Unit III Transformers

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Output Equations – Main Dimensions - KVA output for single and three phase transformers – Window space factor – Overall dimensions – Operating characteristics – Regulation – No load current – Temperature rise in Transformers – Design of Tank -Methods of cooling of Transformers.

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3 3

INTRODUCTION:

 A transformer is essentially a static electromagnetic device consisting of two or more windings which link with a common magnetic field.

 The primary is connected to an alternating voltage source, an alternating flux is produced whose amplitude depends on the primary voltage and the no. of turns.

 A transformer is not an energy conversion device, but a device that transforms electrical energy from one or more primary a.c circuits to one or more secondary a.c circuits with changed values of voltage and current.

 The transformer is extremely important as a component in many different types of electric circuits, from small-signal electronic circuits to high voltage power transmission systems.

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IMPORTANT FUNCTIONS OF A TRANSFORMER

:

Changing voltage and current level in an

electric system.

Matching source and load impedances for

maximum power transfer in electronic and

control circuitry.

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5 5

PARTS OF A TRANSFORMER

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7 7

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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 other around the legs or limbs.

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9 9

 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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11 11

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 day

whether carrying any load or not.

 The load varies from time to time & it will be on no

load most of the time.

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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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13 13

 A substation has number of transformers working

in parallel.

 During heavy loads all the transformers are put in

operation & 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 TYPE

1. 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 be easily 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 for repair work.

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15 15 5. Better heat dissipation from

windings.

6. Has longer mean length of core and shorter mean

length of coil turn, hence best suited for EHV

requirements.

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

H

H L L H H L L H H L L H

CORE CORE

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17 17

CROSS SECTION OF SINGLE PHASE TRANSFORMER

W_w

H_w

W_w _W

w

H_w

CROSS SECTION OF CORE TYPE SINGLE PHASE

TRANSFORMER

CROSS SECTION OF SHELL TYPE SINGLE PHASE

TRANSFORMER WINDOW AREA WINDOW AREA WINDOW 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 flux

density & 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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19 19  The induced emf in a transformer, E= 4.44fΦ_mT volt

Emf per turn, E_t= E/T

= 4.44fΦ_m volts

 The window in a single-phase transformer contains one primary and one secondary winding.

 Window space factor K_wis the ratio of conductor area in window to the total area.

K_w= A_c/ A_w

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 The current density is the same for both the windings.

/ δ

 Ampere turns, AT= T

_p

I

_p

= T

_s

I

_s

A

_w

= total window area ;

K

_w

= window space factor = A

_c

/ A

_w

A

_c

= conductor area

= K

_w

A

_w

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21 21

The total copper area in the window:

A

_c

= Copper area of primary winding+ Copper area of

secondary winding

= (no. of primary turns × area of cross section of primary

conductor)+ (no. of secondary turns × area of cross

section of secondary conductor)

_p

I

_p

+ T

_s

I

_s

)/δ

= 1/ δ (AT+AT)

= 2AT/ δ

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On equating equation (1) & (2), we get K_wA_w = 2AT/ δ

Ampere turns, AT = K_wA_w δ/2

KVA OUTPUT OF SINGLE PHASE TRANSFORMER:

Rating in kVA ,Q = V_p I_p x 10 -3 (1Φ) = E_p I_p x 10 -3

= E_p ( T_p I_p )/ T_p x 10 -3 = E_tAT x 10 -3

On substituting the value of E_t & AT,

Q = 4.44 f Φ_m (K_w A_w δ ) /2 x 10 -3 where Φ_m = B_m A_i

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23 23

OUTPUT EQUATION OF THREE-PHASE TRANSFORMERS

 The induced emf in a transformer, E= 4.44fΦ_mT volt  Voltage per turn, E_t= E/T

= 4.44fΦ_m volt

 In case of a three-phase transformer, each window contains two primary and two secondary windings.

 The total copper area in the window:

A_c = 2T_pa_p + 2T_sa_s= ( T_pI_p+ T_s I_s)2/δ = 4AT/ δ

since ap = I_p/ δ and a_s = I_s/ δ

since AT= T_pI_p = T_sI_s ( neglecting magnetizing current) A_w = total window area ;

K_w = window space factor = A_c/ A_w A_c = conductor area = K_w A_w

= 4AT/ δ ampere turns, AT = K_wA_w δ/4

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KVA OUTPUT OF THREE PHASE TRANSFORMER: Rating in kVA ,Q = 3 x V_p I_p x 10 -3 = 3 x E_p I_p x 10 -3 = 3 x E_p ( T_p I_p )/ T_p x 10 -3 = 3 x E_tAT x 10 -3 Q= 3 x 4.44 f Φ_m (K_w A_w δ ) /4 x 10 -3 where Φ_m = B_m A_i = 3.33 f B_m A_i K_w A_w δ x 10 -3 Using the output equation it can also be shown that

E t = K √ kVA where K =√ 4.44 f r x 10 -3

r = Φm / AT

r is a constant for transformer of a given type ,service and

method of connection, since Φm determines the core section and 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)

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Output Equations – Main Dimensions - KVA output for

single and three phase transformers – Window space

factor – Overall dimensions – Operating characteristics – Regulation – No load current – Temperature rise in Transformers – Design of Tank -Methods of cooling of Transformers.

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WINDOW SPACE FACTOR

 Defined as the ratio of copper area in the window of the total window area.

 It depends upon the relative amounts of insulation and copper provided, which in turn depends upon the voltage rating and output of transformer.

Empirical formula:

For below 50kVA  Kw=8/(30+kV) For 50kVA to 200kVA  Kw=10/(30+kV) For 1000kVA Kw=12/(30+kV)

Where Kv is the voltage of high voltage winding in kv

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MAIN DIMENSIONS

(i) Design of core.

(ii) Design of yoke.

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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 for high 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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31 31

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/√2)

2

_{= 0.5d}

2

Let stacking factor, Sf=0.9

a

circumscribing circle

= 0.5d

2

/(Π /4d

2

) = 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 the

circumscribing circle

= Ai/d

2

= 0.45d

2

/d

2

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TWO STEPPED CORE (or) CRUCIFORM CORE:

 In stepped core the dimensions of the steps should be

chosen, such as to occupy maximum area within a

circle.

 The dimensions of the 2 step to give maximum area

for the core in the given area of the circle are

determined below:

Let, a= length of the rectangle.

b= breath of the rectangle.

d= diameter of the circumscribing circle.

Also

d=diagonal of the rectangle.

θ=Angle between the diagonal & the

length of the rectangle.

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35 35 b d θ b a a b d θ b a b a-b/2 a-b/2 b b d θ

CROSS SECTION OF TWO STEPPED CORE

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 The maximum core area for a given d is obtained

when θ is maximum.

Cosθ= a/d; so a=dcosθ

(1)

sin θ= b/d; so b=dsin θ

(2)

 The 2 stepped core can be divided into 3 rectangles.

 The area of three rectangles give the gross core area.

Gross core area, A

_gi

= ab+[(a-b)/2]b+[(a-b)/2]b

= ab+ab-b

2

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37 37

Substitute for a and b in equation (3),

2

(2cosθsinθ-sin

2

θ)

= d

2

(sin2θ -sin

2

θ)

2cosθsinθ =0

d

2

2cosθsinθ = d

2

cos2θ×2

d

2

sin2θ = d

2

cos2θ×2

sin2θ/cos2θ =2

tan2θ= 2; 2θ=tan

-1

2 θ= 1/2 tan

-1

_{2 = 31.72°}

 When θ= 31.72°, the dimensions of the core will give

the maximum area for the core for a specified d.

a=d cosθ ; b=d sin θ

= d cos31.72; = d sin31.72

= 0.85d; = 0.53d

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39 39

Substitute the values of a &b in equation (3) we get,

Gross core area, A

_gi

= 2ab-b

2

= 0.618d

2

Let stacking factor, Sf=0.9

Net core area, Ai=Stacking factor× Gross core

area

circumscribing circle

= Ai/d

2

= 0.56d

2

/d

2

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CHOICE FLUX DENSITY and CURRENT DENSITY ( B_mand δ )

 B_m determines the core area.

 Higher B_m → smaller area → smaller Lmt → saving in the cost of iron and copper.

 But higher B_m increases the iron loss and temp rise.  For Distribution transformer B_m = 1.1 to 1.35 Wb/m2_.  For Power transformer B_m = 1.25 to 1.45 Wb/m2_.

 The area of conductors for the primary and secondary windings determined after choosing a suitable value for δ which depends on the method of cooling.

 Current density value depends on method of cooling and range 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 rectangular

conductors.

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43 43

DESIGN OF WINDING

 The design of winding involves the

determination of

no. of turns & area of cross section

of the conductor

used.

 The

no. of turns

is estimated using

voltage rating &

e.m.f per turns.

 The

area of cross section

is estimated using

rated

current & 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,

T

_LV

= V

_LV

/ E

_t

or AT/I

_LV

where,

V

_LV

= rated voltage of low voltage winding.

I

_LV

= rated current of low voltage winding.

 Number of turns in a high voltage winding,

T

_HV

= T

_LV

× V

_HV

/ V

_LV

where,

V

_HV

= rated voltage of high voltage winding.

 Rated current in a winding =

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45

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 leg

area.

• In case of rectangular yoke ,

depth of yoke = the depth of core.

• In square or stepped ,

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DESIGN OF YOKE

• Area of yoke =depth of yoke x height of yoke

= D

_y

x H

_y

D

_y

= width of largest core stamping = a

H

_y

=(1.15 to 1.25) A

_gi

for transformers using grain

oriented steel

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47 47

OVERALL DIMENSIONS OF A TRANSFORMER

 The main dimensions of a transformer are:

Height of the window (H_w) and Width of the window (W_w)  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(H_y)

Depth of the yoke(D_y)

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;

W

_w

, H

_w

= width and height of the window ( length of the

window);

Hy = height of the yoke;

For

core type

: D = d + W

_w

; D

_y

=a,

W = D+a ;

H = H

_w

+ 2 H

_y

Width over two limbs=D + outer diameter of h.v.windings Width over one limbs=outer diameter of h.v.windings

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49 49 For three phase transformers :

D= d + W_w , D_y=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 = 2W_w+4a ; H = H _w+ 2a

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Output Equations – Main Dimensions - KVA output for single and three phase transformers – Window space factor – Overall dimensions – Operating characteristics – Regulation – No load current – Temperature rise in Transformers – Design of Tank -Methods of cooling of Transformers.

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51

OPERATING CHARACTERISTICS

Resistance of winding

L_mts,L_mtp= length of primary & secondary windings,m;

r_p,r_s= resistance of primary and secondary winding respectively,m

Total I2 R loss in windings

Total resistance (per phase) of transformer referred to primary side

Per unit resistance

p mtp _s _mts p p p s

T L

_{T L}

r

andr

a





2 2 c p p s s

P



I r



I r

2 ₂ 2

.

p c s p p s p s p p s

T

P

I

R

r

and r

r

I

T





_

_





_

_





_

_













p p r p

I R

V





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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 resulting

flux 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

2

3

p s mt x o t c

b

L

AT

f

a

E

L





 



_





_





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53

Note on Reactance:

• Useful flux:

It is the flux that links with both

primary and secondary windings and is responsible

in transferring the energy Electro-magnetically from

primary to secondary side. The path of the useful

flux is in the magnetic core.

• Leakage flux:

It is the flux that links only with the

primary or secondary winding and is responsible in

imparting inductance to the windings. The path of

the leakage flux depends on the geometrical

configuration of the coils and the neighboring iron

masses.

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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 s o mt x t

b

f

AT

L

a

n

E

w

 







_





_





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55

Output Equations – Main Dimensions - KVA output for single and three phase transformers – Window space factor – Overall dimensions – Operating characteristics –

Regulation – No load current – Temperature rise in Transformers – Design of Tank -Methods of cooling of Transformers.

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REGULATION

 On no load the secondary terminal voltage Vp’=Vp. The drop in secondary terminal voltage from no load to full load can be calculated by using the phasor diagram.

At lagging power factor cos Φ,

Assuming that the angle θ between Vp and Vp’ is very small, we have

, 2 2

( cos sin ) ( cos sin )

p p p p p p p p p p V  V  I R



 I X



 I X



 I R





sin

cos

' p p p p p p

V

I

R

I

X

V





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57

REGULATION

The p.u regulation, for full load rated output Q and full load current I_p is :

If the regulation is large and the phase shift between Vp and Vp’ is not justified. For this case:

p p p p p p p p V X I R I V V V





cos sin ,    











_r

cos



_p

sin

2

)

sin

cos

(

2

1 sin

cos



















_r



_p



_p



_r

(58)

Regulation – No load current – Temperature rise in

Transformers – Design of Tank -Methods of cooling of Transformers.

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59

ESTIMATION OF NO LOAD CURRENT OF TRANSFORMER

No load current of a transformer has 2 components:

Magnetizing component –depends 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=2l_c Total length of yoke= 2l_y

l_c=H_w= height of the window l_y=W_w= width of the window

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mmf for core = mmf/metre for max. flux density in core × total length of core

= at_c×2l_c= 2 at_cl_c

mmf for yoke = mmf /metre for max. flux density yoke × total length of yoke

= at_y ×2l_y= 2 at_yl_y

Total magnetizing mmf,AT₀=mmf for core + mmf for yoke + mmf for joints

=2 at_cl_c+ 2 at_yl_y + mmf for joints

Maximum value for magnetizing current = AT₀/ T_p

 If the magnetizing current is sinusoidal, rms value for magnetizing current,

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(62)

NO LOAD CURRENT OF THREE PHASE TRANSFORMER:

Total length of core=3l_c Total length of yoke= 2l_y l_c=H_w = height of the window l_y=W_w= width of the window

mmf for core= mmf/metre for max.flux density in core × total length of core

= at_c×3l_c= 3 at_cl_c

mmf for yoke = mmf /metre for max. flux density in yoke × total length of yoke

= at_y×2l_y= 2 at_yl_y

Total magnetizing mmf,AT₀=mmf for core + mmf for yoke + mmf for joints

(63)

(64)

Regulation – No load current – Temperature rise in

Transformers – Design of Tank -Methods of cooling of Transformers.

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65

DESIGN OF INSULATION

Basic 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.

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TRANSFORMER OIL AS A COOLING MEDIUM

 The specific heat dissipation due to convection of oil

λ

_conv

= 40.3 (θ /H)

¼

_W/m

2

_{- °C ;}

where,

θ = temp difference of the surface relative to the oil, °C H = 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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67

TEMPERATURE RISE IN PLAIN TANKED WALLS

 The transformer core and winding is placed inside a container called tank

 The tank will dissipate the heat by both radiation and convections

 For temperature rise over 400_{C over the ambient temperature} 200C,

The specific heat dissipation are follows,

 Due to radiation 6.0 W/m2_{-°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 loss Temperature rise

sp heat dissipation heat dissipation surface of the k P P Temperature rise S     

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Where,

S_t = heat dissipating surface area of the tank. λ = Specific heat dissipation

Pi = Iron loss; Pc = copper loss

Heat 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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69

Output Equations – Main Dimensions - KVA output for single and three phase transformers – Window space factor – Overall dimensions – Operating characteristics – Regulation – No load current – Temperature rise in

Transformers – Design of Tank -Methods of cooling of

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DESIGN OF TANK WITH COOLING TUBES

 For small transformers,plain walled tank is enough to dissipate the losses.

 The transformers are provided with cooling tubes to increase the heat dissipating area.

 Tubes mounted on vertical side of the tank

 Other hand, the tube will improve the circulation of oil. This

improves the dissipation of loss by convection  The improvement in

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71

Let,

_t

Total loss dissipated by the walls and tubes

= (12.5 S

_t

+ 8.8 xS

_t

)

= (12.5 + 8.8 x) S

_t

(72)

Actual total area of tank walls and tubes

= S

_t

_t

l

_t

)

n

_t

= (1/8.8π d

_t

l

_t

) [ {(P

_i

+ P

_c

)/ θ} –12.5 S

_t

]

 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

(74)

 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.

(75)

75

DIMENSIONS OF TRANSFORMER TANK

H_T L_T W_T D_OC D_OC H D D C3 C₄ C₂

(76)

With reference to the figure we get,

Width of the tank, W_T= 2D+ D_oc+2 C₁(3-phase) = D+ D_oc+2 C₁(1-phase)

length of the tank , L_T= D_oc+2 C₂

Height of the tank , H_T= H+C₃+C₄

The clearance on the sides depends on voltage and power rating of the winding

Voltage KVA rating Clearance in mm

C1 C2 C3 c4

Up to 11KV <1000KVA 40 50 75 375 Up to 11KV 1000-5000KVA 70 90 100 400 11KV to 30KV <1000KVA 75 100 75 450 11KV to 30KV 1000-5000KVA 85 125 100 475

(77)

77

Output Equations – Main Dimensions - KVA output for single and three phase transformers – Window space factor – Overall dimensions – Operating characteristics – Regulation – No load current – Temperature rise in