DO NOT WANT AI SOLUTIONS. Thank you P1-Equipment ratings for the three-bus power system shown below are as follows (all per unit valueslisted were determined utilizing the equipment's Voltage and MVA ratings):Generator G1:Generator G2:550 MVA, 13.8 kV, X" = 0.18 per unit (pos, neg, zero)750 MVA, 18 kV, X" = 0.20 per unit (pos, neg, zero)Transformer T1:500 MVA, 13.8 A/500 YkV X = 0.12 per unitTransformer T2:Each 500 kV line:750 MVA, 18 A/500 Y kV, X = 0.10 per unitX1 = 385 OhmsA three-phase bolted short circuit occurs at bus 2, where the prefault voltage is 525 kV. Prefaultload current is neglected.a) Identify per unit impedance of the G1, G2, T1, T2, and the 500 kV line based upon theproper voltage base and 1000 MVAb) Draw the positive, negative, and zero-sequence reactance diagram in per unit on a 1000-MVA.c) Determine the Thevenin reactance in per-unit at the fault for the positive, negative, andzero sequences.d) Determine the subtransient fault current in per unit and in kA rms, ande) contributions to the fault current from generator G1 and G2T1 1G1750038ΔΥj5012YA22G2

a) Per unit impedances on 1000 MVA base (already calculated):G1: 0.327 p.u.G2: 0.267 p.u.T1: 0.24…

Answered: P1-Equipment ratings for the three-bus…

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter9: Unsymmetrical Faults

Section: Chapter Questions

Problem 9.5P

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Equipment ratings for the five-bus power system shown in Figure 7.15 are as follows: Generator G1:Â Â Â Â 50Â MVA,Â 12kV,Â X=0.2 per unit Generator G2: 100Â MVA, 15 kV, X=0.2 per unit Transformer T1: 50 MVA, 10 kV Y/138kVY,X=0.10 per unit Transformer T2: 100 MVA, 15 kV /138kVY,X=0.10 per unit Each 138-kV line: X1=40 A three-phase short circuit occurs at bus 5, where the prefault voltage is 15 kV. Prefault load current is neglected. (a) Draw the positive-sequence reactance diagram in unit on a 100-MVA, 15-kV base in the zone of generator G2. Determine (b) the ThĂ©venin equivalent at the fault, (c) the subtransient fault current in per unit and in kA rms, and (d) contributions to the fault from generator G2 and from transformer T2.
Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of three identical single-phase transformers, each specified by X1=0.24 (on the low-voltage side), negligible resistance and magnetizing current, and turns ratio =N2/N1=10. The transformer bank is delivering 100 MW at 0.8 p.f. lagging to a substation bus whose voltage is 230 kV. (a) Determine the primary current magnitude, primary voltage (line-to-line) magnitude, and the three-phase complex power supplied by the generator. Choose the line-to-neutral voltage at the bus, Va as the reference Account for the phase shift, and assume positive-sequence operation. (b) Find the phase shift between the primary and secondary voltages.
Equipment ratings for the four-bus power system shown in Figure 7.14 are as follows: Generator G1: 500 MVA, 13.8 kV, X=0.20 per unit Generator G2: 750 MVA, 18 kV, X=0.18 per unit Generator G3: 1000 MVA, 20 kV, X=0.17 per unit Transformer T1: 500 MVA, 13.8/500YkV,X=0.12 per unit Transformer T2: 750 MVA, 18/500YkV,X=0.10 per unit Transformer T3: 1000Â MVA, 20/500YkV,X=0.10 per unit A three-phase short circuit occurs at bus 1, where the prefault voltage is 525 kV. Prefault load current is neglected. Draw the positive-sequence reactance diagram in per unit on a 1000-MVA, 20-kV base in the zone of generator G3. Determine (a) the ThĂ©venin reactance in per unit at the fault, (b) the subtransient fault current in per unit and in kA rms, and (c) contributions to the fault current from generator G1 and from line 1-2.
The per-unit equivalent circuit of two transformers Ta and Tb connected in parallel, with the same nominal voltage ratio and the same reactan of 0.1 per unit on the same base, is shown in Figure 3.43. Transformer Tb has a voltage-magnitude step-up toward the load of 1.05 times that of Ta (that is, the tap on the secondary winding of Tb is set to 1.05). The load is represented by 0.8+j0.6 per unit at a voltage V2=1.0/0 per unit. Determine the complex power in per unit transmitted to the load through each transformer, comment on how the transformers share the real and reactive powers.
Three zones of a single-phase circuit are identified in the figure. The zones are connected by transformers T₁ and T2, whose ratings are also shown. Using base values of 100 kVA and 240 volts in zone 1, draw the per-unit circuit and determine the per-unit impedances and the per-unit source voltage. Then calculate the load current both in per-unit and in amperes. Transformer winding resistances and shunt admittance branches are neglected. Zone 1 Zone 2 Vs = 220/0° volts 3---38 T, 30 KVA 240/480 volts M 0.10 p.u. Xoa Xune = 2 fl T T₂ 20 kVA 460/115 volts Xeg = 0.10 p.u. Zone 3 ww Zload = 0.9 - 10.20
Three zones of a single-phase circuit are identified shown in Figure below. The zones are connected by transformers T1 and T2, whose ratings are also shown. Using base values of 33 kVA and 232 volts in zone 1, Find: 1- Draw the per-unit circuit including the per-unit impedances and the per-unit source voltage. 2- Calculate the load current both in per-unit and in amperes (actual or original value). Vs Zone 1 232.940° Vs G. 38 T₁ 30 KVA 240/480 volts Xeq = 0.10 p.u. Zone 2 Xiine = 4 Ω T₂ 20 KVA 460/115 volts Zload = Xea = 0.10 p.u. Zone 3 u 1+j2.2 Ω 2
Three single-phase two-winding transformers, each rated 3 kVA, 220/110volts, 60 Hz, with a 0.10 per-unit leakage reactance, are connected as athree-phase extended D autotransformer bank, as shown in Figure 3.36 (c).The low-voltage D winding has a 110-volt rating. (a) Draw the positive sequencephasor diagram and show that the high-voltage winding has a479.5-volt rating. (b) A three-phase load connected to the low-voltageterminals absorbs 6 kW at 110 volts and at 0.8 power factor lagging. Drawthe per-unit impedance diagram and calculate the voltage and current atthe high-voltage terminals. Assume positive-sequence operation.
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b) A fault occurs at bus 3 of the network shown in Figure Q4. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, transformer and load are given in Figure Q4. V₁ = 120° p.u. V₂ = 120° p.u. V₂ = 1/0° p.u. V₂= 120° p.u. jXj0.1 p.u. JX2) 0.1 p.u. jX0j0.15 p.u. jXn-j0.2 p.u. 1 JX(2)-j0.2 p.u. 2 jX)=j0.25 p.u. JX20-10.15 p.u. jXa(z)-j0.2 p.u. 4 jX2(0)=j0.2 p.u. jXT(1) j0.1 p.u. jXT(2)=j0.15 p.u. jXT(0)=j0.1 p.u. Figure Q4. Circuit for problem 4b). = jXj0.1 p.u. j0.1 p.u. - JX(2) JXL(0) 10.1 p.u. = (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 3. (ii) Determine the positive sequence fault current for the case when a three- phase-to-ground fault occurs at bus 3 of the network. (iii) Determine the short-circuit fault current for the case when a one-phase- to-ground fault occurs at bus…
b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodal voltages throughout the network are of 1 p.u. and the impedance of the electric arc is neglected. Sequence impedance parameters of the generator, transmission lines, and transformer are given in Figure Q3, where X and Y are the last two digits of your student number. V₁ = 120° p.u. V₂ = 120° p.u. jX(1) j0.1Y p.u. jX2)= j0.1Y p.u. jXko) j0.1X p.u. - 0 jX(1) = j0.2 p.u. 1JX(2) = 0.2 p.u. 2 jX1(0) = j0.25 p.u. jX2(1) j0.2 p.u. V₁=1/0° p.u. jX(2(2) = j0.2Y p.u. jX2(0) = j0.3X p.u. = V₂ = 120° p.u. jXT(1) j0.1X p.u. jXT(2) j0.1X p.u. JX3(1) j0.1Y p.u. JX3(2)=j0.1Y p.u. jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u. 0- = 3 = Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(n) = j0.13 p. u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u. 4 (i) Assuming a balanced excitation, draw the positive, negative and zero sequence Thévenin equivalent circuits as seen from bus 4. (ii)…
FIGURE Q2 shows a four bus three phase power system. The generators, transformers and transmission line per unit reactance are given as follows: Generator G1, G2: XG= 0.05 pu Transformers T1, T2: XT= 0.13 pu Transmission Line 2 – 3: XL = 0.50 pu Bus 1 Bus 2 Bus 3 Bus 4 Line GI T1 G2 FAULT FIGURE Q2 Evaluate the per unit fault current, fault MVA and voltage at all buses if a three phase bolted fault occurs at Bus 2.
b) The single line diagram of a power system is shown in Figure Q2.1 including generator and transformer winding connection and earthing details. The parameters for this system have been calculated on a common 100 MVA base and are given in Table Q2.1. All resistances and shunt susceptances are neglected. This system experiences a single line to ground fault at a point F on line L1. The point F is at a distance d from Bus 4 along the line LI. The total length of the line L1 is 50 km. Note that the location of d is not drawn to scale in Figure Q2.1. The fault current at the fault point F is measured to be 6.106 kA. G1 G1 G2 i) G2 T1 T2 T3 L1 11) 2 Rated voltage (kV) 20 20 20/33 20/33 33/11 11 T1 ΔΕ T2 T3 Figure Q2.1 Table Q2.1 Zero sequence reactance Xo (p.u.) 0.20 0.15 0.12 0.20 0.20 1.20 Positive sequence reactance X₁ (p.u.) 0.30 0.25 0.12 0.20 0.20 0.50 UNIVERSITY OF LIVERPOOL L1 (l-d) 5 Negative sequence reactance X₂ (p.u.) 0.35 0.30 0.12 0.20 0.20 0.50 Determine the zero, positive,…

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