Community Answer

" Base MUA "=100Gencrator reactarce X_(G(0.0))=0.09 (8.0){:[{:X_(G(" new "))=0.09 xx(" MUA(new) ")/(" MVA(old) ")],[X_(a)(" new ")=0.15(8.0)]:}{:[" Trausformil "(T_(1))quadX_(T_(1))(6.0)=0.1(f.0)],[X_(T)" (new) "=0.1 xx((100)/(50))],[x_(T_(1))" (new) "=0.2(rho.0)],[" Trausformer "(T_(2))quadX_(T_(2))(6.0)=0.18.0". "],[X_(T_(2))" (new) "=0.1 xx(100)/(50)],[xT_(2)" (new) "=0.28.0]:}[1]Motos reacance X_(M)=0.08 p.0.{:[X_(M)" (new) "=(0.08)xx((100)/(43.2))^(xx)xx((18)/(20))^(2)],[X_(M(" new ")=0.15(8.0).]:}transmission Line z=(120+j 200)Omega{:[Z(f.0)=Z(" MVA Base ")/((kV_("Bose "))^(2))],[Z(p.0)=(120+j 200)((100))/((200)^(2))],[Z(p.0)=(0.3+j 0.5)p.0.]:}Now we have all quantities of reactances in P.U. on 100 MNAA bose so we can draw P.0. Insedance diagram.2(a)(b) Motor terminal voltage{:[V_(M)=18kV],[" Motot drawing "=45MVA" at "0.80" fowir factor "],[45 xx10^(6)=sqrt3V_(M)I_(M)],[45 xx10^(6)=sqrt3(18 xx10^(3))I_(M)],[I_(M)=(45 xx10^(6))/((sqrt3)(18 xx10^(3)))],[IM=1443.376A],[cos phi=0.8=>phi=cos^(-1)(0.8)],[phi=36.87^(@)],[I=1443.376/_-36.87^(@)A]:}Lagging{:[" Motor side Base MVA "=100MVA],[" Base kV "=2 ... See the full answer