"DETAILED DESIGN OF TRANSFORMERINTRODUCTION: Design of transformer and rotating machine plays important role in manufacturing andmarketing. Most important unit in practical application of electrical network is transformers ofvarious specifications. Starting from distribution to power transformers. Transformers likeinstrumentation transformer is also very important. For fast design in meeting demand for theunits, standard software is necessary.A common program using MATLAB or any application software to provide such design isessential. In view of above facts this project tries to develop a software using MATLAB forcommon optimized design output and provides necessary information about operating parameters.METHODOLOGY:It is a software programming which provides the details about transformersspecifications as per the data fed by user or manufacturer. This program is useful to obtaindesigning details of different types of transformer and also provide the specified data as per“INDIAN ELECTRICITY STANDARD” it also provide the details of minimum cost formanufacturing of a transformer.A software developed using MATLAB takes data input and given data will be made use inproviding detailed design data pertaining to design of any transformer.Software provide an optimal design data for the required dimension. Input parameters likesecondary and primary voltages KVA ratings with frequency and power factor. Certain designspecification can be made us through look-up table using MATLAB and design formulaecommon program is developed.The software developed makes use of rated KVA, number of phases, frequency, type oftransformer, rated voltages, tapping - number of tapping, range, type of connections, temperaturerise and some time limitations for short circuit impedance, no load loss and load loss.Certain data’s like volts per turn, maximum flux density, current density, window space factor,stocking factor is to be fed through look-up table.Asoftware developed using MATLAB provides information on main dimensions of the core,area of the core, overall length of the core, yoke, height of the yoke, details of low voltage andhigh voltage windings etc..This program also provides information about performance like total loss, efficiency, all dayefficiency, regulation and design of cooling tank if necessary. Dept. of Electrical & Electronics Engineering-BIET,Davangere Page 1 DETAILED DESIGN OF TRANSFORMERBLOCK DIAGRAM:NECESSARY RELATIONS DESIGN DATA MATLAB SOFTWARE FORINPUTRELATIONSLOOK UP TABLE DESIGN OUTPUTOUTPUT SELECTION OF DATAS:A set of standard data’s related to design of transformer (both core type and shell type)makes use of standard inputs like voltage rating, current rating, kVA rating gets utilized inobtaining standard dimensions. It makes use of design formula and MATLAB is used insimulating the results.EXPECTED RESULTS: In this project for any given data corresponding design parameters and operatingparameters can be obtained.? Using common program design and performance parameters can be obtained.? This can be made used as standard software and the hand tool to obtain design parametersand performance parameters.? Software available in market is costly (Runs lacks of rupees) may not be affordable toordinary small scale industries.Dept. of Electrical & Electronics Engineering-BIET,Davangere Page 2 DETAILED DESIGN OF TRANSFORMERAPPLICATIONS:? Small scale industries can utilize this software to obtain design data, operating parameterand standard drawing.? Same can be made using in the design of very large transformers. SUMMARY:This project work is an attempt to design the transformer of all the ranges and developdrawing using Auto-CAD. In this project there is a provision for optimization and providesinformation about performance parameters.PROCEDUREFOR THE DESIGN OF TRANSFORMERSThe following points may be noted in designing the transformers:? Specifications of the transformer to be designed. ? Choose the voltage per turn, ? volts.? 2 ? Choose the maximum flux density, ? Wb/?? ? Calculate the net cross-section of the core, ?? ? Find the diameter of the circumscribing circle of the core, D.? Choose the window area ? from ? .? ? ? Find the turns in the low voltage winding (L.V) ,? .2 ? Find the turns in the high voltage winding (H.V) , ? .1 2 2 ? Choose the current density, ? A/? or MA/? .? Find the conductor size of the L.V winding, ? .2 ? Find the conductor size of H.V winding, ?1 Dept. of Electrical & Electronics Engineering-BIET,Davangere Page 3 DETAILED DESIGN OF TRANSFORMER? Find the ratio of cross-section of copper of L.V and H.V windings in window towindow area ? .Check kW.? ? Choose the layout of L.V and H.V winding.? Choose the core frame; the core diameter, window, yoke, overall core size.? Design the low voltage winding.? Design the high voltage winding.? Calculate the % reactance.? Calculate the % resistance.? Calculate the % impedance.? Find the weight of the iron in core and yoke.? Find the core or iron loss.? Find the magnetizing VA.? Find the weight of L.V winding.? Find the weight of H.V winding.? Find the resistance of L.V and H.V winding.? Find the turns ratio of H.V/L.V., i.e.? /? .1 2 ? Calculate the equivalent resistance referred to H.V.? Calculate the copper loss 3I^2r allowing 7% to 10% for stray load loss; find theload loss at 75 degree CelsiusEXAMPLE: Design a 5MVA, three-phase, 50 Hz, 66kV/11kV, delta-deltaconnected transformer. The construction is of core type; temperaturerise of oil 50 degree Celsius, % impedance 7.15%. Total loss not moreDept. of Electrical & Electronics Engineering-BIET,Davangere Page 4 DETAILED DESIGN OF TRANSFORMERth than 50kW.Find the efficiency of transformer on full load, 3/4 loads,and ½ full load at 75degree Celsius. Find the regulation on full load atunity power factor and at 0.8pf lagging. Workout the detaileddimensions of cores, coils.Solution:Voltage per turn:1 1000? = ? ? ? ×? 40 ? ? ? ? ? ? ? ? ? ? ? ? 1 1000= v(5000 × )40 3= 32.27 , choose 32 V per turn.High voltage side: volts per phase=66000V. Turns per phase=66000/32=2064Low voltage side: Volts per phase=11000 V Turns per phase=11000/32=344Core: Use core laminations of cold rolled steel of 0.35mm thickness; mitered coreconstruction 45° cut (CRGO). Choose flux density 1.6 Wb/m^2; specific loss=1.3W/kg; use 9-step core so that core space factor=0.89; take stacking factor=0.92.Cross-sectional area of core:6 ? 10 ? 2? = × ? × 50 = 32 × × 1.6 × 50? ?? ? 4.44 4.44Dept. of Electrical & Electronics Engineering-BIET,Davangere Page 5 DETAILED DESIGN OF TRANSFORMER2 ? ? 2 ? = 0.89 × 0.92 × = 90,090? ?? 4 2 ? = 90090 × 4/0.89 × 0.92 × ? Hence d=374.28mmTake d=375mm.Then ? =90,387mm^2? Take core and yoke of the same sections.With d=375mm, the largest width of the core section and hence the width of thelimb=375*0.93=349 say 350mm approxArea of window ? : ? -3 ? = 3.33? ? ? ? ? ? ? × 10 kVA? ? ? Take kW=0.28; current density ? =3 A/mm^290387 6 -35000 = 3.33 × × ? × 3 × 10 × 1.6 × 0.28 × 10? 6 10 Solving ? =247,200 mm^2. Choose larger ? for H.V transformer.? ? Choose width=375mm and height=1050mm.Then ? =375×1050 or? ? =393,750mm²? Low voltage winding: 1000? = 5000 × × 11000 = 151.5 ?? ? 3 Choose current density ? = 3 A/mm²Area of conductor for L.V winding ? =151.5/3=50.5 mm² L.V turns=344. Choose2 a rectangular conductor, two strips of 45mm thick×6.3mm wide conductor; usepaper insulation. Area of conductor=27.49mm²; area of twoconductors=27.5×2=55mm², then current density= 151.5/55=2.75 Amm². Use disccoils.Dept. of Electrical & Electronics Engineering-BIET,Davangere Page 6 DETAILED DESIGN OF TRANSFORMERInsulation: Use a Bakelite cylinder between the core and L.V 5mm thick and stickof 10mm; the distance between L.V and core diameter =15mm; paper insulationon conductors 0.5 to 0.6mm thick. Spacers between coils (disc of) 10mm.There 344 turns per limb. Split it into 34 sections or discs of 10 turns each andth 35 disc of 4 turns with more insulation. Each conductor is 2mm×6.3mm×4.5mm.With 0.6mm paper insulation around the conductor, the dimensions of the winconductors for one turn of coil are 6.9mm×2mm×5.1mm. There are two such turnslengthwise and 5 such turns widthwise or arrangement is 2× high×5 wide urns perdisc. Size of each disc=13.8mm high×51mm wide. Height of L.V coils over copper= 35 × 13.8 + 34 × 10 = 483 + 340 = 823? ? , where 34 spacers each of10mm are used between discs. Distance between core diameter and L.V =15mmInside diameter of L.V=375+2 × 15 = 405? ? Thickness of L.V coils=51mm Outside diameter of L.V. coils=405 + 2 × 51 = 507? ?Mean diameter of L.V=405+51 = 456? ?Mean length of turn of L.V=? × 456 = 1432? ?High voltage winding: ? = 5000*1000/3*66000=25.25 A? ? High voltage turns per phase=2064Current density chosen ? = 3 ? /mm²Size of conductor a1=25.25/3=8.42mm²Dept. of Electrical & Electronics Engineering-BIET,Davangere Page 7 DETAILED DESIGN OF TRANSFORMERChoose rectangular conductor 4.5mm×2.0mm size; area=8.637mm²Then current density =25.25/8.637=2.92A/mm²Use disc coils just as in case of LV.choose 35disc coils. Each conductordimension with paper insulation is 5.1×2.6mm. Each disc (except the last whichwill have remaining turns) will have 60 turns: 4 conductors high and 15conductors wide. So height of conductors in each disc=5.1×4=20.4mm andwidth=2.6×15=39mm. Spacers between discs=10mm each. Thus height of theH.V.winding over copper =35 × 20.4 + 34 × 10 = 714 + 340 = 1054? ?Distance between L.V.and H.V.=32mmInside diameter of H.V.coils=507+ (2×32)=571mmThickness of H.V.coils=39mmOutside diameter of H.V.coils=571+2×39=649mmMean diameter of H.V.coils 571+39=610mmMean length of turn=? × 610 = 1916? ?Average mean length of L.V.and H.V.=1432+1916/2=1674mmMean height over copper of L.V. and H.V.=823+1054/2=938.5mm? =32 volts; ? =151.5×344 ? ? a=32mm; ? =39mm;? =51mm; (a+? +;? =/3)=62mm1 2 1 2 Percentage reactance:? -7 2 2? × ? × 4? × 10 × ? ? ? ? ? × ? × (? + ? + ) ? ? ? 1 3 %? = ? ? ? ? ? × ? ? ? -7 2? × 50 × 4 × ? × 10 × 1.674 × 151.4 × 344 × 0.062 =0.9835 × 32 Dept. of Electrical & Electronics Engineering-BIET,Davangere Page 8 DETAILED DESIGN OF TRANSFORMER =7.11%Percentage resistance: ? =0.021 ohm/mm²/mm for copper conductorsResistance of L.V.winding per phase= ? ? × ? ? ? ? ?? ? ?2=0.021×1432×344/55×1000=0.188 ohmResistance of H.V.winding per phase =0.021×1916×2064/8.636×1000=9.615 ohm 2 Equivalent resistance referred to H.V.= 9.615+(6 ×0.188) = 9.62+6.77=16.39 ohm25.25×16.39%resistance = ×100% = 0.63%66000Percentage impedence:2 2 2 2%impedence=v%? + %? = v7.11 + 0.63 =7.14% Copper loss : load loss :2 2 Copper loss=3? r=3×25.25 × 16.39 = 31.35 ? ? .Add 10% stray loss=3.135kW. Load loss at 75?=34.485 kWDept. of Electrical & Electronics Engineering-BIET,Davangere Page 9 DETAILED DESIGN OF TRANSFORMERWeight of iron: ? =90,387 mm²? 90,387Weight of iron = (3× 1335 + 2 × 1720) × 7.85 × 1000? ?6 10 =5283kgIron loss = 5283×1.3= 6.87KwFull load loss =34.485+6.87=41.36kWEfficiency on full load at unity pf:5000= ×100%= 99.18%5000+41.36 thOn 3/4 of full load, load loss=(9/16)×34.485=19.4kW On ½ of full load, load loss=(1/4)×34.485=8.62 kWthEfficiency on 3/4 of full load at unity power factor 5000×0.75=×100%=99.3%5000×0.75+6.87+19.4 5000×0.5Efficiency on ½ of full load at unity power factor =×100%=99.38% 5000×0.5+6.87+8.62Core loss current per phase:6870 ? ==0.034 A? 3×66000 Magnetizing VA for ?(=1.6 Wb/m) is 5 VA/kg? ? ? Magnetizing VA=5283×5=26415 VADept. of Electrical & Electronics Engineering-BIET,Davangere Page 10 "