Question

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2 - The three-phase power and line-line ratings of the electric power system shown in Figure 2 are given below. T T2 V Line 2 Vm G M Figure 2 One-line diagram for problem 2 60 MVA 20 kV 50 MVA 20/200 kV 50 MVA 200/20 kV 200 kV 43.2 MVA 18 kV TI: T: Line: M: X=9% X=10% X=10% Z=120+j2002 X=8% a) Draw an impedance diagram showing all impedances in per unit on a 100-MVA base. Choose 20 kV as the voltage base for the generator. (b) The motor is drawing 45 MVA, 0.8 power factor lagging at a line-to-line terminal voltage of 18 KV. Determine the terminal voltage and the internal emf of the generator in per unit and in kV.

Answer 1

Solution't Given single line diagram of the Pow or system and rating of components are as follows - Ti 2 Vm 013 + Line 200 W GOMVA 5 OMVA 2= 120+J2ook 20kv 20kv/zooku x =gt. X= 107 erxzjoogpu or xajo lpu SOMVA 200/20kr R210y. orxajolpo 43.2 muĄ 18 kv x=8%. or x =jo- of po we have to take Base MUA CSB) = LooMNA and Base voltage (VB) = 20kV for generator Generator (G)- old values, GoMUA, 20 kV, X,jo og Pu new values 100MVA, 2oker, X2 -? we know New Base MUA Xpucnew) - Xpucold X old base kvi new baseur old Base MVA so Xpu (new) = 10.ogx 100 6o xebe 20 20 Xpu(new) =jo.15 so 2 Transformereti) old values, somUA, 20/200 kev, Xj0.100 New Values, 100 MUA, 20/200 kV, Xticnews =? Xti cnew) = 30.1% loo eeg Xto (new) =jo. 2 po similarly for Transformer Tz * fnto Xtachew) = jo 2 pu

Line & old values, 200kw, 2= 120+32000 New values are loomua, 200ler, Line Base impidance ZBL ² (2008 10372 1008106 z yoor 2 Sь So Perunit impidance of line Zepuz 120 +5200 ५०० Zlpu = 0.3 + jo. 5 ou Motor & old values, 43.2 MNA, 18 kv, Xijo.08 po New values are bo Mv 6, 2_oky, so Xmcnews =jo.08 x 100 43.2 х 92 Xmcnews ajo is pu (9) so the impidance diagram is jo.15 Vg jo.z! jos? jo.2 Ym ; 0.15 mamat mnm timmt 0.3 figr Reseta Impidance diagram of given power system (6) Given that Motor is drawing USMUA, at 0.8 PF lag at a line-to-line voltage of 18 kv motor current is Im z 1443.3 A V3 x 18x102 6 uslo

Power factor Cosa z of lag so at a cos'08 = -36.86 so Im = 1443.3 L-36.86 A Base current Is 1ooxlob √3x20x103 = 288 6.75 A So per unit Motor current Impo Im 1443.3L-36-86 2886.75 IB Impu z 0.5 2-36.86 PU motor terminal voltage in kv are Per unit motor terminal voltage will be 18kV 18 20 = 0.9 Pu The terminal voltage of geners for inpucan be Calculated as Vyno Impu (jo.2 +0.3+30.5 +30.2) + um Vypu z 0.5L-36.86(0.3+ jo.g) + 0.9 Loº Vypu = 1:31 2 11.82° PU Vg kv = Ugpu x Vo 1.31211.82 x 20 7103 Vyur = 26.2 111.82 kv

Gemera for Internal {mf can be calculated as Egpu Impux jo.1s + Vypu Eyou = 0.5L-36.86 x 10.15 + 1,31211.82 Egpu 1.367 213.89 Po Generator internal emf in kr is Eg Egpu x VB = 1.367 113.89 x 20X103 Eg 227.34 213.89 kv so, terminal voltage of generator in pu is 1.31211.82 and in ku is 2 6.2 211.82 ur and Internal emf of gentre for in po is 1.367213.89 and in uw is 27.34 213.89 kv