Consider the singleline diagram of the power system shown in

Consider the single-line diagram of the power system shown in Fig. 2. Equipment ratings are: G_1 : 750 MVA, 18 kV, x\" = 0.2 pu G2: 750 MVA, 18 kV, x\" = 0.2 pu M_3: 1500 MVA, 20 kV, x\" = 0.2 pu T_1 : 750 MVA, 500V/20 Delta kV, x = 0.10 pu T_2: 750 MVA. 500V/20 Delta kV, x = 0.10 pu T_3: 750 MVA, 500V/20 Delta kV, x = 0.10 pu T_4 : 750 MVA, 500V/20 Delta kV, x = 0.10 pu T_5: 1500 MVA, 500F/20Y kV, x = 0.10 pu Neglecting resistance, transformer phase shift, and magnetizing reactance, draw the equivalent reactance diagram. Use a base of 100 MVA and 500 kV for the 40 Ohm line. Determine the per-unit reactances.

Solution

MVAB=100

KVB=500

PU REACTANCE OF j40 LINE=(j40*100)/(500*1000)=j0.008 pu

new pu reactance of T1=j0.1*(100/750)*(500/500)^2=j0.0133 pu

new pu reactance of T2= j0.1*(100/750)*(500/500)^2=j0.0133 pu

base voltaage for T1 primarry=20 kv

new pu reactance of G1=j0.2*(100/750)*(18/20)^2=j0.0216 pu

new pu reactance of T3=j0.1*(100/750)*(20/20)^2=j0.0133 pu

base voltaage for T3 Secondary=500 kv

pu reactance of j25 line=(j25*100)/(500*1000)=j5*10^-3 pu

pu reactance of j25 line=j5*10^-3 pu

new  pu reactance of T5=j0.1*(100/1500)*(500/500)^2=j6.66*10^-3 pu

base voltage for M3=20 kv

new pu reactance of M3=j0.2*(100/1500)*(20/20)^2=j0.0133 pu

new pu reactance of T4=j0.1*(100/750)*(500/500)^2=j0.0133 pu

Base voltage for T4 secondary=20 kv

new pu reactance of G2=j0.2*(100/750)*(18/20)^2=j0.0216 pu