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IRWIN 9e 11_74

IRWIN 9e 11_74 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 11.74 A standard practice for utilit},r companies is to divide its customers into single-phase users and three-phase users. The utility must provide three—phase users, typically industries, with all tluee phases. However, single—phase users, residential and light commercial, are connected to only one phase. To reduce cable costs, all single—phase users in a neighborhood are connected together. This means that even if the threephase users present perfectlyr balanced loads to the power grid, the single—phase loads will never be in balance, resulting in current flov.r in the neutral connection. Consider the 60H; air—sequence network in Fig. P1134. With a line voltage of 4lﬁf3lJ° V rms, phase a supplies the single—phase users on A Street, phase 11 supplies B Street, and phase r: supplies C 5H Furthermore, the three—phase industrial load, which is connected in delta, is balanced. Find the neutral current. Figure P1134 SOLUTION: IAN ‘ WWW—0° 96: 40010. A/mnc Q40Lo° ? 30ki°° * IZEZ‘ ".45 _ = ~u~ = —420 we ’3“ a q 0 but)“ 0 t) éok Lo *3 0350020 74 MM 44'!) [120° “ 33,95: 36kN (H Z H %—¢ ,, Bék . iszA pp 015’ 334, a 72k Low” (0.5): '72. [600 KVA in WA FWWmea—ﬁSwM 55—w)= Chapter 11: Polyphase Circuits Problem 11.74 2 Irwin. Basic Engineering Circuit Analysis, 91E .- 0 7% u IA 3 V2“ k 439%.“) = 5100 ("60 A mm: 3 (2900)”) I,5= loot—1&0 Pr arms 12: loog‘goU A «ﬁrm ,— Iam ; IA ’r—IAN ;’00 1'60 +1004!) lip, ’“MLW 53 (7 l9://9 A 75m; ’ P 156 3 1151’ 1m =- ’00 5:30 +115” [—22.0 153 3' /95'v26 Ari/903236 A Ymﬁ ’- n "" 0 Ice ‘5 Ic,"' 1m “MOO 1600* 25-0 [19-0 fee: 321.25 005.?” A xmg .— 1"“ lnN 3' ”‘ am ”Ibis” Inc ‘7: ,: ,(g49.§8L’l‘?»1/9)’(/7§v2é[:/96:33) ——@i2.2-5‘/_/03"79 1’th l08~ 763’ 1+ 83. 42°A Problem 1174 Chapter 11: Polyphase Circuits ...
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