Practical Transformer on Load We now consider the deviations from the last two
ideality conditions :
1. The resistance of its windings is zero.
2. There is no leakage flux.
The effects of these deviations become more prominent when a practical transformer is put on load.
Monday, August 01, 2011 Transformers 1Next
(1) Effect of Winding Resistance Current flow through the windings causes a power loss
called I2R loss or copper loss.
This effect is accounted for by including a resistance R1
in the primary and resistance R2 in the secondary
Monday, August 01, 2011 Transformers 2Next
(2) Effect of Flux Leakage The difference between the total flux linking with the
primary and the useful mutual flux Φu linking with both the windings is called the primary leakage flux, ΦL1.
Similarly, ΦL2 represents the secondary leakage flux.
Flux leakage results in energy being alternately stored in and discharged from the magnetic fields with each cycle of the power supply.
It is not directly a power loss, but causes the secondary voltage to fail to be directly proportional to the primary voltage, particularly under heavy loads.
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Leakage flux in a transformer
Monday, August 01, 2011 Transformers 4
(a) Its definition. (b) Its effect accounted for.
• The useful mutual flux Φu is responsible for the transformer action.• The leakage flux ΦL1 induces an emf EL1 in the primary winding.
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Similarly, flux ΦL2 induces an emf EL2 in the secondary.
Hence, we include reactances X1 and X2 in the primary and secondary windings, in the equivalent circuit.
The paths of leakage fluxes ΦL1 and ΦL2 are almost entirely due to the long air paths and are therefore practically constant.
The reluctance of the paths being very high, X1 and X2
are relatively small even on full load.
However, the useful flux Φu remains almost independent of the load.
Monday, August 01, 2011 Transformers 5Next
Monday, August 01, 2011 Transformers 6Next
Equivalent Circuit of a Transformer
Monday, August 01, 2011 Transformers 7
It is merely a representation of the following KVL equations :
1 1 1 1 1 1 1 1 1 1( )I R jI X I R jX V E E
2 2 2 2 2 2 2 2 2 2( )I R jI X I R jX E V V
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Points to Draw Phasor Diagram
1. Resistive voltage drop in phase with current phasor.
2. Inductive voltage drop in quadrature with current.
3. To get V1, add I1Z1 to –E1.
4. Add V2 and I2Z2 , to get E2.
5. Current I1 is vector sum of I0 and I2’.
6. Angle between V1 and I1 gives the power factor angle of the transformer.
Monday, August 01, 2011 Transformers 8
Monday, August 01, 2011 Transformers 9
m
E2
E1
-E1
I1R1
I1X1V1
I1
I0
I1'
I2
V2
I2R2
I2X2
0
1
Ph
aso
r D
iag
ram
fo
r
Pra
ctic
al T
ran
sfo
rmer
on
R
esi
stiv
e L
oa
d
1 1 1 1 1
2 2 2 2
( )
( )
I R jX
V I R jX
2
V E
E
O
I1Z1
I2Z2
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Monday, August 01, 2011 Transformers 10
Pra
ctic
al T
ran
sfo
rmer
on
In
du
ctiv
e L
oa
d
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Monday, August 01, 2011 Transformers 11
Pra
ctic
al T
ran
sfo
rmer
on
C
ap
aci
tive
Lo
ad
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Simplified Equivalent Circuit The no-load current I0 is only about 3-5 % percent of the
full-load current.
The exciting circuit R0-X0 in is shifted to the left of impedance R1-X1.
Monday, August 01, 2011 Transformers 12
Transforming the impedances from the secondary to the primary side.
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Monday, August 01, 2011 Transformers 13
Equivalent resistance and reactance referred to the primary side
2 2
e1 1 2 e1 1 2( / ) and ( / )R R R K X X X K
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Approximate Equivalent Circuit
Monday, August 01, 2011 Transformers 14Next
Example 5 A single-phase, 50-kVA, 4400-V/220-V, 50-Hz transformer
has R1 = 3.45 Ω, R2 = 0.009 Ω, X1 = 5.2 Ω and X2 = 0.015 Ω. Calculate
(a) the Re as referred to the primary,
(b) the Re as referred to the secondary,
(c) the Xe as referred to the primary,
(d) the Xe as referred to the secondary,
(e) the Ze as referred to the primary,
(f) the Ze as referred to the secondary, and
(g) the total copper loss.
Monday, August 01, 2011 Transformers 15Next
Monday, August 01, 2011 Transformers 16
Solution : Full-load primary current,
1
1
kVA 5000011.36 A
4400I
V
Full-load secondary current, 2
2
kVA 50000227.27 A
220I
V
2
1
220 10.05
4400 20
VK
V
(a) 2 2
e1 1 2( / ) 3.45 [0.009/(0.05) ]R R R K 7.05 Ω
(b)
(c)
2 2
e2 1 2 (0.05) 3.45 0.009R K R R 0.0176 Ω
2 2
e1 1 2( / ) 5.2 [0.015/(0.05) ]X X X K 11.2 Ω
(d) 2 2
e2 1 2 (0.05) 5.2 0.015X K X X 0.028 Ω
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Monday, August 01, 2011 Transformers 17
(e) 2 2 2 2
e1 e1 e1 (7.05) (11.2)Z R X 13.23 Ω
(f)
(g) Total copper loss
2 2 2 2
e2 e2 e2 (0.0176) (0.028)Z R X 0.0331 Ω
2 2 2 2
1 1 2 2 (11.36) 3.45 (227) 0.009I R I R 909 W
Alternatively, by considering equivalent resistances, total copper loss
2 2
1 e1 (11.36) 7.05I R 909.8 W
2 2
2 e2 (227.27) 0.0176I R 909 W
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Voltage Regulation
Monday, August 01, 2011 Transformers 18
V2(0) = secondary terminal voltage at no load,
and V2 = secondary terminal voltage at full load.
The voltage regulation of a transformer is defined as the change in its secondary terminal voltage from no load to full load, the primary voltage being assumed constant.
The voltage drop V2(0) - V2 is called the inherent regulation.
2(0) 2
2(0)
( ) Per unit V V
iV
regulation down
2(0) 2
2(0)
% 100V V
V
regulation down
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Monday, August 01, 2011 Transformers 19
2(0) 2
2
( ) Per unit V V
iiV
regulation up
2(0) 2
2
% 100V V
V
regulation up
Normally, when nothing is specified, ‘regulation’ means ‘regulation down’.
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Monday, August 01, 2011 Transformers 20
Exact voltage drop =
2(0) 2 OC OA OG OA AG AF+FGV V
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Approximate Voltage Drop
The secondary terminal voltage at no load,2(0) 2 1 1V E KE KV
Monday, August 01, 2011 Transformers 21
In case of leading power factor,
2 e2 2 e2
Approximate voltage drop, AF AE EF AE BD
cos sinI R I X
2 e2 2 e2
Approximate voltage drop, AF AE EF AE BD
cos sinI R I X
In general,
2 e2 2 e2Approximate voltage drop cos sinI R I X
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Monday, August 01, 2011 Transformers 22
2 2 2 2
2(0)
cos sin% Regulation 100
cos sin
e e
r x
I R I X
V
V V
Condition for Zero Regulation :Possible only if the load has leading power factor.
2 2 2 2cos sin 0e eI R I X 2
2
tan e
e
R
X
Use + sign for lagging power factor and – sign for leading power factor.
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Condition for Maximum Regulation
Monday, August 01, 2011 Transformers 23
2 2 2 2( cos sin ) 0e e
dI R I X
d
2 2 2 2( sin cos ) 0e eI R I X
2
2
tan e
e
X
R
Maximum regulation can occur only for inductive load. The voltage drop is maximum when
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Monday, August 01, 2011 Transformers 24
Example 6
Solution :
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Monday, August 01, 2011 Transformers 25
Example 7
Solution :
2the load voltage, 240 6V 234V
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Example 8 A single-phase, 40-kVA, 6600-V/250-V,
transformer has primary and secondary resistances R1 = 10 Ω and R2 = 0.02 Ω, respectively. The equivalent leakage reactance as referred to the primary is 35 Ω. Find the full-load regulation for the load power factor of
(a) unity,
(b) 0.8 lagging, and
(c) 0.8 leading.
Monday, August 01, 2011 Transformers 26Next
Monday, August 01, 2011 Transformers 27
2 2
e2 1 2 (0.0379) 10 0.02 0.0343R K R R
2 2
e2 e1and (0.0379) 35 0.0502X K X
(a) For power factor, cos = 1; sin = 0. Hence,
2 2 2 2
2(0)
cos sin% Regulation 100
160 0.0343 1 0100
250
e eI R I X
V
2.195 %
Solution : Given : R1 = 10 Ω; R2 = 0.02 Ω; Xe1 = 35 Ω
2
250 the turns-ratio, 0.0379
6600
40000the full-load current, 160 A
250
K
I
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Monday, August 01, 2011 Transformers 28
(b) For power factor, cos = 0.8 (lagging, positive);
(c) For power factor, cos = 0.8 (leading, negative);
sin 0.6
2 e2 2 e2
2(0)
cos sin% Regulation 100
160 0.0343 0.8 160 0.0502 0.6100
250
I R I X
V
0.172 %
2sin 1 cos 0.6
2 e2 2 e2
2(0)
cos sin% Regulation 100
160 0.0343 0.8 160 0.0502 0.6100
250
I R I X
V
3.68 %
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