Designing a 200kVA Distribution Transformer

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A
Design Report
On
Electrical System Design-I
Submitted To
For Partial Fulfillment of the Academic Requirements for the Award of
Bachelor of Technology
In
Electrical Engineering
SUBMITTED BY
Arnab Nandi
Roll No: 16901613025
Reg. No: 131690110525 of 2013-14
UNDER THE GUIDANCE OF
Prof. Probal Mukherjee
Prof. Sandipan Misra
Department of Electrical Engineering
ACADEMY OF TECHNOLOGY
2016-2017

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I undersigned, Arnab Nandi declare that this report on Electrical System Design-I
(EE-782) is the result of my work carried out during the 7th
Semester of the B.Tech
course.
This report has not been previously submitted to any other university / institutions
for any other examination and for any other purpose by any other person. I will not
use this project report in future to use as submission to any other university,
institutions or any publisher. I also promise not to allow / permit any other persons
to copy / publish any part /full material of this report in any form.
ARNAB NANDI
Electrical Engineering,
Academy of Technology
Roll No: 16901613025
Reg. No: 131690110525 of 2013-14

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With profound respect and gratitude, I take this opportunity to convey my heartfelt gratitude
towards the academic and governing board of our University for including this course in our
B.Tech curriculum. I also express my gratitude towards our college for proving us such an
opportunity to improve our skills by assigning us this project.
My special thanks goes to Prof. Probal Mukherjee and Prof. Sandipan Mishra for guiding us
through this course.
I am also thankful Prof. Sandip Saha Chowdhury (H.O.D, Department of Electrical
Engineering) and all professors of Department of Electrical Engineering who have guided and
enlightened us throughout the B.Tech curriculum, and helped us in developing a thorough
knowledge of the various subjects which has been a significant factor for the accomplishment of
this designing project.

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Objective.................................................................................................................................................................5
Assumptions............................................................................................................................................................5
Description..............................................................................................................................................................6
Core Design.............................................................................................................................................................8
Window Dimensions........................................................................................................................................9
Yoke Parameters.............................................................................................................................................10
Weight of Lamination....................................................................................................................................10
Winding Design....................................................................................................................................................11
Calculation of Current Magnitudes ................................................................................................................11
Calculation of Cross Section Area .................................................................................................................11
Calculation of Number of Turns ....................................................................................................................11
Calculation of Length of Secondary Winding................................................................................................12
Calculation of Length of Primary Winding....................................................................................................12
Calculation of Winding Loss..........................................................................................................................12
Tank Design..........................................................................................................................................................13
Cooling Tubes ................................................................................................................................................13
Electrical Parameters ..........................................................................................................................................14
Efficiency Calculation ..................................................................................................................................14
Data Sheet.............................................................................................................................................................15

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To design a distribution transformer meeting the specified conditions
Rated Power (Q) 200 kVA
Primary Voltage 6.6 kV
Secondary Voltage 440 V
Frequency 50 Hz
Phase 3
Phasor Group Dy
Type Core type
Type of Cooling Oil Immersed Natural Cooled
Maximum Temperature Rise 35o
C
Parameter Symbol Value Unit
Voltage/turns Factor Kt 0.5 (mΩ)1/2
Stacking Factor Kf 0.9
Yoke cross-section factor Ky 1.15
Peak flux density in core Bm 1.3 Tesla
Density of Si-steel σFe 7600 kg/m3
Core loss density ρFe 2 W/kg
Winding space factor Kw 0.27
Window aspect ratio α 3
Current density in winding δ 3.2 A/mm2
Resistivity of copper ρ 2.1 µΩcm
Density of copper σCu 8900 Kg/m3

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Electrical transformer is a static device which transforms electrical energy from one circuit to another without
any direct electrical connection and with the help of mutual induction between two windings. It transforms
power from one circuit to another without changing its frequency but may be in different voltage level.
Most transformers used in commercial and industrial facilities fall in the middle ground Because these
transformers represent the majority, we should evaluate them carefully before choosing a unit for a specific
project and/or application.
There are three main parameters in choosing a transformer:
 It must have enough capacity to handle the expected loads (as well as a certain amount of overload)
 Considerations should be given to possibly increasing the capacity to handle potential load growth
 The funds allocated for its purchase be based on a certain life expectancy (with consideration to an
optimal decision on initial, operational, and installation costs.)
Both capacity and cost relate to a number of factors that you should evaluate. These include:
 Application of the unit
 Choice of insulation type (liquid-filled or dry type)
 Choice of winding material (copper or aluminum)
 Possible use of low-loss core material
 Regulation (voltage stability)
 Life expectancy
 Temperature considerations
 Losses (both no-load and operating losses)
 Shielding
 Accessories
The main parts of a transformer are:
 Laminated core
 Windings
 Insulating Materials
 Transformer oil
 Oil Tank
 Conservator
 Cooling tubes
Figure 1: Section through a Distribution Transformer

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Figure 2: Working Principle of Ideal Transformer Figure 3: Per phase Voltage and Current Relation
Figure 2: Isometric Projection of 200kVA Transformer
ARNAB NANDI ©

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The r.m.s value emf generated in transformer core is given as:
E=4.44fNAeBm where Ae = effective cross-section area of limb
N = number of turns in the winding
Bm = peak flux density in core = 1.3 Tesla
 E/N = 4.44fAeBm = et -----(i) is the voltage induced per turn
Again, et = Kt √Q -----(ii) where Kt =voltage/turns factor = 0.5 (mΩ)1/2
Q= kVA rating of transformer = 200kVA
 et = 0.5* √200 = 7.07 V/turns
Substituting et in (i) we get Ae =
7.07
4.44∗50∗1.3
= 244.9 cm2
= KfAg where Ag is the gross-core area
 Ag = 0.02449/0.9 = 0.0272m2
 Ag= a1dc + b1b2 + c1c2
= dc
2 cos θ1 + dc
2 cos θ1(sinθ2 – sin θ1) + a2 dc(cosθ1 – sin θ2)
For maximum condition:
𝜕Ag
𝜕θ1
=
𝜕Ag
𝜕θ2
= 0
 Ag = 0.668dc
2 . Solving the above two equations we get θ2 = π/4 and θ1= 25o
Figure 5: Elevation of Transformer Core
Figure 6: Cross-section of Core Limb
From geometry, we get:
a1 = dc cosθ1
a2= dc sinθ1
b1=dc cos θ1
b2= dc (sinθ2 – sin θ1)
c1=a2
c2=dc (cosθ1 – sin θ2)
Ag= a1 a2 + b1b2 + c1c2
ARNAB NANDI ©

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Figure 7: Elevation of Cross-Section of 3-stepped core
Substituting θ1 and θ2 Substituting dc
a1 = 0.906dc a1 = 18.282 cm
a2= 0.424dc a2= 8.56 cm
b1=0.707dc b1= 14.266 cm
b2= 0.283dc b2= 5.71 cm
c1=0.42dc c1= 8.56 cm
c2= 0.199dc c2= 4.01 cm
Ag = 0.668dc
2
 dc = √
Ag
0.668
= √
272
0.668
= 20.1788 cm
Throughput equation of transformer is given as:
Q*103
= 3.33 Kf Kw f Bm δ Ag Aw
 Aw =
Q∗103
3.33Kf Kw f Bm δ Ag
=
200∗103
3.33∗0.9∗0.2732∗50∗1.3∗0.0272∗3.2∗106
= 561.68 cm2
= Window Area
 Window Dimensions:
Let: Hy = Height of Yoke
Hw = Height of Window
Ww= Width of Window
α = Hw / Ww = Aspect Ratio = 3
Then Aw = Hw *Ww = α Ww
2
 561.68 cm2
= α Ww
2
 Ww = 13.68 cm
 Hw = 41.049 cm
Figure 8: Vertical and Horizontal Cross-Section of Core
ARNAB NANDI ©

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 Yoke Parameters:
Yoke Factor Ky =
Yoke Area
Ag
= 1.15 Yoke Area= Ky* Ag = a1* Hy
 Hy = 17.11 cm
Volume of Yoke = 2* Hy * Kf*D where D= 3dc +2Ww = 87.8964 cm
 Volume of Yoke = 2707.033 cm3
 Volume of Limb = 3* Hw* Ag* Kf = 30146.38 cm3
 Mass of Core = 229.78 kg
 Core loss= 2W/kg
Total loss = 459.561W = 0.4595 kW
% loss =
0.4595∗100
200
= 0.229%
 Weight of Lamination
Step-1
SEGMENT LENGTH L(cm) WIDTH (cm) STACK (cm) COUNT MASS (kg)
1 Hw + Hy = 58.159 a1= 18.282 a2= 8.56 3 186.76
2 d= 33.86 Hy= 17.11 a2 = 8.56 2 40.43
3 2d- a1 = 22.0756 Hy= 17.11 a2 = 8.56 1 22.115
Step-2
1 Hw + Hy = 58.159 b1= 14.266 b2= 5.71 3 97.214
2 d= 33.86 Hy= 17.11 b2= 5.71 2 45.254
3 2d- b1=53.46 Hy= 17.11 b2= 5.71 1 35.72
Step-3
1 Hw + Hy = 58.159 c1= 8.56 c2= 4.01 3 40.964
2 d= 33.86 Hy= 17.11 Hy= 17.11 2 31.78
2 2d- c1= 59.16 Hy= 17.11 Hy= 17.11 1 27.76
Total Mass of Lamination 527.997
Figure 9: Segments of Individual Laminations

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 Calculation of Current Magnitudes
Secondary Current Is =
Q∗103
√3∗Vs
=
200∗103
√3∗440
= 262.43 A
Primary Current Ip =
Q∗103
Vp
=
200∗103
6.6∗103 = 10.10 A
 Calculation of Cross-section Area
Secondary Cross sectional area: asec =
Is
δ
= 82 mm2
Primary Cross sectional area: apri =
Ip
δ
= 3.156 mm2
 Calculation of Number of Turns
Number of turns in secondary winding: Ns=
Vs
√3∗et
= 32.664 ≈ 33
Number of turns in primary winding: Np=
Vp
et
= 933.52 ≈ 934
Figure 10: Arrangement of Windings on Limb
Figure 11: Cross-Section through the Windings
ARNAB NANDI ©

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 Calculation of Length of Secondary Winding
Let: ls = number of layers in secondary winding = 2
ps= number of parallel sections = 2
tis= insulation thickness = 0.2mm
ts= thickness of secondary conductor = 2mm
tf = thickness of former = 3mm
tps = inter-layer insulation thickness = 0.5mm
Thickness of secondary winding: tsw = ps* ls*( ts+2 tis) + 0.5*( ls-1) = 10mm
Inner diameter of secondary winding: IDsec = dc +2tf = 207.788 mm
Outer diameter of secondary winding: ODsec = IDsec +2tsw = 227.788 mm
Mean diameter of secondary winding : MDsec = 217.788 mm
Length of secondary winding: Ls = π * MDsec * ps * (Ns/ ls) = 45.157 m
Mass/phase = ls* ws* ts*density *(10-6
) = 17.686 kg
 Calculation of Length of Primary Winding
Thickness of primary winding: tpw = 0.5mm
Thickness of duct: td = 10mm
Inner diameter of primary winding: IDpri = ODsec +2td = 247.788 mm
Outer diameter of primary winding: ODpri = IDpri +2tpw = 248.788 mm
Mean diameter of primary winding : MDpri = 248.288 mm
Length of primary winding: Lp= π * MDpri * Np = 727.7 m
 Calculation of Winding Loss
Let resistivity ρ= 20nΩm.
Resistance/phase of secondary winding: Rs = 0.00513 Ω/ph
Resistance/phase of primary winding: Rp = 0.005 Ω/ph
Loss in secondary winding = 3*Is
2
*Rs = 1.06 kW
Loss in secondary winding = 3*Ip
2
*Rp = 1.53 W

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Let CH = clearance along height = 50 cm
CL = clearance along length = 5cm
CW = clearance along width = 5cm
d = Ww + dC = 33.8588 cm
dPo = ODpri = 24.8788 cm
Tank height: H = Hw + Hy + 2CH = 175.31 cm
Tank length: L = 2d + dPo +2CL = 102.59 cm
Tank width: W= dPo + 2Cw = 34.8788 cm
 Cooling Tubes
Heat transfer area: St = 2H(L+W) = 4.819 m2
Convection coefficient: λC = 6W/m2
/o
C
Radiation coefficient: λR = 6.5W/m2
/o
C
Heat transfer factor: θT = (λC + λR)*St = 60.2375 W/o
C
Temperature rise: ∆TM = (core+copper loss) / θT
= 24.8 o
C
Tube factor: ζ = 30%
Area of Tubes: Stu =
(
loss
∆TM
)−θT
(1+ζ )∗ λC
= 2.29m2
Let height of tube h=170.31 cm and diameter φ = 5cm
Number of tubes = Stu / (π *h* φ) = 8.75 ≈ 9
NOT TO SCALE
Figure 12: Cross section of Core and Winding Figure 13: Schematic diagram of Core and Tank
ARNAB NANDI ©

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Inductance of primary winding: Lp = μo NP
2
MDpri∗ (
𝑡𝑝
3
+
𝑡𝑑
2
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑝𝑟𝑖𝑚𝑎𝑟𝑦
) = 3.683 mH/m
Inductance of secondary winding: Ls = μo Ns
2
MDsec∗ (
𝑡𝑠
3
+
𝑡𝑑
2
ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦
) = 6.7913 μH/m
Let: R1 = Primary winding resistance/ph =Rp/3
X1 = Primary winding reactance/ph = jω(Lp/3)
R2’= Secondary winding resistance/ph referred to primary = (Np
2
/3Ns
2
)Rs
X2’= Secondary winding reactance/ph referred to primary = jω(Np
2
/3Ns
2
)Ls
Total Resistance/ph = Rp/3 + (Np
2
/3Ns
2
)Rs = 1.3715 Ω
Total Inductance/ph = Lp/3 + (Np
2
/3Ns
2
)Ls = 3.04 mH
Primary side impedance: Zp = Vp/3Ip = 217.8217Ω
% Resistance: εR = (1.3715/217.8217)*100 = 0.6296%
% Reactance: εX = 0.4384%
Core loss current/ph: Icore=
PFe
3𝑉𝑝
= 0.023 A
Primary Magnetizing current/ph: Imag = 3.933A
 Efficiency Calculation
Total winding Loss: PCu = 1034.57 W
Total core Loss: PFe = 459.561 W
Full load Loss: PL = PCu + PFe = 1494.131 W
Assume operating power factor: cosφ = 0.8
 Efficiency at full load: η =
𝑄cosφ
𝑄cosφ+PL
= 99.074%
 Maximum efficiency Load fraction = √
PFe
PCu
= 66.684%
 Maximum efficiency at unity power factor: ηmax = 99.5425 %
Figure 14: Equivalent circuit of Transformer referred to Primary

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SPECIFICATIONS
Parameter Symbol Value Unit
kVA Rating Q 200 kVA
Frequency f 50 Hz
Phases φ 3
Primary Vector Y Star
Secondary Vector ∆ Delta
Type Core
Primary Voltage VP 6600 Volt (V)
Secondary Voltage VS 400 Volt (V)
ASSUMPTIONS
Parameter Symbol Value Unit
Volts/turn factor Kt 0.5 (mΩ)1/2
Stacking factor Kf 0.9 Hz
Yoke factor Ky 1.15
Peak flux density Bm 1.3 Tesla
Density of Steel σFe 7600 kg/m3
Core loss density ρFe 2 W/kg
Winding space factor Kw 0.27
Window aspect ratio α 3
Current density in winding δ 3.2 A/mm2
Resistivity of copper ρ 2.1 µΩcm
Density of copper σCu 8900 Kg/m3
CORE DESIGN
Parameter Symbol Value Unit
Volts/turn Et 7.07 Volt
Net core area Ai 244.9 cm2
Gross core area Ag 272 cm2
Diameter of Circle dc 20.1788 cm
First step width a1 18.282 cm
Second step width b1 14.266 cm
Third step width c1 8.56 cm
First step stack a2 8.56 cm
Second step stack b2 5.71 cm
Third step stack c2 4.01 cm

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WINDOW DESIGN
Parameter Symbol Value Unit
Window area Aw 561.6 cm2
Window width Ww 13.68 cm
Window height Hw 41.09 cm
YOKE DESIGN
Parameter Symbol Value Unit
Gross yoke cross-section area Agy 318.5 cm2
Volume of yoke VY 2707.033 cm3
Depth of yoke DY 18.281 cm2
OVERALL CORE DIMENSIONS
Parameter Symbol Value Unit
Limb-to-limb distance d 33.86 cm
Total width of core W 34.88 cm
Total height of core H 41.09 cm
Mass of core MCore 229.78 kg
Core loss PCore 459.56 Watt (W)
Mass of lamination Mlami 527.997 kg
WINDING DESIGN
Parameter Symbol Value Unit
Secondary turns/phase NS 33
Primary turns/phase NP 934
Secondary current IS 262.43 A
Primary current IP 10.10 A
Secondary area aS 82 mm2
Primary area aP 3.156 mm2
Secondary turns/layer nS 16.5
Number of parallel sections lS 2
Width of section WS 22 mm
Thickness of section tS 2 mm
Secondary internal diameter IDS 207.788 mm
Secondary outer diameter ODS 227.788 mm
Primary internal diameter IDP 247.788 mm
Primary outer diameter ODP 248.788 mm

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EFFICIENCY CALCULATION
Parameter Symbol Value Unit
Total winding loss PCu 1034.57 W
Total iron loss PFe 459.561 W
Full load loss PFL 1494.131 W
Operating power factor Cosφ 0.8
Efficiency at full load η 99.047 %
Maximum efficiency load fraction ξ 66.648 %
Maximum efficiency at UPF ηmax 99.5425 %
TANK DESIGN
Parameter Symbol Value Unit
Height clearance CH 50 cm
Length clearance CL 2 cm
Width clearance CW 5 cm
Tank Height HTk 175.31 cm
Tank Length LTk 102.59 cm
Tank Width WTk 34.8788 cm
Heat dissipation area ST 4.819 m2
Convection coefficient λC 6 W/m2
/o
C
Radiation coefficient λR 6.5 W/m2
/o
C
Allowed temperature rise ∆T 35 o
C
Maximum temperature rise ∆TM 24.8 o
C
Number of cooling tubes ntu 9
ELECTRICAL PARAMETERS
Parameter Symbol Value Unit
Free space permeability µo 4π*10-7 H/m
Primary winding inductance LP 3.683 mH/m
Secondary winding inductance LS 3.6836.7913 µH/m
Equivalent resistance/phase ReqP 1.3715 Ω
Equivalent inductance/phase LeqP 3.04 mH
Percentage resistance εR 0.6296 %
Percentage reactance εX 0.4384 %
Percentage impedence εZ 0.76722 %
Core loss current/phase IO 0.023 A
Fault current If 2.283 kA
Open circuit current IOC 6.813 A
Primary magnetizing current Im 3.933 A
Peak mmf in yoke Hy 235 AT/m
Peak mmf in limb HL 395 AT/m
Peak mmf in core Hm 550.72 AT