By using the Bus admittance matrix that mentioned below, 20 j50 Ybus 10+ j20 10+ j20 26 j52 10+ j30 16+j32 10+ j30 16+j32 26 j62 (i) Draw the system and mark the admittance values on the network. (ii) If the scheduling loads are S2 = 400+j250 MVA. P3 -200 MW. Slack bus voltage V₁-1.05 P.U, V3 = 1.04 P. U. determine the phasor values of V2 and V3 by using Newton-Raphson method. Perform one iteration.
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- Equipment ratings for the five-bus power system shown in Figure 7.15 are as follows: Generator G1:    50 MVA, 12kV, X=0.2 per unit Generator G2: 100 MVA, 15 kV, X=0.2 per unit Transformer T1: 50 MVA, 10 kV Y/138kVY,X=0.10 per unit Transformer T2: 100 MVA, 15 kV /138kVY,X=0.10 per unit Each 138-kV line: X1=40 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 unit on a 100-MVA, 15-kV base in the zone of generator G2. Determine (b) the Thévenin equivalent at the fault, (c) the subtransient fault current in per unit and in kA rms, and (d) contributions to the fault from generator G2 and from transformer T2.The sample large power system network data's are given below, The total number of buses is 5 Three-phase short circuit fault subjected at the bus 5 The initial voltage of the faulted bus is 1.0 p.u The Zbus matrix element Z55 is 0.704 p.u Fault impedance Zf= 0.33 p.u Fault current (If )in p.u ..........Write current equations and determine the admittance matrix of the network Determine the load voltage VL at the 5th Bus-Bar, the current of the Load IL and the current of each generator.
- Refer to the 5-bus data given below. Sbase = 400 MVA, the base voltage is 15.0 kV for buses 1 and 3, and 345 kV for buses 2, 4, 5. Note that 1-5 and 3-4 are 15/345 kV transformers.a) Draw a one-line diagram of the power system.b) Find the elements Y23, Y43, Y11 Y33 of the system bus admittance matrix.2. For the 3-bus network shown in figure, the impedances indicated are in per unit. a) Draw pu admittance diagram and obtain the bus admittance matrix Ybus for the network. b) Find the source voltages Eai and Ecz so that buses 1 and 2 have the voltages V = 120°, V2 = 1.05490° X= 0.20 pu X = 0.20 pu X= 0.36 pu Xo = 0.36 pu Xo 0.15 pu X p= 0.36 pu X- 0.36 pu X - 0.3 pu Load %3Dpls solve in less than 30 min pls
- Following figure shows the one-line diagram of a two bus system. Take bus 1 as slack bus, bus 2 as load (PQ) bus. Neglect the shunt charging admittance. Obtain the bus admittance matrixYBUs and find V₂ and 62, power flows and line losses using FDLF method. All the values are given in per unit on 100MVA base. Use a tolerance of 0.001 for power mismatch. 1 Z12= 0.12+10.16 Slack bus V₁ 1.0/0⁰ pu 2 PL2=1.0pu Q12=0.5puQ2. Figure Q2 shows the single-line diagram. The scheduled loads at buses 2 and 3 are as marked on the diagram. Line impedances are marked in per unit on 100 MVA base and the line charging susceptances are neglected. a) Using Gauss-Seidel Method, determine the phasor values of the voltage at load bus 2 and 3 according to second iteration results. b) Find slack bus real and reactive power according to second iteration results. c) Determine line flows and line losses according to second iteration results. d) Construct a power flow according to second iteration results. Slack Bus = 1.04.20° 0.025+j0.045 0.015+j0.035 0.012+j0,03 3 |2 134.8 MW 251.9 MW 42.5 MVAR 108.6 MVARA network consisting of a set of generator and load buses is to be modeled with a DC power flow, for the sake of conducting a contingency analysis. The initial flows calculated with the DC power flow give the following information: f°2-4 = - 65.3 MW and fº4-5 = 13.6 MW. The following values of LODF and PTDF factors are given: PTDF54,2-4 = -0.2609, LODF2-4,4-5 = -0.6087. Calculate the contingency flow on line 2-4 due to outage of line 4-5. Select one: O a. -75.5MW O b. None of these O c. -68.85MW O d. -73.58MW O e. 75.5MW O f. -61.75MW
- 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) For a 4-bus 3-phase, 100 kV, 50 Hz power system, the following Y-bus and corresponding Z-bus matrix are obtained using bases of 100 MVA and 100 kV. [- j28.333 j5 j6.667 j10 j5 - j28.333 j10 j6.667 Ybus p.u. j6.667 j10 - j16.667 j0 j10 j6.667 j0 - j16.667 [ j0.0903 j0.0597 j0.0719 j0.0780 Zbus j0.0597 j0.0903 j0.0780 j0.0719 pu. j0.0719 j0.0780 j01356 j0.0743 Lj0.0780 j0.0719 j0.0743 j0.1356] A symmetrical three-phase fault occurs at bus 4. By assumming prefault voltages to be at nominal values and prefault current to be zero, determine fault currents (in A) at bus 4 and voltages (in V) at all buses during fault if the fault is: (i) solid (bolted) fault. (ii) through reactance of 10 Q to the ground. (iii) Give comment based on your observation on the differences between results obtained from parts (i) and (ii) above.Figure below shows one-line diagram of a simple three bus power system with generation at bus 1. Bus 1 is considered as slack bus. A load consisting of 250 MW and 110 MVAR is taken from bus 2. A load consisting of 128 MW and 35 MVAR is taken from bus 3. Line impedances are marked in per unit on a 100 MVA base. Line susceptances are neglected. G1 0.01 +10.03 V₁ = 1.040° 0.02 +0.04 Select one: O a. None of these O b. 0.9245-j0.025 O c. 0.9245+j0.025 O d. -0.9245-j0.025 e. 0.9638-j0.03 0.0125+j0.025 ·0 P2 Q2 P3 Q3 Start with flat initial estimates of ₂0) = 1 + j0 & V3⁰) = 1 + j0, and keeping |V₂| = 1 pu, find V₂(¹)