unit-3

T ra

Unit 3Unit 31

Introduction

Constituents of transformer:i. Magnetic Circuitii. Electric Circuitiii. Dielectric Circuitiv. Other accessories

2

Classification or Types

3

Constructional Details

4


5


6


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 figureThe 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


10

Shell Type

Core Type


Ww

Hw

11

Transformer WindingTransformer Winding

Types of transformer windings are, Concentric Sandwich

Disc Cross over Helical

12


Ww

Hw

13


14

InsulationsInsulations

Dry type Transformers:VarnishEnamelLarge size Transformers:Unimpregnated paperCloths

Immersed in Transformer oil – insulation & coolent

15

Comparison between Core & Shell Type

Description Core Type Shell TypeConstruction Easy to assemble &

DismantleComplex

Mechanical Strength

Low High

Leakage reactance Higher Smaller

Cooling Better cooling of Winding

Better cooling of Core

Repair Easy HardApplications High Voltage & Low

outputLow Voltages & Large Output

16

Classification on ServiceDetails Distribution

transformerPower Transformer

Capacity Upto 500kVA Above 500kVAVoltage rating 11,22,33kV/440V 400/33kV;220/11kV

…etc.,Connection Δ/Y, 3φ, 4 Wire Δ/Δ; Δ/Y, 3φ, 4 WireFlux 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

Window in a 1φ transformer contains one primary & one secondary winding.


)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


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,


)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,


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,


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

QKE t

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 K1φ Shell Type 1.0 to 1.21φ Core Type 0.75 to 0.853φ Shell Type 1.2 to 1.33φ Core Type

Distribution Xmer0.45 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 Volumeii. Total Weightiii. Total Costiv. 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


c

im

c

im

AAB2

2A

ABr

)2(rB2A

A

B2AAr

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


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,whereAgcAgcC

ci

ccciiit

mti Ll


MgcMgcC cciit mti Ll

Differeniating Ct with respect to β, 2/3

cc2/1

iit Mgc21)(Mgc

21C

dd 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


ci

ccii

CCGcGc

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

EfficiencyTotal 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 PxPxQxQ

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-QPxPxQx

0d

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

ppx

GGor

GpGpx

GpGp

PP

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 toWidthDepth

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

5.02

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

45.02

2

d

d

64.0

4

5.02

2

d

d


Useful ratio in design – Core area factor,

41


45.0d

d45.0dA

K

2

2

2i

C

Circle bingCircumscri of Square

area Core Net

Let,a – Length of the rectangleb – breadth of the rectangled – 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,0

ddAgi

sindbdbsin

cosdadacos

Two stepped core can be divided in to 3 rectangles.Referring to the fig shown,

43


θ

d

b

b

a

a

(a-b)/2

(a-b)/2

22 bab2babab

b2

)ba(2ab

b2

bab2

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


01

1

22

22gi

72.312tan21

2tan2

22tan

22cos2sin

2sin2cos2cossin2d2cos2d

0cossin2d2cos2dd

dA

d53.0)72.31sin(dbsindbdbsin

d85.0)72.31cos(dacosdadacos

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


79.0d

4

d618.0

71.0d

4

d56.0

2

2

2

2

circle bingCircumscri of Areaarea core Gross

circle bingCircumscri of Areaarea core Net

Core area factor,

Ratios of Multi-stepped Cores,

46


56.0d

d56.0dA

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 Areaarea core Net

circle bingCircumscri of Areaarea core Gross

Flux density decides,Area of cross section of the coreCore loss

Choice of flux density depends on,Service condition (DT/PT)Material used

Cold rolled Silicon Steel-Work with higher flux densityUpto 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

Wa

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,whereVVT

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

HC

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

SPP

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 =


tt SS 5.125.66

)1(SX8.85.12XS8.8S5.12 ttt

tubes and by wallsdissipated loss Total

tt XS8.8XS1001355.6

convectionby tubesby dissipated Loss

)1(SSS tubesand s tank wallof area totalActual ttt XX


)2()1(

)8.85.12()1(

)8.85.12(surface gdissipatin of mper dissipated Loss

area Totaldissipated losses Totalsurface gdissipatin of mper dissipated Loss

2

2

XX

XSXS

t

t

5.12PP8.8

PP)8.85.12(

)8.85.12(PP have, we(3), and (1) From

)3(PPP losses, TotalDissipated Loss

loss Total tubescoolingr with Transforme

in rise eTemperatur

ci

ci

ci

ciloss

t

t

t

SX

SX

XS

5.12PP

8.81 ci

tSX


)6(5.12PP8.8

1each tube of Area

tubesof area Total tubes,ofnumber Total

,

)5(5.12PP8.8

15.12PP8.8

1 tubescooling of area Total

ci

cici

ttt

t

t

tt

t

t

ttt

Sld

n

n

ldtubesofareaSurfacetubesofDiameterd

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


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.


Voltage kVA RatingClearance in mm

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

11kV – 33kV 1000-5000kVA 85 125 100 475


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

unit-3

Documents

transformer windingwwhw

phase transformer

core shell type

core areaampere

core type0

window area window space

core type power xmer0

core type distribution