Transcript Document
EXAMPLE
A pole mounted 100kVA distribution transformer has the following
characteristics
R1 = 1.56W
R2 = 0.005W
X1 = 4.66W
X2 = 0.016W
Volts Ratio = 6600 / 230
(a) 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
V1 = 6600V, voltage turns ratio = 6600 / 230 = 28.7
Therefore 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 =
RO =
XM =
I0 - Im
V1 / I0
V1 / Im
(b) A T/F at 100kVA
= 0.251 - j0.9680 Amps
= 6600 / 0.251 =
26.3kW
= 6600 / 0.9680 =
6.82kW
For HV side,
IFL1
= kVA / V1
= 100 000 / 6 600
= 15.15 Amps
For LV side,
IFL2
= kVA / V2
= 100 000 / 230
= 434.78 Amps
Load current = 15.15
where = cos-1 pf
= cos-1 0.8
= -36o
Therefore Load Current = 15.15 -36.9o
E1 = V1 - ILZL
E1 = V1 - IL( RS + XS )
= ( 6600 + j0 ) – 15.15 -36.9o ( 5.68 + j17.84 )
= ( 6600 + j0 ) – { [15.15 -36.9o ] [ 18.72 o ] }
= 6600 – { 283.6 o }= 6600 – 230.7 – j164.6
= 6369.3 – j164.6 V
(6371.4 -18 o V)
E2 = V2 = E1 / a
= 6371.4 / 28.7 = 222 V
Regulation = (VNL - VFL)
/ VFL
= 100 x (230 - 222) / 222
= 3.6%
(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
=
100 x [ ( 6371) ( 15.15 ) ( cos –36.9 ) ]
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[( 6371 ) ( 15.15 ) ( cos –36.9 ) + 1660 + 1300 ]
=
96.3%
N.B. a transformer only has two losses, no load losses and cu losses.