The load flow data for the sample power system are given below. The voltage magnitude at bus 2 is to be maintained at 1.04 p.u.Determine the set of load flow equations at the end of first iteration by using N-R method. Impedance for sample system Bus Code 1-2 1-3 2-3 Schedule of generation and loads: Impedances 0.08 + j0.24 line charging admittance 0.0 0.02 + j0.06 0.0 0.06 + j0.18 0.0 Bus Code Assured Voltages Generation Load MW MVAR MW MVAR 1 1.06 + j0.0 0 0 0 2 1.0 + j0.0 0.2 0.0 0.0 0.0 3 1.0 + j0.0 0 0 0.6 0.25
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- Consider three ideal single-phase transformers (with a voltage gain of ) put together as three-phase bank as shown in Figure 3.35. Assuming positive-sequence voltages for Va,Vb, and Vc find Va,Vb, and VC. in terms of Va,Vb, and Vc, respectively. (a) Would such relationships hold for the line voltages as well? (b) Looking into the current relationships, express IaIb and Ic in terms of IaIb and Ic respectively. (C) Let S and S be the per-phase complex power output and input. respectively. Find S in terms of S.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…1. FIGURE 52 shows the one-line diagram of a simple three-bus power system with generation at bus I. The voltage at bus l is V1 = 1.0L0° per unit. The scheduled loads on buses 2 and 3 are marked on the diagram. Line impedances are marked in per unit on a 100 MVA base. For the purpose of hand calculations, line resistances and line charging susceptances are neglected a) Using Gauss-Seidel method and initial estimates of Va 0)-1.0+)0 and V o)- ( 1.0 +j0, determine V2 and V3. Perform two iterations (b) If after several iterations the bus voltages converge to V20.90-j0.10 pu 0.95-70.05 pu determine the line flows and line losses and the slack bus real and reactive power. 2 400 MW 320 Mvar Slack 0.0125 0.05 300 MW 270 Mvar FIGURE 52
- Write down the effect of ‘Load Tap Changer (LTC)’ on voltage stability and analyze it using P-U (nose) curve .The one-line diagram of a simple power system is shown in Figure below. The neutral of each generator is grounded through a current-limiting reactor of 0.25/3 per unit on a 100-MVA base. The system data expressed in per unit on a common 100-MVA base is tabulated below. The generators are running on no-load at their rated voltage and rated frequency with their emfs in phase. G Stark Item Base MVA Voltage Rating X' x² 20 kV 20 kV 20/220 kV 20/220 kV 100 0.05 0.15 0.15 0.10 0.10 220 kV 0.125 0.125 0.30 0.15 0.25 025 0.7125 0.15 100 100 0.15 0.05 0.10 0.10 0.10 100 0.10 100 100 Lu La 220 kV 0.15 220 kV 0.35 100 A balanced three-phase fault at bus 3 through a fault impedance Zf= jo.I per unit. The magnitude of the fault current in amperes in phase b for this fault is: Select one: A. 345.3 B. 820.1 C. 312500 3888888 产产6. For a three bus power system assume bus 1 is the swing with a per unit voltage of 1.020 , bus 2 is a PQ bus with a per unit load of 2.0 + j0:5, and bus 3 is a PV bus with 1.0 per unit generation and a 1.0 voltage setpoint. The per unit line impedances are j0.1 between buses 1 and 2, j0.4 between buses 1 and 3, and j0.2 between buses 2 and 3. Using a flat start, use the Newton-Raphson approach to determine the first iteration phasor voltages at buses 2 and 3.
- Q2. 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 MVARFor the system shown in figure, voltages V2, V3 and angles 82, 83 are calculated using Newton-Raphson method. shunt line charging admittances are neglected. All the values are given in per unit on 100MVA base. Calculate the complex power flows S12 and S32 in actual units. 1) Z12 = 0.01+ j0.02 Z3 = 0.02+ j0.04 PL G QL Slack bus V2 = 0.96L -1.67° V3 = 0.884 - 5.48° Vi=1.0/0° (estimated time to answer this question: 13 minutes)Select the base voltages for one section and calculate the base voltages for the rest of the sections. Remember that the voltages given are line-line. Calculate the current and impedance for all sections of the System. Assume a base complex power of 100 MVA.
- What is load curveIn a given system of base power of 250 MW, and bus 3 is taken as reference. The per-unit reactances are X12 = 0.2 p.u., X13 = X23 = 0. 1 p.u. The power flow in the system is given as: PF12 = 50 MW, PF13 = 150 MW, PF23 = 50 MW. Based on readings from 2 meters (not including M12 meter), M13 = 145 MW (not calibrated), and M23 = 50 MW (well calibrated), deduce the flow on line 1-2. Select one: O a. PF12 = - 47.5 MW. O b. None of these O c. PF12 = 67.5 MW. O d. PF12 = 47.5 MW. O e. PF12 =- 57.5 MW. O f. PF12 = 57 MW.A 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