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Setp by step design of transformer

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Steps involved in Transformer Design

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Setp by step design of transformer

  1. 1. 1 Inductor and Transformer Design Design Problem # 1 Design a (250 VA) 250 Watt isolation transformer with the following specifications using core geometry 𝑲 𝒈 approach. Input voltage, 𝑽𝒊 = 230 V Output voltage, 𝑽 𝒐 = 230 V Output Power, 𝑷 𝒐 = 250 Watts Frequency, 𝒇 = 50 Hz Efficiency, 𝜼 = 95 % Regulation, 𝜶 = 5 % Flux density, 𝑩 𝒂𝒄 = 1.6 T Design Steps: - Various steps involved in designing this transformer are: Step # 1: Calculation of total power Total power, 𝑃𝑡 = 𝐼𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 + 𝑂𝑢𝑡𝑝𝑢𝑡 𝑃𝑜𝑤𝑒𝑟 = 𝑃𝑂 𝜂 + 𝑃𝑂 = 250 0.95 + 250 = 513.16 𝑤𝑎𝑡𝑡𝑠. Step # 2: Calculation of electrical condition Electrical conditions, 𝐾𝑒 = 0.145𝐾𝑓 2 𝑓2 𝐵 𝑚 2 × 10−4 = 0.145 × 4.44 × 502 × 1.62 × 10−4 = 1.83 Step # 3: Calculation of core geometry Core geometry, 𝐾𝑔 = 𝑃𝑡 2𝐾 𝑒 𝛼 = 513.16 2×1.83×5 = 𝟏𝟖. 𝟎𝟒 cm5 Step # 4: Selection of transformer core For the core geometry calculated in step # 3, the closest lamination number is 𝑬𝑰 − 𝟏𝟓𝟎. For 𝑬𝑰 − 𝟏𝟓𝟎 lamination, Magnetic path length (MPL) = 22.9 cm Core weight =2.334 Kg
  2. 2. 2 Copper weight = 853 gm Mean length turn (MLT) = 22 cm Iron area, 𝐴 𝑐 = 13.8 cm2 Window area, 𝑊𝑎 = 10.89 cm2 Area product, 𝐴 𝑝 = 𝐴 𝑐 × 𝑊𝑎 = 150 cm2 Core geometry, 𝐾𝑔 = 28.04 cm5 Surface area, 𝐴 𝑡 = 479 cm2 Step # 5: Calculation of primary number of turns Primary number of turns, 𝑁𝑝 = 𝑉𝑖×104 𝐾 𝑓 𝐵 𝑎𝑐 𝑓𝐴 𝑐 = 513.16×104 4.44×1.6×50×13.8 = 𝟒𝟕𝟎 turns Step # 6: Calculation of current density Current density, 𝐽 = 𝑃𝑡×104 𝐾 𝑓×𝐾 𝑢×𝐵 𝑎𝑐×𝑓×𝐴 𝑝 = 513.16×104 4.44×0.4×1.6×50×150 = 240.78 A/cm2 Step # 7: Calculation of input current Input current, 𝐼𝑖 = 𝑖𝑛𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑖𝑛𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 = 𝑃𝑜/𝜂 𝑉𝑖 = ( 250 0.95 ) 230 = 1.144 A Step # 8: Calculation of cross-sectional area (bare) of conductor for primary winding Bare conductor cross-sectional area, 𝐴 𝑤𝑝(𝐵) = 𝐼 𝑖 𝐽 = 1.144 240.78 = 0.00475 cm2 Step # 9: Selection of wire from wire table The closest Standard Wire Gauge (SWG) corresponding to the bare conductor area calculated in step # 8 is 21 SWG. For 21 SWG conductor, 𝐴 𝑤𝑝(𝐵) = 0.00519 cm2 i.e. 0.519 mm2 Resistance for 21 SWG conductor is 33.2 Ω Km = 332 𝜇Ω 𝑐𝑚 Step # 10: Calculation of primary winding resistance Resistance of primary winding, 𝑅 𝑝 = 𝑀𝐿𝑇 × 𝑁𝑝 × 332 × 10−6 = 22 × 470 × 332 × 10−6 = 3.433 Ω
  3. 3. 3 Step # 11: Calculation of copper loss in primary winding Primary winding copper loss 𝑃𝑝 = 𝐼 𝑝 2 × 𝑅 𝑝 = 1.1442 × 3.433 = 4.493 Watts Step # 12: Calculation of secondary winding turns Number of turns in the secondary winding 𝑁𝑠 = 𝑁 𝑝×𝑉𝑠 𝑉𝑖 [1 + 𝛼 100 ] = 469×230 230 [1 + 5 100 ] = 493 Step # 13: Calculation of bare conductor area for secondary winding Cross-sectional area of bare conductor for secondary winding 𝐴 𝑤𝑠(𝐵) = 𝐼 𝑖 𝐽 = 1.087 240.78 = 0.00451 cm2 = 0.451 mm2 Step # 14: Selection of conductor size required for secondary winding From the wire table, the closest cross-sectional area (i.e. next to) is found by choosing the conductor size as 21 SWG. For 21 SWG wire, bare conductor area is 0.519 mm2 , for which resistance/cm is 332 𝜇Ω/𝑐𝑚. Step # 15: Calculation of secondary winding resistance Secondary winding resistance 𝑅 𝑠 = 𝑀𝐿𝑇 × 𝑁𝑠 × 332 × 10−6 = 22 × 493 × 332 × 10−6 = 3.601 Ω Step # 16: Calculation of copper in secondary winding Copper loss in secondary winding, 𝑃𝑠 = 𝐼 𝑜 2 × 𝑅 𝑠 = 1.0872 × 3.601 = 4.255 Watts Step # 17: Calculation of total copper loss Total copper loss, 𝑃𝑐𝑢 = 𝑃𝑝 + 𝑃𝑠 = 4.493 + 4.255 = 8.747 Watts Step # 18: Calculation of voltage regulation Voltage regulation, 𝛼 = 𝑃𝑐𝑢 𝑃𝑜 = 8.747 250 = 0.035 = 3.5 % Step # 19: Calculation of Watts per Kg (W/K) Watts/Kg, 𝑊 𝐾 = 0.000557𝑓1.68 𝐵𝑎𝑐 1.86 = 0.000557 × 501.68 × 1.61.86 = 0.9545
  4. 4. 4 Step # 20: Calculation of core loss Core loss, 𝑃𝑓𝑒 = 𝑊 𝐾 × 𝑊𝑡𝑓𝑒 × 10−3 = 0.9545 × 2334 × 10−3 = 2.23 Watts Step # 21: Calculation of total loss Total loss in the transformer, 𝑃Σ = 𝑃𝑐𝑢 + 𝑃𝑓𝑒 = 8.748 + 2.23 = 10.978 Watts Step # 22: Calculation of Watts/unit area Watts/unit area, 𝜓 = 𝑃Σ 𝐴 𝑡 = 10.978 479 = 0.023 Watts/cm2 Step # 23: Calculation of temperature rise Temperature rise, 𝑇𝑟 = 450𝜓0.826 = 450 × 0.0230.826 = 19.95 0 C Step # 24: Calculation of window utilization factor Window utilization factor, 𝐾 𝑢 = 𝐾 𝑢𝑝 + 𝐾 𝑢𝑠 = 𝑁 𝑝×𝐴 𝑤𝑝(𝐵) 𝑊𝑎 + 𝑁 𝑠×𝐴 𝑤𝑠(𝐵) 𝑊𝑎 = 470×0.00519+493×0.00519 10.89 = 0.46

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