Transformer Design | Core Design | Full Design | EE 3220 Electrical Machine Design EE-3220 Core Design Window Dimensions Yoke Design Overall Dimensions of Frame Low Voltage Winding High Voltage Winding Resistance Leakage Reactance Regulation Losses Core Loss Efficiency No Load Current Tank Project on Transformer Design

1. Transformer Design
Course on Electrical Machine Design
Designed By
Md. Jikrul sayeed Hossain (1503048)
Tanbin islam rohan (1503099)
Lubaina saffana siddiqua (1503022)
Mamun sarker (1503106)
Group: 25

2. Problem on Transformer Design
Design a 65 KVA, 11 KV/400 V, 50 Hz, 3 Phase, core type, delta/star distribution Transformer, 5%
tapping is used in the HV side. (Choose cooling system)
i
Objectives
 To Learn About The Procedure of Transformer Designing
 To Gather Knowledge on The Basic Requirements to
Design a Transformer
 To minimize the cost of Designing

3. Design Requirements
Self
Coolling
HV
6.6KV
DELTA
LV
433V
STAR
3 PHASE
50HZ
5%
Tap
75
KVA
1

4. Core Design
2

5. Core Design
The value of k is taken from the table, [7.2 in A course in Electrical Machine Design. AK. Sawhney]
K=0.45 for 3-phase core type distribution transformer.
Voltage per turn Et =K 𝑄 =0.45 75 =3.90 V Therefore,
Flux in the core, Φm =
𝐸𝑡
4.44×𝑓
=
3.90
4.44×50
= 0.01757 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=
0.01757
1.0
= 0.01757 = 17.57×103 mm2
Using a cruciform core, Ai =0.56d2
Diameter of circumscribing circle, d =
17.57×103
0
.
56
=177.13 mm≈177 mm
Reference widths of laminations:
a=0.85d=0.85×177 =150.45 mm
b=0.53d=0.53×177=93.81 mm
The nearest standard dimensions are: a =151 mm and b =94 mm
3
Type K
Single phase shell type 1-1.2
Single phase core type 0.75-0.85
Three phase shell type 1.3
Three phase core
type(distribution)
0.45
Single phase shell
type(power)
0.6 to 0.7
Area percentage
of circumscribing
circle
Square Cruciform Three
Stepped
Four
step
Net core area,Ai 0.45 0.56 0.6 0.62

6. Window Dimensions
4

7. Window Dimensions
The window space factor for the required rating transformer is Kw = 10/(30+KV)= =
10
30
+
6
.
6
= 0.273 ≈ 0.27
The current density in the windings is taken 2.5 A/mm2
Output of transformer: Q =3.33fBmKwδAwAi× 10−3
Therefore, Window Area, Aw=
75
3.33×50×1×0.273×2.5×106×0.0×10−3
=37.56 m2
=37.56× 103 mm2
Taking the ratio Height to Width of window as 2.7:
Hw×Ww = 37.56 × 103 mm2 or, 2.7×Ww² = 37.56 × 103
therefore width of window, Ww=118 mm &
Window height, Hw =318 mm
Area of window provided, Aw =118 × 318
= 37524mm2
= 0.0375 m2
Distance between adjacent core center, D = Ww+d =118 + 177
=295 mm
5
For a distribution Transformer
with rating 65KVA, Kw =
10/(30+KV)
For distribution , large power
transformers,self oil cooled type
𝜹 = 2.2 to 3.2 A/mm2

8. Yoke Design
6

9. Yoke Design
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 =1.2×17.57×103 =21.084×103 mm2
Gross area of Yoke =21.084×103 /0.9=23.426×103 mm2
Taking the section of Yoke as rectangular,
Depth of Yoke, Dy=a=151mm
Therefore, height of Yoke, Hy =
23.426×10
3
151 ≈ 155 mm
7

10. Overall
Dimension of
frame
8

11. Overall Dimension of Frame
Height of frame, H=Hw+2Hy=318+2*155=628 mm
Width of frame, W=2D+a=2×305 +151=761 mm
Depth of frame, Dy=a=151 mm
9

12. Low voltage
winding
10

13. Low Voltage Winding Continued…
Secondary voltage = 433 V (Connection= Star)
Secondary Phase Voltage ,Vs=
433
3
=250V
Number of turns per phase, Ts =
𝑉𝑠
𝐸𝑡
=
250
3.90
=64.10≈64
Secondary Phase Current, Is=
75×1000
3×250
=100 A
Current density 2.5 A/mm2 is Used.
Area of Secondary Conductor as=100/2.5=40 mm2
Using a bare conductor of 13×3.2 mm,
Area of bare conductor , as=41.6 mm2
Current density in secondary winding , δs=100/41.6=2.40 A/mm2
The conductor are paper covered. The increase in dimension on account of account of paper
covering is 0.5 mm.
Therefore Dimension of insulated conductor = 13×3.7 mm2
11

14. Low Voltage Winding
Using 3 layers for the winding.
Helical winding is used. Therefore, space has to be provided (
𝑇𝑠
3
+ 1)=(
64
3
+ 1) =22.3= 23 (approx)
turns along the axial depth.
Axial depth of L.V winding, Lcs =23×13=299 mm
The Height of window is 486 mm. This leaves a clearance of (318-299)/2=9.5 mm of each side of the
windings. [Which fulfill the minimum requirement 6 mm]
Radial depth of low voltage winding,
bs=no. of layers x radial depth of conductor + insulator betn layers
=3×3.2+2×0.5 = 10.6 mm
Diameter of circumscribing circle, d =177 mm
Using pressboard wraps 1.5 mm thick as insulation between l.v. winding and core.
Inside diameter of l.v. winding=117+1.5×2=120 mm
Outside diameter of l.v. winding= 120+10.6 =130.6 mm 12

15. High voltage
winding
13

16. High Voltage WindingContinued…
Primary Line voltage =6600V (Connection: Delta)
Primary Phase Voltage, Vp= 6600 V
Therefore, no. of turns per phase, Tp=Vp ×
𝑇𝑠
𝑉𝑠
=
6600×64
250
=1689.6 = 1690 (approx)
As ±5% tapping is to be provided, therefore the no. of turns is increased to Tp=1.05×1690=1775
The voltage per coil is about 1500 V, therefore using 15 coils,
Voltage per coil= 6600/15=440 V
Turns per coil= 1775 /15=118
Using 14 normal coils of 120 turns and one reinforced coil of 95 turns.
Total H.V turns provided, Tp =14 × 120 + 1×95=1775
Taking 24 layers per coil. Turns/layer=120/24= 5 (approximately)
Maximum voltage between layers = 2×5×3.9 =39 V
H.V. winding phase current, Ip =(75×1000)/(3×6600)=3.78 A
As the current is below 20A,
cross-over coils are used H.V winding taking a current density of 2.1 A/mm2
Area of H.V. conductor, ap =3.78/2.1=1.8 mm2 14

17. High Voltage WindingContinued…
Diameter of bare conductor =
4
𝜋
× 1.8 =1.51 mm
Using paper covered conductors, the nearest standard conductor size:
Bare diameter =1.5 mm
insulated diameter=1.58 mm with fine covering.
Modified area of the conductor, ap=
𝝅
𝟒
× 𝟏. 𝟓 𝟐
=1.76 mm2
Actual value of current density used, δp=
𝟑.𝟕𝟖
𝟏.𝟕𝟔
=2.13 A/mm2
Axial depth of one coil =5x 1.58 =7.9 mm
The space betn adjacent coils are 5 mm in height.
Axial length of H.V winding:
Lcp =no. of coils x axial depth of each coil + depth of spacers
=15×7.9 + 15×5=193.5 mm
The height of window is 318 mm & therefore the space left betn winding & window is (318-193.5)/2
=62.25 mm.
The clearance left on each side is 62.25 mm, which is sufficient for 6.6kV transformer.
15

18. High Voltage Winding
The insulation used between layers is 0.3 mm thick paper
Radial depth of H.V coil, bp=15 × 1.58 + 14×0.3=27.9
Thickness of insulation between h.v. and l.v. windings= 5+0.9kV=5+0.9*6.6=10.94 mm
Inside diameter of H.V. winding
=Outside diameter of L.V winding + 2 x thickness of insulation.
=130.6+2×10.94≈152.48 mm
Outside diameter of H.V. winding
De = Inside diameter of H.V. winding + 2 x Radial depth of H.V coil
=152.48+2×27.9=208.08 mm
Clearance betn windings=D – Outside diameter of H.V. winding
=295– 208.08=86.92 mm (Which is for enough)
16

19. Resistance &
Leakage
Reactance
17

20. Resistance & Leakage ReactanceContinued…
Mean diameter of primary winding =
𝟏𝟓𝟐.𝟒𝟖+𝟐𝟎𝟖.𝟎𝟖
𝟐
= 180.28 mm
Length of mean turn of primary winding, Lmtp=𝝅×180.28×10-3 =0.566 m
Resistance of primary winding at 75℃, rp=
𝑻𝒑×⍴×𝐋 𝐦𝐭𝐩
𝒂 𝒑
=
𝟏𝟕𝟕𝟓×𝟎.𝟎𝟐𝟏×𝟎.𝟓𝟔
𝟏.𝟖
=11.6 Ω
Mean Diameter of Secondary winding=
𝟏𝟐𝟎+𝟏𝟑𝟎.𝟔
𝟐
=125.3≈125mm
Length of mean turn of Secondary winding, Lmts=𝝅×125×10-3
=0.3927 m
Resistance of Secondary winding at 75℃, rs=
𝑻𝒔×⍴×𝐋 𝐦𝐭𝐬
𝒂 𝒔
=
𝟔𝟒×𝟎.𝟎𝟐𝟏×𝟎.𝟑𝟗𝟐𝟕
𝟒𝟏.𝟔
=0.01268Ω
Therefore, Total Resistance referred to primary side ,
Rp=11.6+
𝟏𝟕𝟕𝟓
𝟔𝟒
𝟐
× 𝟎. 𝟎𝟏𝟐𝟔𝟖=21.35 Ω
P.U. resistance of transformer εr=
𝑰𝒑×𝑹𝒑
𝑽𝒑
=
𝟑.𝟕𝟖 ×𝟐𝟏.𝟑𝟓
𝟔𝟔𝟎𝟎
= 0.012
R
e
s
i
s
t
a
n
c
e
18

21. Resistance & Leakage Reactance
Mean diameter of windings =
120+208.08
2
=164.04 mm
Length of mean turn, Lmt = 𝜋 ×164.04×10-3=0.515 m
Height of winding, Lc =(Lcp +Lcs) /2 =(193.5 +299)/2
= 246.25 mm
Therefore leakage Resistance referred to primary side ,
Xp=2×𝜋×50×4× 𝜋 ×10-7 ×17752×
0.515
0.2462
× 10.94 +
10.6+ 27.9
3
× 10−3
=61.85 Ω
P.U. leakage resistance of transformer εΦ=
3.78 ×61.85
6600
=0.035
P.U Impedance εs = 0.012 2 + 0.035 2=0.037
R
e
a
c
t
a
n
c
e
19

22. Voltage
Regulation
20

23. Voltage Regulation
P.U. regulation ε= εr 𝐜𝐨𝐬𝚽 +
εssinΦ
Therefore, P.U. regulation at unity power factor ε=εr =0.012
P.U regulation at 0.8 p.f. lagging = 0.012*0.8 + 0.037*0.6
=0.0318
21

24. Losses
22

25. Losses
LOSSES:
I2R loss at 75℃ = 3 × 𝐼𝑝2 × 𝑅𝑝=3 × 3.782 × 21.35=915.17 W
Taking stray loss 15% above.
Total I2R loss, 𝑃𝑐=1.5 X 915.17 =1372.755 W
CORE LOSS:
Taking density laminations as 7.6X103 kg/m3
Weight of 3 limbs =3×0.3×0.01757×7.6×103=107.388 kg
Corresponding to flux density 1 Wb/ m2, specific core loss 1.2 W/kg
Core loss in limbs=107.388×1.2=128.8656 W
Weight of two Yokes = 2X0.761X0.021084X7.6X103=243.88 kg
Corresponding to flux density 0.833 Wb/m2, specific core loss 0.85 W/kg
Core loss in Yoke=243.88X0.85=207.298 W
Total core loss, Pi=128.87 + 207.3 =336.17 W
23

26. Efficiency
24

27. Efficiency
Total losses at full load = 336.17 +1372.755 =1708.925 W
Efficiency at full load unity P.f. =
𝟕𝟓𝟎𝟎𝟎
𝟕𝟓𝟎𝟎𝟎+𝟏𝟕𝟎𝟖.𝟗𝟐𝟓
× 𝟏𝟎𝟎% =97.77 %
For maximum efficiency x=
𝑷𝒊
𝑷𝒄
= 𝟎. 𝟒𝟗4
Thus maximum efficiency occurs at 49.4% of full load. This is a good figure for distribution
transformer.
25

28. No load
current
26

29. No load Current
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
𝟐×𝟏𝟔𝟗𝟎
= 30.59×10-3 A
Loss component of no load current, IL =
𝟑𝟑𝟔.𝟏𝟕
𝟐×𝟔𝟔𝟎𝟎
= 36.02×10-3 A
No load current, Io = 𝟑𝟎. 𝟓𝟗 × 𝟏𝟎
^ − 𝟑 𝟐 + 𝟑𝟔. 𝟎𝟐 × 𝟏𝟎^
− 𝟑 𝟐
= 47.26×10-3 A
No load current as a percentage of full load current =
𝟒𝟕.𝟐𝟔×𝟏𝟎
−
𝟑
𝟑.𝟕𝟖
×100%
= 1.25%
Allowing for joints etc the no load current will be about 2.5% of full load current
27

30. Tank Design
28

31. Tank Design
Height over Yoke, H = 628 mm
Allowing 50 mm at the base & 150 mm for oil,
Height of oil level =628+50+150=828 mm
Allowing 200 mm Height for leads etc,
Height of Tank Ht=828 + 200 =1028mm ≈ 1.028 m
Assuming a clearance of 40 mm along the width on each side,
Width of the Tank, Wt=2D+De+2l=2 × 295 + 208.08 + 2 × 40
=878.08 mm≈0.87 m
Length of the tank, Lt=De+2b=208.08+2X50 =308.08 mm≈0.308 m
Total loss dissipating surface of tank =2 0.87 + 0.308 × 1.028 = 2.422 m2
Specific loss dissipation due to radiation & convention is 12.5 w/m2℃
Temperature rise =
1708.925
2.422×12.5
= 56.44 ℃
This is not bellow 35℃ and therefore plain tank is not sufficient for cooling and tubes are required.
29

32. Tank With
Tubes
30

33. Tank With Tube
Area of tubes= x* 2.422= 2.422x
Loss dissipating surface= (1+x)*St= 2.422*(1+x)
Loss dissipated=
12.5+8.8𝑥
𝑥+1
w/m2 - ℃
Temperature rise with tubes,
Ɵ=
𝑃𝑖+𝑃𝑐
𝑆𝑡(12.5+8.86)
=
1708.925
2.422(12.5+8.86𝑥)
Pi + Pc = 1708.925, x= 0.494
Therefore, Ɵ = 41.88 ℃
Total area of tubes =
1
8.8
(
𝑃𝑖+𝑃𝑐
Ɵ
− 12.5 𝑆𝑡) =1.2 m2
Let, dt = 0.05 m
Lt = 1.00 m
Hence, Number of tubes nt =
1
8.8𝜋∗𝑑𝑡
∗
𝐿𝑡
(
𝑃𝑖+𝑃𝑐
Ɵ
− 12.5 𝑆𝑡) = 7.62
But, It can’t be a fraction so nt = 8.
31

34. Design Sheet
32

35. Final Voltage Current Rating
33
Line Volage
HV: 6.6
kV
LV: 433
V
Phase
voltage
HV: 6.6
kV
LV: 250
V
Line
current
HV: 6.55
A
LV:100
A
Phase
current
HV: 3.78
A
LV: 100
A

36. Design Sheet (Continued…)
1 Material --- 0.35mm thick 92 grade
2 Output Constant K .45
3 Voltage per turn Et 3.90 V
4 Circumscribing circle diameter d 177 mm
5 No. of steps --- 2
6 Dimensions a 151 mm
b 94 mm
7 Net iron area Ai 17.57 × 103 mm2
8 Flux density Bm 1.0 Wb/m2
9 Flux Φm 0.01757 Wb
10 Weight 107.388 kg
11 Specific iron loss 1.2 W/kg
12 Iron loss 128.8656 W
CORE:
34

37. Design Sheet (Continued…)
1 Depth of Yoke Dy 151 mm
2 Height of Yoke Hy 155 mm
3 Net Yoke area 21.084x103 mm2
4 Flux density 0.833 Wb/m2
5 Flux 0.01757 Wb
6 Weight 244 kg
7 Specific iron loss 0.85 W/kg
8 Iron loss 207.298 W
YOKE:
WINDOWS:
1 Number 2
2 Window space factor Kw 0.27
3 Height of window Hw 318 mm
4 Width of window Ww 118 mm
5 Area of window Aw 0.0375 m2 35

38. Design Sheet (Continued…)
1 Distance betn adjacent limbs D 295 mm
2 Height of Frame H 628 mm
3 Width of Frame W 761 mm
4 Depth of window Dy 151 mm
1 Betn L.V. winding & Core Press board wraps 1.5mm
2 Betn L.V. winding & H.V. winding Bakelized paper 5mm
3 Width of duct betn L.V & H.V. 5mm
INSULATION:
FRAME:
36

39. Design Sheet (Continued…)
Sl no. Properties L.V. H.V.
1 Type of winding Helical Cross-over
2 Connections Star Delta
3 Conductor Dimensions bare 13x3.2 mm2 Diameter=1.5 mm
insulated 13x3.7 mm2 Diameter=1.58mm
Area 41.6 mm2 1.76 mm2
No. in parallel None None
4 Current Density 2.5 A/mm2 2.13 A/mm2
5 Turns per phase 64 1690 (1775 at ±5%
tapping)
6 Coils total number 3 3x15
per core leg 1 15
7 Turns Per coil 64 14 of 120 turns, 1 of 95
Per layer 22 5
8 Number of layers 3 24
9 Height of winding 318 mm 193.5 mm
10 Depth of winding 299 mm 27.9 mm
11 Insulation Betn layers 0.5 mm press board 0.3mm paper
Betn coils 5mm spacers
12 Coil
Diameters
Inside 120 mm 152.48 mm
Outside 130.6 mm 208.08 mm
13 Length of mean turn 0.566 m 0.3927 m
14 Resistance at 75℃ 11.6 Ω 0.01268 Ω
WINDINGS:
37

40. Design Sheet
TANK:
1. Dimensions Height Ht 1.028 m
Length Lt 0.308 m
Width Wt 0.87 m
1. Tubes 8
1. Temperature rise --- 56.44 ℃
1. Impedance P.U. Resistance --- 0.035
P.U. Reactance --- 0.012
P.U. Impedance --- 0.037
1. Losses Total Core loss --- 336.17 W
Total copper loss --- 1372.755 W
Total losses at full
load
--- 1708.925 W
Efficiency at full
load & unity p.f.
--- 97.77 %
38

41. 39
Thank You all !