Transformer 2 a Ironcored Transformer d Tapped windings

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Transformer -2

Transformer -2

(a) Iron-cored Transformer (d) Tapped windings B A (b) Ferrite –cored Transformer (e) Auto.

(a) Iron-cored Transformer (d) Tapped windings B A (b) Ferrite –cored Transformer (e) Auto. Transformer (c) Multiple -windings B A F E D C C B H G F E D C A

Construction of transformer Laminated steel-core transformer

Construction of transformer Laminated steel-core transformer

Primary windings of a 30 k. VA, 6000 V/230 V transformer has a resistance

Primary windings of a 30 k. VA, 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 I 1 R 1 10 V 1 6000/230 V Xe 23 V 1 ’ R 2 I 2 0. 016 V 2 ’ 30 k. VA V 2

Equivalent resistance of primary and secondary referred to primary is I 1 R e

Equivalent resistance of primary and secondary referred to primary is I 1 R e V 1 Full-load current at primary Xe I 2 6000/230 V V 1 ’ V 2 ’ 30 k. VA V 2

Losses in transformer on load categorized into two 1. I 2 R losses in

Losses in transformer on load categorized into two 1. I 2 R losses in primary and secondary windings, namely I 12 R 1+I 22 R 2. 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. Therefore total loss or R 1 e=equivalent resistance of primary and secondary referred to primary R 2 e=equivalent resistance of primary and secondary referred to secondary

Note: p. f= power factor The wire losses can be expressed as Divided by

Note: p. f= power factor The wire losses can be expressed as Divided by I 2

It is also possible to express Or

It is also possible to express Or

The primary and secondary windings of a 500 k. VA transformer have a resistances

The primary and secondary windings of a 500 k. VA 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 k. W , 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 Coil Wire loss at primary PW 1=I 12 R 1=75. 82 x 0. 42=2415 W Coil Wire loss at secondary PW 2=I 22 R 2=12502 x 0. 0011=1720 W

PW = PW 1 + PW 2 = 2415 + 1720 = 4. 135

PW = PW 1 + PW 2 = 2415 + 1720 = 4. 135 k. W Total loss PL= PW + PC = 4. 135 + 2. 9 = 7. 035 k. W Output power on full load Pout = 500 x 0. 8 = 400 k. W Therefore Pin = Pout + PL = 400+7. 035 = 407. 035 k. W

(b) Since the wire loss varies as square of the current , thus Losses

(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. 034 k. W Total Loss on half-load PL= PC+PW/2=2. 9+1. 034=3. 934 k. W Output power on half-load Pout/2= 400/2 = 200 k. W Input power on half-load Pin/2= Pout/2+PL=200+3. 934 = 203. 934 k. W

(c) Full-load I 2 R loss is PW = 4. 135 k. W Let

(c) Full-load I 2 R loss is PW = 4. 135 k. W Let n= fraction of full-load appearance power at which it is maximum efficiency Total I 2 R loss is = n 2 x 4. 135 k. W=2. 9 Therefore n=0. 837 Output at maximum efficiency is= 0. 837 x 500= 418. 5 k. WA Output power at power factor 0. 8= 418. 5 x 0. 8 = 334. 8 k. WA Since the core and I 2 R are equaled, then total loss is PL= 2 x 2. 9 = 5. 8 k. W

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