Transformer Design

Introduction
Transformer is static device which converts the voltage or current from one level to
another level keeping the power and frequency constant. The transformer are required to
transport the power from the generating station to load center which is far away from the
generating station. Its based on the faraday’s law of electromagnetic induction which states
whenever the coil experiences the changing flux and emf is induced .

𝒅∅
𝒆 = −𝑵
𝒅𝒕
Where N is number of turns in the coil . The negative sign implies that the nature of the
induced emf is such that it opposes the cause which produced it.

When a sinusoidal voltage is applied to the primary winding of a practical transformer on an iron
core with the secondary winding open, a small current In called the exciting current flows. The
major component of this current is called the magnetizing current , which corresponds to the
current through the magnetizing susceptance Bm . The magnetizing current produces the flux in
the core. The much smaller component of Ih+e which accounts for losses in the iron core, leads
the magnetizing current by 90° . The core losses occur due , first, to the fact that the cyclic
changes of the direction of the flux in the iron require energy which is dissipated as heat and is
called hysteresis loss . The second loss is due to the fact that circulating currents are induced in
the iron due to the changing flux, and these current produce an I2R loss in the iron called eddy-
current loss . Hysteresis loss is reduced by the use of certain high grades of alloy steel for the
core and Eddy current loss is reduced building the core with laminated sheet When the
secondary is open then small amount of primary current will flow to establish the flux in core .
The Equivalent circuit consist only magnetising inductance and resistance for the core loss
component .The no load condition and its phasor diagram is shown in below figure
In a well -designed transformer the maximum flux density in the core occurs at the knee of the
B-H or saturation curve of the transformer. So, flux density is not linear with respect to field
intensity . The magnetizing current cannot be sinusoidal if it is to produce sinusoidally varying
flux required for inducing sinusoidal voltages e1 and e2 when the applied voltage is sinusoidal .

The exciting current Inwill have a third harmonic content as high as 40% and lesser amounts of
higher harmonics..

Load Component Current

When the secondary circuit of the Transformer is connected to the load then the load current
develops another flux which opposes the main flux hence to establish the constant flux in
side the core primary will draw the load component current .
Impedance Transfer
If we define the ratio of the primary voltage to secondary voltage is ‘a’ then the load
transferred from secondary to primary will be
𝑽𝟏
𝑰𝟏
𝒁𝟏
= = 𝒂𝟐 ∴ 𝒁𝟏 = 𝒂𝟐 𝒁𝟐
𝒁𝟐 𝑽𝟐
𝑰𝟐
The Exact Equivalent circuit of the single phase transformer is shown in Below Fig

Where equivalent resistance and reactance referred the primary side are
 𝑹𝟎𝟏 = 𝑹𝒑 + 𝒂𝟐 𝑹𝒔 𝒂𝒏𝒅 𝑿𝟎𝟏 = 𝑿𝒑 + 𝒂𝟐 𝑿𝒔

Emf Equation
Its based on the faraday’s law of electromagnetic induction which states whenever the coil
experiences the changing flux and emf is induced .
𝒅∅
𝒆 = −𝑵
𝒅𝒕
Where N is number of turns in the coil . The negative sign implies that the nature of the
induced emf is such that it opposes the cause which produced it.
Let the
 =m Cost

𝒅𝐦 𝐂𝐨𝐬𝐭
𝒆 = −𝑵 = 𝐍𝐦𝐒𝐢𝐧𝐭 = 𝟐𝛑𝐟𝐍𝐦𝐒𝐢𝐧𝐭
𝒅𝒕
Hence the rms value of the induced emf
𝟐𝛑𝐟𝐍𝐦𝐒𝐢𝐧𝐭
𝒆= = 𝟒. 𝟒𝟒𝒇𝑵
√𝟐
So the induced emf lags the main flux by 900
Phasor Diagram
The complete phasor diagram of the transformer for the lagging load representing all
components will be as shown below

2
I1
E2
I2/a
-E1
V2
I2
V1

Where
𝑬𝟐 = 𝑽𝟐 + 𝑰𝟐 (𝑹𝟐 + 𝒋𝑿𝟐 )
𝑰𝟐
𝑰𝟏 = + 𝑰𝒏
𝒂
𝑽𝟏 = 𝑬𝟏 + 𝑰𝟏 (𝑹𝟏 + 𝒋𝑿𝟏 )
Types of Transformer
According two the construction the transformer are two types one is core type whereas other
is shell type . In the core type the core is inside and the winding are outside . But in shell type
the windings are in side and the core is out side as shown in below figure

Core Type Shell Type

According to the service

The transformer are two types one is power transformer which are placed to transport the
power from generating station to Grid and another is distribution transformer which are
located near to the load center . The design criteria for the power and the distribution
transformer will be different
Power transformer
1. Load on the transformer will be at or near the full load through out the period of operation.
When the load is less, the transformer, which is in parallel with other transformers, may be put
out of service.
2. Generally designed to achieve maximum efficiency at or near the full load. Therefore iron loss
is made equal to full load copper loss by using a higher value of flux density. In other words,
power transformers are generally designed for a higher value of flux density.
3. Necessity of voltage regulation does not arise .The voltage variation is obtained by the help of
tap changers provided generally on the high voltage side. Generally Power transformers are
deliberately designed for a higher value of leakage reactance, so that the short-circuit current,
effect of mechanical force and hence the damage is less.
Distribution transformer
1. Load on the transformer does not remain constant but varies instant to instant over 24 hours a
day
2. Generally designed for maximum efficiency at about half full load. In order that the all day
efficiency is high, iron loss is made less by selecting a lesser value of flux density. In other
words distribution transformers are generally designed for a lesser value of flux density. Since
the distributed transformers are located in the vicinity of the load, voltage regulation is an
important factor.
3. Generally the distribution transformers are not equipped with tap changers to maintain a
constant voltage as it increases the cost, maintenance charges etc., Thus the distribution
transformers are designed to have a low value of inherent regulation by keeping down the value
of leakage reactance.
According to Phase
Single phase and three phase transformer are available .The transformers may be
inherently 3-phase,having three primary windings and three secondary windings mounted on a 3-
legged core. The same result can be achieved by using three single-phase transformers connected
together to form a 3-phase transformer bank. Generally in the large power station three single
phase transformer are connected to build three phase transformer mainly due to
transportation and reliability . The open delta connection of the three phase transformer can
handled three phase system in derated state

Output Equation
It gives the relationship between electrical rating and physical dimensions of the machines.
Let
V1 = Primary voltage say LV
V2 = Secondary voltage say HV
I1 = Primary current
I2 = Secondary current
N1= Primary no of turns
N2= Secondary no of turns
I
a1 = Sectional area of LV conductors (m2) = 1

I
a1 = Sectional area of HV conductors (m2) = 2

2
 = Permissible current density (A/m )
Q = Rating in KVA
We place first half of LV on one limb and rest half of LV on other limb to reduce leakage flux.
So arrangement is LV insulation then half LV turns then HV insulation and then half HV turns.

For 1-phase core type transformer

H L Window
L H H L L H
V V V V V V V V

1-phase core type transformer with

concentric windings
Given by
Q = V1 I1  103 KVA
= 4.44 fm N1 I1  103 KVA  V1  4.44 fm N1 
= 4.44 fAi B m N 1 I 1  10 3 KVA -----------(1) (  m  Ai Bm )
Where
f = frequency
m = Maximum flux in the core
Ai = Sectional area of core
Bm = Maximum flux density in the core

Window Space Factor

Actual Cu Section Area of Windings in Window

Kw 
Window Area ( Aw )
a N  a2 N 2
 1 1
Aw
( I /  ) N1  ( I 2 /  ) N 2
 1 ( a1  I /1  & a2  I / 2  )
Aw
I N I N
 1 1 2 2
Aw

2I1 N1
 ( For IdealTransformer I1N1  I 2 N 2 )
Aw
So
 K w Aw 
 N1 I 1                ( 2)
 2 
Put equation value of N1I1 form equation (2) to equation (1)

 K w Aw
Q  4.44 f A i Bm  10  3 KVA
2
Q  2.22 f A i Bm  K w Aw  10 3 KVA          (3) Window

For 1- phase shell type transformer

LV LV
Window Space Factor HV HV

a1 N1  a2 N 2 LV LV
Kw 
Aw
HV HV
( I /  ) N1  ( I 2 /  ) N 2
 1 ( a1  I /1  & a2  I / 2  ) LV LV
Aw

1-phase shell type transformer with

sandwich windings
 I 1 N1  I 2 N 2

Aw
2I N
 1 1 ( For Ideal Transformer I1 N1  I 2 N 2 )
Aw
So
K w Aw
N 1 I1               ( 4)
2
Put equation value of N1I1 form equation (4) to equation (1)
 K w Aw
Q  4.44 f A i Bm  10  3 KVA
2
Q  2.22 f A i Bm  K w Aw  10 3 KVA          (5)
Note it is same as for 1-phase core type transformer i.e. equ (3)

For 3-phase core type transformer

Rating is given by
Q = 3 V1I1 103 KVA
= 3  4.44 fm N1 I1  10 KVA
3
 V1  4.44 fm N1 
= 3  4.44 fAi Bm N1 I1  10 3
KVA-----------(6) (  m  Ai Bm )

Window

H L L H H L L H H L L H
V V V V V V V V V V V V

3-phase core type transformer with concentric

windings
Window Space Factor

Actual Cu Section Area of Windings in Window

Kw 
Window Area ( Aw )
 2(a1 N1  a2 N 2 )

Aw
2  ( I1 /  ) N1  ( I 2 /  ) N 2 
 ( a1  I /1  & a2  I / 2  )
Aw
2 ( I1 N 1  I 2 N 2 )

Aw
2  2 I1 N 1
 ( For Ideal Transformer I1 N1  I 2 N 2 )
Aw
So
 K w Aw
N 1 I1               (7 )
4
Put equation value of N1I1 form equation (7) to equation (6)
 K w Aw
Q  3  4.44 f A i Bm  10  3 KVA
4 Windo
Q  3.33 f A i B m  K w Aw  10 3 KVA      (8)

For 3- phase shell type transformer

Window Space Factor

a1 N1  a2 N 2
Kw 
Aw
( I /  ) N1  ( I 2 /  ) N 2
 1 ( a1  I /1  & a2  I / 2  )
Aw
I N I N
 1 1 2 2
Aw
2I N
 1 1 ( For Ideal Transforme r I1 N1  I 2 N 2 )
Aw
So 3-phase shell type transformer
K w Aw with sandwich windings
N1 I 1               (9 )
2
Put equation value of N1I1 form equation (9) to equation (6)
 K w Aw
Q  3  4.44 f A i Bm  10  3 KVA
2
Q  6.66 f A i Bm  K w Aw  10 3 KVA          (10)

Choice of magnetic loading (Bm)

 Value of flux density in transformer determines core area and yoke area hence the
computational of flux density is very important and crucial part of the design
 Normally flux density is chosen at the knee point of the magnetization curve
  Magnetic material used for the transformer core are hot rolled silicon steel or cold
rolled grain oriented silicon steel
 Choice of flux density effects on the performance parameters such no load current ,
losses , temperature raise and behavior during short circuit
 Higher value of flux density reduce core cross-section which further reduces the
copper material ie reduces the cost and weight
 But higher value of flux density increases iron loss and temperature raise
 Flux density to be chosen depends upon service condition . For distribution
transformer main design aspect is to have high all day efficiency for which
requirements is low iron loss ie low value of flux density
(1) Normal Si-Steel 0.9 to 1.1 T
(0.35 mm thickness, 1.5%—3.5% Si)
(2) HRGO 1.2 to 1.4 T
(Hot Rolled Grain Oriented Si Steel)
(3) CRGO 1.4 to 1.7 T
(Cold Rolled Grain Oriented Si Steel)
(0.14---0.28 mm thickness)

Choice of Electric Loading  

 The conductor cross-section for the l.v. and h.v. Winding are determine after
choosing the suitable value of the current density
 Temperature raise is high if the higher value of the current density is chosen
 current density is significant for the I2R loss hence the load at which maximum
efficiency depends on it
 The level of the I2R loss for the power and distribution transformer are different so
the current density will be different
 This depends upon cooling method employed

(1) Natural Cooling: 1.5---2.3 A/mm2

AN Air Natural cooling
ON Oil Natural cooling
OFN Oil Forced circulated with Natural air cooling

(2) Forced Cooling : 2.2---4.0 A/mm2

AB Air Blast cooling
OB Oil Blast cooling
OFB Oil Forced circulated with air Blast cooling

(3) Water Cooling: 5.0 ---6.0 A/mm2

OW Oil immersed with circulated Water cooling
OFW Oil Forced with circulated Water cooling
Core Construction:

(b) E-I type

(a) U-I type

(c) U-T type 45o

(d) L-L type

(e) Mitred Core Construction (Latest)

EMF per turn:

We know
V1  4.44 fm N1              (1)
V1
So EMF / Turn Et   4.44 fm            (2)
N1
and

Q = V1 I1 103 KVA (Note: Take Q as per phase rating in KVA)

= 4.44 f m N1 I 1  10 3
KVA
 E t N 1 I 1  10 3 KVA        (3)
In the design, the ratio of total magnetic loading and electric loading may be kept constant
Magnetic loading = m
Electric loading = N1 I1
m m
So  cons tan t ( say " r " )  N 1 I 1  put in eqution (3)
N 1 I1 r
m
Q  Et  103 KVA
r
Et
Or Q  Et  10 3 KVA using equation (2)
4.44 f r
2
E t  ( 4 .44 f r  10 3 )  Q

Or Et  K t Q Volts / Turn

3
Where Kt  4.44 f r 10 is a constant and value depend upon service type , construction
and cooling system For same cooling / current density / same type – three phase distribution
transformer will be more heated than power transformer ( distribution transformer are
continuously loaded) so the value of K will be more in case of power transformer Similarly for
same cooling system and current density / for Shell type cooling system is more effective
compared to core type so larger K can be chosen in shell type compared to core type

Kt = 0.6 to 0.7 for 3-phase core type power transformer

Kt = 0.45 for 3-phase core type distribution transformer
Kt = 1.3 for 3-phase shell type transformer

Kt = 0.75 to 0.85 for 1-phase core type transformer

Kt = 1.0 to 1.2 for 1-phase shell type transformer

Estimation of Core X-sectional area Ai

 The core section of the core type transformer may be rectangular , square or stepped.
 Shell type transformers use cores with rectangular cross section. For the rectangular
core the ratio of depth to width 1.4 to 2.0.
 Square or stepped core When circular coils are required for high voltage transformers ,
square and stepped cores are used . Circular coils are preferred because of their superior
mechanical characteristics.
 As the size of the transformer increases, it becomes wasteful to use rectangular cores.
Square cores are used and the surrounding circle, representing the inner surface of the
tubular form carrying the windings, is called the circumscribing circle. Even now a lot of
useful space is wasted and the length of the mean turn increases causing higher I2R
losses.
 With larger transformers, cruciform cores, with better utilization of the space, are used. It
should be born in mind that two different types of laminations are used in cruciform
cores. With still larger transformers, further step sizes are introduced to utilize the core
  even more effectively. However, larger step sizes → larger number of lamination sizes →
higher labor cost.

Core Cross-section
Small core type of transformer gas rectangular core limbs with rectangular coils as shown
below figure however for the larger power transformer economic use of the core material
required so the circular core limb is needed since the circular core has less periphery
than rectangular core which reduced the length of copper . The circular core needs
large no of different size of laminated sheet . Hence the two step or multistep core are
generally used .

We know
Et  K t Q           (1)
Et  4.44 f m
Or Et  4.44 f Ai Bm           ( 2)
Et
So Ai            (3)
4.44 f Bm

a
Two Step Core:
Two Step core is also called cruciform Core . b
Let d is the diameter of the circumscribing circle then
Here a = d cos and b - dsin
Gross Area of the iron Ag=2ab-b2 = 2d2SinCos-d2sin2
Hence for Maximum Gross area =0 Ө
0 b
Which gives = 31.45
Hence a=dCos31.450 =0.81d and b= dSin31.450 =0.53d
Maximum Gross iron area =2ab-b2=0.618d2
Let stacking factor K=0.9
Then Net iron Area = 0.9Ag= 0.9x0.618d2=0.56d2 2-Step
 Ratio
.
= = 0.71 𝑎𝑛𝑑 =
.
= 0.79

Now the core may be following types

d d/√2

d = Diameter of circumscribe circle

For Square core
d d
Gross Area    0 .5 d 2
2 2
Let stacking factor
Ki  0.9
Actual Iron Area
Ai  0.9  0.5d 2
 0.45 d 2 (0.45 for square core and take ‘K’ as a general case
 K d2
Ai
So Ai  K d 2 Or d
K
Window Space Factor (Kw)
The window space factor is defined as the ration of the copper area in the window to total
window area. It depends upon the relative amounts of insulation and copper provided ,which
is in turn depends upon the voltage rating and the output of the transformer . The following
empirical formulas are used to determine the windo space factor

Kw=Window space factor

8
Kw  for upto 10 KVA
30  HigherKV
10
Kw  for upto 200 KVA
30  HigherKV
12
Kw  for upto 1000 KVA
30  HigherKV
For higher rating Kw = 0.15 to 0.20

Window Dimension
The leakage reactance is affected by the distance between adjacent limbs . if this distance
is relatively small width of the winding is limited this must be counter balance by
increasing the height of the window or winding ,thus windings are long and thin . This
arrangement give low value of the leakage reactance. If the height is limited then width of
the window must increase to accommodate windings this results short and wide coils
resulting high value of leakage . Hence the height and width of the window can be
adjusted for the suitable arrangement of windings as well as low value of the leakage

The area of the window depends upon the total copper area and window space factor

Area of the window 𝐴 =

2𝑎 𝑇
𝐴 = 𝑓𝑜𝑟 𝑠𝑖𝑛𝑔𝑙𝑒 𝑝ℎ𝑎𝑠𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝐾
4𝑎 𝑇
𝐴 = 𝑓𝑜𝑟 𝑇ℎ𝑟𝑒𝑒 𝑝ℎ𝑎𝑠𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
𝐾
Area of the Window 𝐴 = ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑤𝑖𝑛𝑑𝑜𝑤 𝑥 𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑤𝑖𝑛𝑑𝑜𝑤 = 𝐻 𝑥𝑊
𝐻
Assuming suitable value for ratio 𝑊 the height and width of the window will be
calculated

Width of the window for optimum output

Let D is the Distance Between the Adjcent Limbs Then D must includes width of iron
,winding, insulation , clearance and ducts for cooling D= width of iron + width of copper
+ width of insulation + width of ducts + width of clearance .let m is the width occupied by
insulation ,ducts and clearance

Then width of iron and copper 𝐷 = 𝐷 − 𝑚 . So width occupied by bare conductor in

window is =𝐷 − 𝑑
Let S be the volt ampere (VA) per unit height of the transformer winding then
𝑆 = 𝐸 𝑇 𝐼 where Th is the number of turn per unit height
𝐸 = 4.44𝑓∅ = 4.44𝑓𝐵 𝐴 = 4.44𝑓𝐵 𝑘 𝑑
For a fixed value of frequency and flux density Et is proportional to d2
Now 𝐼𝑇 = 𝑚𝑚𝑓 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑎𝛿𝑇
Where a is area of the conductor and  is the current density. Now
𝑎𝑇 = height of the conductor x width of the conductor = width of the conductor ( as height
is unity )
𝐼𝑇 = 𝑚𝑚𝑓 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 ℎ𝑒𝑖𝑔ℎ𝑡 = 𝑎𝛿𝑇 𝑖𝑠 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙 𝑡𝑜 𝛿 𝑥𝑤𝑖𝑑𝑡ℎ 𝑜𝑓 𝑐𝑜𝑝𝑝𝑒𝑟 𝑖𝑛 𝑤𝑖𝑛𝑑𝑜𝑤
Hence 𝐼𝑇 𝑖𝑠 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑎𝑙 𝑡𝑜 𝐷 − 𝑑 𝑓𝑜𝑟 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
∴ 𝑆 ∝ 𝑑 (𝐷 − 𝑑)
Hence to obtain maximum output for the given value of 𝐷 , S is differentiated with respect
to d and = 0 .Which gives the relation 𝐷 = 1.5𝑑
The value of the m depends the rating and cooling system of the transformer . For the
normal design m=0,2d
Hence for the maximum output per unit height the relation between D and d will be as
𝐷 = 1.7𝑑
Hence width of window for the maximum output is 0.7d

Design of yoke
The area of the yoke is taken as 15-20% larger than that of core for the transformer using
hot rolled silicon steel . This reduces the value of the dlux density in yoke which reduces
iron losses and magnetizing current . For the transformer made of cold rolled grain oriented
sheet the yoke area is equal to core area

The section of the yoke can be either rectangular or stepped . if the yoke is rectangular
the depth of the yoke is equal to depth of the core . This depth of core is equal to width
of largest stamping size when square and stepped core are used

For rectangular yoke : Area of Yoke = Depth of Yoke x Height of yoke Dyx Hy where
Dy is equal to width of largest stamping

Window Dimension

Three Phase Core Type

Single Phase Core Type
For Single Phase Core Transformer
D= d+Ww
Hw= H+2Hy and W=D+a
Width over two limb = D + outer diameter of h.v. windings
Tank dimension is slightly greater than width over two limbs and Height of core + 2Hy
to accommodate insulation and other cither clearance

For Three Phase Core Transformer

D= d+Ww
Hw= H+2Hy and W=2D+a
Width over three limb = 2D + outer diameter of h.v. windings
Tank dimension is slightly greater than width over two limbs and Height of core + 2Hy
to accommodate insulation and other cither clearance

Position of windings relative to core

Low voltage winding are placed near to core which reduce the insulation requirements
. Also if high voltage winding is kept inner side of the core then insulation requirement
of h,v. winding and core will be more which also increases the mean length of the
copper .
Also it will be more convenient to keep taping from high voltage winding

Ratio of Iron loss to copper loss

Copper loss/m3 = (ρl/a) (current)2 /(la) = (ρl/a) (aδ) 2 /la = ρδ2 W/m3 = ρδ2 /density W/Kg
Taking ρ =0.021 x 10-6 and density = 8.9 x 10 3 Kg/m3
Copper loss/Kg = specific copper loss = pc = 2.36 x 10-12 δ 2 W/Kg .
Then total copper loss = Wc = pcGc where Gc weight of copper , Kg
In addition to the above , we must add stray losses which may be 5 to 25 % of copper loss.
The total iron loss / Kg = specific iron loss (pi) can be found from the iron loss curves. Then the
total iron loss = Wi = pi Gi Where Gi = weight of iron .
Ratio of Iron loss / copper loss = pi Gi / pcGc

Design of insulation
i) Electrical insulation: depends on the operating voltage
ii) Eddy current loss in the conductors and tank walls
iii) Mechanical considerations: high mechanical forces during fault
iv) Thermal considerations: depends on cooling

Major insulation : between windings and core (grounded)

Minor insulation ; between turns and layers Materials : cotton thread, cotton tape, leatheriod
paper, millinax paper etc
Insulation thickness requied for winding to core and between h.v. and lv. = (5+ 0.9 KV )
mm This thickness includes the width of ducts between windings . Generally width of
ducts is 0.6 mm for the small capacity and 7.5- 12 mm for the large capacity transformer

Thickness of the Major insulation of HV winding upto 33 KV ( all dimension are in

mm )
Thickness of the Major insulation of l.v. winding upto 33 KV ( all dimension are in
mm )

Transformer Windings : Materials used as transformer winding Copper, Aluminum

 High conductivity
 Low-temperature coefficient
 High thermal conductivity
 Light weigh

Types of Windings

 Cylindrical winding with circular conductors

 Crossover winding with circular or rectangular conductors
 Continuous disc type winding with rectangular conductors
 Helical winding

Cylindrical winding with circular conductor

 individual cylindrical layers are placed axially one over the other.
 used in transformer with high current and low voltage rating
 Rectangular conductor are used for the voltage upto 500 V and 10 – 600A
 Circular conductor are used for a current rating of 80 A
Uses of Cylindrical Windings
Cylindrical windings are low voltage windings used up to 6.6 kV for kVA up to 600-750, and
current rating between 10 to 600 A.

Continuous disc type winding with rectangular conductors

 used in high capacity transformer.

 conductors can be single-stripe or strands connected in parallel.
 stranded conductors of rectangular or circular cross-section can be used.
 separated from each other by press-board spacers.
Helical winding

 wound in axial direction along a screw line with a certain inclination.

 Spacers are provided to provide support and also to provide path for oil flow. –
 In transformer having high current rating, a no. of strands of conductors may be
connected in parallel in each turn
 The advantage of Double Helical Winding is that it reduces eddy current loss in
conductors
 In Disc-Helical Windings, parallel strips are placed side by side in a radial direction to
cover the entire radial depth of the winding.
Cross-over winding

These windings are used in high voltage windings of small transformers. The conductors are
paper-covered round wires or strips. The windings are divided into several coils to reduce
voltage between adjacent layers. These coils are axially separated by 0.5 to 1 mm, with the
voltage between adjacent coils kept within 800 to 1000 V.

Transposition of windings

 Multilayer windings are transposed after regular interval

 It help to develop the equal voltage in all coils and minimize circulating current
Estimation of core loss and core loss componet of No load current I c:
Volume of iron in core = 3*L*Ai m3
Weight of iron in core = density * volume
= i * 3*L*Ai Kg
i = density of iron (kg/m3)
=7600 Kg/m3 for normal Iron/steel
= 6500 Kg/m3 for M-4 steel
From the graph we can find out specific iron loss, p i (Watt/Kg ) corresponding to flux density Bm
in core.
So
Iron loss in core =pi* i * 3*L*Ai Watt
Similarly
Iron loss in yoke = py* i * 2*W*Ay Watt
Where py = specific iron loss corresponding to flux density B y in yoke

Total Iron loss Pi =Iron loss in core + Iron loss in yoke

Core loss component of no load current

Ic = Core loss per phase/ Primary Voltage
P
Ic  i
3V1
Estimation of magnetizing current of no load current I m:
Find out magnetizing force H (atcore, at/m) corresponding to flux density B m in the core and atyoke
corresponding to flux density in the yoke from B-H curve
Bm  atcore / m, Bc  at yoke / m 
So
MMF required for the core = 3*L*atcore
MMF required for the yoke = 2*W*atyoke

We account 5% at for joints etc

So total MMF required = 1.05[MMF for core + MMF for yoke]

Peak value of the magnetizing current

Total MMF required
I m , peak 
3N1
RMS value of the magnetizing current
I m, peak
I m, RMS 
2
Total MMF required
I m, RMS 
3 2 N1
 V1=-E1

Estitmation of No load current and phasor diagram: Io

Ic
No load current Io
Ф0
2 2
Io  Ic  Im

No load power factor

Ic Im
Cos o 
Io

The no load current should not exceed 5% of the full the load current.

Estimation of no of turns on LV and hv winding E2

V
Primary no of turns N1  1 No load phasor diagram
Et
V
Secondary no of turns N2  2
Et

Estimation of sectional area of primary and secondary windings

Q  103
Primary current I1 
3V1
Q  103 N
Secondary current I2  OR 1 I1
3V2 N2
I1
Sectional area of primary winding a1 

I2
Sectional area of secondary winging a2 

Where  is current the density.
Now we can use round conductors or strip conductors for this see the IS codes and ICC (Indian
Cable Company) table.

Determination of R1& R2 and Cu losses:

Let Lmt = Length of mean turn

Resistance of primary winding
L N (m)
R1, dc , 75 o  0.021  10  6 mt 1 2
a1 ( m )
R1, ac, 75o  (1.15 to 1.20) R1, dc, 75o
Resistance of secondary winding
 Lmt N 2 ( m)
R2, dc , 75 o  0.021  10  6
a2 ( m 2 )
R2, ac, 75o  (1.15 to 1.20) R2, dc, 75o
Copper loss in primary winding  3I12 R1 Watt
Copper loss in secondary winding  3 I R2
2
2 Watt
Total copper loss  3I R  3I R2
2
1 1
2
2

 3I ( R  R )
2
1 1
'
2
 3 I12 R p
R01  R p  R1  R2'
Where
 Total resis tan ce referred to primary side

Note: On No load, there is magnetic field around connecting leads etc which causes additional
stray losses in the transformer tanks and other metallic parts. These losses may be taken as 7% to
10% of total cu losses.

Determination of EFFICIENCY:

Output Power
Efficiency 
Input Power

Output Power

Output Power  Losses
Output Power
  100 %
Output Power  Iron Loss  Cu loss

Estimation of leakage REACTANCE:

Assumptions
1. Consider permeability of iron as infinity that is MMF is needed only for leakage flux path
in the window.
2. The leakage flux lines are parallel to the axis of the core.
Consider an elementary cylinder of leakage flux lines of thickness dx at a distance x as shown in
following figure.

MMF at distance x

N1 I1
Mx  x
b1
Permeance of this elementary cylinder
 A
 o
L
Lmt dx a b2
b1
 o (Lc =Length of winding)
Lc dx
Lc
 1 L 1 x
 S  & Permeance  
 o A S
Leakage flux lines associated with elementary cylinder
dx  M x  Permeance
NI L dx
 1 1 x  o mt
b1 Lc
Flux linkage due to this leakage flux
d x  No of truns with which it is associated d x N1I1=N2I2
N1 x N1I1 L dx
  x  o mt
b1 b1 Lc
2
L x
  o N mt I1   dx
1
2 x
Lc  b1  MMF Distribution
Flux linkages (or associated) with primary winding
b 2
L 1
x 2 L b 
   o N mt I 1    dx  o N1 mt I1  1 
'
1 1
2

Lc 0  b1  Lc  3 
Flux linkages (or associated) with the space ‘a’ between primary and secondary windings
Lmt
 o   o N12 I1a
Lc
We consider half of this flux linkage with primary and rest half with the secondary winding. So
total flux linkages with primary winding

 1   1'  o
2
L b a
 1   o N12 mt I1  1  
Lc  3 2 
Similarly total flux linkages with secondary winding

 2   2'  o
2
L b a
 2   o N 22 mt I 2  2  
Lc  3 2 
Primary & Secondary leakage inductance
 L b a
L1  1   o N 12 mt  1  
I1 Lc  3 2 
 2 Lmt  b2 a 
L2    o N 22
  
I2 Lc  3 2 
Primary & Secondary leakage reactance
L b a
X 1  2fL1  2fo N12 mt  1  
Lc  3 2 
Lmt  b2 a 
X 2  2fL2  2fo N 22   
Lc  3 2 
Total Leakage reactance referred to primary side
L b b 
X 01 X P  X 1  X 2'  2f o N 12 mt  1 2  a 
Lc  3 
Total Leakage reactance referred to secondary side
L  b  b2 
X 02  X S  X 1'  X 2  2f o N 22 mt  1  a
Lc  3 
It must be 5% to 8% or maximum 10%
Note:-How to control XP?
If increasing the window height (L), Lc will increase and following will
decrease b1, b2& Lmt and so we can reduce the value of XP.

Calculation of VolTage Regulation of transformer:

I 2 Ro 2Cos 2  I 2 X o 2 Sin2
V .R.   100
E2
R Cos2 X Sin2
 o2  100  o 2  100
E2 / I 2 E2 / I 2
 % Ro 2Cos2  % X o 2 Sin2

Transformer Tank Design:

Width of the transformer (Tank)

Wt=2D + De + 2b
Where De= External diameter of HV winding
b = Clearance width wise between HV and tank
Depth of transformer (Tank)
lt= De + 2a
Where a= Clearance depth wise between HV and tank
Height of transformer (Tank)
Ht= H + h
Where h=h1 + h2= Clearance height wise of top and bottom
 W

hy

L H
Ww
(D-d)

D D
a

lt
De

b a b

Wt

h
W

Ht
H

h
Optimum design:
Optimum transformer design is a complex process that involves minimizing one or more of the
following quantities: Total weight, Total volume, Total cost, and Total loss. The ratio r=фm/AT
is a controlling factor for these quantities. A higher value of results in higher flux, which
requires a larger cross section. This increases the weight, volume, and cost of iron, and also
increases iron loss. A lower value of AT decreases the weight, volume, and cost of copper, and
also decreases copper losses
 Design for maximum efficiency ( that occur for load current in which iron loss equal to
core losses )
 Design for minimum cost ( that occur when cost of iron is equal to cost of copper
 For minimum losses current density chosen for both primary and secondary should
be equal

Variation of output and losses with linear dimension

Consider Transformer A with its Linear Dimension x times greater than Transformer B
From Output equation
Q  3.33 f A i Bm  K w Aw  103 KVA
4
Output Varies x
Surface Area Varies : x2
Losses Varies: x3
Temperature Raise : Qc/S where S is surface area
Therefore the temeperture raise linearly vary with x

Cooling Transformer
No transformer is truly an 'ideal transformer' and hence each will incur some losses, most of
which get converted into heat. If this heat is not dissipated properly, the excess temperature in
transformer may cause serious problems like insulation failure.
The temperature rise is given by the relationship 𝜃 = . So the temperature rise can be
decreased and brought within limits by two means.
 Increasing the S heat dissipating area
 Decreasing the cooling coefficient
Since the temperature rise is inversely proportional to surface are of the tank . Thus if the
area of the tank is increased the temperature can be controlled within limit. The surface
areaof the tanks can be increased by adopting fins , Corrugation as well as tubes .In high
power transformer forced cooling methods are used .

It is obvious that transformer needs a cooling system. Transformers can be divided in two types
as
 dry type transformers and
 oil immersed transformers.

Different cooling methods of transformers are

Cooling methods for Dry type Transformers
 Air Natural or Self air cooled transformer This method of transformer cooling is
generally used in small transformers (upto 3 MVA). In this method the transformer is
allowed to cool by natural air flow surrounding it.
 Air Blast For transformers rated more than 3 MVA, cooling by natural air method is
inadequate. In this method, air is forced on the core and windings with the help of fans or
blowers. The air supply must be filtered to prevent the accumulation of dust particles in
ventilation ducts. This method can be used for transformers up to 15 MVA.
Cooling methods for Oil Immersed Transformers
 Oil Natural Air Natural (ONAN)

Air cooling is not sufficient and effective for the medium and large size transformer. Oil
has main advantage of high thermal conductivity and high coefficient of volume expansion
with temperature . Hence all transformer are immersed in oil and heat generated in windings
and core are dissipated to the oil by conduction and oil transfer heat by convection .
During convection the heat is transferred to tank wall by convection .During this processes
heated oil gets cool and fall back to the bottom .There fore natural thermal head is
created which further transfer heat from heated part to wall

Temperature rise of tank with Plain wall

Surface area of 4 vertical side of the tank (Heat is considered to be dissipated from 4 vertical
sides of the tank) St= 2(Wt + lt) Ht m2 (Excluding area of top and bottom of tank). The
specific heat dissipation by material of transformer tank is 6 Watt/m2-0C and 6.5
Watt/m2-0C by means of radiation and convection respectively
o o
 = Temp rise of oil (35 C to 50 C)
12.5St  =Total full load losses ( Iron loss + Cu loss)= Total Full load Loss
Total full load losses
So temp rise in o C 
12.5 St
If the temp rise so calculated exceeds the limiting value, the suitable no of cooling tubes or
radiators must be provided

Temperature rise of tank with provision of tubes :

Let xSt= Surface area of all cooling tubes then Losses to be dissipated by the transformer walls
and cooling tube= Total losses . The specific heat dissipation by material of transformer
tank is 6 Watt/m2-0C and 6.5 Watt/m2-0C by means of radiation and convection
respectively. Its is found that due to provision of tubes the convection will be improved by
35%.
12.5 St  6.5 *1.35 xSt   12.5 St  8.8 xSt   Total losses
For Tank Wall For Tubes
2 0 2 0
6 Watt/m - C -Raditon +6.5 Watt/m - C 2 0
6.5*1.35 W  8 . 8 Watt/m - C
2 0
W=12.5 Watt/m - C (  35% more) Convection only
So from above equation we can find out total surface are of cooling tubes (xS t)
Normally we use 5 cm diameter tubes and keep them 7.5 cm apart
At= Surface area of one cooling tube
 d tube ltube , mean
Hence
xS
No of cooling tubes  t
At
 7.5 Cm

d= 5 Cm

Tank and Arrangement of Cooling tubes

Cooling methods used for oil immersed Transformer
Weight of TRANFORMER:

Let

Wi = Weight of Iron in core and yoke (core volume* density + yoke volume* density) Kg
Wc= Weight of copper in winding (volume* density) Kg
(density of cu = 8900 Kg/m3)
Weight of Oil
= Volume of oil * 880 Kg
Add 20% of (Wi+Wc) for fittings, tank etc.
Total weight is equal to weight of above all parts.
Design for maximum efficiency ( that occur for load current in which iron loss equal to
core losses )

Design for minimum cost ( that occur when cost of iron is equal to cost of copper)
Area of the core affected by weight of copper and iron
1. A 25 KVA ,11000/433 V , 50 hz three phase delta star core type transformer has full load
copper loss 720Watt and over all core dimensions are
For limb For Yoke
Cross-section area A = 0.01035m2 Cross-section area A= 0.01216m2
Height H = 0.3m Length = 0.624m
Flux density Bm =1web/m2 Flux density Bm =0.833Web/m2
The magnetizing and specific core loss characteristics are given below table
Bm(Web/m2) 0.8 0.9 1 1.1
Iron loss (W/kg) 0.85 1.05 1.2 1.4
Mmf(A/m) 75 100 120 125
Determine no load current ,Efficiency at full load and identify whether it is distribution or
power transformer
2. Why Large Transformers requires elaborate cooling system? Show that by increasing linear
dimension of transformer n times , the KVA rating is increased by n 4 times and losses to be
dissipated per unit surface area is increased by n times

3. The tank of 575KVA On type transformer has dimensions length , width and height are 105
cm , 65 cm and 150 cm respectively . The full load losses is 6.5 KW .Design suitable
arrangement of 5 cm diameter tubes spaced 7.5 cm and average length of each tubes are
127.5 cm and temperature rise of the wall is limited at 35 0C .
Assume Watt/m2-0C due to convection =6.5, Watt/m2-0C due to radiation =6.5 and
improvement in convection due to provision of tubes =35%

4. Show the path of leakage flux in three phase core type transformer having concentric
cylindrical winding of equal length. What simplifying assumption are made in the derivation
of the formula for leakage reactance calculation .

5. A 750 ,6600V , 50hz three phase delta star core type transformer has following data : width
of l.v. winding 30mm , width of the high voltage winding 25mm , width of the duct between
high voltage and low voltage winding 15mm , height of winding 0.4m , length of the mean
turn 1.5m , h,v winding turns 217 and l.v winding turns is 15 . Estimate leakage reactance of
the transformer referred to h.v side .
6. A transformer has 1500W and 1100W hysterysis and eddy current losses at normal voltage
and frequency . Calculate the both losses when voltage and frequency are raised by 15%
7. Calculate the approximate overall dimension for a 200KVA ,6600/440V ,50 hz 3- phase core
type transformer . The following data may be assumed :Emf per turn =10V , maximum flux
density Bm=1.3Web/m2 , current density  = 2.5amp/mm2 , window space factor =0.3 ,
overall height = overall width , Stacking factor =0.9 .
Use three step cores.
For three step core
Width of the largest stamping =0.9d
Net iron area =0.9d2 where d is diameter of the circumscribing circle
8. Full Load at 0.8 lagging P.F. regulation of power transformer was estimated from its
designed data to be 2.5% against the desire value 2%. What adjustment are necessary in
design to obtain require regulation. Justify your procedure

9. A 15 MVA transformer has iron loss of 80KW and copper loss of 120 KW at full load . The
tank dimensions are 3.5X3X1.4 meter .The transformer oil is cooled by three liter of oil per
second passed through cooling coil .Estimate average temperature rise of tank if the
difference of temperature of water at inlet and outlet is 150C . The specific heat dissipation
from the tank wall is 10W/m2-0C