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Transformer -2 (a) Iron-cored Transformer (d) Tapped windings B A (b) Ferrite –cored Transformer (e) AutoTransformer (c) Multiplewindings F E D C C B H B A G F E D C A Construction of transformer Laminated steel-core transformer Primary windings of a 30 kVA, 6000 V/230 V transformer has a resistance of 10 , the secondary windings has a resistance of 0.016 . The total reactance of the transformer referred to primary is 23 . Calculate the voltage regulation of the transformer when it supplies a the full-load current at power factor of 0.8 lagging I1 R1 10 V1 6000/230 V Xe 23 V1’ R2 I2 0.016 V2’ 30 kVA V2 Equivalent resistance of primary and secondary referred to primary is 2 V1 6000 Re R1 R2 10 0.016 21 V 230 2 2 cos2 0.8 I1 Re sin 2 0.6 Full-load current at primary 6000/230 Xe I2 V V1 V1 V2 ’ ’ 30 KVA 30000 I1 5A V 6000 kVA Per unit voltageregulation I1 Re cos2 X e sin 2 V1 521 0.8 23 0.6 0.0255 6000 2.55 % V2 Losses in transformer on load categorized into two 1. I2R losses in primary and secondary windings, namely I12R1+I22R2. 2. Core losses due to hysteresis and eddy currents Usually variation between no-load and full-load can be negligible thus, the total core loss, PC , is assumed to be constant all the time. 2 2 P I R I Therefore total loss C 1 1 2 R2 or PC I12 R1e PC I 22 R2e R1e=equivalent resistance of primary and secondary referred to primary R2e=equivalent resistance of primary and secondary referred to secondary output power output power Efficiency input power output power losses I 2V2 p. f . I 2V2 p. f . Pc I12 R1 I 22 R2 Note: p.f= power factor The wire losses can be expressed as Efficiency Divided by I2 V R2e R2 R1 2 V1 2 N R2 R1 2 N1 I 2V2 cos. I 2V2 cos. Pc I 22 R2e V2 cos. Efficiency V2 cos. Pc / I 2 I 2 R2e 2 It is also possible to express output power input power losses Efficiency input power input power Or losses 1 input power The primary and secondary windings of a 500 kVA transformer have a resistances of 0.42 and 0.0011 respectively. The primary and secondary voltages are 6600 V and 400 V respectively and the core loss is 2.9 kW , assuming the power factor of the load to be 0.8. Calculate the efficiency on (a) fullload and (b) half-load. (c) assuming the power factor 0.8, find output which the efficiency of the transformer is maximum. (a) Primary current on full-load Secondary current on full-load 500103 I1 75.8 A 6600 500103 I2 1250A 400 Coil Wire loss at primary PW1=I12R1=75.82 x 0.42=2415W Coil Wire loss at secondary PW2=I22R2=12502 x 0.0011=1720W PW = PW1 + PW2 = 2415 + 1720 = 4.135kW Total loss PL= PW + PC = 4.135 + 2.9 = 7.035kW Output power on full load Pout = 500 x 0.8 = 400 kW Therefore Pin = Pout + PL = 400+7.035 = 407.035 kW losses 7.035 1 1 0.9827 per unit Input 407.035 98.27% (b) Since the wire loss varies as square of the current , thus Losses on half-load PW/2=4.135/22=4.135/4=1.034kW Total Loss on half-load PL= PC+PW/2=2.9+1.034=3.934kW Output power on half-load Pout/2= 400/2 = 200kW Input power on half-load Pin/2= Pout/2+PL=200+3.934 = 203.934kW Loss 3.934 1 1 0.9807 per unit Input 203.934 98.07% (c) Full-load I2R loss is PW = 4.135kW Let n= fraction of full-load appearance power at which it is maximum efficiency Total I2R loss is = n2 x 4.135 kW=2.9 Therefore n=0.837 Output at maximum efficiency is= 0.837 x 500= 418.5kWA Output power at power factor 0.8= 418.5 x 0.8 = 334.8 kWA Since the core and I2R are equaled, then total loss is PL= 2 x 2.9 = 5.8 kW Maximum 1 Loss 5.8 1 0.983 per unit 98.3% Input 334.8 5.8