T ra
Unit 3Unit 31
Introduction
Constituents of transformer:i. Magnetic Circuit
ii. Electric Circuit
iii. Dielectric Circuit
iv. Other accessories
2
Classification or Types
3
Constructional Details
4
Constructional Details
5
Constructional Details
6
Constructional Details
7
Constructional DetailsConstructional 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 CoreTransformer Core
Core type Construction Shell Type Construction
9
Stepped Core Construction
Transformer CoreTransformer Core
10
Shell Type
Core Type
Transformer CoreTransformer Core
Ww
Hw
11
Transformer WindingTransformer Winding
Types of transformer windings are, Concentric Sandwich
Disc Cross over Helical
12
Transformer WindingTransformer Winding
Ww
Hw
13
Transformer WindingTransformer Winding
14
InsulationsInsulations
Dry type Transformers:VarnishEnamel
Large size Transformers:Unimpregnated paperCloths
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 ServiceDetails Distribution
transformerPower 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 Cooled17
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=4.44fφm18
Window in a 1φ transformer contains one primary & one secondary winding.
Output Equation of Transformer
)3(aI
a
Iwindingsthebothinsameis)(DensityCurrent
)2(AK A window,in area ConductorAA
factor,K Space Window
Window of area TotalWindow in area Conductor
factor,K Space Window
s
s
p
p
wwc
w
cw
w
19
Output Equation of Transformer
s
sp
pI
a;I
a
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
)5(AT2
ATAT1
ITIT1
IT
IT
aTaT
aTaT
conductorsec of section-X ofareaX turnssec of .No
conductorpry of section-X ofareaX turnspry of .No
sspp
ss
pp
sspp
sspp
20
Therefore, equating (2) & (5),
kVA rating of 1φ transformer is given by,
Output Equation of Transformer
)6(AK21
AT
AT2AK
ww
ww
3wwm
3wwm
3t
t3-
ppp
p
-3pp
-3pp
10AKf22.2
10AK21
.f44.4
)6(10ATE
TE
E),1(from10ITT
E
10IE10IV Q
21
We know that,
Three phase transformer:
Each window has 2 primary & 2 Secondary windings.
Total Cu. Area in the window is given by,
Output Equation of Transformer
immi
mm ABand
AB
kVA10KAABf22.2Q 3wwim
Ww
Hw
4AK
AT
AKAT4
),7(&)2(Compare
)7(AT4
A
aT2aT2A
ww
ww
c
ssppc
22
kVA rating of 3φ transformer,
Output Equation of Transformer
kVA10KAABf33.3
10AK41
.f44.43
10ATE
10ITT
E3
10IE3Q
3wwim
3wwm
3t
3-pp
p
p
3pp
23
EMF per Turn
Design of Xmer starts with the section of EMF/turn.
Let,
f44.410rQ
f44.410rQ
10f44.4rQ
10r
f44.4
10)AT(f44.4
10ITf44.4
10IVQ
ATr
loadingElectric toloadingmagnetic Specific of Ratio
3
m
3
32m
3mm
3m
3ppm
3pp
m
24
Q10rf44.4
f44.4Q
.10rf44.4f44.4f44.410rQ
f44.4
f44.4E,t.k.w
3
33
mt
QKEt
EMF per Turn
310rf44.4K,where K depends on the type, service condition & method of construction of transformer.
25
EMF per Turn
Transformer Type Value of K
1φ Shell Type 1.0 to 1.2
1φ Core Type 0.75 to 0.85
3φ Shell Type 1.2 to 1.3
3φ Core Type Distribution Xmer
0.45 to 0.5
3φ Core Type Power Xmer
0.6 to 0.7
26
Optimum Design
Transformer may be designed to make one of the following quantitites as minimum.
i. Total Volume
ii. Total Weight
iii. Total Cost
iv. Total Losses
In general, these requirements are contradictory & it is normally possible to satisfy only one of them.
All these quantities vary with
ATr m
27
Optimum DesignDesign for Minimum Cost
Let us consider a single phase transformer.
kVA10KAABf22.2Q 3wwim
wwc3
cim AKAkVA10AABf22.2Q
Assuming that φ & B are constant, Ac.Ai – Constant
Let,
In optimum design, it aims to determining the minimum value of total cost.
2A
AK21
AT
ABAT
r
cww
imm
m
28
)1(M2 AA ic
Optimum DesignDesign for Minimum Cost
c
im
c
im
AAB2
2A
ABr
)2(rB2A
A
B2
A
Ar
mc
i
m
c
i
β is the function of r alone [δ & Bm – Constant]
From (1) & (2),
29
M
A&MA ci
2M AA ic
30
Optimum DesignDesign 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 Coppercciicit GcGcCCC
31
.lyrespectivecopperandironoftscosspecificc&c,where
AgcAgcC
ci
ccciiit
mti Ll
Optimum DesignDesign for Minimum Cost
MgcMgcC cciit mti Ll
Differeniating Ct with respect to β, 2/3
cc2/1
iit Mgc2
1)(Mgc
2
1C
d
d mti Ll
For minimum cost, 0Cdd
t 2/3
cc2/1
ii Mgc21
)(Mgc21 mti Ll
1ccii gcgc mti Ll
i
cccii A
Agcgc mti Ll
32
ccciii AgcAgc mti Ll
Optimum DesignDesign for Minimum Cost
ci
ccii
CC
GcGc
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
For minimum weight of transformer,
Weight of iron = Weight of Conductor
For minimum loss,Iron loss = I2R loss in conductor
c
i
c
i
c
c
i
ig
gG
GorgG
gG
ci GG
c2
i PP x
Optimum DesignDesign 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.
Efficiency,
Condition for maximum efficiency is,
33
c2
i PxPxQ
xQ
x
0d
d x x
c2
i
c2
c2
c2
i
cc2
i
cc2
i
c2
i
cc2
i
PxP
PxPxxQPxPxQ
x2xPQPxPxQ
xQ2xPQQPxPxQ
PxPxQ
xQ2xPQ-QPxPxQ
x
0
d
d2
x
Variable losses = Constant losses
for maximum efficiency
34
Optimum DesignDesign for Minimum Loss and Maximum
Efficiency
i
c2
c
i
cc
ii2
cc
ii
c
i
p
px
G
Gor
Gp
Gpx
Gp
GpP
P
Design of Core
Core type transformer : Rectangular/Square /Stepped cross section
Shell type transformer : Rectangular cross section
35
For core type distribution transformer & small power transformer for moderate & low voltages
Rectangular coils are used. For shell type,
36
Design of CoreRectangular Core
24.1 toWidth
Depth
32 toCoreofDepth
limbCentralofWidth
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 CoreSquare & 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 CoreSquare & Stepped Core
Let, d - diameter of circumscribing circle
a – side of square
Diameter,
Gross core area,
Let the stacking factor, Sf=0.9.
Net core area,
39
Design of CoreSquare Core
2
22 222
da
aaaad
2
22
5.0
2
dA
dasquareofAreaA
gi
gi
22 45.05.09.0 ddAi
Gross core area includes insulation area
Net core area excludes insulation area
Area of Circumscribing circle is
Ratio of net core area to Area of Circumscribing circle is
Ratio of gross core area to Area of Circumscribing circle is
40
2
4d
58.0
4
45.0
2
2
d
d
64.0
4
5.0
2
2
d
d
Design of CoreSquare Core
Useful ratio in design – Core area factor,
41
Design of CoreSquare Core
45.0d
d45.0
d
A
K
2
2
2i
C
Circle bingCircumscri of Square
area Core Net
Let,a – Length of the rectangleb – 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.
42
Design of CoreStepped Core or Cruciform Core
θ
d
b
b
a
a
(a-b)/2
(a-b)/2
The max. core area for a given ‘d’ is obtained by the max value of ‘θ’
For max value of ‘θ’,
From the figure,
0d
dAgi
sindbd
bsin
cosdad
acos
Two stepped core can be divided in to 3 rectangles.Referring to the fig shown,
43
Design of CoreStepped Core or Cruciform Core
θ
d
b
b
a
a
(a-b)/2
(a-b)/2
22 bab2babab
b2
)ba(2ab
b2
bab
2
baab
giAarea, core Gross
On substituting ‘a’ and ‘b’ in the above equations,
222gi
222gi
2gi
sind2sindA
sindsincosd2A
)sind()sind)(cosd(2A
For max value of ‘θ’,
0d
dAgi
i.e.,
Therefore, if the θ=31.720, the dimensions ‘a’ & ‘b’ will give maximum area of core for a specified ‘d’.
44
Design of CoreStepped Core or Cruciform Core
01
1
22
22gi
72.312tan2
1
2tan2
22tan
22cos
2sin
2sin2cos2
cossin2d2cos2d
0cossin2d2cos2dd
dA
d53.0)72.31sin(dbsindbd
bsin
d85.0)72.31cos(dacosdad
acos
0
0
Gross core area,
The ratios,
45
22
2gi
2gi
2gi
d56.0d618.09.0
,9.0
d618.0A
)d53.0()d53.0)(d85.0(2A
bab2A
i
i
f
A
area Core Gross X factor StackingAarea, core Net
Sfactor, stackingLet
Design of CoreStepped Core or Cruciform Core
79.0d
4
d618.0
71.0d
4
d56.0
2
2
2
2
circle bingCircumscri of Area
area core Gross
circle bingCircumscri of Area
area core Net
Core area factor,
Ratios of Multi-stepped Cores,
46
Design of CoreStepped Core or Cruciform Core
56.0d
d56.0
d
A
K
2
2
2i
C
Circle bingCircumscri of Square
area Core Net
Ratio 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
Core area factor,KC
0.45 0.56 0.6 0.62
circle bingCircumscri of Area
area core Net
circle bingCircumscri of Area
area core Gross
Flux density decides,
Area of cross section of the core
Core loss
Choice of flux density depends on,
Service condition (DT/PT)
Material usedCold 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 densityDistribution transformer : 1.1 to 1.4 wb/m2
Power Transformer : 1.2 to 1.5 wb/m2
47
Design of CoreChoice of Flux Density in Core
48
dWwd
D
Hw
Hy
Hy
H
Dy
W
a
Overall DimensionsSingle phase Core Type
49
b
Ww
Hw
a
a
2a
Depth Over
winding
Overall DimensionsSingle phase Shell Type
50D
dWwd
Hw
Hy
Hy
H
Dy
a
dWwd
Hw
Hy
Hy
H
Dy
W
a
D
Overall DimensionsThee phase Core Type
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.
51 windingLV of Current RatedI
windingLV of voltage RatedV where,
T winding,LV in turns of No.
LV
LV
LVt
LVLV I
AT)or(
E
V
52
Design of Winding
windingHV of voltage Rated
T winding,HV in turns of No.
HV
LV
HVLVHV
V,where
V
VT
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.
Methods of cooling:1. Air Natural (AN)-upto 1.5MVA2. Air Blast (AB) 3. Oil natural (ON) – Upto 10 MVA4. Oil Natural – Air Forced (ONAF)5. Oil Forced– Air Natural (OFAN) – 30 MVA6. Oil Forced– Air Forced (OFAF)7. Oil Natural – Water Forced (ONWF) – Power plants8. Oil Forced - Water Forced (OFWF) – Power plants
Cooling of transformers
Specific heat dissipation due to convection is,
The average working temperature of oil is 50-600C.
For
The value of the dissipation in air is 8 W/m2.0C. i.e, 10 times less than oil.
Cooling of transformersTransformer Oil as Cooling Medium
m surface, gdissipatin ofHeight
oil, torelative surface theof difference eTemperatur - where,
./3.40
0
024
1
H
C
CmWHconv
,15.0&200 mtoHC
.. W/m100 to80 02 Cconv
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,
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)
Cooling of transformersTemperature rise in plain walled tanks
t
ci
S
PP
5.12
tankof surface
gdissipatinHeat
nDissipatio
heat Specificlosses Total
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.
Let, Dissipating surface of tank – St
Dissipating surface of tubes – XSt
Loss dissipated by surface of the tank by radiation and convection =
Design of tanks with cooling tubes
tt SS 5.125.66
)1(SX8.85.12XS8.8S5.12 ttt
tubes and by walls
dissipated loss Total
tt XS8.8XS100
1355.6
convectionby tubes
by dissipated Loss
)1(SSS tubesand s tank wallof area totalActual ttt XX
Design of tanks with cooling tubes
)2()1(
)8.85.12(
)1(
)8.85.12(surface gdissipatin of mper dissipated Loss
area Total
dissipated losses Totalsurface gdissipatin of mper dissipated Loss
2
2
X
X
XS
XS
t
t
5.12PP
8.8
PP)8.85.12(
)8.85.12(
PP have, we(3), and (1) From
)3(PPP losses, Total
Dissipated Loss
loss Total
tubescoolingr with Transforme
in rise eTemperatur
ci
ci
ci
ciloss
t
t
t
SX
SX
XS
5.12
PP
8.8
1 ci
tSX
Design of tanks with cooling tubes
)6(5.12PP
8.8
1each tube of Area
tubesof area Total tubes,ofnumber Total
,
)5(5.12PP
8.8
15.12
PP
8.8
1 tubescooling of area Total
ci
cici
ttt
t
t
tt
t
t
ttt
Sld
n
n
ldtubesofareaSurface
tubesofDiameterd
tubesofLengthlLet
SSS
Standard diameter of cooling tube is 50mm & length depends on the height of the tank.
Centre to centre spacing is 75mm.
D D
WT
HT
C4
C3
C1DocC2
LT
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
Voltage kVA RatingClearance in mm
C1 C2 C3 C4
Up to 11kV <1000kVA 40 50 75 375
Upto 11 kV1000-
5000kVA70 90 100 400
11kV – 33kV <1000kVA 75 100 75 450
11kV – 33kV1000-
5000kVA85 125 100 475
Design of tanks with cooling tubes
Estimation of No-load Current
No-load Current of Transformer:Magnetizing Component
Depends on MMF required to establish required flux
Loss ComponentDepends on iron loss
65
Total Length of the core = 2lc
Total Length of the yoke = 2ly
Here, lc=Hw=Height of Window
ly= Ww=Width of Window
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 CurrentNo-load current of Single phase Transformer
lClC
ly
ly
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.
No-load current,
67
Estimation of No-load CurrentNo-load current of Single phase Transformer
22m0 III l
68
lClClC
ly
ly
Estimation of No-load CurrentNo-load current of Three phase Transformer