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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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 smallsignal electronic circuits to high voltage power transmission systems. 3 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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. 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 other around 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 crosssectional 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 day whether 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 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 1.

Easy in design and construction. 2. Has low mechanical strength due to nonbracing of windings. 3. Reduction of leakage reactance is not easily possible. 4. The assembly can be easily dismantled for repair work.

SHELL TYPE 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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5.

Better heat dissipation from 5. windings.

6.

Has longer mean length of core and shorter mean length of coil turn, hence best suited for EHV requirements.

6.

Heat is not easily dissipated from windings, since it is surrounded by core. It is not suitable for EHV requirements.

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HIGH VOLTAGE AND LOW VOLTAGE WINDING OF A TRANSFORMER

LOW VOLTAGE WINDING

HL

L H

H L

HIGH VOLTAGE WINDING

L H

H

CORE

L

L H

CORE

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CROSS SECTION OF SINGLE PHASE TRANSFORMER

Ww

Ww Hw WINDOW

Ww Hw

WINDOW

WINDOW

AREA

AREA

AREA

CORE

CORE

CROSS SECTION OF CORE TYPE SINGLE PHASE TRANSFORMER

CROSS SECTION OF SHELL TYPE SINGLE PHASE TRANSFORMER

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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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 The induced emf in a transformer, E= 4.44fΦmT volt Emf per turn, Et = E/T = 4.44fΦm volts  The window in a single-phase transformer contains one primary and one secondary winding.  Window space factor Kw is the ratio of conductor area in window 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, δ= Ip/ ap = Is / as  Area of cross section of primary conductor, ap = Ip/ δ and  Area of cross section of primary conductor, as = Is / δ  Ampere turns, AT= TpIp = TsIs Aw = total window area ; Kw = window space factor = Ac / Aw Ac = conductor area = Kw Aw

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The total copper area in the window: Ac = 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) Ac = Tpap + Tsas = Tp Ip /δ+ Ts Is /δ (since ap = Ip/ δ & as = Is / δ) = ( TpIp+ Ts Is)/δ = 1/ δ (AT+AT) = 2AT/ δ

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On equating equation (1) & (2), we get KwAw = 2AT/ δ Ampere turns, AT = KwAw δ/2 KVA OUTPUT OF SINGLE PHASE TRANSFORMER: Rating in kVA ,Q = Vp Ip x 10 -3 (1Φ) = 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 Φm (Kw Aw δ ) /2 x 10 -3 where Φm = Bm Ai Q = 2.22 f Bm Ai Kw Aw δ x 10 -3 22 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

OUTPUT EQUATION OF THREE-PHASE TRANSFORMERS

 The induced emf in a transformer, E= 4.44fΦmT volt  Voltage per turn, Et = E/T = 4.44fΦm 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/δ = 4AT/ δ since ap = Ip/ δ and as = Is / δ since AT= TpIp = TsIs ( neglecting magnetizing current) Aw = total window area ; Kw = window space factor = Ac / Aw Ac = conductor area = Kw Aw = 4AT/ δ ampere turns, AT = KwAw δ/4 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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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 Φm (Kw Aw δ ) /4 x 10 -3 where Φm = Bm Ai = 3.33 f Bm Ai Kw Aw 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. 24 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

Value of K • • • • •

( 1.0 to 1.2) 1.3 (0.75 to 0.85) (0.6 to 0.7) 0.45

for single phase shell type for three-phase shell type (power) for single phase core type for three phase core type (power) for three-phase core type (distribution)

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

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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 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

SQUARE

CORE

CORE

STEPPED CORE

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SQUARE CORE: a Let d= diameter of the circumscribing circle. d a a Also d=diagonal of the square a= side of the square a 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 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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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 excluding insulation area. Area of circumscribing circle = Π /4d2 The ratio, Net core area/ Area of circumscribing circle = 0.45d2/(Π /4d2) = 0.58 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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The ratio, Gross core area/ Area of circumscribing circle = 0.5d2 /(Π /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 the circumscribing circle = Ai/d2 = 0.45d2 /d2 = 0.45

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

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b

b a-b/2

d

θ

b

a b

d

θ

b

b

a-b/2 a

CROSS SECTION OF TWO STEPPED CORE

a

d θ

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b

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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, Agi = ab+[(a-b)/2]b+[(a-b)/2]b = ab+ab-b2 = 2ab-b2 (3)

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Substitute for a and b in equation (3), Agi = 2(dcosθ)(dsinθ)-(dsinθ)2 = 2d2cosθsinθ-d2sin2θ = d2(2cosθsinθ-sin2 θ) = d2(sin2θ -sin2 θ) = d2sin2θ -d2sin2 θ (4)  To get maximum value of θ, differentiate Agi w.r.t θ , and equate to zero, i.e.,d/dθ Agi = 0  On differentiating equation (4) w.r.t θ we get, d/dθ Agi = d2cos2θ×2- d22cosθsinθ Put d/dθ Agi = 0 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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So d2cos2θ×2- d22cosθsinθ =0 d22cosθsinθ = d2cos2θ×2 d2 sin2θ = d2cos2θ×2 sin2θ/cos2θ =2 tan2θ= 2; 2θ=tan-12 θ= 1/2 tan-12 = 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 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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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 /(Π /4d2)= 0.71

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The ratio, Gross core area/ Area of circumscribing circle = 0.618d2 /(Π /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 δ )

 Bm determines the core area.  Higher Bm → smaller area → smaller Lmt → saving in the cost of iron and copper.  But higher Bm increases the iron loss and temp rise.  For Distribution transformer Bm = 1.1 to 1.35 Wb/m2.  For Power transformer Bm = 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. 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 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. IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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

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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 leg area. • In case of rectangular yoke , depth of yoke = the depth of core. • In square or stepped , depth of core =width of largest stampings IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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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 grain oriented 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 the window); Hy = height of the yoke; For core type: D = d + Ww ; Dy=a, W = D+a ; H = Hw + 2 Hy 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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For three 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 IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/ EE 2355/DEM/VER 1.0

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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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OPERATING CHARACTERISTICS

Resistance of winding Lmts,Lmtp = length of primary & secondary windings,m; rp,rs= resistance of primary and secondary winding respectively,m

Tp Lmtp

Ts Lmts rp   andrp   ap as Total I2 R loss in windings Pc  I p 2 rp  I s 2 rs Total resistance (per phase) of transformer referred to primary side

 Is Pc R p  2  rp    Ip Ip  Per unit resistance

r 

2

2

  Tp   and .rs  rp    rs  Ts  

I p Rp Vp 51

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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

bp  bs  AT Lmt   x  2 f o a   Et Lc  3  52

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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. 53

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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

bp  bs   f o AT Lmt  x  a   n Et w  6 

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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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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,

V p  (V p,  I p R p cos   I p X p sin  ) 2  ( I p X p cos   I p R p sin  ) 2 Assuming that the angle θ between Vp and Vp’ is very small, we have

V p  V p'  I p R p cos   I p X p sin  56

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

 

V p  V p, Vp



I p R p cos   I p X p sin  Vp

   r cos    p sin  If the regulation is large and the phase shift between Vp and Vp’ is not justified. For this case:

1    r cos    p sin   ( p cos    r sin  ) 2 2

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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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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=2lc Total length of yoke= 2ly lc=Hw = height of the window ly=Ww= width of the window 59

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mmf for core = mmf/metre for max. flux density in core × total length of core = atc ×2lc= 2 atclc mmf 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 + mmf for 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 60

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NO LOAD CURRENT OF THREE PHASE TRANSFORMER:

Total length of core=3lc Total 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 62

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

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

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

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.  MATERIALS : cotton thread, cotton tape, leatheroid paper, 65

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

TRANSFORMER OIL AS A COOLING MEDIUM

 The specific heat dissipation due to convection of oil

λconv = 40.3 (θ /H) ¼ W/m2 - °C ; where, θ = temp difference of the surface relative to the oil, °C H = height of the dissipating surface, m. Average values   200 C.and .H  0.5to1m conv  80to100W / m 2 0 C

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

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 400C 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. total.loss sp.heat.dissipation  heat.dissipation.surface.of .the.tan k PP Temperature.rise  i c 12.5  St Temperature.rise 

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

Where, St = 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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IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/EE 2355/DEM/VER 1.0

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

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 loss dissipation by convection = loss dissipated by 35% of tube surface area.

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

Let, Dissipating surface of the tank = St Dissipating surface of the tubes = xSt Loss dissipated by the tank surface = (6+6.5) St =12.5 St Loss dissipated by the tubes = (135/100 x 6.5)× x St by 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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IFETCE/EEE/M.SUJITH/III YEAR/VI SEM/EE 2355/DEM/VER 1.0

Actual total area of tank walls and tubes = St + x St = St(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, θ= Total loss/ Total Loss dissipated Total loss, Ploss = Pi + Pc Hence, θ = ( Pi + Pc)/ (12.5 + 8.8x) St 12.5 + 8.8x = ( Pi + Pc)/ θ St x = [{( Pi + Pc)/ θ St }-12.5] (1/8.8) 72

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

Total area of cooling tubes = x St = (1/8.8) [ {(Pi + Pc)/ θ} – 12.5 St ] St The total number of tubes = nt = Total tube area / Area of each tube = Total tube area /(π dt lt ) nt = (1/8.8π dt lt) [ {(Pi + Pc)/ θ} –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. 73

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

 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 diameter of the coil.

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

DIMENSIONS OF TRANSFORMER TANK C4 HT

H

D

D

C3

WT C2 LT DOC

DOC

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

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 winding Voltage

KVA rating

Clearance in mm C1

C2

C3

c4

Up to 11KV