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ECE421/521 POWER SYSTEM ANALYSISHomework #32.7a2.7b2.7c2.82.92.10a2.10b2.14a2.14b2.15a2.15b2.16a2.16bTotal5215’5’5’10’10’5’5’10’10’10’10’10’5’100’42110’10’10’10’10’10’10’5’5’10’10’10’10’100’+20’2.7(a)I1=VZ1=200∠0°0.8+j5.6=5-j35 A (1’/2’)I2=VZ2=200∠0°8-j16=5+j10 A (1’/2’)S1=VI1*=200*5+j35=1000W+j7000var (1’/2’)S2=VI2*=200*5-j10=1000W-j2000var (1’/2’)S3=5000*0.8+5000*0.6=4000W+j3000var (1’/2’)(b)S=S1+S2+S3=6000W+j8000var (1’/2’)I3=S3*V3*=4000-j3000200=20-j15 A I=I1+I2+I3=5-j35+5+j10+20-j15=30-j40=50∠-53.13° A (2’/4’)PF=cos-53.13°=0.6 lagging (2’/4’)(c)θ'=cos-11=0° Q'=Ptanθ'=0 var Qc=8000-0=8000 var (1’/2’)Sc=-j8 kvar (1’/2’)Z=|V|2Sc*=2002j8000=-j5 Ω (1’/2’)C=1062π60*5=531 μF (1’/2’)I=S*V*=6000200=30∠0° A (1’/2’)2.8I12=V1-V2Zl=500∠16.26°-585∠0°0.7+j2.4=42+j56=70∠53.1° A (1’/1’)I21=V2-V1Zl=585∠0°-500∠16.26°0.7+j2.4=-42-j56=70∠-126.9° A (1’/1’)S12=V1I12*=500∠16.26°×70∠-53.1°=35000∠-36.84° VA=28 kW-j21 kvar (2’/2’)S21=V2I21*=585∠0°×70∠126.9°=40950∠126.9° VA=-25 kW+33 kvar (2’/2’)From the above results, P1 is positive and P2 is negative, machine 1 is delivering 28kW real power and machine 2 is receiving 25kW real power. Q1 is negative and Q2 is positive, machine 1 is receiving 21kvar reactive power and machine 2 is delivering 33kvar reactive power. (2’/2’)Sl=S12+S21=3 kW+12 kvar (2’/2’)2.9Run the following codes, (2’/2’)%HW3 2.9E1 = 120;a1 = -5;E2 = 100;a2 = 0;R = 1;X = 7;Z = R +1i*X;E1 = (0.75*E1:1:E1)';k = length(E1);a1 = ones(k,1)*a1;a1r = a1*pi/180;a2 = ones(k,1)*a2;a2r = a2*pi/180;V1 = E1.*cos(a1r) + 1i*E1.*sin(a1r);V2 = E2.*cos(a2r) + 1i*E2.*sin(a1r);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(' |V| 1 Q-1 Q-2 Q-L ')disp(Result1)plot(E1, Q1, E1, Q2, E1, QL)xlabel('Source #1 Voltage Magnitude, V')ylabel('Q, var')text(115, 300, 'Q1'), text(115, -250, 'Q2'), text(115, 0, 'QL')The result is, (2’/2’)ECE521HW3_2_9 |V| 1 Q-1 Q-2 Q-L 90.0000 -130.7167 145.7984 15.0817 91.0000 -119.4291 131.7447 12.3156 92.0000 -107.8616 117.6910 9.8294 93.0000 -96.0140 103.6372 7.6233 94.0000 -83.8864 89.5835 5.6971 95.0000 -71.4788 75.5298 4.0510 96.0000 -58.7912 61.4760 2.6848 97.0000 -45.8236 47.4223 1.5987 98.0000 -32.5760 33.3686 0.7926 99.0000 -19.0484 19.3148 0.2664 100.0000 -5.2408 5.2611 0.0203 101.0000 8.8468 -8.7926 0.0541 102.0000 23.2144 -22.8464 0.3680 103.0000 37.8620 -36.9001 0.9618 104.0000 52.7895 -50.9538 1.8357 105.0000 67.9971 -65.0076 2.9896 106.0000 83.4847 -79.0613 4.4234 107.0000 99.2523 -93.1151 6.1373 108.0000 115.2999 -107.1688 8.1311 109.0000 131.6275 -121.2225 10.4050 110.0000 148.2351 -135.2763 12.9588 111.0000 165.1227 -149.3300 15.7927 112.0000 182.2903 -163.3837 18.9066 113.0000 199.7379 -177.4375 22.3004 114.0000 217.4655 -191.4912 25.9743 115.0000 235.4731 -205.5449 29.9281 116.0000 253.7606 -219.5987 34.1620 117.0000 272.3282 -233.6524 38.6758 118.0000 291.1758 -247.7061 43.4697 119.0000 310.3034 -261.7599 48.5436 120.0000 329.7110 -275.8136 53.8974 (3’/3’)As the voltage magnitude of V1 increase, the reactive power delivered from machine 1 increases and the reactive power delivered from machine 2 decreases. As V1=|V2|, Q1=Q2. So, the reactive power flows from higher terminal voltage to lower terminal voltage. (3’/3’)2.10(a)Run the following code, (1’/2’)%HW3 2.10wt = 0:0.05:2*pi;van = 2500*cos(wt);vbn = 2500*cos(wt-120*pi/180);vcn = 2500*cos(wt-240*pi/180);Z = 250;delta = 36.87*pi/180;ian = 2500*cos(wt - delta)/Z;ibn = 2500*cos(wt-120*pi/180-delta)/Z;icn = 2500*cos(wt-240*pi/180-delta)/Z;pa = van.*ian/1000;pb = vbn.*ibn/1000;pc = vcn.*icn/1000;p = pa+pb+pc;plot(wt,pa,'*',wt,pb,'-',wt,pc,'+',wt,p,'o')xlabel('Time, s')ylabel('instantaneous power, kW')text(0.3, 25, 'pa'), text(2.4, 25, 'pb')text(4.5, 25, 'pc'), text(6, 28, 'pa+pb+pc') (2’/4’)The plot shows that pc lags pb 120 degree and pb lags pa 120 degree. The sum of them is a constant value 30kVA. (2’/4’)(b)cosθ=cos36.87°=0.8 (1’/2’)P3?=3VpIpcosθ=3×25002×102×0.8=30kVA (2’/4’)The result agrees to the total power obtained in (a). (2’/4’)2.14(a)1horsepower=746watts P1=15×74693.25%=12kVA S1=12+j12×0.80.6=12kW+j16kvar (2’/1’)S2=6kW+j0kvar (2’/1’)S3=0kW-j16kvar (2’/1’)S3?=S1+S2+S3=18kW=18∠0° kVA (1’/0.5’)Total system kW: P=18kWTotal system kvar: Q=0kvarSystem PF: PF=1 (1’/0.5’)Supply current per phase: Ip=S3?*3Vp*=180003×207.853=50∠0° A (2’/1’)(b)If the capacitor bank is switched off,S'3?=S1+S2=18kW+j16kvar=24∠41.6° kVA (4’/2’)New system PF: PF'=0.75 lagging (2’/1’)New supply current per phase: I'p=S3?'*3Vp*=24000∠-41.6°3×207.853=66.7∠-41.6° A (4’/2’)2.15(a)S1=60kW+j660kvar (2’/2’)S2=240kW-j180kvar (2’/2’)S3=60kW+j0kvar (2’/2’)S3?=S1+S2+S3=360kW+j480kvar=600∠53.1° kVA (1’/1’)System PF: PF=0.6 lagging (1’/1’)Supply current per phase: Ip=S3?*3Vp*=600∠-53.1° 3×12.473=27.7∠-53.1° A (2’/2’)(b)To improve the overall PF from 0.6 to 0.8,The new total kvar is: Q'3?=360×0.60.8=270kvar (1’/1’)The kvar been compensated per phase: Qc=480-2703=70kvar (1’/1’)Zc=|Vp|2Sc*=(124703)2j70000=-j740.5Ω (2’/2’)C=1062π×60×740.5=3.6μF (2’/2’)New line current: IL=Ip=S3?'*3Vp*=360-j270 3×12.473=20.8∠-36.9° A (4’/4’)2.16(a)The Δ-connected resistive load is transformed into equivalent Y-connection,Z2=183=6Ω (1’/1’)The phase voltage is,V1=346.413=200V (1’/1’)The total impedance in single phase is,Z=j3+12×612+6=4+j3Ω (2’/2’)The current per phase is,I=V1Z=2004+j3=32-j24A (2’/2’)The three phase power supplied is,S=3V1I*=3×200×32+j24=19.2kW+14.4kvar (2’/2’)(b)The line-to-neutral voltage of phase a at the combined load terminals is,Van=200-j3×32-j24=128-j96=160∠-36.9° V (3’/6’)The line-to-line voltage of phase a at the combined load terminals is,VaL=3∠30°Van=277∠-6.9° V (2’/4’) ................
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