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Unit 3
1
Introduction
Constituents of transformer:
i.
Magnetic Circuit
ii.
Electric Circuit
iii. Dielectric Circuit
iv. Other accessories
2
Classification or Types
Transformers
Based on
transformer
Ratio
Based on Core
Core
Shell
Type
Type
Step up
Step Down
Based on
Service
Distribution
Power
Transformer
Transformer
3
Constructional Details
4
Constructional Details
5
Constructional Details
6
Constructional Details
7
Constructional Details
The requirements of magnetic material
are,
 High permeability
 Low reluctance
 High saturation flux density
 Smaller area under B-H curve
For small transformers, the laminations
are in the form of E,I, C and O as shown in
figure
The percentage of silicon in the steel is
about 3.5. Above this value the steel
becomes very brittle and also very hard to
cut
8
Transformer Core
Core type Construction
Shell Type Construction
9
Transformer Core
Stepped Core Construction
10
Transformer Core
Ww
Hw
Core Type
Shell Type
11
Transformer Winding
 Types of transformer windings are,
 Concentric
 Sandwich
 Disc
 Cross over
 Helical
12
Transformer Winding
Hw
Ww
13
Transformer Winding
14
Insulations
Dry type Transformers:
Varnish
Enamel
Large size Transformers:
Unimpregnated paper
Cloths
Immersed in Transformer oil – insulation & coolent
15
Comparison between
Core & Shell Type
Description
Core Type
Shell Type
Construction
Easy to assemble & Dismantle
Complex
Mechanical Strength
Low
High
Leakage reactance
Higher
Smaller
Cooling
Better cooling of Winding
Better cooling of Core
Repair
Easy
Hard
Applications
High Voltage & Low output
Low Voltages & Large Output
16
Classification on Service
Details
Distribution transformer
Power Transformer
Capacity
Upto 500kVA
Above 500kVA
Voltage rating
11,22,33kV/440V
400/33kV;220/11kV…etc.,
Connection
Δ/Y, 3φ, 4 Wire
Δ/Δ; Δ/Y, 3φ, 4 Wire
Flux Density
Upto 1.5 wb/m2
Upto 1.7 wb/m2
Current Density
Upto 2.6 A/mm2
Upto 3.3 A/mm2
Load
100% for few Hrs, Part loadfor
some time, No-load for few Hrs
Nearly on Full load
Ratio of Iron Loss to Cu loss
1:3
1:1
Regulation
4 to 9%
6 to 10%
Cooling
Self oil cooled
Forced Oil Cooled
17
Output Equation of Transformer
 It relates the rated kVA output to the area of core & window
 The output kVA of a transformer depends on,
 Flux Density (B) – related to Core area
 Ampere Turns (AT) – related to Window area
 Window – Space inside the core – to accommodate primary &
secondary winding
Let,
T- No. of turns in transformer winding
f – Frequency of supply
Induced EMF/Turn , Et=E/T=4.44fφm
18
Output Equation of Transformer
 Window in a 1φ transformer contains one primary & one secondary
winding.
Conductor area in Window
Window Space factor,Kw 
Total area of Window
Ac
Window Space factor,Kw 
Aw
 Conductor area in w indow ,Ac  K w A w  (2)
Current Density() is same in both the w indings
Ip
I
   s  (3)
ap a s
19
Output Equation of Transformer
Ip
Is
ap 
; as 


If we neglect magnetizing MMF, then (AT)primary = (AT)secondary
AT=IpTp=IsTs
 (4)
Total Cu. Area in window, Ac=Cu.area of pry wdg + Cu.area of sec wdg
No. of pry turns X area
 No. of sec turns X area



 

of
X
section
of
pry
conductor
of
X
section
of
sec
conductor

 

 Tp a p  Ts a s
 Tp a p  Ts a s
 Tp
Ip

 Ts
Is

1

TpIp  TsIs 

1
2 AT
 AT  AT 
 (5)



20
Output Equation of Transformer
Therefore, equating (2) & (5),
KwAw 
AT 
2 AT

1
K w A w   ( 6)
2
kVA rating of 1φ transformer is given by,
Q  VpIp  10-3  EpIp  10-3

Ep
Tp
TpIp  10- 3
 E t  AT  103
E

from
(
1
),
E

t

T 

 (6)
1
 4.44fm . K w A w   103
2
 2.22  fm  K w A w   103
21
Output Equation of Transformer
m
and m  Bm A i
We know that, Bm 
Ai
 Q  2.22  f Bm Ai A w K w 103 kVA
Three phase transformer:
 Each window has 2 primary & 2 Secondary
windings.
 Total Cu. Area in the window is given by,
A c  2Tp a p  2Ts a s
4 AT
Ac 

Hw
Ww
 (7)
4 AT
Compare(2) & (7),
 KwAw

K w A w
AT 
4
22
Output Equation of Transformer
kVA rating of 3φ transformer,
Q  3 EpIp  103
3
Ep
Tp
TpIp  10-3
 E t  AT  103
1
 3  4.44  fm .  K w A w   103
4
 3.33 f Bm A i A wK w   103 kVA
23
EMF per Turn
 Design of Xmer starts with the section of EMF/turn.
Let, Ratio of Specific magnetic loading r  m

to Electric loading
Q  VpIp  103


AT
 4.44 f m TpIp  103
 4.44 f m ( AT)  103
m
 103
r
Q r
Q  r  103
2
m 

3
4.44 f  10
4.44 f
 4.44 f m
m 
Q  r  103
4.44 f
24
EMF per Turn
w.k.t , E t  4.44 f m
Q  r  103
Q
3
 4.44 f
 4.44 f  4.44 f  r  10 .
4.44 f
4.44 f
 4.44 f  r  103  Q
Et  K  Q
where, K  4.44 f  r  103
 K depends on the type, service condition & method of construction of
transformer.
25
EMF per Turn
Transformer Type
1φ Shell Type
Value of K
1.0 to 1.2
1φ Core Type
3φ Shell Type
3φ Core Type Distribution Xmer
3φ Core Type Power Xmer
0.75 to 0.85
1.2 to 1.3
0.45 to 0.5
0.6 to 0.7
26
Optimum Design
 Transformer may be designed to make one of the following
quantitites as minimum.
i.
ii.
iii.
iv.
Total Volume
Total Weight
Total Cost
Total Losses
 In general, these requirements are contradictory & it is
normally possible to satisfy only one of them.
 All these quantities vary with
r 
m
AT
27
Optimum Design
Design for Minimum Cost
Let us consider a single phase transformer.
 Q  2.22  f Bm Ai A w K w 103 kVA
Q  2.22  f Bm Ai Ac  103 kVA
 Ac  Kw Aw 
Assuming that φ & B are constant, Ac.Ai – Constant
Let, A A  M2  (1)
c i
In optimum design, it aims to determining the minimum value of
total cost. r  m
AT
m  B m A i
Ac
1
AT  K w A w  
2
2
28
Optimum Design
Design for Minimum Cost
Bm A i
2 Bm A i
r

Ac
Ac
2
A i 2 Bm
r
Ac 
Ai


r  
Ac
2 Bm
 (2)
β is the function of r alone [δ & Bm – Constant]
From (1) & (2),
M
Ai  M  & Ac 

 A A  M 
2
c i
29
Optimum Design
Design for Minimum Cost
Let, Ct - Total cost of transformer active materials
Ci – Cost of iron
Cc – Cost of conductor
pi – Loss in iron/kg (W)
pc – Loss in Copper/kg (W)
li – Mean length of flux path in iron(m)
Lmt – Mean length of turn of transformer winding (m)
Gi – Weight of active iron (kg)
Gc – Weight of Copper (kg)
gi – Weight/m3 of iron
gc – Weight/m3 of Copper
Ct  Ci  Cc  c iGi  c cGc
30
Optimum Design
Design for Minimum Cost
Ct  ci g ili Ai  ccg c Lmt Ac
where, ci & cc  specificcos ts of iron and copperrespectively.
M
C t  c i giliM   c c gc Lmt

Differeniating Ct with respect to β,
d
1
1
1/ 2
C t  ci g ili M()
 c c g c Lmt M3 / 2
d
2
2
For minimum cost,
d
Ct  0
d
1
1
c i giliM() 1/ 2  c c gc Lmt M 3 / 2
2
2
c i gili  c c gc Lmt 1
Ac
c i gili  c c gc Lmt
Ai
31
Optimum Design
Design for Minimum Cost
c i gili Ai  c c gc Lmt A c
ci G i  c c G c
Ci  C c
Hence for minimum cost, the cost of iron must be equal to the cost of
copper.
Similarly,
For minimum volume of transformer,
Volume of iron = Volume of Copper
Gi
gi

Gc
gc
or
Gi
Gc

gi
gc
For minimum weight of transformer,
Weight of iron = Weight of Conductor Gi  Gc
For minimum loss,
Iron loss = I2R loss in conductor
Pi  x Pc
2
32
Optimum Design
Design for Minimum Loss and Maximum Efficiency
Total losses at full load = Pi+Pc
At any fraction x of full load, total losses = Pi+x2Pc
If output at a fraction of x of full load is xQ.
xQ
Efficiency, x 
xQ  Pi  x 2 Pc
Condition for maximum efficiency is,


d x
0
dx
xQ  Pi  x 2 Pc Q - Q  2xPc xQ
dx

0
2
2
dx
xQ  Pi  x Pc


xQ  P  x P Q  Q  2xP xQ
xQ  P  x P   Q  2xP x
2
i
c
c
2
i
c
c
xQ  Pi  x 2 Pc  xQ  x 2 Pc  x 2 Pc
Pi  x 2 Pc
33
Optimum Design
Design for Minimum Loss and Maximum Efficiency
Variable losses = Constant losses
Pi
Pc

pi G i
pcG c
pi G i
x 
pcG c
2
Gi
2 pc
or
x
Gc
pi
for maximum efficiency
34
Design of Core
 Core type transformer : Rectangular/Square /Stepped cross
section
 Shell type transformer : Rectangular cross section
35
Design of Core
Rectangular Core
 For core type distribution transformer & small power
transformer for moderate & low voltages
Depth
 1.4 to 2
Width
Rectangular coils are used.
 For shell type,
Width of Central lim b
 2 to 3
Depth of Core
36
Design of Core
Square & Stepped Core
 Used when circular coils are required for high voltage
distribution and power transformer.
 Circular coils are preferred for their better mechanical
strength.

 Circle representing the inner surface of the tubular form
carrying the windings (Circumscribing Circle)
37
Design of Core
Square & Stepped Core
 Dia of Circumscribing circle is larger in Square core than
Stepped core with the same area of cross section.
 Thus the length of mean turn(Lmt) is reduced in stepped core
and reduces the cost of copper and copper loss.
 However, with large number of steps, a large number of
different sizes of laminations are used.
38
Design of Core
Square Core
Let, d - diameter of circumscribing circle
a – side of square
Diameter, d  a 2  a 2  2a 2  2  a
d
a
2
 d 

Gross core area, Agi  Area of square  a  
 2
Agi  0.5d 2
Let the stacking factor, Sf=0.9.
2
2
Net core area, Ai  0.9  0.5d 2  0.45d 2
39
Design of Core
Square Core
 Gross core area includes insulation area
 Net core area excludes insulation area
 2
Area of Circumscribing circle is d
4
Ratio of net core area to Area of Circumscribing circle is
0.45d 2
 0.58
 2
d
4
Ratio of gross core area to Area
of Circumscribing circle is
2
0.5d
40
 0.64
 2
d
4
Design of Core
Square Core
Useful ratio in design – Core area factor,
KC 
Net Core area
Squareof Circum scribing Circle
A i 0.45d 2
 2 
 0.45
2
d
d
41
Design of Core
Stepped Core or Cruciform Core
Let,a – Length of the rectangle
b – breadth of the rectangle
d – diameter of the circumscribing circle and diagonal
of the rectangle.
θ – Angle b/w the diagonal and length of the rectangle.
 The max. core area for a given ‘d’ is
obtained by the max value of ‘θ’
 For max value of ‘θ’,
b
(a-b)/2
d
b
θ
a
dA gi
d
(a-b)/2
a
0
 From the figure,
a
cos    a  d cos
d
b
sin    b  d sin 
d
42
Design of Core
Stepped Core or Cruciform Core
b
Two stepped core can be divided in to 3 rectangles.
Referring to the fig shown,
ab ab
Gross core area,Agi  ab  
b  
b
2
2

 

2(a  b)
 ab 
b
2
 ab  ab  b 2  2ab  b 2
(a-b)/2
d
b
θ
(a-b)/2
a
On substituting ‘a’ and ‘b’ in the above equations,
A gi  2(d cos)(d sin )  (d sin ) 2
A gi  2d 2 cos sin   d 2 sin 2 
A gi  d 2 sin 2  d 2 sin 2 
For max value of ‘θ’,
dA gi
d
0
43
a
Design of Core
Stepped Core or Cruciform Core
i.e.,
dA gi
 d 2 2 cos 2  d 2 2 sin  cos  0
d
d 2 2 cos 2  d 2 2 sin  cos 
2 cos 2  sin 2
sin 2
2
cos 2
tan 2  2
2  tan1 2 
1
  tan1 2   31.720
2
Therefore, if the θ=31.720, the dimensions ‘a’ & ‘b’ will give
maximum area of core for a specified ‘d’.
a
cos    a  d cos  a  d cos(31.720 )  0.85d
d
b
sin    b  d sin   b  d sin(31.720 )  0.53d
d
44
Design of Core
Stepped Core or Cruciform Core
Gross core area,
A gi  2ab  b 2
A gi  2(0.85d)(0.53d)  (0.53d) 2
A gi  0.618d 2
Let stacking factor,S f  0.9,
Net core area,Ai  Stacking factor X Gross Core area
Ai  0.9  0.618d 2  0.56d 2
The ratios,
Net core area
0.56d 2

 0.71

Area of Circum scribing circle
d2
4
Gross core area
0.618d 2

 0.79
 2
Area of Circum scribing circle
d
4
45
Design of Core
Stepped Core or Cruciform Core
Core area factor,
KC 
Net Core area
Squareof Circum scribing Circle
A i 0.56d 2
 2 
 0.56
2
d
d
Ratios of Multi-stepped Cores,
Ratio
Net core area
Area of Circum scribing circle
Gross core area
Area of Circum scribing circle
Core area factor,KC
Square Core
Cruciform Core
3-Stepped Core
4-Stepped Core
0.64
0.79
0.84
0.87
0.58
0.71
0.75
0.78
46
0.45
0.56
0.6
0.62
Design of Core
Choice of Flux Density in Core
 Flux density decides,
 Area of cross section of the core
 Core loss
 Choice of flux density depends on,
 Service condition (DT/PT)
 Material used
Cold rolled Silicon Steel-Work with higher flux density
 Upto 132 kV : 1.55 wb/m2
 132 to 275kV: 1.6 wb/m2
 275 to 400kV: 1.7 wb/m2
Hot rolled Silicon Steel – Work with lower flux density
 Distribution transformer : 1.1 to 1.4 wb/m2
 Power Transformer : 1.2 to 1.5 wb/m2
47
Overall Dimensions
Single phase Core Type
Hy
Hw
H
Hy
a
W
Dy
d
Ww
D
d
48
Overall Dimensions
Single phase Shell Type
a
Hw
2a
a
Ww
b
Depth
Over
winding
49
Overall Dimensions
Thee phase Core Type
Hy
Hy
a
Hw
H
Hw
Hy
Hy
Dy
Dy
H
a
W
d
Ww
D
d
Ww
D
d
50
Design of Winding
 Transformer windings: HV winding & LV winding
 Winding Design involves:
 Determination of no. of turns: based on kVA rating & EMF per turn
 Area of cross section of conductor used: Based on rated current and Current density
 No. of turns of LV winding is estimated first using given data.
 Then, no. of turns of HV winding is calculated to the voltage rating.
No. of turns in LV winding,TLV 
VLV
AT
(or)
Et
I LV
where,VLV  Rated voltageof LV winding
I LV  Rated Current of LV winding
51
Design of Winding
No.of turns in HV winding,THV  TLV 
VHV
VLV
where, VHV  Rated voltageof HV winding
52
Cooling of transformers
 Losses in transformer-Converted in heat energy.
 Heat developed is transmitted by,
 Conduction
 Convection
 Radiation
 The paths of heat flow are,
 From internal hot spot to the outer surface(in contact with oil)
 From outer surface to the oil
 From the oil to the tank
 From tank to the cooling medium-Air or water.
Cooling of transformers
 Methods of cooling:
1.
2.
3.
4.
5.
6.
7.
8.
Air Natural (AN)-upto 1.5MVA
Air Blast (AB)
Oil natural (ON) – Upto 10 MVA
Oil Natural – Air Forced (ONAF)
Oil Forced– Air Natural (OFAN) – 30 MVA
Oil Forced– Air Forced (OFAF)
Oil Natural – Water Forced (ONWF) – Power plants
Oil Forced - Water Forced (OFWF) – Power plants
Cooling of transformers
Transformer Oil as Cooling Medium
 Specific heat dissipation due to convection is,
1
  4
conv  40.3  W / m 2 .0 C
H
where, - T emperatur
e differenceof thesurface relativeto oil,0C
H  Height of dissipating surface,m
 The average working temperature of oil is 50-600C.

 200C & H  0.5 to 1m,
 For
conv  80 to100 W/m2 .0 C.
 The value of the dissipation in air is 8 W/m2.0C. i.e, 10 times less
than oil.
Cooling of transformers
Temperature rise in plain walled tanks
 Transformer wall dissipates heat in radiation & convection.
 For a temperature rise of 400C above the ambient temperature of 200C, the heat
dissipations are as follows:



Specific heat dissipation by radiation,rad=6 W/m2.0C
Specific heat dissipation by convection, conv=6.5 W/m2.0C
Total heat dissipation in plain wall 12.5 W/m2.0C
 The temperature rise,

P P
T otallosses
 i c
Specific heat Heat dissipating  12.5 St

 

Dissipatio
n
surface
of
tank

 

St – Heat dissipating surface
 Heat dissipating surface of tank : Total area of vertical sides+ One half area of top
cover(Air cooled) (Full area of top cover for oil cooled)
Design of tanks with cooling tubes
 Cooling tubes increases the heat dissipation
 Cooling tubes mounted on vertical sides of the transformer would not
proportional to increase in area. Because, the tubes prevents the
radiation from the tank in screened surfaces.
 But the cooling tubes increase circulation of oil and hence improve
the convection
 Circulation is due to effective pressure heads
 Dissipation by convection is equal to that of 35% of tube surface
area. i.e., 35% tube area is added to actual tube area.
Design of tanks with cooling tubes
Let, Dissipating surface of tank – St
Dissipating surface of tubes – XSt
Loss dissipated by surface of the tank by radiation and convection = 6  6.5St  12.5St
Loss dissipatedby 
135
 XSt  8.8XSt
  6.5 
tubesby convection
100
Total loss dissipated
  12.5St  8.8XSt  12.5  8.8XSt  (1)
by wallsand tubes 
Actual totalarea of tank walls and tubes  St  XSt  St (1  X )
Design of tanks with cooling tubes
T otallosses dissipated
Loss dissipated per m of dissipating surface 
T otalarea
St (12.5  8.8 X ) (12.5  8.8 X )
2
Loss dissipated per m of dissipating surface 

 (2)
St (1  X )
(1  X )
2

T otalloss
 
T ransforme
r with coolingtubes
Loss Dissipated
T emperature rise in
T otallosses, Ploss  Pi  Pc
 (3)
Pi  Pc
From (1) and (3), we have, 
St (12.5  8.8 X )
Pi  Pc
(12.5  8.8 X ) 
 St
Pi  Pc
8.8 X 
 12.5 
 St

1  Pi  Pc

X
 12.5 
8.8   St

Design of tanks with cooling tubes

1  Pi  Pc
1  Pi  Pc



T otalarea of coolingtubes 
 12.5  St 
 12.5St   (5)


8.8   St
8.8  


Let, lt  Length of tubes
d t  Diam eterof tubes
 Surfacearea of tubes   d t lt
T otalarea of tubes
Area of each tube
1
 Pi  Pc

nt 
 12.5St   (6)

8.8 d t lt  

T otalnumber of tubes,nt 
 Standard diameter of cooling tube is 50mm & length depends on the
height of the tank.
 Centre to centre spacing is 75mm.
C4
HT
D
D
C3
WT
LT
C2
Doc
C1
Design of tanks with cooling tubes
 Dimensions of the tank:
Let, C1 – Clearance b/w winding and tank along width
C2 - Clearance b/w winding and tank along length
C3 – Clearance b/w the transformer frame and tank at the bottom
C4 - Clearance b/w the transformer frame and tank at the top
Doc – Outer diameter of the coil.
Width of the tank, WT=2D+ Doc +2 C1 (For 3 Transformer)
= D+ Doc +2 C1 (For 1 Transformer)
Length of the tank, LT= Doc +2 C2
Height of the tank, HT=H+C3+ C4
Design of tanks with cooling tubes
 Clearance on the sides depends on the voltage &
power ratings.
 Clearance at the top depends on the oil height above
the assembled transformer & space for mounting the
terminals and tap changer.
 Clearance at the bottom depends on the space
required for mounting the frame.
Design of tanks with cooling tubes
Clearance in mm
Voltage
kVA Rating
C1
C2
C3
C4
Up to 11kV
<1000kVA
40
50
75
375
Upto 11 kV
1000-5000kVA
70
90
100
400
11kV – 33kV
<1000kVA
75
100
75
450
85
125
100
475
11kV – 33kV 1000-5000kVA
Estimation of No-load Current
No-load Current of Transformer:
Magnetizing Component
Depends on MMF required to establish required flux
Loss Component
Depends on iron loss
65
Estimation of No-load Current
No-load current of Single phase Transformer
ly
Total Length of the core = 2lc
Total Length of the yoke = 2ly
lC
lC
Here, lc=Hw=Height of Window
ly= Ww=Width of Window
ly
MMF for core=MMF per metre for max. flux density in core X Total length of
Core
= atc X 2lc= 2 atc lc
MMF for yoke=MMF per metre for max. flux density in yoke X Total length of
yoke
= aty X 2ly= 2 aty ly
Total Magnetizing MMF,AT0=MMF for Core+MMF for Yoke+MMF for joints
= 2 atc lc +2 aty ly +MMF for joints
The values of atc & aty are taken from B-H curve of transformer steel.
66
Estimation of No-load Current
No-load current of Single phase Transformer
Max. value of magnetizing current=AT0/Tp
If the magnetizing current is sinusoidal then,
RMS value of magnetising current, Im=AT0/√2Tp
If the magnetizing current is not sinusoidal,
RMS value of magnetising current, Im=AT0/KpkTp
The loss component of no-load current, Il=Pi/Vp
Where, Pi – Iron loss in Watts
Vp – Primary terminal voltage
Iron losses are calculated by finding the weight of cores and
yokes. Loss per kg is given by the manufacturer.
2
2
No-load current, I 0  I m  Il
67
Estimation of No-load Current
No-load current of Three phase Transformer
ly
lC
lC
lC
ly
68