1. Assume that a 2.5-kV 1-phase circuit feeds a load of 450 kW at a lagging load power factor and the load current of 215 A.If it is desired to improve the power factor determine the following. a. The new corrected p.f after installing a shunt capacitor unit with a rating of 350kvar. none 0.992 lag 0.9 lag 0.952 lag
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- Under balanced operating conditions, consider the three-phase complex power delivered by the three-phase source to the three-phase load. Match the following expressions, those on the left to those on the right. (i) Realpower, P3 (a) (3VLLIL)VA (ii) Reactive power, Q3 (b) (3VLLILsin)var (iii) Total apparent power, S3 (c) (3VLLILcos)W (iv) Complex power, S3 (d) P3+jQ3 Note that VLL is the rms line-to-line voltage, IL is the rms line current, and is the power-factor angle.The voltage v(t)=359.3cos(t)volts is applied to a load consisting of a 10 resistor in parallel with a capacitive reactance XC=25. Calculate (a) the instantaneous power absorbed by the resistor, (b) the instantaneous power absorbed by the capacitor. (c) the real power absorbed by the resistor, (d) the reactive power delivered by the capacitor, and (e) the load power factor.A single-phase source is applied to a two-terminal, passive circuit with equivalent impedance Z=3.045, measured from the terminals. The source current is i(t)=22cos(t)kA. Determine the (a) instantaneous power, (b) real power, (c) reactive power delivered by the source, and (d) source power factor.
- Two balanced three-phase loads that are connected in parallel are fed by a three-phase line having a series impedance of (0.4j2.7) 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 at the load end of the line is 2203V. 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, and (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-phase comp lex power absorbed by the line and loads.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.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.
- The instantaneous power absorbed by the load in a single-phase ac circuit, for a general R LC load under sinusoidal-steady-state excitation. is (a) Nonzero constant (b) Zero (c) Containing double-frequency componentsFigure 3.32 shows the oneline diagram of a three-phase power system. By selecting a common base of 100 MVA and 22 kV on the generator side, draw an impedance diagram showing all impedances including the load impedance in per-unit. The data are given a follows: G:90MVA22kVx=0.18perunitT1:50MVA22/220kVx=0.10perunitT2:40MVA220/11kVx=0.06perunitT3:40MVA22/110kVx=0.064perunitT4:40MVA110/11kVx=0.08perunitM:66.5MVA10.45kVx=0.185perunit Lines I and 2 have series reactances of 48.4 and 65.43, respectively. At bus 4, the three-phase load absorbs 57 MVA at 10.45 kV and 0.6 power factor lagging.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) False
- Consider a single-phase load with an applied voltage v(t)=150cos(t+10)volts and load current i(t)=5cos(t+50)A. (a) Determine the power triangle. (b) Find the power factor and specify whether it is lagging or leading. (c) Calculate the reactive power supplied by capacitors in parallel with the load that correct the power factor to 0.9 lagging.A single phase, unity power factor load takes 90 Amps at 10 kV. Using a base voltage of 65 kV and base kilovolt amperes of 4475 kVA. Find the base current, base impedance, base power, per unit current and per unit voltage of the load.The three-phase power and line-line ratings of the electric power system shown in Figure-1 are given below: G1:82 MVA 30 kV X=10% T1:72MVA 30/300 kV X=11% T2:72MVA 300/30 kV X=11% M:62MVA 25 kV X=9% 300 kV Z= 89+j179 Apply the knowledge of per unit system on above network and draw an impedance diagram showing all impedances in per unit when Sb =100MVA and Vb = 30 kv