For the three-bus system shown in figure below: • The line from bus 1 to bus 2 has an imped- ance Z = 0.05+j0.1 The other two line (1-3 and 2-3) both have an impedance Z = j0.25 The load at bus 3 is consuming 150 MW and 87 Mvar wwwwww The generator at bus 1 has a voltage set- point of 1.03 per-unit о Bus 1 Bus 2 4-j12 -4+j8j4 Ybus = -4+j8 4j12 j4 j4 j4 -j8. Bus 3 Given the Y-bus above, with bus 1 as slack bus and buses 2 and 3 as PQ buses, write the real and reactive power balance equations for bus 3 only, in terms of 02, 03, V2 and V3. You do not need to find the Jacobian or solve the equations. Include the load P and Q.
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- t Figure below shows a power system where load at bus 5 is fed by generators at bus 1 and bus 4. The generators are rated at 100MVA; 11 KV with subtransient reactance of 25%. The transformers are rated each 100 MVA, 11/112 KV and have a leakage reactance of 8%. The lines have an inductance of 1 mH/phase/km. Line L1 is 100 km long while lines L2 and L3 are each of 50 km in length. Load For a 3-phase fault at bus 5, the fault current is equal to 2530A. What the value of the change in the fault current if the value of the base voltage is changed to .10 kV (i.e. new value/old value) 1.0426 0.9833 1.0245 1.0166 0.9751 1The 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Three zones of a single-phase circuit are identified shown in Figure below. The zones are connected by transformers T1 and T2, whose ratings are also shown. Using base values of 33 kVA and 232 volts in zone 1, Find: 1- Draw the per-unit circuit including the per-unit impedances and the per-unit source voltage. 2- Calculate the load current both in per-unit and in amperes (actual or original value). Vs Zone 1 232.940° Vs G. 38 T₁ 30 KVA 240/480 volts Xeq = 0.10 p.u. Zone 2 Xiine = 4 Ω T₂ 20 KVA 460/115 volts Zload = Xea = 0.10 p.u. Zone 3 u 1+j2.2 Ω 2
- Description In the particular case of figure below derive both the critical clearing angle and the critical clearing time. P, = Pmaz sin d Pm A1 do der Smar A generator having H = 6.R MJ/MVA is delivering power of 1.0 per unit to an infinite bus through a purely reactive network when the occurrence of a fault reduces the generator output power to zero. The maximum power that could be delivered is 2.5S per unit. When the fault is cleared, the original network conditions again exist. Determine the critical clearing angle and critical clearing time. (Roll=PQRS)In the single line diagram shown below, generators G₁ and G₂ have a leakage reactance of 0.26 per unit on a 66.5MVA rating at 11kV, and transformers T₁ and T₂ have a voltage ratio of 11/145kV and a leakage reactance of 0.125 per unit on a 75MVA rating. Choosing 100MVA and 132kV (at the lines side) as base quantities, find the reactance of G₁, G₂, T₁ and T2 in per unit to the new base quantities. 66.5MVA 11kV 0.26pu 66.5MVA 11kV 0.26pu (G1 G2 75MVA 11/145kV 0.125pu T1 T2 75MVA 11/145KV 0.125pu 132kV LinesPER-UNIT SYSTEM TUTORIAL Tutorial 3 Draw an impedance diagram for the electric power system shown in Figure 3 showing all impedances in per unit on a 100-MVA base. Choose 20 kV as the voltage base for generator. The three-phase power and line-line ratings are given below. X = 9% X = 16% X = 20% X = 9% X = 120 Q S = 48MW + j64MVar G1: 90MVA 20kV T1: 80MVA 20/200kV T2: 80MVA 200/20kV G2: 90MVA 18kV Line: 200kV Load: 200kV
- Q2\ The one-line diagram of a simple power system is shown in figure below. All impedances are expressed in per unit (pu) on a common MVA base. All resistances and shunt capacitances are neglected. The generators are operating on no load at their rated voltage. A three-phase fault occurs at bus 1 through a fault impedance of Zf = j0.08 per unit. Using Thevenin's theorem obtain the impedance to the point of fault and the fault current in per unit. Determine the bus voltages and line currents of generators during fault. X₁ = = 0.1 XT-0.1 3 1 to ojo XL=0.2 2 040 X₁ = 0.1Figure below shows a power system where load at bus 5 is fed by generators at bus 1 and bus 4. The generators are rated at 100MVA; 11 KV with subtransient reactance of 25%. The transformers are rated each at 100 MVA, 11/112 KV and have a leakage reactance of 8%. The lines have an inductance of 1 mH/phase/km. Line L1 is 100 km long while lines L2 and L3 are each of 50 km in length. 1- For a 3-phase fault at bus 5, the fault current is equal to 2530A. What the value of the change in the fault current if the value of the base voltage is changed to 10 kV (i.e. new value/old value). 2- For a 3-phase fault at bus 5, the fault current is equal to 2530A. What the value of the change in the fault current if the length of L1 is reduced to 50 km (i.e. new value/old value).Figure below shows a power system where load at bus 5 is fed by generators at bus 1 and bus 4. The generators are rated at 100MVA; 11 KV with subtransient reactance of 25%. The transformers are rated each at 100 MVA, 11/112 KV and have a leakage reactance of 8%. The lines have an inductance of 1 mH/phase/km. Line L1 is 100 km long while lines L2 and L3 are each of 50 km in length. He Load 10 For a 3-phase fault at bus 5, the fault current is equal to 2530A. What the value of the change in the fault current if the length of L1 is reduced to 50 km (i.e. new .value/old value) 1 O 1.0174 0.9965 1.0062 1.0222 0.9867
- 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…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 MVARFigure 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)