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Output Equations Main Dimensions - KVA output forsingle and three phase transformers Window spacefactor Overall dimensions Operating characteristics Regulation No load current Temperature rise inTransformers Design of Tank -Methods of cooling ofTransformers.
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INTRODUCTION:
A transformer is essentially a static electromagnetic
device consisting of two or more windings which link witha common magnetic field.
The primary is connected to an alternating voltagesource, an alternating flux is produced whose amplitudedepends on the primary voltage and the no. of turns.
A transformer is not an energy conversion device, but adevice that transforms electrical energy from one ormore primary a.c circuits to one or more secondary a.ccircuits with changed values of voltage and current.
The transformer is extremely important as a componentin many different types of electric circuits, from small-signal electronic circuits to high voltage powertransmission systems.
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IMPORTANT FUNCTIONS OF A TRANSFORMER:
Changing voltage and current level in anelectric system.
Matching source and load impedances for
maximum power transfer in electronic andcontrol circuitry.
Electrical isolation.
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PARTS OF A TRANSFORMER
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TYPES OF CORES
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CLASSIFICATION:
Based on construction,
i. Core type and
ii. Shell type
Based on application,
i. Distribution and
ii. Power transformers.
CORE TYPE:
The magnetic core is built of laminations to form a
rectangular frame.
The windings are arranged concentrically with each otheraround the legs or limbs.
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The top and bottom horizontal portion of the core are
called yoke.
The yokes connect the 2 limbs and have a cross-
sectional area >or =to that of limbs.
Each limb carries one half of primary and secondary.
The 2 windings are closely coupled together to reduce
the leakage reactance.
The low voltage winding is wound near the core and the
h.v winding is wound away from the core in order to
reduce the amount of insulating materials required.
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SHELL TYPE:
The windings arte put around the central limb & the
flux path is completed through the 2 side limbs.
Central limb carries total mutual flux.
Side limbs form a part of a parallel magnetic circuit& carry half the total flux.
The cross-sectional area of the central limb is twice
that of each side limbs.
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DISTRIBUTION TRANSFORMER:
These are transformers upto 200kVA(or 500kVA)
are used to step down distribution voltage to a
standard service voltage.
They are kept in operation all the 24 hours a daywhether carrying any load or not.
The load varies from time to time & it will be on no
load most of the time.
So copper loss is more compared to core loss.
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Distribution transformers are designed with less
iron loss and have a maximum efficiency at a load
much lesser than the full load.
It should have good regulation to maintain the
variation of supply voltage within limits. So it is
designed with small value of leakage reactance.
POWER TRANSFORMER:
Used in sub-stations and generating stations & have
ratings above 200kVA.
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A substation has number of transformers working
in parallel.
During heavy loads all the transformers are put inoperation & during light loads some of them are
disconnected.
So power transformers should be designed to have
a maximum efficiency at or near full load.
Designed to have a greater leakage reactance to
limit fault current.
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COMPARISON OF CORE TYPE AND SHELL TYPE:
CORE TYPE SHELL TYPE1. Easy in design and
construction.
2. Has low mechanical
strength due to non-
bracing of windings.
3. Reduction of leakage
reactance is not easily
possible.
4. The assembly can beeasily dismantled for
repair work.
1. Comparatively complex.
2. High mechanical
strength .
3. Reduction of leakage
reactance is not highly
possible.
4. The assembly cannot be
easily dismantled forrepair work.
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5. Better heat dissipation from
windings.
6. Has longer mean length of
core and shorter mean
length of coil turn, hence
best suited for EHVrequirements.
5. Heat is not easily
dissipated from windings,since it is surrounded by
core.
6. It is not suitable for EHV
requirements.
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LOW VOLTAGE WINDING HIGH VOLTAGE
WINDING
HH L HL H L HLH HL L
CORECORE
HIGH VOLTAGE AND LOW VOLTAGE WINDING OF A TRANSFORMER
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CROSS SECTION OF SINGLE PHASE TRANSFORMER
Ww
Hw
Ww Ww
Hw
CROSS SECTION OF CORE
TYPE SINGLE PHASE
TRANSFORMER
CROSS SECTION OF SHELL
TYPE SINGLE PHASE
TRANSFORMER
WINDOW
AREA
WINDOW
AREAWINDOW
AREA
CORE CORE
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OUTPUT EQUATION OF SINGLE-PHASE TRANSFORMERS
The equation which relates the rated kVA output of
a transformer to the area of core and window is
called output equation.
In transformers the output kVA depends on fluxdensity & ampere turns.
The flux density is related to the core area and the
ampere-turns is related to the window area.
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The induced emf in a transformer, E= 4.44fmT voltEmf per turn, Et = E/T
= 4.44fm volts
The window in a single-phase transformer contains oneprimary and one secondary winding.
Window space factor Kw is the ratio of conductor area inwindow to the total area.
Kw = Ac / Aw
Conductor area in window, Ac = Kw Aw
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The current density is the same for both the windings.
So current density, = Ip/ ap = Is / as
Area of cross section of primary conductor, ap = Ip/ and
Area of cross section of primary conductor, as = Is /
Ampere turns, AT= TpIp = TsIsAw = total window area ;
Kw = window space factor = Ac / AwAc = conductor area = K w Aw
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The total copper area in the window:
Ac = Copper area of primary winding+ Copper area ofsecondary winding
= (no. of primary turns area of cross section of primaryconductor)+ (no. of secondary turns area of crosssection of secondary conductor)
Ac = Tpap + Tsas= Tp Ip /+ Ts Is / (since ap = Ip/ & as
= Is / )= ( TpIp+ Ts Is)/= 1/ (AT+AT)
= 2AT/
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On equating equation (1) & (2), we get
KwAw = 2AT/
Ampere turns, AT = KwAw /2
KVA OUTPUT OF SINGLE PHASE TRANSFORMER:
Rating in kVA ,Q = Vp Ip x 10-3
(1) = Ep Ip x 10 -3
= Ep ( Tp Ip )/ Tp x 10-3
= Et AT x 10-3
On substituting the value of Et & AT,
Q = 4.44 f m (Kw Aw ) /2 x 10 -3
where m = Bm AiQ = 2.22 f Bm Ai Kw Aw x 10 -3
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OUTPUT EQUATION OF THREE-PHASE TRANSFORMERS
The induced emf in a transformer, E= 4.44fm
T volt
Voltage per turn, Et = E/T
= 4.44fm volt In case of a three-phase transformer, each window contains
two primary and two secondary windings.
The total copper area in the window:
Ac = 2Tpap + 2Tsas = ( TpIp+ Ts Is)2/= 4AT/
since ap = Ip/ and as = Is / since AT= TpIp = TsIs ( neglecting magnetizing current)
Aw = total window area ;Kw = window space factor = Ac / AwAc = conductor area = Kw Aw
= 4AT/ ampere turns, AT = KwAw /4
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KVA OUTPUT OF THREE PHASE TRANSFORMER:
Rating in kVA ,Q = 3 x Vp Ip x 10-3
= 3 x Ep Ip x 10 -3
= 3 x Ep ( Tp Ip )/ Tp x 10-3
= 3 x Et AT x 10-3
Q= 3 x 4.44 f m (Kw Aw ) /4 x 10 -3
where m = Bm Ai= 3.33 f Bm Ai Kw Aw x 10 -3
Using the output equation it can also be shown that
E t = K kVAwhere K = 4.44 f r x 10 -3
r = m / ATr is a constant for transformer of a given type ,service andmethod of connection, since m determines the core sectionand AT fixes the total copper area.
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Value of K
( 1.0 to 1.2) for single phase shell type
1.3 for three-phase shell type (power)
(0.75 to 0.85) for single phase core type
(0.6 to 0.7) for three phase core type (power)
0.45 for three-phase core type (distribution)
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Output Equations Main Dimensions - KVA output forsingle and three phase transformers Window spacefactor Overall dimensions Operating characteristics Regulation No load current Temperature rise inTransformers Design of Tank -Methods of cooling ofTransformers.
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MAIN DIMENSIONS
(i) Design of core.
(ii) Design of yoke.(iii)Design of winding
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DESIGN OF CORES
The core section of the core type transformer may be
rectangular, square or stepped.
Shell type transformers use cores with rectangular cross
section.
In core type transformers with rectangular core the ratio of
depth to width of the core is 1.4 to 2. In shell type transformers with rectangular core the width of
the central limb is 2 to 3 times the depth of the core.
When circular coils are required forhigh voltage transformers
,square and stepped cores are used .
Circular coils are preferred because of their superior
mechanical characteristics.
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CROSS-SECTION OF TRANSFORMER CORES
RECTANGULAR
CORE
SQUARE
CORE
STEPPED CORE
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SQUARE CORE:
Let d= diameter of the
circumscribing circle.
Also d=diagonal of the square
a= side of the square
Diameter of the circumscribing circle,d = a2+a2= 2a2 = 2a
Side of the square, a= d/2
Gross area of the square, Agi = area of the square=a2 =
(d/2)2 = 0.5d2
Let stacking factor, Sf=0.9
a
a
aad
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Net core area, Ai=Stacking factor Gross core area
= 0.9 0.5d2
= 0.45 d2
Gross core area is the area including the insulation
area.
Net core area is the area of iron alone excludinginsulation area.
Area of circumscribing circle = /4d2
The ratio, Net core area/ Area of
circumscribing circle
= 0.45d2/( /4d2) = 0.58
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The ratio, Gross core area/ Area of
circumscribing circle
= 0.5d2 /( /4d2) = 0.64
Core area factor is the ratio of net core area and
square of the circumscribing circle.
Core area factor= Net core area/square of thecircumscribing circle
= Ai/d2 = 0.45d2 /d2
= 0.45
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b
d
b
a
a
b
d
b
a
b
a-b/2
a-b/2
b
bd
CROSS SECTION OF TWO
STEPPED CORE
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The maximum core area for a given d is obtained
when is maximum.
Cos= a/d; so a=dcos (1)
sin = b/d; so b=dsin (2)
The 2 stepped core can be divided into 3 rectangles.
The area of three rectangles give the gross core area.
Gross core area, Agi = ab+[(a-b)/2]b+[(a-b)/2]b
= ab+ab-b2
= 2ab-b
2
(3)
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So d2cos22- d22cossin =0
d22cossin = d2cos22
d2 sin2 = d2cos22
sin2/cos2 =2
tan2= 2; 2=tan-12
= 1/2 tan-12 = 31.72When = 31.72, the dimensions of the core will give
the maximum area for the core for a specified d.
a=d cos ; b=d sin
= d cos31.72; = d sin31.72
= 0.85d; = 0.53d
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Substitute the values of a &b in equation (3) we get,
Gross core area, Agi = 2ab-b2
= 0.618d2
Let stacking factor, Sf=0.9
Net core area, Ai=Stacking factor Gross core
area
= 0.9 0.618d2
= 0.56d2
The ratio, Net core area/ Area of
circumscribing circle= 0.56d2 /( /4d2)= 0.71
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The ratio, Gross core area/ Area of
circumscribing circle= 0.618d2 /( /4d2) = 0.79
Core area factor= Net core area/square of the
circumscribing circle
= Ai/d2 = 0.56d2 /d2
= 0.56
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CHOICE FLUX DENSITY and CURRENT DENSITY ( Bmand )
Bm determines the core area. Higher Bm smaller area smaller Lmt saving in
the costof iron and copper.
Buthigher Bm increases the iron loss and temp rise.
For Distribution transformerBm =1.1 to 1.35 Wb/m2
. For Power transformerBm =1.25 to 1.45 Wb/m
2.
The area of conductors for the primary and secondarywindings determined after choosing a suitable value for which depends on the methodof cooling.
Current density value depends on method of cooling andrange is 1.1 to 2.2 A/mm2
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TYPES OF WINDINGS
Cylindrical winding with circular conductors.Crossover winding with circular or rectangular
conductors.
Continuous disc type winding with rectangularconductors.
Helical winding.
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DESIGN OF WINDING
The design of winding involves the determination of
no. of turns & area of cross section of the conductorused.
The no. of turns is estimated using voltage rating &
e.m.f per turns.
The area of cross section is estimated using ratedcurrent & current density.
Usually the no. of turns of L.V winding is estimated
first using the given data & it is corrected to the
nearest integer.
Then the no. of turns of H.V winding are chosen to
satisfy the voltage rating of the transformer.
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Number of turns in a low voltage winding,
TLV
= VLV
/ Et
or AT/ILV
where,
VLV = rated voltage of low voltage winding.
ILV = rated current of low voltage winding.
Number of turns in a high voltage winding,
THV = TLV VHV / VLV
where,
VHV = rated voltage of high voltage winding.Rated current in a winding =
kVA per phase 103 /voltage rating of the winding
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DESIGN OF YOKE
The purpose of the yoke is to connect the legs
providing a least reluctance path. In order to limit the
iron loss in the yoke, operating flux density is
reduced by increasing the yoke area.
Generally yoke area is made 20% more than the legarea.
In case of rectangular yoke ,
depth of yoke = the depth of core.
In square or stepped ,depth of core =width of largest stampings
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DESIGN OF YOKE
Area of yoke =depth of yoke x height of yoke= Dy x Hy
Dy= width of largest core stamping = a
Hy=(1.15 to 1.25) Agi for transformers using grainoriented steel
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OVERALL DIMENSIONS OF A TRANSFORMER
The main dimensions of a transformer are:
Height of the window (Hw) and
Width of the window (Ww)
Other important dimensions are:
Width of the largest stamping(a)
Diameter of the circumscribing circle(d)
Distance between the core centres(D)
Height of the yoke(Hy)
Depth of the yoke(Dy)
Overall height of transformer frame(H)
Overall width of transformer frame(W)
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OVERALL DIMENSIONS
a = width of the largest stamping ;
d = diameter of the circumscribing circle;
D = distance between centres of adjacent limbs;
Ww, Hw = width and height of the window ( length of thewindow);
Hy = height of the yoke;
Forcore type: D = d + Ww ; Dy=a,
W = D+a ;
H = Hw + 2 HyWidth over two limbs=D + outer diameter of h.v.windings
Width over one limbs=outer diameter of h.v.windings
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Forthree phase transformers :
D= d + Ww , Dy=a,
H=Hw+2Hy ;
W=2D+a;
Width over 3 limbs=2D+outer diameter of h.v.winding
Width over one limb = outer diameter of h.v.winding
For Single phase shell type :
Dy = b ;
Hy = a ;W = 2Ww+4a ;
H = H w+ 2a
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Leakage reactance of winding
The estimation of leakage reactance is the primarily the
estimation of the distribution of leakage flux and the resultingflux leakages of the primary and secondary windings
The distribution of leakage flux depends upon the geometrical
configuration of the coils and the neighboring iron masses and
permeability of the latter
Leakage reactance of core type transformer
Leakage reactance of sandwich coils
Leakage reactance of core type transformer
Per unit leakage reactance
23
p smtx o
t c
b bLATf aE L
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Note on Reactance:
Useful flux: It is the flux that links with both
primary and secondary windings and is responsiblein transferring the energy Electro-magnetically fromprimary to secondary side. The path of the usefulflux is in the magnetic core.
Leakage flux: It is the flux that links only with theprimary or secondary winding and is responsible inimparting inductance to the windings. The path ofthe leakage flux depends on the geometrical
configuration of the coils and the neighboring ironmasses.
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Leakage reactance of sandwich coils
The idealized flux distribution in shell type transformers
Each of n coils is sandwiched between two coils of L.v.winding.
Per unit reactance
6
p so mtx
t
b bf LATa
n E w
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Output Equations Main Dimensions - KVA output forsingle and three phase transformers Window spacefactor Overall dimensions Operating characteristics Regulation No load current Temperature rise inTransformers Design of Tank -Methods of cooling of
Transformers.
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REGULATION
The p.u regulation, for full load rated output Q and full load
current Ip is :
If the regulation is large and the phase shift between Vp and Vp
is not justified. For this case:
p
pppp
p
pp
V
XIRI
V
VV
sincos,
sincos pr
2
)sincos(2
1
sincos rppr
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Output Equations Main Dimensions - KVA output forsingle and three phase transformers Window spacefactor Overall dimensions Operating characteristics Regulation No load current Temperature rise inTransformers Design of Tank -Methods of cooling of
Transformers.
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ESTIMATION OF NO LOAD CURRENT OF
TRANSFORMER
No load current of a transformer has 2 components:
Magnetizing componentdepends on the mmf required
to establish the desired flux.
Loss component depends on the iron losses.
NO LOAD CURRENT OF A SINGLE PHASE TRANSFORMER:
Total length of core=2lc
Total length of yoke= 2ly
lc=Hw = height of the window
ly=Ww= width of the window
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mmf for core = mmf/metre for max. flux density in core
total length of core
= atc 2lc= 2 atclcmmf for yoke = mmf /metre for max. flux density yoke
total length of yoke
= aty 2ly = 2 atyly
Total magnetizing mmf,AT0=mmf for core + mmf for yoke + mmffor joints
=2 atclc + 2 atyly + mmf for joints
Maximum value for magnetizing current = AT0 / Tp If the magnetizing current is sinusoidal, rms value for
magnetizing current,
Im= AT0/2Tp
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NO LOAD CURRENT OF THREE PHASE TRANSFORMER
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NO LOAD CURRENT OF THREE PHASE TRANSFORMER:
Total length of core=3lcTotal length of yoke= 2ly
lc=Hw = height of the window
ly=Ww= width of the window
mmf for core= mmf/metre for max.flux density in core
total length of core
= atc 3lc= 3 atclc
mmf for yoke = mmf /metre for max. flux density in yoke
total length of yoke
= aty 2ly = 2 atyly
Total magnetizing mmf,AT0=mmf for core + mmf for yoke + mmf
for joints
=3 atclc + 2 atyly + mmf for joints
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DESIGN OF INSULATIONBasic consideration in design the insulation
ELECTRICAL INSULATION: Depends on the operating voltage, eddy
current loss in the conductors and tank walls.
MECHANICAL CONSIDERATIONS: depends on the capable to with
stand mechanical Stresses during fault .
THERMAL CONSIDERATIONS: depends on Safe operating of
temperature values and types of cooling employed
Insulation of transformers divided in to four types
Major , Minor , insulation relative to tank , insulation between phases
MAJOR INSULATION : Between windings and core (grounded).
MINOR INSULATION :Between turns, coils and layers.
MATERIALS : cotton thread, cotton tape, leatheroid paper,
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TRANSFORMER OIL AS A COOLING MEDIUM
The specific heat dissipation due to convection ofoil
conv = 40.3 ( /H) W/m2 - C ;where, = temp difference of the surface relative to the oil, CH = height of the dissipating surface, m.
Average values 0
2 0
20 . . 0.5 1
80 100 /conv
C and H to m
to W m C
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TEMPERATURE RISE IN PLAIN TANKED WALLS
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TEMPERATURE RISE IN PLAIN TANKED WALLS
The transformer core and winding is placed inside a containercalledtank
The tank will dissipate the heat byboth radiation andconvections
For temperature rise over 400C over the ambient temperature200C,
The specific heat dissipation are follows,
Due to radiation 6.0 W/m
2
-C and Due to convection 6.5 W/m2-c
Thus a total of 12.5 W/m2- C is taken.
..
. . . . . . . tan
.12.5
i c
t
total lossTemperature rise
sp heat dissipation heat dissipation surface of the k
P PTemperature rise
S
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Where,
St = heat dissipating surface area of the tank.
= Specific heat dissipation
Pi = Iron loss; Pc = copper lossHeat dissipating surface of the tank =Total area of vertical sides+
1/2 area of top cover.
The area of bottom of the tank should be neglected as it has very
little cooling effect. Transformers rated for larger outputs must be provided with
means to improve the conditions of heat dissipation. This achieved
by providing cooling tubes and radiators.
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Output Equations Main Dimensions - KVA output forsingle and three phase transformers Window spacefactor Overall dimensions Operating characteristics Regulation No load current Temperature rise inTransformers Design of Tank -Methods of cooling of
Transformers.
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DESIGN OF TANK WITH COOLING TUBES
For small transformers,plain walled tank is enough todissipate the losses.
The transformers areprovided with cooling tubes to increase
the heat dissipating area.
Tubes mountedon vertical side of the tank
Other hand, thetube will improve the circulation of oil. This
improves the dissipation of lossby convection
The improvement in
loss dissipation by convection = loss dissipated by 35% of tube surface area.
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Let,
Dissipating surface of the tank = St
Dissipating surface of the tubes = xStLoss dissipated by the tank surface = (6+6.5) St=12.5 St
Loss dissipated by the tubes = (135/100 x 6.5) x Stby convection = 8.8 x St
Total loss dissipated by the walls and tubes
= (12.5 St + 8.8 xSt)
= (12.5 + 8.8 x) St
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Actual total area of tank walls and tubes
= St + x St= S
t(1+ x)
Loss dissipated per m2 of dissipating surface
=Total loss dissipated/ Total area
=(12.5 + 8.8x) St / St(1+ x)
= (12.5 + 8.8x)/(1+x)
Temperature rise in transformer with cooling tubes,= Total loss/ Total Loss dissipated
Total loss, Ploss = Pi + PcHence,
= ( Pi + Pc)/ (12.5 + 8.8x) St12.5 + 8.8x = ( Pi + Pc)/ Stx = [{( Pi + Pc)/ St }-12.5] (1/8.8)
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Total area of cooling tubes = x St= (1/8.8) [ {(Pi + Pc)/ } 12.5 St ] St
The total number of tubes = nt
= Total tube area / Area of each tube
= Total tube area /( dt lt )
nt = (1/8.8dtlt) [ {(Pi + Pc)/}12.5 St ]
The arrangement of the tubes on tank side walls should be
made uniformly with a spacing of usually 75 mm.
The standard diameter of the cooling tubes is 50 mm and the
length of the tube depends on the height of the tank.
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The dimensions of the tank are decided by the
dimensions of the transformer and the clearance
required on all sides.Let
C1= clearance between winding & tank along the width
C2= clearance between winding & tank along the length
C3= clearance between the transformer frame & the tank at the
bottom
C4= clearance between the transformer frame & the tank at the
top.
Doc= outer diameterof the coil.
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With reference to the figure we get
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With reference to the figure we get,
Width of the tank, WT= 2D+ Doc+2 C1(3-phase)
= D+ Doc+2 C1(1-phase)
length of the tank, LT= Doc+2 C2
Height of the tank, HT= H+C3+C4
The clearance on the sides depends on voltage and power rating
of the windingVoltage KVA rating Clearance in mm
C1 C2 C3 c4
Up to 11KV
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Output Equations Main Dimensions - KVA output forsingle and three phase transformers Window spacefactor Overall dimensions Operating characteristics Regulation No load current Temperature rise inTransformers Design of Tank -Methods of cooling of
Transformers.
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METHODS OF COOLING OF TRANSFORMERS
The losses developed in a transformer are converted
into heat energy & cause heating of correspondingtransformer parts.
The heat dissipation in a transformer occurs by
conduction, convection & radiation.
The path of heat flow in a transformer are:
From the internal most heated spots of a given
part(core or winding) to their outer surface in
contact with the oil by conduction.
From the outer surface of a transformer part to the
oil that cools it by convection.
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From the oil to the walls of the cooler by
convection (Eg. wall of a tank)
From the walls of the cooler to the coolingmedium air or water by both convection &
radiation.
The various methods of cooling transformers are:
Air natural(AN)
Air blast(AB)
Oil natural(ON)
Oil natural- Air forced(ONAF)
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Oil natural- Water forced(ONWF)
Forced circulation of oil(OF)
Oil forced-Air natural(OFAN)
Oil forced-Air forced(OFAF)
Oil forced - Water forced(OFWF)
The choice of cooling method depends upon the,size,
type of application &
type of conditions obtaining at the site where the
transformer is installed.Natural cooling is suitable up to 10MVA
The forced oil and air circulation -30MVA