Short Notes: Transformers

# Short Notes: Transformers | Short Notes for Electrical Engineering - Electrical Engineering (EE)

``` Page 1

Impact of dimensions on various parameters of Transformer
KVA Rating ? (Core Dimension)
4

Voltage Rating ? (Core Dimension)
2
Current Rating ? (Core Dimension)
2

Core Loss ? Core Volume
Induced EMF in a Transformer
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN

? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively.
? F is the flux in the transformer and F m is maximum value of flux.
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary
should be such that primary and secondary flux should oppose each other.
? Also, primary current enters the positive terminal of primary winding as primary absorbs
power and secondary current leaves the positive terminal of secondary winding as
secondary delivers power and this way we can mark emf polarities.

Exact equivalent circuit

Page 2

Impact of dimensions on various parameters of Transformer
KVA Rating ? (Core Dimension)
4

Voltage Rating ? (Core Dimension)
2
Current Rating ? (Core Dimension)
2

Core Loss ? Core Volume
Induced EMF in a Transformer
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN

? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively.
? F is the flux in the transformer and F m is maximum value of flux.
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary
should be such that primary and secondary flux should oppose each other.
? Also, primary current enters the positive terminal of primary winding as primary absorbs
power and secondary current leaves the positive terminal of secondary winding as
secondary delivers power and this way we can mark emf polarities.

Exact equivalent circuit

Exact equivalent circuit w.r.t. primary

2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
;
? Approximately Equivalent Circuit

01 1 2
R = R R
?
?
01 1 2
X = X X
?
?
Tests Conducted on a Transformer
(i) Open Circuit Test
o Conducted on LV side keeping HV side open circuited
o Equivalent Circuit

Page 3

Impact of dimensions on various parameters of Transformer
KVA Rating ? (Core Dimension)
4

Voltage Rating ? (Core Dimension)
2
Current Rating ? (Core Dimension)
2

Core Loss ? Core Volume
Induced EMF in a Transformer
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN

? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively.
? F is the flux in the transformer and F m is maximum value of flux.
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary
should be such that primary and secondary flux should oppose each other.
? Also, primary current enters the positive terminal of primary winding as primary absorbs
power and secondary current leaves the positive terminal of secondary winding as
secondary delivers power and this way we can mark emf polarities.

Exact equivalent circuit

Exact equivalent circuit w.r.t. primary

2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
;
? Approximately Equivalent Circuit

01 1 2
R = R R
?
?
01 1 2
X = X X
?
?
Tests Conducted on a Transformer
(i) Open Circuit Test
o Conducted on LV side keeping HV side open circuited
o Equivalent Circuit

2
1
1 0 0
c
V
P = V I cos =
R
? -------- (i)
0
I = I
o
0
10
P
cos =
VI
?
o Calculate
2
00
sin = 1 - cos ??
o
2
1
1 0 0
m
V
Q = V I sin =
X
? ------- (ii)
Calculate
c
R from (i) &
m
X from (ii)
(ii) Short Circuit Test
o Conducted on HV side keeping LV side short circuited
o Equivalent Circuit

o
01
R &
01
X are equivalent winding resistance & equivalent leakage reactor referred to
HV side.
2
sc 01
P = I R from this equation, we can calculate
01
R
o
sc
01
sc
V
Z =
I
&
22
01 01 01
X = Z R ?
o We obtain
01
R ,
01
X & full load copper losses from this test.
Losses on Transformers
o Copper Loss

22
Cu 1 1 2 2
P = I R I R ?

22
1 01 2 02
= I R  I R ?
Where
1
I = primary current

2
I = secondary current
Page 4

Impact of dimensions on various parameters of Transformer
KVA Rating ? (Core Dimension)
4

Voltage Rating ? (Core Dimension)
2
Current Rating ? (Core Dimension)
2

Core Loss ? Core Volume
Induced EMF in a Transformer
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN

? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively.
? F is the flux in the transformer and F m is maximum value of flux.
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary
should be such that primary and secondary flux should oppose each other.
? Also, primary current enters the positive terminal of primary winding as primary absorbs
power and secondary current leaves the positive terminal of secondary winding as
secondary delivers power and this way we can mark emf polarities.

Exact equivalent circuit

Exact equivalent circuit w.r.t. primary

2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
;
? Approximately Equivalent Circuit

01 1 2
R = R R
?
?
01 1 2
X = X X
?
?
Tests Conducted on a Transformer
(i) Open Circuit Test
o Conducted on LV side keeping HV side open circuited
o Equivalent Circuit

2
1
1 0 0
c
V
P = V I cos =
R
? -------- (i)
0
I = I
o
0
10
P
cos =
VI
?
o Calculate
2
00
sin = 1 - cos ??
o
2
1
1 0 0
m
V
Q = V I sin =
X
? ------- (ii)
Calculate
c
R from (i) &
m
X from (ii)
(ii) Short Circuit Test
o Conducted on HV side keeping LV side short circuited
o Equivalent Circuit

o
01
R &
01
X are equivalent winding resistance & equivalent leakage reactor referred to
HV side.
2
sc 01
P = I R from this equation, we can calculate
01
R
o
sc
01
sc
V
Z =
I
&
22
01 01 01
X = Z R ?
o We obtain
01
R ,
01
X & full load copper losses from this test.
Losses on Transformers
o Copper Loss

22
Cu 1 1 2 2
P = I R I R ?

22
1 01 2 02
= I R  I R ?
Where
1
I = primary current

2
I = secondary current

1
R = primary winding resistance

2
R = secondary winding resistance

22
12
01 1 2 02 2 1
21
NN
R = R R ; R = R R
NN
? ? ? ?
??
? ? ? ?
? ? ? ?

o Core Loss
(i)  Hysteresis Loss

x
n n m
P = K B f
X = 1.6

m
B = maximum value of flux density

1.6
n n m
P = K B f

m
V
B
f
?
V = applied voltage
f = frequency

?
??
??
??
??
1.6
1.6 0.6
nhh
V
P = K f = K V f
f

If V is constant & f is increased,
h
P decreases
(ii)  Eddy Current Loss

2 2
e e m
P = K B f

m
V
B
f
?

2
22
e e e
V
P = K f = K V
f
??
??
??
??

Core loss =
c e n
P = P P ?

Page 5

Impact of dimensions on various parameters of Transformer
KVA Rating ? (Core Dimension)
4

Voltage Rating ? (Core Dimension)
2
Current Rating ? (Core Dimension)
2

Core Loss ? Core Volume
Induced EMF in a Transformer
?
?
?
?
??
??
11
22
1 1 m
2 2 m
d
EN
dt
d
EN
dt
E (rms) 4.44fN
E (rms) 4.44fN

? Where E1 and E2 are emf in primary and secondary windings of Transformer respectively.
? F is the flux in the transformer and F m is maximum value of flux.
? The polarity of emf is decided on basis of Lenz Law as currents in primary and secondary
should be such that primary and secondary flux should oppose each other.
? Also, primary current enters the positive terminal of primary winding as primary absorbs
power and secondary current leaves the positive terminal of secondary winding as
secondary delivers power and this way we can mark emf polarities.

Exact equivalent circuit

Exact equivalent circuit w.r.t. primary

2 2 2
1 1 1
2 2 2 2 L L
2 2 2
N N N
R = R ;  X = X ;  Z = Z
N N N
? ? ? ? ? ?
? ? ?
? ? ? ? ? ?
? ? ? ? ? ?
;
? Approximately Equivalent Circuit

01 1 2
R = R R
?
?
01 1 2
X = X X
?
?
Tests Conducted on a Transformer
(i) Open Circuit Test
o Conducted on LV side keeping HV side open circuited
o Equivalent Circuit

2
1
1 0 0
c
V
P = V I cos =
R
? -------- (i)
0
I = I
o
0
10
P
cos =
VI
?
o Calculate
2
00
sin = 1 - cos ??
o
2
1
1 0 0
m
V
Q = V I sin =
X
? ------- (ii)
Calculate
c
R from (i) &
m
X from (ii)
(ii) Short Circuit Test
o Conducted on HV side keeping LV side short circuited
o Equivalent Circuit

o
01
R &
01
X are equivalent winding resistance & equivalent leakage reactor referred to
HV side.
2
sc 01
P = I R from this equation, we can calculate
01
R
o
sc
01
sc
V
Z =
I
&
22
01 01 01
X = Z R ?
o We obtain
01
R ,
01
X & full load copper losses from this test.
Losses on Transformers
o Copper Loss

22
Cu 1 1 2 2
P = I R I R ?

22
1 01 2 02
= I R  I R ?
Where
1
I = primary current

2
I = secondary current

1
R = primary winding resistance

2
R = secondary winding resistance

22
12
01 1 2 02 2 1
21
NN
R = R R ; R = R R
NN
? ? ? ?
??
? ? ? ?
? ? ? ?

o Core Loss
(i)  Hysteresis Loss

x
n n m
P = K B f
X = 1.6

m
B = maximum value of flux density

1.6
n n m
P = K B f

m
V
B
f
?
V = applied voltage
f = frequency

?
??
??
??
??
1.6
1.6 0.6
nhh
V
P = K f = K V f
f

If V is constant & f is increased,
h
P decreases
(ii)  Eddy Current Loss

2 2
e e m
P = K B f

m
V
B
f
?

2
22
e e e
V
P = K f = K V
f
??
??
??
??

Core loss =
c e n
P = P P ?

Efficiency

? ?
? ?
2
i Cu,FL
x KVA cos
=
x KVA cos P x P
?
?
? ? ?

cos = power factor ?

i
P = iron loss

Cu,FL
P = Full load copper losses
KVA = Power rating of Transformer
For maximum efficiency,

i
Cu,FL
P
x =
P

Voltage Regulation of Transformer
Regulation down
NL FL
NL
VV
100
V
?
??
Regulation up
NL FL
FL
VV
100
V
?
??
Equivalent circuit with respect to secondary
K = Transformation Ratio
2
1
N

N
?
2
V ?
2
V
?
?
Approximate Voltage Regulation

? ?
2 02 2 02 2
2
I R cos X sin
VR =
V
? ? ?

```

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## Short Notes for Electrical Engineering

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