b) A fault occurs at bus 4 of the network shown in Figure Q3. Pre-fault nodalvoltages throughout the network are of 1 p.u. and the impedance of the electricarc is neglected. Sequence impedance parameters of the generator,transmission lines, and transformer are given in Figure Q3, where X and Y arethe last two digits of your student number.jX(1) j0.1Y p.u.jX2)= j0.1Y p.u.jXko) = j0.1X p.u.V₁ = 120° p.u. V₂ = 120° p.u.(i)(ii)0jX(1) = j0.2 p.u.1 jx(2) j0.2 p.u. 2jX1(0) = j0.25 p.u.jXT(1)jXT(2)종 3j0.1X p.u. JX3(1)j0.1Y p.u.j0.1X p.u. JX3(2)j0.1Y p.u.jXT(0) j0.1X p.u. JX3(0)=j0.15 p.u.0=x = 1,jX2(1) j0.2Y p.u. V₁=1/0° p.u.jX(2(2) = j0.2Y p.u.jX2(0) = j0.3X p.u.=V3 = 120° p.u.Figure Q3. Circuit for problem 3b).For example, if your student number is c1700123, then:y = 7==jXa(r) = j0.13 p.u., jXa(z) = j0.13 p. u., and jXa(o) = j0.12 p. u.Assuming a balanced excitation, draw the positive, negative and zerosequence Thévenin equivalent circuits as seen from bus 4.4Determine the positive sequence fault current for the case when a three-phase-to-ground fault occurs at bus 4 of the network.

x=1y=7Note: reactance of element right to bus 4 is not given. Hence we will not consider that in…

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. 10.av

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. 10.av

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

Chapter3: Power Transformers

Section: Chapter Questions

Problem 3.52P

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Three single-phase, two-winding transformers, each rated 450MVA,20kV/288.7kV, with leakage reactance Xeq=0.10perunit, are connected to form a three-phase bank. The high-voltage windings are connected in Y with a solidly grounded neutral. Draw the per-unit equivalent circuit if the low-voltage windings are connected (a) in with American standard phase shift or (b) in Y with an open neutral. Use the transformer ratings as base quantities. Winding resistances and exciting current are neglected.
Three single-phase two-winding transformers, each rated 3kVA,220/110volts,60Hz, with a 0.10 per-unit leakage reactance, are connected as a three-phase extended autotransformer bank, as shown in Figure 3.36(c). The low-voltage winding has a 110 volt rating. (a) Draw the positive-sequence phasor diagram and show that the high-voltage winding has a 479.5 volt rating. (b) A three-phase load connected to the low-voltage terminals absorbs 6 kW at 110 volts and at 0.8 power factor lagging. Draw the per-unit impedance diagram and calculate the voltage and current at the high-voltage terminals. Assume positive-sequence operation.
b) A fault occurs at bus 2 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. JX20 /0.1X p.u. jXa2) 0.1X p.u. JX20 j0.2Y p.u. V,= 120° p.u. V, 120° p.u. V, 120° p.u. jX4-70.2X p.u. jX2 j0.2X p.u. jX o 0.2Y p.u. jXncay J0.25 p.u. jXna J0.25 p.u. 3 jXno0.3 p.u. jXTu) /0.2Y p.u. jXra j0.2Y p.u. - j0.2Y p.u. Xp-10.1X p.u. jXa j0.1X p.u. jXp0)- j0.05 p.u. 0 Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXac1) = j0.22 p.u., jXac2) = j0.22 p.u., and jXaco) = j0.23 p. u. X-2 Y=8 (iv) Determine the short-circuit fault current for the case when a phase-to- phase fault occurs at bus 2.
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)…
Q2. The single-line diagram of a simple three-bus power system is shown in Figure-2. Each generator is represented by an emf behind the sub-transient reactance. All impedances are expressed in per unit on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage with their emfs in phase. A three-phase fault occurs at bus 3 through a fault impedance of Zf = j0.19 per unit. (i) Using Th'evenin's theorem, obtain the impedance to the point of fault and the fault current in (ii) Determine the bus voltages per unit. ) j0.05 j0.075 j0.75 2 j0.30 j0.45 Figure-2: Single line diagram of the power system network for Q2 3
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…
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 in Figure. The zones are connected bytransformers T1 and T2, whose ratings are also shown.Using base values of 30 kVA and 240 volts in zone 1:a. draw the per-unit circuit and determine the per-unit impedances and the per-unitsource voltage.b. calculate the load current both in per-unit and in amperes. Transformer windingresistances and shunt admittance branches are neglected.
Description In the particular case of figure below derive both the critical clearing angle and the critical clearing time. P, = Pmaz sin d Pm A1 do der Smar A generator having H = 6.R MJ/MVA is delivering power of 1.0 per unit to an infinite bus through a purely reactive network when the occurrence of a fault reduces the generator output power to zero. The maximum power that could be delivered is 2.5S per unit. When the fault is cleared, the original network conditions again exist. Determine the critical clearing angle and critical clearing time. (Roll=PQRS)
For the circuit shown, if a symmetrical three-phase fault occurs in node 2, determinea. The short-circuit current in all three phases. b. The voltage at node 1 during the fault.c. The current in generator 1 during the fault. Neglect the prefault currents and consider the reference voltage of the system at 10√3 ??.
6. A 3 phase fault occurs at point F as shown in figure. Determine the critical clearing anglefor the system. The generator is delivering 1.0 p.u power under prefault conditio
8. A 1000-kV and a 500-kVA, 1-phase transformers are connected to the same bus-bars on the primary side. The secondary e.m.fs at no-load are 500 and 510 V respectively. The impedance voltage of the first transformer is 3.4% and of the second 5%. What cross-current will pass between them when the secondaries are connected together in parallel? Assuming that the ratio of resistance to reactance is the same in each, what currents will flow in the windings of the two transformers when supplying a total load of 1200 kVA. [(a) 290 A (b) 1577 and 900 A]