Transformer Design Ppt

DESIGING OF DISTRIBUTION TRANSFORMER BY MATLAB PROGRAMMING

ABSTRACT The 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

implementation.

have

chosen

MATLAB

for

programming

Transformer construction

Core Windings Insulting materials Insulating oil Tank Fittings and accessories

Core

Different grades of CRGO core and their losses at different Flux densities Flux density (T)

M3 (W/kg)

M4 (W/Kg)

23MOH (W/Kg)

1.5

0.70

0.84

0.67

1.6

0.85

1.00

0.77

1.7

1.09

1.20

0.92

In 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 construction

Stepped core

Window

Core structure

MS

W/H = Window height W/W = window width C/L = Center limb distance MS = maximum step

Winding materials Copper Aluminum

Maximum Current density for Aluminum

1.5 A/sq mm

Current density for copper winding

3.0 A/Sq mm

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.

Why we use Rectangular Strip for LV conductor ?

10

10 mm

Area of LV conductor = 80 sq mm Diameter for round conductor = 10 mm

10 mm

8 mm

Length for rectangular conductor = 10 mm Width for rectangular conductor = 8mm

8 mm

Reduced Winding space factor

Why parallel conductor in LV ? 10 mm 8mm

RαL Surface length = 2x (10+8) = 36 mm

10 mm 4mm

Surface length = 2 x 2 x (10+4) = 56 m 4mm

Skin effect is reduced as surface resistance is inc

Windings Insulation Major insulation It 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 Insulation This 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.

coi l

Transformer insulation construction

Limits for Winding clearances Clearance between

Minimum (mm)

Maximum (mm)

Limb to LV coil (CLLV )

2.5

3.0

LV coil to HV coil (CLVHV )

8

11

Limb to Limb (CLL )

6

10

Yoke to LV coil (CYLV )

5

10

Yoke to HV coil (CYHV )

16

25

Spacer between HV coil

6

10

HV winding to Tank Side walls

30

LV Winding Assembly

LV coil

CYLV CLLV

Front view Top view

HV winding Assembly

Spacer between HV coil to Coil

CYHV

Front view Top view

CLVHV

BASIC STRUCTURE OF DESIGN PROCEDURE START ENTER THE SPECIFICATIONS OF DTR WITH ALL NECESSARY LIMITATIONS PERFORM THE CALCULATION S IS THE CALCULATED RESULTS ARE MATCHED WITH SPECIFICATIONS YES PRINT THE REQUIRED OUTPUT DATA STOP

MAKE NECESSARY CHANGES WITH IN LIMITS NO

PROCEDURE FOR DESIGNING Step 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.

FORMULAE FOR DESIGNING To calculate of EMF per turn (Et ):

E

t

K

KVA

Where KVA = Rating of the Transformer K is a constant and varies from 0.31 – 0.39 for Aluminum winding transformer 0.41 – 0.45 for copper winding transformer To calculate number of LV turns

Where VLV = LV phase voltage Et = EMF per turn

Calculation of gross core area (Ag ) and core diameter (Do ) :

A

g

E



2.22 X

t

X 100

B

m

XCUF

sqcm

Where Bm = maximum flux density (1.5 T – 1.7 T) CUF = core utilization factor (0.95 – 0.97)

D

o



A

g

X4

 X K1

X 10mm

Where K1 = constant and depends on number of steps used in core. Number of steps varies from 6 – 14 depending on core diameter K1 varies from 0.92 – 0.96 with an increment of 0.005.

LV coil Design Calculation of LV current :

I

LV

KVAX 1000  A 3 X V LV

Where VLV = LV phase Voltage

Calculation of Area of LV Conductor :

ALV 

I CD

LV

Sqmm LV

Design of LV conductor Selection of number of parallel conductors Step 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 conductor Step1: 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)

Flow chart for calculation of dimensions of LV conductor AEPC Limits of wec=x lec= y WEC=AEPC/y Is WEC <x Yes LEC= y Display LEC & WEC

No

y= y+0.1

MATLAB Program AEPC=40 LEC=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); Out put length of Lv conductor=12.40 mm width of Lv conductor=3.23 mm

Calculation of dimensions of LV coil Axial 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 =( ρ X MLc)/ ALV LV winding loss (PLV ):

Ohms

MATLAB program for LV coil Design 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;

Why HV coils Are more than One? 1833.3 V

Layer insulation

11000 V

1833.3 V 1833.3 V 1833.3 V 1833.3 V 1833.3 V Single coil per limb

Six coils per limb

Coil voltage

Layer Voltage

For Single coil

11000 V

42 x 3.5 x 2= 294 V

For six

1833.3 V

7 x 3.5 x 2 =

Calculation of window height WH = HLV + 2 CYLV mm

Calculation of HV coil details: Number of HV turns ( NHV ) : NHV = NLV x 44 HV 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 ):

A

DHV =

HV



X4

mm

Inner diameter of HV coil (IDHV ): IDHV = ODLV + 2x CLVHV mm Height 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 coils Calculate Turns per Layer (TPL): TPL =( H HV / (D HVI +0.05) ) -1 Calculate number of layers (NL): NL = TPC/ TPL Outer diameter ( OD HV ): OD HV = 2 x ((NL x D HVI )+((NL-1) x Layer insulation)) mm Mean 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 =( ρ X MLc)/ AHV

Ohms

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

Full load loss (FLL): FLL= P LV + P HV Watts Calculation of core dimensions center limb distance (CL) = OD window width (WW) = CL –MS Where MS = Maximum step

HV

+C

LL

mm

Calculation of weight of core :

W c    3 XWH    4 XCL    2 XMSX 0.86 X 7.65 X10 X Ag XCUF 3

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 density B.F = Building factor (1.20 – 1.24)

Efficiency

% 

KVACOS% L 2

KVA COS % L% L  FLL NLL

Where FLL and NLL are in KW

Percentage resistance (%R) :

FLL( KW ) %R  100 KVA

 100

Percentage reactance (%X) :

RT 

%X   

LV

 RT 3

HV



 C LVHV  

6

2

7.91 50 10  I LV  N LV  MD W    0.95

V

LV

 MH W

Where RT

LV

RT

HV

C

LVHV

= radial thickness of LV coil in cm = radial thickness of HV coil in cm = clearance between LV and HV coils cm

MD

W

= mean diameter of windings in mm

MH

W

= mean height of windings in mm

 100

Percentage Impedance (%Z):

%Z

%R

2

 %X

2

For distribution transformer the impedance voltage should be 4.5% with ±10% tolerance

% Impedance

Short circuit current

5 % impedance

20 times rated current

4 % impedance

25 times rated current

3 % impedance

33.5 times rated current

2 % impedance

50 times rated current

Main dimensions of tank Length of tank = (2x CL) + OD Width of tank = OD

HV

HV

+60 mm

+ 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

conclusion 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

impedance.

variable

inputs

and

with

close

proximity

of