G.M 5. The following is the system data for a load flow solution:The line admittancesBus CodeAdmittance1-22-j81-j40.666-j2.6641-j42-j81-32-32-43-4Neglect line charging admittances.The schedule of active and reactive powers:Bus CodePVRemarks11.06Slack1+ j01+ j01+ j00.50.2PQ30.40.3PQPQ40.30.1Determine the voltages attheendof the1st iteration by using G-Smethod

From the information drawing the single line diagram- Now calculate elements of Y busY11=sum of…

Answered: Bus Code V Remarks 1 1.06 Slack 1+ j0…

Bus Code V Remarks 1 1.06 Slack 1+ j0 1+ j0 1+j0 0.5 0.2 PQ PQ 3 0.4 0.3 4 0.3 0.1 PQ Determine the voltages at the end of the 1st iteration by using G-S method

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

Chapter2: Fundamentals

Section: Chapter Questions

Problem 2.38P

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The figure below shows the one-line diagram of a four- bus power system. The voltages, the scheduled real power and reactive powers, and the reactances of transmission lines are marked at this one line diagram (The voltages and reactances are in PU referred to 100 MW base. The active power P2 in MW is the last three digits (from right) of your registration number (i.e for the student that has a registration number 202112396, P2 =396). [10] Starting from an estimated voltage at bus 2, bus 3, and bus 4 equals V2 (0) = 1.15<0°, V3 = 1.15 < 0°, V4 1.1< 0°. 1- Specify the type of each bus and known & unknown quantities at each bus. 2- Find the elements of the second row of the admittance matrix (i.e. [Y21 Y22 Y23 Y24]). 3- Using Gauss-Siedal fınd the voltage at bus 2 after the first iteration. 4- Using Newton-Raphson, calculate: |- The value of real power (P2), at bus 2 after the first iteration. Il- The second element in the first row of the Jacobian matrix after the first iteration. 2 P2…
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…
5.10 The zero sequence current of a generator for line to ground fault is j2.4 p.u. Then the current through the neutral during the fault is (a) j2.4 p.u. (b) j0.8 p.u. (d) j0.24 p.u. (c) i7.2 p.u.
400Ω A k=0.9 600Ω B k=0.8 66Ω 600Ω
In a 7-bus system, the double line fault is occurred at bus -5. During fault the positive sequence potential difference between the bus-4 and bus-1is found to be 0.0175. what is the line impedance which is connected between bus-4 and bus-1 if the current flowing through transmission line is -j 0.148? Select one: O a. -jo.182 O b. +j0.1182 O c. -jo.1182 O d. +j0.182
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.
Q2\ The one-line diagram of a simple power system is shown in figure below. All impedances are expressed in per unit (pu) on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage. A three-phase fault occurs at bus 1 through a fault impedance of Zf = j0.08 per unit. Using Thevenin's theorem obtain the impedance to the point of fault and the fault current in per unit. Determine the bus voltages and line currents of generators during fault. X₁ = = 0.1 XT-0.1 3 1 to ojo XL=0.2 2 040 X₁ = 0.1
b) The one-line diagram for a simple power system is given in Figure 1(b). į. Draw the impedance diagram of the system. i. Construct the bus admittance matrix, Xeus and bus impedance matrix, Zous of the system. I. Evaluate the fault current, bus voltages and line current during the flow when a balanced three-phase fault occurs on bus 1. bus 1 bus 2 j65 N G G 25 MVA 25 MVA 13.8 kV 15% 13.2/69 kV 11% 25 MVA 69/13.2 kV 11% 15 MVA 13 kV 15% Figure 1(b) m
Derive an expression of per unit impedance of a given base MVA and base KV in term of new base MVA and new base KVA
The system values are given below. The bus 1 voltage after fault = 1.5 p.u The bus 2 voltage after fault = 1.2 p.u The line admittance between bus 1 and bus 2 (Y12 ) is = 0.8 p.u The post fault current current flow between bus 1 and 2 is ..............
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)…