How did we get this result? PG₁ = 15 KWStep2b)Part AThe single-line diagram of the hole system will be,V-460/20VG₁IGI=-WW m1.4j1.6PG₁√3VLL COS15×10³√3×460×0.8The active power supplied to the generator will be,PG1√3VLLIL COSso the current supplied by generator 1 will be,IG₁www m0.8jl- 36.86=23.533 – 36.86° AOn applying KVL in the above figure will be,V₁ + IG₁ Z1 − Vg₁=0162VL=VG₁ - IG₁Z₁460 22 592G2100 Problem 7. 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 phasebetween generator G1 and the load, and (0.8+j1) per phase between generator G2 and the load. Ifgenerator G1 supplies 15 kW at 0.8 power 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.c) The real and reactive power supplied by generator G2.

Answered: Image /qna-images/answer/d52a5c52-91fb-458c-8ca7-3819781c950f.jpg

PG₁ = 15 KW Step2 b) Part A The single-line diagram of the hole system will be, VG₁ IGL = -WW 1.4 PG₁ √3VLL COS 15×10³ √3x460x0.8 j1.6 www m 0.8 jl The active power supplied to the generator will be, PG1 √3VLLIL COS so the current supplied by generator 1 will be, IG₁ 162 - 36.86 =23.533 – 36.86° A On applying KVL in the above figure will be, G2

PG₁ = 15 KW Step2 b) Part A The single-line diagram of the hole system will be, VG₁ IGL = -WW 1.4 PG₁ √3VLL COS 15×10³ √3x460x0.8 j1.6 www m 0.8 jl The active power supplied to the generator will be, PG1 √3VLLIL COS so the current supplied by generator 1 will be, IG₁ 162 - 36.86 =23.533 – 36.86° A On applying KVL in the above figure will be, G2

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)

10th Edition

ISBN:9781337399128

Author:Russell E. Smith

Publisher:Russell E. Smith

Chapter7: Alternating Current, Power Distribution, And Voltage Systems

Section: Chapter Questions

Problem 7RQ

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