A Project on Transformer Design.pdf

Project on Transformer Design
Course no.: EE-3220
Group No.: 16
Name Roll
Md. Ataur Rahman 1703002
Shetu Mohanto 1703018
Md. Nafis Jawad 1703026
Md. Sabbir Mahamud 1703072

• Design a 150 KVA, 11KV/0.415KV, 50Hz, 3-phase, core type,
delta/star and ONAN cooling-based distribution transformer.
Use cylindrical and crossover type windings, and 5% tapping at
HV side. Also ensure that the impedance voltage is below 4%.

• The value of k is taken from the previous table 𝑘 = 0.45 for 3-phase core type
distribution transformer.
• Voltage per turn, 𝐸𝑡 = 𝑘 𝑄 = 0.45 150 = 5.511𝑉
• Therefore, Flux in the core, 𝜙𝑚 =
𝐸𝑘𝑡
4..44∗𝑓
=
5.511
4..44∗50
= 0.02482𝑊𝑏
• Hot rolled silicon steel grade 92 is used. The value of flux density 𝐵𝑚 is assumed as
1.0𝑊𝑏/𝑚2
• ∴Net iron Area , 𝐴𝑖 =
𝜙𝑚
𝐵𝑚
=
0.02482
1
=0.02482 𝑚2
=24.82 ∗ 103
𝑚𝑚2

• Using a cruciform core, cross section area of the core,
𝐴𝑖 = 0.56 ∗ 𝑑2 [𝑐𝑜𝑟𝑒 𝑎𝑟𝑒𝑎 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 = 0.5 𝑑 = 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑢𝑚𝑠𝑐𝑟𝑖𝑏𝑖𝑛 𝑐𝑖𝑟𝑐𝑙𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒]
• Diameter of the circumscribing circle,
𝑑 =
𝐴𝑖
0.56
=
24.82∗103
0.56
= 211𝑚𝑚
• Reference width laminations,
𝑎 = 0.85 ∗ 𝑑 = 0.85 ∗ 210.52 = 178.94𝑚𝑚 ≈ 179𝑚𝑚
𝑏 = 0.53 ∗ 𝑑 = 0.53 ∗ 210.52 = 111.57𝑚𝑚 ≈ 112𝑚𝑚

• For transformer rating in between 50 𝑡𝑜 200 𝐾𝑉𝐴 ,the equation for window spacing factor,
𝐾𝑤 =
10
30 + 𝐾𝑉 𝑟𝑎𝑡𝑖𝑛𝑔 𝑜𝑓 𝐻𝑇 𝑠𝑖𝑑𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟
=
10
30+11
= 0.244
• Assuming the current density of the conductor 𝛿 = 2 𝐴/𝑚𝑚2
,
• We can find the window area from the KVA equation of the Transformer
𝑄 = 3.33 ∗ 𝑓 ∗ 𝐵𝑚 ∗ 𝐾𝑤 ∗ 𝛿 ∗ 𝐴𝑤 ∗ 𝐴𝑖 ∗ 10−3
⟹ 𝐴𝑤 =
𝑄
3.33∗𝑓∗ 𝐵𝑚∗ 𝐾𝑤∗ 𝛿∗𝐴𝑖∗10−3
=
150
3.33∗50∗1∗0.244∗ 2∗106 ∗0.02482∗10−3
∴ 𝐴𝑤 = 0.0744𝑚2
• Taking the ratio of Height to width 3.28 [𝑏𝑒𝑡𝑤𝑒𝑒𝑛 2 𝑡𝑜 4] for the window dimension,
𝐻𝑤 ∗ 𝑊
𝑤 = 74,400 𝑚𝑚2
• Width of the window, 𝑊
𝑤 =
74400
3.28
= 150.607𝑚𝑚 ≈ 150.6𝑚𝑚
• Height of the window, 𝐻𝑤 =
74400
𝑊𝑤
= 494.02𝑚𝑚 ≈ 494𝑚𝑚
• Distance between two adjacent limb center ,
𝐷 = 𝑊
𝑤 + 𝑑 = 150.6 + 211 = 361.6𝑚𝑚 𝑑 = 𝐷𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒 𝑐𝑖𝑟𝑐𝑢𝑚𝑠𝑐𝑟𝑖𝑏𝑖𝑛𝑔 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒

• The cross sectional area of yoke is 15 𝑡𝑜 25% greater than that of limb. Considering 20% larger,
𝐹𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 𝑦𝑜𝑘𝑒 =
𝐹𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 𝑙𝑖𝑚𝑏 ∗ 𝑙𝑖𝑚𝑏 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎
𝑦𝑜𝑘𝑒 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎
=
1∗ 𝑙𝑖𝑚𝑏 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎
1.2∗𝑙𝑖𝑚𝑏 𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎
= 0.833𝑊𝑏/𝑚2
• Net area of yoke = 1.2 ∗ 24820 𝑚𝑚2
= 29784 𝑚𝑚2 = .02978 𝑚
• Gross area of yoke =
29784
0.9
= 33093.33 𝑚𝑚2
• Taking the cross section of the yoke as rectangular,
∴The depth of yoke 𝐷𝑦 = 𝑎 = 179 𝑚𝑚
• Height of the yoke 𝐻𝑦 =
33093.33
179
= 184.87 𝑚𝑚 ≈ 185 𝑚𝑚

• Height of Frame ∶ 𝐻𝑤 + 2 ∗ 𝐻𝑦 = 431 + 2 ∗ 185 = 801 𝑚𝑚
• Width of frame: 2 ∗ 𝐷 + 𝑎 = 2 ∗ 361.6 + 179 = 902.2 𝑚𝑚
• Depth of frame, 𝐷𝑦 = 𝑎 = 179 𝑚𝑚

Low Voltage(L.V.) Winding
(Cylindrical Winding)

• Secondary voltage 415𝑉, star connected.
• Phase voltage 𝑉
𝑠 =
𝑙𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒
3
=
415
3
= 239.6 ≈ 240 𝑉
• No of turn per phase 𝑇𝑠 =
𝑉𝑠
𝐸𝑡
=
239
5.511
= 43.476 ≈ 44
• Secondary phase current, 𝐼𝑠 =
𝐾𝑉𝐴
3∗𝑉𝑠
=
150∗1000
3∗239.6
= 208.68 A
• Assuming current density 2.1 𝐴/𝑚𝑚2,
∴Area of the conductor =
208.68
2.1
= 99.37 𝑚𝑚2
.
• As the current rating is greater than 𝟐𝟎𝑨, it is not advisable to use circular cross section conductor.
• Taking 15 ∗ 7 𝑚𝑚 (𝑤𝑖𝑑𝑡ℎ ∗ 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠) strip (rectangular) type bare conductor.
∴Conductor area, 𝐴𝑠 = 105 𝑚𝑚2
• Current density in secondary winding,
𝐼𝑠
𝐴𝑠
=
208.68
105
= 1.98𝐴/𝑚𝑚2
That is less than our assumed current density value.
• The conductor is paper insulated (0.25 𝑚𝑚 in each side).
• The increase in dimension on account of paper covering is 0.5 𝑚𝑚.
• So, Dimension of insulated conductor = 15.5 ∗ 7.5 = 116.25 𝑚𝑚2

• Taking 2 − 𝑙𝑎𝑦𝑒𝑟 cylindrical winding, 22 𝑇𝑢𝑟𝑛𝑠 per layer,
∴Axial depth of L.V. winding = 22 ∗ 15.5 = 341 𝑚𝑚.
• The height of window is 431 mm. So, the winding leaves
( 431−341)
2
= 45 𝑚𝑚 clearance in each side.
• Considering 1.5 𝑚𝑚 thick Bakelite cylinders between the core and the L.V. winding , using 0.5 𝑚𝑚 pressboard
cylinder between layers of the L.V. winding,
• Radial depth of low voltage winding,
𝑏𝑠 = 𝑛𝑜. 𝑜𝑓 𝑙𝑎𝑦𝑒𝑟𝑠 ∗ 𝑟𝑎𝑑𝑖𝑎𝑙 𝑑𝑒𝑝𝑡ℎ 𝑜𝑓 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟 + 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑜𝑟 𝑏𝑒𝑡𝑛 𝑙𝑎𝑦𝑒𝑟𝑠
= 2 ∗ 7.5 + 0.5
= 15.5 𝑚𝑚
• Diameter of circumscribing circle ,𝑑 = 211 𝑚𝑚
• The inner diameter of of L.V. winding = 211 + (2 ∗ 1.5) = 214 𝑚𝑚
• The outer diameter of L.V. winding = 214 + 2 ∗ 15.5 = 245 𝑚𝑚

High Voltage(H.V.) Winding
(Crossover Winding)

• 𝑃𝑟𝑖𝑚𝑎𝑟𝑦 𝑙𝑖𝑛𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑝ℎ𝑎𝑠𝑒 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 11 𝐾𝑉
• No of turns per phase 𝑇𝑝 =
𝑉𝑝∗𝑇𝑠
𝑉𝑠(𝑝𝑒𝑟 𝑝ℎ𝑎𝑠𝑒)
=
11000∗44
239.6
= 2020
• As 5% tapping is to be provided so, the no. of turn has to be,= 1.05 ∗ 2020 = 2121
• Taking no. of coil in the H.V. side =14 ,
• 7 coils of 169 Turns , each of them has 13 layers , so there is 13 turns per layer.
• 6 coils of 140 turns, each of them has 14 layers , so there is 10 turns per layer.
• 1 coil of 98 turns, it has 14 layer , so there is 7 turns per layer.
• So the overall no of turns = 7 ∗ 169 + 6 ∗ 140 + 1 ∗ 98 = 2121.
• In the turn combinations, the maximum no of turns in the coils is 169. the impressed voltage on those coils is
169 ∗ 5.511 = 931.36 𝑉 which is below 1000 𝑉. The impressed voltage on the rest coils will also be less
than 1000 𝑉.
• In the turn combinations, the maximum voltage between two layers will be
= 2 ∗ 𝑛𝑜. 𝑜𝑓 𝑡𝑢𝑟𝑛𝑠 𝑝𝑒𝑟 𝑙𝑎𝑦𝑒𝑟 ∗ 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑝𝑒𝑟 𝑡𝑢𝑟𝑛
= 2 ∗ 13 ∗ 5.511
= 143.286; 𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑎𝑙𝑙𝑜𝑤𝑎𝑏𝑙𝑒 𝑙𝑖𝑚𝑖𝑡.
• So, the inter layer voltage difference for other coils also will be less than 143.286 V.

• The phase current in H.V. side,
𝐼𝑝 =
150000
3∗11000
= 4.54 𝐴
• As the current is below 20 𝐴, circular cross-sectional conductor can be used . Taking 2.4 𝐴/𝑚𝑚2
current density,
∴Area of H.V. conductor=
4.54
2.4
= 1.892 𝑚𝑚2
• Diameter of bare conductor=
4∗1.892
𝜋
= 1.5520 𝑚𝑚
• Taking diameter of bare conductor , 𝑑 = 1.55 𝑚𝑚.
• Modified area of bare conductor =
𝜋
4
∗ 𝑑2 = 1.887 𝑚𝑚2
• The actual current density will be =
4.54
1.887
= 2.4 𝐴/𝑚𝑚2
• Insulated conductor diameter will be, 2.05 𝑚𝑚. [ .25 𝑚𝑚 𝑖𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑐𝑜𝑣𝑒𝑟𝑖𝑛𝑔]

• Total axial depth of coils,
= (7 ∗ 13 ∗ 2.05 + 6 ∗ 10 ∗ 2.05 + 1 ∗ 7 ∗ 2.05)
= 323.9 𝑚𝑚
• Total no. of coil = 14, so there is 13 inter coil spacing.
• Considering 3.5 𝑚𝑚 inter coil spacing,
• Total axial depth will be of H.V. = 323.9 + 13 ∗ 3.5 = 369.4 𝑚𝑚
• The axial depth takes
369.4
494
∗ 100% = 74.77% of the window height which is below 75% that leaves
494−369.4
2
= 62.3 𝑚𝑚 spacing between coil and yoke on both side.

• From 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 7.22 the thickness of insulation between H.V. & L.V. winding = 5 + 0.9 × 11 = 14.9 𝑚𝑚,
this includes the width of oil duct also.
• The insulation between H.V. & L.V. winding is a 5𝑚𝑚 thick Bakelite paper cylinder. The H.V. winding is
wound on a former 5 𝑚𝑚 thick and the duct is 5𝑚𝑚 wide, space making the total insulation between
H.V. & L.V. winding 15𝑚𝑚.
• Considering 0.5 𝑚𝑚 thick paper insulation between layers,
• Maximum Radial depth of H.V. winding = 14 ∗ 2.05 + 13 ∗ 0.5 = 35.2 𝑚𝑚
• 𝐼𝑛𝑠𝑖𝑑𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝐻𝑉 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 𝑂𝑢𝑡𝑠𝑖𝑑𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝐿. 𝑉. 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 + 𝐼𝑛𝑠𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑑𝑒𝑝𝑡ℎ 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝐻. 𝑉. & 𝐿. 𝑉.
= 245 + 2 ∗ 15
= 275 𝑚𝑚
• 𝑂𝑢𝑡𝑠𝑖𝑑𝑒 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑜𝑓 𝐻. 𝑉. 𝑤𝑖𝑛𝑑𝑖𝑛𝑔 = 275 + 2 ∗ 35.2 = 345.4 𝑚𝑚
• 𝐶𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑙𝑖𝑚𝑏 = 𝐷 – 2 ∗ 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝐻. 𝑉. 𝑤𝑖𝑛𝑑𝑖𝑛𝑔
= 361.6 – 345.4
= 16.2 𝑚𝑚

➢ Resistance of primary (H.V.) side :
• Mean diameter of primary winding for 140 turns =
275 + 345.4
2
= 310.2 𝑚𝑚
• Length of mean turn of 140 turns of winding = 310.2 ∗ 𝜋 ∗ 10−3
= 0.9745 𝑚
• Resistance of 6 coils having 140 turns at 75 − 𝑑𝑒𝑔𝑟𝑒𝑒 Celsius ,
𝑇𝑝 =
𝑛𝑜.𝑜𝑓 𝑐𝑜𝑖𝑙∗ (𝑛𝑜.𝑜𝑓 𝑡𝑢𝑟𝑛 ∗ 𝑚𝑒𝑎𝑛 𝑙𝑒𝑛𝑔𝑡ℎ ∗ 𝑟𝑒𝑠𝑖𝑠𝑡𝑖𝑣𝑖𝑡𝑦)
𝑐𝑟𝑜𝑠𝑠 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑏𝑎𝑟𝑒 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑜𝑟
= 6 ∗
140∗ 0.9745∗ 0.021
1.887
= 9.11 𝑜ℎ𝑚
• Radial depth of 169 turns ,
• Radial depth = 13 ∗ 2.05 + 12 ∗ 0.5 = 32.65 𝑚𝑚
• Mean diameter of primary winding for 169 turns =
275+275+2∗32.65
2
= 307.65 𝑚𝑚
• Length of mean turn of primary winding = 307.65 ∗ 𝜋 ∗ 10−3 = 0.9665 𝑚
• Resistance of 7 coils having 169 turns at 75 − 𝑑𝑒𝑔𝑟𝑒𝑒 Celsius
= 7 ∗
169 ∗ 0.021 ∗ 0.9665
1.887
= 12.72 𝑜ℎ𝑚

• Radial depth = 14 ∗ 2.05 + 13 ∗ 0.5 = 35.2 𝑚𝑚
• Outer diameter of H.V. winding = 275 𝑚𝑚 + 2 ∗ 35.2 = 345.4 𝑚𝑚
• Mean diameter of winding =
275 + 345.4
2
= 310.2 𝑚𝑚
• Length of mean turn = 310.2 ∗ 𝜋 ∗ 10−3 = 0.975 𝑚
• Resistance of turn = 1 ∗
98∗0.021∗0.975
1.887
= 1.063 𝑜ℎ𝑚
• Total primary winding resistance = 9.11 + 12.72 + 1.063 𝑜ℎ𝑚 = 22.89 𝑜ℎ𝑚
• Mean diameter of secondary winding =
214 + 245
2
= 229.5 𝑚𝑚
• Length of mean turn of secondary winding = 229.5 ∗ 𝜋 ∗ 10−3 = 0.721 𝑚
• Resistance of secondary winding at 75-degree Celsius ,
𝑅𝑠 =
0.721 ∗ 0.021 ∗ 44
105
= 6.34 ∗ 10−3
𝑜ℎ𝑚
• Total Resistance referred to primary = 22.89 + (
2121
44
)2
∗ 6.34 ∗ 10−3
= 37.62 𝑜ℎ𝑚
• P.U. resistance of transformer =
𝐼𝑝∗𝑅𝑝
𝑉𝑝
=
37.62 ∗ 4.54
11000
= 0.0155

Leakage Resistance
Calculation

• Mean diameter of windings =
214 + 345.4
2
= 279.7 𝑚𝑚
• Length of mean turn = 279.7 ∗ 𝜋 ∗ 10−3 = 0.88 𝑚
• Height of winding =
341+369.4
2
= 355.2 𝑚𝑚 = 0.3552 𝑚
• Leakage reactance of transformer referred to primary side,
𝑋𝑝 = 2 ∗ 𝜋 ∗ 50 ∗ 4 ∗ 𝜋 ∗ 10−7
∗ 21212
∗
0.88
0.3552
∗ (15 +
35.2+15.5
3
) ∗ 10−3
= 140.36 𝑜ℎ𝑚
• P.U. leakage reactance of transformer =
4.54∗ 140.36
11000
= 0.058
• P.U. impedance of transformer = 0.0582 + 0.01552 = 0.06

• Per unit regulation, Ꜫ = 0.0155 ∗ cos Φ + 0.058 ∗ sin Φ
• So, per unit regulation at 𝑢𝑛𝑖𝑡𝑦 𝑃. 𝐹. = 0.0155
• At zero P.F. lagging = 0.058 [Φ = 90°]
• At 0.8 P.F. lagging = 0.0155 ∗ 0.8 + 0.058 ∗ 0.6 = 0.0472

• 𝐼2𝑅 loss at 75-degree Celsius , 3 ∗ 𝐼𝑝
2
∗ 𝑅𝑝 = 3 ∗ 4.542 ∗ 37.62 = 2326.23 𝑊𝑎𝑡𝑡
• Total loss including 15% stray load loss = 1.15 ∗ 2326.23 = 2675.16 𝑊𝑎𝑡𝑡
• Taking density laminations as 7.6 ∗ 103 𝑘𝑔/𝑚3
• 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 3 𝑙𝑖𝑚𝑏𝑠 = 3 ∗ 𝑖𝑟𝑜𝑛 𝑎𝑟𝑒𝑎 ∗ 𝑤𝑖𝑛𝑑𝑜𝑤 ℎ𝑒𝑖𝑔ℎ𝑡 ∗ 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑙𝑎𝑚𝑖𝑛𝑎𝑡𝑖𝑜𝑛𝑠
= 3 ∗ 0.02482 ∗ 0.494 ∗ 7.6 ∗ 103
= 279.55 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 = 279.55𝑋1.2 = 335.46 𝑊
• Weight of two yokes = 2 ∗ 0.9022 ∗ 0.02978 ∗ 7.6 ∗ 103
= 408.4 𝑘𝑔
• Corresponding to 0.833 𝑊𝑏/𝑚2 flux density in the yoke, Specific core loss = 0.85W
• Core loss in Yoke = 408.4𝑋0.85 = 347.14 𝑊
• Total core loss, 𝑃𝑖 = 335.46 + 347.14 = 682.6 𝑊

• Total losses at full load : 682.6 + 2675.16 = 3357.76 𝑤𝑎𝑡𝑡
• Efficiency at full load unity P.F. = (
150∗103
3357.76 + 150∗103) ∗ 100%
•
= 97.81%
• For maximum efficiency, (𝑥2) ∗ 𝑐𝑜𝑝𝑝𝑒𝑟 𝑙𝑜𝑠𝑠 = 𝑐𝑜𝑟𝑒 𝑙𝑜𝑠𝑠
⇒ 𝑥 =
682.6
2675.16
∴ 𝑥 = 0.505
• So maximum efficiency occurs at 50.5% of the full load . This is a good figure for distribution transformer.

• Corresponding to flux densities of 1 𝑊𝑏/𝑚2
& 0.833 𝑊𝑏/𝑚2
in core & yoke respectively, with the help
of B-H curve 𝑎𝑡𝑐 = 120 𝐴/𝑚 & 𝑎𝑡𝑦 = 80 𝐴/𝑚.
• So, total magnetizing m.m.f. = 3 × 120 × 0.494 + 2 × 80 × 0.9022 = 322.192 𝐴
• So, magnetizing m.m.f. per phase, 𝐴𝑇𝑜 =
322.192
3
= 107.4 𝐴
• Magnetizing current 𝐼𝑚 =
𝐴𝑇𝑜
2∗𝑇𝑝
=
102.173
2∗2121
= 0.0358 𝐴
• Loss component of no load current =
𝑐𝑜𝑟𝑒 𝑙𝑜𝑠𝑠
3∗𝑉𝑝
=
682.6
3∗11000
= 0.021 𝐴
• No load current = 0.03582 + 0.0212 = 0.0415 𝐴
• No load current as a percent of full load current =
0.0415
4.15
∗ 100% = 1%
• Allowing for joints etc. the no load current will be about 2 − 2.5% of full load current.

• Impedence referred to primary = 37.622 + 140.362 = 145.31 𝑜ℎ𝑚
• Percentage of input voltage that requires to flow rated current on the secondary side at short circuit,
145.31∗4.54
11000
∗ 100% = 5.9% [ which lies in standard limit 5 − 10%].

• Height over yoke , 𝐻 = 801 𝑚𝑚
• Allowing 50 𝑚𝑚 at the base & 150 𝑚𝑚 for oil,
• Height of oil level = 801 + 50 + 150 = 1001 𝑚𝑚
• Allowing another 200 𝑚𝑚 height for leads and bushing
Height of Tank, 𝐻𝑡 = 1001 + 200 = 1201 𝑚𝑚
• The height of tank is taken as 1.201 𝑚 .
• Width of the Tank,
𝑊𝑡 = 2𝐷 + 𝐷𝑒 + 2𝑙 = 𝟐 × 361.6 + 345.4 + 𝟐 × 𝟒𝟎 = 1148.6 𝑚𝑚 [ 𝑙 = 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒 𝑤𝑖𝑡ℎ 𝑡𝑎𝑛𝑘]
• The width of tank is taken as 1.148 𝑚 .
• The clearance used is approximately 50 𝑚𝑚 on each side.
• Length of the tank,
𝐿𝑡 = 𝐷𝑒 + 2𝑏 = 345.4 + 2 ∗ 70 = 385.4 𝑚𝑚 [𝑏 = 𝑐𝑙𝑒𝑎𝑟𝑎𝑛𝑐𝑒]
• The length of tank is taken as 0.385 𝑚
• Total loss dissipating surface of tank
= 𝟐 ∗ (1.148 + 0.385 ) × 1.201 = 3.68 𝑚2
• Total specific loss dissipation due to radiation & convection is 12.5 𝑊/𝑚2
℃

• Temperature rise =
3357.56
3.68∗12.5
= 72.99 °𝐶. This is over 35℃, therefore plain tank alone is not sufficient
for cooling & so tubes are required.
• Let the area of tubes be 𝑥 ∗ 3.68
• Loss dissipating surface= (1 + 𝑥) ∗ 𝑆𝑡 = 3.68 ∗ (1 + 𝑥)
• Loss dissipated=
12.5+8.8𝑥
𝑥+1
W/m2 - ℃
• So, specific loss dissipation =
3357.56
3.68∗ 1+𝑥 ∗35
=
26.1
1+𝑥
⇒
26.1
1+𝑥
=
12.5+8.8∗𝑥
1+𝑥
∴ 𝑥 = 1.55
• Area of tubes needed = 1.55 × 3.68 = 5.7 𝑚2
• So, dissipating area of each tube = 𝜋 × 0.05 × 1.55 = 0.243 𝑚2
• So number of tubes will be provided = 5.7/0.243 ≈ 23.46
• Arrangement of tubes : 𝐴𝑙𝑜𝑛𝑔 𝑙𝑒𝑛𝑔𝑡ℎ – 2 𝑟𝑜𝑤𝑠 – 3 & 2 𝑡𝑢𝑏𝑒𝑠

• Core:
1 Material --- 0.35mm thick 92 grade
2 Output Constant K 0.45
3 Voltage per turn Et 5.511V
4 Circumscribing circle
diameter
d 211 mm
5 No. of steps --- 2
6 Dimensions a 178.94 mm
b 111.57 mm
7 Net iron area Ai 24.82 × 103
mm2
8 Flux density Bm 1.0 Wb/m2
9 Flux Φm 0.02482 Wb
10 Weight 279.55 kg
11 Specific iron loss 1.2 W/kg
12 Iron loss 335.46 W

• Yoke:
• Window:
1 Depth of Yoke Dy 179 mm
2 Height of Yoke Hy 184.87 mm
3 Net Yoke area 29.784x103 mm2
4 Flux density 0.833 Wb/m2
5 Flux 0.025 Wb
6 Weight 408.4 kg
7 Specific iron loss 0.85 W/kg
8 Iron loss 347.14 W
1 Number 2
2 Window space
factor
Kw 0.244
3 Height of window Hw 494 mm
4 Width of window Ww 150.6 mm
5 Area of window Aw 0.0744 m2
6 Height to width ratio 3.28

• Frame:
• Insulation:
1 Distance betn adjacent
limbs
D 361.6 mm
2 Height of Frame H 801 mm
3 Width of Frame W 902.2 mm
4 Depth of window Dy 179 mm
1 Betn L.V. winding & Core Press board wraps 1.5mm
2 Betn L.V. winding & H.V.
winding
Bakelite paper 5mm
3 Width of duct betn L.V &
H.V.
5mm

• Winding:
Sl no. Properties L.V. H.V.
1 Type of winding Cylindrical Cross-over
2 Connections Star Delta
3 Conductor Dimensions bare 15x7 mm2 Diameter=1.55 mm
insulated 15.5x7.5 mm2 Diameter=2.05mm
Area 116.25 mm2 1.887 mm2
No. in parallel None None
4 Current Density 1.98 A/mm2 2.4 A/mm2
5 Turns per phase 44 2020 (2121 at ±5%
tapping)
6 Coils total number 3 3x14
per core leg 1 14
7 Turns Per coil 44 7 of 169, 6 of 140, 1 of 98
Per layer 22 13,10,7
8 Number of layers 2 13,14,14
9 Height of winding 341 mm 369.4 mm
10 Depth of winding 15.5 mm 35.2 mm
11 Insulation Betn layers 0.5 mm press board 0.5mm paper
Betn coils 3.5mm spacers
12 Coil Diameters Inside 214mm 275mm
Outside 245mm 345.4mm
13 Length of mean turn 229.5mm 310.2mm
14 Resistance at 75℃ 0.00634Ω 22.89Ω

• Tank:
1. Dimensions Height Ht 1.201m
Length Lt 0.385
Width Wt 1.148m
2. Tubes 8
3. Temperature rise --- 72.99 ℃
4. Impedance P.U. Resistance --- 0.0155
P.U. Reactance --- 0.058
P.U. Impedance --- 0.06
5. Losses Total Core loss --- 682.6 W
Total copper loss --- 2675.16 W
Total losses at full
load
--- 3357.76 W
Efficiency at full
load & unity p.f.
--- 97.81%

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