Example In the power system shown in Fig. three phase fault occurs at P and the faulty line was opened a little later. Find the power output equations for the pre-fault, during fault and post-fault conditions. 0.16 X₁ = 0.28 3 G |E| = 1.25 p.u. 0.16 0.16 0.24 0.16 38 0.24 |V|= 1.0 p.u. 0.16 Values marked are p.u. reactances Fig
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- Calculate the fault current in each phase for a L-G fault occurs at machine-2 for the system shown in below fig.below.AIII the reactances are given in p.u on 100 MVA base. E- 16 *- j0.2 Line X2 = j0.22 X x2 j0.11 Xp = j0.33 - j0.16 Xe = 0.15 X2 0.17 Xo-j0.06 x, = x2 = j0.10 X1 =X2 = Xo = 0.10 None -j2.8622 -j2.6228 -j2.2668A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is ignored. A fault takes place. reducing the maximum power transferable to 0.5 pu whereas before the fault, this power was 2.0 pu and after the clearance of the fault, it is 1.5 pu. By the use of equal area criterion, determine the critical clearing angle.Which of the following statements is/are true? A In purely resistive circuit, the current lags voltage. (B In purely capacitive circuit, the voltage lags current. In purely resistive circuit, the current is in phase with voltage. D In purely inductive circuit, the current leads voltage.
- A 400 V, 3-phase alternator is subjected to an unsymmetrical fault. The sequence currents are Ia1= -j10 A, Ia2 = j6 A and Iao = j4 A. The sequence impedances are Z1 = j10 ohm, Z2 = j5 ohm and z0 = j7.5 ohm. Find the fault current and the type of fault. Also find the terminal voltages (phase and line) of the generator.A parallel connection of a RL branch with a C branch is connected across a 100V AC mains. At first R = 10 ohms, L = 20mH and frequency of 1000rad / s, the current measured is 2.2361A at 89.44 leading power factor. A) Determine the initial capacitance. B) However, a fault occurs on the capacitor branch making its capacitance 20% lower and a resistance of 5 ohms is detected. If this faulty circuit is rerun but at a frequency of 500 rad / s, determine the new current that will flow through the circuit.Voltage sags due to short circuit and earth faults are the cause for the vast majority of equipment problems. Electronics equipment become more susceptible to voltage sag, hence the companies experience production stoppages. Voltage sag analysis is very complex issue. Based on voltage sag randon factors describe on it.
- A generator operating at 50Hz delivers 1 pu power to an infinite bus through a transmission circuit inwhich resistance is ignored. A fault takes place reducing the maximum power transferable to 0.5 puwhereas before the fault, this power was 2.0pu and after the clearance of the fault, it is 1.5 pu. By theuse of equal area criterion, determine the critical clearing angle.Calculate the fault current in each phase for a L-G fault occurs at machine-2 for the system shown in below fig.below.All the reactances are given in p.u on 100 MVA base. E= 16º xj = j0.2 Line X2 = ]0.22 X, = x, = j0.11 Xp = j0.33 xe = 30.15 - j0. 16 X = J0.17 Xn = j0.06 X, = x2 = j0.10 X =x2 = Xg = 0.10 -j2.8622 -j2.6228 None -j2.2668 ele eleReconsider Problem 3.29. If Va,VbandVc are a negative-sequence set, how would the voltage and current relationships change? (a) If C1 is the complex positive-sequence voltage gain in Problem 3.29 and (b) if C2 is the negative sequence complex voltage gain, express the relationship between C1andC2
- In a power system with negligible resistance, the fault current at a point is 800 p. u. The series reactance to be included at the fault point to limit the short circuit to 5.00 p. u. is (a) 3.00 p.u (b) 0.200 p.u () 0.125p.u8 (d)0.075 p.uQ3/ For the system shown in figure. The generator is supplying 1.0 pu MW power under prefault condition. When three-phase fault occur at point F. What should the value of fault reactance must be to make the critical clearing angle equal to 58.9. Sc j0.1 j0.1 10.21 BE j0.38 38 j0.21 IEgl-12 pu 10.19 F 10.19 IVI-1.00 pu j0.1 j0.1 Q4/ Figure below shows the one-line diagramCalculate the fault current in each phase for a L-G fault occurs at primary of transformer-2 for the system shown in below fig.All the reactaces are given in p.u on 100 MVA base E= 1/0º x = j0.2 Line X =x, = j0.11 Xp = j0.33 X2 = ]0.22 xe = 30.15 - j0.16 X2 = J0.17 Xo = j0.06 X, = x2 = j0.10 X1 =x2 = xg = 0.10 -j2.7831 -j2.3781p.u None -j2.2668 p.u ele