Solve problems 5.2 in detail and thank you 5.1 Determine the three zone settings for the relay Rab in the system shown in Figure 5.26. Thesystem nominal voltage is 138 kV, and the positive sequence impedances for the variouselements are given in the figure. The transformer impedance is given in ohms as viewedfrom the 138 kV side. Assume that the maximum load at the relay site is 120 MVA, andselect a CT ratio accordingly. The available distance relay has zone 1 and zone 2 settingsfrom 0.2 to 10 2, and zone 3 settings from 0.5 to 40 2, in increments of 0.1 2. The angleof maximum torque can be adjusted to 75° or 80°. Remember that the zone 3 of the relaymust back up the line BC, as well as the transformer.B(2+j50)A(3 + j40)++(0+j9)Rabwww.wwwFigure 5.26 System for problem 5.15.2 Consider the system shown in Figure 5.13. Assume that F is at the end of the line. If themaximum load in the line is 10 MVA, what are the zone 1 and zone 2 settings for this line?You may assume that the available distance relay is the same as that in problem 5.1.

Answered: Image /qna-images/answer/61aed5c6-cd9f-48e4-a4d0-f1ca94698dda.jpg

Answered: 5.1 Determine the three zone settings…

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

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.20MCQ

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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)…
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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…
plz help, explain in detail and correct. need typed solution
The base current and base voltage of a 345 kV system are chosen to be 3000A and300kV respectively. Determine the base impedance and the per-unit voltage for thesystem.
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