Unit III Transformers

7/28/2019 Unit III Transformers

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IFETCE/EEE/M.SUJ ITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0


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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

28IFETCE/EEE/M.SUJ ITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0


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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


= 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


= 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

IFETCE/EEE/M.SUJ ITH/III YEAR/VI SEM/EE 2355/DEM/VER 1.0


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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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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


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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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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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

Unit III Transformers

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Transcript of Unit III Transformers