformer is operating at full load 08 power factor lagging and 240 V

CONTENTS 31 (c) Find the primary voltage and voltage regulation when the transformer is operating at full-load 08 power factor leading (d) Verify your answers by running the trans program in MATLAB and obtain the transformer efficiency curve (a) Referring the secondary impedance to the primary side, the transformer equivalent impedance referred to the high voltage-side is Z e1 = 02 + j045 + ( ) 2400 2 (0002 + j00045) = 04 + j09 Ω 240 + I 2 I = 625 3687 1 I 0 I c Z e1 = 040 + j090 Ω I m V 1 1000 Ω j1500 Ω V 2 = 2400 FIGURE 20 Equivalent circuit referred to the high-voltage side (b) At full-load 08 power factor lagging S = 150 primary current is I 2 = S V = 150000 3687 = 625 3687 A 2400 3687 kva, and the referred V 1 = 2400 + (04 + j09)(625 3687 ) = 2453933 07 V The voltage regulation is VR = 2453933 2400 2400 (c) At full-load 08 power factor leading S = 150 primary current is I 2 = S 100 = 2247% V = 150000 +3687 = 625 +3687 A 2400 3687 kva, and the referred V 1 = 2400 + (04 + j09)(625 +3687 ) = 2387004 144 V

32 CONTENTS The voltage regulation is VR = 2453933 2400 2400 100 = 0541% (d) Entering trans at the MATLAB prompt, result in P erc e n t 96 94 92 90 88 86 84 Transformer Efficiency, PF = 08 82 20 40 60 80 100 120 140 160 Output Power, kw FIGURE 21 Efficiency curve of Problem 35 TRANSFORMER ANALYSIS Type of parameters for input Select ---------------------------- ------ To obtain equivalent circuit from tests 1 To input individual winding impedances 2 To input transformer equivalent impedance 3 To quit 0 Select number of menu --> 2 Enter Transformer rated power in kva, S = 150 Enter rated low voltage in volts = 240 Enter rated high voltage in volts = 2400

CONTENTS 33 Enter LV winding series impedance R1+j*X1 in ohm=0002+j*0045 Enter HV winding series impedance R2+j*X2 in ohm=02+j*045 Enter lv within quotes for low side shunt branch or enter hv within quotes for high side shunt branch -> hv Shunt resistance in ohm (if neglected enter inf) = 1000 Shunt reactance in ohm (if neglected enter inf) = 1500 Shunt branch ref to LV side Shunt branch ref to HV side Rc = 10000 ohm Rc = 1000000 ohm Xm = 15000 ohm Xm = 1500000 ohm Series branch ref to LV side Series branch refto HV side Ze = 00040 + j 00090 ohm Ze = 04000 + j 090 ohm Hit return to continue Enter load kva, S2 = 150 Enter load power factor, pf = 08 Enter lg within quotes for lagging pf or ld within quotes for leading pf -> lg Enter load terminal voltage in volt, V2 = 240 Secondary load voltage = 240000 V Secondary load current = 625000 A at -3687 degrees Current ref to primary = 62500 A at -3687 degrees Primary no-load current = 2949 A at -3369 degrees Primary input current = 65445 A at -3673 degrees Primary input voltage = 2453933 V at 070 degrees Voltage regulation = 2247 percent Transformer efficiency = 94249 percent Maximum efficiency is 95238 %occurs at 288 kva with 08pf The efficiency curve is shown in Figure 21 The analysis is repeated for the fullload kva, 08 power factor leading 36 A 60-kVA, 4800/2400-V single-phase transformer gave the following test results: 1 Rated voltage is applied to the low voltage winding and the high voltage winding is open-circuited Under this condition, the current into the low voltage winding is 24 A and the power taken from the 2400 V source is 3456 W 2 A reduced voltage of 1250 V is applied to the high voltage winding and the low voltage winding is short-circuited Under this condition, the current flowing into the high voltage winding is 125 A and the power taken from the 1250 V source is 4375 W (a) Determine parameters of the equivalent circuit referred to the high voltage side

34 CONTENTS (b) Determine voltage regulation and efficiency when transformer is operating at full-load, 08 power factor lagging, and a terminal voltage of 2400 V (c) What is the load kva for maximum efficiency and the maximum efficiency at 08 power factor? (d) Determine the efficiency when transformer is operating at 3/4 full-load, 08 power factor lagging, and a terminal voltage of 2400 V (e) Verify your answers by running the trans program in MATLAB and obtain the transformer efficiency curve (a) From the no-load test data R clv = V 2 2 = (2400)2 P 0 3456 = 1666667 Ω I m = I c = V 2 R clv = 2400 166667 = 144 A I0 2 I2 c = (24) 2 (144) 2 = 192 A X mlv = V 2 I m = 2400 192 = 1250 Ω The shunt parameters referred to the high-voltage side are R C HV = ( ) 4800 2 1666667 = 6666667 Ω 2400 X mhv = From the short-circuit test data ( ) 4800 2 1250 = 5000 Ω 2400 Z e1 = V sc I sc = 1250 125 = 100 Ω R e1 = P sc (I sc ) 2 = 4375 (125) 2 = 28 Ω

CONTENTS 35 X e1 = Z 2 e1 R2 e1 = (100) 2 (28) 2 = 96 Ω (b) At full-load 08 power factor lagging S = 60 primary current is I 2 = S V = 60000 3687 = 125 3687 A 4800 3687 kva, and the referred V 1 = 4800 + (28 + j96)(125 3687 ) = 584829 7368 V The voltage regulation is VR = 584829 4800 4800 Efficiency at full-load, 08 power factor is η = 100 = 21839% (60)(08) 100 = 85974 % (60)(08) + 3456 + 4375 (c) At Maximum efficiency P c = P cu, and we have P cu S 2 fl P c S 2 maxη Therefore, the load kva at maximum efficiency is P c 3456 S maxη = S fl = 60 = 53327 kva P cu 4375 and the maximum efficiency at 08 power factor is η = (53327)(08) 100 = 86057 % (53327)(08) + 3456 + 3456 (d) At 3/4 full-load, the copper loss is P cu3/4 = (3/4) 2 (4375) = 24609375 W Therefore, the efficiency at 3/4 full-load, 08 power factor is η = (45)(08) 100 = 85884 % (45)(08) + 3456 + 24609 (e) The commands trans display the following menu

36 CONTENTS TRANSFORMER ANALYSIS Type of parameters for input Select ---------------------------- ------ To obtain equivalent circuit from tests 1 To input individual winding impedances 2 To input transformer equivalent impedance 3 To quit 0 Select number of menu --> 1 Enter Transformer rated power in kva, S = 60 Enter rated low voltage in volts = 2400 Enter rated high voltage in volts = 4800 Open circuit test data ---------------------- Enter lv within quotes for data referred to low side or enter hv within quotes for data referred to high side -> lv Enter input voltage in volts, Vo = 2400 Enter no-load current in Amp, Io = 24 Enter no-load input power in Watt, Po = 3456 Short circuit test data ------------------------ Enter lv within quotes for data referred to low side or enter hv within quotes for data referred to high side -> hv Enter reduced input voltage in volts, Vsc = 1250 Enter input current in Amp, Isc = 125 Enter input power in Watt, Psc = 4375 Shunt branch ref to LV side Shunt branch ref to HV side Rc = 1666667 ohm Rc = 6666667 ohm Xm = 1250000 ohm Xm = 5000000 ohm Series branch ref to LV side Series branch refto HV side Ze = 70000 + j 240000 ohm Ze = 28000 + j 96000 ohm Hit return to continue Enter load KVA, S2 = 60 Enter load power factor, pf = 08 Enter lg within quotes for lagging pf or ld within quotes for leading pf -> lg Enter load terminal voltage in volt, V2 = 2400 Secondary load voltage = 2400000 V Secondary load current = 25000 A at -3687 degrees

CONTENTS 37 Current ref to primary = 12500 A at -3687 degrees Primary no-load current = 1462 A at -5313 degrees Primary input current = 13910 A at -3856 degrees Primary input voltage = 5848290 V at 737 degrees Voltage regulation = 21839 percent Transformer efficiency = 85974 percent Maximum efficiency is 86057%occurs at 5333kVA with 08pf The efficiency curve is shown in Figure 22 The analysis is repeated for the 3/4 full-load kva, 08 power factor lagging 76 78 80 82 84 86 88 P erc e n t 10 15 20 25 30 35 40 45 50 55 60 Output Power, kw Transformer Efficiency, PF = 08 FIGURE 22 Efficiency curve of Problem 36 37 A two-winding transformer rated at 9-kVA, 120/90-V, 60-HZ has a core loss of 200 W and a full-load copper loss of 500 W (a) The above transformer is to be connected as an auto transformer to supply a load at 120 V from 210 V source What kva load can be supplied without exceeding the current rating of the windings? (For this part assume an ideal transformer) (b) Find the efficiency with the kva loading of part (a) and 08 power factor The two-winding transformer rated currents are: I LV = 9000 90 = 100 A

38 CONTENTS I HV = 9000 120 = 75 A The auto transformer connection is as shown in Figure 23 + 90V 100A 210V 120V 75A 175A + 120V FIGURE 23 Auto transformer connection for Problem 37 (a) With windings carrying rated currents, the auto transformer rating is S = (120)(175)(10 3 ) = 21 kva (b) Since the windings are subjected to the same rated voltages and currents as the two-winding transformer, the auto transformer copper loss and the core loss at the rated values are the same as the two-winding transformer Therefore, the auto transformer efficiency at 08 power factor is η = (21000)(08) 100 = 96 % (21000)(08) + 200 + 500 38 Three identical 9-MVA, 72-kV/416-kV, single-phase transformers are connected in Y on the high-voltage side and on the low voltage side The equivalent series impedance of each transformer referred to the high-voltage side is 012 + j082 Ω per phase The transformer supplies a balanced three-phase load of 18 MVA, 08 power factor lagging at 416 kv Determine the line-to-line voltage at the high-voltage terminals of the transformer The per phase equivalent circuit is shown in Figure 24 The load complex power is S = 18000 3687 kva I 1 = 18000 3687 = 833333 3687 A (3)(72)

CONTENTS 39 Z e1 = 012 + j082 + I 1 V 1 V 2 = 7200 V FIGURE 24 The per phase equivalent circuit of Problem 38 V 1 = 72 0 + (012 + j082)(833333 3687 )(10 3 )= 77054 362 kv Therefore, the magnitude of the primary line-to-line supply voltage is V 1LL = 3(77054) = 13346 kv 39 A 400-MVA, 240-kV/24-KV, three-phase Y- transformer has an equivalent series impedance of 12 + j6 Ω per phase referred to the high-voltage side The transformer is supplying a three-phase load of 400-MVA, 08 power factor lagging at a terminal voltage of 24 kv (line to line) on its low-voltage side The primary is supplied from a feeder with an impedance of 06 + j12 Ω per phase Determine the line-to-line voltage at the high-voltage terminals of the transformer and the sending-end of the feeder The per phase equivalent circuit referred to the primary is is shown in Figure 25 The load complex power is + I 1 Z f = 06 + j12 Z e1 = 12 + j6 + V s V = 138564 kv V 1 FIGURE 25 The per phase equivalent circuit for Problem 39 2 S = 400000 3687 kva

40 CONTENTS and the referred secondary voltage per phase is V 2 = 240 3 = 138564 kv I 1 = 400000 3687 (3)(138564) = 96225 3687 A V 1 = 138564 0 + (12 + j6)(96225 3687 )(10 3 ) = 143 157 kv Therefore, the line-to-line voltage magnitude at the high voltage terminal of the transformer is V 1LL = 3(143) = 24769 kv V s = 138564 0 + (18 + j72)(96225 3687 )(10 3 ) = 144177 179 kv Therefore, the line-to-line voltage magnitude at the sending end of the feeder is V 1LL = 3(143) = 24972 kv 310 In Problem 39, with transformer rated values as base quantities, express all impedances in per unit Working with per-unit values, determine the line-to-line voltage at the high-voltage terminals of the transformer and the sending-end of the feeder With the transformer rated MVA as base, the HV-side base impedance is Z B = (240)2 400 = 144 Ω The transformer and the feeder impedances in per unit are Z e = Z f = (12 + j6) = 0008333 + j004166 pu 144 (06 + j12) = 0004166 + j000833 pu 144 I = 1 3687 1 0 = 1 3687 pu V 1 = 10 0 + (0008333 + j004166)(1 3687 = 103205 157 pu

CONTENTS 41 Therefore, the line-to-line voltage magnitude at the high-voltage terminal of the transformer is (103205)(240) = 24769 kv V s = 10 0 + (00125 + j005)(1 3687 = 10405 179 pu Therefore, the line-to-line voltage magnitude at the high voltage terminal of the transformer is (10405)(240) = 24972 kv 311 A three-phase, Y-connected, 75-MVA, 27-kV synchronous generator has a synchronous reactance of 90 Ω per phase Using rated MVA and voltage as base values, determine the per unit reactance Then refer this per unit value to a 100- MVA, 30-kV base The base impedance is Z B = (KV B) 2 MV A B = (27)2 75 = 972 Ω X pu = 9 = 0926 pu 972 The generator reactance on a 100-MVA, 30-kV base is ( ) ( ) 100 27 2 X punew = 0926 = 10 pu 75 30 312 A 40-MVA, 20-kV/400-kV, single-phase transformer has the following series impedances: Z 1 = 09 + j18 Ω and Z 2 = 128 + j288 Ω Using the transformer rating as base, determine the per unit impedance of the transformer from the ohmic value referred to the low-voltage side Compute the per unit impedance using the ohmic value referred to the high-voltage side The transformer equivalent impedance referred to the low-voltage side is ( ) 20 2 Z e1 = 09 + j18 + (128 + j288) = 122 + j252 Ω 400 The low-voltage base impedance is Z B1 = (20)2 40 = 10 Ω Z pu1 = 122 + j252 10 = 0122 + j0252 pu

42 CONTENTS The transformer equivalent impedance referred to the high-voltage side is ( ) 400 2 Z e2 = (09 + j18) + (128 + j288) = 488 + j1008 Ω 20 The high-voltage base impedance is Z B2 = (400)2 40 = 4000 Ω Z pu2 = 488 + j1008 4000 = 0122 + j0252 pu We note that the transformer per unit impedance has the same value regardless of whether it is referred to the primary or the secondary side 313 Draw an impedance diagram for the electric power system shown in Figure 26 showing all impedances in per unit on a 100-MVA base Choose 20 kv as the voltage base for generator The three-phase power and line-line ratings are given below G 1 : 90 MVA 20 kv X = 9% T 1 : 80 MVA 20/200 kv X = 16% T 2 : 80 MVA 200/20 kv X = 20% G 2 : 90 MVA 18 kv X = 9% Line: 200 kv X = 120 Ω Load: 200 kv S = 48 MW +j64 Mvar T 1 T 2 G 1 G 2 1 Line 2 Load FIGURE 26 One-line diagram for Problem 313 The base voltage V BG1 on the LV side of T 1 is 20 kv Hence the base on its HV side is ( ) 200 V B1 = 20 = 200 kv 20 This fixes the base on the HV side of T 2 at V B2 = 200 kv, and on its LV side at ( ) 20 V BG2 = 200 = 20 kv 200

CONTENTS 43 The generator and transformer reactances in per unit on a 100 MVA base, from (369)and (370) are ( ) 100 G: X = 009 = 010 pu 90 ( ) 100 T 1 : X = 016 = 020 pu 80 ( ) 100 T 2 : X = 020 = 025 pu 80 ( ) ( ) 100 18 2 G 2 : X = 009 = 0081 pu 90 20 The base impedance for the transmission line is The per unit line reactance is Z BL = (200)2 100 = 400 Ω Line: X = The load impedance in ohms is ( ) 120 = 030 pu 400 Z L = (V L L) 2 S L(3φ) = (200)2 = 300 + j400 Ω 48 j64 The load impedance in per unit is Z L(pu) = 300 + j400 400 = 075 + j10 pu The per unit equivalent circuit is shown in Figure 27 j02 j03 j025 j01 G 1 075 + j10 j0081 G 2 FIGURE 27 Per unit impedance diagram for Problem 311

44 CONTENTS G 1 2 3 4 T 1 T 2 Line 1 220 kv T 4 M T 3 5 Line 2 6 110 kv Load FIGURE 28 One-line diagram for Problem 314 314 The one-line diagram of a power system is shown in Figure 28 The three-phase power and line-line ratings are given below G: 80 MVA 22 kv X = 24% T 1 : 50 MVA 22/220 kv X = 10% T 2 : 40 MVA 220/22 kv X = 60% T 3 : 40 MVA 22/110 kv X = 64% Line 1: 220 kv X = 121 Ω Line 2: 110 kv X = 4235 Ω M: 6885 MVA 20 kv X = 225% Load: 10 Mvar 4 kv -connected capacitors The three-phase ratings of the three-phase transformer are Primary: Y-connected 40MVA, 110 kv Secondary: Y-connected 40 MVA, 22 kv Tertiary: -connected 15 MVA, 4 kv The per phase measured reactances at the terminal of a winding with the second one short-circuited and the third open-circuited are Z P S = 96% 40 MVA, 110 kv / 22 kv Z P T = 72% 40 MVA, 110 kv / 4 kv Z ST = 12% 40 MVA, 22 kv / 4 kv Obtain the T-circuit equivalent impedances of the three-winding transformer to the common MVA base Draw an impedance diagram showing all impedances in per unit on a 100-MVA base Choose 22 kv as the voltage base for generator The base voltage V B1 on the LV side of T 1 is 22 kv Hence the base on its HV side

CONTENTS 45 is V B2 = 22 ( ) 220 = 220 kv 22 This fixes the base on the HV side of T 2 at V B3 = 220 kv, and on its LV side at ( ) 22 V B4 = 220 = 22 kv 220 Similarly, the voltage base at buses 5 and 6 are V B5 = V B6 = 22 Voltage base for the tertiary side of T 4 is ( ) 110 = 110 kv 22 ( ) 4 V BT = 110 = 4 kv 110 The per unit impedances on a 100 MVA base are: ( ) 100 G: X = 024 = 030 pu 80 ( ) 100 T 1 : X = 010 = 020 pu 50 ( ) 100 T 2 : X = 006 = 015 pu 40 ( ) 100 T 3 : X = 0064 = 016 pu 40 The motor reactance is expressed on its nameplate rating of 6885 MVA, and 20 kv However, the base voltage at bus 4 for the motor is 22 kv, therefore ( ) ( ) 100 20 2 M: X = 0225 = 027 pu 6885 22 Impedance bases for lines 1 and 2 are Z B2 = (220)2 100 = 484 Ω Z B5 = (110)2 100 = 121 Ω

46 CONTENTS 1 j020 j025 j015 4 j016 j035 j006 j018 j03 j027 E g I 012 G j10 I m M Em FIGURE 29 Per unit impedance diagram for Problem 314 Line 1 and 2 per unit reactances are ( ) 121 Line 1 : X = = 025 pu 484 ( ) 4235 Line 2 : X = = 035 pu 121 The load impedance in ohms is Z L = (V L L) 2 S L(3φ) The base impedance for the load is = (4)2 j10 = j16 Ω Z BT = (4)2 100 = 016 Ω Therefore, the load impedance in per unit is Z L(pu) = j16 = j10 pu 016 The three-winding impedances on a 100 MVA base are ( ) 100 Z P S = 0096 = 024 pu 40 ( ) 100 Z P T = 0072 = 018 pu 40 ( ) 100 Z ST = 0120 = 030 pu 40

CONTENTS 47 The equivalent T circuit impedances are Z P = 1 (j024 + j018 j030) = j006 pu 2 Z S = 1 (j024 + j030 j018) = j018 pu 2 Z T = 1 (j018 + j030 j024) = j012 pu 2 The per unit equivalent circuit is shown in Figure 29 315 The three-phase power and line-line ratings of the electric power system shown in Figure 30 are given below V g G T 1 T 2 1 Line 2 V m M FIGURE 30 One-line diagram for Problem 315 G 1 : 60 MVA 20 kv X = 9% T 1 : 50 MVA 20/200 kv X = 10% T 2 : 50 MVA 200/20 kv X = 10% M : 432 MVA 18 kv X = 8% Line: 200 kv Z = 120 + j200 Ω (a) Draw an impedance diagram showing all impedances in per unit on a 100-MVA base Choose 20 kv as the voltage base for generator (b) The motor is drawing 45 MVA, 080 power factor lagging at a line-to-line terminal voltage of 18 kv Determine the terminal voltage and the internal emf of the generator in per unit and in kv The base voltage V BG1 on the LV side of T 1 is 20 kv Hence the base on its HV side is ( ) 200 V B1 = 20 = 200 kv 20 This fixes the base on the HV side of T 2 at V B2 = 200 kv, and on its LV side at ( ) 20 V Bm = 200 = 20 kv 200

48 CONTENTS The generator and transformer reactances in per unit on a 100 MVA base, from (369)and (370) are ( ) 100 G: X = 009 = 015 pu 60 ( ) 100 T 1 : X = 010 = 020 pu 50 ( ) 100 T 2 : X = 010 = 020 pu 50 ( ) ( ) 100 18 2 M: X = 008 = 015 pu 432 20 The base impedance for the transmission line is The per unit line impedance is Line: Z BL = (200)2 100 = 400 Ω Z line = ( 120 + j200 ) = 030 + j05 pu 400 The per unit equivalent circuit is shown in Figure 31 j015 V g j02 + 03 + j05 j020 + V m j015 E g E m FIGURE 31 Per unit impedance diagram for Problem 315 (b) The motor complex power in per unit is S m = 45 and the motor terminal voltage is 3687 100 V m = 18 20 0 = 045 3687 pu = 090 0 pu

CONTENTS 49 I = 045 3687 090 0 = 05 3687 pu V g = 090 0 + (03 + j09)(05 3687 = 131795 1182 pu Thus, the generator line-to-line terminal voltage is V g = (131795)(20) = 26359 kv E g = 090 0 + (03 + j105)(05 3687 = 1375 1388 pu Thus, the generator line-to-line internal emf is E g = (1375)(20) = 275 kv 316 The one-line diagram of a three-phase power system is as shown in Figure 32 Impedances are marked in per unit on a 100-MVA, 400-kV base The load at bus 2 is S 2 = 1593 MW j334 Mvar, and at bus 3 is S 3 = 77 MW +j14 Mvar It is required to hold the voltage at bus 3 at 400 0 kv Working in per unit, determine the voltage at buses 2 and 1 V 1 V 2 V 3 j05 pu j04 pu FIGURE 32 One-line diagram for Problem 316 S 2 S 3 S 2 = 1593 MW j334 Mvar = 01593 j0334 pu S 3 = 7700 MW + j140 Mvar = 07700 + j0140 pu V 3 = 400 0 = 10 0 pu 400 I 3 = S 3 077 j014 V3 = 10 0 = 077 j014 pu V 2 = 10 0 + (j04)(077 j014) = 11 1626 pu

50 CONTENTS Therefore, the line-to-line voltage at bus 2 is V 2 = (400)(11) = 440 kv I 2 = S 2 01593 + j0334 V2 = 11 1626 = 0054 + j0332 pu I 12 = (077 j014) + (0054 + j0332) = 0824 + j0192 pu V 1 = 11 1626 + (j05)(0824 + j0192) = 12 3687 pu Therefore, the line-to-line voltage at bus 1 is V 1 = (400)(12) = 480 kv 317 The one-line diagram of a three-phase power system is as shown in Figure 33 The transformer reactance is 20 percent on a base of 100-MVA, 23/115-kV and the line impedance is Z = j66125ω The load at bus 2 is S 2 = 1848 MW +j66 Mar, and at bus 3 is S 3 = 0 MW +j20 Mar It is required to hold the voltage at bus 3 at 115 0 kv Working in per unit, determine the voltage at buses 2 and 1 V 1 V 3 V 2 j66125 Ω S 2 S 3 FIGURE 33 One-line diagram for Problem 317 S 2 = 1848 MW + j66 Mvar = 1848 + j0066 pu S 3 = 0 MW + j200 Mvar = 0 + j020 pu V 3 = 115 115 0 = 10 0 pu I 3 = S 3 V3 = j02 = j02 pu 10 0 V 2 = 10 0 + (j05)( j02) = 11 0 pu