Basic Engineering Circuit Analysis
11th Edition
ISBN: 9781118539293
Author: J. David Irwin, R. Mark Nelms
Publisher: WILEY
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Two 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 voltages
A 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.
Under balanced operating conditions, consider the 3-phase complex power delivered by the 3-phase source to the 3-phase load. Match the following expressions, those on the left to those on the right. 0) Real power, Py (ii) Reactive power, Qu (ii) Total apparent power Sy (iv) Complex power, Sy (a) (V3 VLL. IL)VA (b) (V3 Vu sng) var (e) (V3 VLL. IL cos ø) W (d) Py +Qy Note that VLL is the rms line-to-line voltage, IL is the ms line current, and o is the power-factor angle.
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- 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.arrow_forwardA 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.arrow_forwardUnder 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.arrow_forward
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