EEE471HW2solp1

# EEE471HW2solp1 - 3 15 = = Ω-= = o L j j j Z 7 12 1 12 67...

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EEE471/ 591 GSO 4 th Ed. Chapter 3 Homework Problem 3.4 Single phase, 100KVA, 2400-240V, 60 Hz Transformer, Load = 80 [email protected] pf lagging at 230 V V p 100 KVA 2400:240V I L 80KVA @ 0.8 pf lagging At 230 V V L =230 / 0 o Find: V p , Z L , Z L referred to primary, P p , Q A I o L 9 . 36 / 8 . 347 8 . cos / 23 . 80 1 - = - = - V V o o P 0 / 2300 240 2400 0 / 230 = = = - = = o o o L L I V Z 9 . 36 / 661 . 0 9 . 36 / 8 . 347 0 / 230 = = o o L primary Z 9 . 36 / 1 . 66 ) 9 . 36 / 661 (. 240 2400 ) ( 2 KVA j S o ) 48 64 ( 9 . 36 / 80 + = = a b d c

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o I 9 . 36 / 83 . 20 8 . cos / 10 ) 240 ( 50000 ' 1 - = - = - V j V o o H 2448 ) 5 . 2 1 ( 9 . 36 / 83 . 20 0 / 2400 = + - + = V j V o o S 14 . 1 / 2490 ) 5 . 4 2 ( 9 . 36 / 83 . 20 2400 = + - + = KVA j I V S o o S S S 96 . 31 85 . 40 9 . 36 / 83 . 20 ( 14 . 1 / 2490 * + = = = Problem 3.19 (Repeat Problem 3.14 in per unit) VA B = 50 V B = 2400 VA A I B 83 . 20 2400 50000 = = I’ = 1/-36.9 o pu = = 2 . 115 50000 ) 2400 ( 2 B Z pu Z o o TU 2 . 68 / 0234 . 2 . 115 2 . 68 / 69 . 2 = = pu j Z Z o T L 66 / 0427 . 2 . 115 5 . 4 2 = + = + V pu V o o o HU 2448 68 . / 02 . 1 ) 2 . 68 / 0234 (. 9 . 36 / 1 1 = = - + = V pu V o o o SU 2490 14 . 1 / 038 . 1 ) 66 / 0427 (. 9 . 36 / 1 1 = = - + = KVA j pu j S o o S 32 88 . 40 640 . 818 . ) 9 . 36 / 1 ( 14 . 1 / 038 . 1 + = + = = Load 50 KVA @ .8 pf lag 240 V V S 1+j2 V H 1+j2.5 V X I’ 50 KVA 2400:240V 240 /0 o V Find: V H , V S , S S a b Problem 3.14 c a b C
Problem 3.22 Find: I S KVA S B 15 3 = kV kV B 480 . 3 277 . 3 = = = 36 . 15 015 . ) 48 (. 2 B Z A I B 18 ) 48

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Unformatted text preview: (. 3 15 = = Ω-= +-+ = o L j j j Z 7 . 12 / 1 . 12 67 . 1 45 ) 33 . 8 15 )( 10 30 ( o o LU Z 7 . 12 / 788 . 36 . 15 7 . 12 / 1 . 12-=-= A pu I o o o S 7 . 12 / 9 . 22 7 . 12 / 27 . 1 7 . 12 / 788 . 1 = =-= Z Y 30+j10Ώ Z Δ = 45 – j25 15 – j 8.33Ώ 3 φ I S I S V o / 3 277 + I S Z LU 1/0 o-Problem 3.33 a c b c’ a’ b’ A B C HV Y LV Δ V X V Y V Z V A V’ A For ASA Standard V C V B V A V a V c V b V’ A a) z x b a A V V V V nV +-=-= therefore: z a = and x b = V c V a V b b) High leads low by 90 o z x b c A V V V V nV +-=-= therefore: x b = and z a =...
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## This note was uploaded on 02/21/2010 for the course EEE ??? taught by Professor Farmer during the Spring '10 term at ASU.

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EEE471HW2solp1 - 3 15 = = Ω-= = o L j j j Z 7 12 1 12 67...

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