Question

34 A 5kVA, 415/230 single phase transformer has the following resistances and leakage reactances: R1 = 0.2002; x1 = 0.8102; and R2 = 0.0792; X2 = 0.130 Q- Determine the current fault level at the consumer terminals of the transformer.

Please show clear working out, so I understand better.

Answer 1

- Vence = 230V primary 3 Seconday, VRASE= 415V Side - side. · ZONE-1 ZONE - 2 UVA)b = 5KVA primary side or Zane - I Base impidence. Zk= Ke i MVAB ZB= (0.415) ? 5 mVA ::. 1000 So zB = 34.4452 pey unit nufstanie and reactance R1 (pu) = n(real) oro. :: 21 .34.445 ..14 cpu) = 5.806*16* piu f 19 6P14) = x (real) Old 2. 3B .34.005 x4 Cpu) = 23.515741032.4. perunit veltage. Vi (PU) = vieraal) VB Micpy) = 415V Vi (PM) = 1pu Iov

10. secondany side ör' Zone - Zb= (@:230) 10.78.2 5 M1A/1000 12 CPU) = 0.07 = 661641ā=> pu.' «() = OB = (xs-KIC 4: V24p.4) = VzCreal) - Bove . VB 2300 - 1pu SOV = 1 P14 let. consumer terminal ör' load terminal veltage is refference veltage: Iz = V2 (pu) Lo V2 = I lopu per unit eanincitent ckt. Cho need to reffering the tramsformer in either side when per Unit value syften used) um in mo M. SM B jxz a ti T.4 е с y short, Ч(р) us drait - during I pecult 12=0

wing KUL in uresh 00 - -Ti (pw+ IC 14+b + jcxq+ke)) to 50 .: (te) +504+x2) I = 118 3 1 6*6.806+ 6616) + (23:515112.84% I= 1.18 103 37.895/ 70.2650_) I= 36-3881|S-70866° pour raginitude of current Il'I = 26.3881 poll this current is facilt current. :: If= I = 26 388). Pių - pault level at load sto Vici prepaut) * IF ste: 114*26:388) :sta 26.3881 pie seal value st (real) = squ*(V4) st (real) - 26.3881*5 KVA ist (real) = 1210 9405 KVA Answers