DESIGING OF DISTRIBUTION TRANSFORMERBYMATLAB PROGRAMMING

ABSTRACTThe concept of optimization and performance of three phase distribution transformer will be achieved by using the computer technology. The computer aided design eliminates the tedious and time consuming hand calculations thereby realizing the designers from numerical drudgery to enable the designer time to grapple with physical and logical ideas thereby accelerating the design process.In this project for designing of transformer we are using the MATLAB software because MATLAB is using everywhere in Electrical fields. MATLAB program facilitates us to make design simple and accurate. As our project requires lot of mathematical calculations we have chosen MATLAB for programming implementation.

Transformer constructionCore Windings Insulting materialsInsulating oilTank Fittings and accessories

Core Different grades of CRGO core and their losses at different Flux densitiesIn our design we select the flux density from 1.5 T to 1.7 T (max), depending on the rated no load loss.Three phase core type and shell type constructionStepped core

Flux density(T)M3 (W/kg)M4 (W/Kg)23MOH (W/Kg)1.50.700.840.671.60.851.000.771.71.091.200.92

Window Core structureW/H = Window heightW/W = window widthC/L = Center limb distanceMS = maximum stepMS

Winding materialsCopper Aluminum Choice of using copper or aluminum as winding material generally depends upon the end-users. Most of our power utilities in India desire transformers up to 250 KVA/11 KV with aluminum windings, because of its wide availability and economy in cost. But owing to some limitation in its inherent properties, higher rating transformers are made with copper winding.

MaximumCurrent density for Aluminum1.5 A/sq mmCurrent density for copper winding3.0 A/Sq mm

Why we use Rectangular Strip for LV conductor ?Area of LV conductor = 80 sq mmDiameter for round conductor = 10 mm 10Reduced Winding space factorLength for rectangular conductor = 10 mmWidth for rectangular conductor = 8mm8 mm8 mm10 mm10 mm

Why parallel conductor in LV ?10 mm8mm4mm4mm10 mmR LSurface length = 2x (10+8) = 36 mmSurface length = 2 x 2 x (10+4) = 56 mmSkin effect is reduced as surface resistance is increased.

Major insulationIt is well known to the transformer designer that the main and most important insulation consists of the insulation between the high voltage coil and the low voltage coil in the same phase and from the LV coil to ground. In Medium and Low Voltage transformers, the insulation material used between High and low voltage coils are pressboard and cooling ducts, which are also used for cooling.

Minor InsulationThis category is normally for the insulation between the adjacent turns in a coil and between different sections in the same coil. Synthetic enamel covered and DPC wires are normally used in Medium and Low voltage transformers.

A Pressboard represents a thick insulation paper made of extremely pure cellulose fiber, suitably treated at the wet stage of manufacturing process and then compacted at very high pressure.

Windings Insulation

Transformer insulation constructioncoil

Limits for Winding clearances

Clearance betweenMinimum (mm)Maximum (mm)Limb to LV coil (CLLV )2.53.0LV coil to HV coil (CLVHV )811Limb to Limb (CLL )610Yoke to LV coil (CYLV )510Yoke to HV coil (CYHV )1625Spacer between HV coil610HV winding to Tank Side walls30

CYLV LV coilCLLVFront viewTop viewLV Winding Assembly

Spacer between HV coil to CoilCYHVCLVHVFront viewTop viewHV winding Assembly

BASIC STRUCTURE OF DESIGN PROCEDURESTARTENTER THE SPECIFICATIONS OF DTR WITH ALL NECESSARY LIMITATIONSPERFORM THE CALCULATION SIS THE CALCULATED RESULTS ARE MATCHED WITH SPECIFICATIONSMAKE NECESSARY CHANGES WITH IN LIMITS PRINT THE REQUIRED OUTPUT DATASTOPNOYES

PROCEDURE FOR DESIGNINGStep 1: start the design by calculating Et . Step 2: Calculate the core area and core diameter.Step 3: Calculate the LV winding details.Step 4: Calculate the HV winding details.Step 5: Calculate the dimensions of core.Step 6: Calculate the main dimensions of tank.

To calculate of EMF per turn (Et ):FORMULAE FOR DESIGNINGWhere KVA = Rating of the TransformerK is a constant and varies from0.31 0.39 for Aluminum winding transformer 0.41 0.45 for copper winding transformerTo calculate number of LV turnsWhere VLV = LV phase voltageEt = EMF per turn

Calculation of gross core area (Ag ) and core diameter (Do ) :Where Bm = maximum flux density (1.5 T 1.7 T)CUF = core utilization factor (0.95 0.97)Where K1 = constant and depends on number of steps used in core.Number of steps varies from 6 14 depending on core diameterK1 varies from 0.92 0.96 with an increment of 0.005.

LV coil DesignCalculation of LV current : Where

VLV = LV phase VoltageCalculation of Area of LV Conductor :

Design of LV conductorSelection of number of parallel conductorsStep 1: Fix boundary area of each parallel conductor. (aepc)Step2: Calculate number of parallel conductors.(NPC)

Step3: Round off to next integer NPC= Fix(npc)(for ex: 3.41 to 4)Step4: Calculate new area of each parallel conductor,

Calculation of dimensions of conductorStep1: Specify maximum limit of conductor thickness.Setp2: Select starting range of Conductor length.Step4: By using conductor area calculate the actual dimensions of conductor.(LEC and WEC)

AEPCLimits of wec=xlec= yWEC=AEPC/yIs WEC

AEPC=40LEC=5;WEC=AEPC/LEC;while (WEC>3.25) LEC=LEC+0.1; WEC=AEPC/LEC;End;fprintf (Length of Lv conductor= %3.2f mm \n\n, LEC);fprintf (Width of Lv conductor= %3.2f mm \n\n, WEC);MATLAB Programlength of Lv conductor=12.40 mm

width of Lv conductor=3.23 mmOut put

Calculation of dimensions of LV coilAxial Height (HLV ): HLV = (TPL+1) x (nA x (LECI +0.1)) mm

Inner diameter of LV coil(IDLV): IDLV = Do + 2 CLLV mm

Outer diameter of LV coil (ODLV):ODLV = IDLV+ 2x ((2x nR x WECI) + Layer insulation) mm

Mean diameter(MDLV):MDLV = (IDLV + ODLV )/2 mm

Mean length of Conductor (MLc ):MLc = MDLV x 3.14 x NLV x 10-3 M

Resistance (RLV ): RLV =( X MLc)/ ALV Ohms

LV winding loss (PLV ): PLV = 3 I2LV RLV Watts

NL=2;TLP=NLV/NL;%enter 1 for transposition and 0 for no transposition%TRANS=1;IDLV=DO+2*CLLV;if (TRANS==1) HLV=((TLP+2)*(LECI+0.1))*a;end;if (TRANS==0) HLV=(TLP+1)*a*(LECI+0.1);ODLV=IDLV+2*((2*b*WECI)+LI);MLV=mean([IDLV,ODLV]);%calculation of LV winding Resistance%if WM==1 R=0.021;else R=0.034;end;RLV=(MLV*pi*0.001*TLP*2*R)/ALVN;MATLAB program for LV coil Design

Single coil per limbSix coils per limb11000 V1833.3 V1833.3 V1833.3 V1833.3 V1833.3 V1833.3 VWhy HV coils Are more than One?Layer insulation

Coil voltageLayer VoltageFor Single coil11000 V42 x 3.5 x 2= 294 VFor six coils1833.3 V7 x 3.5 x 2 = 49 V

Calculation of HV coil details:Calculation of window heightWH = HLV + 2 CYLV mmNumber of HV turns ( NHV ) :NHV = NLV x 44HV Current (IHV ):IHV = (KVA x 1000)/(3 x 11000)Area of HV conductor (AHV): AHV = IHV / CDHV where CDHV = Current density on HV side.HV conductor Diameter ( DHV ):DHV = mm

Inner diameter of HV coil (IDHV ):IDHV = ODLV + 2x CLVHV mmHeight of HV coil (HHV ):HHV = (WH-2CYHV -(n-1)SCC)/CPL

Select number HV coils and should be even.Calculate turns per coil :TPC = N HV /Number of HV coilsCalculate Turns per Layer (TPL):TPL =( H HV / (D HVI +0.05) ) -1Calculate number of layers (NL):NL = TPC/ TPLOuter diameter ( OD HV ):OD HV = 2 x ((NL x D HVI )+((NL-1) x Layer insulation)) mmMean diameter(MDHV):MDHV = (IDHV + ODHV )/2 mm

Mean length of Conductor (MLc ):MLc = MDHV x 3.14 x NHV x 10-3 M

Resistance (RHV ): RHV =( X MLc)/ AHV Ohms

HV winding loss (PLV ): PHV = 3 I2HV RHV Watts

Full load loss (FLL):FLL= P LV + P HV WattsCalculation of core dimensionscenter limb distance (CL) = OD HV + C LL mmwindow width (WW) = CL MSWhere MS = Maximum step

Calculation of weight of core :Calculation of no load loss (NLL):NLL = W c x (W/KG) x B.F watts

Where W/KG = core loss per kg at specified flux densityB.F = Building factor (1.20 1.24)

Efficiency Where FLL and NLL are in KWPercentage resistance (%R) :

Percentage reactance (%X) :WhereRT LV = radial thickness of LV coil in cmRT HV = radial thickness of HV coil in cmC LVHV = clearance between LV and HV coils cmMD W = mean diameter of windings in mmMH W = mean height of windings in mm

Percentage Impedance (%Z):For distribution transformer the impedance voltage should be 4.5% with 10% tolerance

% ImpedanceShort circuit current5 % impedance20 times rated current4 % impedance25 times rated current3 % impedance33.5 times rated current2 % impedance50 times rated current

Main dimensions of tankLength of tank = (2x CL) + OD HV +60 mm

Width of tank = OD HV + 45 (LV connection side) + 50 (HV side) mm

Height of tank = Overall height of core + bottom clearance 35-50mm)+ top clearance (100 mm)

MATLAB Program

Output

Computer aided design has various advantages over manual design. It is error free, economical and saves valuable time. In computer aided design programme, very few inputs are available from the customer specifications. The rest of the inputs are estimated by the designer on the basis of performance parameters. Since programme discussed is on performance based, we may run the programme with different values of variable inputs and with close proximity of impedance.conclusion