12 he data in table below for the system shown, if there is a 3phase fault occurs at bus 3,find the fault ent using bus impedance matrix. MVA voltage 33kv clement XI (p.u) G1 100 0.2 G2 100 33kv 0.2 TI 100 33/220 kv 0.05 T2 100 331220 kv 0.05 LI 100 220 kv 0.1 12 100 220 kv 0.1 L3 100 220 kv 0.1 Ti 2) Tz (G) Le 3) 3.
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Q: b) The one-line diagram for a simple power system is given in Figure 1(b). į. Draw the impedance…
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Q: Q1/The data in table below for the system shown, if there is a 3phase fault occurs at bus 3,find the…
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Q: A single line diagram of a power system is shown in Fig.3.1. All the sequence reactances of the…
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- 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 ..............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. +j0.182The LLLG fault occurs at bus-8with pre-fault voltage as 1.02 p.u. If the positive sequence impedance at bus-8 is j0.15 and positive sequence current is -j1.125. which one of the following values will be the value of zero sequence voltage at bus 8? Select one: a. 1.3125 O b. 0 O c. 0.7587 O d. 0.3125
- 1 Xine = j0.30 %3D G M X" = j0.15 X" = j0.20 Xiine = j0.30 Xiine = j0.30 %3D 1. Find the Thévenin equivalent looking into bus 3. 2. Calculate the subtransient fault current for a bolted three-phase fault at bus 3. Pre-fault voltage is VF 1Z°, and pre-fault current is neglected. 3. Distribute the fault current between the motor and generator.The one line diagram of a simple 3-bus power system shown below. All impedances are expresses in per unit on a common base. A three phase fault occurs at bus 3 through a fault impedance Z =j 0.19 pu, calculate the per unit fault current. A. 1.5 p.u. B.2.5 p.u. G₁ j 0.05 j0.75 j 0.3 3 C. 5.0 p.u. D. 3.0 p.u. G₂ j 0.075 j0.45 2Q3: - A three-phase fault occurs at bus-1 of the network shown in figure below. The pre-fault voltage is 1.05 PU and the pre-fault load current is neglected. (45%) i) determine the bus impedance matrix. ii) calculate the sub-transient fault current and the contribution to the fault current from the transmission line. T1 T2 T.L j0.1 PU 10.1 PU j0.105 PU G jXdg j0.15 PU M jXdm $0.2 PU
- j0.2 For the following power system the impedance matrix is given below. Initially all bus voltages are 1 pu. Then a three-phase fault occurs at bus (2). j0.25 j0.2 (3) (1) j0.1 : (a) What is the fault current? (2) j0.125 j0.1 (b) What are the post-fault bus voltages? 3. 0.0625 0.02083 0.02083' Zbus 0.02083 0.07044 0.02282 0.02083 0.02282 0.060915.10 The zero sequence current of a generator for line to ground fault is /2.4 p.u. Then the current through the neutral during the fault is (a) j2.4 p.u. (c) j7.2 p.u. (b) j0.8 p.u. (d) j0.24 p.u.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…
- Q1. Given Zo = 0.3L60°, Z¡ = 0.17L80° and, Z2 = 0.45L120°. Compute the fault current and voltages for a Double Line-to-Ground Fault. Note that the sequence impedances are in per- unit. This means that the solution for current and voltage will be in per-unit. [1] Note: Formulas for the reference For Double line to ground fault the positive, negative and zero sequence circuits are connected in parallel The sequence networks are interconnected, as shown in Fig. 8.9 Zo To Because the sequence currents sum to one node, it follows that Vo I, =-(I, +1,) The current I, is the voltage drop across Z, in series with the parallel combination of Z, and V, I = Z, + Z, +Z, V2 Fig 8.9Thevenin equivalents of the zero, positive, and negative sequence circuits, respectively, seen from the faulty busbar of a power system are Za0 = j0.265 p.u., Za1 = j0.45 p.u. and Za2 = j0.38 p.u. given as. Pre-fault voltage Vf = 120° p.u. and fault impedance Zf = j0.012 p.u. given as. According to this; a) In case of single phase-to-earth short-circuit fault, with component and phase currents Calculate the phase voltages at the fault point. b) Repeat the steps in step a for the phase-to-phase short circuit fault that will occur between phases b-c.The bus impedance matrix of a 3-bus system is: [j0.2295 j0.1494 j0.1506] j0.1494 j0.1954 j0.1046 Lj0.1506 j0.1046 j0.1954] Zbus Draw the reactance diagram, and find the fault current if a 3-ph fault occurs at bus 1. Also find the voltage at each bus during the fault and the current contributions to the fault from buses 2 and 3.