Heat transfer analysis of a 6- Nodal triangular element under steady state

8/8/2019 Heat transfer analysis of a 6- Nodal triangular element under steady state

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Abstract

The UEL Subroutine in Abaqus can be used for various user de-fined applications. In this project, we program a user element for a

steady state heat transfer case using Fortran. This user element is

used subsequently to analyze a part geometry which is subjected to

varying boundary conditions of temperature and heat transfer coeffi-

cient along its surface. Part geometry was meshed using the Abaqus

6-Nodal element and then a set of user elements were generated using

the input file. Finally, the accuracy of the temperature analysis was

compared with the results generated by Abaqus.

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Contents

1 Task Formulation 4

2 Introduction 5

3 Formulation of governing equation 6

4 Shape Functions 9

5 Part Geometry 11

6 Results 12

6.1 Temperature distribution across the element . . . . . . . . . . 126.2 Comparison of nodal temperatures . . . . . . . . . . . . . . . 12

7 Conclusion 14

Appendices 15

A Fortran code 15

B Input File 23

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1 Task Formulation

Heat transfer computation under steady-state condition, using the 6 nodaltriangular isotropic element formulation.

Tasks:

1. Write a UEL program for ABAQUS for the problem above.

2. Consider the heat convection boundary condition in the UEL program.

3. Analysis of temperature distribution in plates and discs.

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2 Introduction

In a steady state analysis, the temperature is not a function of time.Hencethe whole analysis is done in one step even though the number of incrementsused to finish one step may vary. Since, it is an isotropic element, the heattransfer in both x and y directions are equal. These are important consider-ations since they influence the final governing equations to be formulated.

A Two-Dimensional heat transfer element in space is always under theinfluence of forces. The forces are in the form of Internal heat generated,Flux, and Force due to convection. Hence suitable boundary conditions haveto be applied in order to deduce the appropriate governing equation.

A schematic diagram of force function is as shown below:

Figure 1: Types of boundary conditions for a two dimensional conductionand convection.

1. S1: Surface on which the temperature is known.

2. S2: Surface on which heat flux acts.

3. S3: Surface on which convection has to be considered.

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3 Formulation of governing equation

Consider a differential element of a solid body that has a small dimensionin z direction. Thus now, the whole domain is reduced to two dimensions xand y.We consider for the time being that the differential element is in theinterior of the body. Hence, only the lateral convection is considered but notthe edge convection.Under the assumptionof steady state conditions, we obtain the equation

qxtdy+qytdx+Qtdydx =

qx +

qxy

dx

tdy+

qy +

qyy

dy

tdx+2h(TTa)dydx

(1)

Figure 2: Two Dimensional case of conduction and convection

where

1. t=thickness

2. h=heat transfer coefficient

3. Ta=ambient temperature of the surrounding fluid

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Utilizing Fouriers law,

qx = kxTx

(2)

qy = kyT

y(3)

After substituting flux values and simplifying, the equation is of theform:

x

tkx

T

x

+

y

tky

T

y

+ Qt = 2h(T Ta)

(4)

For Finite element formulation, we describe the temperature distribu-tion by :

T(x, y) =

Mi=1

Ni(x, y)Ti = [N][T]

(5)

Considering equation (1)and applying Galerkins formulation and con-sidering first two integrals and manipulating it mathematically, we ob-tain

A

Ni(x, y) x

tkxTx

+ y

tky Ty

+ Qt 2h(T Ta) dA = 0

(6)

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Applying Green Gauss theorom, and substituting and re-ordering theterms, we get the final equation,

A

kx

N

x

T N

x

+ ky

N

y

T N

y

{T}tdA+2h A

[N]TN{T}dA

=A

[N]TtdA + 2hTa

[N]TdA

S

qsns[N]TdS

(7)

which is of the formk(e)

{T} =

f(e)Q

+f(e)h

+f(e)g

(8)

In our specific task, the Internal heat generated is zero and there is noflux surrounding the part. Hence the governing equation reduces to theform of

kA

Nx

TNx

+Ny

TNy

{T}tdA + 2h A

[N]TN{T}dA

= 2hTa

[N]TdA

(9)

Thus, equation (9) is the governing equation for the given task.Thefurther coding in fortran for the user element is done for this governingequation.

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4 Shape Functions

Shape functions are important factors in any analysis for appropriateinterpolation of field variables.Its imperative to find the right position-ing of the element in the co-ordinate system. In most of the cases, thebest way to do this is to write shape functions for the element in thenatural co-ordinates, when the co-ordinate varies from 1 to +1 andthe co-ordinate varies from 1 to +1.In the case of a 6-node triangular element, selecting the natural co-ordinate system would create complications during the coding for thesurface convetion forcing functions since on two edges of the element,both the co-ordinates and would vary and neither of them would

be constant.Hence a better way to write shape functions for the element would bein terms of the area co-ordinates, where the two-dimensional system isexpressed in terms of three co-ordinates , and which vary from 0to 1 with the condition

+ + = 1

(10)

Figure 3: Element expressed in Area Co-ordinates

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Shape functions are :

(a) N1=(2-1)

(b) N2=(2-1)

(c) N3=(2-1)

(d) N4=4

(e) N5=4

(f) N6=4

The gauss points for integration choses are

(a) (2/3,1/6,1/6)(b) (1/6,2/3,1/6)

(c) (1/6,1/6,2/3)

The weight for the integration points is 1/6

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5 Part Geometry

Figure 4: Part Geometry for Analysis

The part geometry to be analysed is as shown. A noticable factor hereis that the heat transfer coefficient of the surrounding fluid is varying attwo sections along the surface. Hence, the edge number of the elementon specific surfaces should be considered in order to program for thesurface convection factor. Additionally, the mesh must be in a way thatin no element, both the edges are exposed to surface convection. Allthese factors are taken into consideration and the geometry is meshed.

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6 Results

6.1 Temperature distribution across the element

Figure 5: Comparison of Abaqus and UEL temperature contour

The above figure establishes the fact that the UEL results when com-pared with the Abaqus results are pretty accurate. The red region in-dicates regions of high temperature and blue indicates low temperatureregions. The result is quite logical since the temperature at the upperedge is 10000 with the heat transfer coefficient of 10000Wm2K1 andat the lower edge, the surrounding temperature is 3000 with a heattransfer coefficient of 20Wm2K1.

6.2 Comparison of nodal temperatures

In order to have an idea of the accuracy of the results, the followingtable shows the values of temperatures at a specific node.The valuesare observed for nodes on different edges and surfaces.

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1- Comparison of Nodal temperature values

Node Number Abaqus UEL Accuracy in %348 991.682 999.915 0.83177 500.13 500 0.0229 408.701 404.998 0.9

331 920.821 913.406 0.871 316.962 302.382 4.5939 389.319 395.474 1.5

The above table shows that there is a small deviation in the tempera-ture values of Abaqus and UEL.

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7 Conclusion

The deviations in the results of abaqus and user element is in mostof the cases around 0.5% to 1.25%. In a few regions , i.e right upperregion of the geometry, the maximum deviation is around 5.02%. Thefactors influencing the accuracy in case of a six nodal quadratic elementmay be the use of a B bar element by abaqus, repeated integration atthe gauss points while calculating the temperatures developed due tosurface convection. The change of gauss points also has no dominantchange in the results obtained.A better accuracy was obtained at theinner nodes than at the outer nodes where the geometry was exposedto surface convection.

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Appendices

A Fortran code

C 6 POINTSSUBROUTINE UEL( RHS,AMATRX,SVARS,ENERGY,

NDOFEL,NRHS, NSVARS,1 PROPS, NPROPS,COORDS,MCRD,NNODE,U, DU, V, A,

JTYPE, TIME, DTIME,2 KSTEP, KINC, JELEM,PARAMS,NDLOAD,JDLTYP,

ADLMAG,PREDEF,3 NPREDF,LFLAGS,MLVARX,DDLMAG,MDLOAD,PNEWDT,JPROPS,NJPROP,

4 PERIOD)C

INCLUDE ABA PARAM. INC C

DIMENSION RHS(MLVARX, ) ,AMATRX(NDOFEL,NDOFEL),

1 SVARS( ) ,ENERGY( 7 ) ,PROPS( ) ,COORDS(MCRD,NNODE) ,

2 U(NDOFEL) ,DU(MLVARX, ) ,V(NDOFEL) ,A(NDOFEL) ,TIME( 2 ) ,

3 PARAMS( ) ,JDLTYP(MDLOAD, ) ,ADLMAG(MDLOAD, ) ,

4 DDLMAG(MDLOAD, ) ,PREDEF( 2 ,NPREDF,NNODE) ,LFLAGS(4) ,

5 JPROPS( )C

DIMENSION PHI ( 6 ) , PHIX( 6 ) , PHIY( 6 ) , PHIC( 6 ) ,PHIE( 6) ,

1 GPX( 3 ) ,GPY( 3 ) ,IFACE( 7 )C

PARAMETER (ZERO=0.D0,FOURTH=0.25D0,HALF=0.5D0,ONE=1.D0,TWO=2.D0)

PARAMETER (FOUR=4.D0,THREE=3. D0, SIX =6 .D0)

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DATA IFACE/ 1 ,4 , 2 ,5 , 3 ,6 , 1 /

CC MATERIAL PROPERTY DEFINITIONC

THICK = PROPS(1)COND = PROPS( 2 )

CC INITIALIZATION (NRHS=1)C

DO 6 K1=1,NDOFELRHS(K1 ,NRHS)=ZERODO 4 K2=1,NDOFEL

AMATRX(K2 , K1)=ZERO4 CONTINUE6 CONTINUE

CIF (LFLAGS( 3 ) . EQ. 4 ) RETURN

C SteadyS t a t e ANALYSISIF (LFLAGS(1) .EQ.31) THEN

CC DETERMINE GAUSS POINT LOCATIONSC

CALLGSPT(GPX,GPY)C

C ASSEMBLE AMATRX AND RHSC

DO 300 K=1,3C LOOP THROUGH GAUSS PTS

C=GPX(K)E=GPY(K)CALL DER(C , E ,GPX,GPY,GWEI, PHI , PHIX, PHIY,

PHIC, PHIE1 ,DXDC,DXDE,DYDC,DYDE,AJACOB,COORDS

,MCRD,NNODE)C

DTDX=ZERODTDY=ZEROT =ZERO

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C

DO I=1, 6DTDX=U( I ) PHIX( I )+DTDXDTDY=U( I ) PHIY( I )+DTDYT =U( I ) PHI( I )+T

END DO

CWE=(ONE/SIX ) (ONE/SIX) (ONE/SIX )GWE=WEAJACOB

CDO KI=1,6

C LOOP OVER NODES

RHS( KI ,NRHS) = RHS( KI ,NRHS) 1 GWE(COND(PHIX(KI) DTDX +

PHIY(KI) DTDY) )

DO KJ=1,6AMATRX(KI , KJ)= AMATRX(KI , KJ) + GWE

1 (COND(PHIX(KI) PHIX( KJ) +PHIY(KI) PHIY(KJ) ) )

END DO

END DO

300 CONTINUEC

IF (JDLTYP(1 ,1) .EQ.1) THENC ZETA=1CE=0C E=1CC

HCOEF123=PROPS( 3 )SINKT1=ADLMAG( 1 , 1 )

CDO KI=1,3

C LOOP THROUGH GAUSS PTSC=GPX( KI )E=1C

CC

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CALL DER(C , E ,GPX,GPY,GWEI, PHI , PHIX,

PHIY,PHIC,PHIE1 ,DXDC,DXDE,DYDC,DYDE,AJACOB,COORDS,MCRD,NNODE)

CT=ZERO

DO I=1, 6T=U( I ) PHI( I )+T

END DO

CDS=SQRT(DXDEDXDE + DYDEDYDE)

C

DO KJ=1,3C LOOP THROUGH NODES

RHS( IFACE( KJ) ,NRHS) = RHS( IFACE(KJ),NRHS)

1 THREEWEDSPHI(IFACE(KJ) ) HCOEF123(TSINKT1)

DO KK=1,6C LOOP THROUGH NODES

AMATRX(IFACE (KJ) ,KK)= AMATRX(IFACE (KJ) ,KK) +

1 THREEWEDSPHI(IFACE(KJ) ) HCOEF123PHI(KK)END DO

END DO

END DO

CEND IF

CC END IFC

IF (JDLTYP(1 ,1) .EQ.2) THEN

CC=0

CHCOEF123=PROPS( 3 )SINKT1=ADLMAG( 1 , 1 )

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C

DO KI=1,3C LOOP THROUGH GAUSS PTSE=GPY( KI )

CC

CALL DER(C , E ,GPX,GPY,GWEI, PHI , PHIX,PHIY,PHIC,PHIE

1 ,DXDC,DXDE,DYDC,DYDE,AJACOB,COORDS,MCRD,NNODE)

CT=ZERO

DO I=1, 6T=U( I ) PHI( I )+T

END DO

CDS=SQRT(DXDEDXDE + DYDEDYDE)

CDO KJ=3,5

C LOOP THROUGH NODESRHS( IFACE( KJ) ,NRHS) = RHS( IFACE(KJ)

,NRHS)

1 THREEWEDSPHI(IFACE(KJ) ) HCOEF123(TSINKT1)DO KK=1,6

C LOOP THROUGH NODESAMATRX(IFACE (KJ) ,KK)= AMATRX(IFACE (

KJ) ,KK) +1 THREEWEDSPHI(IFACE(KJ) )

HCOEF123PHI(KK)END DO

END DO

END DO

CEND IF

CC END IFC

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C

IF (JDLTYP(1 ,1) .EQ.3) THENCE=0

CHCOEF123=PROPS( 3 )SINKT1=ADLMAG( 1 , 1 )

CDO KI=1,3

C LOOP THROUGH GAUSS PTSC=GPX( KI )

C

CCALL DER(C , E ,GPX,GPY,GWEI, PHI , PHIX,

PHIY,PHIC,PHIE1 ,DXDC,DXDE,DYDC,DYDE,AJACOB,

COORDS,MCRD,NNODE)C

T=ZERODO I=1, 6

T=U( I ) PHI( I )+TEND DO

C DS=SQRT(DXDEDXDE + DYDEDYDE)C

DO KJ=5,7C LOOP THROUGH NODES

RHS( IFACE( KJ) ,NRHS) = RHS( IFACE(KJ),NRHS)

1 THREEWEDSPHI(IFACE(KJ) ) HCOEF123(TSINKT1)

DO KK=1,6C LOOP THROUGH NODES

AMATRX(IFACE (KJ) ,KK)= AMATRX(IFACE (KJ) ,KK) +

1 THREEWEDSPHI(IFACE(KJ) ) HCOEF123PHI(KK)

END DO

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END DO

END DOCEND IF

CEND IF

RETURN

END

CC

SUBROUTINE GSPT(GPX,GPY)INCLUDE aba param . inc

DIMENSION GPX( 3 ) ,GPY( 3 )C

PARAMETER(ONE=1.D0 ,TWO=2 .D0 ,THREE=3.D0 , SIX=6 .D0)

CC GPX: X COORDINATE OF GAUSS PTC GPY: Y COORDINATE OF GAUSS PTCC GPX(1 )=ONE/TWOC GPY(1 )=ONE/TWO

C GPX( 2 ) =ZEROC GPY(2 )=ONE/TWOC GPX(3 )=ONE/TWOC GPY( 3 ) =ZERO

GPX( 1 )=TWO/THREEGPY( 1 )=ONE/SIXGPX( 2 )=ONE/SIXGPY( 2 )=TWO/THREEGPX( 3 )=ONE/SIXGPY( 3 )=TWO/THREERETURN

ENDCC

SUBROUTINE DER(C , E ,GPX,GPY,GWEI, PHI , PHIX, PHIY,PHIC,PHIE,

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1 DXDC,DXDE,DYDC,DYDE,AJACOB,COORDS,MCRD,NNODE

)INCLUDE aba param . inc DIMENSION PHI ( 6 ) , PHIX( 6 ) , PHIY( 6 ) , PHIC( 6 ) ,PHIE

( 6) ,1 COORDS(MCRD,NNODE)

CPARAMETER(ZERO=0.D0 ,FOURTH=0 .25D0 ,HALF=0 .5 D0 ,

ONE=1.D0 ,TWO=2.D0 )PARAMETER(FOUR=4.D0 ,THREE=3 .D0 )

CC INTERPOLATION FUNCTIONS

CP H I( 1) = C(TWOCONE)P H I( 2) = E(TWOEONE)PHI( 3) = (ONECE) (TWO(ONECE)ONE)PHI ( 4 ) = FOURCEPHI ( 5 ) = FOUR(ONECE) EPHI ( 6 ) = FOURC(ONECE)

CC DERIVATIVES WRT TO CC

PHIC( 1 ) = FOURCONEPHIC ( 2 ) = ZEROPHIC( 3 ) = FOURC+FOURETHREEPHIC( 4 ) = FOUREPHIC( 5) = FOUREPHIC( 6 ) = FOUR(ONETWOCE)

CC DERIVATIVES WRT TO EC

PHIE( 1 ) = ZEROPHIE( 2 ) = FOUREONE

PHIE( 3 ) = FOURC+FOURETHREEPHIE( 4 ) = FOURCPHIE( 5 ) = FOUR(ONECTWOE)PHIE( 6) = FOURC

C

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DXDC=ZERO

DXDE=ZERODYDC=ZERODYDE=ZERO

CDO 3 I = 1,6

DXDC=DXDC+COORDS( 1 , I ) PHIC( I )DXDE=DXDE+COORDS( 1 , I ) PHIE( I )DYDC=DYDC+COORDS( 2 , I ) PHIC( I )DYDE=DYDE+COORDS( 2 , I ) PHIE( I )

3 CONTINUEC

C CALCULATION OF JACOBIANC

AJACOB=(DXDCDYDEDXDEDYDC)CC DERIVATIVES WRT TO X AND YC

DO 5 I = 1,6PHIX( I ) =(PHIC( I )DYDEPHIE( I ) DYDC) /AJACOBPHIY( I ) =(PHIE( I ) DXDCPHIC( I ) DXDE) /AJACOB

5 CONTINUERETURN

END

B Input File

HeadingPre pri nt , echo=NO, model=NO, h is to ry=NO, co nt ac t=

NO

Node1 , 1 7 . 5 , 101.6999972 , 1 6 . 5 , 101.6999973 , 1 6 . 5 , 129.4 , 1 7 . 5 , 130.

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5 , 0 . , 101.699997

6 , 0 . , 129.7 , 0 . , 130.8 , 1 7 . 5 , 100.9 , 1 6 . 5 , 100.699997

10 , 1 . , 85.1999969. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 87 6 , 0 .5 55 60 85 71 , 8 9. 54 84 16 1

1877 , 0 . , 8 8 . 69 9 99 6 91878 , 2 . 2 0 7 7 0 3 5 9 , 97 . 0 39 9 7 8

1 87 9 , 0 .7 05 30 79 61 , 9 6. 34 90 44 81880 , 1 .7 26 14 40 8 , 96 .33 718 871881 , 5 .1 17 66 76 8 , 98 .24 401 861882 , 5 .7 83 60 79 6 , 98 .95 977 781883 , 4 .8 15 17 26 7 , 99 .19 019 321884 , 3 .9 06 20 13 6 , 96 .98 712 921885 , 4 .4 78 42 21 6 , 97 .68 341 061886 , 3 .8 38 31 21 5 , 99 .29 848 481887 , 0 . , 9 6 . 69 9 99 6 9

USER ELEMENT,NODES=6,TYPE=U1,PROP=4,COORDINATES=2,

VAR=111 ,Element , type=U1 , ELSET=UEL

1 , 1 , 2 , 27 , 536 , 537 , 5382 , 2 7 , 46 , 1 , 539 , 540 , 5383 , 4 6 , 27 , 2 8 , 539 , 541 , 5424 , 2 8 , 45 , 4 6 , 543 , 544 , 5425 , 4 5 , 28 , 2 9 , 543 , 545 , 5466 , 2 9 , 44 , 4 5 , 547 , 548 , 546. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .815 , 534 , 396 , 390 , 1834 , 1872 , 1870

816 , 534 , 389 , 527 , 18 71 , 1830 , 1868817 , 535 , 394 , 395 , 18 37 , 1887 , 1879818 , 535 , 527 , 530 , 18 80 , 1852 , 1878

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ELEMENT,TYPE=DC2D6 , ELSET=ALLE

1000 , 1 , 2 , 27 , 53 6 ,537 , 5382000 , 27 , 46 , 1 , 53 9 ,

540 , 5383000 , 46 , 27 , 2 8 , 5 39 ,

541 , 5424000 , 28 , 45 , 4 6 , 5 43 ,

544 , 5425000 , 45 , 28 , 2 9 , 5 43 ,

545 , 5466000 , 29 , 44 , 4 5 , 5 47 ,

548 , 5467000 , 44 , 29 , 3 0 , 5 47 ,

549 , 5508000 , 30 , 43 , 4 4 , 5 51 ,

552 , 5509000 , 43 , 30 , 3 1 , 5 51 ,

553 , 55410000 , 31 , 42 , 43 , 55 5 ,

556 , 55411000 , 42 , 31 , 32 , 55 5 ,

557 , 558. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

814000 , 533 , 528 , 532 , 1865 ,1860 , 1886

815000 , 534 , 396 , 390 , 1834 ,1872 , 1870

816000 , 534 , 389 , 527 , 1871 ,1830 , 1868

817000 , 535 , 394 , 395 , 1837 ,

1887 , 1879818000 , 535 , 527 , 530 , 1880 ,

1852 , 1878

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MATERIAL,NAME=MAT1

CONDUCTIVITY0. 000000 0000000 01 ,SOLID SECTION, ELSET=ALLE,MATERIAL=MAT1UEL PROPERTY,ELSET=UEL1 . , 5 0 . , 1 0 0 0 . , 2 0E l s e t , e l s e t=BOTTOMEDGE2, g e n e r a t e177 , 189 , 2 Sur f ac e , type=ELEMENT, name=BOTTOMEDGEBOTTOMEDGE2, S2E l s e t , e l s e t=RIGHTLOWEREDGE1, g e n e r a t e

2 , 22 , 2

Sur f ac e , type=ELEMENT, name=RIGHTLOWEREDGERIGHTLOWEREDGE1, S1E l s e t , e l s e t=MIDDLEEDGE1

1 9 2 , 2 3 6 , 2 3 8 , 2 4 0 , 2 4 2 , 2 4 4 , 2 4 6 , 2 4 8 , 2 5 0 , 2 5 2 ,2 5 4 , 2 5 6 , 2 5 8 , 2 5 9 , 2 6 1 , 2 6 3 , 2 6 5

2 67 , 2 69 , 2 71 , 2 73 , 2 75 , 27 8 , 2 80 , 2 82 , 2 84 , 2 86 ,2 8 8 , 2 9 0 , 2 9 2 , 2 9 4 , 2 9 6 , 2 9 8 , 3 0 0 , 3 0 2 , 3 0 4 ,3 06 , 3 08

3 1 0 , 3 1 2 , 3 1 4 , 3 1 6 , 3 1 8 , 3 2 0 , 3 2 2 , 3 2 4 , 3 2 6 , 3 2 8 ,330 , 33 2 , 334 , 336 , 338 , 340

3 4 2 , 3 4 4 , 3 4 6 , 3 4 8 , 3 5 0 , 3 5 2 , 3 5 4 , 3 5 6 , 3 5 8 , 3 6 0 ,362 , 36 4 , 366 , 368 , 370 , 3723 7 4 , 3 7 6 , 3 7 8 , 3 8 0 , 3 8 2 , 3 8 4 , 3 8 6 , 3 8 8 , 3 9 0 , 3 9 2 ,

394 , 39 6 , 398 , 400 , 402 , 4044 0 6 , 4 0 8 , 7 5 2 , 7 5 4 , 7 5 6 , 7 5 8 , 7 6 0 , 7 6 2 , 7 6 4 , 7 6 6 ,

7 68 , 7 70 , 7 72 , 774 Sur f ac e , type=ELEMENT, name=MIDDLEEDGEMIDDLEEDGE1, S1El se t , e l s e t=UPPEREDGE3, ge ne ra te

581 , 593 , 2 Sur f ac e , type=ELEMENT, name=UPPEREDGE

UPPEREDGE3, S3E l s e t , e l s e t=SYMMETRICEDGE1

5 94 , 8 00E l s e t , e l s e t=SYMMETRICEDGE2

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35 , 4 9 , 63 , 77 , 91 , 105 , 119 , 13 3 , 147 , 161 ,

17 5 , 190 , 193 , 195 , 201 , 2032 0 6 , 2 2 9 , 4 0 9 , 4 1 1 , 4 1 3 , 4 1 5 , 4 1 7 , 4 1 9 , 4 2 1 , 4 2 3 ,425 , 42 7 , 429 , 431 , 433 , 435

4 3 7 , 4 3 9 , 4 4 1 , 4 4 3 , 4 4 5 , 4 4 7 , 4 4 9 , 4 5 1 , 4 5 3 , 4 5 5 ,457 , 45 9 , 461 , 463 , 465 , 467

4 6 9 , 4 7 1 , 4 7 3 , 4 7 5 , 4 7 7 , 4 7 9 , 4 8 1 , 4 8 3 , 4 8 5 , 4 8 7 ,489 , 49 1 , 493 , 495 , 497 , 499

5 0 1 , 5 0 3 , 5 0 5 , 5 0 7 , 5 0 9 , 5 1 1 , 5 1 3 , 5 1 5 , 5 1 7 , 5 1 9 ,521 , 52 3 , 525 , 527 , 529 , 531

5 3 3 , 5 3 5 , 5 3 7 , 5 3 9 , 5 4 1 , 5 4 3 , 5 4 5 , 5 4 7 , 5 4 9 , 5 5 1 ,553 , 55 5 , 557 , 596 , 610 , 624

6 3 8 , 6 5 2 , 6 6 6 , 6 8 0 , 6 9 4 , 7 0 8 , 7 2 2 , 7 3 6 , 7 7 7 , 7 8 6 ,817

E l s e t , e l s e t=SYMMETRICEDGE31 96 , 7 94 , 7 97 , 798

Sur f ac e , type=ELEMENT, name=SYMMETRICEDGESYMMETRICEDGE1, S1SYMMETRICEDGE2, S2SYMMETRICEDGE3, S3El se t , e l s e t=TOPRIGHTEDGE1, g en er a te

560 , 580 , 2

E l s e t , e l s e t=TOPRIGHTEDGE27 5 0 , Sur f ac e , type=ELEMENT, name=TOPRIGHTEDGETOPRIGHTEDGE1, S1TOPRIGHTEDGE2, S2NSET , NSET=UPPERNODES22 , 1544 , 348 , 1540 , 347 , 1536 , 346 , 1532 , 345 ,

1 52 8 , 3 44 , 1 52 4 , 3 43 , 1 52 0 , 20 S e c t i o n : S e c ti o n1 S o l i d S e c t i o n , e l s e t =ALLE , m a t e r i a l =M a t e r i a l1

, MATERIALS M at e r i al , name=Material 1

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C o n d u c t i v i t y

50 , De nsi t y7830 INTERACTION PROPERTIESF il m P r oper t y , name=IntProp11 0 0 .F il m P r oper t y , name=IntProp22 0 .

STEP: Step1Step , name=Step 1, i nc =10000Hea t T r a ns f er , s t e ad y s t a t e0 . 0 05 , 1 . , 1 e05 , 0 . 0 0 5 , BOUNDARY CONDITIONS INTERACTIONS

I n t e r a c t i o n : I nt1BOUNDARYUPPERNODES, 1 1 , 1 1 , 9 9 9 . 9 1 5DLOADUPPEREDGE3, U3, 1 0 0 0 . ,MIDDLEEDGE1, U1, 5 0 0 .RIGHTLOWEREDGE1, U2 , 4 0 0 .BOTTOMEDGE2, U2 , 3 0 0 . OUTPUT REQUESTS

R e st ar t , write , f r e q u e n c y =0 FIELD OUTPUT: FOutput1

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Output , f i e l d , v a r i a b l e=PRESELECT

Output , h i s t o r y , f r e q u e n c y =0End Step

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Heat transfer analysis of a 6- Nodal triangular element under steady state

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