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Sheet 2 Example 1: A 20 kVA, 2500/250 V, 50Hz, single-phase transformer gave the following test result: Open-circuit test: 250V, 1.5A, and 100 watts measured on the low voltage side Short-circuit test: 100V, 8A, and 320 watts measured on the on high voltage side (a) Compute the parameters of the equivalent circuits and draw the equivalent circuit of the transformer showing all the values. Solution: From the given test results data, we have Req1 Psc 320 2 =5Ω I sc2 8 2 V X eq1 sc 52 = 12 Ω I sc Reh Voc2 2502 =12 Ω Poc 100 2 V X m oc 122 =173Ω I oc Re+h and Xm are low-voltage side values, which can be converted to the high voltage side 2 Reh N Reh * 1 62.5 k Ω N1 2 N X m X m * 1 17.3 k Ω N1 Example 2: A 10 kVA, 2400/240 V, single-phase transformer have the following parameters: R1 = 3Ω, R2 = 15 Ω the leakage reactance are X1 = 0.03 Ω and X2 = 0.15 Ω. Calculate the full-load voltage regulation of the transformer when operated at rated secondary voltage and (a) 0.8 power factor lagging and (b) 0.8 power factor leading. Solution: R eq2 = R2 + R1' = R2 + R1 / α 2 = = 0.03 + 3/ (10)2 = 0.06 Ω X eq2 = X2 + X1' = X2 + X1 / α 2 = = 0.15 + 15/ (10)2 = 0.3 Ω Zeq2 = Req2 + j X eq2 = 0.06 + j 0.3 Ω = 0.3 78.7 0 (a) For a 0.8 power factor, lagging I2 S = 41.7 A V2 Referring to Fig. 10.17 V1 V2 I 2 Z eq 2 240 0 41.7 37 0 X 0.3 78.7 0 = 249.5 + j 8.5 = 249.65 1.950 V1 V2 249.65 240 VR% = X 100 = X 100 = 4% 240 V2 (b) For a 0.8 power factor, leading I 2 = 41.7 37 0 A V1 V2 I 2 Z eq 2 240 0 41.7 37 0 X 0.3 78.7 0 = 234.5 + j 11.5 = 235 2.80 249.65 240 VR% = * 100 = -2.18% 240 Practical Problems 1- A 50 kVA, 4400/220 V transformer has R1 = 3.45 Ω, R2 = 0.009 Ω. The values of reactances are X1 = 5.2 Ω and X2= 0.015 Ω. Calculate for the transformer : a. b. c. d. e. Equivalent resistance as referred to primary Equivalent resistance as referred to secondary Equivalent reactance as referred to both primary and secondary, Equivalent impedance as referred to both primary and secondary, and Total copper loss, first using individual resistances of the two windings and secondly, using equivalent resistance as referred to each side. 2. The primary and secondary winding resistance of a 40 kVA, 6600/250 V singlephase transformer are 10 Ω and 0.05 Ω respectively. The equivalent leakage reactance as referred to the primary winding is 35 Ω. Find the full-load regulation for load power factors of a. unity ; b. 0.8 lagging ; and c. 0.8 leading 3. A 20 kVA, 2200/220 V 50-Hz distribution transformer is tested for efficiency and regulation as follows; d. e. f. g. h. i. j. k. l. O.C. test (L.V. side) : 220 V, 4.2 A, 148 W. S.C. test (H.V. side) : 86 V, 10.5 A, 360 W. Determine: (i) Core loss (ii) Equivalent resistance referred to primary. (iii)Equivalent resistance referred to secondary. (iv) Equivalent reactance referred to primary. (v) Equivalent reactance referred to secondary. (vi) Regulation of transformer at 0.8 power factor lagging current.