b) A fault occurs at bus 2 of the network shown in Figure Q3. Pre-fault nodalvoltages throughout the network are of 1 p.u. and the impedance of theelectric arc 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.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. 3jXno0.3 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.0Figure 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-2Y=8(iv) Determine the short-circuit fault current for the case when a phase-to-phase fault occurs at bus 2.

Answered: Image /qna-images/answer/4c9b19b6-6e77-470b-b5eb-97e4f49ffdb1.jpg

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. jXa0) J0.1X p.u. jXaa-j0.1X p.u. jX 20 j0.2Y p.u. !3! %3D V = 120° p.u. V 120° p.u. V,-120° p.u. jXa)= j0.2X p.u. JX42-J0.2X p.u. JX40 J0.2Y p.u. !! jXaa)= j0.25 p.u.. A Ma- 70.25 p.u. 3 jXno-J0.3 p.u. %3! jXr j0.2Y p.u. jXra /0.2Y p.u. jXro /0.2Y p.u. jX- /0.1X p.u. jXe= j0.1X p.u. X0) = 10.05 p.u. 0 %3D %3D 0. Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(1) = j0.22 p. u., jXa(2) = j0.22 p.u., and jXaco)= j0.23 p.u. %3D %3D 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 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. jXa0) J0.1X p.u. jXaa-j0.1X p.u. jX 20 j0.2Y p.u. !3! %3D V = 120° p.u. V 120° p.u. V,-120° p.u. jXa)= j0.2X p.u. JX42-J0.2X p.u. JX40 J0.2Y p.u. !! jXaa)= j0.25 p.u.. A Ma- 70.25 p.u. 3 jXno-J0.3 p.u. %3! jXr j0.2Y p.u. jXra /0.2Y p.u. jXro /0.2Y p.u. jX- /0.1X p.u. jXe= j0.1X p.u. X0) = 10.05 p.u. 0 %3D %3D 0. Figure Q3. Circuit for problem 3b). For example, if your student number is c1700123, then: jXa(1) = j0.22 p. u., jXa(2) = j0.22 p.u., and jXaco)= j0.23 p.u. %3D %3D X=2 Y=8 (iv) Determine the short-circuit fault current for the case when a phase-to- phase fault occurs at bus 2.

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 3.47P: Consider the oneline diagram shown in Figure 3.40. The three-phase transformer bank is made up of...

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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.
In developing per-unit circuits of systems such as the one shown in Figure 3.10. when moving across a transformer, the voltage base is changed in proportion to the transformer voltage ratings. (a) True (b) False
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.
Determine the positive- and negative-sequence phase shifts for the three- phase transformers shown in Figure 3.36.
Consider a bank of this single-phase two-winding transformers whose high-voltage terminals are connected to a three-phase, 13.8-kV feeder. The low-voltage terminals are connected to a three-phase substation load rated 2.0 MVA and 2.5 kV. Determine the required voltage, current, and MVA ratings of both windings of each transformer, when the high-voltage/low- voltage windings are connected (a) Y-, (b) -Y, (c) Y-Y, and (d) -.
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.
Consider the three single-phase two-winding transformers shown in Figure 3.37. The high-voltage windings are connected in Y. (a) For the low-voltage side, connect the windings in , place the polarity marks, and label the terminals a, b, and c in accordance with the American standard. (b) Relabel the terminals a, b, and c such that VAN is 90 out of phase with Va for positive sequence.
A bank of three single-phase transformers, each rated 30MVA,38.1/3.81kV, are connected in Y- with a balanced load of three 1, Y-connected resistors. Choosing a base of 90MVA,66kV for the high-voltage side of the three-phase transformer. spify the base for the low-voltage side. Compute the per-unit resistance of the load on the base for the low-voltage side. Also, determine the load resistance in ohms referred to the high-voltage side and the per-unit value on the chosen base.
Three single-phase two-winding transformers, each rated 25MVA,54.2/5.42kV, are connected to form a three-phase Y- bank with a balanced Y-connected resistive load of 0.6 per phase on the low-voltage side. By choosing a base of 75 MVA (three phase) and 94 kV (line-to-line) for the high-voltage side of the transformer bank, specify the base quantities for the low-voltage side. Determine the per-unit resistance of the load on the base for the low-voltage side. Then determine the load resistance RL in ohms referred to the high-voltage side and the per-unit value of this load resistance on the chosen base.
A Cock-croft Walton type voltage multiplier has 7 stages with capacitance all equal 0.07pF. The supply a frequency of to be supplied is 6.5mA. to transformer secondary voltage is 133kV at 250HZ. If the load current The total voltage ripple of the system is: (a) 15.3kv (b) 12.2kv (c) 9.1kV (d) 10.4kv (e) 1.9kv a. d O b. a OC C O d. e O e. b
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. 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) 0 jX(1) = j0.2 p.u. 1 jx(2) j0.2 p.u. 2 jX1(0) = j0.25 p.u. jXT(1) jXT(2) 종 3 j0.1X 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 zero sequence Thévenin equivalent circuits as seen from…
A short transmission line connects a step-up transformer on the source side with a series impedance of j0.5 ohms (referred to the primary) to a step-down transformer on the load side with a series impedance of 50 ohms (referred to the primary). The turn ratio of the step-up transformer is 1:10 and a no-load primary voltage of 23 (line-to-line) kV. The step-down transformer has a turn ratio of 35:2 and it has a Y-connected, balanced load of 0.490 +j 0.0816 ohms connected to its secondary side. A capacitor bank of -j0.171 ohms is added parallel to the load. Assuming an ideal transmission line, the load voltage (line-to-neutral) would be: O A. 7360 V O B. 6.2 kV O C. 12750 V O D. 4806 V O E. None of the other choices are correct