Short Notes: Transformers

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

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``` 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

No-Load Current ? Core Dimension
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

No-Load Current ? Core Dimension
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

No-Load Current ? Core Dimension
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

o Power reading =
2
1
1 0 0
c
V
P = V I cos =
R
? -------- (i)
o Ammeter reading ?
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.
o Wattmeter reading =
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

No-Load Current ? Core Dimension
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

o Power reading =
2
1
1 0 0
c
V
P = V I cos =
R
? -------- (i)
o Ammeter reading ?
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.
o Wattmeter reading =
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

No-Load Current ? Core Dimension
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

o Power reading =
2
1
1 0 0
c
V
P = V I cos =
R
? -------- (i)
o Ammeter reading ?
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.
o Wattmeter reading =
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
?
?
? ? ?

X  = % loading of Transformer
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
?
No-load voltage
2
V ?
Full-load voltage
2
V
?
?
Approximate Voltage Regulation

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

```

38 docs

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

 1. What is a transformer in electrical engineering?
A transformer is an electrical device that is used to transfer electrical energy between two or more circuits through electromagnetic induction. It consists of two or more coils of wire, known as windings, that are wound around a core made of magnetic material. Transformers are essential in electrical power transmission and distribution systems as they can step up or step down the voltage levels.
 2. How does a transformer work?
A transformer works on the principle of electromagnetic induction. When an alternating current (AC) flows through the primary winding, it creates a changing magnetic field around the winding. This changing magnetic field induces a voltage in the secondary winding, which is connected to the load. The ratio of the number of turns in the primary and secondary windings determines the voltage transformation ratio of the transformer.
 3. What are the different types of transformers?
There are several types of transformers used in electrical engineering. Some common types include: - Power transformers: These are used in power transmission and distribution systems to step up or step down voltages. - Distribution transformers: These are used to step down the voltage for distribution to households and commercial buildings. - Instrument transformers: These are used for measurement and protection purposes, such as current transformers (CTs) and voltage transformers (VTs). - Autotransformers: These have a single winding that serves as both the primary and secondary winding, resulting in a voltage transformation ratio between the input and output. - Isolation transformers: These provide electrical isolation between the input and output circuits, often used for safety purposes.
 4. What are the main components of a transformer?
The main components of a transformer include: - Core: The core is made of a magnetic material, such as laminated iron or steel, and provides a path for the magnetic flux. - Windings: There are two or more windings, including the primary winding and secondary winding, which are made of insulated copper or aluminum wire. These windings are responsible for transferring the electrical energy. - Insulation: Insulation materials, such as varnish or enamel, are used to electrically isolate the windings and prevent short circuits. - Tap changer: Some transformers have a tap changer that allows adjusting the turns ratio to compensate for variations in the input or output voltage. - Cooling system: Transformers may have cooling systems, such as oil or air cooling, to dissipate the heat generated during operation.
 5. What are some common problems with transformers?
Some common problems with transformers include: - Overheating: Transformers can overheat due to excessive loading, poor cooling, or insulation degradation. This can lead to insulation failure and reduced efficiency. - Winding faults: Short circuits or open circuits in the windings can cause a loss of voltage or current, resulting in decreased performance. - Insulation breakdown: Over time, the insulation in a transformer can degrade, leading to breakdowns and potential electrical faults. - Voltage regulation issues: Transformers should provide a stable output voltage, but variations in input voltage or load can cause voltage regulation problems. - Transformer noise: Transformers can produce audible noise due to magnetostriction, vibration, or loose components. Excessive noise can indicate a potential issue.

## Short Notes for Electrical Engineering

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