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EXAMPLE A pole mounted 100kVA distribution transformer has the following characteristics

R 1 = 1.56 W. R 2 = 0.005 W. X 1 = 4.66 W. X 2 = 0.016 W. EXAMPLE A pole mounted 100kVA distribution transformer has the following characteristics. Volts Ratio = 6600 / 230. If the no load current is given by 0.251 - j0.9680, find R O and X M

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EXAMPLE A pole mounted 100kVA distribution transformer has the following characteristics

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  1. R1 = 1.56W R2 = 0.005W X1 = 4.66W X2 = 0.016W EXAMPLE A pole mounted 100kVA distribution transformer has the following characteristics Volts Ratio = 6600 / 230

  2. If the no load current is given by 0.251 - j0.9680, find RO and XM • (b) For a full load current at 0.8pf lag, find the secondary voltage and • voltage regulation • (c) Find the efficiency of the transformer

  3. V1 = 6600V, voltage turns ratio = 6600 / 230 = 28.7Therefore V2 = 230            R2 is going from LV to HV side - it therefore increases in value.                                     Since a2 = (28.7)2                 RS = R1 + (28.7)2 R2                     = 5.68W                 XS = X1 + (28.7)2 X2                     = 17.84W (a) INL = I0 -Im = 0.251 - j0.9680 Amps             RO = V1 / I0 = 6600 / 0.251 = 26.3kW            XM = V1 / Im = 6600 / 0.9680 = 6.82kW (b) A T/F at 100kVA

  4. For HV side, IFL1 = kVA / V1                                               = 100 000 / 6 600                                              = 15.15 AmpsFor LV side, IFL2 = kVA / V2= 100 000 / 230                                              = 434.78 Amps Load current = 15.15 Ðq                                                  where q = cos-1 pf                                                               = cos-1 0.8                                                               = -36o                Therefore Load Current = 15.15 Ð-36.9o

  5. E1 = V1 - ILZLE1 = V1 - IL( RS + XS )= ( 6600 + j0 ) – 15.15 Ð-36.9o ( 5.68 + j17.84 )  = ( 6600 + j0 ) – { [15.15 Ð-36.9o ] [ 18.72 Ð 72.3 o ] } = 6600 – { 283.6 Ð 35.4 o }= 6600 – 230.7 – j164.6 = 6369.3 – j164.6 V (6371.4 Ð -1.48 o V)           E2 = V2 = E1 / a                = 6371.4 / 28.7 = 222 V       Regulation = (VNL - VFL)      / VFL                       = 100 x (230 - 222) / 222                       = 3.6%

  6. (c) Efficiency = Pout / Pin = (Pin - Losses) / Pin                    LOSSES                              No load losses = IO2RO = (0.251)2(26 300)                                                                  = 1.66kW                             Cu Losses = I12RS                                             = (15.15)2(5.68)                                             = 1.3kW         So Efficiency = 100 x [V2I2cosf] / [V2I2cosf + sum of losses]                        For max efficiency PNL = PCu

  7. = 100 x [ ( 6371) ( 15.15 ) ( cos –36.9 ) ] ________________________________________________________________ [( 6371 ) ( 15.15 ) ( cos –36.9 ) + 1660 + 1300 ] = 96.3% N.B. a transformer only has two losses, no load losses and cu losses.

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