EXAMPLE A pole mounted 100kVA distribution transformer has the following characteristics R 1 = 1.56  R 2 =  X 1 = 4.66  X 2 =  Volts Ratio = 6600 / 230

(a)If the no load current is given by j0.9680, find R O and X M (b) For a full load current at 0.8pf lag, find the secondary voltage and voltage regulation (c) Find the efficiency of the transformer

V 1 = 6600V, voltage turns ratio = 6600 / 230 = 28.7 Therefore V 2 = 230 R 2 is going from LV to HV side - it therefore increases in value. Since a 2 = (28.7) 2 R S = R 1 + (28.7) 2 R 2 = 5.68  X S = X 1 + (28.7) 2 X 2 =  (a) I NL = I 0 - I m = j Amps R O = V 1 / I 0 = 6600 / = 26.3k  X M = V 1 / I m = 6600 / = 6.82k  (b) A T/F at 100kVA

For HV side, I FL1 = kVA / V 1 = / = Amps For LV side, I FL2 = kVA / V 2 = / 230 = Amps Load current =  where  = cos -1 pf = cos = -36 o Therefore Load Current =  o

E 1 = V 1 - I L Z L E 1 = V 1 - I L ( R S + X S ) = ( j0 ) –  o ( j17.84 ) = ( j0 ) – { [15.15  o ] [  o ] } = 6600 – {  o }= 6600 – – j164.6 = – j164.6 V (  o V) E 2 = V 2 = E 1 / a = / 28.7 = 222 V Regulation = (V NL - V FL ) / V FL = 100 x ( ) / 222 = 3.6%

(c) Efficiency = P out / P in = (P in - Losses) / P in LOSSES No load losses = I O 2 R O = (0.251) 2 (26 300) = 1.66kW Cu Losses = I 1 2 R S = (15.15) 2 (5.68) = 1.3kW So Efficiency = 100 x [V 2 I 2 cos  ] / [V 2 I 2 cos  + sum of losses] For max efficiency P NL = P Cu

= 100 x [ ( 6371) ( ) ( cos –36.9 ) ] ________________________________________________________________ [( 6371 ) ( ) ( cos –36.9 ) ] = 96.3% N.B. a transformer only has two losses, no load losses and cu losses.