2. A Y-connected balanced three-phase generator with an impedance of 0.4 + j0.3.2 per phase is connected to a Y-connected balanced load with an impedance of 24+/19 per phase. The line joining the generator and the load has an impedance of 0.6+ j0.7 per phase. Assuming a positive sequence for the source voltages and that Van = 120230° V calculate the complex power at the source and at the load.
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- While the instantaneous electric power delivered by a single-phase generator under balanced steady-state conditions is a function of time havi ng two components of a constant and a double-frequency sinusoid, the total instantaneous electric power delivered by a three-phase generator under balanced steady-state conditions is a constant. (a) True (b) FalseQ2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Ztoad 500 236.87° ohm , and the transmission line's impedance is Zine = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.The one-line diagram of a three-phase power system is shown in the above figure. The line impedances are in per- unit on a 100-MVA base, and the line admittances are neglected. Use Gauss-Seidel method in Matlab to analyze the system the tolerance of error is 10-3. Then anwser the questions as follows. V, = 1.05200 0.01+ j0.025 300 + j200 MVA 0.015 + j0.035 150 MW 200 +j150 MVA 0.01+ j0.03 |V = 1.03 The admittance matrix (Y_bus)? 0.02 + jo.04 0.0125 + jo.03 2.
- Q2) A 13.2-kV single-phase generator supplies power to a load through a transmission line. The load's impedance is Zload = 500 236.87° ohm, and the transmission line's impedance is Zline = 60 253.1° ohm. To reduce transmission line losses to 0.0103 of its losses without using the transformers design and use two transformers T1 between the generator and the transmission line and T2 between the transmission line and the load.A 3-phase generator possesses a synchronous reactance of 62 (line reactance between the source and the load) and the excitation voltage Eo = 3KV (source) per phase. Calculate the line-to-neutral voltage ELoad for a balanced resistive load of 8. Also, draw a phasor diagram between three-phase source and load voltages.A balanced Y-connected generator with terminal voltage Vbc 5 200/08 volts is connected to a balanced-D load whose impedance is 10/408 ohms per phase. The line impedance between the source and load is 0.5/808 ohm for each phase. The generator neutral is grounded through an impedance of j5 ohms. The generator sequence impedances are given by Zg0 5 j7, Zg1 5 j15, and Zg2 5 j10 ohms. Draw the sequence networks for this system and determine the sequence components of the line currents.
- two generators supplying a load. Generator I has a no-load frequency of 62.5 Hz and a slope Sp1 of I MW/Hz. Generator 2 has a no-load frequency of 62.0 Hz and a slope sp2 of I MW/Hz. The two generators are supplying a real load totaling 2.5 MW at 0.8 PF lagging. (a) At what frequency is this system operating, and how much power is supplied by each of the two generators? (b) Suppose an additional I-MW load were attached to this power system. What would the new system frequency be, and how much power would Gl and G2 supply now? Generator 1 VT V2 Generator 2 VTí KVAR KVARIn a three-phase system the per-unit phase voltages are 100° , Vôn = 0.8Z180° Van 0.920° , Von = 0.92 - а What are the per-unit symmetrical components V0) and V)? Express in polar notation.Transkrmers phase with each bank with - adjuste 35. Three single-phase transformers, cach of 100 kVA rating, are conected in a closed-delta arrangement. If one of them is taken out, it would be possible to load the bank in such a manner that cach one is loaded to the extent of: A 86.6 kVA C. 57.7KVA B. 66.7 kVA D. 33.33 kVA when load is at unity power factor. 36. Average power factor at which V-V bank separate is A 86.6% C. 66.67% В. 73.3% D. None of these 37. Scott connections are used for: A. single-phase to three-phase transformation C. three-phase to two-phase transformation B. three-phase to single-phase transformation (D Any of the above 38. The current in low voltage winding of a 25 kVA, 6600/400 V, 50 Hz, 3-phase Y-A transformer, at full load is: A 36 A Č. 65,32 A B. 20.78 A D. 40.2 A 39. The current in high voltage winding of a 20 kVA, 2200/220 V, 50 Hz, 3-phase A-A transformer, at full load is: (A. 3.03 A C. jo.5 A B. 5.25 A D. None of above 40. The current in low voltage…
- In the system shown in Figure 1, the transformers are connected star-star with both star points grounded and the generator is connected in star with its star points grounded. The per unit impedances of each element on a 40 MVA base are given in Table 1 and the voltage levels are given in Table 2. Z [p.u.] 0 2 tööt Generator 0.01 +j0.08 p.u. L Figure 1: A section of the distribution system Transformer T1 Line V BASE [KV] 0.04 + j 0.03 + j 0.15000000000000002 0.06000000000000000 Generator Table 1: Sequence impedances (p.u. on 40 MVA base) 3 T2 181. Line 3 10 Table 2: Voltage bases (kV) Load 1 Transformer T2 Load Current A load current of 11.316 kA is flowing with a lagging power factor of 90 %. Convert this to a current vector in per-unit. IL = 0.04 + j 0.06000000000000000A single-phase power system consists of a 450-V 60-Hz generator supplying a load-bank consisting of three loads (L1, L2 and L3) through a transmission line of impedance Z line=0.22+j0.43 as shown in Figure 1. A transformer (T1) with turn ratio N1:N2 is connected in the source end. Another tapping transformer (T2) with turn ratio M1:M2 is connected in the load end with a switch (K1). The switch can be connected to any of the tapping location: a, b or c. The load bank is connected with the system through another switch (K2). This switch can be connected to any of the load: L1, L2 or L3. Only one load can be connected at a time. There is a Control Box (C.B.) in between K1 and K2. It is designed in a way that if K2 is connected with a load (L1, L2 or L3) based on the demand of the consumer, it will trigger K1 to connect with any of the tap of T2 (a, b or c), based on the required power by the load. Consider that the power loss in the CB is negligible Load 1 (L1): Light load, 200W…Two single phase transformers with equal turns have impedance of (0.5+j0.3) ohms and (0.6+j10) ohms with respect to the secondary. If they operate in parallel, how will they share a load of 100 kW at 0.8 p.f. lagging?