Four-bus power system shown in Fig. 1 are as follows: Generator G1: 200 MVA, 7.2 kv,...
Transformer TI : 50 MVA, 10 kV Y/138 kV Y, X=0.10 per unit; Transformer T2: 100 MVA, 15 kV D/138 kV Y, X-0.10 per unit; Each 138-kV line: X1-400 A three-phase short circuit occurs at bus 5, where the prefault voltage is 15 kV. Prefault load current is neglected. (a)Draw the positive-sequence reactance diagram in per-unit on a 100-MVA, 15-kV base in the zone of generator G2. Determine: (b) the The'venin equivalent at the fault, (c) the subtransient fault current...
1) Consider the power system shown in Fig. 1. Use a power base of 500 MVA to calculate the fault current in amperes for a double line-to-ground fault at bus B. G: 500 MVA, 13.8 kv, xa = 0.2 p.M., X2 = 0.2 p.j. and x = 0.1 p.u. G2:600 MVA, 26 kv, xa = 0.15 p.u., X2 = 0.15 p.u. and X, = 0.1 p.u. G3:400 MVA, 13.8 kv, x, = 0.2 p.u., x2 = 0.2 p.u. and x...
Please show all the clearly step Y11 ist j30 and Y44 isnt -j12.85 Consider the 4-bus power system shown in Fig. 1. The system parameters are given below: 50 MVA, 20 kV, X-2090 40 MVA, 20 kV, X-20%, X, = 5% 50 MVA, 20 kV Δ /110 kV Ý, X= 1090 50 MVA, 20 kV MI 10 kV Ý, X= 10% Xi-24.2 Ω Generator G: Motor M: Transformer T1 : Transformer T2 : Transmission line: 3 4 T2 nu)M Fig....
The equipment ratings for a five bus system are given as Generator G1: 50 MVA, 12 kV, Xd ’’=X2=0.20, X0= 0.10 per unit Generator G2: 100 MVA, 15 kV, Xd ’’=0.2, X2=0.23, X0= 0.10 per unit Transformer T1: 50 MVA, 10 kV (Y)/138 kV (Y), X=0.10 per unit Transformer T1: 100 MVA, 15 kV (∆)/138 kV (Y), X=0.10 per unit Each 138 kV line: X1=40 Ohms, X0=100 ohms (1) Draw out the zero-, positive-, and negative- sequence reactance diagrams for the original system using a 100-MVA,...
The ratings of the components shown in the one-line diagram are G1: 25 MVA, 13.8 kV, x-0.15 pu G2:15MVA, 13 kV, x = 0.1 5 pu. TI : 25 MVA, 13.2/69 kV, x-0. I 1 pu T2: 25 MVA, 69/13.2 kV,x-0.220 pu Transmission line: j65 ohms/pha bus 2 BE 165Ω ISMVA e ratings of generator 1 as base valu 25MVA 13.8 kV 1 5% 69113.2 kV13kV 1 1% 13.2169k 1 1% 1- Draw the reactance diagram. 2- Find the Y-bus...
Assuming there is a FAULT at BUS 3, Determine the thevenin equivalent of each series network as viewed from the fault bus. Given: -Prefault voltage is 1.0 per unit -Prefault load currents and delta-wye transformer phase shifts are neglected Synchronous generators G1 1000 MVA 15 kVX"-X2 0.18, Xo 0.07 per unit G2 1000 MVA 15 kV X: X, = 0.20, X,-0.10 per unit G3 500 MVA 13.8 kV X: X,-0.15, X,-0.05 per unit G4 750 MVA 13.8 kV X,-0.30, X,-0.40,...
3.13 A single-line diagram of a three-phase power system is shown in Fig. 3.51. The ratings of the equipment are shown below Generator G: 100 MVA, 11 kV, Xi -X2-0.20 pu, Xo -0.05 pu Generator G2 : 100 MVA, 20 kV, Xi=X2=0.25 pu, Xo=0.03 pu, X,,-0.05 pu Transformer T: 100 MVA, 11/66 kV, Xi -X2-Xo 0.06 pu Transformer T2: 100 MVA, 11/66 kV, Xi-X2 = Xo 0.06 pu Line: 100 MVA, X,-X2 = 0.15 pu, Xo = 0.65 pu A...
The following are the answers: Autumn 2013 T1 G1 5 kV 50 MVA 5/20 kV 50 MVA 20 kV 50 MVA X 0.1 p.u Figure A13 In the system shown, per-unit series equivalent impedances are shown for each element. Before a balanced 3-phase fault occurs in the location shown, the pre-fault current is at its rated value at 90 % lagging power factor flowing from generator G1 The generator is operating at full voltage. How much current in kA is...
A single line diagram of a power system is shown in Fig. 2. The system data with equipment ratings and assumed sequence reactances are given the following table. The neutrals of the generator and A-Y transformers are solidly grounded. The motor neutral is grounded through a reactance Xn 0.05 per unit on the motor base. Assume that Pre-fault voltage is takin as VF-1.0 ,0° per unit and Pre- fault load current and Δ-Y transformer phase shift are neglected In the...
3) The single-line diagram of a three-phase power system is shown in Fig. 1. Equipment ratings are given as follows: G1 1,000 MVA, 15.0 kV, 20.18, o 0.07 pu G2 : 1,000 MVA. 15.0 kV, 攻=エ1 =エ2 = 0.20, ro = 0.10 pu G3 : 500 MVA, 13.8 kV. 1" = 띠 z2 = 0.15, zo 0.05 pu G4 : 750 MVA, 13.8 kV. ェd =ェ1 = 0.30, T2 = 0.40 ro = 0.10 pu Ti : 1,000 MVA. 15.0Δ/765Y...