Question

Name: ID# Q.2 A single phase, 100kVA, 1000/100 V transformer is to be tested to determine its excitation branch component and its series impedances. The following test data have been taken from the high-voltage side of the transformer: Open-circuit test (1000V, 0.6A, 400W); Short-circuit test (50V, 100A, 1800W). (25 marks) a) Find the equivalent circuit of the transformer referred to the high-voltage side? b) Find the equivalent circuit of the transformer referred to the low-voltage side? c) The core losses at rated parameters?

Answer 1

Criren A 1-0, 100KVA 1000/1001 transformer 4oow O.C Test : TOOOV OGA / S.C Test 5ov, 100 1800 W solution branch conponents open circuit To deternine encitation will obtain from test data- have test we So circuit from open Noc 1000V loc = 0.GA Poc 4oo Woo 2 no load power factor Step ① finding 100 Poc Voc. loc 400 400 6oo 1000 x 0.6 cosdo= 0.64 1 de 0-1 (0.67) Hence 0 48. 19° the we Kow that no load so current QS components of Celtrent that has two CS Scanned with CamScanner

and second ore magretising component cursed Loss corporent Curret. core V loc Phasor Pc Do Therefore I'm a loc sindo Toc Les do & Ic = 0.6 x sin (48.19) Hence , I'm o.1453 0.6 x Im 0.4472 Arps 2 1. 0.6 x 0.6f Apps 0.402 Pc determining Rc 8 xm Now 1000 or 0.402 Rc Voc Ic 2487.56 N Rc 10oo R Voc & 0.4472 xm Im 2236.14 n xm- CS Scanned with CamScanner

Now will use series impedances Circuit test data. Short have sic So, from Test هوا Psc = 180ow 100 A Isc 5ov Vsc = Roc. Calculating 1800 100x100 PSC 1800 (100)? RSC = (Isc) 2 RSC 0.18 on 5o Vsc 2 Esc 100 Isc 0.2 n Esc cwe krow Therefore Zsc = Rsch + (xsc.) 2 2 2 RSC + Xsch Esc or 2 2 Esc RSC XSC Hence Xsc J (02) -0.18)? Jo.oy-0.0324 0.087 18 n x sc CS Scanned with CamScanner

so have calculated excitation branch all the poraneters te & series irpedances. the Rc xm o encitation Branch 2481.56 22 36.140 R sc 1 series Xsc 0.087 180 Zsc Oran 0-18 n irpe danu An Parta of the Circuit tray former to high voltage equivalent reffered side on carried Ligh voltage have alkeady the test Since we therefore parameters Side calculated. Is/a 0.180 jo.087180 lp + Sooo00 La Escaoilu avs 2487.564 j2236.14 Vp trans forner Part le circuit of the في ما equivalent reffered voltage side. to 2 a have to divide my we this So tern ratio where cell inpedances to 10 Calculating 100 .. la = 10) CS Scanned with CamScanner

Reffered Rc 2487.56 = 24.8756 on 100 Reffered xm 22.36 14 d 22 36.14 102 Reffered Rsc 0.00185 0.18 102 0.0008 1180 Reffered Xsc = 0.08718 102 Reffered Zsc 0.2 0.002 n 102 so we have all reffered paraneters side. low Voltage circuit diagram reffered Hene equivalent Voltage side to جب ما 0.00181 jo.000871871 alp Ps Escz 0.002 ore 24 87560 2 22.3614 n Vs Parte Core losesses at rated parareters briven Sraled 100 KVA Hence Curreant. on secondary fell load side. CS Scanned with CamScanner

Srated Vooted. Prated Irated Srated Nrated 3 1%XIO 100 Hence Trated =1000 Arps to low KVL on reffered Now have a opplying voltage side. so we Irated ( Rsct .jxsc) Vs + Vela 1000L0 (0.0018 +j 0.0008718) 100 200 or VP/a + 1.8 + J 0.8718 100 UP/a vela 10118 + j 0.8718 J1101.39€ + (0.87183° Lten a Come vela 101.8037 20:49069 VPla 2 2 [Velal (101.8034) wss core 24.8756 Rc walts 416.63 Loss Pcore ** Note Voltage side for low Vprinary ve 07 V secondary CS Scanned with CamScanner

feel free to ask if any doubt regarding the solution and concept thank you