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Transformer Design Project
- 1. 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
- 2. Question
- 3. Project on Transformer Design β’ 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%.
- 4. Core Design
- 5. Project on Transformer Design β’ The value of k is taken from the previous table π = 0.45 for 3-phase core type distribution transformer. β’ Voltage per turn, πΈπ‘ = π π = 0.45 150 = 5.511π β’ Therefore, Flux in the core, ππ = πΈππ‘ 4..44βπ = 5.511 4..44β50 = 0.02482ππ β’ Hot rolled silicon steel grade 92 is used. The value of flux density π΅π is assumed as 1.0ππ/π2 β’ β΄Net iron Area , π΄π = ππ π΅π = 0.02482 1 =0.02482 π2 =24.82 β 103 ππ2
- 6. Project on Transformer Design β’ Using a cruciform core, cross section area of the core, π΄π = 0.56 β π2 [ππππ ππππ ππππ‘πππ = 0.5 π = ππππππ‘ππ ππ π‘βπ πππππ’ππ ππππππ ππππππ ππ π‘βπ ππππ] β’ Diameter of the circumscribing circle, π = π΄π 0.56 = 24.82β103 0.56 = 211ππ β’ Reference width laminations, π = 0.85 β π = 0.85 β 210.52 = 178.94ππ β 179ππ π = 0.53 β π = 0.53 β 210.52 = 111.57ππ β 112ππ
- 8. Project on Transformer Design β’ For transformer rating in between 50 π‘π 200 πΎππ΄ ,the equation for window spacing factor, πΎπ€ = 10 30 + πΎπ πππ‘πππ ππ π»π π πππ ππ π‘βπ π‘ππππ ππππππ = 10 30+11 = 0.244 β’ Assuming the current density of the conductor πΏ = 2 π΄/ππ2 , β’ We can find the window area from the KVA equation of the Transformer π = 3.33 β π β π΅π β πΎπ€ β πΏ β π΄π€ β π΄π β 10β3 βΉ π΄π€ = π 3.33βπβ π΅πβ πΎπ€β πΏβπ΄πβ10β3 = 150 3.33β50β1β0.244β 2β106 β0.02482β10β3 β΄ π΄π€ = 0.0744π2 β’ Taking the ratio of Height to width 3.28 [πππ‘π€πππ 2 π‘π 4] for the window dimension, π»π€ β π π€ = 74,400 ππ2 β’ Width of the window, π π€ = 74400 3.28 = 150.607ππ β 150.6ππ β’ Height of the window, π»π€ = 74400 ππ€ = 494.02ππ β 494ππ β’ Distance between two adjacent limb center , π· = π π€ + π = 150.6 + 211 = 361.6ππ π = π·πππππ‘ππ ππ π‘βπ ππππππ πππππ’ππ πππππππ π‘βπ ππππ
- 9. Yoke Design
- 10. Project on Transformer Design β’ The cross sectional area of yoke is 15 π‘π 25% greater than that of limb. Considering 20% larger, πΉππ’π₯ ππππ ππ‘π¦ ππ π¦πππ = πΉππ’π₯ ππππ ππ‘π¦ ππ ππππ β ππππ ππππ π π πππ‘πππ ππππ π¦πππ ππππ π π πππ‘πππ ππππ = 1β ππππ ππππ π π πππ‘πππ ππππ 1.2βππππ ππππ π π πππ‘πππ ππππ = 0.833ππ/π2 β’ Net area of yoke = 1.2 β 24820 ππ2 = 29784 ππ2 = .02978 π β’ Gross area of yoke = 29784 0.9 = 33093.33 ππ2 β’ Taking the cross section of the yoke as rectangular, β΄The depth of yoke π·π¦ = π = 179 ππ β’ Height of the yoke π»π¦ = 33093.33 179 = 184.87 ππ β 185 ππ
- 12. Project on Transformer Design β’ Height of Frame βΆ π»π€ + 2 β π»π¦ = 431 + 2 β 185 = 801 ππ β’ Width of frame: 2 β π· + π = 2 β 361.6 + 179 = 902.2 ππ β’ Depth of frame, π·π¦ = π = 179 ππ
- 13. Low Voltage(L.V.) Winding (Cylindrical Winding)
- 14. Project on Transformer Design β’ Secondary voltage 415π, star connected. β’ Phase voltage π π = ππππ π£πππ‘πππ 3 = 415 3 = 239.6 β 240 π β’ No of turn per phase ππ = ππ πΈπ‘ = 239 5.511 = 43.476 β 44 β’ Secondary phase current, πΌπ = πΎππ΄ 3βππ = 150β1000 3β239.6 = 208.68 A β’ Assuming current density 2.1 π΄/ππ2, β΄Area of the conductor = 208.68 2.1 = 99.37 ππ2 . β’ As the current rating is greater than πππ¨, it is not advisable to use circular cross section conductor. β’ Taking 15 β 7 ππ (π€πππ‘β β π‘βππππππ π ) strip (rectangular) type bare conductor. β΄Conductor area, π΄π = 105 ππ2 β’ Current density in secondary winding, πΌπ π΄π = 208.68 105 = 1.98π΄/ππ2 That is less than our assumed current density value. β’ The conductor is paper insulated (0.25 ππ in each side). β’ The increase in dimension on account of paper covering is 0.5 ππ. β’ So, Dimension of insulated conductor = 15.5 β 7.5 = 116.25 ππ2
- 15. Project on Transformer Design β’ Taking 2 β πππ¦ππ cylindrical winding, 22 ππ’πππ per layer, β΄Axial depth of L.V. winding = 22 β 15.5 = 341 ππ. β’ The height of window is 431 mm. So, the winding leaves ( 431β341) 2 = 45 ππ clearance in each side. β’ Considering 1.5 ππ thick Bakelite cylinders between the core and the L.V. winding , using 0.5 ππ pressboard cylinder between layers of the L.V. winding, β’ Radial depth of low voltage winding, ππ = ππ. ππ πππ¦πππ β ππππππ ππππ‘β ππ πππππ’ππ‘ππ + πππ π’πππ‘ππ πππ‘π πππ¦πππ = 2 β 7.5 + 0.5 = 15.5 ππ β’ Diameter of circumscribing circle ,π = 211 ππ β’ The inner diameter of of L.V. winding = 211 + (2 β 1.5) = 214 ππ β’ The outer diameter of L.V. winding = 214 + 2 β 15.5 = 245 ππ
- 16. High Voltage(H.V.) Winding (Crossover Winding)
- 17. Project on Transformer Design β’ πππππππ¦ ππππ π£πππ‘πππ = πππππππ¦ πβππ π π£πππ‘πππ = 11 πΎπ β’ No of turns per phase ππ = ππβππ ππ (πππ πβππ π) = 11000β44 239.6 = 2020 β’ As 5% tapping is to be provided so, the no. of turn has to be,= 1.05 β 2020 = 2121 β’ Taking no. of coil in the H.V. side =14 , β’ 7 coils of 169 Turns , each of them has 13 layers , so there is 13 turns per layer. β’ 6 coils of 140 turns, each of them has 14 layers , so there is 10 turns per layer. β’ 1 coil of 98 turns, it has 14 layer , so there is 7 turns per layer. β’ So the overall no of turns = 7 β 169 + 6 β 140 + 1 β 98 = 2121. β’ In the turn combinations, the maximum no of turns in the coils is 169. the impressed voltage on those coils is 169 β 5.511 = 931.36 π which is below 1000 π. The impressed voltage on the rest coils will also be less than 1000 π. β’ In the turn combinations, the maximum voltage between two layers will be = 2 β ππ. ππ π‘π’πππ πππ πππ¦ππ β ππππ’πππ π£πππ‘πππ πππ π‘π’ππ = 2 β 13 β 5.511 = 143.286; π€βππβ ππ ππ π‘βπ πππππ€ππππ πππππ‘. β’ So, the inter layer voltage difference for other coils also will be less than 143.286 V.
- 18. Project on Transformer Design β’ The phase current in H.V. side, πΌπ = 150000 3β11000 = 4.54 π΄ β’ As the current is below 20 π΄, circular cross-sectional conductor can be used . Taking 2.4 π΄/ππ2 current density, β΄Area of H.V. conductor= 4.54 2.4 = 1.892 ππ2 β’ Diameter of bare conductor= 4β1.892 π = 1.5520 ππ β’ Taking diameter of bare conductor , π = 1.55 ππ. β’ Modified area of bare conductor = π 4 β π2 = 1.887 ππ2 β’ The actual current density will be = 4.54 1.887 = 2.4 π΄/ππ2 β’ Insulated conductor diameter will be, 2.05 ππ. [ .25 ππ πππ π’πππ‘πππ πππ£πππππ]
- 19. Project on Transformer Design β’ Total axial depth of coils, = (7 β 13 β 2.05 + 6 β 10 β 2.05 + 1 β 7 β 2.05) = 323.9 ππ β’ Total no. of coil = 14, so there is 13 inter coil spacing. β’ Considering 3.5 ππ inter coil spacing, β’ Total axial depth will be of H.V. = 323.9 + 13 β 3.5 = 369.4 ππ β’ The axial depth takes 369.4 494 β 100% = 74.77% of the window height which is below 75% that leaves 494β369.4 2 = 62.3 ππ spacing between coil and yoke on both side.
- 20. Project on Transformer Design β’ From πππ’ππ‘πππ 7.22 the thickness of insulation between H.V. & L.V. winding = 5 + 0.9 Γ 11 = 14.9 ππ, this includes the width of oil duct also. β’ The insulation between H.V. & L.V. winding is a 5ππ thick Bakelite paper cylinder. The H.V. winding is wound on a former 5 ππ thick and the duct is 5ππ wide, space making the total insulation between H.V. & L.V. winding 15ππ. β’ Considering 0.5 ππ thick paper insulation between layers, β’ Maximum Radial depth of H.V. winding = 14 β 2.05 + 13 β 0.5 = 35.2 ππ β’ πΌππ πππ ππππππ‘ππ ππ π»π π€ππππππ = ππ’π‘π πππ ππππππ‘ππ ππ πΏ. π. π€ππππππ + πΌππ π’πππ‘πππ ππππ‘β πππ‘π€πππ π». π. & πΏ. π. = 245 + 2 β 15 = 275 ππ β’ ππ’π‘π πππ ππππππ‘ππ ππ π». π. π€ππππππ = 275 + 2 β 35.2 = 345.4 ππ β’ πΆππππππππ πππ‘π€πππ π‘π€π ππππππππ‘ ππππ = π· β 2 β ππ’π‘π πππ πππππ’π ππ π». π. π€ππππππ = 361.6 β 345.4 = 16.2 ππ
- 22. Project on Transformer Design β’ Resistance of primary (H.V.) side : β’ Mean diameter of primary winding for 140 turns = 275 + 345.4 2 = 310.2 ππ β’ Length of mean turn of 140 turns of winding = 310.2 β π β 10β3 = 0.9745 π β’ Resistance of 6 coils having 140 turns at 75 β ππππππ Celsius , ππ = ππ.ππ ππππβ (ππ.ππ π‘π’ππ β ππππ πππππ‘β β πππ ππ π‘ππ£ππ‘π¦) ππππ π π πππ‘πππ ππππ ππ ππππ πππππ’ππ‘ππ = 6 β 140β 0.9745β 0.021 1.887 = 9.11 πβπ β’ Radial depth of 169 turns , β’ Radial depth = 13 β 2.05 + 12 β 0.5 = 32.65 ππ β’ Mean diameter of primary winding for 169 turns = 275+275+2β32.65 2 = 307.65 ππ β’ Length of mean turn of primary winding = 307.65 β π β 10β3 = 0.9665 π β’ Resistance of 7 coils having 169 turns at 75 β ππππππ Celsius = 7 β 169 β 0.021 β 0.9665 1.887 = 12.72 πβπ
- 23. Project on Transformer Design β’ Radial depth = 14 β 2.05 + 13 β 0.5 = 35.2 ππ β’ Outer diameter of H.V. winding = 275 ππ + 2 β 35.2 = 345.4 ππ β’ Mean diameter of winding = 275 + 345.4 2 = 310.2 ππ β’ Length of mean turn = 310.2 β π β 10β3 = 0.975 π β’ Resistance of turn = 1 β 98β0.021β0.975 1.887 = 1.063 πβπ β’ Total primary winding resistance = 9.11 + 12.72 + 1.063 πβπ = 22.89 πβπ β’ Mean diameter of secondary winding = 214 + 245 2 = 229.5 ππ β’ Length of mean turn of secondary winding = 229.5 β π β 10β3 = 0.721 π β’ Resistance of secondary winding at 75-degree Celsius , π π = 0.721 β 0.021 β 44 105 = 6.34 β 10β3 πβπ β’ Total Resistance referred to primary = 22.89 + ( 2121 44 )2 β 6.34 β 10β3 = 37.62 πβπ β’ P.U. resistance of transformer = πΌπβπ π ππ = 37.62 β 4.54 11000 = 0.0155
- 25. Project on Transformer Design β’ Mean diameter of windings = 214 + 345.4 2 = 279.7 ππ β’ Length of mean turn = 279.7 β π β 10β3 = 0.88 π β’ Height of winding = 341+369.4 2 = 355.2 ππ = 0.3552 π β’ Leakage reactance of transformer referred to primary side, ππ = 2 β π β 50 β 4 β π β 10β7 β 21212 β 0.88 0.3552 β (15 + 35.2+15.5 3 ) β 10β3 = 140.36 πβπ β’ P.U. leakage reactance of transformer = 4.54β 140.36 11000 = 0.058 β’ P.U. impedance of transformer = 0.0582 + 0.01552 = 0.06
- 26. Regulation
- 27. Project on Transformer Design β’ Per unit regulation, κͺ = 0.0155 β cos Ξ¦ + 0.058 β sin Ξ¦ β’ So, per unit regulation at π’πππ‘π¦ π. πΉ. = 0.0155 β’ At zero P.F. lagging = 0.058 [Ξ¦ = 90Β°] β’ At 0.8 P.F. lagging = 0.0155 β 0.8 + 0.058 β 0.6 = 0.0472
- 28. Loss Calculation
- 29. Project on Transformer Design β’ πΌ2π loss at 75-degree Celsius , 3 β πΌπ 2 β π π = 3 β 4.542 β 37.62 = 2326.23 πππ‘π‘ β’ Total loss including 15% stray load loss = 1.15 β 2326.23 = 2675.16 πππ‘π‘ β’ Taking density laminations as 7.6 β 103 ππ/π3 β’ ππππβπ‘ ππ 3 πππππ = 3 β ππππ ππππ β π€πππππ€ βπππβπ‘ β ππππ ππ‘π¦ ππ πππππππ‘ππππ = 3 β 0.02482 β 0.494 β 7.6 β 103 = 279.55 ππ β’ The flux density in the limbs is 1 ππ/π2 & corresponding to this density, specific core loss is 1.2π/ππ β’ Core loss in limbs = 279.55π1.2 = 335.46 π β’ Weight of two yokes = 2 β 0.9022 β 0.02978 β 7.6 β 103 = 408.4 ππ β’ Corresponding to 0.833 ππ/π2 flux density in the yoke, Specific core loss = 0.85W β’ Core loss in Yoke = 408.4π0.85 = 347.14 π β’ Total core loss, ππ = 335.46 + 347.14 = 682.6 π
- 30. Efficiency
- 31. Project on Transformer Design β’ Total losses at full load : 682.6 + 2675.16 = 3357.76 π€ππ‘π‘ β’ Efficiency at full load unity P.F. = ( 150β103 3357.76 + 150β103) β 100% β’ = 97.81% β’ For maximum efficiency, (π₯2) β ππππππ πππ π = ππππ πππ π β π₯ = 682.6 2675.16 β΄ π₯ = 0.505 β’ So maximum efficiency occurs at 50.5% of the full load . This is a good figure for distribution transformer.
- 32. No Load Current
- 33. Project on Transformer Design β’ Corresponding to flux densities of 1 ππ/π2 & 0.833 ππ/π2 in core & yoke respectively, with the help of B-H curve ππ‘π = 120 π΄/π & ππ‘π¦ = 80 π΄/π. β’ So, total magnetizing m.m.f. = 3 Γ 120 Γ 0.494 + 2 Γ 80 Γ 0.9022 = 322.192 π΄ β’ So, magnetizing m.m.f. per phase, π΄ππ = 322.192 3 = 107.4 π΄ β’ Magnetizing current πΌπ = π΄ππ 2βππ = 102.173 2β2121 = 0.0358 π΄ β’ Loss component of no load current = ππππ πππ π 3βππ = 682.6 3β11000 = 0.021 π΄ β’ No load current = 0.03582 + 0.0212 = 0.0415 π΄ β’ No load current as a percent of full load current = 0.0415 4.15 β 100% = 1% β’ Allowing for joints etc. the no load current will be about 2 β 2.5% of full load current.
- 35. Project on Transformer Design β’ Impedence referred to primary = 37.622 + 140.362 = 145.31 πβπ β’ Percentage of input voltage that requires to flow rated current on the secondary side at short circuit, 145.31β4.54 11000 β 100% = 5.9% [ which lies in standard limit 5 β 10%].
- 36. Tank Design
- 37. Project on Transformer Design β’ Height over yoke , π» = 801 ππ β’ Allowing 50 ππ at the base & 150 ππ for oil, β’ Height of oil level = 801 + 50 + 150 = 1001 ππ β’ Allowing another 200 ππ height for leads and bushing Height of Tank, π»π‘ = 1001 + 200 = 1201 ππ β’ The height of tank is taken as 1.201 π . β’ Width of the Tank, ππ‘ = 2π· + π·π + 2π = π Γ 361.6 + 345.4 + π Γ ππ = 1148.6 ππ [ π = πππππππππ π€ππ‘β π‘πππ] β’ The width of tank is taken as 1.148 π . β’ The clearance used is approximately 50 ππ on each side. β’ Length of the tank, πΏπ‘ = π·π + 2π = 345.4 + 2 β 70 = 385.4 ππ [π = πππππππππ] β’ The length of tank is taken as 0.385 π β’ Total loss dissipating surface of tank = π β (1.148 + 0.385 ) Γ 1.201 = 3.68 π2 β’ Total specific loss dissipation due to radiation & convection is 12.5 π/π2 β
- 38. Project on Transformer Design β’ Temperature rise = 3357.56 3.68β12.5 = 72.99 Β°πΆ. This is over 35β, therefore plain tank alone is not sufficient for cooling & so tubes are required. β’ Let the area of tubes be π₯ β 3.68 β’ Loss dissipating surface= (1 + π₯) β ππ‘ = 3.68 β (1 + π₯) β’ Loss dissipated= 12.5+8.8π₯ π₯+1 W/m2 - β β’ So, specific loss dissipation = 3357.56 3.68β 1+π₯ β35 = 26.1 1+π₯ β 26.1 1+π₯ = 12.5+8.8βπ₯ 1+π₯ β΄ π₯ = 1.55 β’ Area of tubes needed = 1.55 Γ 3.68 = 5.7 π2 β’ So, dissipating area of each tube = π Γ 0.05 Γ 1.55 = 0.243 π2 β’ So number of tubes will be provided = 5.7/0.243 β 23.46 β’ Arrangement of tubes : π΄ππππ πππππ‘β β 2 πππ€π β 3 & 2 π‘π’πππ
- 39. Design Sheet
- 40. Project on Transformer Design
- 41. Project on Transformer Design β’ 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 Γ 103 mm2 8 Flux density Bm 1.0 Wb/m2 9 Flux Ξ¦m 0.02482 Wb 10 Weight 279.55 kg 11 Specific iron loss 1.2 W/kg 12 Iron loss 335.46 W
- 42. Project on Transformer Design β’ Yoke: β’ 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
- 43. Project on Transformer Design β’ Frame: β’ 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
- 44. Project on Transformer Design β’ 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 Β±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 75β 0.00634β¦ 22.89β¦
- 45. Project on Transformer Design β’ Tank: 1. Dimensions Height Ht 1.201m Length Lt 0.385 Width Wt 1.148m 2. Tubes 8 3. Temperature rise --- 72.99 β 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%
- 46. Thank You