design of an transformer , Exercises for Electric Machines
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arjun-khatiwada

design of an transformer , Exercises for Electric Machines

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Design a 400 KVA, 66/ 6.6 KV, 50 Hz, 3-phase delta/star core type oil immersed natural cooled distribution transformer. Maximum temperature rise not to exceed 45 ºC.
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TRIBHUVAN UNIVERSITY

INSTITUTE OF ENGINEERING

CENTRAL CAMPUS, PULCHOWK

Report of Machine design

on

Transformer Design

Submitted By

Submitted To

Er. Purushottam Khadka

Lecturer

Department of electrical

Engineering

Date of Submission: 2074/11/21

Arjun Khatiwada (072/BEL/305)

Group: A

Department of Electrical

Engineering

072BEL305

2

5. Question: Design a 400 KVA, 66/ 6.6 KV, 50 Hz, 3-phase delta/star core type oil

immersed natural cooled distribution transformer. Maximum temperature rise

not to exceed 45 ºC.

Design of Core:-

Voltage per turn, Et= k√��

For 3-phase core type distribution transformer, constant for voltage per turn, k=0.45

So, Et=0.45*√400 = 9 volts/turns

Now Фm = Et

4.44∗�� =

9

4.44∗50 =0.0405 weber

Using the hot rolled silicon steel (grade 92) for yoke and limb.

Bm=1.35 wb/m 2

Area of iron, Ai= Фm / Bm =0.0405/1.35 =0.03 m 2=30*103mm2

Using three stepped cruciform core with stacking factor 0.9

Ai=0.6*d 2

d = √ Ai

0.6 =√

30∗103

0.6 = 223.6����

Now the dimensions of three stepped core,

Width of the largest Stamping (a) = 0.9d= ( 0.9 * 223.6) mm= 201.24 mm

Width of the middle stamping (b) = 0.7d= (0.7 * 223.6)mm= 156.52 mm

Width of the smallest stamping(c) = 0.42d= (0.42 * 223.6)mm= 93.91 mm

Design Of Window:-

For 400 KVA rating transformer,

Window Space Factor (kw) = 10

30+���� =

10

30+66 =0.1041

Taking current density, δ =2.5 A/mm2=2.5*106 A/m2 [Distribution Transformer]

The KVA rating of the transformer is given by,

Q=3.33*f*Bm*δ*Kw*Aw*Ai*10-3 KVA

Solving for Aw

072BEL305

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Aw = Q

3.33 ∗ f ∗ Bm ∗ δ ∗ kw ∗ Ai ∗ 10−3

= 400

3.33 ∗ 50 ∗ 1.35 ∗ 2.5 ∗ 106 ∗ 0.1041 ∗ 0.03 ∗ 10−3

= 227.92 * 103mm2

Taking ratio of height to width of window as 2.5 i.e.

Hw/Ww = 2.5 [ Value must be between 2.5 and 3.5]

Hw/Ww=2.5 , Hw=2.5*Ww

Also Aw=Hw*Ww , 2.5 (Ww)2=Aw

Ww= √ Aw

2.5 = 301.94 mm

Hw = 2.5 * Ww

= 2.5* 301.94

= 752.625 mm

Design of Yoke:-

Taking stacking factor (Ki=0.9)

As we have taken the yoke is formed by hot rolled steel,

Taking net area of yoke 20% greater than area of core

Flux density in the yoke = 1.35

1.2 =1.125 Wb/m2

Area of the yoke, Ay = 1.2 *Agi= 1.2 * Ai /Ki =1.2*30*10 3/0.9=40*103mm2

Taking yoke as rectangular,

Ay=Dy*Hy

Dy = a = 201.24 mm

Hy= ����

�� = 40*103/201.24 =198.76 mm

072BEL305

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Overall Dimension Of Frame:-

Distance between Two Adjacent Limb Centre(D):

D = WW + d = ( 301.94 + 223.6)mm

= 525.54 mm

Overall height of the Frame(H)

H = Hw + 2*Hy = 752.625 + 2*198.76

= 1150.145 mm

Overall width of the frame(W)

W= 2*D + a = (2 *525.54+ 201.24) mm

= 1252.32 mm

The detailed figure for three phase core type transformer with three stepped core with dimensions is

given below:

072BEL305

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Design of LV Winding(secondary winding):-

Secondary line voltage Vs = 6.6kV=6600V ; connection=star

Secondary Phase voltage, Vsp = 6600/√3 = 3810.51V

No. of turns per phase, Ts = = ����

Et =

3810.51

9 = 423.39 ≈ 424 turns

Phase current on secondary Is = 400∗1000

3∗3810.51 = 34.99 A

Current density δ =2.5 A/mm2

Area of a secondary conductor, as = 34.99

2.5 = 13.996 mm2

From table, we use a bare conductor of width and thickness of 4.5 mm and 3.2 mm respectively.

Area of bare conductor, as =13.9 mm2

Actual Current density in secondary winding �� = 34.99

13.9 = 2.517 A/mm2

The conductors are paper insulated. The increase in dimension on account of paper covering is 0.5mm.

Dimension of the insulated conductor = 5.0 mm*3.7 mm

Choosing 3 layers for the winding, taking 145 turns per layer for first 2 layers and 134 turns for 3rd

layer

Ts=145*2+134=424 turns

Helical winding is used. Therefore, space has to be provided for (145+1) = 146 turns

Axial depth of l.v. winding, Lcs = 146*5.0 = 730 mm

The height of the window is 752.625. This leaves a clearance of (752.625- 730)/2=11.3125 mm on

each side of the winding. (This value must lie between 8 & 15 mm)

Using 0.5mm pressboard cylinders between layers

Radial depth of low voltage winding, bs = 3*3.7+2*0.5

= 12.1 mm

Diameter of circumscribing circle, d = 223.6 mm

Using pressboard of 1.5mm as insulation between LV and core,

Inside diameter of low voltage winding = 223.6 + 2*1.5

= 226.6 mm.

072BEL305

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Outside diameter of low voltage winding = 226.6 + 2*12.1 =250.8 mm

Design of HV winding(primary winding):-

Primary Line voltage = 66 kV ; connection = delta

Primary Phase voltage (Vp)=66 kV

No. of turns per phase, Tp Ts Vs

Vp * =

66000

6600/√3 *424 = 7343.895 ≈ 7344

Phase current on HV side, Ip = 400∗1000

3∗66000 =2.02 A

As the current is less than 20 A .we use cross-over coils as H.V. windings.

Choosing current density in HV winding 10% more than LV winding

, δ =(1.1*2.5)A/mm2 = 2.75 A/mm2

Cross-sectional area of conductor on HV , ap = ����

δ =2.02/2.75=0.7345 mm2

Diameter of conductor =√ ap∗4

�� =√

0.7345∗4

�� =0.967 mm

Taking the standard size from the table,

Diameter of bare conductor = 0.950 mm

Diameter with fine covering =1.125 mm

Modified cross-sectional area (ap) = ��d2/4 = �� ∗ 0.952/4 = 0.7088 mm2

Actual current density δ= 2.02/0.7088= 2.85 A/mm2

The voltage per coil is about 1500V.

Using 44 coils, voltage per coil = 66000

44 = 1500��

Turns per coil = 7344

44 = 166.91 turns

Taking 168 turns in 43 coils and 120 turns in 44th coil,

Total HV coils = 170*43 +34 = 7344

Let the number of layers per coil be x .

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No. of turns per layer per coil = 1.125∗168

�� for 43 coils

= 1.125∗120

�� for reinforced coil

Now, we have to maintain this relationship

Total height of the window ≤ window height – clearance

1.125∗168

�� *43 +

1.125∗120

�� +43*5 ≤ 752.625- 40*2

8262

�� ≤ 457.625

So, x ≥ 18.05.

Taking 21 layers per coil in first 43 coils, turns per coil per layer = 168

21 = 8

Taking 20 layers per coil in 44th, turns per coil per layer = 120

20 = 6

Maximum voltage between layers = 2*8*9V =144 V, which is below 300V

Axial depth of each coil for first 43 coils = 8*1.125= 9 mm

Axial depth of each coil for 44th coil = 6*1.125 = 6.75 mm

The spacers used between adjacent coils are 5mm in height.

Total axial depth of HV winding, Lcp = 9*43+6.75+5*43 = 608.75 mm

Therefore, clearance left on each side of winding = 752.625−608.75

2 = 71.9375 mm (This value

should be greater than clearance for LV winding and also> 15 mm)

Let the insulation between the layer be Bakelite paper of width = 0.3mm

Radial depth of HV coil )( pb = 21*1.125+20*0.3 = 29.625 mm

Thickness of insulation between HV and LV winding = 5+0.9*KV=5+0.9*66=64.6 mm

But to adjust the radial clearance we use insulation of thickness = 40 mm

Inside diameter of HV winding

= outside diameter of LV + 2*thickness of insulation between HV and LV

= 250.8+2*40

=330.8 mm

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Outer diameter of HV winding

= Inside diameter of HV winding + 2*radial depth of HV coil

= 330.8+2*29.625

= 390.05 mm

Clearance between windings of two adjacent limbs

= D-Outer diameter of HV winding

= 525.54-390.05

= 135.49mm

Operating Characteristics:-

Calculation of Resistance

Resistivity of copper at C75 mmm /021.0)( 2

Mean diameter of primary winding = 330.8+390.5

2 =360.65 mm

Length of mean turn of primary winding (Lmtp)= ��*360.65 =1.133 m

Resistance of primary winding,

rp= ��∗����∗��������

���� =

0.021∗7344∗1.133

0.7088 = 246.52 Ω

Mean diameter of secondary winding = 226.6+250.8

2 = 238.7 mm

Length of mean turn of secondary winding (Lmts)= ��*238.7 =0.75 m

Resistance of Secondary Winding

rp= ��∗����∗��������

���� =

0.021∗424∗0.75

13.996 = 0.477 Ω

Now,

Total resistance referred to primary side

072BEL305

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RP = rp + rs*( ����

���� )2 = 246.52 + 0.477*(7344/424)2=389.624Ω

Per unit voltage drop in the resistance of transformer

Er = ����∗����

���� =

2.02∗389.624

66000 = 0.0107 pu

Leakage reactance

Mean diameter of Winding

= (inside diameter of LV + outside diameter of HV)/2

=(226.6+390.5)/2

=308.55 mm

Length of mean turn DLmt *)(  = Π *328.55

=0.969 m

Height of the winding )(Lc

= (axial depth of LV winding + axial depth of HV winding)/2

=(730+608.75)/2 =0.669 m

Leakage reactance(Xp)  

  

  

  

  

3 *****2 2

sp bb a

Lc

Lmt Tpf

= 2π ∗ 50 ∗ 4π ∗ 10−7 ∗ 73442 ∗ 0.969

0.669 ∗ (40 +

12.1+29.625

3 )*10-3

= 1658.86Ω

Per unit voltage drop in leakage reactance

Ex= ����∗����

���� =

2.02∗1658.86

66000 = 0.0507 pu

Per unit impedance

Ε= (0.01072 + 0.05072)1/2 = 0.0518 pu

072BEL305

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Regulation:-

 CosSin rx 

Per unit regulation at unity pf = 0.0507*0+0.0107*1=0.0107 pu

Per unit regulation at 0.8 pf = 0.0507*0.6+0.0107*0.8=0.0389 pu

Calculation of Losses:

Copper loss

Copper loss= 3*Ip 2*Rp = 3*2.02

2*389.624= 4769.465 W

Taking 15% as stray loss

Total copper loss = 1.15 *4769.465=5484.88 W

Core loss

Taking density of laminations for HRS = 7.6*103 kg/m3

Weight of three limbs

= 3* height of window* iron area*density

= 3*752.625*30*103 *10-9 *7.6*103 kg

= 514.795 kg

Flux density in the limb Bm = 1.35 Wb/m 2 and corresponding to this density, specific iron loss in

limb= 2.1 W/Kg

Core loss in the limb = 2.1* 514.795

= 1081.0695W

Weight of two yoke

= 2*iron area of yoke (Ayi) * width of frame (W) * density

= 2*36*103 *1252.32*10-9 *7.6*103

= 685.27 Kg

As HRS is used for so taking area of yoke be 20% larger than area of limb

Bm for yoke = 1.125 Wb/m 2

072BEL305

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Corresponding to Bm= 1.125 Wb/m 2,specific core loss in yoke= 1.4 W/Kg

Core loss in yoke = 1.4*685.27

= 959.378 W

Total loss

Total core loss (iron loss Pi) = (1081.0695 + 959.378) = 2040.447 W

Total Loss at full load =iron loss + copper loss =2040.447+ 5484.88 = 7525.327 W

Calculation of Efficiency :

At unity pf, efficiency= 400∗103

400∗103+7525.327 * 100% = 98.15 %

At 0.8 pf, efficiency= 400∗103∗0.8

400∗103∗0.8+7525.327 * 100% = 97.70 %

For maximum efficiency:

X2 * Pc = Pi

X= √ Pi

���� = √

2040.447

5484.88 = 0.61

Calculation of No load current

From graph the values of magnetizing mmf

Atc =1000 A/m

Aty =190 A/m

Total magnetizing mmf = 3*ATC *LC + 2 * ATY*LY

Length of core, Lc= Hw = 752.625 mm = 0.752 m

Length of yoke, Ly = W = 1252.32 mm =1.252 m

Total magnetizing mmf = 3*ATC *LC + 2 * ATY*LY

= 3*1000*0.752+ 2*190*1.252

=2731.76

Magnetizing mmf per phase = 2731.76/3 =910.58 AT

072BEL305

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Magnetizing current Tp

ATperphase Im

*2 )( 

Im = 910.58

√2∗7344 = 0.0876A per phase

Loss component of current Vp

coreloss I l

*3 

Il= 7525.327

3∗66∗103 = 0.038A per phase

Total no load current 22

lm III 

Io = (0.0876 2+0.0382)1/2 = 0.0955 A

No-load current as % of full load current is

= 0.0955

2.02 *100 % =4.73 %

Cooling system

Tank Design:

For transformer of 66Kv and less than 1000KVA the required specification are:

Clearance along the height, h=500mm

Clearance along the width, b=75mm

Clearance along length, l=100mm

Height of tank(Ht)= H+h

= 1.15 + 0.5 =1.65 m

Width of tank(Wt)=2D+De+2b=

= 2*0.525+0.39+2*0.075=1.59 m

Length of tank(Lt)=De+2l

= 0.39+ 2*0.1= 0.59 m

Here, we use Tank having 1.65 m height,1.59 m width and 0.59 m length.

072BEL305

13

Loss dissipating surface area of tank

St = 2(Lt+Wt)Ht+ Lt∗Wt

2 = 2(0.59+1.59)*1.65+

0.59∗1.321

2

=7.584 m2

Now,

θmax= ����+����

����∗12.5 =

7525.327

7.584∗12.5

θmax= 79.38 oC

The maximum temperature rating of transformer is 79.38 ˚C.

Now for the given temperature of 45 ˚C,

Let the area of tubes = xSt

Assume that the heat dissipated by tubes by convection increases by 35%

xSt = 1

8.8 [ Pi+Pc

ᶿ -12.5St] =[7525.327/45 -12.5*7.584]/8.8

=8.23 m2

The diameter of tube is taken 50 mm.

The average height of the tube is 1.6 mm. Area of each tube = π×0.05*1.6 = 0.2513 m2

So the number of tubes = 8.23/0.2513 =32.75

≈ 33 tubes

Arrangement of tubes

Applying only one row and column keeping certain mm gap between two tubes. 12 tubes along

width of front face and 11 tubes along width of back face with spacing of 125 mm in front and 140

mm in back face. Along breadth, 5 tubes are used in each sides with spacing of 110 mm.

Total number of tubes = 12+11+2*5= 33

The arrangement of tubes can be as shown below.

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Fig:- Arrangements of tube outside the tank

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DESIGN SHEET

kVA 400 Phase 3 Frequency 50 Hz Delta / Star

Line Voltage L.V. 6600 V

H.V. 66000 V

Phase Voltage L.V. 3810.51 V

H.V. 66000 V

Line Current L.V. 34.99 A

H.V. 3.49 A

Phase Current L.V.34.99 A

H.V. 2.02 A

Type – Core Type of Cooling – Oil immersed

Core

Material 0.35mm thick 92 grade

Output constant K 0.45

Voltage per turn Et 9 V/turn

Circumscribing circle diameter d 223.6 mm

No of steps 3

Dimensions

a 201.24 mm

b 156.52 mm

c 93.91 mm

Net iron area Ai 30000 mm2

Flux density Bm 1.35 Wb/mm2

Flux Øm 0.0405 Wb

Weight 514.795 Kg

Specific iron loss 2.1 W/Kg

Iron loss 1081.0695 W

Yokes

Depth of yoke Dy 201.24 mm

Height of yoke Hy 198.76 mm

Net yoke area 36000 mm2

Flux density 1.125 Wb/mm2

Flux Ø 0.03375 Wb

Weight 685.27 Kg

Specific iron loss 1.4 W/Kg

Iron loss 959.378 W

072BEL305

16

Window

Number 2

Window space factor Kw 0.1041

Area of window Aw 227920 mm2

Width of window Ww 301.94 mm

Height of Window Hw 752.625 mm

Frame

Distance between adjacent limbs D 525.54 mm

Height of frame H 1150.145 mm

Width of frame W 1252.32 mm

Depth of frame Dy 201.24 mm

Windings L.V H.V

Types of winding Helical Cross over

Connections Star Delta

Conductors

Dimensions

Bare

width = 4.5 mm

thickness =

3.2 mm

Diameter = 0.95 mm

Insulated width = 5.0 mm

thickness =

3.7 mm

Diameter = 1.125 mm

New current density 2.517 A/mm2 2.85 A/mm2

No of turns per phase 424 7344

Coils

1 44

No of layers

3

21 layers for first 43 coils and 20

layers for 44th coil

Turns

Per coil

424 168 turns for 43 coils

120 turns for 44th coil

Per layer

145 turns for 2

layers , 134 turns

for 3rd layer

8 for first 43 coils

6 for 44th coil

072BEL305

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Height of windings 730 mm 608.75 mm

Depth of windings 12.1 mm 29.625 mm

Insulation

Between layers 0.5mm 0.3mm

Between coils 5mm

Coil diameters

Inside 226.6 mm 330.8 mm

Outside 250.8 mm 390.05 mm

Length of mean turns 750 mm 1133 mm

Resistance at 75 ºC 0.477 Ω 246.52 Ω

Insulation

Between Lv and Hv windings Bakelite paper 40 mm

Between Lv winding and core Press board wraps 1.5mm

Tank

Dimensions

height of tank Ht 1650 mm

width of tank Wt 1590 mm

length of tank Lt 590 mm

max temp oC 79.38 ℃

Impedance

P.U Resistance drop 0.0107 pu

P.U Reactance drop 0.0507 pu

P.U Impedance drop 0.0518 pu

Losses

Total copper loss 5484.88 W

Total core loss 2040.447 W

Total losses at full load 7525.327 W

Efficiency at full load and Unity pf

Efficiency at full load and 0.8 pf

Ratio of iron loss and copper loss at full

98.15 %

97.70 %

0.61

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