3
3
Design of Transformer
Design of Transformer
3.1 Introduction
3.1
introduction
Transformers are
are static
static electromagnetic
electromagnetic devices
devices with
with two
two or
or more
more windings
windings and
and aa common
common
Transformers
magnetic
circuit.
They
are
one
of
the
primary
power
apparatus
used
in
the
power
system
magnetic circuit. They are one of the primary power apparatus used in the power system
network.
The
transformers
change
the
AC
voltage
and
the
associated
current
from
one
network. The transformers change the AC voltage and the associated current from one
level
to
another
level,
maintaining
power
balance
between
input
and
output
side,
without
level to another level, maintaining power balance between input and output side, without
change in
in frequency.
frequency. Hence,
Hence, transformers
transformers perform
perform AC
AC energy
energy transformation
transformation rather
rather than
than
change
energy
conversion
from
one
or
more
primary
side
circuit/s
to
one
or
more
secondary
side
energy conversion from one or more primary side circuit/s to one or more secondary side
circuit/s.
circuit/s.
The main
main parts
parts of
of the
The
the transformer
transformer are
are as
as
follows:
follows:
Coolant
• • Core
conservator
• • Yoke
Bushing
• • Winding
When the MVA rating is high, in addition
to the above parts, transformer is equipped
with
• • Bushing
• • Coolant conservator
The different parts of the transformer in
general are represented in Fig. 3.1.
The key functions and materials used for
different parts of the transformer are stated
in Table. 3.1.
Transformers are classified on the basis
of the following aspects.
Yoke
Core surrounded
by winding
Winding
Fig. 3.1 | Parts of a transformer
fig.
3.2 Design of Transformer
Table 3.1 | Functions of various parts of the transformer
Part
Material
Function
Core
Cold-rolled grain-oriented steel
• Creation of closed magnetic
circuit for passage of flux
between primary and secondary
windings
• Windings are made around the
core
Yoke
Cold-rolled grain-oriented steel
• It acts as the top and bottom of
core, providing support to it
Winding
Copper
• Carries the current and used for
transfer of power
Bushing
Polymer or porcelain
• Prevention of flashover due to
connection of high voltage to
earthing point
Coolant conservator
Steel
• Provides storage space for
coolant used for circulation in
cooling the transformer
••
••
••
••
Construction (shape of magnetic medium)
Number of phases
Winding/turns/voltage ratio
Application (loading condition)
The classification of transformers is represented in Fig. 3.2. In addition, instrument
transformers such as current transformers and potential transformers are also available.
The detailed classification based on the aforementioned aspects is given in the following
section.
3.1.1 Based on Voltage Ratio
•• Step-up transformer: This type of transformer increases the voltage, with
corresponding decrease in current. They have a larger number of turns in secondary
side, i.e., (T2 / T1 ) > 1, where T1 and T2 are the number of turns in primary and
secondary sides, respectively. A step-up transformer is shown in Fig. 3.3.
•• Step-down transformer: This type of transformer decreases the voltage, with
the corresponding increase in current. They have a smaller number of turns in
secondary side, i.e., T2/T1 < 1, where T1 and T2 are the number of turns in Primary
and secondary sides respectively. A step-down transformer is shown in Fig. 3.4.
3.1.2 Based on Construction
•• Core-type transformer: In core-type transformer, windings encircle the core as
shown in Fig. 3.5(a) and (b). The primary and secondary windings are wound
around the limbs. It therefore creates a single magnetic circuit. The low-voltage
winding is wound inside, near the core and the high-voltage winding is wound
Introduction
Introduction
Introduction 3.3
3.3
3.3
Introduction
3.3
Single
Single
Single
Three
Three
Three
Number
Number
Number
of
ofof
phases
phases
phases
Distribution
Distribution
Distribution
Application
Application
Application
Core
Core
Core
Transformer
Transformer
Transformer
types
types
types
Construction
Construction
Construction
Shell
Shell
Shell
Power
Power
Power
Voltage
Voltage
Voltage
ratio
ratio
ratio
Step
Step
Step
down
down
down
Step
Stepup
up
up
Step
fig.
fig.3.2
3.2
3.2 ||| Classification
Classification
Classificationofof
oftransformers
transformers
transformers
fig.
Fig.
VVV
2 22
VVV
1 11
TTT
2 22
TTT
1 11
fig.
fig.3.3
3.3
3.3 |||| Step-up
Step-up
Step-uptransformer
transformer
transformer
fig.
Fig.
3.3
Step-up
transformer
VVV
1 11
VVV
2 22
TTT
1 11
TTT
2 22
fig.
fig.3.4
3.4
3.4 |||| Step-down
Step-down
Step-downtransformer
transformer
transformer
fig.
Fig.
3.4
Step-down
transformer
outside,
low-voltage
away
from the core.
The reason
for
winding
windingencircling
iswound
wound
woundthe
inside,
inside,
near
nearthe
the
thewinding,
core
coreand
and
andthe
the
thehigh-voltage
high-voltage
high-voltage
winding
winding
iswound
wound
wound
winding
isis
inside,
near
core
winding
isis
this
winding
arrangement
is
that
the
quantity
of
insulation
materials
required
is
outside,
outside,
encircling
encircling
the
the
low-voltage
low-voltage
winding,
winding,
away
away
from
from
the
the
core.
core.
The
The
reason
reason
for
for
outside, encircling the low-voltage winding, away from the core. The reason for
less.
In
order
to
have
reduced
leakage
reactance,
two
windings
can
be
coupled
this
this
winding
winding
arrangement
arrangement
is
is
that
that
the
the
quantity
quantity
of
of
insulation
insulation
materials
materials
required
required
is
this winding arrangement is that the quantity of insulation materials required isis
together
tightly,
with
each
limb
carrying
one
half
of
the
primary
and
one
half
of
the
less.
less.
In
In
order
order
to
to
have
have
reduced
reduced
leakage
leakage
reactance,
reactance,
two
two
windings
windings
can
can
be
be
coupled
coupled
less. In order to have reduced leakage reactance, two windings can be coupled
secondary.
together
togethertightly,
tightly,
tightly,with
with
witheach
each
eachlimb
limb
limbcarrying
carrying
carryingone
one
onehalf
half
halfof
of
ofthe
the
theprimary
primary
primaryand
and
andone
one
onehalf
half
halfof
of
of
together
the
thesecondary.
secondary.
secondary.
the
3.4 Design of
of Transformer
3.4
3.4 Design
Design of Transformer
Transformer
(a)
(a)
(b)
(b)
fig. 3.5 || (a)
(a) Single-phase core-type
core-type transformer. (b)
(b) Three-phase core-type
core-type transformer
Fig.
fig. 3.5
3.5 | (a) Single-phase
Single-phase core-type transformer.
transformer. (b) Three-phase
Three-phase core-type transformer
transformer
• Shell-type transformer: In shell-type transformer, the core encircles the winding as
••• Shell-type
transformer:
In shell-type
transformer,
the core
the
as
Shell-type
transformer:
shell-type
transformer,
core encircles
encircles
the winding
winding
as
shown in Fig.
3.6(a) andIn(b).
The primary
and thethe
secondary
windings
are placed
shown
in
Fig.
3.6(a)
and
(b).
The
primary
and
the
secondary
windings
are
placed
shown
in
Fig.
3.6(a)
and
(b).
The
primary
and
the
secondary
windings
are
placed
together alternatively, giving rise to sandwich type of winding with high-voltage
together
alternatively, giving
rise
to
type
winding
with
high-voltage
together
rise
to sandwich
sandwich
type of
of The
winding
with
high-voltage
windingsalternatively,
sandwiched giving
between
low-voltage
windings.
central
limb
carries the
windings
sandwiched
between
low-voltage
windings.
The
central
limb
carries
the
windings
sandwiched
between
low-voltage
windings.
The
central
limb
carries
the
total mutual flux, while the other limbs carry half the total flux. This creates
a double
total
mutual
flux,
while
the
other
limbs
carry
half
the
total
flux.
This
creates
aa double
total
mutual
flux,
while
the
other
limbs
carry
half
the
total
flux.
This
creates
double
magnetic circuit.
magnetic
magnetic circuit.
circuit.
(a)
(a)
(b)
(b)
fig. 3.6 | (a) Single-phase shell-type transformer. (b) Three-phase shell-type transformer
fig. 3.6
3.6 || (a)
(a) Single-phase
Single-phase shell-type
shell-type transformer.
transformer. (b)
(b) Three-phase
Three-phase shell-type
shell-type transformer
transformer
Fig.
Introduction 3.5
The comparison between core-type and shell-type transformers is represented in Table 3.2.
Table 3.2 | Comparison between core-type and shell-type transformers
Feature\Type
Core type
Shell type
Basic construction
Simple, as the core is
­encircled by the windings
Complex, as the core encircles
maximum part of the windings
Number of magnetic circuit
Single
Double
Number of limbs (for single- Two
phase type)
Three
Type of coil preferred
Cylindrical
Multilayer disc or sandwich
type
Leakage reactance and
­possibility of reduction
Comparatively high and
cannot be reduced
Comparatively low and can be
reduced
Maintenance and repairs
Easier to maintain and
repair as winding can be
removed or dismasted for
repairs
Difficult to maintain and repair
as for dismasting, a large
number of laminations are
required to be removed
Cooling and heat
­dissipation
Effective, due to distribution Difficult, as mainly the core can
of windings
be effectively cooled rather than
the windings where more heat is
produced
Electromagnetic forces expe- Large
rienced under fault or short
circuit conditions
Small
Applications
Suited for low- to mediumvoltage applications
Suited for high-voltage
­applications
3.1.3 Based on Application
•• Distribution transformer: This type of transformer is used for stepping down the
transmission voltage to distribution voltage or stepping down the distribution
voltage to a standard voltage as per the service requirements of an industrial
or commercial load. The rating of this transformer is usually below 250 kVA.
The type of connection used is three-phase four-wire delta-star. The period of
operation is 24 hours throughout the day. Hence, the design should be done in a
way that the efficiency is maximum even at half-load condition, leading to a good
all day efficiency. The flux density is usually less than 1.7 Wb/m2. The ratio of iron
loss to copper loss is 1:3. As this type of transformer is located in the vicinity of
the load, the voltage regulation must be good, which can be satisfied by having
a small leakage reactance. Usually, the percentage impedance is in the range of
4–18%. The voltage regulation is in the range of 4–9%. The type of cooling used is
self-oil cooling.
3.6 Design of Transformer
•• Power transformer: This type of transformer is used for either stepping up the voltage
or stepping down the voltage in generating stations and substations at every end of
power transmission line. The rating of this transformer is usually above 250 kVA.
The type of connection used is three-phase three-wire-delta-delta or star-delta. The
period of operation is during heavy loading periods and is disconnected during
light loading periods. Hence, the design is in such a way that efficiency is maximum
at full or near full load. The flux density is usually between 1.5 and 1.7 Wb/m2. The
ratio of iron loss to copper loss is 1:1. The voltage regulation is not an important
factor as this transformer is not located in the vicinity of the load. Hence, the leakage
reactance can be made high, used for limiting the short circuit current, leading to less
mechanical forces. Usually, the percentage impedance is in the range of 6–18%. The
voltage regulation is in the range of 6–10%. The type of cooling used is forced air or
forced oil cooling.
3.1.4 Based on Number of Phases
•• Single-phase transformer: This type of transformer has a single-phase primary and a
single-phase secondary placed with either core or shell-type construction. It is used
for low-power applications.
•• Three-phase transformer: This type of transformer has three-phase primary and
three-phase secondary placed with either core or shell-type construction. It is used
for high-power applications.
The three-phase transformer can be made by connecting three single-phase transformers
into a three-phase transformer bank or can be an inherently constructed three-phase
transformer, with a common magnetic circuit for all three phases. The comparison between
three-phase transformer bank and three-phase transformer is shown in Table 3.3.
Table 3.3 | Comparison between three-phase transformer bank and three-phase transformer
Feature/Type
Three-phase transformer bank
Three-phase transformer
Cost
Costly, due to large volume
Cheaper, due to small
volume
Efficiency
Less
High
Installation and
maintenance
Difficult
Easy
Faults and clearance
Easier to clean, as any single
unit in which fault has
occurred can be replaced
Difficult to clean, as it is a
single unit
The three-phase transformer connections can be of star or delta type. In star connection,
the phase voltage is 1 / 3 times the line voltage; hence, the number of turns is 1 / 3 times
the number of turns in delta connection. But the phase current in star connection is 3 times
the phase current in delta connection, hence requiring large area of cross-section. The amount
of insulation used in star connection is smaller compared to delta connection. In star
connection, neutral grounding/earthing can be done easily, reducing effect of faults. For
high voltages, cost of star connection is less. Moreover, four-wire connection with neutral
high voltages, cost of star connection is less. Moreover, four-wire connection with neutral
(fourth wire) can be easily done. But, for delta connection, under unbalance condition,
response is better than star connection.
Apart from the above, there is another type of transformer with three winding
(primary, secondary and tertiary), as shown in Fig. 3.7 to operate at high,
medium and
Introduction
3.7
low voltages.
high voltages,
R cost of star connection is less. Moreover, four-wire connectionr with neutral
(fourth wire) can be easily done. But, for delta connection, under unbalance condition,
I0
response is better than star connection.
Apart from the above, there is another type of transformer with three winding
(primary, secondary and tertiary), as shown in Fig. 3.7 to operate at high, medium and
Stabilizing
low voltages.
or tertiary
winding
Y
R
B
I0
Primary side
y
r
b
Secondary side
fig.
Fig. 3.7 | Three-winding transformer
Stabilizing
The reasons for
are
as follows:
tertiary
Y the use of tertiary winding or
y
(fourth wire) can be easily done. But, for winding
delta connection, under unbalance condition,
1. Ability
anconnection.
additional load with a small voltage different frombprimary and
B to supply
response
is better
than star
secondary
Apart
from the above, there
another type ofSecondary
transformer
with three winding
Primaryisside
side
2. To secondary
provide voltage
for compensating
powerat factor
correctionand
at
(primary,
and tertiary),
as shown incapacitors
Fig. 3.7 tofor
operate
high, medium
different voltages fig. 3.7 | Three-winding transformer
low voltages.
3. reasons
As stabilizing
for limiting
short
or fault current, when connected in
The
for thewinding
use of tertiary
winding
arecircuit
as follows:
deltafor the use of tertiary winding are as follows:
The reasons
1.
to supply
an additional
with
a small voltage
different
from primary and
4. Ability
For indicating
voltage
levels in load
testing
transformer
used for
high voltages
1.
Ability
to
supply
an
additional
load
with
a
small
voltage
different
from
and
secondary
5. To reduce zero sequence impedance, making easy flow of fault current
easyprimary
to ground
secondary
2. To
to provide voltage for compensating capacitors for power factor correction at
The
possible
arrangements
of windings
in three for
winding
transformer
are shown
2. different
To provide
voltage
for compensating
capacitors
power
factor correction
at
different
voltages
in Fig.
3.8(a)
and
(b).
different
voltages
3. As
stabilizing
winding for limiting short circuit or fault current, when connected in
3. delta
As stabilizing winding for limiting short circuit or fault current, when connected in
LV winding
MV winding
delta
4. For
indicating voltage levels
in testing transformer used for high
voltages
4. To
Forreduce
indicating
voltage levels
in testing
transformer
used
for high
voltages
5.
zero sequence
impedance,
making
easy flow
of fault
current
easy to ground
MV winding making easy flow of fault LV
winding
5. To reduce zero sequence impedance,
current
easy to ground
The different possible arrangements
of windings in three windingHV
transformer
are shown
winding
HV
winding
The3.8(a)
different
arrangements of windings in three winding transformer are shown
in Fig.
andpossible
(b).
in Fig. 3.8(a) and (b).
(a)
LV winding
MV winding
MV winding
Core
LV winding
Core
HV winding
(b)
HV winding
fig. 3.8 | Winding arrangements of three-winding transformer: (a) LV, MV and HV;
(b) MV, LV and HV
Core
(a)
Core
(b)
fig.
3.8
transformer:(a)
(a)LV,
LV,MV
MVand
andHV;
HV;
Fig.
3.8| |Winding
Windingarrangements
arrangementsof
of three-winding
three-winding transformer:
(b)
LV and
and HV
HV
(b) MV,
MV, LV
3.8 Design of Transformer
3.2 Specifications of a Transformer
The transformer specifications are their rating and performance expectations. A designer
needs to know the following important specifications with regards to the design of the
transformer.
1. Rating/Capacity/Output-in kVA
2. Voltage–Voltage rating of HV and LV windings with or without tap changers and
tappings
3. Frequency – f in Hz
4. Number of phases – one or three
5. Type of winding connection in case of three-phase transformers – star-star, star-delta,
delta-delta, delta-star with or without grounded neutral
6. Rating – continuous or short time
7. Cooling – natural or forced
8. Type – core or shell, power or distribution
9. Efficiency, per unit impedance, location (i.e., indoor, pole or platform mounting,
etc.), temperature rise, etc.
3.3 Design of Transformer
The design of transformer includes determination of its main dimensions namely the
overall width and height based on the geometry of windows, distance between limbs and
diameter of circumscribing circle for various types of core. Design also includes design of
yoke, core and winding. Estimation of no load current, iron loss and copper loss facilitates
the temperature rise calculations based on which cooling systems are designed. Calculation
of resistances and leakage reactances are used for the determination of the efficiency and
regulation.
Hence, the design of transformer can be split into following four steps:
1.
2.
3.
4.
Determining output equation relating output with machine dimensions
Determining overall dimensions
Design of cooling apparatus
Estimation of quantities such as no load current, iron losses, resistance and leakage
reactance and so on.
3.3.1 Output Equation of Single-phase Transformer
The kVA rating of a single-phase transformer is given as follows:
Q(kVA) = Vp I p ×10−3 (or) Vs I s × 10−3 (3.1)
 Ep I p ×10−3 [ ∵ Terminal voltage, Vp  Induced voltage, Ep ] (3.2a)
(or)
 Es I s ×10−3 [ ∵ Vs  Es ] (3.2b)
where p → primary, V and E → voltage, s → secondary and I → current.
Design of Transformer 3.9
Design of Transformer 3.9
Design
of Transformer 3.9
The voltage induced in a transformer winding is given as follows:
The voltage induced in a transformer winding is given as follows:
Ex = 4.44fφmTx
(3.3)
The voltage induced in a transformer
winding
is given as follows:
Ex = 4.44fφ
Tx
(3.3)
m
where
Ex = 4.44fφmTx(3.3)
where
f – frequency (Hz)
fwhere
–
frequency
(Hz)
φm – Flux
f – (Wb)
frequency (Hz), φm – Flux (Wb), T – number of turns, x – can be p or s, denoting
φm––number
Fluxor(Wb)
T
of turns
primary
secondary.
T
number
x –– can
be p of
or turns
s, denoting primary or secondary
x – can be p or s, denoting primary or secondary
3.3.2 Output Equation of Single-phase Transformer (Core-type)
3.3.2 output equation of single-phase Transformer (Core-type)
The
cross-section
a 1φ core-type
transformer Transformer
is shown in Fig.
3.9. The arrangement of the
3.3.2
output of
equation
of single-phase
(Core-type)
The cross-section
of a 1φ core-type
transformer is shown in Fig.
3.9. The arrangement of the
winding
is shownof
in Fig.
3.10.
The
cross-section
1φ core-type
transformer is shown in Fig. 3.9. The arrangement of the
winding
is shown inaFig.
3.10.
winding is shown in Fig. 3.10.
Window area
Aw area
Window
Aw
Hw
Hw
Ww
Ww
Iron area
Ai area
Iron
Ai
fig. 3.9 | Cross-section of 1φ core-type transformer
Fig.
fig. 3.9 | Cross-section of 1φ core-type transformer
H
H
L
L
V
V
V
V
L
L
V
V
H
H
V
V
a1
a1
a2
a2
H
H
V
V
L
L
V
V
L
L
H
H
V
V
V
V
fig. 3.10 | Winding arrangement of 1φ core-type transformer
Fig.
fig. 3.10 | Winding arrangement of 1φ core-type transformer
The output of the transformer in kVA is related to its main dimensions. Hence, the output
output of the
is related
toarea
its main
dimensions.
Hence,
the output
canThe
be determined
intransformer
terms of areainofkVA
core/iron
(Ai),
of window
(Aw) and
window
space
can
be
determined
in
terms
of
area
of
core/iron
(A
),
area
of
window
(A
)
and
window
space
factor (Kw). Hence, the following steps are followed.
w
ii
w
factor (Kw
).
Hence,
the
following
steps
are
followed.
w
3.10 Design of Transformer
In case of single-phase transformers, there is one primary and secondary winding. Half
of primary and half of secondary are placed on one limb and the other half of each of the
winding is placed on the other limb. Each limb has half of the primary and secondary
windings as shown in Fig. 3.10. Hence, the total copper area in window is given as follows:
Ac = Total copper area of [ primary winding + secondary windiing ]
= [ number of turns in primary × area of primary conductor ]
+ [ number of turns in secondary × area of secondary conductor ]
= Tp Ap + Ts As (3.4)
Assuming same current density, δ in both primary and secondary windings, we also get
δ=
⇒
Ap =
Ip
Ap
=
Is
(3.5)
As
Ip
I
and As = s (3.6)
δ
δ
Substituting Eqs. (3.5) and (3.6) in Eq. (3.4), we get
Ac = Tp
(
Assuming,
Ip
+ Ts
Is
δ
δ
1
= Tp I p + Ts I s (3.7)
δ
)
TpIp = Ts Is = AT(3.8)
where AT is the magnetic mmf.
Substituting Eq. (3.8) in Eq. (3.7), we get
Ac =
AT + AT 2 AT
=
(3.9)
δ
δ
We know that the ratio of copper area (Ac) in window to the total window area (Aw) is
window space factor, which is given by
copper area in window
A
= c (3.10)
total window area
Aw
Kw =
⇒
Ac = K w Aw (3.11)
Equating, Eqs. (3.9) and (3.11), we get
2AT
= K w Aw
δ
⇒
AT =
δ K w Aw
(3.12)
2
Output Equation of Single-phase Transformer
(Shell-type) 3.11
Design of Transformer
From Eq. (3.2a),
3
Q = Ep I p × 10−
Q = Ep I p ×10−3
−33
10−
=
×10
(3.13)
=E
Ett ×
×T
Tpp ×
× II pp ×
(3.13)
Ep
(3.14)
[where Volt/turn, Et = Ep ⇒ Ep = Et × Tp ]
[where Volt/turn,
Et = Tp ⇒ Ep = Et ×Tp ] (3.14)
Tp
Substituting T I = AT from Eq. (3.8) in Eq. (3.13), we get
Substituting TppIpp= AT from Eq. (3.8) in Eq. (3.13), we get
Q = Et × AT × 10−3
Q = Et × AT × 10−3
Substituting Eq. (3.12) in the above equation, we get
Substituting Eq. (3.12) in the above equation, we get
δK A
Q = Et × δ K w Aw × 10−3
Q = Et × w2 w ×10−3
2
Substituting Ep / Tp from Eq. (3.3) in the above equation, we get
Substituting Ep / Tp from Eq. (3.3) in the above equation, we get
δK A
Q = 4.44× f ×φm × w w × 10−3
δ K 2A
Q = 4.44× f ×φm × w w ×10−3
2
= 2.22 f φmδ K w Aw × 10−3
(3.15)
= 2.22 f φm δ K w Aw ×10−3 (3.15)
Substituting φm = Bm (maximum flux density) × Ai (area of core or area of iron) in the above
Substituting
equation, weφget
m = Bm (maximum flux density) × Ai (area of core or area of iron) in the above
equation, we get
3
(3.16)
Q = 2.22 f Bmδ K w Aw Ai × 10−
Q = 2.22 f Bm δ K w Aw Ai ×10−3 (3.16)
Equation (3.16) is the output equation of a single-phase core-type transformer and it relates
Equation
the output
equation ofand
a single-phase
the output(3.16)
kVA is
rating
to its dimensions
parameters.core-type transformer and it relates
the output kVA rating to its dimensions and parameters.
3.3.3 output
equation
of single-phase
Transformer
(shell-type)
3.3.3 Output
Equation
of Single-phase
Transformer
(Shell-type)
The
The cross-section
cross-section of
of the
the core
core of
of 1φ
1φ shell-type
shell-type transformer
transformer is
is shown
shown in
in Fig.
Fig. 3.11.
3.11. The
The
arrangement
of
winding
is
shown
in
Fig.
3.12.
arrangement of winding is shown in Fig. 3.12.
Window area
Aw
Hw
Window area
Aw
Iron area
Ai
Ww
Hw
Ww
fig. 3.11 | Cross-section of 1φ shell-type transformer
Fig.
3.12
3.12 Design
Design of
of Transformer
Transformer
HV
LV
LV
HV
fig. 3.12
3.12 || Winding
Fig.
Winding arrangement
arrangement of
of 1φ
1φ shell-type
shell-type transformer
transformer
Here, both the windings are wound over the central limb and so each window has all the
Here, both the windings are wound over the central limb and so each window has all the
primary and secondary windings.
primary and secondary windings.
Hence, the total copper area in the window is given by
Hence, the total copper area in the window is given by
A = T A +T A
Acc= TppApp+ TssAss
2IT
=2IT
= δ
δ
2 AT
=2 AT = K w Aw
= δ = K w Aw
δ
The aforementioned equation is same as Eqs. (3.9) and (3.11) and so the output equation
equationis issame
sameasasthe
Eqs.
(3.9) and
(3.11)of
and
the output
equation
for The
a 1φaforementioned
shell-type transformer
output
equation
1φso
core-type
transformer
for
a
1φ
shell-type
transformer
is
same
as
the
output
equation
of
1φ
core-type
transformer
and is given by
and is given by
Q = 2.22 f Bmδ K w Aw Ai × 10−3
Q = 2.22 f Bm δ K w Aw Ai ×10−3
The above equation is the output equation of a single-phase shell-type transformer.
The above equation is the output equation of a single-phase shell-type transformer.
3.3.4 output equation of Three-phase Transformer
3.3.4
of Three-phase
The kVAOutput
rating ofEquation
a three-phase
transformer isTransformer
given by
The kVA rating of a three-phase transformer is given
by
Q (kVA) = 3 Vp I p × 10−3 (or ) 3Vs I s × 10−3
(3.17)
−3
−3
Q (kVA) = 3 Vp I p ×10
(or) 3Vs Is ×10 (3.17)
 3Ep I p × 10−3 [∵ Vp  Ep ]
(3.18a)
 3Ep I p ×10−3 [∵ Vp  Ep ] (3.18a)
(or)
 3Es I s × 10−3 [∵ Vs  Es ]
(3.18b)
(or)
 3Es I s ×10−3 [∵ Vs  Es ] (3.18b)
where p – primary, V, E – voltage, s – secondary, I – current.
where p – primary, V, E – voltage, s – secondary, I – current.
3.3.5 output equation of Three-phase Transformer (Core-type)
The cross-section Equation
of a 3φ core-type
transformer
is shown in (Core-type)
Fig. 3.13. The arrangement of
3.3.5 Output
of Three-phase
Transformer
winding
is shown of
in aFig.
The
cross-section
3φ 3.14.
core-type transformer is shown in Fig. 3.13. The arrangement of
winding is shown in Fig. 3.14.
Design of Transformer
Output Equation
Equation of
of Single-phase
Single-phase Transformer
Transformer
(Shell-type) 3.13
Output
(Shell-type)
3.13
Window
Window area
area
A
Aw
Window
Window area
area
A
Aw
w
w
H
Hww
H
Hww
Iron
Iron area
area
A
Aii
W
Www
W
Www
fig.
Fig.
fig. 3.13
3.13 || Cross-section
Cross-section of
of 3φ
3φ core-type
core-type transformer
transformer
H
H
L
L
V
V
V
V
L
L
H
H
H
H
L
L
L
L
H
H
H
H
L
L
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
V
a1
a
1
a2
a
2
L
L
H
H
V
V
V
V
fig.
Fig.
fig. 3.14
3.14 || Winding
Winding arrangement
arrangement of
of 3φ
3φ core-type
core-type transformer
transformer
Each
Each limb
limb carries
carries both
both the
the primary
primary and
and secondary
secondary windings
windings of
of one
one phase.
phase. So,
So, in
in each
each
window,
there
are
two
primary
and
two
secondary
windings.
There
are
two
windows,
window, there are two primary and two secondary windings. There are two windows, and
and
in
in each
each window,
window, the
the total
total copper
copper area
area in
in the
the window
window is
is given
given by
by
A
(3.19)
2TppA
2Ts sA
AAc c=
==2T
2T
AApp+
++2T
2T
AAs s(3.19)
(3.19)
c
p
p
s
s
Following
Following the
the same
same steps
steps as
as used
used in
in the
the derivation
derivation of
of output
output equation
equation of
of single-phase
single-phase
transformer,
substituting
Eq.
(3.6)
in
Eq.(3.19),
we
get
(3.19), we
we get
get
transformer, substituting Eq. (3.6) in Eq.
Eq.(3.19),
II pI
I
AAcc=
= 2Tpp pp+
+ 2TssIIsss
A
c =22TT
p δ +22TT
s δ
δδ
δδ
2
=22 TpI I p++TTsII s (3.20)
=
(3.20)
= δδ T
Tpp I pp + Tss I ss
(3.20)
δ
((
))
3.14 Design of Transformer
Substituting Eq. (3.8) in Eq. (3.20), we get
2
4 AT
Ac = ( AT + AT ) =
(3.21)
δ
δ
Similarly, equating Eqs. (3.21) and (3.11), we get
4AT
= K w Aw
δ
⇒
AT =
δ K w Aw
(3.22)
4
From Eq. (3.18a),
Q = 3Ep I p ×10−3
Substituting Eq. (3.14) in the above equation, we get
Q = 3Et ×Tp × I p ×10−3 (3.23)
Substituting Eq. (3.8) in Eq. (3.23), we get
Q = 3Et × AT ×10−3
Substituting Eq. (3.22) in the above equation, we get
Q = 3Et ×
δ K w Aw
×10−3
4
Substituting Ep/Tp from Eq. (3.3) in the above equation, we get
Q = 3 × 4.44 × f ×φm ×
δ K w Aw
×10−3
4
= 3.33 f φm δ K w Aw ×10−3 (3.24)
Substituting φm = Bm (maximum flux density) × Ai (area of core) in Eq. (3.24), we get
Q = 3.33 f Bm δ K w Aw Ai ×10−3 (3.25)
Equation (3.25) is the output equation of a three-phase core-type transformer.
3.3.6 Output Equation of Three-phase Transformer (Shell-type)
The cross-section of a 3φ shell-type transformer is shown in Fig. 3.15. The arrangement of
winding is shown in Fig. 3.16.
This transformer has six windows and each window carries LV and HV windings of one
phase only.
Copper area in each window is given by
Ac = Tp Ap + Ts As
(3.26)
Design of(Shell-type)
Transformer
3.15
OutputEquation
EquationofofThree-phase
Three-phaseTransformer
Transformer
(Shell-type) 3.15
3.15
Output
Windowarea
area
Window
AAw w
HHw w
Windowarea
area
Window
AAw w
HHw w
WWw w
WWw w
Windowarea
area
Window
AAw w
HHw w
LV HV
HV
LV
HV LV
LV
HV
LV HV
HV
LV
HV LV
LV
HV
LV HV
HV
LV
Windowarea
area
Window
AAw w
Iron
Iron
area
area
AAi i
HHw w
WWw w
WWw w
Windowarea
area
Window
AAw w
HHw w
HV LV
LV
HV
Windowarea
area
Window
AAw w
HHw w
WWw w
WWw w
fig.
3.15
Cross-section
3φ
shell-type
transformer
fig.
Fig.3.15
3.15|||Cross-section
Cross-sectionofof
of3φ
3φshell-type
shell-typetransformer
transformer
fig.
3.16
Winding
fig.
Fig.3.16
3.16|||Winding
Winding
arrangement
of
3φ
shell-type
arrangement
arrangementofof3φ
3φshell-type
shell-type
transformer
transformer
transformer
Following the same steps as used in the derivation of output equation of single-phase
Followingthe
thesame
samesteps
stepsas
asused
usedininthe
thederivation
derivationofofoutput
outputequation
equationofofsingle-phase
single-phase
Following
transformer, substituting Eq. (3.6) in Eq. (3.19), we get
transformer,
substituting
Eq.
(3.6)
in
Eq.
(3.19),
we
get
transformer, substituting Eq. (3.6) in Eq. (3.19), we get
Ip
I
Ac = TpI pI p + TIssI ss
A
=
T
+
T
Acc= Tpp δ+ Ts s δ
δδ
δδ
Tp I p + Ts I s
=TTppI pI p++TTs sI sI s (3.27)
δ
(3.27)
==
(3.27)
δδ
Substituting
Eq.
(3.8)
in
Eq.
(3.20),
we
get
SubstitutingEq.
Eq.(3.8)
(3.8)in
inEq.
Eq.(3.20),
(3.20),we
weget
get
Substituting
+ AT
AT +
AT
AT
+ AT
A =AT
AAccc==
δ
δδ
2AT
AT
2
AT
2
=
(3.28)
(3.28)
(3.28)
==
δδδ
3.16 Design of Transformer
Similarly, equating Eqs. (3.21) and (3.11), we get
2AT
= K w Aw
δ
⇒
AT =
δ K w Aw
(3.29)
2
From Eqn. (3.18a),
Q = 3Ep I p ×10−3
Substituting Eqn. (3.14) in the above equation, we get
Q = 3Et ×Tp × I p ×10−3 (3.30)
Substituting Eqn. (3.8) in Eq. (3.30), we get
Q = 3Et × AT ×10−3
Substituting Eq. (3.22) in the above equation, we get
Q = 3Et ×
δ K w Aw
×10−3
2
Substituting Ep/Tp from Eq. (3.3) in the above equation, we get
Q = 3 × 4.44 × f ×φm ×
δ K w Aw
×10−3
2
= 6.66 f φm δ K w Aw ×10−3 (3.31)
Substitution φm = Bm (maximum flux density) × Ai (area of core) in Eq. (3.31), we get
Q = 6.66 f Bm δ K w Aw Ai ×10−3 (3.32)
The above equation is the output equation of a three-phase shell-type transformer.
3.4 Volt Per Turn of Winding
The rating of a transformer per phase is given as
Q (kVA) = VI ×10−3 = 4.44 f φm TI ×10−3
= 4.44 f φm AT × 10-3 [ ∵ TI = AT](3.33)
Here, the term, φm is called the magnetic loading and AT is called electric loading.
The magnetic loading, φm = BmAi gives a measure of cross-section of iron core. As the
magnetic loading increases, the flux density Bm and hence the core loss increases which
reduces the efficiency of operation.
Similarly, the electric loading AT gives a measure of the cross-section of the winding.
Increase in electric loading increases the number of turns, resistance and hence results in
increased copper losses reducing the efficiency.
Choice of Flux Density 3.17
So, higher values of φm and AT will only reduce the efficiency of the transformer. For
economical design of transformer, the ratio of φm and AT is maintained constant, which
depends on the type of the transformer.
φ
φ
The ratio,
x = m ⇒ AT = m (3.34)
AT
x
Substituting Eq. (3.34) in Eq. (3.33), we get
Q = 4.44 f
⇒
φm =
2
φm
×10−3
x
x ×10 3
× Q
4.44 f
Now, Volt per turn of winding is given by
Et = 4.44 f φm
= 4.44 f
x ×10 3
× Q
4.44 f
⇒
Et = 4.44 f x ×10 3 × Q
⇒
Et = K Q
where
K = 4.44 f x×10 3
Here, x and hence K are constants. Their value depends on type, construction and service
conditions. Table 3.4 gives the various values of K for different types of transformers.
Table 3.4 | Values of K for various types of transformers
Type of transformers
K
Single-phase core type
0.75–0.85
Three phase core type( power)
0.6–0.7
Three phase core type( distribution)
0.45
Single phase shell type
1.0–1.2
Three phase shell type
1.3
3.5 Choice of Flux Density
The output equation and the emf equation indicate that the choice of flux density Bm affects
the core area Ai. Higher flux density reduces the core area. This will reduce the diameter of the
circumscribing circle, thereby reducing the length of mean turn of winding. Thus, there will be
3.18 Design of Transformer
a huge saving in the cost of copper and iron. But, a higher flux density, while saving copper and
iron, will also result in increased magnetizing current leading to saturation, increased iron losses
requiring elaborate cooling arrangements. The choice of Bm also depends on the type and service
conditions of the transformer. It has already been pointed out that a distribution transformer
should be designed for lower iron losses giving good all day efficiency. Hence, for distribution
transformers, a comparatively lower flux density is assumed. The permissible values of Bm for
various types of transformers are given in Table 3.5. The variation depends on the rating of the
transformer as well, higher values for higher rating.
Table 3.5 | Values of flux density for various types of transformers
Type of transformers
Value of flux
density
Distribution transformer-hot rolled silicon steel
1.1 to 1.4 Wb/m2
Power transformer-hot rolled silicon steel
1.2 to 1.5 Wb/m2
Transformers using cold rolled grain oriented steel
Up to 132 kV
For 275 kV
For 400 kV
1.55 Wb/m2
1.6 Wb/m2
1.7 Wb/m2
3.6 Choice of Current Density
The choice of specific electric loading, i.e., the current density (δ), will affect the transformer
dimensions and efficiency. A higher value of current density will result in smaller dimensions
of transformer. But with increased electric loading heating, losses and hence efficiency
decreases. Higher value of current density will result in excessive temperature rise, resulting
in damage in insulation due to local heating. Even with improved methods of cooling and
with well-ventilated winding, the choice of current density is limited by the local heating.
The type of transformer also affects the choice of current density as the requirement of copper
and iron losses is different for power and distribution transformers. Table 3.6 gives the range
of permissible values of current density for different types of transformers.
Table 3.6 | Values of current density for various types of transformers
Type of transformers
Value of current
density (A/mm2)
Standard distribution transformer and
small power transformer
1.5 to 2.6
Medium and large power transformer
2.4 to 3.4
Large power transformers
5.4 to 6.3
3.7 Design of Core
The core forms the magnetic circuit of the transformer along with the yoke. The core is built
up of silicon steel laminations of 0.3–0.5 mm thickness with varnish insulation in between.
Design of Core
3.19
The core section of a small size transformer is rectangular with circular or rectangular coils.
For higher sizes, circular cores with circular coils are preferred over rectangular coils as the
former occupies less space and uses lesser amount of copper. Also, the mechanical forces
developed under short circuit conditions, change the shape of square or rectangular sections
and damage the coil and insulation. This is not so in circular section. Hence, a circular core
section and circular coils are generally used. The different types of core laminations used for
core-type and shell-type transformers are represented in Fig. 3.17.
Shell-type Laminations
“E-I” Laminations
Core-type Laminations
“E-E” Laminations
“L” Laminations
“U-I” Laminations
fig. 3.17 | Types of core laminations used for core-type and shell-type transformers
Fig.
As a laminated core needs to be used in a transformer, a larger number of laminations of
different sizes are required to build a circular core. It also becomes difficult to arrange them
together to get a circular section. So, it is a general practice to approximate a circular core as
a stepped core with infinite number of steps. However, the maximum number of steps can
be restricted for practical reasons with minimum being one. A stepped core also enables the
use of circular coil. It is also called as cruciform type of core if the number of steps are two.
Figure 3.18 gives the types of cores and Fig. 3.19 shows the core sections of multi-stepped,
square and four-stepped cruciform cores. The shape of the core has to be selected based
on minimum wastage of useful space and for minimum length of copper. The imaginary
circle passing through the corners of the core section is known as circumscribing circle or
Cruciform
(2 step)
Cruciform(2step)
Types of core
Stepped
3
step
3step
Square
4
step
4step
Rectangle
fig. 3.18 | Types of cores
Fig.
3.20
3.20 Design
Design of
of Transformer
Transformer
fig.
Fig.
fig. 3.19
3.19 || Core
Core sections
sections of
of multi-stepped,
multi-stepped, square
square and
and four-stepped
four-stepped cruciform
cruciform cores
cores
circum
circum circle.
circle. The
The dimensions
dimensions for
for obtaining
obtaining the
the maximum
maximum area
area in
in the
the given
given circum
circum circle,
circle, for
for
various
various number
number of
of steps,
steps, are
are given
given in
in Fig
Fig 3.19.
3.19.
Since
Since the
the core
core is
is prepared
prepared by
by stacking
stacking the
the laminations
laminations together
together which
which are
are insulated
insulated from
from
each
each other
other by
by aa thin
thin coating
coating of
of varnish,
varnish, assembled
assembled core
core includes
includes the
the area
area of
of insulation
insulation also.
also.
The
The stacking
stacking factor
factor that
that relates
relates the
the gross
gross core
core section
section with
with the
the net
net core
core section
section is
is defined
defined as
as
Net core section,
A
Net
Aii =i K =
Netcore
coresection,
section, A
Ks (Stacking
factor) factor)
(Stacking
factor)
Gross section,
core section, =
AgiKss(Stacking
Gross
Agi
Grosscore
core section, A
gi
(3.35)
(3.35)
(3.35)
TheThe
stacking factor
usually varies
from 0.850.85
to 0.9.0.9.
The stacking
stacking factor
factor usually
usually varies
varies from
from 0.85 to
to 0.9.
In
order
to
determine
the
utilization
of
the
available space
space of transformer
transformer core, the
the ratios,
In
In order
order to
to determine
determine the
the utilization
utilization of
of the
the available
available space of
of transformer core,
core, the ratios,
ratios,
net
core
area
to
circumscribing
circle’s
area
and
gross
core
area
to
circumscribing
circle’s
area
net
net core
core area
area to
to circumscribing
circumscribing circle’s
circle’s area
area and
and gross
gross core
core area
area to
to circumscribing
circumscribing circle’s
circle’s area
area
are
determined.
are
are determined.
determined.
3.7.1 Square
Core
3.7.1
3.7.1 square
square Core
Core
Square
Square cross
cross section
section
dd
aa
fig.
fig. 3.20
Fig.
3.20 ||| Square
Square core
core
From
From Fig.
Fig. 3.20,
3.20, the
the gross
gross core
core area
area for
for square
square core
core is
is given
given by
by
From
Fig.
3.20,
the
gross
core
area
for
square
core
is
given
by
A
= aa×
(3.36
Agi
×aa,,
(3.36 a)
a)
gi =
Agi = a× a, (3.36a)
where
where aa –– side
side of
of square.
square.
where
ais–observed
side of square.
And
it
And it is observed from
from Fig.
Fig. 3.20
3.20 that
that
And it is observed from Fig. 3.20 that
diagonal
of
square
diagonal of square =
= diameter
diameter of
of circumscribing
circumscribing circle
circle
diagonal of square
=
diameter
of
circumscribing
circle
=
d
=d
=d
Design of Core
Also, for any square, diagonal (d) =
3.21
2a
d
a= d
(3.36 b)
a = 2 (3.36b)
2
Substituting
Eq.
(3.36
b)
in
Eq.
(3.36
a),
we
get
Substituting Eq. (3.36b) in Eq. (3.36a), we get
Hence,
Hence,
d2
Agi = d 2 = 0.5d22
(3.36 c)
Agi = 2 = 0.5d (3.36c)
2
We know that
We know that
Net core area,
A = Stacking factor × gross core area
Net core area,
Aii = Stacking factor × gross core area
A = K × Agi
Aii = Kss × Agi
Assuming stacking factor = 0.9 and substituting Eq. (3.36 c) in the aforementioned equation,
Assuming stacking factor = 0.9 and substituting Eq. (3.36c) in the aforementioned equation,
we get
we get
A = 0.9 × 0.5d 2
Aii = 0.9× 0.5d 2
= 0.45d 2
= 0.45d 2
To determine
determine the
the utilization
utilization of
of available
available space
space in
in transformer,
transformer, area
area of
of circumscribing
circumscribing
To
circle is
is required
required to
to be
be found.
found. It
It is
is given
given by
by
circle
π d22
Acc =
= πd
A
cc
44
where dd –– diameter
diameter of
of circle.
circle. Hence,
Hence, the
the ratio
ratio of
of net
net core
core area
area to
to circumscribing
circumscribing circle’s
circle’s area
area
where
is
given
by
is given by
Ai
0.45d22
A
i =
= 0.45d2 =
=00..58
58
Acc
A
ππdd2
cc
44
And, the
the ratio
ratio of
of gross
gross core
core area
area to
to circumscribing
circumscribing circle’s
circle’s area
area is
is given
given by
by
And,
Agi 0.5d22
A
gi = 0.5d = 0.64
Acc = π d22 = 0.64
A
πd
cc
44
3.7.2 Stepped
steppedCore
Core
The stepped core has two regions, region 1 and region 2, as shown in Fig. 3.21.
2
d
θ
1
b
2
fig. 3.21 || Stepped core
Fig.
a
3.22 Design of Transformer
From Fig. 3.21, the gross core area is given by
Agi = Area of region 1 + Area of region 2
where
Area of region 1 = ( a×b)
Area of region 2 = ( a − b) × b
Hence,
Agi = ab + ab − b 2
= 2 ab − b 2 (3.37)
It is required to express the terms ‘a’ and ‘b’ in terms of ‘d’, which is the diameter of
Agi
A
circumscribing circle, in order to find the ratios, i and
.
Acc
Acc
From Fig. 3.21,
cos θ =
b
a
and sin θ =
d
d
a = d cos θ and b = d sin θ (3.38)
⇒
Substituting Eq. (3.37) in Eq. (3.38), we get
2
Agi = 2× d sin θ × d cos θ − (d sin θ)
= 2d 2 sin θ cos θ − d 2 sin 2 θ
Agi = d 2  sin 2θ − sin 2 θ 


[ ∵ 2 sin θ cos θ = sin 2θ ] (3.39)
The maximum area for a given d is obtained when θ is maximum. So, to get maximum θ,
Agi is differentiated with respect to θ and equated to zero.
dAgi
dθ
=0
Hence, on differentiating Eq. (3.39) with respect to θ and equating it to zero, we get
dAgi
dθ
= d 2 [ 2 cos 2θ − 2 sin θ cos θ ] = 0
⇒
2 cos 2θ = 2 sin θ cos θ
⇒
2 cos 2θ = sin 2θ
⇒
sin 2θ
=2
cos 2θ
Design of Core
Design of Core
⇒
⇒
⇒
⇒
tan 2θ = 2
tan 2θ = 2
2θ = tan−1 (2) and θ = 31.72° (3.40)
2θ = tan−1 (2) and θ = 31.72°
(3.40)
2 (stepped) (Cruciform)
Hence, with
withθ =
θ=
, we
maximum
Hence,
3131
.72.72
°, °we
getget
maximum
area
area
of for
corea for
a given
d. Hence,
of
core
given
d. Hence,
aa==ddcos
cosθθ==0.85d
0.85d
and
b
=
d
sin
θ
=
0.53d
and
b = d sin θ = 0.53d
Using Eq. (3.40) in Eq. (3.39), we get
Using Eq. (3.40) in Eq. (3.39), we get
100
50
80
A = 0.618 d 2
Agi = 0.618gid 2
Net core area,
Net core area,
3.23
3.23
50
80
Ai = 0.9× Agi
Ai = 0.9 × Agi
=0.56 d 2
= 0.56 d 2
 ∵ stacking factor is
  ∵ stacking factor is 
 assumed
to be 0.9  

 assumed to be 0. 9 


0.56 d 22
Ai
Ai = 0.56 d = 0.71
Acc = π d 22 = 0.71
Acc
πd
4
4
2
A
Agi
0..618
618dd 2 = 0.79
gi = 0
=
= 0.79
2
A
π
Acc
πdd 2
cc
44
100
85
3 (stepped)
40
66
85
87
Hence, it
it is
Hence,
is proved
proved that
that the
the area
area ofof
circumscribing
circle
can
be
more
effectively
circumscribing circle can be more effectively
used by
by increasing
used
increasing the
the number
number of
ofsteps
stepsof
ofthe
the
core.
Table
3.7
shows
the
ratio
of
net,
gross
core. Table 3.7 shows the ratio of net, gross core
core area
to area
the area
of circumscribing
circle
area
to the
of circumscribing
circle
for
for
various
cores
represented
in
Fig.
3.22.
various cores represented in Fig. 3.22.
87
100
4 (stepped)
34
57
74
Fig.
fig. 3.22
3.22 || Types
Types of
of stepped
stepped cores
cores
Table 3.7
| 3.7
Ratio|ofRatio
net, gross
area
to area
the area
of area
circumscribing
circle forcircle
various
cores cores
Table
of net,core
gross
core
to the
of circumscribing
for various
ratio
Ratio
square core
core
Square
2-stepped
3-stepped
4-stepped
/area of the circumscribing circle, Acc
A
Agi
gi/area of the circumscribing circle, Acc
Ai/area of the circumscribing circle, Acc
Ai/area of the circumscribing circle, Acc
0.64
0.64
0.58
0.58
0.79
0.79
0.71
0.71
0.84
0.84
0.75
0.75
0.87
0.87
0.78
0.78
Note that the values in the second row of Table 3.7 can be obtained by multiplying the
Note that the values in the second row of Table 3.7 can be obtained by multiplying the
respective values in the first row. This is because, Ai = Agi × Stacking factor. Stacking factor
respective values in the first row. This is because, Ai = Agi × Stacking factor. Stacking factor
is assumed as 0.9.
is assumed as 0.9.
3.24 Design of Transformer
3.8 Design of Yoke
The yoke of a transformer connects the legs and provides the least reluctance path. The yoke
portion does not have windings and so the iron loss in yoke can be minimized by increasing
the yoke area, thereby reducing the operating flux density. Generally, the yoke area is made
20% more than the leg area. The yoke sections can be rectangular or cruciform. Figure 3.23
shows the yoke for rectangular and cruciform types of leg. A cruciform leg can also have a
rectangular yoke as shown in Fig. 3.23.
3.23
Dy
a
a
Hy
a
a
a
a
Rectangular leg and yoke
Cruciform leg and yoke
Rectangular yoke and
cruciform leg
fig. 3.23 | Yoke section for various types of legs
Fig.
For a rectangular yoke, the depth Dyy of the yoke will be same as the depth of the core.
This will be equal to the width of the largest core stamping, ‘a’. The height of the yoke Hyy can
in general be taken as 1–1.5 times the width of the leg.
Area
of yoke,
Ay A
=yD=y D
×yH×y H
=y1.15
to 1.20
times
AgiA
(for
Hot Rolled Silicon Steel)
Area
of yoke,
= 1.15
to 1.20
times
gi (for Hot Rolled Silicon Steel)
Ay A
=yA=giA(for
Cold
Rolled
Silicon
Steel)
(for
Cold
Rolled
Silicon
Steel)
gi
3.9 Window and Core Proportions
The area of the window can be determined from the output equation of the transformer.
kVA
inin
kVA
QQ
Area
windowAAw==
Area
of of
window
w
33
K ××
.22
.33
.66f δf B
δ BmAAK
2.222
oror
3.333
oror
6.666
1010
m i i ww
If WIfwW
is the
width of the window and Hw is the height of the window, then Aw = Ww × Hw
w is the width of the window and Hw is the height of the window, then Aw = Ww ×
in
order
to
limit
the the
leakage
reactance
of the
transformer,
Hw H
is made
more than the Ww. In
Hw in order to limit
leakage
reactance
of the
transformer,
w is made more than the Ww.
practice,
the
ratio
H
/W
lies
between
2.5
and
3.5.
w
In practice, the ratiowH /W
lies between 2.5 and 3.5.
w
w
3.10 Overall
3.10 overall Dimensions
Dimensions
After
After designing
designing the
the core,
core, yoke
yoke sections
sections and
and the
the window
window sections,
sections, the
the overall
overall dimensions
dimensions
could
be
computed
as
follows.
could be computed as follows.
3.10.1 Single-phase
3.10.1 single-phase Core-type
Core-type Transformer
Transformer
Figure
Figure 3.24
3.24 shows
shows the
the overall
overall dimensions
dimensions of
of aa single-phase
single-phase core-type
core-type transformer,
transformer, where
where H
H
–– overall
height,
W
–
overall
width,
D
–
distance
between
the
adjacent
limbs,
d
–
diameter
overall height, W – overall width, D – distance between the adjacent limbs, d – diameter of
of
Overall
Overall Dimensions
Dimensions 3.25
3.25
circumscribing
circumscribing circle,
circle, W
Www –– width
width of
of the
the window,
window, H
Hww –– height
height of
of the
the window,
window, H
Hyy –– Height
Height of
of
yoke,
D
–
depth
of
yoke.
yoke, Dyy – depth of yoke.
H
+ 2H
H=
=H
Hw
w + 2Hyy
D
+a
D=
=W
Ww
w+a
W
W=
=D
D+
+ aa
D
=
a
Dyy = a
Horizontal cross section
Width of
one limb
D
Ww
Hy
Hw
H
Hy
a
d
W
a
d
Width of two limbs
Vertical cross section
Fig.
fig. 3.24
3.24 || Overall
Overall dimensions
dimensions of
of single-phase
single-phase core-type
core-type transformer
transformer
3.10.2
3.10.2 Three-phase
Three-phase Core-type
Core-type Transformer
Transformer
Figure
Figure 3.25
3.25 shows
shows the
the overall
overall dimensions
dimensions of
of aa three-phase
three-phase core-type
core-type transformer,
transformer, where
where
D
D=
=W
Www +
+ dd
H
H=
=H
Hww +
+H
Hyy
W=
= 2D
2D +
+ aa
W
3.10.3 single-phase Shell-type
shell-type Transformer
Transformer
3.10.3 Single-phase
In shell-type
shell-typetransformer
transformershown
shown
Fig.
3.26,
section
be rectangular
in shape.
In
in in
Fig.
3.26,
thethe
corecore
section
will will
be rectangular
in shape.
Only
Only
the
central
limb
carries
windings
and
so
the
flux
in
the
central
limb
is
twice
that
the
the central limb carries windings and so the flux in the central limb is twice that of the of
outer
outer
limbs.
Hence,
the
area
of
cross-section
of
the
central
limb
is
twice
that
of
the
other
limbs.
limbs. Hence, the area of cross-section of the central limb is twice that of the other limbs.
3.26
3.26 Design
Design of
of Transformer
Transformer
Horizontal cross section
D
Hww
Www
d
Hyy
Hww
H
Hyy
a
d
W
a
d
a
d
Width of three limbs
Vertical cross section
fig. 3.25 | Overall dimensions of a three-phase core-type transformer
Fig.
w
Horizontal
cross
section
a
Www
Vertical
cross
section
b
b
2a
Www
a
Hww H
a
fig. 3.26
3.26 || Overall
Overall dimensions
dimensions of
of aa single-phase
single-phase shell-type
shell-type transformer
transformer
Fig.
Overall Dimensions
Dimensions 3.27
3.27
Overall
The dimensions
dimensions are
are given
given by
by
The
Gross
Grosscore
coresection
sectionofofthe
thecentral
centrallimb
limb==2a2a××b b
2Ww++4a4a
WW==2W
w
2H
HH==HH
++2H
HH
++2a2a
y=
ww
y=
ww
Over all
all width,
width,
Over
Overall height,
Note: The
Note:
Theratio
ratiob/2a
b/2a=
=2.5
2.5
3.10.4 Three-phase
Three-phase shell-type
Shell-type Transformer
Transformer
3.10.4
Figure 3.27
3.27 shows
shows the
the overall
overall dimensions
dimensions of
of aa three-phase
three-phase core-type
core-type transformer.
transformer. The
The
Figure
dimensions
are
given
by
dimensions are given by
Gross
Grosscore
coresection
sectionofofthe
thecentral
centrallimb
limb==2a2a××b b
2Ww++4a4a
WW==2W
w
H
=
3H
4Hy
= 3Hw +
H = 3Hw w++4H
4a4a
y = 3Hw +
Overall width,
width,
Overall
Overall height,
Three phase core type
w
Horizontal
section
Depth of winding
a
b
ww
ww
a
2a
Vertical
section
Hw
a
H
fig. 3.27 | Overall dimensions of a three-phase shell-type transformer
Fig.
3.28 Design of Transformer
3.11
Design of Windings
The applied voltage V1 is approximately equal to the voltage induced:
V1 = E1 = 4.44 φm f T1 = Et T1
Number of primary turns (or turns/phase),
T1 =
=
V1
in case of single-phase transformers
Et
V1ph
Et
in case of three-phase transformers
Number of secondary turns (or turns/phase),
T2 =
=
V2
in case of single-phase transformers
Et
V2ph
Et
in case of three-phase transformers
Primary current (or current/phase),
I1 =
kVA ×10 3
in case of single phase transformers
V1
=
kVA ×10 3
in case of three-phase transformers
3V1ph
Cross-sectional area of primary winding conductor
I
a1 = 1 mm 2
δ
Secondary current (or current/phase)
I2 =
kVA ×10 3
in case of single-phase transformers
V2
=
kVA ×10 3
in case of three-phase transformers
3V2ph
Cross-sectional area of secondary winding conductor
I
a2 = 2 mm 2
δ
Knowing the number of turns and cross-sectional area of the primary and secondary
winding conductors, number of turns/layer in a window height of Hw and number of layers
in a window width of Ww can be found out.
Design
Design of
of Windings
Windings 3.29
3.29
The
The various
various types
types of
of windings
windings used
used in
in transformers
transformers are
are shown
shown in
in Fig.
Fig. 3.28.
3.28.
Winding
types
Cylindrical
Helical
Continuous
disc type
Layer
fig. 3.28 | Different types of windings used in transformers
Fig.
3.11.1 Type
Type and
and Choice
Choice of
of Windings
Windings
3.11.1
Differenttypes
typesof of
windings
namely,
cylindrical,
layer
and continuous
disc
Different
windings
namely,
cylindrical,
helical,helical,
layer and
continuous
disc windings
windings
areused
generally
usedupon
depending
uponand
thecurrent
voltage
and current
ratings.
are
generally
depending
the voltage
ratings.
The various
typesThe
of
various types
of windings
used
in transformer
construction
are explained as
windings
frequently
used infrequently
transformer
construction
are explained
as follows.
follows.
Cylindrical winding
Cylindrical winding
The cylindrical winding may use circular, rectangular or strip conductors. It is generally
The cylindrical winding may use circular, rectangular or strip conductors. It is generally
wound with many layers. The layered cylindrical winding with sheet conductors is shown
wound with many layers. The layered cylindrical winding with sheet conductors is shown
in Fig. 3.29(a).
in Fig. 3.29(a).
Helical
Helical winding
winding
This
This type
type of
of winding
winding may
may be
be of
of single
single or
or double
double helical
helical type.
type. They
They are
are usually
usually wound
wound around
around
aa bakelite
cylinder.
In
a
single
helical
type,
the
turns
are
wound
in
the
axial
direction.
bakelite cylinder. In a single helical type, the turns are wound in the axial direction. There
There
is
is only
only one
one turn
turn in
in each
each winding
winding layer.
layer. If
If the
the winding
winding consists
consists of
of many
many strands
strands in
in parallel,
parallel,
they
they are
are called
called double
double helical
helical winding.
winding. Helical
Helical windings
windings are
are usually
usually used
used for
for low-voltage
low-voltage
winding
where
the
number
of
turns
is
less
and
current
is
high.
Since
the
current
winding where the number of turns is less and current is high. Since the current is
is high,
high, aa
large
number
of
strips
are
used
in
parallel
and
are
placed
side
by
side
in
a
radial
direction.
large number of strips are used in parallel and are placed side by side in a radial direction.
These
These parallel
parallel connected
connected conductors
conductors can
can be
be transposed
transposed at
at regular
regular intervals
intervals to
to equalize
equalize the
the
resistance
and
leakage
reactance
of
each
conductor.
A
single
helical
winding
resistance and leakage reactance of each conductor. A single helical winding is
is shown
shown in
in
Fig.
Fig. 3.29(b).
3.29(b).
Continuous
Continuous disc
disc winding
winding
The
The disc
disc winding
winding consists
consists of
of number
number of
of flat
flat coils
coils or
or discs
discs connected
connected in
in series
series or
or parallel.
parallel. The
The
discs
discs are
are formed
formed by
by single
single or
or number
number of
of strips
strips in
in parallel
parallel wound
wound spirally.
spirally. These
These windings
windings are
are
used
used for
for high
high rated
rated transformers.
transformers. A
A simple
simple disc
disc winding
winding is
is shown
shown in
in Fig.
Fig. 3.29(c).
3.29(c).
Layer winding
This type of winding consists of several cylindrical layers wound concentrically and are
connected in series as shown in Fig. 3.29(d). They are used for windings rated above 100 kV,
with the number of layers dependent on the voltage rating. Generally, the outer layers are
made shorter to distribute the winding capacitance uniformly.
Once the choice of winding is properly made, then it is designed in such a way that it
fits properly with the calculated window dimensions accommodating provision for ducts,
insulation and clearances. Table 3.8 gives a guideline about the choice of windings based on
the voltage ratings.
3.30 Design
Design of
of Transformer
Transformer
3.30
12 11 10 9 8 7 6 5 4 3 2 1
Low
voltage
winding
Core
leg
(a)
(b)
Layer
Winding
Spacer
Transposition
(b)
(c)
(d)
fig.
Fig. 3.29
3.29 || Different
Different types
types of
of windings
windings used
used inin transformers:
transformers: (a)
(a) Continuous
Continuous disc
disc type
type winding,
winding,
(b)
Helical
winding,
(c)
Layer
winding
and
(d)
Cylindrical
winding
(b) Helical winding, (c) Layer winding and (d) Cylindrical winding
Table 3.8 | Choice of windings based on the voltage ratings of transformers
Table 3.8 | Choice of windings based on the voltage ratings of transformers
Type of transformers
kVa
Type of transformers
kVA
Transformers for
1.6–100
Transformers for rural
1.6–100
rural supply
supply
Distribution
100–1000
Distribution small
100–1000
transformer,
transformer,
small
power transformer
power transformer
Transformers
10,000–20,000
Transformers
10,000–20,000
for
secondary for
secondary transmission
transmission
Transformers
Transformers for
for
primary
primary transmission
transmission
Transformers
Transformers at
at
generating
generating stations
stations
Voltage in kV
Winding
Voltage in kV
Winding
HV – 3.3 to 11
Foil, crossover
HV – 3.3 to 11
Foil, crossover
LV – 0.44
Helical
LV – 0.44
Helical
HV – 11 to 33
Foil, crossover, disc, multilayer
HV – 11 to 33
Foil, crossover, disc,
helical
multilayer helical
LV – 0.44
Helical, multilayered helical
LV – 0.44
Helical, multilayered helical
HV – 33 to 66
Disc
HV – 33 to 66
Disc
LVLV
– 11
Disc
or helix
– 11
Disc
or helix
45,000–15,00,00
– 132
to to
440
or multi-layered
helix
45,000–15,00,00HV
HV
– 132
440 DiscDisc
or multi-layered
helix
LVLV
– 11,33,66
DiscDisc
or helix
– 11,33,66
or helix
72,000–35,00,00
– 132
to to275
or multi-layered
helix
72,000–35,00,00HV
HV
– 132
275 DiscDisc
or multi-layered
helix
LVLV
– 1122
– 1122
DiscDisc
or helix
or helix
Design of Windings 3.31
Example 3.1: Determine the main dimensions of a 350 kVA, 3φ, 50 Hz Y/Δ, 11,000/3300
V, core-type distribution transformer. Assume distance between core
centres as twice the width of core.
Solution: Given,
Q = 350 kVA
Transformer rating,
Type = 3φ , Y-Δ
Voltage rating = 11000/3300V
For a 3φ core-type distribution transformer volt per turn,
Et = 0.45 kVA
(refer Table 3.4)
(1)
Et = 0.45 350 = 8.4
E
8.4
= 0.038 Wb
φm = t =
4.44f ( 4.44)(50)
φm
(Assuming 1.2 T flux density)
Bm
0.038
Ai =
= 0.032 m 2
1.2
Ai =
For a three-stepped core,
Width of largest stamping,
Ai = 0.6 d 2 = 0.032 m 2
d=
0.032
= 0.23 m
0.6
a = 0.9d
a = (0.9)(0.23)
a = 0.21 m
Depth of yoke,
Dy = a = 0.21 m
We know that
kVA = 3.33 f δ Ai Bm Aw kw × 10−3(2)
Let us assume δ as 2.5 A/mm2
and
kw =
=
10
30 + kVin
10
= 0.24
30 + 11
3.32 Design of Transformer
Substituting the determined values in Eq. (2), we get
350 = (3.33) (50) (2.5 × 106) (0.032) (1.2) Aw (0.24) × 10−3
Aw = 0.09 m2
Given that the distance between the core centre, D = 2a
Ww + d = 2a
Ww = 2a - d = 0.19 m
Hw =
Overall width,
Aw
0.09
=
= 0.47 m
Ww 0.19
W = 2D + a = 2(Ww + d) + a
= 2(0.19 + 0.23) + 0.21
= 1.05 m
Overall height,
H = H w + 2 H y = 0.89 m
Width = 0.21 m
W = 1.05 m
H = 0.89 m
a = 0.21m
Example 3.2: Estimate the main dimensions of a 3φ, Δ-Y core-type transformer
rated for 300 kVA, 6600/440 V, 50 Hz. A three-step core with a diameter
of circumscribing circle of 0.25 m and a leg spacing of 0.4 m is to be
designed. The volts per turn is 8.5 volts, the current density is 2.5 A/
mm2, the window space factor is 0.28 and the stacking factor is 0.9. Also
find the conductor dimensions of primary and secondary windings.
Solution: Given
Transformer rating = 300 kVA
Voltage rating and type = 6600/440V, Δ/Y
Diameter of circumscribing circle of
core = 0.25 m
Leg spacing = 0.4 m
Volts/turn, Et = 8.5 V
Current density, δ = 2.5 A/mm2
Window space factor, kw = 0.28
Stacking factor, ki = 0.9
Primary voltage = 6000 V (Δ connected)
6600
Et
6600
=
= 776
8.5
Number of primary turns =
Design of Windings 3.33
Number of secondary turns =
Secondary phase voltage
Et
440 / 3
8.5
254
=
= 30
8.5
=
Primary winding current,
I Lp =
=
kVA rating of transformer
3VLp ×10−3
300
( 3 )(6600)10−3
= 26.24 A
Primary is Δ connected. So phase current of primary
=
26.24
3
= 15.15 A
Area of conductors of primary conductor,
ap =
Ip
δ
15.15
=
2.5
= 6.06 mm 2
Q
Line current of secondary side =
3VLs ×10−3
Since the secondary side is star connected, phase current and line current are the same.
Area of cross-section of secondary conductor,
I
393.65
as = s =
δ
2.5
= 157.5 mm 2
Where,
Is =
Q
( 3 )(440×10−3 )
= 393.65 A
Copper area in window,
(
Ac = 2 apTp + as Ts
)
= 2 (6.06 ×776 + 157.5× 30)
= 18855.12 mm 2
3.34 Design of Transformer
Aw =
Window area,
=
Area of circumscribing circle,
Acc =
Copper area
window space factor
18855.12
= 67339.71 mm 2
0.28
πd2
4
2
=
π (0.25)
4
= 0.049 m 2
We know that for a three-stepped core,
Agi = 0.84 Acc
Gross iron area,
= (0.84)(0.049)
= 0.041 m 2
Ai = Agi ×stacking factor
Net iron area,
= (0.041)(0.9)
= 0.0369 m 2
Given leg spacing = 0.45 m (∵ D = Ww + d = 0.45)
Width of window Ww = D − d = 0.45 − 0.25
= 0.20 m
Height of window = H w =
=
Width of largest stamping,
Aw
Ww
0.067
= 0.335 m
0.20
a = 0.84 d = 0.21 m
Overall width of transformer,
W = 2D + a
= 2(0.45) + 0.21
= 1.11 m
Overall height of transformer,
H = Hw + 2Hy
= 0.335 + 2(0.21)
= 0.755 m
Design of Windings 3.35
The main dimensions are
W = 1.11m
H = 0.755 m
a = 0.21m
Example 3.3: A three-phase, 50 Hz oil cooled core-type transformer has the following
dimensions. Distance between core centres = 0.2 m, height of window
= 0.24 m, diameter of circumscribing circle = 0.14 m. The flux density
in the core = 1.25 Wb/m2, current density in conductor = 2.5 A/mm2.
Assume a window space factor of 0.2 and core area factor of 0.56. The
core is two stepped. Estimate the kVA rating of transformer.
Solution:
Given
f = 50 Hz
Bm = 1.25 Wb/m 2
D = 0.2 m
H w = 0.24 m
δ = 2.5 A/mm 2
kw = 0.2
d = 0.14 m
Ai = 0.56 d 2
Q = 3.33 f Bm Ai kw Aw δ × 10−3(1)
Ai = 0.56 d2 = (0.56) (0.14)2
For a 3φ core-type transformer,
Area of window,
= 0.0109 m2
D = d + Ww
Ww = D - d = 0.2 - 0.14
= 0.06 m
Aw = H w ×Ww = 0.24 × 0.06
=0.0144 m 2
Q = 3.33 × 50 × 1.25 × 0.0109 × 0.2 × 0.0144 ×
2.5 × 106 × 10−3
Q = 16.33 kVA
Example 3.4: Calculate the core and window area of a 125 kVA, 2000/400V, 50 Hz,
1φ, shell-type transformer. Find also the amount of copper and iron
required. Flux density is 1.1 T, current density is 2.2 A/mm2, Volt/turn
is 11.2 volts and window space factor is 0.33. Take specific gravity of
copper as 8.9 g/cm3 and that of iron as 7.8 g/cm3. Assume a rectangular
core with stamping of 7 cm width.
3.36 Design of Transformer
Solution: Given
kVA rating, Q = 125
Volt and type of transformer = 2000/400 V 1φ and shell
Flux density, Bm = 1.1 T
Current density, δ = 2.2 A/mm2
Volt/turn = 11.2 V
Window space factor = 0.33
Width of stamping, a = 7 cm
Specific gravity of copper = 8.9 g/cm3
Specific gravity of iron = 7.8 g/cm3
Et = 4.44 f φm = 11.2V
Volts per turn,
φm =
Et
11.2
=
4.44 f ( 4.44)(50)
= 0.05 Wb
Ai =
φm 0.05
=
1.1
Bm
= 0.045 m 2
We know that kVA rating of a 1φ shell-type transformer is given by
Q = 2.22 f δ Ai Bm Aw kw × 10−3
125 = (2.22) (50) (2.2 × 106) (0.045) (1.1) A
Aw = 0.03 m2
(1)
w (0.33) × 10
−3
The stampings are of 7 cm width. Hence,
a = 7 cm
2a = 14 cm
b=
Agi
2a
⇒ b=
=
Ai
(ks )(2a)


∵ Agi = Ai 

ks 
0.045
(0.9)(0.14)
= 0.36 m
Area of window,
Aw = 0.03 m2
Let the ratio of height of window to width of window is 3. The window is represented in
Fig. 3.30.
Hence,
Hw
=3
Ww
Aw = 0.03
2
= 3 Ww
Design of Windings 3.37
c
y
a
x
d
z
2a
0
b
2a
e
fig. 3.30
Fig.
0.003
.03
==0.01.1mm
33
0.01.1==0.03.3mm
HH
3×
ww==3 ×
WW
ww==
The whole
whole window
window in
in this
this shell-type
shell-type transformer
transformer is
is assumed
assumed to
to be
be filled
filled with
with both
both LV
LV
The
and
HV
windings.
and HV windings.
So,
So,
height
heightofofwinding
winding==height
heightofofwindow
window
And, width
width of
of LV
LV and
and HV
HV winding
winding =
= width
width of
of window
window
And,
Weight
Weightofofcopper
copper==Volume
Volumeofofcopper
copper××density
density
= (Area of copper × length of copper in bcdeb)
=×(A
rea of copper × length of copper in bcdeb)
density
× density
= Aw kw× length of mean turn × density
(2)
= Aw kw× length of mean turn × density (2)
Length of mean turn of winding = 2(2 a + Ww ) + 2 (b + Ww )
Where,
Where,
Length of mean turn of winding==2 (214
(2 a++10W)w
2 (+
b+
+)2+
(32
10W
) w)
==132
cm+ 10) + 2(32 + 10)
2 (14
From Eq. (2),
From Eq. (2),
= 132 cm
Weight of copper = 0.03 × 10 4 ×0.33 × 132×8.9× 10−3
= 116.4 kg
-3
Weight
of
copper
0.03
× 10 4of
×0.33
×8.9×
Weight of iron ==2×
volume
iron×
×132
density
of10
iron
= 116.4
A kg
= 2× i × mean core length × density
Weight of iron = 2×2volume of iron ×density of iron
0.A
045
i × mean
2×
 2 (10length
==2×
×10 4 ×core
+ 7 ) +×2 (density
30 + 7 )

2
2
0.045−3
=×27×.8 ×10 ×10 4 ×  2 (10 + 7 ) + 2 (30 + 7 )
2
Weight of iron = 379 kg
×7.88 ×10−3
Weight of iron = 379 kg
3.38 Design of Transformer
Example 3.5: Determine the dimensions of the core and window for a 125 kVA, 50 Hz,
single-phase oil immersed core-type transformer. Assume flux density
as 1.23 T, current density as 2.5 A/mm2 and window space factor as 0.32.
Take emf per turn as 17 V. Design a cruciform core with window height
as three times the window width.
Solution: Given
Emf per turn = 17 V
Flux density, Bm = 1.23 T
Rating = 125 kVA
Current density, δ = 2.5 A/mm2
Type = single phase
Window height, Hw = 3Ww
Window space factor, kw = 0.32
We know that
Volt per turn Et = 4.44 f φm
(1)
From Eq. (1),
φm =
Main flux,
=
Et
4.44 f
17
( 4.44)(50)
= 0.0765 Wb
Also, net iron area,
Ai =
φm
(2)
Bm
Ai =
0.0765
= 0.062 m 2
1.23
For a cruciform core, net iron area,
Ai = 0.56d2
(3)
From Eq. (3), the diameter of the circumscribing circle is given by
d=
=
Width of the largest stamping,
Ai
0.56
0.062
= 0.332 m
0.56
a = 0.85d (4)
a = (0.85)(0.332) = 0.282 m
∵
width of transformer = width of largest stamping
width of transformer = 0.282 m (app)
We know that
Width of smallest stamping,
b = 0.53d(5)
Design of Windings 3.39
Design of Windings 3.39
b = (0.53)(0.332) = 0.175 m
b = (0.53)(0.332) = 0.175 m
H
(6)
Hyy=
=(1
(1to
to1.5)a
1.5)a(6)
Height of
of yoke,
yoke,
Height
HHyy==(1)
(1)(0.282)
(0.282)==0.282
0.282m
m
−3 (7)
Q=
=2.22
2.22ffδδAAiBBmAAwkkw×
×10
10−3
Q
(7)
i m w w
The
The kVA
kVA rating,
rating,
((
))
125==((22.22
50)) 22.5
×10
1066 ((00.062
Aww((00.32
10−−33
)(11.23
125
.22)()(50
.5 ×
.062)(
.23))A
.32))××10
125×1033
125 × 10 = 0.0184 m22
= 0.0184 m
AAww==
1066
66.77
.77 ××10
Hww=
=3W
3Wwwand
andAAww=
=H
HwwW
Www(8)
H
(8)
∵
∵
The main dimensions in window and core are shown in Fig. 3.31.
The main dimensions in window and core are shown in Fig. 3.31.
D
w = 0.688 m
a
0.292
Hw = 0.23
Ww = 0.078
H = 0.79 m
Hy = 0.292
d
0.332
fig. 3.31
Fig.
Overall height of transformer,
Overall height of transformer,
Aw = 3Ww22 = 0.0184
Aw = 3Ww = 0.0184
0.0184
Ww = 0.0184 = 0.078 m
Ww =
= 0.078 m
3
3
H w = 3Ww = (3)(0.078) = 0.234 m
H w = 3Ww = (3)(0.078) = 0.234 m
H = Hw + 2H y
H = Hw + 2H y
= 0.23 + 2 (0.28)
= 0.23 + 2 (0.28)
= 0.79 m
= 0.79 m
3.40 Design of Transformer
Overall width of transformer,
W = Ww + d + a
= 0.078 + 0.33 + 0.28
= 0.688 m
a = 0.282 m
Overall depth of transformer,
The main dimensions are
H = 0.79 m
W = 0.688 m
a = 0.282 m
Example 3.6: Find the dimensions of the core and yoke of a single-phase coretype transformer of rating 250 kVA, 50 Hz. Given the Volt per turn as
14 V, maximum flux density as 1.2 Wb/m2, the window space factor
as 0.3 and current density as 2.8 A/mm2. Design a square core with
distance between the adjacent limbs equal to 1.5 times the width of the
transformer. Assume a stacking factor of 0.8 and the flux density in the
yoke is 80% flux density in the core.
Solution: Given
kVA rating, Q and type = 250 kVA and 1φ
Frequency, f = 50 Hz
Volt/turn, Et = 14 V
Maximum flux density, Bm = 1.2 Wb/m2
Window space factor, kw = 0.3
We know that
Current density, δ = 2.8 A/mm2
Distance between adjacent limbs, D = 1.5a
Stacking factor = 0.8
Flux density in yoke, Bmy = 80% of Bm in core
Volt/turn,
Et = 4.44 f φm
Et = 4.44 f Bm Ai
(1)
From Eq. (1), area Ai can be written as
Et
4.44 f Bm
14
=
( 4.44)(50)(1.2)
Ai =
= 0.0525 m 2
For a square core,
Ai = 0.45d2(2)
0.0525 = 0.45d 2
0.0525
d=
= 0.341 m
0.45
Design of Windings 3.41
Width of largest stamping,
a = 0.85d = 0.289 m
Given that distance between adjacent limbs,
∴ Width of window,
D = 1.5a = (1.5)(0.289)
= 0.4335 m
Ww = D - d = 0.4335 - 0.341
= 0.0925 m
For a single-phase core-type transformer,
Q = 2.22 f Bm Ai Aw kw δ × 10−3(3)
250 = (2.22)(50)(1.2)(0.0525) Aw × 0.3 × 2.8 ×106 ×10−3
250 = 5874.12 Aw
Aw =
250
= 0.0425 m 2
5874.12
Height of window,
Aw
Ww
0.0425
=
0.0925
= 0.4594
Hw =
Given that
Bmyoke = 80% Bmcore
Bmy = 0.8 Bmc
φm
φ
= 0.8 m
Ay
Agi
Ay =
1
Agi
0.8
Ay = 1.25 Agi
Ay = (1.25)(0.0656) = 0.082 m2
Where,
⇒
Ai
= Agi
Stacking factor
Agi =
Ai
0.0525
=
0.8
0.8
= 0.0656 m 2 (4)
3.42 Design of Transformer
Depth of yoke,
Dy = a = 0.289 m
Height of yoke,
Hy =
Overall width,
W = D + a = 0.43 35 + 0.289
0.082
= 0.283 m
0.289
= 0.7225 m
H = Hw + 2Hy
Overall height,
= 0.4594 + (2)(0.283)
= 1.0254 m
The main dimensions of transformer are
W = 0.72 m
H = 1.0254 m
a = 0.289 m
Example 3.7: Find the main dimensions of a three-phase, 200 kVA, 6600/600 volts,
core-type transformer. Assume a three-stepped, cruciform core with
area factor of 0.62 and window space factor of 0.25. The current density
is 2.8 A/mm2 and the emf per turn is 9 V. Assume the height of window
as three times the width of window and the maximum flux density as
1.25 Wb/m2.
Solution: Given
Rating, Q and type= 200 kVA and 3φ
Voltage = 6600/600 V
Window space factor kw = 0.25
We know that
Volt per turn,
Net iron area,
Maximum flux density = 1.25 Wb/m2
Current density = 2.8 A/mm2
Volt/turn = 9 V
Hw
=3
Ww
Et = 4.44 f Bm Ai
Ai =
=
(1)
Et
4.44 f Bm
9
( 4.44)(50)(1.25)
= 0.0324 m 2
For a four-stepped core,
Ai = 0.62d2
(2)
Design of Windings 3.43
0.0324 = 0.62d 2
d=
0.0324
= 0.2286 m
0.62
Width of largest stamping for a four-stepped section,
a = 0.92d(3)
⇒
a = 0.210 m
kVA of a 3φ transformer is
Q = 3.33 f Bm Ai Aw kw δ × 10−3
(4)
200 = (3.33) (50) (1.25) (0.0324) Aw (0.25) (2.8 × 106)
× 10−3
Window area,
Aw =
200
4.72×10 3
= 0.0423 m 2
Given that
Height of window H w
=
=3
Width of window Ww
Area,
Aw = ( H w )(Ww ) = 0.0423 m 2
2
Aw = 3 Ww
= 0.0423
0.0423
= 0.11874
3
H w = 0.1188 × 3 = 0.3564 m
Ww =
We know that Hy = a and Dy = a
Overall height,
H = H w + 2H y
= 0.3564 + 2 (0.210)
= 0.7764 m
Overall width,
W = 2D + a = 2Ww + 2d + a
= (2)(0.1188) + 2 (0.2286) + 0.210
= 0.9048 m
W = 0.9048 m
H = 0.7764 m
a = 0.210 m
The main dimensions in window and core are shown in Fig. 3.32.
3.44 Design of Transformer
0.28
0.22
0.22
0.210
a
0.7764
0.3564
0.1188
0.21
0.22
0.9048
fig. 3.32
3.32
Fig.
Example 3.8: 3.8: Determine
Determine the
the main
main dimensions
dimensions of
of aa 100
100 kVA,
kVA, 2200/480
2200/480 V,
V, 50
50 Hz,
Hz, singlesingleExample
phase
core-type
transformer.
Find
also
the
number
of
turns
and
crossphase core type transformer. Find also the number of turns and crosssectional
area
of
primary
and
secondary
conductors.
Assume
Volt/turn
sectional area of primary and secondary conductors.
Assume Volt/turn
as 7.5
7.5 V,
V, maximum
maximum flux
flux density
density as
as 1.2
1.2 Wb/m
Wb/m22.. Ratio
Ratio of
of effective
effective crosscrossas
sectional
area
of
core
to
square
of
diameter
of
circumscribing
circle
is
sectional area of core to square of diameter of circumscribing circle is
0.6,
ratio
of
height
to
width
of
window
is
2,
window
space
factor
is
0.28
0.6, ratio of height to width of window
is 2, window space factor is 0.28
and current
current density
density is
is 2.5
2.5 A/mm
A/mm22..
and
Solution: Given
Solution:
Given
2 2
Rating,
Current
density,
δ=
A/mm
Rating, Q
Q and
and type
type =
= 100
100 kVA
kVA and
and 1φ
1φ
Current
density,
δ=
2.52.5
A/mm
Voltage = 2200/480 V
Window
space
factor,
0.28
Window
space
factor,
kwkw
==
0.28
2 2
Volt/turn = 7.5
AiA=i =
0.6d
0.6d
Bm =
= 1.2
1.2 Wb/m
Wb/m22
B
m
Hw
=
2
Hw
W
=2
w
Ww
Volt per turn,
Et = 4.44 f Bm Ai = 7.5
Volt per turn,
Et = 4.44 f Bm Ai = 7.5
Et
Net iron area,
Ai =
E
A4i .44
= f Bmt
Net iron area,
4.44 f Bm
7.5
=
7.5
.44)(50)(1.2)
( 4=
( 4.44)(250)(1.2)
= 0.02815 m
= 0.02815 m 2
Given
Ai = 0.6 d2
(1)
Given
Ai = 0.6 d2 (1)
0.02815 = 0.6 d2
0.02815 = 0.6 d2
⇒
d = 0.2166
m m
⇒
d = 0.2166
Design of Windings 3.45
Let us assume a three-stepped core, and the width of largest stamping,
a = 0.9d(2)
a = 0.1949
Q = 2.22 f Bm Ai Aw kw δ × 10−3
100 = (2.22) (50) (1.2) (0.02815) Aw (0.28) (2.5 × 106)
× 10−3
Given
Aw = 0.03809 m2
Hw
=2
Ww
and
Hence,
Aw = 0.03809 m2
( H w )(Ww ) = 0.03809
2Ww2 = 0.03809
Ww = 0.138 m
Hw = 0.276 m
Depth of yoke = a = 0.1949 m
Let the area of yoke is 20% greater than the area of core section.
Ay = 1.2
Ai
0.02815
= 1.2×
0.9
0.9
= 0.03753 m2
∵
Dy = 0.1949 m
and
Ay = 0.03753 m2
Hy =
Ay
Dy
=
0.03753
0.1949
= 0.1925 m
Overall width,
W = D + a = W w+ d + a
= 0.138 + 0.2166 + 0.1949
= 0.5495 m
Overall height,
H = Hw + 2Hy = 0.276 + 2 × 0.1925
= 0.661 m
Number of secondary turns,
T2 =
Secondary voltage
Volt per turn
3.46 Design of Transformer
=
Number of primary turns,
T1 =
=
I1 =
Primary current,
=
480
= 64
7.5
Primary voltage
Volt per turn
2200
= 293
7.5
kVA ×1000
primary voltage
100 ×1000
= 45.45 A
2200
Area of cross-section of primary winding,
I
a1 = 1
δ
45.45
=
= 18.18 mm 2
2.5
Secondary current,
I2 =
kVA ×1000
secondary voltage
=
100 ×1000
= 208.34 A
480
Area of cross-section of secondary winding,
I
a2 = 2
δ
208.34
=
=83.34 mm 2
2.5
3.12 Resistance Calculation
The resistances of HV and LV windings are to be calculated in order to estimate the copper
loss in the transformer. Based on the losses and hence the temperature rise, the cooling
system will be designed. The area of cross-section of HV and LV coils, the number of turns
and the layout of the winding are used to calculate the mean length of LV and HV windings.
Then the resistance of each winding can be calculated as follows:
Resistance of the primary winding/phase,
Rp =
ρLmt Tp
ap
Reactance Calculation 3.47
Resistance of the secondary winding/phase,
Rs =
ρLmt Ts
as
Mean length of turn of the primary winding,
Lmt = π × mean diameter of the winding
Mean length of turn of the secondary winding,
Lmt = π × mean diameter of the winding
Where resistivity of copper at 60°C, r = 2.1 × 10−6 ohm-cm or 2.1 × 10−8 ohm-m or 0.021
ohm/m/mm2
Lmtp and Lmts are the length of mean turns of primary and secondary windings, respectively.
Tp, Ts are the corresponding number of turns
ap, as are the area of cross-sections of the conductors
Total resistance per phase referred to primary side,
 Tp 2
R1 = Rp +   Rs
 Ts 
3.13
Reactance Calculation
The reactance of the windings primarily involves estimation of the leakage flux. Reactance
calculations are required to make an estimate the regulation of the transformer to be
designed.
(a) Useful flux
The flux that links with both primary and secondary windings and is responsible in
transferring the energy electromagnetically from primary to secondary side is called
the useful flux. The path of the useful flux is in the magnetic core.
(b) Leakage flux
The flux that links only with the primary or secondary winding and is responsible
in imparting inductance to the windings is called the leakage flux. The path of
the leakage flux depends on the geometrical configuration of the coils and the
neighbouring iron masses.
(c) Reactance
1. Leakage reactance = 2πfl × inductance = 2πf × (Flux linkage/current)
2. Flux linkage = flux × number of turns
3. Flux = (mmf or AT)/Reluctance = AT × permeance ( ∧ )
4. Permeance ( ∧ ) = 1/Reluctance =
aµ0 µr
l
where a = area over which the flux is established, l = length of the flux path.
If xp and xs are the leakage reactances of the primary and secondary windings, then the
total leakage reactance of the transformer referred to primary winding
3.48 Design of Transformer
 Tp 2
X p = xp + xs′ = xp + xs  
 Ts 
Similarly, the leakage reactance of the transformer referred to secondary winding
 T 2

Xs = xp′ + xs = xp  s  + xs
 Tp 
Estimation of the leakage flux or reactance is always difficult, on account of the complex
geometry of the leakage flux path and great accuracy is unobtainable. A number of
assumptions are to be made to get a usable approximate expression. Validity or the accuracy
of the expression is checked against test data.
3.13.1 Leakage Reactance of a Core-type Transformer with ­Concentric
LV and HV Coils of Equal Height or Length
The following are the assumptions considered for the leakage reactance of the core-type
transformer.
••
••
••
••
••
••
••
••
Effect of magnetizing current is neglected.
Reluctance and effect of saturation of iron are neglected.
All the mmf is assumed to be used to overcome the reluctance of coil height.
Leakage flux distribution in coil and in the space between the LV and the HV coils is
assumed to be parallel to the leg axis.
Equal axial lengths for both primary and secondary windings.
The flux paths are parallel to the winding along the axial height.
Ip Tp = Is Ts = Total mmf
Length of mean turn of both windings is equal.
Let,
bp and bs = Radial depth of primary and secondary windings
Tp and Ts = Number of primary and secondary turns per phase for three phase
Ip and Is = Primary and secondary currents per phase for three phase
Lmtp/Lmts = Mean length of turn of primary or secondary windings, respectively
Lmt = Mean length of primary and secondary windings considered together
L0 = Circumference of the insulation portion or duct or both between LV and HV coils
Lc = Axial height or length of both LV and HV coils
The total flux linkage of the primary or secondary winding is due to
•• Leakage flux inside the primary or secondary winding and
•• Leakage flux in between the LV and HV coils
To determine the flux linkage due to the flux inside the coil, consider an elemental strip
dx at a distance ‘x’ from the edge of the LV winding.
Then the flux linkage of the LV winding due to the flux φx in the strip,
Ψx = φx × number of turns linked by φx
Ψx = ampere turns producing φx × permeance of the strip × number of turns linked by φx
Reactance Calculation 3.49
 I pTp x   Lmtp dxµ0   Tp x 

 

×
ψx = 
×

 bp  
Lc
  bp 
Considering that the mean length of the strip is approximately equal to Lmtp.
Therefore, the total flux linkage due to the flux inside the coil is given by
bp
 I pTp x   Lmtp dxµ0   Tp x 

 
×

ψx = ∫ 
×
 bp  
Lc
bp 



0
ψx =
I pTp2 µ0 Lmtp bp
3 Lc
If one half of the flux φ0, in between the LV and HV windings, is assumed to be linking
with each winding, then the flux linkage of the primary winding due to half of the flux φ0 in
between LV and HV windings is
1
ψ0 = φ0 ×Number of turns linked by φ0
2
= ampere turns producing φ0 × permeance of the
duct × number of turns linked by flux φ0
L aµ
1
ψ0 = ×Tp I p × 0 0 ×Tp
2
Lc
Therefore, the total flux linkage of the primary winding is
λ = ψ + ψ0 = I pTp2 ×
µ0  Lmtp bp L0 a 


+
Lc  3
2 
Assuming that Lmtp and L0 to be equal, the total flux linkage of the primary winding
becomes, with the assumption that Lmtp = L0
λ = ψ + ψ0 = I pTp2 ×
µ0 Lmtp  bp a 
 + 
Lc  3 2 
Therefore, leakage reactance of the primary/phase,
 flux linkage 
xp = 2π f ×

 current 
I pTp2 ×
x p = 2π f
µ0  Lmtp bp L0 a 


+
2 
Lc  3
Ip
3.50 Design of Transformer
xp = 2π fTp2 ×
µ0  Lmtp bp L0 a 


+
Lc  3
2 
xp = 2π f µ0 T12
Lmt 
b1 + b2 

 a +
Lc 
3 
The leakage reactance of the secondary/phase,
xs = 2π f µ0 Ts 2
Lmt 
b1 + b2 

 a +
3 
Lc 
The total reactance is therefore given by
X p = xp + xs′ = 2π f µ0 Tp2
bp + bs 
Lmt 

 a +
3 
Lc 
Example 3.9: Calculate the leakage reactance of a 50 Hz transformer with the
following data.
Number of primary and secondary turns = 510 and 30
Mean length of primary and secondary turns = 1.25 m and 1.05 m
Thickness of each winding = 0.025 m
Width of duct = 0.015 m
Height of each winding = 0.61 m
Solution:
We know that the leakage reactance referred to primary side is given by
X p = 2π f µ0 Tp2
Lmt 
b1 + b2 

 a +
3 
Lc 
From the given data,
a = 0.015 Lc = 0.61m
b = 0.025 f = 50 Hz
Lmt =
1.25 + 1.05
= 1.15 m
2
Hence,
(
)
2
X p = (2π )(50) 4π ×10−7 (510)
×
0.025 
1.15 

0.015 + 0.025 +
3 
0.61 
X p = 6.1 Ω
Reactance Calculation 3.51
Example 3.10: A 600 kVA, 6600/400 volts, 50 Hz, three-phase core-type transformer
has the following data.
Width of LV and HV winding = 3 cm
Width of duct between LV and HV winding = 2 cm
Height of LV and HV winding = 40 cm
Length of mean turn = 1.5 m
Number of turns of HV winding = 220
Estimate the leakage reactance of the transformer referred to HV side.
Solution: Given
f = 50 Hz
a = 0.02 m
T1 = 220
b1 = 0.03 m
Lmt = 1.5 m
b2 = 0.03 m
Lc = 0.4 m
L 
b + b 2 
X p = 2π f µ0 Tp2 mt  a + 1

3 
Lc 
0.03 + 0.03 
2 1.5 
= (2π )(50) 4π ×10−7 (220)

0.02 +

0.4 
3
(
)
X p = 2.866 Ω
Example 3.11: Calculate the % regulation at full load, 0.8 pf lag for a 500 kVA,
6000/400 V Δ-Y . 3 φ, 50 Hz core-type transformer having cylindrical
coils of equal length with the following data.
Height of coils = 48 cm
Thickness of LV and HV coils = 2.53 cm and 1.65 cm
Insulation between LV and HV coil = 1.4 cm
Mean diameter of coil = 28 cm
Volt/turn = 8 V
Full load copper loss = 3.8 kW
Solution:
Number of primary turns,
T1 =
6000
= 750
8
Primary current,
Ip =
kVA ×10 3
3V1
=
500 ×10 3
= 27.7 A
3 × 6000
3.52 Design of Transformer
Rp =
Primary resistance,
copper loss
(3) I p2
=
3.8 ×10 3
(3)(27.7 )2
Rp = 165Ω
Reactance of primary winding,
Lmt 
b1 + b2 
 (1)
 a +
Lc 
3 
X p = 2π f Tp2 µ0
Lmt = Length of mean term of both primary and
secondary winding
= π × 28
From Eq. (1),
= 87.96 cm
(0.0165 + 0.0253) 
2  0.8796 
X p = (2π )(50) 4π ×10−7 (750) 
0.014 +

 0.48 

3
(
)
X p = 11.35 Ω
And,
% Regulation =
=
I p Rp cos φ + I p X p sin φ
V1
×100
(27.7 )(1.65)(0.8) + (27.7 )(11.35)(0.6)
6000
×100
%Regulation = 3.75
3.14 Flowchart for Overall Design of Transformer
Example 3.12: Design a 100 kVA, 3φ, 50 Hz, 11 KV/440 V, Δ-Y distribution transformer,
self-oil cooled with temperature rise not to exceed 40°C. The no load
loss is 350 W and copper loss and stay load loss on full load is 2000 W.
The percentage impedance is 4.5%.
Calculate also, the no load current, efficiency at 60°C at full load, 75% of
load at unity power factor and regulation at 75°C at unity power factor.
Solution:
Volt/Turn:
Emf/turn = Et = k kVA ,
where k = 0.45 for 3φ core-type distribution transformer
= 0.45 100 = 4.5 volts
Flowchart for Overall Design of Transformer 3.53
Flowchart for Overall Design of Transformer 3.53
Start
Get the input data:
- Number of phases
- Type of transformer
- Distance between core centres
- kVA rating
- Frequency
- Voltage ratio
- Type of core and details of core dimensional relations, width
of largest stamping, stacking factor, window-width depth
relation, core width depth relation
- Flux density (specific magnetic loading) and current density
(specific electric loading)
- Window space factor
- Type of core, yoke
Calculate emf per turn
Calculate net iron area,
gross cross sectional
core area
Core design
Calculate diameter of
circumscribing circle
Calculate window area
Window design
Calculate window
height and width
Calculate net and gross
area of yoke
Calculate flux density
in yoke
Yoke design
Calculate depth and
height of yoke
Determine distance
between adjacent core
center
Overall design
Determine overall
width, overall height
and depth of frame
Determine the number
of turns of hv and lv
winding
Determine resistance,
and reactance
Determine height of lv
and hv coil
Print all values
Stop
fig. 3.33
3.33 || Flowchart
Flowchart for
for overall
overall design
design procedure
procedure
Fig.
3.54
3.54 Design
DesignofofTransformer
Transformer
(or)
(or)
1
1
kVA
×kVA
1000×1000
Et = Et =
40
Number
40 Number of legsof legs
=
1 × 1000
100 ×1000
1 =100
= 4.5 V/turn
= 4.5 V/turn
3
40 3
40
Specificmagnetic
magneticloading:
loading:
Specific
Let
us
choose
B
=
1.35Wb/m
Wb/m22. .Material
Materialfor
forcore
corecan
canbe
behot
hotrolled
rolledsteel
steellaminations.
laminations.
Let us choose Bmm= 1.35
Specificcurrent
currentloading:
loading:
Specific
Let
Letcurrent
currentdensity
density δδ be
be2.5
2.5A/mm
A/mm22=
=2.5
2.5×
×10
1066A/m
A/m22
Cross-section
Cross-sectionof
ofcore:
core:
Et =f B4.44
Et = 4.44
m Aif Bm Ai
Netiron
ironarea,
area,
Net
Et
E
4.5 4.5
=
Ai = Ai =t
=
4.44
50 ×1.35
f
B
4.44f Bm 4.44
50 ××
1.35
m × 4.44
2
= 0.015015
m 2 = 150.15
= 0.015015
m 2 = 150.15
cm 2 cm
Gross
Grosscore
coresection
sectionarea,
area,
A
Ai
Agi = Agi = i
Stacking
factor factor
Stacking
150.15150.15
m2 m2
= = 166.83
= 166.83
0.9 0.9
Let us choose a three-stepped core as shown in Fig. 3.34(a).
Let us choose a three-stepped core as shown 2in Fig. 3.34(a).
Ai = 0.6 d
Ai = 0.6 d2
Diameter of the circumscribing circle,
Diameter of the circumscribing circle,
150.15
Ai
d=
=
0.6
0.6
Ai
150.15
d=
=
0.6
0.6
=15.82
cm  16 cm
= 15.82 cm  16 cm
Let
d =16 cm.
Let
d =16 cm. a = 0.9d = 14.4 cm
b = 0.7d = 11.2 cm
a = 0.9d = 14.4 cm
c = 0.42d = 6.72 cm
b = 0.7d = 11.2 cm
Design of window:
c=
0.42d
=A6.72
cm
−3
Q=
3.33
f Bm
i Aw kw δ × 10
Design of window:
copper area in window
Window space factor =
Q = 3.33
kw δ × 10−3
Totalf B
window
m Ai Awarea
=
Let kw be 0.25.
copper area in window
Window space factor =
Total window area
c
b
a
16 = d
fig. 3.34(a)
Fig.
Flowchart for Overall Design of Transformer 3.55
Let kw be 0.25.
Aw =
Windwo area,
100
(3.33)(50)(1.35)(150.15)(0.25)(2.5×106 )×10−3
= 4.74 ×10−6 m 2 = 474 cm 2
Let the ratio of window height of window width be 2.5.
Hw
= 2.5
Ww
Aw = H wWw = 474
(Ww )(2.5)Ww = 474
a=
2.5 Ww 2 = 474
Ww = 13.8 cm
H w = 34.4 cm
The window is shown in Fig. 3.34(b).
Design of yoke:
14.4
= 13.8
D = 29.62
= (1.2)(150.15)
= 180.18 cm2
Gross yoke area =
180.18
= 200.2 cm 2
0.9
Flux density in yoke =
1.35
=1.125 Wb/m 2
1.2
Assuming yoke to be rectangular,
Depth of yoke,
Dy = a = 14.4 cm
Height of yoke,
Hy =
Ay
Wy
=
200.2
14.4
= 13.9 cm  14 cm
Overall dimension of core:
Distance between adjacent core centre = D = Ww + d
= 13.8+15.82 = 29.62 cm
W = 2D + a = (2× 29.62) + 14.4
=73.64 cm
Overall height,
D = 29.62
W = 73.64
Net iron area of yoke = 1.2Ai
Overall width,
hw = 34.4
Ww
H = H w +2 H y
Fig. 3.34(b)
Hw= 34.4
3.56 Design of Transformer
= 34.4 + (2×14) = 62.4 cm
Dy = a = 14.24 cm
Depth of frame,
Design of windings:
LV winding:
Line voltage on LV side = 440 V, Y connected
4400
V
Phase voltage =
3
254 254
Number of turns/phase =
=
= 56.4  57 turns
4.5
Et
100 ×1000
= 131.23 A
3 × 254
Current perphase,
I2 =
Current density,
δ = 2.5 A/mm 2
I
Area of conductors of LV winding = 2
δ
131.23
=
= 52.49 mm 2
2.5
HV winding:
HV side line voltage = 11,000 V, Δ connected
Phase voltage = 11,000 V
Number of turns,
T1 =
V1
×T2
V2
11, 000
× 57 = 2468.5
254
 2469
=
Current in HV winding =
100 ×1000
= 3.03 A
3 ×11,000
Area of conductors of HV winding =
3.03
= 1.212 mm 2
2.5
Diameter of conductor =
4
×1.212 = 1.242 mm
π
Total copper area in window = 2 ( a1T1 + a2T2 )
= 2 ((52.49× 57 ) + (1.212× 2469))
Flowchart for Overall Design of Transformer 3.57
= 2 (2991.93 + 2992.44)
= 11968.7 mm 2
= 119.687 cm 2
Window space factor,
kw =
copper area
window area
=
119.68
= 0.252
474
Inside and outside diameter of LV winding:
Number of turns = 57  58
Area of conductors = 53.33 mm2
Let us choose copper rectangle conductors of size 6.3 × 4.5 mm (2 strips)
Let a paper insulation be used for these conductors. Let us take the inverse in dimension
because of insulation be 0.5 mm.
Diameter of insulated conductor = (6.3 + 0.5)×( 4.5 + 0.5)
= 6.8 × 5 = 34 mm
Let us choose two layers.
Turns per layer = 29
Width of conductor is 6.8 mm along the height of the window = 5 + 5 =10 mm along the
width of the window.
For two layers, the conductor dimension is 6.8 mm along the height of window and with
two conductors 5 + 5 =10 mm per paper width wise.
Helical winding is already used.
Therefore, space required is 28 + 1 times along the axial length.
Axial length of winding = 29 × 6.8 = 197.2 mm = 19.72 cm
Height of window = 34.4 cm
Clearence =
34.4 − 19.72
= 7.34 cm
2
on each side of winding and will be filled with insulation
Using 0.5 mm paper cylinders between layers,
Radial depth of LV winding = (2×10) + 0.5
= 20.5 mm
Diameter of circumscribing circle = 15.82 cm
Using 1.5 mm thick, pressboard insulation between LV winding and core,
(
Inside diameter of LV winding = 15.82 + 2 ×1.5×10−1
= 16.12 cm
)
3.58 Design of Transformer
Outside diameter = 16.12 + (2× 2.05)
= 20.22 cm
Mean diameter = 18.17 cm
Mean length of turn = π × 18.17 = 57.08 m
Inside and outside diameter of HV winding:
Maximum number of turns = 2468
Size of conductor = 1.212 mm 2 round conductors,
With insulation, size of conductor = 1.375 (using enamel coating) .
Let us choose number of coils for HV winding such that the volt/coil is nearly 1500 V.
If 8 disc coils are used,
Volt/coil =
11, 000
= 1375.
8
Turns/coil =
2468
= 308.5
8
= 310 coils  315
7 coils of 315 turns can be used and end coil of 263 turns can be used.
For normal coils, we can we 15 layers, turns per layer 21.
Maximum voltatge between layers = 2× 21× 4.5
= 189 V
(This is much below the normal voltage of 300 V between layers)
Axial depth of normal coil = 21 × 1.375 = 28.875 mm
Axial depth of end coil = 15 layers, 18 turns per layer
= 18 × 1.375 = 24.75 mm
Let the spaces used between adjacent coils is 5 mm in height.
Axial length of HV winding = (7 × 28.875) + (1 × 24.75) + (7 × 5)
= 261.875 mm
Height of window = 344 mm
Clearance on either side =
344 − 261.875
= 41.06 mm
2
Insulation used between layers is 0.3 mm thick paper,
Radial depth of HV coil = (15 × 1.375) + (14 × 0.3)
= 24.825 mm
Total insulation between LV and HV winding = 15 mm
(bakelite paper = 5 mm, oil duct = 5 mm, kV insulation = 5 mm)
Flowchart for Overall Design of Transformer 3.59
Inside diameter of HV winding = 202.5 + (2×15)
= 232.2 mm
Outsidediameter of HV winding = 232.2 + (2× 24.825)
= 281.85 mm
Clearance between two HV windings of adjacent lines = D − 281.85 = 14.35 mm (This is
sufficient for 11 kV)
Resistance:
232.2 + 281.85
Mean diameter of HV winding =
2
= 257.025 mm
Length of mean turn of HV winding = π × 257.025
= 807.47 mm
= 0.80747 m
Lm 1 T1 0.021 ×.80747 × 2468
R1 = ρ
=
a1
1.212
= 34.52 Ω
161.2 + 202.2
Mean diameter of LV winding =
2
= 181.7 mm
Length of mean turn of LV = π × 181.7
= 570.83 mm = 0.57083 m
R2 =
ρLm 2 T2
=
a2
= 0.0126 Ω
0.021 × 0.57083 × 56
`
53
Total resistance referred to primary side,
Rep = R1 +
R2
k2
= 34.52 +
0.0126
 56 2

 2468 
= 58.9 Ω
(3.03) (58.9)
I R
P.u resistance = 1 1 =
V1
11, 000
= 0.016
Leakage resistance:
Mean diameter of windings =
181.7 + 257.025
= 219.36 mm
2
3.60 Design of Transformer
Length of mean turn = π × 219.36 = 689.13 mm
Height of HV coil:
Let us use a bakelite cylinder between LV winding and core of 5 mm thick. Therefore, the net
distance between LV winding and core = 10 mm.
There are 8 coils on LV side with 15 layers and 21 turns/layer.
Height of HV winding over copper = (Height of HV winding in each disc × number
of disc) + Height of spacer × number of spacer
= (21 × diameter of conductor including
insulation × number of disc) + Height of
spacer × number of spacers
= (21 × 1.375 × 8) + (10 × 7) = 301 mm
Height of LV coil:
Number of turns of LV coils = 58
This can be arranged in 2 layers with 29 turns in each layer.
The conductor size for LV is 6.8 × 10 mm.
Height of LV over copper = 29 × 6.8 = 197.2 mm
197.2 + 301
= 249.1 mm
2
Ampere turns = (131.23 A) (57 turns) = 7480.11
Mean height of coil =
p.u reactance =

2π f µ0 Lmt ( AT ) 
 a + b1 + b2  p.u


L0 Et
3 
a = distance between LV and HV coil = 15 mm
b1 = radial depth of LV winding = 20.5 mm
b2 = radial depth of HV winding = 24.823 mm
b + b2
a+ 1
= 0.03108
3
=
2π(50)( 4π ×10−7 )(689.13) (7480.11) (0.03108)
(0.2491)( 4.5)
p.u reactance = 0.0564
p.u impedance = 0.016 2 + 0.0564 2 = 0.05862 p.u
3.15
No Load Current of a Transformer
Determination of the no load current enables to calculate the core losses of the transformer.
The no load current, I0, represented as phasor diagram in Fig. 3.35 is the vectorial sum of the
magnetizing current, Im, and core loss or working component current Ic as given by
2
I0 = Im
+ I c2
No Load Current of a Transformer
3.61
Im produces flux φm in the magnetic circuit and No
Ic supplies
the of
noa Transformer
load losses of
the
Load Current
3.61
transformer.
IThus,
flux
φm in
the magnetic
circuit and=IcV
supplies
the no load losses of the transformer.
load
input
to the transformer
m producesno
1(I0 cos φ0) = V1 Ic
Thus,
no load input to the transformer =
cos φlosses
1(I0load
0) = V(as
1 Ic the output is zero)
=VNo
= No load losses (as the output is zero)
and
input = output + losses
and
input = output + losses
Since the copper loss under no load condition is almost negligible, the no load losses can
Since be
thetaken
copper
load
condition is almost negligible, the no load losses can
entirely
as loss
due under
to coreno
loss
only.
entirely be taken as due to core loss only.
core loss
Thus, the core loss component of the no load current, I c =core loss , for single-phase
Thus, the core loss component of the no load current, I c =
, for single-phase
V
V11
core loss/phase
transformers and I c =core loss/phase , for three-phase transformers.
, for three-phase transformers.
transformers and I c =
V1ph
V1ph
RMS value of magnetizing current,
RMS value of magnetizing current,
Peak magnetizing ampere turns
Im =
Peak magnetizing ampere turns 2T
1
Im =
2T1
where T1 is the number of turns of primary.
where
of primary.
TheTmagnetic
circuitof
ofturns
a transformer
consists of both iron and air path. The iron path is
1 is the number
magnetic
circuitand
of athe
transformer
of both
iron and joints
air path.
The iron
path
is
dueThe
to legs
and yokes
air path isconsists
due to the
unavoidable
created
by the
core
due
to legs of
and
yokes and
air path
is dueIf to
joints created
by the core
composed
stampings
of the
different
shapes.
allthe
theunavoidable
joints are assumed
to be equivalent
to
composed
different
shapes.
If all
the for
joints
assumed to
be equivalent
an air gap of
of stampings
length, lg, of
then
the total
ampere
turns
theare
transformer
magnetic
circuitto
is
an
air to
gap
length,
, then the
ampere
turns
for
the
transformer
magnetic
circuit
is
equal
AToffor
iron +lg800,000
lgBtotal
.
Therefore,
m
equal to AT for iron + 800,000 lgBm. Therefore,
Total AT = AT for iron + 800,000 lg Bm
Total AT = AT for iron + 800,000 lg Bm
AT for iron + 800,000 lg Bm
I m = AT for iron + 800,000 lg Bm
Im =
2T1
2T1
Vp
Ic
I0
φº
90º
flux φ
Im
fig.
Fig. 3.35
3.35 || No
No load
load current
current and
and its
its components
components
The
The no
no load
load current
current and
and its
its components
components of
of aa transformer
transformer in
in terms
terms of
of mmf,
mmf, number
number of
of
turns,
power
and
voltage
are
derived
for
singleand
three-phase
transformers
as
turns, power and voltage are derived for single- and three-phase transformers as follows.
follows.
3.62 Design of Transformer
3.15.1 Single-phase Transformer
Magnetizing current
The peak value of magnetizing current in terms of mmf and number of turns is given by
I m(pk) =
AT0
(3.41)
Tp
where AT0 – Total magnetizing mmf under no load condition, Tp – Number of turns in
primary.
In Eq. (3.41),
AT0 = mmf of core + mmf of yoke + mmf of joints (3.42)
= ATc + ATy + ATj
In the above equation,
(ATc) mmf of core = mmf per metre for maximum flux density in core ×
length of path traced by flux in core
= atc× 2lc = 2atc hw [ ∵ lc = hw (height of window)]
(ATy) mmf of yoke = mmf per metre for maximum flux density in yoke ×
length of path traced by flux in yoke
= aty× 2ly = 2aty ww [ ∵ ly = ww (width of window)]
atc, aty – mmf per metre for maximum flux density in core and yoke, respectively
2lc, 2ly – Total length of core and yoke, respectively
ATj = 800000 lgBm
(As all joints are assumed to be equivalent to an airgap (lg))
From Eq. (I), the rms value of magnetizing current for sinusoidal current is given by
I m (rms) =
AT0
2Tp
(3.43)
If the magnetizing current is non-sinusoidal, its rms values are given by
I m (rms) =
AT0
(3.44)
kpeak Tp
where kpeak – peak factor
Core loss or iron loss current
The core loss or iron loss component of no load current is given by
I c (I i or I l ) =
Pi or Pc
Vp
where Pi (Pc) – iron loss or core loss, Vp – Primary voltage.
3.15.2 Three-phase Transformer
The magnetizing current derived for single-phase transformer holds good for three-phase
transformer too, but the length of the core is 3lc.
No Load Current of a Transformer
3.63
3.15.3 Magnetizing Volt – Ampere and Alternate Expression
for ­Magnetizing Current
The magnetizing Volt – Ampere (VA)M is expressed as the product of voltage induced
primary of transformer winding and magnetizing current.
Voltage induced in primary of transformer winding
= 4.44 f φm TP
= 4.44 f Bm Ai TP [ ∵ φm = Bm Ai ]
Magnetizing current =
AT0
2TP
Hence, magnetizing volt ampere,
( VA)M = 4.44 f Bm Ai Tp ×
AT0
2TP
(3.45)
Expressing mmf in terms of mmf per metre and length of iron path, we get
AT0 = ati × li (3.46)
Substituting Eq. (3.46) in Eq. (3.45), we get
( VA)m =
4.44 f Bm Ai ati li
2
The Magnetizing Volt Ampere per kg can be to found out by determining the ratio of
Magnetizing Volt Ampere to the weight of iron, which is given by
Wi = density × volume
We know that the density of iron is 7.8 × 103 kg/m3 and volume is given by product of
area of iron (Ai) and length of iron (li). Hence, using these values in the above equation, we
get
Wi = 7.8 ×10 3 × Ai × li (3.48)
Hence, magnetizing volt ampere per kg, obtained by dividing Eq. (3.47) by Eq. (3.48) is
(VA)m / kg =
4.44 f Bm ati
2 ×7.8 ×10 3
= 0.4025 f Bm ati ×10−3
 0.4 fBm ati ×10−3
Magnetizing current,
Im =
Magnetizing VA/kg × weight of core
number of phases × voltage/phase
3.64 Design of Transformer
Example 3.13: Find the no load current of a 400 V, 50 Hz, 1φ core-type transformer
with the following data.
Length of mean magnetic path = 200 cm
Gross core section = 100 cm2
Joints have 0.1 mm equivalent to air gap,
maximum flux density = 0.7 Tesla,
Specific core loss = 0.5 watts per kg at 50 Hz and 0.7 Tesla
Ampere turns = 2.2 per cm
Stacking factor = 0.9
Density of steel = 7.5 × 103 kg/m3.
Solution:
We know that
Total mmf = AT0 = 2 ati li + 2 aty ly + mmf for joints
= 2ati li + 2 aty ly +
B
ljoints
µ0
= (Amp. turns/m)(length of mean magnetic path )
+
B
0.1×10−3
µ0
AT0 = (2.2) (200) + (800000) (0.7) × 0.1 × 10−3
AT0 = 496
Ai = gross iron area × stacking factor
Net iron area,
= 100 × 0.9 × 10-4
= 90 × 10-4 m2
Eph = 4.44 f Bm Ai Tph = 400 V
Emf per phase,
Tph =
400
( 4.44)(50)(0.7 )(90 ×10−4 )
Tph = 286
Magnetizing current,
I m (rms) =
=
AT0
2Tph
496
( 2 )(286)
= 1.226 A
Total weight of core = volume of core ×density of steel
= net core section × mean length
×density of steel
= 90 ×10−4 × 2×7.5×10 3 = 135 kg
No Load Current of a Transformer
3.65
Total core loss = specific loss × weight of core
= 0.5 × 135
= 67.5 W
67.5
= 0.168 A
400
Core loss current,
Ic =
No load current,
2
I0 = Im
+ I c2
= 1.226 2 + 0.168 2
I 0 = 1.237 A
Example 3.14: Find the no load current of a 200/100 V, 1 kVA, 50 Hz single-phase
transformer with the following data. Gross cross-sectional area of core
= 27 cm2, effective magnetic core length = 0.5 m, weight of core =
7.5 kg, maximum flux density = 1.1 T, magnetizing mmf = 200 AT/m
and specific core loss = 2 W/kg.
Solution: Given,
Rating, Q = 1 kVA, frequency, f = 50 Hz, gross core area, Agi = 27 cm2, core length = 0.5 m,
weight of core = 7.5 kg, maximum flux density, Bm = 1.1 T, mmf = 200 AT/m and specific
core loss = 2 W/kg
We know that
⇒
Emf per phase = 4.44 f Bm Ai Tph = 200
( 4.44) f Bm (ks Agi )Tph = 200
Hence,
Turns per phase,
Tph =
Also, we know that
=
Im =
200
( 4.44)(50)(1.1)(0.9× 27 ×10−4 )
200
= 337
0.593
AT0
2Tph
Let us assume stacking factor ki to be 0.9.
Im =
=
AT0
2Tph
=
magentizing mmf/m × core length
(200)(0.5)
2Tph
2Tph
3.66 Design of Transformer
Substituting Tph in the above equation, we get
Im =
100
2 × 337
= 0.209 A
core loss Loss/kg × weight of core
=
v1
v1
2×7.5
=
= 0.075 A
200
Core loss current I c =
2
No load current I 0 = I m
+ I c2
I 0 = 0.2092 + 0.0752
= 0.0436 + 5.625×10−3
I 0 = 0.22186 A
Example 3.15: A 6600 V, 50 Hz, 1φ transformer has a sheet steel core. The crosssectional area of iron is 24 × 10−3m2. The maximum flux density is
1.2 T and the mean length of iron path is 2.5 m. There are 4 joints
in the core and each joint takes 40% as much reactive mmf, as it is
required per metre of the core. Find the active, reactive components
of no load current, if magnetizing mmf/m is 232, specific core loss is
1.76 W/kg and density is 7.5 × 10−3 kg/m3.
Solution:
The magnetizing or reactive component of current,
Im =
ATiron + ATjoints
2 T1
(1)
AT for core = mmf for core × length of iron paths
= 232 × 2.5
AT for 1 joint = 40% AT/m for core
40
× 232× 2.5
100
ATjoints = 928
AT for 4 joints = 4 ×
Number of primary turns T1 =
V1
6600
=
4.44 f φm ( 4.44)(50) 24 ×10−3 (1.2)
(
(∵ φm = Bm Ai )
T1 =
6600
= 1032
6.393
)
Transformer Losses 3.67
From Eq. (1), the magnetizing component of current is given by
Im =
(232)(2.5) + 928
= 1.0334 A
( 2 )(1032)
Active or core loss component of no load current,
core loss
V1
specific core loss× volume of core ×density
=
66000
Ic =
1.76 × 24 ×10−3 × 2.5×7.5×10 3
6600
= 0.12 A.
=
2
No load current,
2
I0 = Im
+ I c2 = (1.033) + 0.122
I 0 = 1.0399 A
3.16
No load line current = 3 ×1.0399 = 1.801 A
Transformer Losses
The loss in the transformer has two components, namely load losses and no load losses, as
mentioned in the previous section.
No load losses
The no load losses that occur in the core or iron part of the transformer occur at all times
when the primary is energized, irrespective of whether the secondary is loaded or not.The
I2R loss occurring on the primary due to no load current is negligible. The core or iron loss,
which is constant, has two components, namely hysteresis and eddy loss.
Hysteresis loss: This loss is due to magnetization and demagnetization of the iron core,
when it is subjected to alternating magnetic field. This loss is directly proportional to the flux
density Bm, volume V of the core and frequency f of the alternating field.
Hysteresis loss,
Ph = Kh V Bm1.6 f,
where Kh is a constant whose value depends on the type of the magnetic material used
Eddy current loss: Like hysteresis loss, the eddy current loss also occurs in the core of the
transformer. The alternating magnetic flux produces eddy current in the core which results
in loss of power called eddy loss. To reduce this eddy current and hence the eddy loss, the
core is laminated in to thin sheets, which are insulated from each other by enamel coating.
This loss is proportional to the volume of the core, square of flux density Bm, frequency f and
also the square of the thickness of laminations, t.
3.68 Design of Transformer
Eddy loss Pe = Ke V Bm2 f2 t2,
where Ke is a constant whose value depends on the resistivity of the core material used.
Copper loss: These are the losses occurring in the two windings of the transformer on load
conditions. Since these losses vary with the load, they are called load losses.
Calculation of total core loss
Total core loss = core loss in legs + core loss in yokes
The core loss can be estimated at design stage by referring to graph of core loss/kg versus
flux density.
Core loss in leg = loss/kg in leg × weight of leg in kg
= loss/kg in leg × volume of the leg (Ai Hw) ×
density of steel or iron used
Total core loss = Number of legs × loss/kg in leg × volume of the
leg (Ai Hw) × density of steel or iron used
Core loss in yoke = loss/kg in yoke × volume of yoke (Ay × mean
length of the yoke) × density of iron used
The density of iron or steel used for the transformer core lies between 7.55 and 7.8 g/cm3.
Calculation of copper loss
Since this is the I2R loss in the windings, it can be calculated from the resistance and current
rating of the winding.
3.17 Effects of Change in Frequency in Parameters
of the Transformer
The input frequency of the transformer is one of the determining factors of the performance of
the transformer. When the input frequency is changed, the transformer iron loss and hence the
efficiency change. Similarly, the winding resistance, leakage reactance and voltage are affected
by the change in input frequency. The frequency change effects are discussed as follows.
1. Effect on core loss: Let us examine the effect of frequency variation on core loss if the
voltage remains constant.
2
We have the voltage, E = 4.44fBmAiT and eddy current loss, Pe = Ke f 2 Bm
If the voltage E remains constant, then the product (f Bm) remains constant. Thus,
eddy current loss will remain constant as long as E is kept constant, no matter the
(
x
frequency is changed. But the hysteresis loss is proportional to the product f Bm
and is given as
Hysteresis loss,
)
x
Ph = K h f Bm
The exponent x can take the value from 1.6 to 2.2 depending on the magnetic
characteristics of the core material.
Optimum Design 3.69
If x is assumed to be 2, then
Ph = K h ( f Bm ) Bm
= K h K Bm
(∵ E is constant, f Bm is constant and equal to K )
= Kh K
K
K2
= Kh
f
f
Hence, hysteresis loss will decrease with increase in frequency.
2. Effect on winding resistance: The winding resistance increases due to skin effect
when the frequency is increased. This increase is negligible, when the change in
frequency is very small.
3. Effect on leakage reactance: Since the leakage reactance Xl = 2πfl, the leakage
reactance changes linearly with the increase in frequency.
4. Effect on voltage: Along with the change in frequency, if the flux density is also
changed, then a new voltage rating is possible. For the new voltage rating, the iron
loss and no load current can be calculated.
Table 3.9 gives the summary of effects of frequency variation on various parameters
discussed above.
Table 3.9 | Summary of effects of frequency variation on various parameters
Parameter
Effect of change in frequency
Core loss
Decreases with increase in frequency
with voltage maintained constant
Voltage
Increases with increase in frequency
Leakage reactance
Increases with increase in frequency
Resistance
Increases with increase in frequency
3.18 Optimum Design
A transformer design may be optimized to minimize overall volume, weight, cost and losses,
etc. These requirements are generally conflicting and so it is possible to satisfy any one of the
above at one time.
We know that, from Eq. (3.16), the output equation of single-phase transformer is given by
Q (kVA) = 2.22 f Bm δ K w Aw Ai ×10−3
= 2.22 f Bm δ Ac Ai
[ ∵ K w Aw = Ac ]
In the above equation, on assuming f, Bm and δ to be constant, the left out product of
Ac Ai is constant and is assumed to be given by
Ac Ai = y 2 (3.49)
3.70 Design of Transformer
We know that, from Eq. (3.34),
x=
where
φm
(3.50)
AT
φm = Bm Ai (3.51)
And from Eq. (3.12),
AT =
=
δ K w Aw
2
δAc
(3.52)
2
Substituting Eqs. (3.52) and (3.51) in Eq. (3.50), we get
x=
Bm Ai 2Bm Ai
=
δ Ac
δ Ac
2
On simplifying,
⇒
Ai
δx
=
= z (Assumption ) (3.53)
Ac 2Bm
The assumption is made such that z is a function of x, as Bm and δ are assumed to be constants.
Using Eq. (3.53) in Eq. (3.49) to determine Ai and Ac, we get
Substituting Ai = z Ac in Eq. (3.49), we get
zAc × Ac = y 2
⇒
zAc 2 = y 2
⇒
Ac =
Substituting Ac =
y
z
(3.54)
Ai
in Eq. (3.49), we get
z
Ai
× Ai = y 2
z
⇒
Ai 2 = zy 2
⇒
Ai = y z (3.55)
The total cost of transformer is given by
CT = ci + Cc (3.56)
where Ci – total cost of iron, Cc – total cost of copper conductor
Optimum Design 3.71
And,
Ci = ci gi li Ai (3.57)
Cc = cc gc lmt Ac (3.58)
where Ci – specific cost of iron, Cc – specific cost
Substituting Eqs. (3.57) and (3.58) in Eq. (3.56), we get
CT = ci gi li Ai + cc gc lmt Ac (3.59)
Substituting Eqs. (3.54) and (3.55) in Eq. (3.59), we get
CT = ci gi li y z + cc gc lmt
y
z
(3.60)
In order to determine the minimum cost, differentiating the above equation with respect to
z, and equating to zero we get
⇒
y
y
dCT 1
1
= ci gi li
− cc gc lmt −3/2 = 0
dz
2
z 2
z
ci gi li
y
z
= cc gc lmt
y
z
3/2
On simplifying the above equation, we get
c g l
ci gi li = c c mt
z
Substituting Eq. (3.53) in the above equation, we get
Ac
Ai
ci gi li = cc gc lmt
⇒
ci gi li Ai = cc gc lmt Ac
⇒
ci Gi = cc Gc (3.61)
[ ∵ Gi = gi li Ai and Gc = gc lmt Ac ]
⇒
Ci = Cc (3.62)
Similarly, from Eqs. (3.61) and (3.62),
For minimum volume of transformer,
Volume of iron = Volume of conductor
c
Gi
= c
Gc
ci
For minimum weight of transformer,
Weight of iron = Weight of conductor
3.72 Design of Transformer
GG
GG
i =
i=
cc
For minimum
minimum losses
losses of
of transformer
transformer (maximum
(maximum efficiency),
efficiency),
For
Iron
loss
Copper
loss
Iron
loss
==
Copper
loss
Pi P=i =
x 2xP2cPc
3.19 Cooling
Cooling of
of Transformer
Transformer
3.19
The losses developed in the transformer core and the winding cause heating of the
The losses developed in the transformer core and the winding cause heating of the
transformer parts. Based on the thermal gradient, the heat is transferred to the cooling
transformer parts. Based on the thermal gradient, the heat is transferred to the cooling
medium depending on the method of cooling. In very small transformer of a few kVA rating,
medium depending on the method of cooling. In very small transformer of a few kVA rating,
the cooling surface of the core and winding is sufficient to be cooled by normal circulation
the cooling surface of the core and winding is sufficient to be cooled by normal circulation
of atmospheric air. As the capacity of the transformer increases, the losses also increase, so
of atmospheric air. As the capacity of the transformer increases, the losses also increase, so
it is required to provide better cooling arrangement. The various cooling methods can be
it is required to provide better cooling arrangement. The various cooling methods can be
grouped as follows (Figs. 3.36 and 3.37).
grouped as follows (Figs. 3.36 and 3.37).
Types of
cooling
Oil cooling (Oil
immersed
transformer)
Air cooling (Dry
type transformer)
Natural
cooling
Air
natural
Air
forced
Air
natural
Forced cooling
Air
blast
Forced
cooling
Natural cooling
Oil
natural
Air
natural
Oil
natural
Air
forced
Oil
natural
Water
forced
Oil
forced
Air
Natural
Oil
Forced
Air
forced
Oil
forced
Water
forced
fig.
Fig. 3.36 | Types of cooling
Air
Air Natural
Natural (AN)
(AN)
•
This
method
•• This method uses
uses the
the ambient
ambient air
air as
as the
the cooling
cooling medium.
medium.
•
The
natural
circulation
of
surrounding
air
•• The natural circulation of surrounding air is
is utilized
utilized to
to carry
carry away
away the
the heat
heat generated
generated
by
natural
convection.
by natural convection.
•
•• A
A sheet
sheet metal
metal enclosure
enclosure is
is used
used to
to protect
protect the
the windings
windings against
against mechanical
mechanical
damage.
damage.
•
•• This
This method
method is
is used
used for
for small
small low-voltage
low-voltage transformers
transformers up
up to
to 1.5
1.5 MVA.
MVA.
Air Forced Air Natural (AFAN)
Air Forced Air Natural (AFAN)
•
•• In
In this
this method,
method, air
air is
is circulated
circulated through
through the
the transformer
transformer with
with the
the help
help of
of aa fan
fan and
and
cooled
in
a
heat
exchanger
by
natural
circulation
of
air.
cooled in a heat exchanger by natural circulation of air.
•
•• This
This method
method is
is not
not commonly
commonly used.
used.
CoolingofofTransformer
Transformer 3.73
3.73
Cooling
Main tank
Main tank
Pump
Pump
Radiator
tank
Oil flow
Radiator
tank
Oil flow
(a)
(b)
Main tank
Main tank
Fan
Tube
Fan
(c)
(d)
Radiator
tank
Transformer plain tank
Tank with ribs and
corrugations
(e)
fig.Fig.
3.37
| Cooling
methods:
(a)(a)
OFAN,
(b)(b)
OFAF,
(c)(c)
ONAF,
(d)(d)
ONAN
with
tubes
and
3.37
| Cooling
methods:
OFAN,
OFAF,
ONAF,
ONAN
with
tubes
(e)
ONAN
and (e) ONAN
3.74 Design of Transformer
Air Blast (AB)
•• In this method, the transformer is cooled by a continuous blast of cool air forced
through the cores and the windings.
•• The air blast is produced by external fans. The improvement in heat dissipation
caused by air blast allows higher specific loadings to be used in air-type transformers.
Oil Natural Air Natural (ONAN)
•• Oil is a better conductor of heat than air.
•• Oil has a high co-efficient of volume expansion with temperature.
•• Oil and air are naturally circulated in this cooling
Oil Natural Air Forced (ONAF)
•• In this method, the oil circulating under natural circulation, transfers the heat to the
tank walls.
•• The transformer tank is made hollow and air is blown through the hollow space to
cool the transformer.
•• However, the normal way of cooling the transformers by air blast is to use radiation
tanks of corrugated or elliptical tubes separated from the transformer tank and cooled
by air blast produced by fans.
Oil Natural Water Forced (ONWF)
•• In this method, copper cooling coils are mounted above the transformer and below
the surface of oil.
•• Water is circulated through the cooling coils to cool the transformer.
•• This method proves to be cheap where a natural water head is already available.
Oil Forced Air Natural (OFAN)
•• In large transformers, the natural circulation of oil is insufficient for cooling the
transformer and forced circulation is employed.
•• Oil is circulated by a motor-driven pump from the top of a transformer tank to an
external cooling plant where the oil is cooled.
Oil Forced Air forced (OFAF)
•• In oil natural–air forced method, the oil circulating under natural heat transfers heat
to tank walls.
•• The air is blown through the empty space to cool the transformer.
•• The oil is cooled in external heat exchanger using air blast produced by fans.
•• Therefore, mixed cooling conditions are used the transformer working with oil
natural air natural conditions upto 50% rating and oil forced air forced conditions at
higher loads.
Oil Forced Water Forced (OFWF)
•• In this method, the pressure of oil is kept higher than that of water.
•• This cooling method is suitable for banks of transformers, but from the system
reliability considerations not more than say three tanks should be connected in one
cooling pump circuit.
•• OFWF is used for transformers designed for hydroelectric plants.
Temperature Rise in Tank with Tubes 3.75
3.20 Temperature Rise and Design of Cooling System
Tanks with plain walls are used for smaller rating transformers up to 20–30 kVA rating, as
they have sufficient surface area to dissipate the heat. But with the increase in the rating
of the transformer, maintaining the temperature rise within the safe working limit is not
possible with simple plain walled tanks. Hence, tubes and radiators are used. The tank
walls, tubes and radiators dissipate heat by radiation and convection. The modes of heat
transfer from various regions of transformer are given in Table 3.10.
Table 3.10 | Types of heat transfer from various regions of transformer
From region
To region
Type of heat transfer
Core and winding
Outer surface, yoke
Conduction
Outer surface, yoke
Oil
Convection
Oil
Tank or cooler walls
Convection
Tank or cooler walls
Cooling medium (air
or water)
Convection and radiation
3.21 Temperature Rise in Plain-walled Tank
The temperature rise should not exceed 55oC for natural cooling, 60oC for forced air cooling
and 65oC for forced water cooling and 45oC for oil cooling. The cooling system dissipates
6 W by radiation and 6.5 W by convection per m2 of tank surface per oC. Generally for
heat dissipation, the bottom surface is not considered. Similarly, the top surface has fittings,
bushings, etc. and is not considered to dissipate heat. So, if the cooling surface of the
transformer tank neglecting the top and bottom surfaces is taken as St m2, then 12.5St watts
per m2 per oC will be dissipated.
Hence, for a plain-walled tank, the temperature rise,
θ=
Total loss in watts
12.5 St
3.22 Temperature Rise in Tank with Tubes
As the capacity of the transformer increases, the losses and the temperature rise increases.
In order to keep the temperature rise within limits, air may have to be blown over the
transformer. This is not advisable as the atmospheric air containing moisture, oil particles, etc.,
may affect the insulation. To overcome the problem of atmospheric hazards, the transformer
is placed in a steel tank filled with oil. The oil conducts the heat from core and coil to the tank
walls. From the tank walls, the heat goes dissipated to the surrounding atmosphere due to
radiation and convection. Further as the capacity of the transformer increases, the increased
losses demand a higher dissipating area of the tank or a bigger sized tank. This calls for more
space, more volume of oil and increases the cost and transportation problems. To overcome
these difficulties, the dissipating area is to be increased by using auxiliary radiator tanks or
by using tubes without increasing the size of the tank.
3.76 Design of Transformer
Let the cooling surface of the tank be St m2. This will dissipate 12.5St W/m2/oC. By
provision of tubes, the total surface area becomes x St m2.
Convection from tube walls = x St 6.5 × 1.35 W = x St 8.8 W ( ∵ 1.35 is due to 35%
increase in heat transfer due to convection by syphoning action of the tubes)
Total dissipation = 12.5 St + x St 8.8 W
The number of tubes can be determined by knowing the surface area of each tube. The
diameter of tubes normally used is 5 cm and spaced 7 to 7.5 cm centre to centre apart.
3.23 Design of Tank with Tubes
Dissipating surface area of tank = St
Dissipating surface area of tubes = xSt
Total loss dissipation in surface area of tank = 12.5 St W/°C
Total loss dissipation in surface area of and cooling tubess = (12.5 + 8.8 x )St W/°C
Total loss dissipition by tubes due to convection = 1.35× 6.5 xSt
= 8.8 xSt W/°C
Total surface area of walls of tank and tubes = St + x St
= (1 + x ) St
Total loss dissipation by in surface area of tank and cooling tubes
Total surface area of wallls of tank and tubes
(12.5 + 8.8 x ) St
=
(1 + x )St
Total loss dissipation =
Temperature rice with tubes,
(θ ) =
=
Total loss
Specific heat dissipation × surface area
Pi + Pc
(12.5 + 8.8 x )St
From the above equation, area of tubes can be found as follows.
θ=
⇒
Pi + Pc
12.5St + 8.8 xSt
12.5 St + 8.8 x St =
Pi + Pc
θ
Design of Tank with Tubes 3.77
⇒
8.8 xSt =
xSt =
Total area of tubes,
Pi + Pc
− 12.5 St
θ

1  Pi + Pc

− 12.5St 


8.8  θ
Length of a tube = lt
Diameter of a tube = dt
Radius of a tube = rt
Area of a tube =πdt lt
Total number of tubes,
Total area of tubes
Area of a tube
 Pi + Pc

1

=
− 12.5St 

8.8 π dt lt  θ
nt =
The dimensions of tank depends on the type and capacity of transformer, voltage rating
and electrical clearance to be provided between the transformer and the tank, clearance to
accommodate the connections and taps, clearance for base and oil above the transformer, etc.
The clearance between different parts depends on rating of the transformer.
Let D = distance between adjacent limbs
De = external diameter of outer winding
Cw = width wise clearance between the outer winding and the tank wall
Cl = length wise clearance between the tank wall and the outer winding
Ch = height wise clearance between the tank wall and the outer winding
The various clearance values for a typical transformer of rating 1000 to 5000 kVA and 11
to 33 kV voltage rating is given in Table 3.11.
Table 3.11 | Clearance values for a typical transformer
Clearances
Values
Cw
Cl
Ch
40 to 80 mm
50 to 125 mm
400 to 600 mm
The height wise clearance includes the clearance of 50 to 60 mm at the base, clearance of
150 to 250 mm above oil and space of about 200 to 250 mm for leads. Figure 3.38 shows the
transformer main dimensions with tank along with various clearances.
3.23.1 Flow Chart for Design of Cooling System
The design procedure for design of the cooling system is given below as a flowchart in
Fig. 3.39. This involves the calculation of number of cooling tubes required and suitably
arranging them around the tank.
3.78 Design of Transformer
Ch
Ht
D
D
wt
C1
Lt
Cw
Fig. 3.38 | Transformer main dimensions with tank along with clearances
Example 3.16: Design a cooling system for a 3 φ, delta/star core type oil immersed
transformer of rating 200 kVA, 6000/400 V, 50 Hz. Allowable
temperature rise for the tank walls is 50°C. Dimensions of tank are
125 cm height, 100 cm length and 50 cm width, Allowable losses is
5 kW. What will be temperature rise without the cooling arrangement?
Solution: Given
Transformer rating, Q and type = 200 kVA and 3φ
Voltage rating = 6600/400 V
Allowable temperature rise = 50°C
Allowable losses = 5 kW
Tank size = 1.25 × 1 × 0.5 m
We know that
Losses = 12.5 St θ + 8.78 At θ
Design of Tank with Tubes 3.79
Start
Read kVA rating, voltage ratio, type of the transformer,
core + copper loss = total loss, temperature rise limit, specific
heat dissipation due to radiation, convection, convection
improvisation percentage, tube dimension and limb, frame,
winding dimension
Tubes
Calculate dissipating
surface of plain tank
Determine total losses
Find x, the increase in dissipating
surface area, due to tubes
and specific loss dissipation
Determine area of tube
Determine total
dissipating area
Determine number of
tubes
Determine spacing
number of rows and
columns
Print all values
Stop
fig. 3.39 | Flowchart for design of tank with tubes in transformer
Fig.
3.80 Design of Transformer
3.80 Design of Transformer
Dissipating surface of tank, neglecting the top and bottom surfaces is given by
Dissipating surface of tank, neglecting the top and bottom surfaces is given by
St = 2Ht (Lt + Wt) = 2 ×1.25 (1 + 0.5)
St = 2Ht (Lt + Wt) = 2 ×1.25 (1 + 0.5) = 3.75 m2
= 3.75 m2
5000 = (12.5)(3.75)(50) + (8.78) At (50)
5000 = (12.5)(3.75)(50) + (8.78) At (50)
5000 = 2343.75 + 439 At
5000 = 2343.75 + 439 At
⇒
At = 6.05 m2 2
⇒
At =
6.05 m
Let the diameter of tube be 5 cm and the average height of tube is 105 cm.
Let the diameter of tube be 5 cm and the average height of tube is 105 cm.
Dissipating area of each tube a = π dl = π (0.05)(1.05) = 0.1649 m 2
Dissipating area of each tube at t= πdl = π (0.05)(1.05) = 0.1649 m 2
6.05
Number of tubes = 6.05 = 36
0.1647
= 36
Number of tubes =
0.1647
If the tubes are placed 7 cm apart from centre to centre, then the number of tubes on 100
theand
tubes
are side
placed
from centre
centre, then the number of tubes on
cm If
side
50 cm
are 712cm
andapart
6 as shown
in Fig.to3.40.
100 cm side and 50 cm side are 12 and 6 as shown in Fig. 3.40.
Total tubes = (2 ×12) + (2 × 6) = 36
Total tubes = (2 ×12) + (2 × 6) = 36
30
29
28
27
26
25
24
23
22
21
20
19
31
18
32
17
33
16
50 cm
34
15
35
14
36
13
100 cm
1
2
3
4
5
6
7
7 cm
fig. 3.40
Fig.
8
9
10
11
12
Design of Tank with Tubes 3.81
Specific heat of dissipation in case of plain walled tank = 12.5/W/m 2 / °C
5000
12.5× 3.75
= 106.66°C
So, temperature rise in case of plain walled tank =
To have this temperature rise within 50°C, additional surface is provided in are form
of 36 tubes, to dissipate heat.
Example 3.17: Design a suitable cooling system for a 500 kVA, 6600/440 V 50 Hz, 3φ
transformer with a total full load loss of 7 kW. The transformer main
dimensions are 1 m in height, 0.96 m in length and 0.47 m in breadth.
Use cooling tubes of diameter 50 mm to limit the average temperature
rise to 35°C. Use clearance of 50, 14 and 13 cm on the height, length
and width sides.
Solution: Given
Transformer rating, Q and type = 500 kVA and 3φ
Voltage rating = 6600/440 V
Tank size = 1 × 0.96 × 0.47 m
Clearances = 50 × 14 × 13 cm
Diameter of cooling tube = 50 mm
Allowable temperature rise = 35°C
Total size of transformer including clearances on all sides,
H t × Lt ×Wt = (100 + 50)×(96 + 14)×( 47 + 13)
= 150 ×110 × 60 cm
Total dissipating surface neglecting top and bottom surfaces
= 2 ( H t × Lt ) + ( H t ×Wt )
= 2 (1.5×1.1) + (1.5× 0.6)
= 5.1 m 2
Total losses = 12.5St θ + 8.8 At θ
7000 = (12.5)(5.1)(35) + (8.8)At (35)
Area of tubes,
At = 15.6 m2
Let the average height of each tube = 0.9 Ht = 0.9 × 150 = 135 cm
Dissipating surface area of each tube,
at = πDl
3.82 Design of Transformer
3.82 Design of Transformer
= (π )(0.05)(1.35)
= (π )(0.05)(1.35)
= 0.212 m 2
= 0.212 m 2
At
Number of tubes required = A
at t
Number of tubes required =
at
15.6
= = 15.6
0.212
0.212
= 73.6
 74
= 73.6
 74
If the tubes are placed 7.5 cm apart, then the number of tubes that can be placed along
If the tubes are placed 7.5 cm apart, then the number of tubes that can be placed along 110
110
60
110 cm and 60 cm side are
and
, respectively.
110
60
,7.respectively.
cm and 60 cm side are
7and
.5
5
7.5
7.5
i.e.,
i.e., They
They are
are 15
15 and
and 8,
8, respectively.
respectively.
If
+ 8)==46.
46.
If 15
15 and
and 88 tubes
tubes are
are provided,
provided, then
then the
the total
total numbers
numbers of
of tubes
tubes used
used are
are 2(15
2(15+8)
Since
74
tubes
are
to
be
used
to
dissipate
the
heat,
an
additional
row
of
46
tubes
can
Since 74 tubes are to be used to dissipate the heat, an additional row of 46 tubes can be
be
used,
the total
total number
number of
of tubes
tubes to
to 46
46 ×
× 22 =
used, which
which increases
increases the
= 92
92 tubes
tubes that
that is
is much
much more
more
than
than 74.
74.
Hence,
Hence, we
we can
can select
select 13
13 tubes
tubes that
that can
can be
be provided
provided on
on 110
110 cm
cm and
and 66 tubes
tubes can
can be
be provided
provided
on
60
cm.
on 60 cm.
Total
number
of of
tubes
==
2 [13
++
6] 6]
++
2 [13
+ 6] ==76.
Total
number
tubes
2 [13
2 [13+6]
76.
Hence,
Hence, instead
instead of
of 74
74 tubes,
tubes, 76
76 can
can be
be used
used as
as shown
shown in
in Fig.
Fig. 3.41.
3.41.
2
1
2
13
1
2
13
2
6
6
1
1
60 cm
110 cm
1
1
1
2
13
1
2
13
7 cm
fig. 3.41
Fig.
5 cm
Design of Tank with Tubes 3.83
Example 3.18: Design a cooling system for a natural oil cooled transformer of rating
1200 kVA, with its main dimensions as 0.6 × 1.5 × 1.8 m length,
breadth and height, respectively. The full load loss is 12 kW. Take
loss dissipation due to radiation and convection as 6 W/m2/°C and 6.5
W/m2/°C and 35% improvement in convention is to be provided by
tubes. Allowable temperature rise is 40°C. Use cooling tubes of 5 cm
diameter and 1 m length.
Solution: Given
Rating of transformer = 1200 kVA
Main dimensions = 0.6×1.5×1.8 m
Full load loss = 12 kW
Allowable temperature rise = 40oC
Improvement in convection = 35%
Loss dissipation due to radiation and convection = 6 W/m2/oC and 6.5 W/m2/oC
Total heat dissipating surface = 2( H t Lt + Wt H t )
= 2× (1.8 × 0.6) + (1.5 × 1.8)
St = 7.56 m 2
Loss dissipation due to radiation and convection = (6 + 6.5)St
= 12.5St
Heat dissipating area of tubes = At
Given that 40% improvement in dissipation due to convection is to be provided by tubes of
area At.
Loss dissipated due to cooling tubes = 6.5×
135
× At = 8.77 At
100
Total losses dissipating area by tank and tubes = 12.5 St + 8.77 At
Total losses = 12.5 St θ + 8.77 At θ
12 kW = (12.5St + 8.77 At )θ
θ=
12×1000
12.5St + 8.77 At
40 =
12×1000
(12.5)(7.56) + 8.77 At
At = 23.43 m 2
Total area of tubes,
At= 23.43 m2
3.84 Design
Design of Transformer
3.84
of Transformer
Area of each tube = a = πDl = (π)(5×10−2)(1) = 0.157 m2
Area of each tube = att = πDl = (π)(5×10−2)(1) = 0.157 m2
A
23.43
Total number of tubes = A t = 23.43 = 149
Total number of tubes = att = 0.157 = 149
at
0.157
Let 7.5 cm be the centre to centre distance between cooling tubes.
Let 7.5 cm be the centre to centre distance between cooling tubes.
150
60
Number of
of tubes
tubes on
on 150
150 cm
cm and
and 60
60 cm
cm sides
sides are
are 150 and
and 60 ,, respectively.
respectively.
Number
77..55
77..55
So, 20
20 and
and 88 tubes
tubes can
can be
be placed.
placed.
So,
Number of
of tubes
tubes in
in each
each row
row =
= 2(20
2(20 +
+ 8)
8) =
= 56
56
Number
If two
two rows
rows of
of tubes
tubes are
are placed,
placed, total
total number
number of
of tubes
tubes will
will be
be 22 ×
× 56
56 =
= 112
112
If
Since we
we have
have calculated
calculated only
only 150
150 tubes,
tubes, the
the placement
placement of
of tubes
tubes can
can be
be 20,
20, 16,
16, 20
20 its
its widthwise
widthwise
Since
and 8,
8, 4,
4, 88 along
along lengthwise
lengthwise as
as shown
shown in
in Fig.
Fig. 3.42.
3.42.
and
Hence, the
the total
total will
will be
be 22 (20
(20 +
+ 8)
8) +2(16
+2(16 +
+ 4)
4) +
+ 2(20
2(20 +
+ 8)
8) =
= 56
56 +
+ 40
40 +
+ 56
56 =
= 152
152 tubes.
tubes.
Hence,
20
2
1
25
2
2
1
2
1
1
3
2
3
60 cm
6
4
7
7
8
8
8
1
150 cm
1
1
2
3
1
1
4
17
2
18 19
15
2
20
16
20
fig.
Fig. 3.42
Example 3.19:
A 3φ,
3.19: A
3φ,10
10MVA,
MVA,33/6.6
33/6.6 kV
kV 50
50 Hz,
Hz, transformer
transformer has the following values on
initial design. Length of transformer is 160 cm, height of transformer
is 250 cm and the width is 78.5 cm. Use clearances of 50 cm, 11.5 cm and
11.5 cm, respectively, on the height, width and length for designing
the tank. Total iron loss is 25 kW and the copper loss is 100 kW.
Design of Tank with Tubes 3.85
Calculate the temperature rise of transformer without cooling tubes.
Calculate the number of tubes to limit the temperature rise to 50°C.
Also, comment on the choice of cooling system and whether the
cooling tubes are advisable in a practical case.
Solution:
Given
Transformer rating, Q and type = 10 MVA Total iron loss = 25 kW
Total copper loss = 100 kW
and 3φ
Tank size = 1.60 × 2.50 × 0.785 m
Voltage rating = 33,000/6600 V
Clearances = 50 × 11.5 × 11.5 cm
Allowable temperature rise = 50oC
We know that
Tank size = Transformer size + clearances
Height × width × length of tank = ( H + 50)×(W + 11.5)×(L + 90)
H t ×Wt × Lt = 300 × 90 × 250 cm
H t (Lt + Wt )
Heat dissipeting surface area of tank, St = 2H
= 2× 3 (2.5 + 0.9)
= 20.4 m 2
Without cooling tubes,
Total losses = 12.5 St θ
125×10 3 = (12.5)(20.4)θ
θ=
125×10 3
= 490°C
(12.5)(20.4)
The temperature rise without cooling tubes is 490°C. To limit this to 50°C, the dissipating
surface area has to be increased by providing cooling tubes. Let At be the area of all cooling
tubes.
125 kW = 12.5 St θ + 8.78 At θ
125×10 3 = (12.5)(20.4)(50) + (8.78)( At )(50)
At = 255.694 m 2
Let the average height of tube is 0.9 Ht = 0.9 × 300 = 270 cm and the diameter of each
tube is 5 cm.
Dissipating area of each tube = πDl = (π)(0.05)(2.7) = 0.424 m2
Clearances = 50 × 14 × 13 cm
3.86 Design of Transformer
3.86 Design of Transformer
3.24 Mechanical
Forces
3.24
mechanical forces
Mechanical forces are present in transformer due to leakage flux interaction, near the
Mechanical forces are present in transformer due to leakage flux interaction, near the
windings carrying current.
windings carrying current.
The forces can be classified into axial forces and radial forces, based on resolving the
The forces can be classified into axial forces and radial forces, based on resolving the
leakage flux component direction.
leakage flux component direction.
The leakage flux is represented in Fig. 3.43(a). The axial leakage field and radial leakage
The leakage flux is represented in Fig. 3.43(a). The axial leakage field and radial leakage
field are represented in Fig. 3.43(b) and 3.43(c), respectively.
field are represented in Fig. 3.43(b) and 3.43(c), respectively.
Fr
Fr
Leakage
flux
(a)
(b)
Fa
(c)
fig.
Fig. 3.43
3.43 || (a)
(a) Leakage
Leakage flux;
flux; (b)
(b) Axial
Axial leakage
leakage field;
field; (c)
(c) Radial
Radial leakage
leakage field
field
Generally,
Fleming’s
hand
rule
is used
to determine
direction
of forces.
Generally,
Fleming’s
leftleft
hand
rule
is used
to determine
thethe
direction
of forces.
The
cause
and
effect
of
the
axial
and
radial
forces
are
represented
in
Table
3.12.
The cause and effect of the axial and radial forces are represented in Table 3.12.
Computer-aided Design of Transformer 3.87
Table 3.12 | Cause and effect of the axial and radial forces
Feature\Type
Axial forces
Radial forces
Nature of occurrence
(cause)
• Due to interaction of radial
component of leakage flux
Effect
• Compression caused in windings
• Compression in inner
can become tensile if an
winding. Tensile force in
asymmetry exists in axial magnetic
outer winding
field
• Due to interaction of axial
component of leakage flux
The magnitude of force acting on a conductor is proportional to the current flow in it
and magnetic field intensity due to current in neighbouring conductors. Generally, as the
magnetic field is proportional to the current, force is proportional to the square of current.
The effect of large mechanical forces can be reduced by bracing (reinforcing) of windings.
3.25 Computer-aided Design of Transformer
Manual calculations are done for the design of transformer so far. The design of the transformer
starts from the output equation based on the specifications and other parameters of the
transformer. The different variables and parameters are interrelated and they have a nonlinear relationship with the performance of the machine. Taking all these in to consideration,
the design has to be done to give best performance. For a particular application, several sets of
design need to be carried out to find the optimum one. Hence, many iterations and parameter
changes have to be done. They can be easily done with the help of digital computers. Hence,
computers have become an integral part of the design process of electrical machines. Sample
computer programmes for the design of transformers are given in the following sections.
Example Program 1: Determine the main dimensions of a 350 kVA, 3φ, 50 Hz Y/Δ,
11000/3300V, core-type distribution transformer. Assume
distance between core centres as twice the width of core.
Solution:
Matlab program
disp(‘DESIGN OF OVERALL DIMENSION OF TRANSFORMER\n’);
disp(‘Enter the following values: ‘)
KVA=input(‘KVA rating: ‘);
disp(‘voltage per turn calculation give k :\n1 0.75-0.85 for 1phase
core\n2 0.45 for 3phase core\n3 1-1.2 for 1phase shell\n4 1.3 for
3phase\n’ );
k=input(‘k: ‘);
Et=k*sqrt(KVA);
f=input(‘Line frequency: ‘);
m=input(‘Number of phases: ‘);
3.88 Design of Transformer
Bm=input(‘Flux density: ‘);
Ki=input(‘Stacking factor: ‘);
disp(‘The values of peak flux per pole, net iron area and Gross iron
area calculated are’);
PHlm=Et/(4.44*f);
Ai=PHlm/Bm%net iron area;
Agi=Ai/Ki %gross core section area;
c=input(‘Enter the type of core:\n1)Square\n2)Stepped\n3)3-Stepped\
n4)4-Stepped \n’);
switch c
case 1
ct=0.45;
d=sqrt(Ai/ct) % dia od circumsribing circle
a=sqrt(0.5)*d %width of the largest stamping
case 2
ct=0.56;
d=sqrt(Ai/ct)
a=0.85*d %dimensions
b=0.53*d
case 3
ct=0.6;
d=sqrt(Ai/ct)
a=0.42*d %dimensions
b=0.7*d
c=0.9*d
case 4
ct=0.62;
d=sqrt(Ai/ct)
a=0.36*d %dimensions
b=0.36*d
c=0.78*d
r=0.92*d
end;
disp(‘WINDOW DESIGN OF TRANSFORMER\n’)
KV=input(‘Primary winding voltage: ‘);
deltap=input(‘Primary current density: ‘);
Kw=input(‘Enter window space factor: ‘)
Aw=(KVA*1000)/(3.33*Bm*f*Kw*deltap*Ai) %window area
ratiohw=input(‘Ratio - height to width of window in the range of 2-4:
‘); %Range 2-4
disp(‘The window width and window height are’);
Ww=sqrt(Aw/ratiohw) %window width
Hw=Ww*ratiohw%window height
Computer-aided Design of Transformer 3.89
disp(‘YOKE DESIGN OF TRANSFORMER\n’)
ratioyl=input(‘ratio - area of yoke to limbs: ‘);
Dy=input(‘Depth of yoke: ‘);
disp(‘The flux density in yoke,yoke area, gross yoke area and height
of yoke are calculated as’);
FDy=Bm/ratioyl%flux density in yoke
Ay=ratioyl*(PHlm/Bm) %yoke area
Agy=Ay/Ki %gross area of yoke
Hy=Agy/Dy%height of yoke
disp(‘OVERALL DIMENSION OF TRANSFORMER\n’)
disp(‘The distance betwn core centers,height, width and depth of
transformers are obtained as’);
D=d+Ww%dist between adjacent core centres
H=Hw+2*Hy%height of frame
W=2*D+Dy%width of frame
Df=Dy%depth of frame
C program
#include<stdio.h>
#include<math.h>
int main() {
printf(“DESIGN OF OVERALL DIMENSION OF TRANSFORMER\n”);
printf(“Enter the following values: \n”);
printf(“KVA rating: “);
double KVA;
scanf(“%lf”,&KVA);
double k;
printf(“voltage per turn calculation give k :\n1 0.75-0.85 for 1phase
core\n2 0.45 for 3phase core\n3 1-1.2 for 1phase shell\n4 1.3 for
3phase\n”);
scanf(“%lf”,&k);
double Et=k*sqrt(KVA);
double f;
printf(“\nLine frequency: “);
scanf(“%lf”,&f);
double m;
printf(“\nNumber of phases: “);
scanf(“%lf”,&m);
doubleBm;
printf(“\nFlux density: “);
scanf(“%lf”,&Bm);
double Ki;
3.90 Design of Transformer
printf(“\nStacking factor: “);
scanf(“%lf”,&Ki);
printf(“\nThe values of peak flux per pole, net iron area and Gross
iron area calculated are “);
doublePHlm=Et/(4.44*f);
printf(“\n%lf”,PHlm);
double Ai=PHlm/Bm; //net iron area
printf(“\n%lf”,Ai);
doubleAgi=Ai/Ki; //gross core section area
printf(“\n%lf”,Agi);
printf(“\nEnter the type of core:\n1)Square\n2)Stepped\n3)3-Stepped\
n4)4-Stepped \n”);
int core;
scanf(“%d”,&core);
double d,a,b,c,ct,r;
switch (core)
{
case 1:
ct=0.45;
d=sqrt(Ai/ct); // dia of circumsribing circle
a=sqrt(0.5)*d; //width of the largest stamping
break;
case 2:
ct=0.56;
d=sqrt(Ai/ct);
a=0.85*d; //dimensions
b=0.53*d;
break;
case 3:
ct=0.6;
d=sqrt(Ai/ct);
a=0.42*d; //dimensions
b=0.7*d;
c=0.9*d;
break;
case 4:
ct=0.62;
d=sqrt(Ai/ct);
Computer-aided Design of Transformer 3.91
a=0.36*d; //dimensions
b=0.36*d;
c=0.78*d;
r=0.92*d;
break;
default:
printf(“invalid”);
}
printf(“WINDOW DESIGN OF TRANSFORMER\n”);
double KV;
printf(“\nPrimary winding voltage: “);
scanf(“%lf”,&KV);
doubledeltap;
printf(“\nPrimary current density: “);
scanf(“%lf”,&deltap);
double Kw;
printf(“\nEnter Window space factor: “);
scanf(“%lf”,&Kw);
double Aw=(KVA*1000)/(3.33*Bm*f*Kw*deltap*Ai); //window area
doubleratiohw;
printf(“\nRatio - height to width of window in the range of 2-4: “);
scanf(“%lf”,&ratiohw);
doubleWw=sqrt(Aw/ratiohw); //window width
doubleHw=Ww*ratiohw; //window height
printf(“\nThe window width and window height are %lf, %lf”,Ww,Hw);
printf(“YOKE DESIGN OF TRANSFORMER\n”);
doubleratioyl;
printf(“\nratio - area of yoke to limb: “);
scanf(“%lf”,&ratioyl);
doubleDy;
printf(“\nDepth of yoke: “);
scanf(“%lf”,&Dy);
doubleFDy=Bm/ratioyl; //flux density in yoke
double Ay=ratioyl*(PHlm/Bm); //yoke area
doubleAgy=Ay/Ki; //gross area of yoke
doubleHy=Agy/Dy; //height of yoke
3.92 Design of Transformer
printf(“\nThe flux density in yoke,yoke area, gross yoke area and
height of yoke are calculated as %lf, %lf, %lf, %lf \n”,FDy, Ay, Agy,
Hy);
printf(“OVERALL DIMENSION OF TRANSFORMER\n”);
double D=d+Ww; //dist between adjacent core centres
double H=Hw+2*Hy; //height of frame
double W=2*D+Dy; //width of frame
doubleDf=Dy; //depth of frame
printf(“The distance betwn core centers, height, width and depth of
transformers are obtained as %lf, %lf, %lf, %lf \n”,D,H,W,Df);
return 0;
}
Example Program 2: Computer program for overall design of transformer
Solution:
Matlab program
disp(‘CORE DESIGN OF TRANSFORMER\n’);
disp(‘Enter the following values: ‘)
KVA=input(‘KVA rating: ‘);
disp(‘voltage per turn calculation give k :\n1 0.75-0.85 for 1phase
core\n2 0.45 for 3phase core\n3 1-1.2 for 1phase shell\n4 1.3 for
3phase\n’ );
k=input(‘k: ‘);
Et=k*sqrt(KVA);
f=input(‘Line frequency: ‘);
m=input(‘Number of phases: ‘);
Bm=input(‘Flux density: ‘);
Ki=input(‘Stacking factor: ‘);
disp(‘The values of peak flux per pole, net iron area and Gross iron
area calculated are’);
PHlm=Et/(4.44*f);
Ai=PHlm/Bm%net iron area;
Agi=Ai/Ki %gross core section area;
c=input(‘Enter the type of core:\n1)Square\n2)Stepped\n3)3-Stepped\
n4)4-Stepped \n’);
switch c
case 1
Computer-aided Design of Transformer 3.93
ct=0.45;
d=sqrt(Ai/ct) % dia od circumsribing circle
a=sqrt(0.5)*d %width of the largest stamping
case 2
ct=0.56;
d=sqrt(Ai/ct)
a=0.85*d %dimensions
b=0.53*d
case 3
ct=0.6;
d=sqrt(Ai/ct)
a=0.42*d %dimensions
b=0.7*d
c=0.9*d
case 4
ct=0.62;
d=sqrt(Ai/ct)
a=0.36*d %dimensions
b=0.36*d
c=0.78*d
r=0.92*d
end;
disp(‘WINDOW DESIGN OF TRANSFORMER\n’)
KV=input(‘Primary winding voltage: ‘);
deltap=input(‘Primary current density: ‘);
Kw=input(‘Enter window space factor: ‘)
Aw=(KVA*1000)/(3.33*Bm*f*Kw*deltap*Ai) %window area
ratiohw=input(‘Ratio - height to width of window in the range of 2-4:
‘); %Range 2-4
disp(‘The window width and window height are’);
Ww=sqrt(Aw/ratiohw) %window width
Hw=Ww*ratiohw%window height
disp(‘YOKE DESIGN OF TRANSFORMER\n’)
ratioyl=input(‘ratio - area of yoke to limbs: ‘);
Dy=input(‘Depth of yoke: ‘);
disp(‘The flux density in yoke,yoke area, gross yoke area and height
of yoke are calculated as’);
FDy=Bm/ratioyl%flux density in yoke
Ay=ratioyl*(PHlm/Bm) %yoke area
Agy=Ay/Ki %gross area of yoke
Hy=Agy/Dy%height of yoke
3.94 Design of Transformer
disp(‘OVERALL DIMENSION OF TRANSFORMER\n’)
disp(‘The distance betwn core centers,height, width and depth of
transformers are obtained as’);
D=d+Ww%dist between adjacent core centres
H=Hw+2*Hy%height of frame
W=2*D+Dy%width of frame
Df=Dy%depth of frame
disp(‘LOW VOLTAGE WINDING DESIGN OF TRANSFORMER\n’);
Vls=input(‘Secondary line voltage: ‘);
c1=input(‘Type of connection: \n1.Star\n2.Delta:\n’);
switch c1
case 1
Vsp=Vls/sqrt(3) %secondary phase voltage - star
case 2;
Vsp=Vls;%secondary phase voltage - delta
end;
disp(‘The no of turns per phase and current per phase of LV winding
are’);
Ts=round(Vsp/Et) %no of turns per phase
Isp=(KVA*1000)/(3*Vsp) %secondary current per phase
delta=input(‘Enter secondary current density: ‘)
disp(‘The area of conductor of LV winding is’);
as=Isp/delta %area of secondary conductor
disp(‘Let us choose copper rectangular conductors and paper insulation
for these conductors.’);
x=input(‘Width of conductor along height of window(mm): ‘);
y=input(‘Width of conductor along width of window(mm): ‘);
z=input(‘Increase in dimensions because of insulation(mm): ‘);
x1=x+z
y1=y+z%dimension with covering
ly=input(‘Number of layers: ‘);
disp(‘Using helical winding: ‘)
Ts1=round((Ts/ly)+1) %turns along axial length
Lcs=Ts1*x1 %axial length of lv winding
cls=(Hw*1000-Lcs)/2 %clearance
if(cls<6)
{
disp(‘clearance is <6. Min limit not satisfied’)
}
end;
cly=input(‘Enter thickness of pressboard cylinders(mm): ‘);
bs=2*ly*y1+cly %radial depth of lv winding
Computer-aided Design of Transformer 3.95
lvi=input(‘Enter thickness of insulation between Lv winding and
core(mm): ‘)
disp(‘The inside , outside, mean dia of LV winding and ita mean
length of turn are’);
Idl=d*1000+2*lvi %inside diameter
Odl=Idl+2*bs%outside diameter
Mdl=(Idl+Odl)/2 %Mean diameter
Mlt=pi*Mdl%Mean length of turn
disp(‘HIGH VOLTAGE WINDING DESIGN OF TRANSFORMER\n’);
Vlp=input(‘primary line voltage: ‘);
c1=input(‘Type of connection: \n1.Star\n2.Delta:\n’);
switch c1
case 1
Vpp=Vlp/sqrt(3) %primary phase voltage - star
case 2;
Vpp=Vlp;%primary phase voltage - delta
end;
disp(‘The no of turns per phase and current per phase of LV winding
are’);
Tp=round(Ts*(Vpp/Vsp)) %no of turns
Ipp=(KVA*1000)/(3*Vpp) %primary current per phase
delta=input(‘primary current density: ‘)
disp(‘Let us choose copper round conductors\n’)
disp(‘The area of conductor of HV winding,Dia of conductor,Total
copper area in window and window space factor are’);
ap=Ipp/delta %area of primary conductor
dp=sqrt((4*ap)/pi) %diamater of conductor
Acw=2*(ap*Tp+as*Ts) %Total copper area in window
Kw=Acw/(Aw*10000) %window space factor
ca=input(‘Number of coils in HV winding: ‘); %volt/coil should not
exceed 1500V
ta=input(‘Number of turns in each coil of HV winding: ‘);
tec=Tp-ca*ta %Number of turns in end coil
ly=input(‘Number of layers of normal coil: ‘);
tly=input(‘Turns per layer of normal coil: ‘);
Mxvly=2*tly*(Vlp/Tp) %Max voltage between layers
sci=input(‘Size of conductor with insulation: ‘);
adn=tly*sci%axial depth of normal coil
lye=input(‘Number of layers of end coil: ‘);
tlye=input(‘Turns per layer of end coil: ‘);
ade=tlye*sci%axial depth of end coil
sp=input(‘Height of Spaces used between adjacent coils: ‘);
3.96 Design of Transformer
Lcp=ca*adn+ade+ca*sp%axial length of HV winding
Cl=(Hw-Lcp)/2 %clearance
cly=input(‘Thickness of insultaion between layers: ‘);
bp=ly*sci+(ly-1)*cly%radial depth of HV coil
T=5+((0.9*Vlp)/1000)
Idh=ca*adn+2*T %Inside diameter
Odh=Idh+2*ade%Outside diameter
Mdh=(Idh+Odh)/2 %Mean diameter
disp(‘RESISTANCE DESIGN OF TRANSFORMER\n’);
Lmtp=(pi*Mdh)/1000 %length of mean turn in HV winding
rop=input(‘Resistivity of material in HV winding: ‘);
disp(‘The Resistane of conductor of LV winding is’);
Rp=(Tp*Lmtp*rop)/ap%resistance in HV side
Lmts=(pi*Mdl)/1000 %length of mean turn in Lv winding
ros=input(‘Resistivity of material in LV winding: ‘);
disp(‘The Resistane of conductor of HV winding, Resistance referred
to HV side and Per unit resistance are’);
Rs=(Ts*Lmts*ros)/as %resistance in HV side
Ref=Rp+(((Tp*Tp)/(Ts*Ts))*Rs) %Resistanereffered to primary
ep=(Ipp*Ref)/Vlp%per unit resistance
disp(‘LEAKAGE REACTANCE DESIGN OF TRANSFORMER\n’);
Dm=(Odl+Odh)/2 %Mean diameter of windings
Lmt=(pi*Dm)/1000 %Length of mean turn of winding
Lc=(Lcp+Lcs)/2 %Mean axial length of winding
disp(‘The leakage Reatance referred to HV side and Per unit impedance
are’);
Xp=(2*pi*f*4*pi*10e-7*Tp*Tp*Lmt*(T+(bp+bs)/3))/Lc%Leakage reactance
epx=(Ipp*Xp)/Vpp%per unit leakage reactance
epi=sqrt((ep*ep)+(epx*epx)) %per unit impedance
C program
#include<stdio.h>
#include<math.h>
int main()
{
const double pi=3.14;
printf(“DESIGN OF OVERALL DIMENSION OF TRANSFORMER\n”);
printf(“Enter the following values: \n”);
printf(“KVA rating: “);
Computer-aided Design of Transformer 3.97
double KVA;
scanf(“%lf”,&KVA);
double k;
printf(“voltage per turn calculation give k :\n1 0.75-0.85 for 1phase
core\n2 0.45 for 3phase core\n3 1-1.2 for 1phase shell\n4 1.3 for
3phase\n”);
scanf(“%lf”,&k);
double Et=k*sqrt(KVA);
double f;
printf(“\nLine frequency: “);
scanf(“%lf”,&f);
double m;
printf(“\nNumber of phases: “);
scanf(“%lf”,&m);
doubleBm;
printf(“\nFlux density: “);
scanf(“%lf”,&Bm);
double Ki;
printf(“\nStacking factor: “);
scanf(“%lf”,&Ki);
printf(“\nThe values of peak flux per pole, net iron area and Gross
iron area calculated are “);
doublePHlm=Et/(4.44*f);
printf(“\n%lf”,PHlm);
double Ai=PHlm/Bm; //net iron area
printf(“\n%lf”,Ai);
doubleAgi=Ai/Ki; //gross core section area
printf(“\n%lf”,Agi);
printf(“\nEnter the type of core:\n1)Square\n2)Stepped\n3)3-Stepped\
n4)4-Stepped \n”);
int core;
scanf(“%d”,&core);
double d,a,b,c,ct,r;
switch (core)
{
case 1:
ct=0.45;
d=sqrt(Ai/ct); // dia of circumsribing circle
a=sqrt(0.5)*d; //width of the largest stamping
break;
case 2:
ct=0.56;
d=sqrt(Ai/ct);
3.98 Design of Transformer
a=0.85*d; //dimensions
b=0.53*d;
break;
case 3:
ct=0.6;
d=sqrt(Ai/ct);
a=0.42*d; //dimensions
b=0.7*d;
c=0.9*d;
break;
case 4:
ct=0.62;
d=sqrt(Ai/ct);
a=0.36*d; //dimensions
b=0.36*d;
c=0.78*d;
r=0.92*d;
break;
default:
printf(“invalid”);
}
printf(“WINDOW DESIGN OF TRANSFORMER\n”);
double KV;
printf(“\nPrimary winding voltage: “);
scanf(“%lf”,&KV);
doubledeltap;
printf(“\nPrimary current density: “);
scanf(“%lf”,&deltap);
double Kw;
printf(“\nEnter Window space factor: “);
scanf(“%lf”,&Kw);
double Aw=(KVA*1000)/(3.33*Bm*f*Kw*deltap*Ai); //window area
doubleratiohw;
printf(“\nRatio - height to width of window in the range of 2-4: “);
scanf(“%lf”,&ratiohw);
doubleWw=sqrt(Aw/ratiohw); //window width
doubleHw=Ww*ratiohw; //window height
printf(“\nThe window width and window height are %lf, %lf”,Ww,Hw);
printf(“YOKE DESIGN OF TRANSFORMER\n”);
Computer-aided Design of Transformer 3.99
doubleratioyl;
printf(“\nratio - area of yoke to limb: “);
scanf(“%lf”,&ratioyl);
doubleDy;
printf(“\nDepth of yoke: “);
scanf(“%lf”,&Dy);
doubleFDy=Bm/ratioyl; //flux density in yoke
double Ay=ratioyl*(PHlm/Bm); //yoke area
doubleAgy=Ay/Ki; //gross area of yoke
doubleHy=Agy/Dy; //height of yoke
printf(“\nThe flux density in yoke,yoke area, gross yoke area and
height of yoke are calculated as %lf, %lf, %lf, %lf \n”,FDy, Ay, Agy,
Hy);
printf(“OVERALL DIMENSION OF TRANSFORMER\n”);
double D=d+Ww; //dist between adjacent core centres
double H=Hw+2*Hy; //height of frame
double W=2*D+Dy; //width of frame
doubleDf=Dy; //depth of frame
printf(“The distance betwn core centers, height, width and depth of
transformers are obtained as %lf, %lf, %lf, %lf \n”,D,H,W,Df);
printf(“\nLOW VOLTAGE WINDING DESIGN OF TRANSFORMER\n”);
doubleVls;
printf(“\nSecondary line voltage: “);
scanf(“%lf”,&Vls);
int c1;
printf(“\nType of connection: \n1.Star\n2.Delta:\n “);
scanf(“%d”,&c1);
doubleVsp;
switch (c1)
{
case 1:
Vsp=Vls/sqrt(3); //secondary phase voltage - star
break;
case 2:
Vsp=Vls;//secondary phase voltage - delta
break;
}
doubleTs=round(Vsp/Et); //no of turns per phase
printf(“\nThe no of turns per phase and current per phase of LV
winding are: %lf”,Ts);
doubleIsp=(KVA*1000)/(3*Vsp); //secondary current per phase
double deltas;
3.100 Design of Transformer
printf(“\nEnter secondary current density: “);
scanf(“%lf”,&deltas);
double as=Isp/deltas; //area of secondary conductor
printf(“\nThe area of conductor of LV winding is %lf”,as);
printf(“\nLet us choose copper rectangular conductors and paper
insulation for these conductors.”);
double x;
printf(“\nWidth of conductor along height of window(mm): “);
scanf(“%lf”,&x);
double y;
printf(“\nWidth of conductor along width of window(mm): “);
scanf(“%lf”,&y);
double z;
printf(“\nIncrease in dimensions because of insulation(mm): “);
scanf(“%lf”,&z);
double x1=x+z;
double y1=y+z; //dimension with covering
doublely;
printf(“\nNumber of layers: “);
scanf(“%lf”,&ly);
printf(“\nUsing helical winding: “);
double Ts1=round((Ts/ly)+1); //turns along axial length
doubleLcs=Ts1*x1; //axial length of lv winding
doublecls=(Hw*1000-Lcs)/2; //clearance
if(cls<6)
{
printf(“\nclearance is <6. Min limit not satisfied”);
}
doublecly;
printf(“\nEnter thickness of pressboard cylinders(mm): “);
scanf(“%lf”,&cly);
doublebs=2*ly*y1+cly; //radial depth of lv winding
double lvi;
printf(“\nEnter thickness of insulation between Lv winding and
core(mm): “);
scanf(“%lf”,&lvi);
printf(“\nThe inside , outside, mean dia of LV winding and its mean
length of turn are”);
doubleIdl=d*1000+2*lvi; //inside diameter
doubleOdl=Idl+2*bs; //outside diameter
doubleMdl=(Idl+Odl)/2; //Mean diameter
doubleMlt=pi*Mdl; //Mean length of turn
printf(“\nThe inside , outside, mean dia of LV winding and its mean
length of turn are %lf, %lf, %lf, %lf”,Idl,Odl,Mdl,Mlt);
printf(“\nHIGH VOLTAGE WINDING DESIGN OF TRANSFORMER\n”);
Computer-aided Design of Transformer 3.101
doubleVlp;
printf(“\nprimary line voltage: “);
scanf(“%lf”,&Vlp);
int c2;
printf(“\nType of connection: \n1.Star\n2.Delta:\n”);
scanf(“%d”,&c2);
doubleVpp;
switch (c2)
{
case 1:
Vpp=Vlp/sqrt(3); //primary phase voltage - star
break;
case 2:
Vpp=Vlp;//primary phase voltage - delta
break;
default:
printf(“\ninvalid”);
}
printf(“\nThe no of turns per phase and current per phase of LV
winding are”);
doubleTp=round(Ts*(Vpp/Vsp)); //no of turns
doubleIpp=(KVA*1000)/(3*Vpp); //primary current per phase
printf(“\nThe no of turns per phase and current per phase of LV
winding are %lf, %lf”,Tp,Ipp);
printf(“\nLet us choose copper round conductors\n”);
doubleap=Ipp/deltap; //area of primary conductor
doubledp=sqrt((4*ap)/pi); //diamater of conductor
doubleAcw=2*(ap*Tp+as*Ts); //Total copper area in window
double Kw1=Acw/(Aw*10000); //window space factor
printf(“\nThe area of conductor of HV winding,Dia of conductor,Total
copper area in window and window space factor are %lf, %lf, %lf,
%lf”,ap,dp,Acw,Kw1);
doubleca,ta;
printf(“\nNumber of coils in HV winding: “); //volt/coil should not
exceed 1500V
scanf(“%lf”,&ca);
printf(“\nNumber of turns in each coil of HV winding: “);
scanf(“%lf”,&ta);
doubletec=Tp-ca*ta; //Number of turns in end coil
doublelys,tly;
printf(“\nNumber of layers of normal coil: “);
scanf(“%lf”,&lys);
printf(“\nTurns per layer of normal coil: “);
scanf(“%lf”,&tly);
doubleMxvly=2*tly*(Vlp/Tp); //Max voltage between layers
doublesci,lye,tlye;
3.102 Design of Transformer
printf(“\nSize of conductor with insulation: “);
scanf(“%lf”,&sci);
doubleadn=tly*sci; //axial depth of normal coil
printf(“\nNumber of layers of end coil: “);
scanf(“%lf”,&lye);
printf(“\nTurns per layer of end coil: “);
scanf(“%lf”,&tlye);
doubleade=tlye*sci; //axial depth of end coil
double sp;
sp=printf(“\nHeight of Spaces used between adjacent coils: “);
scanf(“%lf”,&sp);
doubleLcp=ca*adn+ade+ca*sp; //axial length of HV winding
double Cl=(Hw-Lcp)/2; //clearance
doubleclyt;
printf(“\nThickness of insultaion between layers: “);
scanf(“%lf”,&clyt);
doublebp=lys*sci+(lys-1)*clyt; //radial depth of HV coil
double T=5+((0.9*Vlp)/1000);
doubleIdh=ca*adn+2*T; //Inside diameter
doubleOdh=Idh+2*ade; //Outside diameter
doubleMdh=(Idh+Odh)/2; //Mean diameter
printf(“\nThe radial depth, inner diameter, outer diameter, mean
diameter of the high are %lf, %lf, %lf, %lf”,bp,Idh,Odh,Mdh);
printf(“\nRESISTANCE DESIGN OF TRANSFORMER\n”);
doubleLmtp=(pi*Mdh)/1000; //length of mean turn in HV winding
doublerop;
printf(“\nResistivity of material in HV winding: “);
scanf(“%lf”,&rop);
printf(“\nThe Resistance of conductor of LV winding is”);
doubleRp=(Tp*Lmtp*rop)/ap; //resistance in LV side
doubleLmts=(pi*Mdl)/1000; //length of mean turn in LV winding
doubleros;
printf(“\nResistivity of material in LV winding: “);
scanf(“%lf”,&ros);
doubleRs=(Ts*Lmts*ros)/as; //resistance in HV side
double Ref=Rp+(((Tp*Tp)/(Ts*Ts))*Rs); //Resistance referred to
primary
double ep=(Ipp*Ref)/Vlp; //per unit resistance
printf(“\nThe Resistance of conductor of HV winding, Resistance referred
to HV side and Per unit resistance are %lf, %lf, %lf”,Rs,Ref,ep);
printf(“\nLEAKAGE REACTANCE DESIGN OF TRANSFORMER\n”);
doubleDm=(Odl+Odh)/2; //Mean diameter of windings
Computer-aided Design of Transformer 3.103
doubleLmt=(pi*Dm)/1000; //Length of mean turn of winding
doubleLc=(Lcp+Lcs)/2; //Mean axial length of winding
doubleXp=(2*pi*f*4*pi*10e-7*Tp*Tp*Lmt*(T+(bp+bs)/3))/Lc; //Leakage
reactance
doubleepx=(Ipp*Xp)/Vpp; //per unit leakage reactance
double epi=sqrt((ep*ep)+(epx*epx)); //per unit impedance
printf(“\nThe leakage Reactance referred to HV side and Per unit
impedance are %lf, %lf”,epx,epi);
return 0;
}
Example Program 3: Computer program for design of cooling system of transformer
with total loss specified
Solution:
Matlab program
KVA=input(‘KVA rating: ‘);
P=input(‘allowable losses: ‘);
t=input(‘allowable temperature rise: ‘);
l=input(‘length of tank (including clearance): ‘);
w=input(‘width of tank (including clearance): ‘);
h=input(‘height of tank (including clearance): ‘);
ht=input(‘average height of tube: ‘);
dt=input(‘diameter of tube: ‘);
St=2*h*(l+w) %dissipating surface area of tank
At=(P-(12.5*St*t))/(8.78*t) %total heat dissipating area of tubes
at=pi*dt*ht%area of each tube
Nt=round(At/at) %number of tubes
C program
#include<stdio.h>
#include<math.h>
int main()
{
const double pi=3.14;
doubleKVA,P,t,l,w,h,ht,dt;
KVA=printf(“\nKVA rating: “);
printf(“\n\enter the following parameters: “);
KVA=printf(“\nKVA rating: “);
scanf(“%lf”,&KVA);
3.104 Design of Transformer
P=printf(“\nallowable losses: “);
scanf(“%lf”,&P);
t=printf(“\nallowable temperature rise: “);
scanf(“%lf”,&t);
l=printf(“\nlength of tank (including clearance): “);
scanf(“%lf”,&l);
w=printf(“\nwidth of tank (including clearance): “);
scanf(“%lf”,&w);
h=printf(“\nheight of tank (including clearance): “);
scanf(“%lf”,&h);
ht=printf(“\naverage height of tube: “);
scanf(“%lf”,&ht);
dt=printf(“\ndiameter of tube: “);
scanf(“%lf”,&dt);
double St=2*h*(l+w); //dissipating surface area of tank
printf(“\ndissipating surface area of tank is %lf”,St);
double At=(P-(12.5*St*t))/(8.78*t); //total heat dissipating area of
tubes
printf(“\ntotal heat dissipating area of tubes is %lf”,At);
double at=pi*dt*ht; //area of each tube
printf(“\nArea of each tube is %lf”,at);
doubleNt=round(At/at); //number of tubes
printf(“\nTotal number of tubes is %lf”,Nt);
return 0;
}
Review Questions
Multiple-choice Questions
1. Magnetic material used in large transformer is
.
(a) cast steel
(b) hot-rolled silicon steel
(b) cold-rolled grain-oriented steel
(d) either (b) or (c)
2.
type of cylindrical windings using circular conductors are employed in
transformers.
(a) Multi-layered
(b) double layered (c) triple layered
(d) single layered
3. The main disadvantage of cylindrical winding is
.
(a) high copper loss
(b) poor mechanical strength
(c) more eddy current loss
(d) none of the above
4. Helical windings are used in
transformers.
(a) distribution
(b) power
(c) shell-type
(d) none of these
5. Disc windings are primarily used in
capacity transformers.
(a) low
(b) medium
(c) high
(d) any of these
Review Questions 3.105
6. Yokes with rectangular cross-section are used for
(a) small
(b) medium
transformers.
(c) large
kVA.
7. Power transformers have ratings above
(a) 50
(b) 100
(d) any of these
(c) 250
8. Cold-rolled grain-oriented steel has
steel.
(d) 500
in comparison to hot-rolled silicon
(a) better finish
(b) improved space factor
(c) much better magnetic properties
(d) all of the above
.
9. Typical value of window space factor is
(a) 0.5
(b) 0.6
(c) 0.1
(d) 0.3
10. The reason for using multi-step core is used in a transformer is to
.
(a) increase the efficiency
(b) decrease the cost of core material
(c) decrease the cost of copper
(d) increase the output
.
11. Tappings of a transformer are provided
(a) at the middle of h.v. side
(b) at the neutral end of h.v. side
(c) at the phase end of h.v. side
(d) at the phase end of l.v. side
12. For transformer laminations, which type of silicon steel is preferred?
(a) cold rolled
(b) hot rolled
(c) grain-oriented
(d) any of these
13. The primary and secondary windings are interlaced in a transformer for
(a) reduced cost
(b) uniform heating
(c) easiness of coil making
(d) reduced leakage reactance
.
14. Helical winding with rectangular strip conductors is generally used for
(a) h.v. coils
(b) both l.v. and h.v. coils
(c) l.v. coils
(d) none of the above
.
15. The cross-over coils in transformers are generally used for
(a) h.v. winding
(b) l.v. winding
(c) both l.v. and h.v. winding
(d) none of the above
.
16. The cylindrical winding in transformers is generally not used beyond
(a) 6.6 kV
(b) 3.3 kV
(c) 66 kV
(d) 33 kV
17. The percentage of silicon in the transformer core steel is
(a) 2 to 3%
(b) 3 to 4%
.
(c) 4 to 5%
(d) 5 to 6%
18. The transformer core laminations have thickness in the range of
(a) 3.5 to 5 mm
(b) 2 to 3 mm
(c) 0.35 to 0.5 mm
19. The core laminations are prepared using
transformers.
.
(d) 0.035 to 0.05 mm
for large capacity power
(a) cold-rolled grain-oriented silicon steel (b) cold-rolled silicon steel
(c) hot-rolled silicon steel
.
(d) any one of the above
3.106 Design of Transformer
20. The core section in a large capacity power transformer is
.
(a) square
(b) rectangular
(c) multi-stepped
(d) Any of the above
21. Tap changer is normally provided on
transformer.
(a) distribution
(b) step up
(c) instrument
(d) high voltage
22. A three-phase power transformer is generally of
type.
(a) berry
(b) core
(c) shell
(d) toroidal
23. In comparison with power transformer, a distribution transformer has
.
(a) low percentage impedance and high copper iron loss ratio
(b) high percentage impedance and high copper iron loss ratio
(c) high percentage impedance and low copper iron loss ratio
(d) low percentage impedance and low copper iron loss ratio
24. What are the effects of making limb section lower than the yoke section?
(a) economy in copper usage
(b) decreased iron losses
(c) decreased magnetizing current
(d) all of the above
25. Circular coils are preferred in transformers because of
.
(a) of its superior mechanical stability under short circuit conditions
(b) it becomes wasteful to employ rectangular coils
(c) both (a) and (b) above
(d) it is easier to make circular coils.
26. The reason for single-phase shell-type distribution transformer to have sandwich
type winding is
.
(a) to reduce the leakage reactance
(b) to save copper
(c) to improve the voltage regulation
(d) both (a) and
(c) above.
27. What is the effect of increase in the height of the window in comparison to the width
in a transformer?
(a) cost of copper will be reduced
(b) voltage regulation will decrease
(c) efficiency will decrease due to increase in copper losses
(d) both (a) and (b)
28.
causes hum in a transformer.
(a) Magnetostriction
(b) Vibrations developed by laminations depending upon the tightness of clampings
(c) Cushions and paddings
(d) all of the above
29. Humming noise in a transformer can be decreased by
.
(a) using lower flux densities
(b) tightening the clampings of laminations
Review Questions 3.107
(c) using suitable cushions, padding and oil barriers
(d) all of the above
30. If all the dimensions of a transformer is doubled, its iron loss will be
compared to iron loss with the original dimensions.
(a) half
(b) double
(c) four times
(d) 8 times
31. Total copper area accommodated in the window of gross area 300 cm2 of a particular
transformer is 0.48 times, the net area of iron in the core of 200 cm2. Then, the window
space factor is
.
(a) 0.72
(b) 0.28
(c) 0.32
(d) 0.48
32. If the net iron area of a three stepped core is 240 cm2, then the diameter of the
circumscribing circle is
.
(a) 25 cm
(b) 18 cm
(c) 20 cm
(d) 15 cm
33.
to reduce hysteresis lossin a transformer.
(a) Core may be laminated
(b) Silicon steel may be used as the core material.
(c) Core may be constructed with any permanent magnet material such as alnico
(d) Core may be impregnated with varnish.
34. Which of the following relation must be satisfied if a transformer having constant flux
and constant current density is designed for minimum cost?
(a) iron loss = copper loss
(b) weight of iron = weight of copper
(c) weight of iron/weight of copper = specific cost of copper/specific cost of iron
(d) weight of iron/weight of copper = specific cost of iron/specific cost of copper
35. The criteria for a transformer to be designed for the minimum volume is
.
(a) iron loss = copper loss
(b) volume of iron = volume of copper
(c) weight of iron = weight of copper
(d) the volume of iron is minimum
36. Choice of higher core flux density in a transformer leads to
.
(a) increased overall size
(b) reduced magnetizing current
(c) reduced iron losses
(d) reduction in overall cost
37. With rise in voltage, the window space factor of a transformer
.
(a) decreases
(b) increases
(c) remains constant
(d) decreases or increases depending upon whether it is a distribution or power
transformer
38. If the total area occupied by the insulating material in the window of a transformer is
.
72%, then the window space factor is
(a) 0.28
(b) 0.72
(c) 0.32
(d) 0.25
3.108 Design of Transformer
39. If the gross area of window is 750 cm2 for a particular transformer, then the approximate
height of the window is
.
(a) 25 cm
(b) 75 cm
(c) 50 cm
(d) 30 cm
40. If for a single phase, 6600/400V, core-type transformer, the e.m.f.per turn is 6 V, then
the number of turns in h.v. winding is
.
(a) 1106
(b) 1122
(c) 1100
(d) 1089
41. If for a single phase, 6600/400V, core-type transformer, the e.m.f. per turn is 6 V, then
.
the number of turns in l.v. winding is
(a) 65
(b) 66
(c) 67
(d) 68
42. If for a single phase, 6600/400V, core-type transformer, the e.m.f. per turn is 6 V, then
the number of turns in h.v. winding with + 5% tapping is
.
(a) 1064
(b) 1178
(c) 1068
(d) 1072
43. If for a three phase, 11000/440V, delta/star, core-type transformer, e.m.f. per turn is
.
10.8V, then the number of turns per phase in l.v. winding is
(a) 23
(b) 23
(c) 24
(d) 40
44. If for a three phase, 11000/440V, delta/star, core-type transformer, e.m.f. is per turn is
10.8V, then the number of turns per phase in h.v. winding is
.
(a) 1018
(b) 600
(c ) 1000
(d) 1038
45. If for a three phase, 11000/440V, delta/star, core-type transformer, e.m.f. is per turn
is 10.8V, then the number of turns per phase in h.v. winding with - 5% tapping is
.
(a) 1050
(b) 1092
(c) 1080
(d) 1070
46. Eddy currents are reduced in high silicon steel as it provides
.
(a) increases resistivity
(b) reduces resistivity
(c) short circuits
(d) none of the above
47. In order to reduce the eddy current losses within the conductor, the thickness of the
rectangular conductor selected for l.v. and h.v. winding should not be greater than
.
(a) 2 mm
(b) 2.5 mm
(c) 3 mm
(d) 3.5 mm
48. Stacking factor will be minimum for
typeofcore.
(a) four-stepped
(b) three-stepped
(c) square
(d) cruciform
49. The usual values of current densities for medium and large power transformers are
.
(a) 1.5 to 2.6 A/mm2
(b) 2.4 to 3.4 A/mm2
(c) 1.5 to 2 A/mm2
(d) 1 to 2.6 A/mm2
50. If a 200/400V transformer has a secondary winding resistance of 0.5Ω, the total
resistance referred to primary is
.
(a) 0.125Ω
(b) 0.5Ω
(c) 1Ω
(d) 2Ω
51. If the frequency of supply voltage to the primary of a two-winding transformer is
doubled, then the induced emf is
.
(a) unaltered
(b) doubled
(c) halved
(d) none of these
Review Questions 3.109
52. Leakage reactance of a transformer is
.
(a) directly proportional to number of turns
(b) directly proportional to square of number of turns
(c) inversely proportional to number of turns
(d) inversely proportional to square of number of turns
53. Large value of flux density can be adopted while designing
.
(a) distribution transformer
(b) welding transformer
(c) large capacity power transformer
(d) current transformer.
54. Lower value of window space factor will be adopted in design of
.
(a) 400 kVA, 11/0.4 kV, distribution transformer
(b) 20,000 kVA, 66/11 kV, power transformer
(c) 20,000 kVA, 33/11 kV, power transformer
(d) 100 kVA, 11/0.4 kV, distribution transformer
55. Larger value of current density can be adopted for transformer employing
cooling.
(a) oil-forced water forced
(b) oil-natural air forced
(c) oil-immersed self-cooled
(d) any of the above
56. While designing a transformer if increased window height is adopted, it may result into
(a) poor voltage regulation
(b) reduced leakage reactance
(c) increased leakage reactance
(d) both (a) and (b) above
57. If the thickness of laminations is t, then the eddy current losses are proportional to
.
(a) t4
(b) t2
(c) t3
(d) t
58. If wider window is adopted in designing a transformer, it may result in to
(a) reduced leakage reactance
(b) good voltage regulation
(c) increased leakage reactance
(d) both (b) and (c) above
.
59. In the design of a transformer, the usual value of the ratio of window height to
window width used is
.
(a) 5
(b) 3
(c) 2
(d) 4
60. In a transformer, the emf per turn is determined, in terms of its kVA output rating (Q)
.
from the relation
(a) Et = KQ
(b) Et = k
Q
(c) Et = kQ3/2
(d) Et = k/Q
61. An iron cored transformer is working at a maximum flux density of 0.8 Wb/m2. Its core
is replaced by silicon steel core, working at a maximum flux density of 1.2 Wb/m2. If
the total flux is to remain the same, what is the reduction in volume expressed as of the
original volume? The frequency and voltage per turn are the same in both the cases.
(a) 33%
(b) 9%
(c) 22%
(d) 11%
3.110 Design of Transformer
62. If the total losses of a transformer during its design is 500 W at 50% full load, then the
total copper losses of the same transformer at 1.25 times full load will be
.
(a) 500 W
(b) 625 W
(c) 1250 W
(d) 3125 W
63. Typical value of no load current expressed as percentage of full load current in
transformer is
.
(a) 10%
(b) 15%
(c) 3%
(d) 8%
64. In a transformer, iron losses and full load copper losses are 900 and 1600 W, respectively.
The ratio of load for maximum efficiency in terms of full load is
.
(a) 0.56
(b) 0.85
(c) 0.75
(d) 1.0
65. Magnetic couplings are present closer in a transformer to ensure
.
(a) high efficiency
(b) good regulation
(c) good regulation and high efficiency
(d) good regulation and high power factor
66. The useful flux in a transformer links
.
(a) only l.v.turns
(b) only h.v. turns
(c) both l.v. and h.v. turns
(d) none of the above
67. Under which of the following conditions, the hysteresis loss in a transformer remains
unaffected?
(a) When both frequency and flux density are increased by 10%
(b) When flux density is increased by 10%
(c) When thickness of lamination is increased by 10%
(d) When frequency is increased by 10%
68. Maintaining the same thickness but selecting a higher silicon content core material for
a transformer reduces eddy current loss due to
.
(a) decrease in resistivity
(b) increase in resistivity
(c) decrease in malleability
(d) both (a) and (b) above
.
69. In order to have minimum copper loss in the transformer windings,
(a) the primary and secondary currents should be equal
(b) the current densities in primary and secondary windings must be the same
(c) copper losses should be equal to iron losses
(d) none of the above
70. Transformer core is laminated to decrease
.
(a) stray loss
(b) eddy current loss
(c) copper loss
(d) hysteresis loss
71. In transformers, with rise in supply frequency,
.
(a) copper loss remains unaffected but efficiency increases.
(b) copper loss decreases but efficiency increases
Review Questions 3.111
(c) copper loss increases but efficiency decreases
(d) both copper loss and efficiency remain unaffected
72. In transformers, with increase in supply frequency, the iron losses
.
(a) decreases
(b) increases
(c) remain unaffected
(d) becomes zero
73. Under which situation(s), a transformer can be slightly over loaded?
(a) if the ambient temperature is much below the designed value
(b) if the supply frequency is increased
(c) both (a) and (b) above
(d) none of the above
74.
transformer is designed for good all day efficiency.
(a) Distribution transformer
(b) Current transformer
(c) High voltage transformer
(d) Power transformer
.
75. Mechanical forces in a transformer are developed due to
(a) interaction of current flowing in the winding and leakage flux around it
(b) vibrations
(c) gap between laminations
(d) none of these
76. The overload capacity in a transformer depends on
.
(a) supply frequency
(b) core size
(c) both (a) and (b)
(d) none of these
transformer, use of higher leakage reactance is permitted.
77. In
(a) current transformer
(b) instrument transformer
(c) distribution transformer
(d) power transformer
78. In a transformer having a higher leakage reactance leads to an advantage of
.
(a) reducing the magnetizing current
(b) improving the voltage regulation
(c) limiting the inrush current during a short circuit
(d) none of the above
79. The leakage reactance of a transformer depends on
.
(a) configuration of the winding
(b) number of turns
(c) frequency
(d) all the above
.
80. The leakage reactance of a transformer is
(a) proportional to square of number of turns
(b) directly proportional to number of turns
(c) inversely proportional to number of turns
(d) proportional to inverse square of number of turns
3.112 Design of Transformer
81. Addition of cooling tubes to the transformer tank improves heat dissipation capacity
because of
.
(a) additional dissipation by convection
(b) additional dissipation by radiation
(c) additional cooling surface
(d) all the above
82. A conservator tank along with the main tank of a transformer is mostly adopted
.
(a) to prevent formation of sludge in the main tank
(b) to improve the cooling
(c) to keep the oil in reserve
(d) to facilitate the periodical check up of the oil
83.
can be adopted for transformer cooling.
(a) Animal oil
(b) Vegetable oil
(c) Mineral oil
(d) Any oil
84. Heat dissipation by means of radiation in oil immersed transformers with cooling
tubes is
.
2
(a) about 6.0 W/m /°C
(b) about 2 W/m2/°C
(c) about 10 W/m2/°C
(d) about 50 W/m2/°C
85. The heat dissipation capacity of transformer exceeding 50 kVA rating is increased by
providing
.
(a) fins
(b) tubes
(c) radiator tanks
(d) corrugations
(e) all of these
86. Oil used in cooling of transformer should have
.
(a) low viscosity
(b) low dielectric strength
(c) low flash point
(d) none of the above
.
87. Transformer oil should be devoid of
(a) sulphur
(b) moisture
(c) acids
(d) all the above
88. Transformer oil should possess
.
(a) high flash point
(b) high dielectric strength
(c) high viscosity
(d) both (a) and (b) above
89. How much heat can be dissipated by natural means from the plain walled tank of a
transformer?
(a) 8.78W/m2/°C
(b) 3.72 W/m2/°C (c) 6 W/m2/°C
(d) 12.5 W/m2/°C
Answers
1. (b) 2. (a) 3. (b) 4. (b) 5. (c) 6. (a) 7. (c) 8. (d) 9. (d) 10. (c) 11. (b)
12. (d) 13. (d) 14. (c) 15. (a) 16. (a) 17. (c) 18. (c) 19. (a) 20. (c) 21. (a) 22. (b)
Review Questions 3.113
23. (c)
34. (c)
45. (b)
56. (b)
67. (c)
78. (c)
89. (d)
24. (d)
35. (b)
46. (a)
57. (b)
68. (b)
79. (d)
25. (c)
36. (d)
47. (a)
58. (c)
69. (b)
80. (a)
26. (d)
37. (a)
48. (c)
59. (b)
70. (b)
81. (d)
27. (d)
38. (a)
49. (b)
60. (b)
71. (a)
82. (a)
28. (a)
39. (c)
50. (a)
61. (a)
72. (a)
83. (c)
29. (d)
40. (b)
51. (b)
62. (d)
73. (c)
84. (a)
30. (d)
41. (d)
52. (b)
63. (c)
74. (a)
85. (e)
31. (c)
42. (c)
53. (c)
64. (c)
75. (a)
86. (a)
32. (c)
43. (c)
54. (b)
65. (d)
76. (b)
87. (d)
33. (b)
44. (d)
55. (a)
66. (c)
77. (d)
88. (d)
Short Type Questions
Difficulty level – Easy
1. What are the major components of a transformer?
Refer Section 3.1.
2. What are the functions of various parts of transformer?
Refer Table 3.1.
3. Explain the classification of transformers.
Refer Sections 3.1.1 to 3.1.4.
4. Compare core and shell-type transformer.
Refer Table 3.2.
5. Compare three-phase transformer bank and three-phase transformer.
Refer Table 3.3.
6. Mention the reasons for the use of tertiary winding in transformer.
Refer Section 3.1.4.
7. Mention the specifications of transformer.
Refer Section 3.2.
8. What are the steps involved in the design of transformer?
Refer Section 3.3.
9. Give the output equation of single phase core-type and shell-type transformer.
Refer Sections 3.3.2 and 3.3.3.
10. Give the output equation of three phase core-type and shell-type transformer.
Refer Sections 3.3.5 and 3.3.6.
11. Provide the expression for volt per turn of winding.
Refer Section 3.4.
12. What are the values of ‘K’ for various types of transformers?
Refer Table 3.4.
13. Explain about choice of flux density in a transformer.
Refer Section 3.5.
14. What are the values of flux densities for various types of transformers?
Refer Table 3.5.
3.114 Design of Transformer
15. Explain about choice of current density in a transformer.
Refer Section 3.6.
16. What are the values of current densities for various types of transformers?
Refer Table 3.6.
17. What are the types of laminations used in core and shell-type transformers?
Refer Fig. 3.17.
18. What are types of cores used in transformer?
Refer Fig. 3.18.
19. Define stacking factor.
Refer Section 3.7.
20. Provide the ratios of net and gross area to area of circumscribing circle for different
types of cores.
Refer Table 3.7.
21. Explain about design of yoke.
Refer Section 3.8.
22. Provide the expression for area of window.
Refer Section 3.9.
23. Give the expressions for overall dimensions of single phase core and shell-type
transformers.
Refer Sections 3.10.1 and 3.10.3.
24. Give the expressions for overall dimensions of three phase core and shell-type
transformers.
Refer Sections 3.10.2 and 3.10.4.
25. What are the different types of windings used in transformer?
Refer Fig. 3.28.
26. Explain about the choice of windings for different types of transformer.
Refer Table 3.8.
27. Provide the expression for resistance of winding.
Refer Section 3.12.
28. Provide the expression for reactance of winding.
Refer Section 3.13.
29. Give the expression for no load current in transformer.
Refer Section 3.15.
30. Provide the expression for magnetizing volt-ampere in a transformer.
Refer Page 3.63.
31. Provide the alternate expression for magnetizing current in a transformer.
Refer Page 3.63.
Review Questions 3.115
32. Mention the different losses in a transformer.
Refer Section 3.16.
33. Provide a brief summary on effects of variation of frequency on core loss, voltage,
leakage reactance and resistance of transformer.
Refer Table 3.9.
34. Provide the condition for obtaining minimum volume of transformer.
Refer Section 3.18.
35. Provide the condition for obtaining minimum weight of transformer.
Refer Section 3.18.
36. Provide the condition for obtaining minimum losses of transformer.
Refer Section 3.18.
37. What are the various types of cooling methods for a transformer?
Refer Fig. 3.36.
38. Mention the heat transfer methods in various regions of transformer.
Refer Table 3.10.
39. Mention the typical clearance values used in providing cooling arrangements of a
transformer.
Refer Table 3.11.
40. Explain about the cause and effect of axial and radial force in a transformer.
Refer Table 3.12.
Difficulty level – Medium to hard
1. What are the magnetic materials used for the magnetic frame?
(i) Hot-rolled silicon steel; (ii) Cold-rolled grain-oriented silicon steel.
2. Which magnetic material used for the magnetic frame has better magnetic properties?
Cold-rolled grain-oriented silicon steel.
3. Mention the reason for using five limb construction for three phase core-type
transformer.
Five limb construction is used so as to reduce the height of the assembled transformer
and thereby to overcome the transportation limitations.
4. What are the effective requirements of a properly designed core of a transformer?
The key requirements of a properly designed core include that of having minimum
iron losses and no load current.
5. What are the essential properties of transformer steel?
The transformer steel should exhibit
(i) high permeability
(ii) high resistivity
(iii) low coercive force
3.116 Design of Transformer
6. Provide the limiting values of flux density used with hot-rolled steel and cold-rolled
grain-oriented steel.
(i) hot-rolled steel – 1.4 Tesla
(ii) cold-rolled grain-oriented steel – 1.7 Tesla
7. Why transformers employ stepped core?
The following are the reasons for employing stepped core:
(i) Low voltage and high voltage coils are circular, providing better utilization of
space
(ii) For reducing the mean length of low voltage and high voltage turns, resulting in
saving of copper material.
8. Mention the merits of using comparatively higher flux density in the core.
(i) Decrease in core and yoke cross-sectional area for same output.
(ii) Reduction in mean length of low voltage and high voltage turns resulting in
saving of copper material.
(iii) Reduced overall size and weight of the transformer.
(iv) Decrease in overall cost of the transformer.
9. Why low voltage and high voltage windings employ circular coils?
(i) To attain better mechanical strength.
(ii) To lessen the mean length of low voltage and high voltage turns, leading to copper
saving.
10. Mention the types of windings generally used for low voltage winding.
(i) Helical winding.
(ii) Cylindrical winding with rectangular conductors.
11. What are the types of winding that are generally used for high voltage winding?
(i) Cylindrical winding with circular conductor.
(ii) Continuous disc type winding.
(iii) Cross-over winding.
12. Mention the demerits of sandwich winding.
(i) Difficulty in manufacturing due to requirement of high labour.
(ii) Less stability under short circuit conditions.
(iii) Difficulty in insulating different coils from each other and from the yoke.
13. Why are stranded conductors used for the winding in transformer?
Stranded conductors are used for the winding in transformer to reduce the eddy
current losses within the conductor.
14. Mention the insulating materials that are used in transformers.
• Press board
• Porcelain
• Cable paper
• Insulating varnish
Review Questions 3.117
• Varnished silk
• Bakelite
• Transformer oil
15. What are the essential properties of insulating materials used in transformer?
(i) Sufficient mechanical strength to withstand the stresses.
(ii) High dielectric strength to prevent break down.
(iii) Good oil absorbing capability.
16. Mention the reason for using different current densities in low voltage (l.v.) and high
voltage (h.v.) windings.
L.v. winding is an inner winding, placed on the core, in which cooling is less compared
to h.v. winding, which is the outer winding.Thus, l.v. winding should be designed
with a comparatively lesser current density than the h.v. winding.
17. Why stranded conductors are required to be transposed?
It is essential to transpose the stranded conductors, to equalize the reactance of the
strands.
18. Define interleaved winding.
Interleaved winding is a type of winding in which l.v. and h.v. coils are arranged
alternately one over the other along the height of the limb.
19. What is the necessity for providing tappings in a transformer?
(i) To maintain practically constant voltage at consumer’s end,
(ii) To control active and reactive power over a transmission line interconnecting two
generating stations,
(iii) To provide voltage regulation.
20. Provide the percentage tappings, generally employed in transformers.
2.5% and 5%
21. Mention the advantages of providing tapping in h.v. winding.
(i) Precise voltage regulation, because of presence of larger number of turns.
(ii) Ease of providing tappings, as the h.v. winding is an outer winding.
(iii) Relatively lower current is interrupted, while the taps are changed.
22. What are the different methods of changing the tappings in a transformer?
(i) On load tap changing
(ii) Off load tap changing
23. What are the properties required to be attained in designing the windings of
transformer?
(i) Improved electrical performance characteristics
(ii) Sufficient mechanical strength
(iii) Proper ventilation
24. In which type of transformer leakage reactance can be varied over a wide range?
Shell-type transformer with sandwich winding.
3.118 Design of Transformer
25. Why power transformers are designed to have maximum efficiency at or near fullload
whereas distribution transformers are designed to have maximum efficiency at loads
quite lower than the full load?
Power transformers are designed to have maximum efficiency at or near full load
whereas distribution transformers are designed to have maximum efficiency at loads
quite lower than the full load, since power transformers are allowed to work at or
near full load and switched off during light load hours, but distribution transformers
are required to work all the time including light load hours.
26. Why circular coils are always preferred over rectangular coils for winding a
transformer?
Circular coils have the following advantage over rectangular coils:
• Higher mechanical stability under short circuit conditions, due to the presence
of radial forces and there is no tendency for the coil to change its shape due to
mechanical forces
• Proper utilization of space
• Ease of winding
27. Why is the area of yoke of a transformer usually kept 15–20% more than that of core?
By maintaining the area of yoke 15–20% more than the area of core, the yoke flux
density is reduced, and hence reducing the iron losses in the yoke portion.
28. Why are mittred joints preferred in transformer core?
The mittred joints make the flux lines to flow along the direction of grain orientation,
thereby reducing the iron loss and the no load current.
28. Why l.v. winding is placed first on the core and then h.v. winding in case of a coretype
transformer?
By placing l.v. winding near the core, the insulation thickness reduces resulting into
ease in insulation. It further reduces the length of mean turn of conducting material.
29. What are the types of transformer according to service conditions?
(i) Distribution transformer
(ii) Power transformer
30. Why sandwiched winding is preferred for a distribution transformer?
Sandwich winding is preferred to reduce the leakage reactance in order to have a
good voltage regulation.
31. In case of a power transformer, a higher leakage reactance is not considered a
disadvantage – Justify.
Since higher leakage reactance can limit the rush of current during a short circuit, a
power transformer of a higher leakage reactance is not considered a disadvantage.
32. Why a stepped core is used in transformer?
Stepped core is used to approximate the core section near to a circle, therefore reduces
the length of mean turn leading to reduction in the amount of copper.
33. Justify the use of cross-over coils for winding h.v. side of a transformer.
Review Questions 3.119
The h.v. winding has larger number of turns, which when employed with a cross-over
winding can be easily divided into a number of coils depending upon the voltage
rating. Insulation spacers can then separate every coil, which assists in free circulation
of oil.
34. Mention the reason for use of stranded conductors instead of a single conductor of
large cross-section.
Stranded conductors are preferred due to reduction in eddy current loss, compared to
a single conductor of large cross-section.
35. Why stepping is not employed in yoke?
Stepping is required in portions where winding is present. As yoke portions do not
carry any winding, it is not stepped.
36. Why an elaborate clamping and tightening arrangement is required in transformer
core?
Clamping and tightening arrangement is required to overcome the large amount of
mechanical forces developed during short circuit conditions.
37. What is the range of efficiency of a commercially available transformer?
The efficiency of a commercially available transformer will be in the range of 94% to
99%.
38. Define a transformer bank.
A transformer bank consists of three independent single phase transformers with
their primary and secondary windings connected either in star or in delta.
39. How does the design of distribution transformer differ from the design of a power
transformer?
• The distribution transformers are designed in such a way that copper loss will be
higher than iron loss, whereas in power transformers the copper loss will be lesser
than the iron loss.
• The distribution transformers are designed to have the maximum efficiency at a
load much lesser than full load, whereas the power transformers are designed to
have maximum efficiency at or near full load.
• The leakage reactance is kept low in a distribution transformer to have better
regulation, whereas in power transformers the leakage reactance is kept high to
limit the short circuit current.
40. What is the necessity of using sheet steel stampings in the construction of electrical
machines?
The stampings reduce the eddy current losses as the stampings are insulated by a thin
coating of varnish; hence when the stampings are stacked to form a core, the resistance
for the eddy current is very high. (This is due to very small area of crosssection of
laminations.)
41. How the insulation is provided in the laminations of core?
• In laminations made of hot-rolled silicon steel, a thin coating of kaolin or varnish is
applied to insulate them.
3.120 Design of Transformer
• In laminations of cold-rolled silicon steel, a phosphate-based coating is applied to
insulate them.
• In high capacity transformers, above 10 MVA rating in addition to phosphate
coating, a coating of kaolin or varnish is applied.
42. Mention the merits of three-phase transformers over single-phase transformers.
Three-phase transformers have the following advantages compared to single-phase
transformers.
(i) Less weight
(ii) Less cost
43. What are the types of winding commonly employed for LV winding?
(i) Helical winding
(ii) Cylindrical winding with rectangular conductors
44. What are the disadvantages of using higher flux density in design of core of
transformer?
(i) Increased magnetizing current and iron losses.
(ii) Higher temperature-rise of transformer.
(iii) Lower efficiency, because of higher no load losses.
(iv) Saturation of magnetic material.
45. Why the cross-section of core is taken less than yoke?
The cross-section of yoke is larger than core so as to reduce flux density by limiting
the no load current and thereby reducing iron losses.
46. Describe the placement of low voltage and high voltage winding in a single phase
coretype transformer.
On each core, half the number of low voltage and half the number of high voltage turns
are placed.
The low voltage winding is first placed on the core and then the high voltage winding is
placed over the low voltage winding.
47. What is the reason to not use large number of steps in a stepped core?
Even though, the large number of steps reduces the amount of copper used, it
increases the labour charge which may make implementation costlier.
48. Why window space factor is lesser for higher voltage rating transformers?
As the voltage rating increases, larger space is required for insulation and clearances;
hence, the window space factor is lower.
49. Mention the effect on leakage reactance in the construction of narrower window
transformer.
In a transformer with narrow window, the limbs are placed closer thereby the leakage
flux is reduced resulting in the reduction of leakage reactance.
50. Explain the requirement of estimation of Ampere Turn in the design of transformer.
Ampere Turn is estimated to determine the magnetizing current for the transformer.
Review Questions 3.121
51. Mention the reason to add the Ampere Turn required by joints in the estimation of the
total A.T. required for transformer.
As a high portion of the total Ampere Turn links the joints, it is necessary to add it to
the total Ampere Turn.
52. Define window space factor.
The window space factor is defined as the ratio of copper area in the window to the
total area of window.
53. Mention the advantages of stepped cores.
For same cross-sectional area, the stepped cores will have lesser diameter of
circumscribing circle when compared to square cores. This causes reduction in length
of mean turn of the winding with subsequent reduction in costs, i.e., volume of copper
and copper loss.
54. Define copper space factor.
The copper space factor is the ratio of conductor area and window area in case of
transformers.
(In case of rotating machines, it is the ratio of conductor area and slot area.)
55. Mention the factors to be considered in choosing the type of winding for a core-type
transformer.
• Current density
• Impedance
• Temperature rise
• Surge voltage
• Short circuit current
56. Mention the reason for use of delta connection of tertiary winding in a transformer.
The unbalance in phase voltage during unsymmetrical faults in primary or secondary
of transformer is compensated by the circulating currents flowing in the closed delta
connection.
57. How the tank dimensions are fixed based on overall dimensions of transformer frame?
• The dimensions of the tank are decided by the dimensions of the transformer frame
and clearance required on all the sides.
• The clearance on the side depends on voltage and power rating of the winding.
• The clearance at the top depends on the oil height above the assembled transformer
and the space for mounting the terminals and tap changing gear.
• The clearance at the bottom depends on the space required for mounting the
transformer frame inside the tank.
58. Define leg spacing.
Leg spacing is the distance between the centres of two adjacent limbs of a transformer.
It is denoted by D.
59. What are the factors affecting the choice of flux density of core in a transformer?
(i) Core area,
(ii) Core loss and
(iii) Magnetizing current
3.122 Design of Transformer
60. Where does the iron loss take place in a transformer?
(i) Magnetic core
(ii) Yoke
61. Where does the copper loss take place in a transformer?
(i) High voltage winding
(ii) Low voltage winding
62. What is the percentage no load current of transformers?
3 to 5% – small transformer
1 to 3% – medium transformer
0.5 to 1% – large transformer
63. Mention the nominal efficiency of transformer.
97 to 98% – small transformer
98 to 99% – medium transformer
98.5 to 99.4% – large transformer
64. Why the efficiency of a transformer is very high?
(i) Mechanical losses are zero (absent), which forms great part of total losses.
(ii) Iron losses are comparatively less due to use of better magnetic material in
magnetic frame.
65. Why working flux density is on the lower side for distribution transformers?
By maintaining the flux density on the lower side, the iron losses are reduced thereby
improving the all day efficiency which is very important in case of a distribution
transformer.
66. What is the effect of variation of frequency on eddy current loss in transformer core if
the voltage is maintained constant?
The eddy current loss will not change with the effect of variation of frequency in
transformer core with the voltage is maintained constant.
67. How the leakage reactance is affected with the variation of frequency?
The leakage reactance variation is linear with the change in frequency.
68. Mention the relation of output and losses in a transformer with variation in linear
dimensions.
• The output increases as the fourth power with linear dimensions.
• The loss increases as the third power with linear dimensions.
69. How the leakage reactance of a transformer can be reduced?
By interleaving the high voltage and low voltage winding, leakage reactance of a
transformer can be reduced.
70. Mention the cause of noise in transformer.
The causes of noise in the transformer are
• Magnetostriction effect
Review Questions 3.123
• Loosening of stampings
• Mechanical forces produced during operation
71. What are the main functions of cooling medium in transformers?
(i) Transfer of heat by convection from the heated surface to the tank surface.
(ii) Creation of good level of insulation between various conducting parts.
72. How much heat is dissipated by radiation and convection?
Heat dissipated by radiation – 6 W/m2/°C
Heat dissipated by convection – 6.5 W/m2/°C
73. Why elaborate cooling arrangement is required for large transformers whereas
natural cooling is sufficient for small transformers?
The losses are high in large transformers, requiring elaborate cooling arrangement to
limit the temperature rise. As losses are less in small transformers, natural cooling is
sufficient.
74. Why conservator tank has to be necessarily used along with the main tank in a
transformer cooling system?
In order to avoid the contact of air and moisture with the oil of the main tank and to
keep it free from sludge formation, conservator tank has to be necessarily used along
with the main tank in a transformer cooling system.
75. Mention the reason for using breather along with the conservator tank.
The reason for using breather along with the conservator tank is that it is required to
allow the air of the conservator to be driven out or drawn in, for the oil level raise or
fall depending upon the load on the transformer.
76. Mention the factors to be considered for selecting the cooling method of a transformer.
The factors are
• kVA rating
• Size
• Application
• The site condition where it will be installed
77. Why transformer oil is used as a cooling medium?
By using transformer oil as a coolant, the heat dissipation by convection is 10 times
more than the convection due to air which is given by the values of Specific heat
dissipation by convection due to air = 8 W/m2-°C
Specific heat dissipation by convection due to oil = 80 to 100 W/m2-°C
78. Why cooling tubes are provided in a transformer?
Cooling tubes are provided in a transformer to increase the heat dissipating area of
the tank as the cooling tubes will improve the circulation of oil. The circulation of
oil is due to more effective pressure heads produced by columns of oil in tubes. The
improvement in cooling is accounted by taking the specific heat dissipation due to
convection as 35% more than that without tubes.
3.124 Design of Transformer
79. Explain about simple cooling and mixed cooling of transformer.
In simple cooling, the transformer is cooled by a particular method of cooling. In
mixed cooling, the transformer is provided with two or more cooling methods and
at a particular point of time, a single cooling method is employed depending on the
amount of heat to be removed.
80. Why in mine applications transformers, oil cooling should not be used?
The oil used for transformer cooling is inflammable (i.e., can easily set on fire), which
can cause leakage of cooling oil leading to fire accidents in mines.
81. What is breather?
The breather is a device fitted in transformer for breathing. In small oil cooled
transformers, some air-gap is provided between the oil level and tank top surface.
When the oil is heated, it expands and air is expelled out of transformer to the
atmosphere through breather. When the oil is cooled, it shrinks and air is drawn from
the atmosphere into the transformer through breather. This action of transformer is
called breathing.
82. Why silica gel is used in breather of transformer?
The silica gel absorbs the moisture when the air is drawn from atmosphere into the
transformer, while cooing is employed.
83. What is conservator in a transformer?
A conservator is a small cylindrical drum fitted just above the transformer main
tank. It is used to allow the expansion and contraction of oil without contact with
surrounding atmosphere.
When conservator is fitted in a transformer, the tank is fully filled with oil and the
conservator is half filled with oil.
84. What are the functions of transformer oil?
Transformer oil behaves as a cooling medium and provides considerable level of
insulation.
85. Why oil immersed transformers employ bushings?
Bushings make necessary connections to external circuit with windings.
86. Why bracing is employed in case of large transformers?
Bracing is required to take care of the large amount of mechanical forces developed
during short circuit conditions.
Long Type Questions
1. Explain the constructional features of the transformer.
2. Derive the output equation of 1φ core-type and shell-type transformers.
3. Derive the output equation of 3φ core-type and shell-type transformers.
4. Obtain the expression for ratio of gross core area to circumscribing circle’s area for
(a) square core and (b) stepped core.
5. Derive the expression for magnetizing current of 1φ transformer.
Review Questions 3.125
6. Derive the expression for magnetizing volt ampere and magnetizing current of 3φ
transformer.
7. Explain about the types of losses in transformer.
8. What are the effects of frequency in parameters of transformer?
9. Obtain the expression for optimum design of transformer based on minimum cost,
minimum volume, minimum weight and maximum efficiency.
10. Explain the various methods of cooling of transformer.
11. Obtain the expression for design of cooling tank with tubes and give the various
clearance values.
Problems
1. Determine the main dimensions of a 300 kVA, 3φ, 50 Hz Y/Δ, 11,000/3300 V, coretype distribution transformer. Assume distance between core centres as twice the
width of core.
2. A three-phase, 50 Hz oil-cooled core-type transformer has the following dimensions.
Distance between core centres = 0.25 m, height of window = 0.30 m, diameter of
circumscribing circle = 0.14 m. The flux density in the core = 1.1 Wb/m2, current
density in conductor = 2 A/mm2. Assume a window space factor of 0.2 and core area
factor of 0.56. The core is two stepped. Estimate the kVA rating of transformer.
3. Find the dimensions of the core and yoke of a single-phase core-type transformer
of rating 200 kVA, 50 Hz. Given the volt per turn as 15 V, maximum flux density as
1.1 Wb/m2, the window space factor as 0.3 and current density as 2.6 A/mm2. Design
a square core with distance between the adjacent limbs equal to 1.45 times the width
of the transformer. Assume a stacking factor of 0.8 and flux density in the yoke is 80%
flux density in the core.
4. Find the no load current of a 250/100 V, 2 kVA, 50 Hz single-phase transformer with
the following data. Cross-sectional area of core = 25 cm2, effective magnetic core
length = 0.45 m, weight of core = 8 kg, maximum flux density = 1.5 T, magnetizing
mmf = 250 AT/m and specific core loss = 1.9 W/kg.
5. Design a suitable cooling system for a 400 kVA, 6000/400V 50 Hz, 3φ transformer
with a total full load loss of 6 kW. The transformer tank is 1 m in height, 0.96 m
in length and 0.47 m in breadth. Use cooling tubes of diameter 50 mm to limit the
average temperature rise to 37°C. Use clearance of 50, 14 and 13 cm on the height,
length and width sides.
6. A 3φ,15 MVA, 30/6.6 kV 50 Hz, transformer has the following values on initial
design. Length of transformer is 160 cm, height of transformer is 250 cm and the
width is 78.5 cm. Use clearances of 50 cm, 11.5 cm and 11.5 cm, respectively, on the
height, width and length for designing the tank. Total iron loss is 20 kW and 110 kW is
the copper loss. Calculate the temperature rise of transformer without cooling tubes.
Calculate the number of tubes to limit the temperature rise to 55°C.