Project On Transformer(Distribution) Design

PROJECT ON TRANSFORMER
DESIGN
EE-3220 ELECTRICAL
MACHINE DESIGN
SUBMITTED BY,
GROUP 02
SAKIB HOSSAIN
ROLL:1703011
MUHAMMED SAJJAD HOSSAIN
ROLL:1703022
RAIAN HABIB HIMEL
ROLL:1703025
MUHAMMED ZUBAIR RAHMAN
ROLL:1703053
DEPT OF EEE,KUET
SUBMITTED TO,
DR.MD.HABIBULLAH
PROFESSOR,DEPT OF
EEE,KUET
MD.ABU SAYED JANNATUL
ISLAM
ASSISTANT PROFESSOR,
DEPT OF EEE,KUET

DESIGN
PROBLEM:
Design a 200 KVA, 33KV/0.415KV,
50Hz, 3-phase, core type, delta/star
and ONAN cooling based
distribution (station type)
transformer. Use 5% tapping at HV
side and also ensure that the
impedance voltage is below 6%.
2

3
INTRODUCTION:
THE RATING OF THIS TRANSFORMER IS 200KVA,33KV/415V ,50HZ 3 PHASE
CORE TYPE TRANSFORMER.THIS IS A DISTRIBUTION(STATION TYPE)
TRANSFORMER.IT IS MAINLY USED TO POWER THE SUBSTATION OF ANY
AREA.AS 415V(3-PHASE) IS GENERATED IN SECONDARY WINDING SO IT IS
THEN GIVEN TO THE INPUT OF SUBSTATION.
THE STATIONARY TRANSFORMER DUTIES MAYBE SUMMARIZED AS
FOLLOWS:
1) SUPPLY TOTAL STATION LOAD(DUE TO OUTAGE OF THE OTHER STATION
TRANSFORMER)AS WELL AS SUPPLYING THE STARTING LOAD OF A UNIT.
2)SUPPLY ITS PROPORTION OF THE STATION LOAD AND THE CMR UNIT
LOAD WHEN ACTING AS REPLACEMENT FOR A UNIT TRANSFORMER.

CORE DESIGN
The value of k is taken from the table as k=0.45 for 3 phase core type distribution
transformer.
Voltage per turn,ET=K 𝑄 =0.45× 200=6.36 V
Therefore Flux in the core, Φm=
𝐸𝑡
4.44×𝑓
=
6.36
4.44×50
=0.0286 wb
Hot rolled silicon steel guard 92 is used.The value of flux density Bm is assumed as 1.0
Wb/m2.
Net iron Area =
0.0286
1.0
=28.6× 103 𝑚𝑚2
Net core area,Ai=0.56d2[for cruciform or 2 stepped core]
Diameter of circumscribing circle,d=
28.6×103
0.56
=225.98mm
Reference widths of Laminations:
a=0.85d=0.85× 225.98 = 192.08𝑚𝑚
b=0.53d=0.53× 225.98 = 119.77𝑚𝑚
The laminations are punched from 750 mm wide plates and the nearest standard
dimentions are:
a≈ 192 𝑚𝑚
b≈ 120𝑚𝑚
TYPE K
SINGLE PHASE
SHELL TYPE
1.0 T0 2.0
SINGLE PHASE
CORE TYPE
0.75 TO 0.85
THREE PHASE
SHELL TYPE
1.3
THREE PHASE
CORE
TYPE(DISTRIBU
TION)
0.45
THREE PHASE
CORE
TYPE(POWER)
0.6 TO 0.7
CHOICE OF FLUX DENSITY
1)FOR HOT
ROLLED
SILICON
STEEL
DISTRIBUTIO
N
TRANSFORM
ER
1.1 TO 1.35
Wb/b2
POWER
TRANSFORM
ER
1.25 TO 1.45
Wb/m2
2)FOR
CRGO
CORE
UPTO 132KV 1.55 Wb/m2
FOR 275 KV 1.66 Wb/m2
FOR 400KV
& GEN.
TRANSF.
1.7 TO 1.75
Wb/m2

W i n d o w
D i m e n t i o n s

Window Dimentions
7
Window space factor for Distribution Transformer Kva rating From 50 to 200 is:
Kw=
10
30+𝑘𝑣
=
10
30+33
=0.159
The current density in the windings is taken 2.1 A/mm2 for 200kva distribution transformer(ONAN
Cooling)
We know,Q=3.33fBmKwδAwAi×10-3[For 3-phase transformer]
Q=200KVA[AS ONAN COOLING IS USED HERE]
Therefore window area,AW=
200
3.33×50×1.0×0.159×2.1×106
×0.0286×10−3=0.126m2=126× 103mm2
Taking the ratio of Height to width as 3.
HW× WW = 126 × 103 mm2
Or,3× WW = 126 × 103
Therefore width of window,WW=204.94mm≈ 205 𝑚𝑚
Height of window,HW=3× 204.94mm ≈ 615𝑚𝑚
Distance between Adjacent Core Center ,D=WW+d=(204.94+225.98)mm
=430.92mm
For distribution and
small power
transformers, self oil
cooled type
δ=1.1 to 2.3 A/mm2
Window Height to
Width ratio is
between
2 to 4

Yoke Design
9
The area of Yoke is taken as 1.3 times that of limb.
Therefore flux density in yoke=1.0/1.3=0.77 Wb/m2
Net Area of Yoke=1.3× 𝐀𝐢
=1.3× 28.6 × 103 mm2
=37.18× 103 mm2
Iron Stacking Factor,Sf = 0.9
Gross Area of Yoke=
37.18×103
0.9
=41.31× 103 𝑚𝑚2
Taking the Section of Yoke As Rectangular
Depth of Yoke,Dy=a=192mm
Therefore,Height of Yoke,Hy=
41.31×103
192
=215.16mm
≈215mm
15 to 25% larger than core

O v e r a a l l
D i m e n t i o n
O F
F r a m e

OVERALL DIMENTION OF FRAME
11
Height Of Frame,H=HW+2HY=615+2× 215 𝑚𝑚
=1045mm
Width Of Frame,W=2D+a
=2× 430.92 + 192 𝑚𝑚
=1053.84mm
≈ 1054 mm
Depth OF Frame,DY=a=192mm

L O W
V O L T A G E
W I N D I N G

Low Voltage Winding
13
Secondary voltage=0.415KV=415V
Secondary Phase Voltage,VS=
415
3
=239.6=240v[Star Connected]
Number of Turns Per Phase,TS=
𝑉𝑠
𝐸𝑡
=
240
6.36
= 37.74 ≈ 38
Secondary Phase Current,IS=
200×1000
3×240
= 277.77𝐴
Current Density, δS=2.1 A/mm2 is Used[For ONAN Cooling δS range 1.1-2.3 A/mm2]
Area Of Secondary Conductor,as=IS/ δS=
277.77
2.1
= 132.27𝑚𝑚2
Using a bare Conductor of 30× 𝟒. 𝟓𝒎𝒎 𝑊𝑖𝑑𝑡ℎ × 𝑇ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠
Area Of Conductor=30× 4.5 = 135 𝑚𝑚2
Current density in Secondary Winding, δS=
277.77
135
=2.06 A/mm2
The conductor is Paper Covered.The increase in dimension on account of paper covering is 0.5 mm
Therefore The Dimentions Of Insulated Conductor=(30+0.5)× 4.5 + 0.5 𝑚𝑚2
=30.5× 5 𝑚𝑚2
As The Kva rating of the transformer is 200kva and Low Voltage Winding is 415 V So Here 3 Layer helical
Winding is Used.

LOW VOLTAGE WINDING
14
Using the 3 layer Helical winding the space has to be provided(
𝑇𝑠
3
+ 1) = (
38
3
+ 1)=14 turns along axial depth.
Axial Depth of Low Voltage Winding,LCS=14× 30.5 = 420𝑚𝑚
Height Of Window is 615 mm.This Leaves a clearance of (615-420)/2 =97.5 mm clearance of each side of the
winding.
Using 0.5 mm pressboard cylinder between layers,
Radial depth of Low Voltage Winding,bS=no. of layers× 𝐫𝐚𝐝𝐢𝐚𝐥 𝐝𝐞𝐩𝐭𝐡 𝐨𝐟 𝐜𝐨𝐧𝐝𝐮𝐜𝐭𝐨𝐫 + 𝐢𝐧𝐬𝐮𝐥𝐚𝐭𝐨𝐫 𝐁𝐞𝐭𝐰𝐞𝐞𝐧 𝐋𝐚𝐲𝐞𝐫𝐬
=3× 5 + 2 × 0.5 = 16𝑚𝑚

LOW VOLTAGE WINDING
15
Diameter Of circumscribing Circle,d=225.98 mm using Pressboard Wraps.
1.5 mm thick as insulation between L.V. winding and Core.
Inside Diameter of L.V. winding=225.98+1.5× 2 𝑚𝑚
=228.98 mm
Outside Diameter Of L.V. Winding=228.98+2× 𝒃𝒔 𝑚𝑚
=228.98+2× 16 𝑚𝑚
=260.98mm

H I G H
V O L T A G E
W I N D I N G

HIGH VOLTAGE WINDING
17
H.V. winding phase current, Ip =(200×1000)/(3×33000)
=2.02 A
AS current is below 20A,
Cross-over coils are used in H.V winding taking a current density of 1.8 A/mm2
Primary Line Voltage,VL=33KV
Primary Phase Voltage,VP=33KV[Delta Connected]
Therfore No. of turns Per Phase,TP=33000×
𝑇𝑠
𝑉𝑠
=33000×
38
240
= 5225
As ±𝟓% tapping are to be provided,Therefore the no.of turns is increased to TP=1.05× 5225
=5486.25
≈ 5486
The voltage Per coil is about 1000V[as Standard range is 800-1000V].therefore using 33 coils,
Voltage per Coil =33KV/33=1KV
Turns per coil=5486/33=166.24≈ 167
Using 32 Normal Coils of 167 turns and 1 reinforced coil of 142 turns
Total H.V. turns Provided,TotalP=32× 162 + 1 × 142 = 5486

18
Taking 28 layers per coil ,(Turns /layer)=167/28= 6 (approximately)
Maximum Voltage Between layers=2×(Turns/Layer)× ET = 2 × 6 ×6.36V
=76.32V
Area of H.V. conductor , ap=(2.02) / 1.8=1.12 mm²
Diameter of bare conductor =√((4/3.1416)×1.12 ) =1.19 mm
Bare diameter =1.180 mm
insulated diameter is 1.380 mm with fine covering.
Modified area of the conductor, ap= (
⺎
4
)× (1.18)2=1.094mm2
Actual value of current density used , δp=2.02/1.094
=1.85 A/ mm2
Axial depth of one coil =6×1.38 mm =8.28 mm
The space between adjacent coils are 5mm in height .

19
LCP=no. of coils x axial depth of spaces
=33×8.28 + 33×5 = 438.24 mm
Winding Covering percentage=
438.24
615
× % =71.26%
The height of window is 615 mm & therefore the space left between winding & window is (615-438.24) =176.76 mm.
Clearance percentage=
176.76
615
=28.74%.
The clearance left on each side is
176.76
2
= 88.38 mm.
The insulation used between layers is 0.3 mm thick paper.
Radial depth of H.V coil, bp=24 × 1.38 + 27×0.3 = 46.74 mm
thickness of insulation between H.V. and L.V. is = 5 + 0.9xkV =5+0.9×33 ≈ 34.7 mm. This includes the width oil ducts
also.
The insulation between H.V. and L.V. winding is a 5 mm thick bakelized paper cylinder .The H.V winding is wound on
a former 5mm thick and the duct is 5 mm wide ,making the total insulation between H.V. and L.V. windings 15 mm.

20
Inside diameter of H.V. winding =Outside diameter of L.V winding + 2 x thickness of insulation.
=260.98+2×15 mm
= 290.98 mm
Outside diameter of H.V. winding,De= Inside diameter of H.V. winding + 2 x Radial depth of H.V coil
= 290.98 + 2×46.74 mm
=384.46 mm
Clearance between windings= D- Outside diameter of H.V. winding
=430.92 – 384.46 mm
=46.46 mm

RESISTANCE CALCULATION:
21
Mean diameter of primary winding =
384.46+290.98
2
=337.72mm
length of mean turn of primary winding Lmtp= 𝜋×337.72×10-3
=1.06 m
Resistance of primary winding at 75℃,rp==
𝑇𝑝×⍴×Lmtp
𝑎𝑝
=
5225×0.021×1.06
1.094
=106.41Ω
Mean Diameter of Secondary winding=
228.98+260.98
2
= 244.98mm
Length of mean turn of Secondary winding Lmts= 𝜋 ×244.98×10-3
=0.77m
Resistance of Secondary winding at 75℃,rs=
𝑇𝑠×⍴×Lmts
𝑎𝑠
=
38×0.021×0.77
132.27
=4.64×10-3 Ω
Therefore Total Resistance referred to primary side ,Rp=106.41+ (
5225
38
)2 × 4.64×10-3
=194.19 Ω
P.U. resistance of transformer,εr=
𝐼𝑝×𝑅𝑝
𝑉𝑝
=
2.02×194.19
33000
=0.012 Ω

Leakage Reactance:
22
Mean diameter of windings =
384.46+228.98
2
𝑚𝑚
= 306.72 mm
Length of mean turn,Lmt = 𝜋 × 306.72 ×10-3
=0.96 m
Height of winding ,Lc =( Lcp +Lcs) /2
=(438.24+420)/2
= 429.12 mm
=0.43 m
Therefore leakage reactance referred to primary side ,
Xp=2× 𝜋 ×50×2× 𝜋 ×10-7 ×(5225)2×
0.96
0.43
× (33+
16+46.74
3
) × 10-3 bs=16mm ,bp=46.74 mm
=648.64 Ω
P.U. leakage reactance of transformer, εx=
IP
×XP
VP
=
2.02×648.64
33000
=0.039

P.U. Impedance:
23
P.U Impedance εs = 0.039 2 + 0.012 2 =0.041 or, 4.1% [%Z=
𝐼𝑝×𝑍𝑝
𝑉𝑝
is Less Than 6% So Design
is Right.]
Regulation:
P.U regulation, ε=εrcosφ+ εxsinφ
So per unit regulation at unity p.f. ε=εr=0.012
At zero p.f. lagging P.U. regulation, ε=εx=0.039
At 0.8 p.f. lagging P.U. regulation, ε = 0.012×0.8 + 0.039×0.6 =0.033

LOSSES:
24
I2R loss at 75℃ = 3 × 𝐼𝑝
2 × 𝑅𝑝=3 × (2.02)2 × 194.19 = 2377 W=2.377KW
Taking stray loss 15% above.
Total I2R loss, Pc = 1.15 X 2377 = 2.73 KW=2733.68 W
Core Loss:
Taking density laminations as 7.6X103 kg/m3
Weight of 3 limbs =3 X 0.615 X 0.0286 × 7.6X103 = 401.03 kg
The flux density in the limbs is 1.0 Wb/m2 and corresponding to this
density, specific core loss is 1.2 W/Kg
Core loss in limbs=401.03 X 1.2 = 481.236 W
Weight of two Yokes = 2 X 1.054 X 0.03718X7.6X103
=595.65 kg
Corresponding to flux density 0.833 wb/m2 in the yoke the specific
core loss = 0.85 W/kg.
Core loss in Yoke= 595.65 X 0.85 = 506.30 W
Total core loss, Pi= 481.236 + 506.30 = 987.536 W

EFFICIENCY:
25
Total losses at full load = 2733.68+ 987.536 =3721.22 W=3.72 KW
Efficiency at full load unity P.f. =
200
200+3.72
× 100% = 98.17%
For maximum efficiency ,𝑥2𝑃𝑐 = 𝑃𝑖
or,𝑥 =
𝑃𝑖
𝑃𝑐
=
987.536
2733.68
=0.601
The maximum efficiency occurs at 60.1% percent of full load. This is
good figure for distribution transformer
www.yourelectricalguide.com

IMPEDANCE VOLTAGE OF
TRANSFORMER:
26
THE IMPEDANCE VOLTAGE SHOULD BELOW 6%. So We Find impedance
voltage of Transformer is 4.1%.
%Z=percentage impedance Voltage.
Rated Voltage at H.V. Winding=33000V
As we know,
%Z=(Impedance Voltage / Rated Primary Voltage) x 100
or,4.1=
Impedance Voltage
33000
× 100
or, Impedance Voltage=0.041 × 33000
=1353V
This means there would be a 1353 voltage drop in the high-voltage winding at
full load due to losses in the windings and core.

NO LOAD CURRENT:
27
Corresponding to flux densities of 1.0 Wb/m2 and 0.833 Wb/m2 in Core And Yoke
Respectively atc =120 A/m and aty=90 A/m.
Total Magnetizing MMF=3×120× 0.615 + 2 × 90 ×1.054=411.12 A
Magnetizing mmf per Phase,AT0=411.12/3=137.04 A
Magnetizing Current,Im=AT0/( 2 𝑇𝑃) =
137.04
2 ×5225
= 18.55 × 10-3A
Loss Component Of no Load Current,IL=
987.536
3×33000
=9.975× 10−3𝐴
No Load Current,I0= (18.55 × 10−3)2+(9.975 × 10−3)2 =21.06× 10−3𝐴
No Load Current as a percentage of full load current=
21.06×10−3
2.02
× 100%
=1.04%
Allowing For Joints etc. the no load Current Will be about 2% of Full Load Current.

TANK:
29
Height over Yoke,H = 1045 mm
Allowing 50mm at the base & 150 mm for oil,
Height of oil level =1045+50+150=1245 mm
Allowing 200mm Height for leads etc
Height of Tank,Ht=1245 + 200 =1445 mm ≈1.45 m
Assuming a clearance of 75 mm along the wide of each side
Width of the Tank,Wt=2D+De+2l=2 ×430.92+ 384.46 + 2 × 75 =1396.3 mm ≈ 1.4 m
The clearance along the length of transformer is greater than that along the wide .This is because
additional space is needed along the length to accommodate tapings etc.The clearance used is
approximately 100mm on each side .
Length of the tank,Lt=De+2b=384.46+2X120 =624.46mm = 0.624m
Total loss dissipating surface of tank =2(1.4 + 0.624) × 1.45 = 5.87m2
Specific loss dissipation due to radiation & convention is 12.5W/m2℃
Temperature rise =
3721.22
5.87×12.5
=50.72℃ .This is over 35℃, therefore plain tank alone is not sufficient for
cooling & so tubes are required .
Voltage
KV
Rating
KVA
Clearance mm
b l
h
11KV OR
LESS
Less than
1000
40 50 450
1000-5000 70 90 420
ABOUT
11KV
&Upto
33KV
Less than
1000
75 100 550
1000-5000 85 125 550

TUBE CALCULATION:
31
Area of the plane tank 𝑆𝑡= 2*(W𝑡 + L𝑡) * H𝑡 =2(1.4+0.624) X1.45=5.87 𝑚2
Let the Tube area be 𝑥𝑆𝑡
Total dissipating surface = 𝑆𝑡(1+x) =5.87 (1+x)
Specific loss dissipation=
3721.22
5.87 1+𝑥 35
=
18.11
1+𝑥
W/m2-℃
Loss dissipated =
12.5+8.8𝑥
1+𝑥
W/m2-℃
Or,
12.5+8.8𝑥
1+𝑥
=
18.11
1+𝑥
Or, x=0.638
Area of tubes =0.638 X 5.87=3.74 𝑚2
Taking the diameter of tubes 100mm and the average height is 1.75m.
We get ,
Wall area of each tube=𝜋𝑑𝑡 𝑙𝑡= 𝜋 X 0.1X1.75=0.55 𝑚2
Total number of tubes to be provided=3.74/0.55=6.8 ≈ 7
Which is preferable for ONAN COOLING SYSTEM.

Design Sheet:
32
200KVA 3-phase
Type:Core
Frequency
50Hz
Delta/Star
Cooling:
ONAN
%impedance:4.1%
Tapping:5%
Line Voltage H.V :33KV Phase Voltage H.V :33KV
L.V :415V L.V :240V
Line Current H.V :3.5A Phase Current H.V :2.02A
L.V :277.77A L.V :277.77A

CORE:
33
1 Material --- 0.35mm thick 92 grade
2 Output Constant K 0.45
3 Voltage per turn Et 6.36V
4 Circumscribing circle diameter d 225.98 mm
5 No. of steps --- 2
6 Dimensions a 192 mm
b 120 mm
7 Net iron area Ai 28.6 × 103 mm2
8 Flux density Bm 1.0 Wb/m2
9 Flux Φm 0.0286 Wb
10 Weight 401.03 kg
11 Specific iron loss 1.2 W/kg
12 Iron loss 481.236 W

34
1 Depth of Yoke Dy 192 mm
2 Height of Yoke Hy 215 mm
3 Net Yoke area 37.18x103 mm2
4 Flux density 0.77 Wb/m2
5 Flux 0.0286 Wb
6 Weight 595.65 kg
7 Specific iron loss 0.85 W/kg
8 Iron loss 506.30 W
1 Number 2
2 Window space factor Kw 0.159
3 Height of window Hw 615 mm
4 Width of window Ww 205 mm
5 Area of window Aw 0.126 m2
YOKE:
WINDOWS:

35
1 Distance betn adjacent limbs D 430.92 mm
2 Height of Frame H 1045 mm
3 Width of Frame W 1054 mm
4 Depth of window Dy 192 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
FRAME:
INSULATION:

36
Sl no. Properties L.V. H.V.
1 Type of winding Helical Cross-over
2 Connections Star Delta
3 Conductor Dimensions bare 30x4.5 mm2 Diameter=1.18 mm
Insulated 30.5x5 mm2 Diameter=1.38mm
Area 135 mm2 1.094 mm2
No. in parallel None None
4 Current Density 2.1 A/mm2 1.8 A/mm2
5 Turns per phase 38 5225(5486 at ±5%
tapping)
6 Coils total number 3 3x33
per core leg 1 33
7 Turns Per coil 38 32 of 167 turns, 1 of
142 turns
Per layer 14 6
8 Number of layers 3 28
9 Height of winding 420 mm 438.24 mm
10 Depth of winding 16 mm 46.74 mm
11 Insulation Betn layers 0.5 mm press board 0.3mm paper
Betn coils 5mm spacers
12 Coil
Diameters
Inside 228.98 mm 290.98 mm
Outside 260.98 mm 384.46 mm
13 Length of mean turn 0.77 m 1.06 m
14 Resistance at 75℃ 0.00464 Ω 106.41 Ω
WINDINGS:

37
1. Dimensions Height Ht 1.45 m
Length Lt 0.624 m
Width Wt 1.4 m
1. Tubes 7
1. Temperature rise --- 50.72 ℃
1. Impedance P.U. Resistance --- 0.012
P.U. Reactance --- 0.039
P.U. Impedance --- 0.041
1. Losses Total Core loss --- 987.536 W
Total copper
loss
--- 2733.68 W
Total losses at
full load
--- 3.72 KW
Efficiency at full
load & unity p.f.
--- 98.17 %
TANK:

Project On Transformer(Distribution) Design

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