This document presents the design of a 55 KVA, 6.6 KV/433 V, 3 phase core type distribution transformer. It includes calculations for the core, winding, and overall dimensions based on design parameters. Core materials, conductor sizes, and insulation thicknesses are selected. Resistance, reactance, regulation and losses are calculated. The transformer is designed to have an efficiency of 97.4% at full load and unity power factor.

1. EE 3220
Electrical Machine Design
Project on: Transformer Design
▫ Submitted To:
▫ Dr. Mohammad Shaifur Rahman
▫ Professor,
▫ Dept. of EEE,KUET
▫ Abu Syed MD. Jannatul Islam
▫ Lecturer,
▫ Dept. of EEE,KUET
▫ Nashrah Afroz
▫ Lecturer,
▫ Dept of EEE,KUET
▫ Submitted By:
▫ 1503015
▫ 1503023
▫ 1503066
▫ 1503103

2. 2
Question
Design a 55 KVA, 6.6 KV/433 V, 50 Hz, 3 Phase, core type,
delta/star distribution Transformer,
5% tapping is used in the HV side. (Choose cooling system)

3. 3
A distribution
transformer or service
transformer is a
transformer that
provides the final voltage
transformation in the
electric power
distribution system,
stepping down the
voltage used in the
distribution lines to the
level used by the
customer.

4. Core design
4

5. 5
The value of k is taken from the table k=0.45 for 3phase core type
distribution transformer.
Voltage per turn Et = K 𝑸 = 0.45 𝟓𝟓 = 3.33V
Therefore Flux in the core, Φm =
𝑬𝒕
𝟒.𝟒𝟒×𝒇
=
𝟑.𝟑𝟑
𝟒.𝟒𝟒×𝟓𝟎
= 0.015 Wb
Hot rolled silicon steel grade 92 is used. The value of flux density Bm
is assumed as 1.0 Wb/m2.
Net iron Area Ai =
𝟎.𝟎𝟏𝟓
𝟏.𝟎
= 0.015 m2 = 15×103 mm2

6. 6

7. 7
The value of k is taken from the table k=0.45 for 3phase core type
distribution transformer.
Voltage per turn Et = K 𝑸 = 0.45 𝟓𝟓 = 3.33V
Therefore Flux in the core, Φm =
𝑬𝒕
𝟒.𝟒𝟒×𝒇
=
𝟑.𝟑𝟑
𝟒.𝟒𝟒×𝟓𝟎
= 0.015 Wb
Hot rolled silicon steel grade 92 is used. The value of flux density Bm
is assumed as 1.0 Wb/m2.
Net iron Area Ai =
𝟎.𝟎𝟏𝟓
𝟏.𝟎
= 0.015 m2 = 15×103 mm2

8. 8
Using a cruciform core, Ai = 0.56d2
Diameter of circumscribing circle, d =
𝟏𝟓×𝟏𝟎 𝟑
𝟎.𝟓𝟔
=163.66 mm
Reference widths of laminations:
a=.85d=.85×163.66=139.1 mm
b=.53d=.53×163.66=86.74mm
Core area factor

9. Window Dimension
9

10. 10
For transformers of ratings between 50 to 200 kVA window space
factor,
Kw =
𝟏𝟎
𝟑𝟎+𝟔.𝟔
= 0.273
The current density in the windings is taken 2.3 A/mm2 output of
transformer.
Q = 3.33fBmKwδAwAi×10-3
55 = 3.33×50×1.0×0.273×2.3× 106 ×Aw ×0.015× 10−3
Therefore Window Area, Aw = 0.0351 m2
For distribution transformers
δ=1.1 to 2.3 A/mm2

11. 11
Taking the ratio of height to width of window as 2.5
Hw×Ww = 35.1× 103mm2 or, 2.5×Ww²=35.1× 103
So width of window, Ww = 118.49mm ≈ 119 mm
Height Hw=296mm
Distance between adjacent core center,
D = Ww + d = 119 + 164 =283 mm
Between 2 to 4

12. Yoke Design
12

13. 13
The area of Yoke is taken as 1.2 times that of limb. Therefore Flux density
in Yoke = 1/1.2 = 0.833 Wb/m2
Net area of Yoke = 1.2×15×103=18×103 mm2
Gross area of Yoke =18×103/0.9=20×103mm2
Taking the section of Yoke as rectangular,
Depth of Yoke, Dy = a = 139 mm
Therefore, height of Yoke, Hy =
𝟐𝟎×𝟏𝟎
𝟑
𝟏𝟑𝟗 ≈ 139mm
15 to 25% larger than core

14. Overall Dimension of Frame
14
Height of frame, H = Hw + 2Hy = 296+2×139 ≈ 574 mm
Width of frame, W = 2D + a = 2×283+139.1 = 705 mm
Depth of frame, Dy = a = 139 mm

15. L.V. Winding
15

16. 16
Secondary voltage = 433 V
Secondary Phase Voltage, Vs =
𝟒𝟑𝟑
𝟑
=250V
Number of turns per phase, Ts =
𝑽𝒔
𝑬𝒕
=
𝟐𝟓𝟎
𝟑.𝟑𝟑
= 75.1
Secondary Phase Current, Is=
𝟎.𝟓𝟓×𝟏𝟎𝟎𝟎
𝟑×𝟐𝟓𝟎
= 73.3 A
Current density 2.3 A/mm2 is Used.
Area of Secondary Conductor ag =73.3/2.3 = 31.87 mm2
Using a bare conductor of 10×3.5 mm
Area of conductor, as = 34.1 mm2
Current density in secondary winding, δs = 73.3/34.1 = 2.15A/mm2
The conductor are paper covered.

17. 17
The increase in dimension on account of account of paper covering is 0.5 mm.
So Dimension of insulated conductor = 10.5×4mm2
Using 3 layer Helical winding. So space has to be provided (25+1) = 26 turns along
the axial depth.
Axial depth of L.V. winding, LCS = 26×10.5 = 273 mm
The Height of window is 296 mm. This leaves a clearance of (296-273)/2 = 11.5 mm
of each side of the windings.
Using 0.5 mm pressboard cylinders between layers, radial depth of low voltage
winding,
bs = no. of layers x radial depth of conductor + insulator betn layers
= 3×4+2×0.5 = 13 mm
Diameter of circumscribing circle ,d =164mm

18. H.V. Winding
18

19. 5
Primary Line voltage = Primary Phase Voltage, Vp= 6.6×103V; Delta
No. of turns per phase, Tp=𝟔𝟔𝟎𝟎 ×
𝑻𝒔
𝑽𝒔
=
𝟔𝟔𝟎𝟎×𝟕𝟓.𝟏
𝟐𝟓𝟎
= 1983
As ±5% tapings are to be provided, Therefore the no. of turns is increased to
Tp= 1.05×1983 = 2082
The voltage per coil is about 1500 V. Using 5 coils, Voltage/coil is 6600/5 = 1320 V.
Turns per coil 2082 /5 = 417
Using 4 normal coils of 440 turns and one reinforced coil of 322 turns,
Total H.V turns provided, Tp = 4 × 440 + 322 = 2082
Taking 24 layers per coil, Turns/coil = 440/24 = 19
Maximum voltage between layers = 2×19×3.33 = 126.54 V, which is below the
allowable limit.

20. 5
H.V. winding phase current, Ip = (55×1000)/(3×6600) = 2.78 A
As the current is below 20A, cross-over coils are used taking a current density of
2.4 A/mm2
Area of H.V. conductor, ap = (2.78 )/2.4=1.16 mm2
Diameter of bare conductor =√(4×1.16/π ) = 1.22 mm, using paper covering conductors.
From table 23.4 (BIS:3454-1966) the nearest standard conductor size has:
Bare diameter = 1.25 mm
Insulated diameter = 1.45 mm with fine covering.
Modified area of the conductor, ap =
𝝅
𝟒
× 𝟏. 𝟐𝟓 𝟐
= 1.23 mm2
Actual value of current density used, δp =
𝟐.𝟕𝟖
𝟏.𝟐𝟑
= 2.26 A/ mm2

21. 21
Axial depth of one coil = 19 x 1.15 = 27.55 mm
The space used betn adjacent coils are 5mm in height.
Axial length of H.V winding:
LCP = no. of coils x axial depth of each coil + depth of spacers
= 5×27.55 + 5×5 = 162.75 mm
The height of window is 296mm & therefore the space left betn winding & window
is (296-162.75) = 133.3 mm.
The clearance left on each side is 78.5 mm which is sufficient for 6.6kV
transformers.
The insulation used between layers is 0.3 mm thick paper.
Radial depth of H.V coil, bp = 24 × 0.715 + 23×0.3 = 24

22. 22
From equation 7.22 the thickness of insulation between H.V. & L.V. winding =
5+0.9×6.6 = 10.94 mm, this includes the width of oil duct also.
The insulation between H.V. & L.V. winding is a 5mm thick bakelized paper
cylinder. The H.V. winding is wound on a former 5mm thick and the duct is
5mm wide, space making the total insulation between H.V. & L.V. winding 15mm.
Inside diameter of H.V. winding= Outside diameter of L.V winding +
2 x thickness of insulation =193+2×15 = 223 mm
Outside diameter of H.V. winding
De= Inside diameter of H.V. winding + 2 x Radial depth of H.V coil
=223+2×24 = 271 mm
Clearance between two adjacent limbs = D- Outside diameter of H.V. winding
=283– 271 = 12 mm

23. Resistance Calculation
23

24. 5
Mean diameter of primary winding =
𝟐𝟐𝟑+𝟐𝟕𝟏
𝟐
= 247 mm
length of mean turn of primary winding Lmtp = π×247×10 = 0.78 m
Resistance of primary winding at 75℃,rp=
𝑻𝒑×⍴×𝐋 𝒎𝒕𝒑
𝒂𝒑
=
𝟏𝟗𝟖𝟑×𝟎.𝟎𝟐𝟏×𝟎.𝟕𝟖
𝟏.𝟐𝟑
= 26.4Ω
Mean Diameter of Secondary winding =
𝟏𝟔𝟕+𝟏𝟗𝟑
𝟐
= 180 mm
Length of mean turn of Secondary winding Lmts = π×180×10-3 = 0.565 m
Resistance of Secondary winding at 75℃, rs=
𝑻𝒔×⍴×𝐋 𝒎𝒕𝒔
𝒂𝒔
=
𝟕𝟔×.𝟎𝟐𝟏×.𝟓𝟔𝟓
𝟑𝟒.𝟏
= 0.0264Ω
Therefore total Resistance referred to primary side ,
Rp = 26.4+
𝟏𝟗𝟖𝟑
𝟕𝟔
𝟐
× 𝟎. 𝟎𝟐𝟓𝟔 = 44.4 Ω
P.U. resistance of transformer εr =
𝑰𝒑×𝑹𝒑
𝑽𝒑
=
𝟐.𝟕𝟖 ×𝟒𝟒.𝟒
𝟔𝟔𝟎𝟎
= 0.019

25. Leakage Reactance
Calculation
25

26. 5
Mean diameter of windings =
𝟏𝟔𝟕+𝟐𝟕𝟏
𝟐
=219 mm
Length of mean turn, Lmt =π×219×10-3=.69m
Height of winding , Lc =( Lcp+Lcs) /2 =(162.75+273)/2
= 220.4 mm
Therefore leakage resistance referred to primary side ,
Xp=2×π×50×4×π×10-7 ×19832×
.𝟔𝟗
.𝟐𝟐
× 𝟏𝟓 +
𝟐𝟒+𝟏𝟑
𝟑
× 𝟏𝟎−𝟑 =133.1 Ω
P.U. leakage reactance of transformer εx=
𝟐.𝟕𝟖×𝟏𝟑𝟑.𝟏
𝟔𝟔𝟎𝟎
=0.056
P.U Impedance, εs = . 𝟎𝟏𝟗 𝟐 + . 𝟎𝟓𝟔 𝟐=0.059

27. Regulation
27
P.U regulation, ε=εrcosφ+ εrsinφ
So per unit regulation at unity p.f. ε=εr=0.019
At zero p.f. lagging ε=εx=0.056
At 0.8 p.f. lagging ε = 0.019×0.8 + 0.056×0.6 = 0.0488

28. Loss Calculation
28

29. 29
I2R loss at 75℃ = 𝟑 × 𝑰𝒑 𝟐
× 𝑹𝒑= 3×2.782×44.4 = 1029.4 W
Total I2R loss including 15% stray load loss, Pc=1.15 X 1029.4 = 1183.84 W
Taking density laminations as 7.6X103kg/m3
Weight of 3 limbs =3X0.296X0.015×7.6X103=101.2 kg
The flux density in the limbs is 1 Wb/m2 & corresponding to this density,
specific core loss is 1.2W/kg
Core loss in limbs = 101.2X1.2=121.5W
Weight of two Yokes = 2X.705X.018X7.6X103=192.9 kg
Corresponding to .833 Wb/m2 flux density in the yoke, Specific core loss = 0.85W
Core loss in Yoke =192.9X0.85=163.9 W
Total core loss, Pi=121.5 + 163.9 =285.5 W

30. Efficiency
30
Total losses at full load = 285.5+1183.84=1469.3 W
Efficiency at full load unity P.f. =
𝟓𝟓𝟎𝟎𝟎
𝟓𝟓𝟎𝟎𝟎+𝟏𝟒𝟔𝟗.𝟑
× 𝟏𝟎𝟎% =97.4%
For maximum efficiency, 𝒙 𝟐 𝑷𝒄 = 𝑷𝒊 or,𝒙 =
𝑷𝒊
𝑷𝒄
=
𝟐𝟖𝟓.𝟓
𝟏𝟏𝟖𝟑.𝟖𝟒
=0.491
The maximum efficiency occurs at 49.1 percent of full load. This is good
figure for distribution transformer.

31. No Load Current
31

32. 32
Corresponding to flux densities of 1 Wb/m2 & 0.833 Wb/m2 in core & yoke
respectively atc= 120 A/m & aty = 80 A/m.
So, total magnetizing m.m.f. = 3×120×0.296+2×80×0.705 = 219.36 A
So, magnetizing m.m.f. per phase, ATo = 219.36/3 = 73.12
Magnetizing current, Im = ATo/ 𝟐 TP =
𝟕𝟑.𝟏𝟐
𝟐×𝟏𝟗𝟖𝟑
= 26.1×10-3 A
Loss component of no load current, IL =
𝟐𝟖𝟓.𝟓
𝟐×𝟔𝟔𝟎𝟎
= 30.5×10-3 A
No load current, Io = 𝟐𝟔. 𝟏 × 𝟏𝟎
^ − 𝟑 𝟐 + 𝟑𝟎. 𝟓 × 𝟏𝟎^
− 𝟑 𝟐 = 40.1×10-3 A
No load current as a percentage of full load current =
𝟒𝟎.𝟏×𝟏𝟎
−
𝟑
𝟐.𝟕𝟖
×100% = 1.44%
Allowing for joints etc the no load current will be about 2.5% of full load current.

33. Tank
33

34. 34
Height over Yoke, H = 139 mm
Allowing 50 mm at the base & 150 mm for oil,
Height of oil level =139+50+150 = 339 mm
Allowing another 200 mm height for leads etc
Height of Tank Ht=339+ 200 =539 mm
The height of tank is taken as 0.6 mm.
Width of the Tank, Wt= 2D+De+2l= 𝟐 × 𝟐𝟖𝟑 + 𝟐𝟕𝟏 + 𝟐 × 𝟒𝟎 = 917 mm
The width of tank is taken as 0.95 m.
The clearance used is approximately 50 mm on each side.
Length of the tank, Lt = De+2b = 271+2X86.74 = 444.5 mm
The length of tank is taken as 0.45 m
Total loss dissipating surface of tank = 𝟐 𝟎. 𝟗𝟓 + 𝟎. 𝟒𝟓 × 𝟎. 𝟔=1.68 m2
Total specific loss dissipation due to radiation & convection is 12.5 W/m2℃

35. 5
Temperature rise =
𝟏𝟒𝟔𝟗.𝟑
𝟏.𝟔𝟖×𝟏𝟐.𝟓
= 69.97 ℃ ≈ 70 ℃
This is over 35℃, therefore plain tank alone is not sufficient for cooling & so
tubes are required.
Let the area of tubes be xSt
So, specific loss dissipation =
𝟏𝟒𝟔𝟗.𝟑
𝟏.𝟔𝟖(𝟏+𝒙)×𝟑𝟓
=
𝟐𝟒.𝟑𝟗
𝟏+𝒙
Or,
𝟐𝟒.𝟑𝟗
𝟏+𝒙
=
𝟏𝟐.𝟓+𝟖.𝟖𝒙
𝟏+𝒙
So, x = 1.35
Area of tubes needed = 1.35×1.68= 2.27 m2
So, dissipating area of each tube = π×0.05×1.35 = 0.212 m2
So number of tubes will be provided = 2.27/0.212 ≈ 10
Arrangement of tubes : Along length – 2 rows – 3 & 2 tubes

36. 36
DESIGN SHEET
KVA 55
Frequency
50 Hz
Delta/
Star
Type of
cooling:
ON
Type :
Core
Phase:
3-ϕLineVolage
HV:66000
LV:433
Phasevoltage
HV:66000
LV:250
Linecurrent
HV:3.2A
lV:473.33A
Phasecurrent
HV:1.8A
lV:473.33A

37. 37
 CORE:
1 Material --- .35 mm thick 92 grade
2 Output Constant K .45
3 Voltage per turn Et 3.33 V
4 Circumscribing
circle diameter
d 163.66 mm
5 No. of steps --- 2
6 Dimensions a 139.1 mm
b 86.74 mm
7 Net iron area Ai 15×103 mm2
8 Flux density Bm 1 Wb/m2
9 Flux Φm 0.015 Wb
10 Weight 101.2 kg
11 Specific iron loss 1.2 W/kg
12 Iron loss 121.5 W

38. 38
 YOKE:
1 Depth of Yoke Dy 139 mm
2 Height of Yoke Hy 139 mm
3 Net Yoke area 18x103 mm2
4 Flux density .833 Wb/m2
5 Flux .015 Wb
6 Weight 193 kg
7 Specific iron loss 0.8 W/kg
8 Iron loss 163.9 W
 WINDOWS:
1 Number 2
2 Window space factor Kw 0.273
3 Height of window Hw 296 mm
4 Width of window Ww 119 mm
5 Area of window Aw 0.035 m2

39. 39
 FRAME:
 INSULATION:
1 Distance btn adjacent limbs D 283 mm
2 Height of Frame H 574 mm
3 Width of Frame W 705 mm
4 Depth of wondow Dy 139 mm
1 Btn L.V. winding & core Press board wraps 1.5mm
2 Btn L.V. winding & H.V. winding Bakelized paper 5 mm
3 Width of duct btn L.V & H.V. 5mm

40. 40 WINDINGS:
Sl no. Properties L.V. H.V.
1 Type of winding Helical Cross-over
2 Connections Star Delta
3 Conductor Dimensions bare 10x3.5 mm2 Diameter = 1.23 mm
insulated 10.5x4 mm2 Diameter = 1.45mm
Area 34.1 mm2 1.23 mm2
No. in parallel None None
4 Current Density 2.15 A/mm2 2.26 A/mm2
5 Turns per phase 75.1 1983
6 Coils total number 3 3x5
per core leg 1 5
7 Turns Per coil 75 4 of 440, 1of 322
Per layer 26 19

41. 41 WINDINGS(cont.):
8 Number of layers 3 24
9 Height of winding 296 162.75
10 Insulation Betn layers .5 mm
pressboard
.3mm paper
Betn coils 5mm spacers
11 Coil
Diameters
Inside 167 mm 223 mm
Outside 193 mm 271 mm
12 Depth of winding 13 mm 24 mm
13 Length of mean turn .78 mm .565 mm
14 Resistance at 75℃ 26.4 Ω 0.0264 Ω
 TANK:
1 Dimensions Height Ht .6 m
Length Lt .45 m
Width Wt .95 m
2 Oil level --- 0.728 mm
3 Tubes 10
4 Temperature rise --- 70℃

42. 42 TANK(cont.):
5 Impedance P.U. Resistance --- 0.019
P.U. Reactance --- 0.056
P.U. Impedance --- 0.059
6 Losses Total Core loss --- 285.5 W
Total copper
loss
--- 1183.84 W
Total losses at
full load
--- 1469.3 W
Efficiency at full
load & u.p.f.
--- 97.4%