500 KVA Transformer

Design Problem for Transformer

Design of 11KV/440V Distribution Transformer

Submitted by: Vinit Khemka (08104EN030) Pushkar Garg (08104EN032) Shivanshu Goswami (08104EN033) Vivek Singh (08104EN034) Shardul Kulkarni (08104EN035)

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Design Problem for Transformer QuestionQuestion- Design a 500KVA, 11KV/ 440V, 3 phase, transformer.     

Connection: delta/star Core cross section: Square Winding scheme: Core type Cooling: oil immersed Type: distribution transformer

Calculations of losses (Iron, Copper); temperature rise; efficiency and regulation at full/half/quarter-loads and unity/ 0.8 leading/ 0.8 lagging power-factors; all day efficiency

Reasons for making above choices: 1.) Star connection is needed at distribution (secondary) end to provide users with a single phase supply for domestic consumption. 2.) Square type cross section is used because of the high rating of transformer. 3.) Core type winding scheme is preferred for high rating transformers. All windings are easily accessible so repair work is easier. From a design point of view, core winding is simpler. 4.) Since at peak load at UPF, the transformer losses are high (5.6 KW), oil immersed cooling is used. 5.)






Design of 500KVA, 11kV/440V, Dyn11, 3-phase, 50 Hz Distribution Transformer





Design Problem for Transformer Core design: Assumed V/N ratio=20 Maximum flux density for CRGO is 1.73 Tesla (knee point: 1.9 Tesla, saturation: 2.03 Tesla) So, Let us choose Bm = 1.7 tesla Assumed space factor = 0.7 Calculating the core area, V= 4.44*Φm *f* N = 4.44*(Bm*Ac)*50*N 2 Ac=20/(4.44*1.7*50)=529.6 cm 2 Ac(actual- including the space factor term)=529.6x.7=370.7 cm Square winding and cuboidal core 2 Therefore, Ac= 19.25*19.25 cm

LV design: LV voltage= 440 V Phase Voltage=440/1.732=254.03 V LV turns = 254/20=13 Current (phase) = 500000/(3*254) =656 A Number of turns =13 2 Assuming current density= 5 A/mm 2 Cross section area required =656/5=131.2 mm (diameter=12.92 mm) Insulation on conductor: 0.5 mm





Design Problem for Transformer 2

For current density of 2.5 A/mm , total conductor area required = 45.45/2.5 =18.18 mm Diameter= 4.81 mm, Height = (diameter of 1 wire +0.5mm conductor covering) * 550 turns=292 cm

2

Core dimensions: Core side Core-LV gap: LV:

2 mm 12.92 mm

LV-HV gap: HV:

1 mm 40 mm

19.25 cm 19.65 cm (LV-inner side) 20.942 cm (LV mean side) 22.234 cm (LV-outer side) 22.434 cm 26.434 cm (HV mean side) 30.434 cm (HV-outer side)

Core Loss: Loss: Weight = volume*density 3 = (96.25*77*19.25)-(2*(38.5*19.25*19.25))*(7.65 * 10^-3 kg/cm ) (density of CRGO) = 114133.25*7.65*10^-3=873.2 kg Watts/kg for built core at 1.7 tesla flux density => 1.56 Watts = 1.56*873.2= 1.36 KW Copper Loss:






Calculating Leakage reactance: Φ=N*L*I

L=Bm *Ac/N*I 6 -6 L=(1.7*131.2)/(13*656*10 )=0.311*10 H -6 X2=ω*L=97.654*10 ohm Calculating voltage regulation at: V.R. =(I2r2cosƟ+I2x2sinƟ)/E Full Load 1.) Unity power factor V.R. = 656*1.56*10^-5=1.03%. 2.) 0.8 Lagging power factor V.R. = 656*1.56*10^-5*.8+656*97.654*10^-6*0.6=0.00818+.0384 =4.66%. 3.) 0.8 Leading Leading power factor V.R. = 656*1.56*10^-5*.8-656*97.654*10^-6*0.6=0.00818-.0384 =-3.02%.







Design Problem for Transformer 3.) 0.8 Leading Leading power factor V.R. = (656*1.56*10^-5*.8-656*97.654*10^-6*0.6)/2 = (0.00818-.0384)/2 =-1.51%.

Quarter Load 1.) Unity power factor V.R. = 656*1.56*10^-5/4=0.26%. 2.) 0.8 Lagging power factor V.R. = (656*1.56*10^-5*.8+656*97.654*10^-6*0.6)/4 = (0.00818+.0384)/4 =1.17%.





Design Problem for Transformer Output = 250 KW 2 Ohmic losses = 4.24 * (0.5) = 1.06 KW Core losses = 1.36 KW Efficiency = 250/(250+1.06+1.36) = 99.04% 2.) 0.8 Lagging/Leading power factor Output = 200 KW 2 Ohmic losses = 4.24 * (0.5) = 1.06 KW Core losses = 1.36 KW Efficiency = 200/(200+1.06+1.36) = 98.8%





Design Problem for Transformer All day efficiency: 12 AM – 06 AM: 200 KW at 0.8 pf, i.e. 250 KVA Ohmic loss for 6 hours = 4.24*(0.5)2 KW = 1.06 KW Energy loss = 6.36 KWh 06 AM – 10 AM: 300 KW at 0.8 pf, i.e. 375 KVA 2 Ohmic loss for 4 hours = 4.24*(0.75) KW = 2.385 KW Energy loss = 9.4 KWh 10 AM – 12 PM: no load






Temperature rise: The energy dissipated in the resistance sums up to the heat emitted by the core and the winding surfaces, and corresponding temperature rise. Energy dissipated by the windings (E) = copper loss(at full load) + iron loss = 5600 W Temperature rise = E/CxS Where, C = specific heat capacity S = Cooling surface area






Contributions 1.) Problem statement analysis: Shardul Kulkarni, Shivanshu Goswami 2.) Core design: Vivek Singh, Pushkar Garg, Shivanshu Goswami, Shardul Kulkarni 3.) LV/HV design: Shardul Kulkarni, Vivek Singh, Pushkar Garg 4.) Loss calculations: Pushkar Garg, Vivek Singh 5.) Leakage reactance & Voltage regulation: Vinit Khemka, Pushkar Garg

500 KVA Transformer

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