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# homework9solutions - Homework 9 Solutions O l l 50480°...

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Unformatted text preview: Homework 9 Solutions O - l l 50480° 1180°+240°+lz90 100 8'4 Y3" l ‘3 a 10010"=501480°+24240°+142 o ‘2’" = l a a7 50.4900 1180°+24120°+lz330 V_ l a * 2.174+jl.985 147214240 50 —l 697—jl.247 2105.14216‘4 V = . e o 0.0397+j2.217 ”09488.97 l840£82.8° selecting a base of 2300 Vand 500 kVA. each resistor has an impedance of 140°pu; Va» :08; be = l.2; V = 1.0 I'll r ‘ voltages are: —- 1 Va,” =;(0.8182.8°+ 1.24120°—4i.4° + l .0z240° + l 80°) = 0.2792 + 10.9453 = O.9857z73.6° 17“” = 31(0.8482.8°+ l .24240°—4i.4° +l.Oél20°+l80°)=~O.l790-j0.1517 = 0234612203o [Note: An angle of 180° is assigned to 17”] P Va”2 = 0234642203" + 30° = 0234642503" Zero-sequence currents are not present due to the absence of a neutral c I:l = Z, /l£0° = 0.9857443.6°pu I pu onnection. a: = Val/140°: 0234642503" The positive direction of current is from the supply toward the load. 9.1 Calculation of per unit reactanccs ﬁxnghmﬂogs gengrators: Gl XI=XJ=0.18 XZ=XJ=0.18 X(,=0.07 02 x.=x;=0.2o x,=x;=0.20 x‘,=0.10 xx 0151:8188) 0018—8118—8881 500 15 500 = 0.2539 = 0.08464 X2 = X: 20.2539 3X” =3X” =0.2539 G4 X: XJ—030[138)[—1—000] 750 =0.3386 X :_O40[1_._38MI__000] Xn_010[1_.38](1_000] 15 750 15 750 =0.4514 :0.1129 Trgngfomgg; 1000 X =0.10 X.=0.10 =0.12 .. .. .. [.— 500 __j XN =0.11[ﬂO-g]=0.1467 =0.24 750 Tr mis ion 'nes: (765)2 g-mm = =585.23Q 1000 Positive/Negative Sequence Zero Sequence 150 5 X , : XI, = 0 =0.08544 ‘8 585.23 ‘ 585.23 20.2563 X11:X71: 40 X11=Xﬁ= loo ' 585.23 ‘ " 585.23 = 0.06835 = 0.1709 ( X: W0068; Xi— 0/700» Slauk : [OWVIQ Vbqat7é) €3,953}, ?//)7c / l jdyV {(12):} 0X46 “:(1 46V «1/ Xia: O'l/0«O772+ 0«O7?Z—for/ ‘ + 0,qq}°V/0,qg;3:0 14“; ' 2 , I Xi»; 0,3/(0, 072634 (aoUOleZlWﬂ/(O’MU 4b,7/5)) :'0.3//(0,0163+ muggy/0'20“) : o'?//(D<<>’L63+ 0, (a 23) ~ 30,20 Ollbge 0.! ) 070’ 17‘ : D/IO’Iq Wﬂj KEN Qﬁuehff \ o( 0557? 0153454an OHM—e 015074 W 0.0%} 0-1—60, [X , OVH‘L'L O,§‘Wl// 0, 4/1729 —— {£170}— \// X71: 0.3//(0,0’L63+{0,07J0+0.1703’)//[o,016§+o,zx)) :9, z//(0,ou3+ ,O‘m/y/o, 3 063) 70/3/{0‘0167j‘ (Mam) :oz2// 0475—7 (QMOX :- 04H)? 1’ R T NWO“ \j‘ 5Wa(e ‘— ...
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