Single phase Transformer

Vivekananda College of Engineering for Women
Department of Electrical & Electronics
Engineering
Electrical Machines – I
[Sub.Code - U15EE413 – Regulation 2015 ]
UNIT – IV - TRANSFORMERS
Present by
Mr. A. Johny Renoald M.E, Ph.D

UNIT – IV
 Transformers - Principle of operation
 Types
 Basic construction
 Equivalent circuit
 Regulation and efficiency
 Auto transformer

Transformer Equation
• Faraday’s Law states that,
– If the flux passes through a coil of wire, a voltage will be induced in
the turns of wire. This voltage is directly proportional to the rate of
change in the flux with respect of time.
If we have N turns of wire,
dt
t
d
Emf
V ind
ind
)
(




dt
t
d
N
Emf
V ind
ind
)
(




Lenz’s Law
8
Dr.B.GOPINATH, Prof / EEE

• For an ac sources,
– Let V(t) = Vm sint
i(t) = im sint
Since the flux is a sinusoidal function;
Then:
Therefore:
Thus:
t
t m 
sin
)
( 


t
N
dt
t
d
N
Emf
V
m
m
ind
ind



cos
sin







m
m
ind
ind fN
N
Emf
V 



 
 2
(max)
m
m
m
rms
ind fN
fN
N
Emf 




 44
.
4
2
2
2
)
(


28

• For an ideal transformer
• In the equilibrium condition, both the input power will be equaled to the output
power, and this condition is said to ideal condition of a transformer.
• From the ideal transformer circuit, note that,
• Hence, substitute in (i)
m
m
fN
E
fN
E




2
2
1
1
44
.
4
44
.
4
1
2
2
1
2
2
1
1 cos
cos
I
I
V
V
I
V
I
V
power
output
power
Input






………………… (i)
2
2
1
1 V
E
and
V
E 

29

a
I
I
N
N
E
E
Therefore 


1
2
2
1
2
1
,
Where, ‘a’ is the Voltage Transformation Ratio; which will
determine whether the transformer is going to be step-up or
step-down
E1 > E2
For a >1
For a <1 E1 < E2
30

Transformer Rating
• Transformer rating is normally written in terms of Apparent
Power.
• Apparent power is actually the product of its rated current and
rated voltage.
2
2
1
1 I
V
I
V
VA 

 Where,
 I1 and I2 = rated current on primary and secondary winding.
 V1 and V2 = rated voltage on primary and secondary winding.
 Rated currents are actually the full load currents in transformer
31

Conservator Tank of a Transformer
• This is a cylindrical tank mounted on
supporting structure on the roof of the
transformer's main tank. When
transformer is loaded, the temperature of
oil increases and consequently the
volume of oil in the transformer gets
increased. Again; when ambient
temperature is increased, the volume of
oil is also increased.
• The conservator tank of a transformer
provides adequate space for expansion of
oil. Conservator tank of transformer also
acts as a reservoir of oil.
32

Silica Gel Breather of Transformer
• Whenever electrical power
transformer is loaded, the
temperature of the transformer
insulating oil increases,
consequently the volume of the oil
is increased.
• The color of silica gel crystal is dark
blue but, when it absorbs moisture;
it becomes pink.
33

Explosion Vent of Transformer
• The purpose of the
explosion vent in a
transformer is to
prevent damage of
the transformer tank
be releasing any
excessive pressure
generated inside the
transformer.
34

Construction of Buchholz Relay
• Buchholz relay in transformer is an oil container housed the
connecting pipe from main tank to conservator tank. It has
mainly two elements. The upper element consists of a float.
The float is attached to a hinge in such a way that it can move
up and down depending upon the oil level in the Buchholz
relay Container.
• One mercury switch is fixed on the float. The alignment of
mercury switch hence depends upon the position of the float.
The lower element consists of a baffle plate and mercury
switch.
• This plate is fitted on a hinge just in front of the inlet (main
tank side) of Buchholz relay in transformer in such a way that
when oil enters in the relay from that inlet in high pressure the
alignment of the baffle plate along with the mercury switch
attached to it, will change.
35

Cont…
36

Practical Transformer (Equivalent Circuit)
V1 = primary supply voltage
V2 = 2nd terminal (load) voltage
E1 = primary winding voltage
E2 = 2nd winding voltage
I1 = primary supply current
I2 = 2nd winding current
I1
’ = primary winding current
Io = no load current
Ic = core current
Im = magnetism current
R1= primary winding resistance
R2= 2nd winding resistance
X1= primary winding leakage reactance
X2= 2nd winding leakage reactance
Rc= core resistance
Xm= magnetism reactance
V1
I1 R
1
X1
RC
Ic
Xm
Im
Io
E1 E2
V2
I1
’
N1: N2
R2
X2
Load
I2
38

R2’ = R2 / K2
X2’ = X2 / K2
V2’ = V2 / K
ZL = ZL / K2
I2’ = I2K 40

Single Phase Transformer
(Referred to Primary)
• Actual Method
2
2
2
2
2
2
1
2 '
' R
a
R
OR
R
N
N
R 









2
2
2
2
2
2
1
2 '
' X
a
X
OR
X
N
N
X 








 a
I
I
aV
V
OR
V
N
N
V
E
2
2
2
2
2
2
1
'
2
1
'
'












V1
I1 R1 X1
RC
Ic
Xm
Im
Io
E1 E2 V2
I2
’ N1: N2
R2
’
X2
’
Load
I2
42

• Approximate Method
2
2
2
2
2
2
1
2 '
' R
a
R
OR
R
N
N
R 









V1
I1 R1
X1
RC
Ic
Xm
Im
Io
E1 E2 V2
I2
’ N1: N2
R2
’ X2
’
Load
I2
2
2
2
2
2
2
1
2 '
' X
a
X
OR
X
N
N
X 









a
I
I
aV
V
OR
V
N
N
V
E
2
2
2
2
2
2
1
'
2
1
'
'












43

2
2
2
2
2
2
1
2 '
' R
a
R
OR
R
N
N
R 









V1
I1
R01 X01
aV2
2
2
2
2
2
2
1
2 '
' X
a
X
OR
X
N
N
X 









'
'
2
1
01
2
1
01
X
X
X
R
R
R




2
2
2
2
1
'
2 ' aV
V
OR
V
N
N
V 









In some application, the excitation branch
has a small current compared
to load current, thus it may be neglected
without causing serious error.
44

(Referred to Secondary)
• Actual Method
2
1
1
1
2
1
2
1 '
'
a
R
R
OR
R
N
N
R 









a
V
V
OR
V
N
N
V 1
1
1
1
2
1 '
' 









I1’ R1’ X1’
RC’
Ic
Xm’
Im
Io
I2 R2
X2
V2
2
1
1
1
2
1
2
1 '
'
a
X
X
OR
X
N
N
X 









a
V1
45

(Referred to Secondary)
2
1
1
1
2
1
2
1 '
'
a
X
X
OR
X
N
N
X 









2
1
02
2
1
02
'
'
X
X
X
R
R
R




I1’
R02 X02
V2
a
V1
2
1
1
1
2
1
2
1 '
'
a
R
R
OR
R
N
N
R 








 a
V
V
OR
V
N
N
V 1
1
1
1
2
1 '
' 









1
1' aI
I 
Neglect the excitation
branch
46

Transformer Losses
• Generally, there are two types of losses;
i. Iron losses :- occur in core parameters
ii. Copper losses :- occur in winding resistance
i. Iron Losses
i. Copper Losses
circuit
open
c
c
c
iron P
R
I
P
P 

 2
)
(
02
2
2
01
2
1
2
2
2
1
2
1
)
(
)
(
,
)
(
)
(
R
I
R
I
P
referred
if
or
P
R
I
R
I
P
P
cu
circuit
short
cu
copper






Poc and Psc will be discusses later in transformer test
49

Transformer Efficiency
• To check the performance of the device, by comparing the
output with respect to the input.
• The higher the efficiency, the better the system.
%
100
cos
cos
%
100
%
100
,
2
2
2
2









cu
c
losses
out
out
P
P
I
V
I
V
P
P
P
Power
Input
Power
Output
Efficiency



%
100
cos
cos
%
100
cos
cos
2
)
(
)
(








cu
c
n
load
cu
c
load
full
P
n
P
nVA
nVA
P
P
VA
VA






Where, if ½ load, hence n = ½ ,
¼ load, hence n = ¼ ,
90% of full load, n =0.9
Where Pcu = Psc
Pc = Poc
50

Power Factor
• Power factor = angle between Current
and voltage, cos 
V
I

 = -ve
V
I

 = +ve
V
I
 = 1
Lagging Leading unity
 Power factor always lagging for real transformer.
51

Voltage Regulation
• The purpose of voltage regulation is basically
to determine the percentage of voltage drop
between no load and full load.
• Voltage Regulation can be determine based
on 3 methods:
a) Basic Definition
b) Short – circuit Test
c) Equivalent Circuit
54

Voltage Regulation (Basic Definition)
• In this method, all parameter are being referred to
primary or secondary side.
• Can be represented in either
– Down – voltage Regulation
– Up – Voltage Regulation
%
100
. 


NL
FL
NL
V
V
V
R
V
%
100
. 


FL
FL
NL
V
V
V
R
V
55

Voltage Regulation (Short – circuit Test)
• In this method, direct formula can be used.
  %
100
cos
.
1
.


V
V
R
V
f
p
sc
sc 
 
If referred to primary side
  %
100
cos
.
2
.


V
V
R
V
f
p
sc
sc 
  If referred to secondary side
Note that:
‘–’ is for Lagging power factor
‘+’ is for Leading power factor
Isc must equal to IFL
56

Voltage Regulation (Equivalent Circuit )
• In this method, the parameters must be referred to primary or
secondary
  %
100
sin
cos
.
1
.
01
.
01
1



V
X
R
I
R
V
f
p
f
p 
 If referred to
primary side
If referred to
secondary side
Note that:
‘+’ is for Lagging power factor
‘–’ is for Leading power factor
j terms ~0
  %
100
sin
cos
.
2
.
02
.
02
2



V
X
R
I
R
V
f
p
f
p 

57

S.no IOC VOC WOC
1. 0.7 230 40
61

S.NO ISC VSC WSC
1. 4.5 15 90
P = VI ; 1KVA, 220/110V ; Ip = 1000/220 = 4.5 Amps
Is = 1000 / 110 = 0.09 Amps
67

Example 1 : A 5 KVA, 500/250 V, 50 Hz, single phase transformer gave the following readings,
O.C. Test : 500 V, 1 A, 50 W (L.V. side open)
S.C. Test : 25 V, 10 A, 60 W (L.V. side shorted)
Determine : i) The efficiency on full load, 0.8 lagging p.f.
ii) The voltage regulation on full load, 0.8 leading p.f.
iii) The efficiency on 60% of full load, 0.8 leading p.f.
iv) Draw the equivalent circuit referred to primary and insert all the values in it.
Solution : In both the tests, meters are on H.V. side which is primary of the transformer. Hence
the parameters obtained from test results will be referred to primary.
From O.C. test, Vo = 500 V, Io = 1 A, Wo= 50 W
... Cos Φo = Woc/Voc Ioc = 50/(500x1) = 0.1
... Iw = Ioc cos = 1 x 0.1 = 0.1 A
and Im = Io sin Φo = 1 x 0.9949 = 0.9949 A
... Ro =Vo /Ic = 500/0.1 = 5000 Ω
and Xo = Vo/Im = 500/0.9949 = 502.52 Ω and Wo = Pi= iron losses = 50 W
75

From S.C. test, Vsc = 25 V, Isc = 10 A, Wsc = 60 W
... Rsc = Wsc /Isc
2 = 60/(10)2 = 0.6 Ω
Zsc = Vsc /Isc = 25/10 = 2.5 Ω
... Xsc= √(2.52 - 0.62) = 2.4269 Ω
(I1) F.L. = VA rating/V1
= (5 x 103 )/500 = 10 A
and Isc = (I1) F.L.
... Wsc = (Pcu) F.L. = 60 W
i) η on full load, cos = 0.8 lagging
76

I
ii) For 60% of full load, n = 0.6 and cos Φ2 = 0.8 leading]
... Pcu = copper loss on new load = n2 x (Pcu) F.L.
= (0.6)2 x 60 = 21.6 W
= 97.103 %
iv) The equivalent circuit referred to primary is shown in the Fig.
77
ii) Regulation on full load, cos Φ2 = 0.8 leading
= - 1.95 %

Example 2 : The open circuit and short circuit tests on a 10 KVA, 125/250 V, 50 Hz,
single phase transformer gave the following results :
O.C. test : 125 V, 0.6 A, 50 W (on L.V. side)
S.C. test : 15 V, 30 A. 100 W (on H.V. side)
Calculate : i) copper loss on full load
ii) full load efficiency at 0.8 leading p.f.
iii) half load efficiency at 0.8 leading p.f.
iv) regulation at full load, 0.9 leading p.f.
Solution : From O.C. test we can weite,
Wo = Pi = 50 W = Iron loss
From S.C. test we can find the parameters of equivalent circuit. Now S.C. test is
conducted on H.V. side i.e. meters are on H.V. side which is transformer secondary.
Hence parameters from S.C. test results will be referred to secondary.
78

Vsc = 15 V, Isc = 30 A, Wsc = 100 W
... Rsc = Wsc/(Isc)2 =10/(30)2 = 0.111Ω
ZSC = Vsc /Isc = 15/30 = 0.5 Ω
... XSC = √(Zsc2 - RSC
2) = 0.4875 Ω
i) Copper loss on full load
(I2) F.L. = VA rating/V2 = (10 x 103)/250 = 40 A
In short circuit test, Isc = 30 A and not equal to full load value 40 A.
Hence Wsc does not give copper loss on full load
... Wsc = Pcu at 30 A = 100 W
Now Pcu α I2
( Pcu at 30 A)/( Pcu at 40 A) = (30/40) 2
100/( Pcu at 40 A) = 900/1600
Pcu at 40 A = 177.78 W
... (Pcu) F.L. = 177.78 W
79

ii) Full load η , cos Φ2 = 0.8
iii) Half load η , cos Φ2 = 0.8
n = 0.5 as half load, (I2) H.L. = 0.5 x 40 = 20
= 97.69%
80

iv) Regulation at full load, cos Φ = 0.9 leading
= -1.8015%
81

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