10.11 A line to line fault occurs between two phases at the remote end of line of the system in problem 10.10. Find (a) fault current (b) line to neutral voltage of the healthy phase at fault point. [(a) 1.545 kA (b) 17.5 kV]
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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…Two alternators rated 30 MVA and 20 MVA have per unit reactance of 0.30 and 0.50 respectively are connected to a common 12-kV bus bar. Solve the fault current if a symmetrical 3-phase fault occurs at the bus bar.estion 9 The fault current level of a circuit breaker used in a transmission line with phase voltage 2.2 kV and breaking capacity 50 MVA is yet ......kA. vered -ked out of Flag question estion 10 The voltage between lines in a transmission line is the maximum value of the sine waveform. yet vered Select one: -ked out of O True Flag question O False
- In the shown below power system, both generators G1 and G2 emf’s are equal.All the system data are shown on the system diagram.a) Draw the system diagram in per unit using (MVA)base = 75 and (KV)base =20 KV in the generators side.b) Calculate the three-phase balanced short circuit current (If) at busbar 3 inAmpere and the fault level (MVA)sc in MVA.c) If the fault at busbar 3 is a line-to-line fault, calculate the fault current inAmpere and the fault level (MVA)sc in MVA.A 3-phase 33 kV, 37.5 MVA alternator is connected to a 33 kV overhead line which develops an earth fault on one phase at the remote end. The positive, negative and zero sequence reactances of the alternator are 0.18, 0.12 and 0.1 per unit of its rating while those of the line are 6.3, 6.3 and 12.6 ohm per conductor. Find (a) fault current (b) line to neutral voltages at alternator terminals. = 13 20KV an (a) I 1.555 kA (b) V [Von = 18 -119.16° kV, V, = 18 119.16° kVJ %3D %3D enA 25 MVA, 13.2 kV alternator with solidly grounded neutral has a subtransient reactance of 0.40 p.u. The negative and zero sequence reactances are 0.20 and 0.3 p.u. respectively. Find the fault current when, (i) A single line to ground fault occurs at the terminals of an unloaded alternator and (ii) ALL fault occurs.
- A hydro powered 50 MVA generator, with a synchronous reactance 0.33 p.u., delivers 40 MW over a transmission line of 0.15 p.u. reactance to an infinite system. A 3-phase short circuit transient fault occurs on the busbar between the generator and transmission line. Given that the generator emf is 1.4 p.u. prior to and during the fault and all reactances are to a 50 MVA base: Sketch the single line schematic diagram of the system. Determine the maximum load angle the generator can swing to without losing stability. i) ii) iii) iv) Sketch the corresponding power-angle curve for the system showing the accelerating and decelerating areas. Calculate the critical clearing angle of the system to keep its stability.Q5: A generator is connected through a transformer to a synchronous motor. Reduced to the same base, the per unit subtransient reactances of the generator and motor are 0.15 and 0.35, respectively, and the leakage reactance of the transformer is 0.1 per uint. A three-phase fault occurs at the terminals of the motor when the terminal voltage of the generator is 0.9 per uint and the output current of the generator is 1 per unit at 0.8 power factor leading. Find the subtransient current in per unit in the fault, in the generator, and in the motor. Use the terminal voltage of the generator as the reference phasor and obtain the solution (a) by using the internal voltages of the machines and (b) by using Thevenin's theorem.Q.3 When a line-to-ground fault occurs, the current in faulted phase 'a' is 100A. The zero-sequence current in phase 'c' is r
- 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 3A 10,000 kVA, 6.9 kV generator has percentage reactance to positive, negative and zero sequence currents 10%, 7% and 3% respectively and its neutral is grounded solidly. The generator is open circuited and excited to its rated voltage. Calculate the line currents (a) when a single line to ground fault develops on phase 'a' of the generator and (b) when the generator phases b and c are shorted.Consider the system shown in the single-line diagram of Figure (3). All reactances are shown in per unit to the same base. Assume that the voltage at both sources is 1 p.u. a Find the fault current due to a bolted- three-phase short circuit at bus 3 b- Find the fault current supplied by each generator and the voltage at each of the buses 1 and 2 under fault conditions 0.06 p.u. 0.2 p.u. 0.04 p.u. 0.25 p.u. 0.2 p.u. 0.2 p.. 0.2 p.u. 0.06 p.u. 0.25 p.u. Figure (3) Single-line diagram ele ver ele 888 ele 0.06 p.u. 0.25 p.u.