Power System Analysis And Design, 6th Edition Test Bank

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Quiz 1ECE 4761.Two balanced three-phase loads are in parallel.a.Load 1 draws 10 kW at 0.8 PF laggingb.Load 2 draws 20 kVA at 0.6 PF leadingThe loads are supplied by a balanced three-phase 480 VLLsource.(a)Draw the power triangle for the combined load.SL= 23.59kVAQL= Q1+ Q2= -8.5 kVARPL= P1 +P2= 22 kW21.12°(b)Determine PF of the combined load.cos21.12° = 0.933 leading(c)Determine the magnitude of the line current from the source.𝐼𝐿=𝑆𝐿√3𝑉𝐿𝐿=23.59×103√3×480= 28.37 𝐴(d)Y-connected inductors are now installed in parallel with the combined load. What value ofinductive reactance is needed in each leg of the Y to make the source power factor unity?𝑄𝐼𝑛𝑑= |𝑄𝐿| = 8.5 × 103𝑉𝐴𝑅 =3(𝑉𝐿𝑁)2𝑋Y𝑋Y=3(480/√3)28.5×103= 27.1 Ω

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Quiz 2ECE 476Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗1.A 60-Hz single–phase, two-wire overhead line has solid cylindrical copper conductors with 2.4cm diameter. The conductors are arranged in a horizontal configuration with 3 m spacing.Calculate the inductance of each conductor due to both internal and external flux linkages inmH/km (40 points).𝐿𝑥= 𝐿𝑦= 2 × 10−7𝐿𝑛 (𝐷𝑟′) 𝐻/𝑚𝐷 = 3𝑚𝑟′= 0.7788 × 𝑟 = 0.7788 × (0.0242) = 9.346 × 10−3𝐿𝑥= 𝐿𝑦= 2 × 10−7𝐿𝑛 (39.346 × 10−3) 𝐻𝑚 (1000𝑚𝑘𝑚) (1000𝑚𝐻𝐻) = 1.154 𝑚𝐻/𝑘𝑚2.The figure below is a completely transposed three-phase overhead transmission line withbundled phase conductors. All conductors have a radius of 2 cm.0.4m0.4m0.4m0.4m12m12ma.Determine the inductance per phase in mH/km (40 points).𝑟′= 0.7788 ∗ 0.02 = 0.0156𝑅𝑏= √(𝑟′)(0.4)(0.4)(√2 × 0.4)4= 0.1938𝑚𝐷𝑒𝑞= √𝐷𝐴𝐵𝐷𝐵𝐶𝐷𝐶𝐴3= √(12)(12)(24)3= 15.12𝑚𝐿 = 0.2𝐿𝑛 (𝐷𝑒𝑞𝑅𝑏) = 0.2𝐿𝑛 ( 15.120.1938) = 0.8714 𝑚𝐻/𝑘𝑚b.Find the inductive line reactance per phase in Ω/km at 60 Hz(20 points).𝑋 = 𝜔𝐿 = 2𝜋60 × 0.8714 × 10−3= 0.3285 Ω/𝑘𝑚

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Quiz 3ECE 476Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗A generator bus (slack bus) supplies a 100 MW, 50 Mvar load through a lossless transmission line withparameters on 100 MVA base below. Also, there is a 10 Mvar shunt capacitor at the load bus.X per unit: 0.2 & B per unit: 0.21.Determine the bus admittance matrix (Ybus) for the system.10 Mvar = 0.1 p.u.Assuming 1 pu bus voltage,𝑌𝑐𝑎𝑝=𝑆𝑉2=0.112= 0.1 𝑝. 𝑢.𝑌11=1𝑗0.2+ 𝑗0.22,𝑌12= −1𝑗0.2,𝑌21= −1𝑗0.2,𝑌22=1𝑗0.2+ 𝑗0.22+ 𝑗0.1𝑌𝑏𝑢𝑠= [−𝑗4.9𝑗5𝑗5−𝑗4.8]2.Calculate the voltage magnitude and angle at the load bus using the Newton-Rapson method. Use theflat-start. Provide the values after 2 iterations. General power balance equations are given below.𝑃𝑖= ∑|𝑉𝑖||𝑉𝑘|(𝐺𝑖𝑘𝑐𝑜𝑠𝜃𝑖𝑘+ 𝐵𝑖𝑘𝑠𝑖𝑛𝜃𝑖𝑘) = 𝑃𝐺𝑖− 𝑃𝐷𝑖𝑛𝑘=1𝑄𝑖= ∑|𝑉𝑖||𝑉𝑘|(𝐺𝑖𝑘𝑠𝑖𝑛𝜃𝑖𝑘− 𝐵𝑖𝑘𝑐𝑜𝑠𝜃𝑖𝑘) = 𝑄𝐺𝑖− 𝑄𝐷𝑖𝑛𝑘=1𝑃2(𝑥) = |𝑉2|(5𝑠𝑖𝑛𝜃2) + 1 = 0𝑄2(𝑥) = |𝑉2|(−5𝑐𝑜𝑠𝜃2) + |𝑉2|2(4.8) + 0.5 = 0𝐽(𝑥) = [5|𝑉2|𝑐𝑜𝑠𝜃25𝑠𝑖𝑛𝜃25|𝑉2|𝑠𝑖𝑛𝜃2−5𝑐𝑜𝑠𝜃2+ 9.6|𝑉2|]𝑥(0)= [01],𝑓(𝑥(0)) = [ 10.3],𝐽(𝑥(0)) = [5004.6]𝑥(1)= [01] − [5004.6]−1[ 10.3] = [ −0.20.9348]𝑓(𝑥(1)) = [0.9348 × 5 sin(−0.2) + 10.9348 × −5 cos(−0.2) + 0.93482(4.8) + 0.5] = [0.07140.1137]𝐽(𝑥(1)) = [5(0.9348) cos(−0.2)5 sin(−0.2)5(0.9348) sin(−0.2)−5 cos(−0.2) + 9.6(0.9348)] = [ 4.5808−0.9933−0.92864.0737 ]

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𝑥(2)= [ −0.20.9348] − [4.58−0.9933−0.92864.0737 ]−1[0.07140.1137] = [−0.22280.9017 ]∴ 𝑉2= 0.9017∠ − 12.765°

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Quiz 4ECE 476Name:‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗‗1.A 1000-MVA 20-kV, 60-Hz three-phase generator is connected through a 1000-MVA 20-kVΔ/138-kV Y transformer to a 138-kV circuit breaker and a 138-kV transmission line. Thegenerator reactances are𝑋𝑑′′=0.15,𝑋𝑑′=0.45 and𝑋𝑑=1.8 pu. The transformer series reactanceis 0.1 pu; transformer losses and exciting current are neglected. A 3-phase short-circuit occurson the line side of the circuit breaker when the generator is operated at rated terminalvoltage and at no-load. Determine the substransient current through the breaker in pu andin kA rms ignoring any dc offset.A.The base current on the high voltage side of the transformer is:𝐼𝑏𝑎𝑠𝑒,𝐻𝑉=𝑆3∅√3𝑉𝐿𝐿=1000√3 × 138 = 4.184 𝑘𝑉𝐴𝐼" =𝐸𝑞𝑋𝑑′′+ 𝑋𝑇𝑅=10.15 + 0.1 = 4 𝑝. 𝑢. = 4 × 4.184 = 16.735 𝑘𝐴2.Determine Ybusin per-unit for the circuit below using a 1000 MVA system base. Per unit valuesfor the generators and transformers are given on their own base so you need to first convertsome of them to the system MVA base.G1G2500 MVA, 13.8 kVX”=0.2 pu1000 MVA, 20 kVX”=0.17 pu500 MVA13.8Δ/500 Y kVX=0.12 pu1000 MVA20Δ/500 Y kVX=0.10 puT1T2ΔYYΔ500 kV LineX = 50Ω12Hint:𝑋𝐺1′′= 0.2 ×1000500= 0.4 𝑝𝑢,𝑍𝑏𝑎𝑠𝑒=𝑉𝑏𝑎𝑠𝑒2𝑆𝑏𝑎𝑠𝑒=50021000= 250 Ω,𝑋𝑙𝑖𝑛𝑒=50250= 0.2 𝑝𝑢A.𝑋𝑇1= 0.12 ×1000500= 0.24 𝑝𝑢𝑌11= −𝑗 (10.4+0.24+10.2),𝑌12= 𝑌21= 𝑗10.2,𝑌22= −𝑗 (10.1+0.17+10.2)𝑌𝑏𝑢𝑠= 𝑗 [−6.5655−8.7]