In a DC optimal power flow problem of 3 bus network. The flow on lines was found to as follow: f12=100 M, MW, f13=150 MW, and f23 =120MW. The lines limits are 140 MW each. and the system base MVA =100MVA. What additional constraint to be added on the solution of the DCOPF
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- 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. -61.75MWThe 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…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=0.5pu
- 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 52What is load curveQ2. 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 MVAR
- For a system consist from 3 buses and bus 1 is the slack bus. The data of the system are:- -0.2 -0.05 Ybus-j-0.2 0.225 -0.025 p.u. -0.05 -0.025 0.075 V₁=1+j0 p.u., P₂+jQ=0.6+j0.25 p.u., P3+jQ3-0.8+j0.5 p.u. 0.25 By using Gauss-Siedal method, the voltage of the bus 2 after two iteration is -0.762-j0.826 -0.762+j0.826 0.762+j0.826 0.762-j0.826 By using Gauss-Siedal method, the voltage of the bus 1 after two iteration is 1+j0 0+j1 -0.3-j0.6 0.3-j0.6The ---_characteristi cs shows the relation between the generated e.m.f at no load and the field current. O a. internal b. non-magnetic C. external d. magnetic2. 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 %3D
- A DC Optimal Power Flow problem is to be solved for a 3-bus network. The per-unit reactances of the lines interconnecting the buses are as follows: X12 = 0.1pu, X13 = 0.12 pu and X23 = 0.2 pu. There is a generator at each bus. The loads at buses 1, 2 and 3 are 150MW, 200mw, and 100MWrespectively. Bus 1 is taken as the reference bus, and SBase = 100 MVA. Which one of the below is a constraint of the DCOPF problem? Select one: O a. None of these O b. -500 0₂-1000 03 = P3 - 110 O c. 1500 0₂-500 03 = P₂ - 220 O d. 1500 0₂-500 03 = P₂ - 200 O e. -500 0₂-1000 03 = P3 - 150 O f. -1500 8₁-1000 03 = P₁ - 150Solve numerical : 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 δ2, power flows and line losses by using Fast decoupled power flow method. All the values are given in per unit on 100MVA base. Use a tolerance of 0.001 for power mismatch.A DC Optimal Power Flow problem is to be solved for a 3-bus network. The per-unit reactances of the lines interconnecting the buses are as follows: X12 = 0.25 pu, X13 = 0.15 pu and X23 = 0.2 pu. There is a generator at each bus. The loads at buses 1, 2 and 3 are 125 MW, 300 MW and 100 MW respectively. Bus 1 is taken as the reference bus, and SBase = 100 MVA. Which one of the below is NOT a constraint of the DCOPF problem? Select one: O a. -400 02 - 666.67 03 = P, - 125 O b. None of these C. -500 02 - 1166.67 03 = P3 - 100 d. 900 02 - 500 03 = P2 - 300