Two balanced three-phase loads that are connected in parallel are fed by a three-phase line
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- A three-phase line with an impedance of (0.2+j1.0)/ phase feeds three balanced three-phase loads connected in parallel. Load 1: Absorbs a total of 150 kW and 120 kvar. Load 2: Delta connected with an impedance of (150j48)/phase. Load 3: 120 kVA at 0.6 PF leading. If the line-to-neutral voltage at the load end of the line is 2000 v (rms), determine the magnitude of the line-to-line voltage at the source end of the line.Two three-phase generators supply a three-phase load through separate three-phase lines. The load absorbs 30 kW at 0.8 power factor lagging. The line impedance is (1.4+j1.6) per phase between generator G1 and the load, and (0.8+j1) per phase between generator G2 and the load. If generator G1 supplies 15 kW at 0.8 poir factor lagging, with a terminal voltage of 460 V line-to-line, determine (a) the voltage at the load terminals. (b) the voltage at the terminals of generator G2, and (c) the real and reactive power supplied by generator G2. Assume balanced operation.A three-phase line, which has an impedance of (2+j4) per phase, feeds two balanced three-phase loads that are connected in parallel. One of the loads is Y-connected with an impedance of (30+j40) per phase, and the other is -connected with an impedance of (60j45) per phase. The line is energized at the sending end from a 60-Hz, three-phase, balanced voltage source of 1203V (rms. line-to-line). Determine (a) the current, real power. and reactive power delivered by the sending-end source: (b) the line-to-line voltage at the load: (C) the current per phase in each load: and (d) the total three-phase real and reactive powers absorbed by each load and by the line. Check that the total three- phase complex power delivered by the source equals the total three-phase power absorbed by the line and loads.
- For the circuit element of Problem 2.3, calculate (a) the instantaneous power absorbed, (b) the real power (state whether it is delivered or absorbed). (c) the reactive power (state whether delivered or absorbed). (d) the power factor (state whether lagging or leading). [Note: By convention the power factor cos() is positive. If | | is greater than 90, then the reference direction for current may be reversed, resulting in a positive value of cos() ].Figure 2.33 gives the general -Y transformation. (a) Show that the general transformation reduces to that given in Figure 2.16 for a balanced three-phase load. (b) Determine the impedances of the equivalent Y for the following impedances: ZAB=j10,ZBC=j20, and ZCA=j25. ZAB=ZAZB+ZBAC+ZCZAZCZA=ZABZCAZAB+ZBC+ZCAZBC=ZAZB+ZBAC+ZCZAZAZB=ZABZBCZAB+ZBC+ZCAZCA=ZAZB+ZBAC+ZCZAZBZA=ZCAZBCZAB+ZBC+ZCAIt is stated that (i) balanced three-phase circuits can be solved in per unit on a per-phase basis after converting - load impedances to equivalent Y impedances. (ii) Base values can be selected either on a per-phase basis or on a three-phase basis. (a) Both statements are true. (b) Neither is true. (c) Only one of the above is true.
- Consider a three-phase Y-connected source feeding a balanced- load. The phasor sum of the line currents as well as the neutral current are always zero. (a) True (b) FalseTwo balanced loads are connected to a 240-kV rms 50-Hz line. Load 1 draws 30 kW at a power factor of 0.6 lagging, while load 2 draws 45 kVAr at a power factor of 0.8 lagging. Assuming the RYB sequence, determine: (a) the complex, real, and reactive powers absorbed by the combined load, (b) the line currents, and (c) the kVAr rating of the three capacitors ∆-connected in parallel with the load that will raise the power factor to 0.9 lagging and the capacitance of each capacitor. Verify the results by creating a MATLAB Simulink model and program. Obtain the plots for the currents and voltagesA 3-phase Y-connected load draws power from a 3-phase Y-connected source. The source voltage, ?=440∠0º?_rms and the load impedance, Z_y=(30+j15) Ω. The line impedance, Z_line= (1+j1)Ω and the system operates at 60 Hz. a) Draw the single-phase equivalent circuit and label all given quantities numerically. b) Calculatethe rms line (phase) current phasor in each phase. c) Calculate the total active or real power supplied by the source. d) Calculate the total active or real power consumed by the load. e) Calculate the total line loss i.e. the total activepower lost due to line. f) Determine the power efficiency of the system.
- A balanced delta connected load of "14+j19" 2- per phase is connected at the end of a three-phase line. The line impedance is "6+18j" 2- per phase. The line is supplied from a three-phase source with a line-to-line voltage of 207.85 Vrms. Taking phase "a" voltage Va as reference, determine the following: (a) Current in phase a. (b) Total complex power supplied from the source. (c) Magnitude of the line-to-line voltage at the load terminal.The one-line diagram of a three-phase power system is as shown in the figure attached. Impedances are marked in per unit on a 100-MVA, 400-kv base. The load at bus 2 is S(sub 2) = 15.93 MW - j33.4 Mvar, and bus 3 is S(sub 3) = 77MW + j14 Mvar. I tis required to hold the voltage at bus 3 at 400 angle 0 degrees kV. Working in per unit, determine the voltage at buses 2 and 1.Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of ZL = (0:4 + j2:7) Ω per phase. One of the loads absorbs 560 kVA at 0.707 power factor lagging, and the other 132 kW at unity power factor. The line-to-line voltage atthe load end of the line is 2; 200p3 V. Compute: a) The line-to-line voltage at the source end of the line.b) The total real and reactive power losses in the three-phase line.c) The total three-phase real and reactive power supplied at the sending end of the line. Check that the total three-phase complex power delivered by the source equals the total three-phasecomplex power absorbed by the line and loads.