HEAT TRANSFER FINITE ELEMENT FORMULATION VIJAYAVITHAL BONGALE DEPARTMENT OF MECHANICAL ENGINEERING MALNAD COLLEGE OF ENGINEERING HASSAN - 573 202. Mobile.

HEAT TRANSFER

FINITE ELEMENT FORMULATION

VIJAYAVITHAL BONGALEDEPARTMENT OF MECHANICAL ENGINEERING

MALNAD COLLEGE OF ENGINEERINGHASSAN - 573 202.Mobile : 9448821954

N I T CALICUT 17/08/2014


What is Finite Element Method?

FEM is a numerical analysis technique for obtaining approximate solutions to a wide variety of engineering problems.


Applications of FEM:

1. Equilibrium problems or time independent problems. e.g. i) To find displacement distribution and stress distribution for a mechanical or thermal loading in solid mechanics. ii) To find pressure, velocity, temperature, and density distributions of equilibrium problems in fluid mechanics.

2. Eigenvalue problems of solid and fluid mechanics. e.g. i) Determination of natural frequencies and

modes of vibration of solids and fluids. ii) Stability of structures and the stability of laminar flows.

3.Time-dependent or propagation problems of continuum mechanics.

e.g. This category is composed of the problems that results when the time dimension is added to the problems of the first two categories.


Similarities that exists between various types of engineering problems:

1. Solid Bar under Axial Load

areasectionalcrossisAand

nt,displacemeaxialisu

modulus,sYoung'theisE,Where

0,x

uAE

x


2. One – dimensional Heat Transfer

areasectionalcrossisAande,temperaturisT

ty,conductivithermaltheisK,Where

equationLaplace0,x

TKA

x

3. One dimensional fluid flow

x

ΦuandareasectionalcrossisAand

,functionpotentialisφ,densitytheisρ

,Where0,x

ΦρA

x


How the Finite Element Method WorksDiscretize the continuum: Divide the

continuum or solution region into elements.Select interpolation functions: Assign nodes

to each element and then choose the interpolation function to represent the variation of the field variable over the element.

Find the Element Properties: Determine the matrix equations expressing the properties of the individual elements. For this one of the three approaches can be used. i) The direct approach ii) The variational approach or iii) the weighted residuals approach.


Assemble the Element Properties to Obtain the System equations: Combine the matrix equations expressing the behavior of the elements and form the matrix equations expressing the behavior of the entire system.

Impose the Boundary Conditions: Before the system equations are ready for solution they must be modified to account for the boundary conditions of the problem.

Solve the System Equations: Solve the System Equations to obtain the unknown nodal values like displacement, temperature etc.

Make Additional Computations If Desired: From displacements calculate element strains and stresses, from temperatures calculate heat fluxes if required.



General Steps to be followed while solving a problem on heat transfer by using FEM:

1. Discretize and select the element type2. Choose a temperature function3. Define the temperature gradient /

temperature and heat flux/ temperature gradient relationships

4. Derive the element conduction matrix and equations by using either variational approach or by using Galerkin’s approach

5. Assemble the element equations to obtain the global equations and introduce boundary conditions

6. Solve for the nodal temperatures7. Solve for the element temperature

gradients and heat fluxesN I T CALICUT 17/08/2014

Step 1.Select element type.

t1 t2

1

x

One –D element

L

2

Step 2.Choose a temperature function.

t1

t2

T = N1t1+ N2 t2

1 2

Temperature Variation along the length of element

ONE DIMENSIONAL FINITE ELEMENT FORMULATION USING VARIATIONAL

APPROACH


We choose, T (x) = N1 t1 + N2 t2 --------------------------- (1)

N1 & N2 are shape functions given by,

)2(L

xN ,

L

x1N 21

In matrix form

)3(t and L

x

L

x1N

2

1

t

t

and

)4( tNT


Step 3. Define the temperature gradient / temperature and heat flux/

temperature gradient relationships

Temperature gradient matrix is given by,

(5)tBdx

dTg

Where matrix B is,

(6)L

1

L

1

dx

dN

dx

dNB 21

The heat flux /temperature gradient relationship is,

(7)gDq x

(5)tBdx

dTg

Where matrix B is,

(6)L

1

L

1

dx

dN

dx

dNB 21

The heat flux /temperature gradient relationship is,

(7)gDq x


The material property matrix is,

)8( xxKD

Step 4. Derive the element conduction matrix and equations

Consider the following equations

(I)0Qdx

TdK

2

2

xx

L

T=TB

S1

q x = + q*x

S2

q x = 0

L

T=TB

S1

Insulated

With T =TB on surface S1,

2xx*x SonConstant

dx

dTKq N I T CALICUT 17/08/2014

Where, Q is heat generated / unit volume*xq is the heat flow / unit area

is positive when heat is flowing into body is negative when heat is flowing out of the bodyis Zero on an insulated boundary

(II)TTA

hP

t

TCρQ

x

TK

x xx

Insulated

h

T2 T

Consider

With the first boundary condition of above equation and /or second boundary condition and /or loss of heat by convection from the ends of 1-D body, we have

3xx SsurfaceonTThdx

dTK N I T CALICUT 17/08/2014

Minimize the following functional :

(Analogous to the potential energy functional Π)

IIIΩΩΩUΠ hqQh

Where,

dVdx

dTK

2

1U

v

2

xx

V

Q dVTQΩ 2S

* dSTqΩq

)9(2

1Ω

2

h

3

dSTThS N I T CALICUT 17/08/2014

Important:

q* and h on the same surface cannot be specified simultaneously because they cannot occur on the same surface

)10(dSTTh2

1dSTqdVTQdV

dx

dTK

2

1

ΩΩΩUΠ

2

SS

*

Vv

2

xx

hqQh

32

We now have the functional given by

Consider the first term,

)11(2

1dV

dx

dTK

2

1

v

2

xx

dVgDg

V

T


Second term gives,

(12)-------dVTQV

dVQNtV

TT

Third term gives,

(13)dSqNtdSTq *T

S

T

S

*

22

Similarily, fourth term gives,

)14(2

1

2

1 22

33

dSTNthdSTThS

TT

S

Substituting equations (11),(12),(13) and (14) in equation (10) we obtainN I T CALICUT 17/08/2014

(15)dSTTtNNttNNth2

1

dS*qNtdVQNttdVBDBt2

1

dSTNth2

1dSqNt

dVgDg2

1Π

3

2

32

S

2TTTT

T

S

TT

V

TT

V

T

2

S

TT*T

S

T

V

T

h

dVQNtV

TT

Equation (15) has to be minimized with respect to and equated to zero

t


)16(0dShTNtdSNNh

dS*qNdVQNtdVBDBt

Π

3 3

2

S S

TT

T

S

T

V

T

V

h

On simplifying,

(17)ffftdSNNhdVBDB hqQS

TT

V 3

(18)tKf

The above equation is of the form,


)19(dSNNhdVBDB3S

TT

V

hC KKK

Where,

Element Conduction matrix is

The first term is conduction part of K and

second term represent convection part of K

And the force matrices have been defined by,

3

2

S

T

h

T

Sq

T

VQ

dShTNf

,dS*qNf

,dVQNf


Consider the conduction part,

(20)-------dVBDBT

VCK

Substituting for B, D and dV in the above equation,

)21(11

11

L

KAKi.e. ,dx

11

11

L

KA

dxAL

1

L

1K

L

1L

1

xxC

L

02

xx

xx

L

0


The convection part is,

dxPdSWhere

(22)21

12

6

hPLdx

L

x

L

x1

L

xL

x1

hPdSNNhKL

0S

T

h

3

Therefore, element conduction matrix is,

)23(21

12

6

hPL

11

11

L

KA xx

K


The force matrix terms will be,

1

1

2

PLhTdShTNf

1

1

2

PL*qdx

L

xL

x1

P*qdS*qNf

,1

1

2

QALdx

L

xL

x1

QAdVQNf

3

2

S

T

h

L

0

T

Sq

L

0

T

VQ


By adding we get,

)24(1

1

2

PLhTPL*qQAL

1

1

2

PLhT

1

1

2

PL*q

1

1

2

QALffff hqQ

Convection force from the end of the element

1 2

h

T00


We have an additional convection term contribution to the stiffness matrix and is,

(25)dSNNhKendS

T

endh

N1 = 0 and N2 = 1 at right end and

(26)10

00hAdS10

1

0hK

endSendh

The convection force from the free end

)26(1

0AhT

)(

)(AhTf

2

1

h

LxN

LxNend


Step 5. Assemble the element equations to obtain the global equations and introduce boundary conditions.

The global structure conduction matrix is

n

e

eKK1

The global force matrix is

n

1e

efF

and global equations are

tKF N I T CALICUT 17/08/2014

Step 6. Solve for the nodal temperatures

Step 7. Solve for the element temperature gradients and heat fluxes.


ONE DIMENSIONAL FINITE ELEMENT FORMULATION

GALERKIN’S APPROACH

Equation representing one dimensional formulation of conduction with convection is given by,

(1)TTA

hPQ

dx

Tdk

2

2

x

Element – 1-D linear 2 noded element with temperature function

T (x) = N1 t1 + N2 t2 --------------------------- (2)

L

xN ,

L

x1N 21

Where,


The residual equations for the equation (I) are

(3)0dxANTTA

hPQ

dx

Tdk i

x

x2

2

x

2

1

Where i = 1,2

(4)0dxNThPdxNQAdxNThPdxNdx

TdkA

2

1

2

1

2

1

2

1

x

x

x

xi

x

xi

x

xii2

2

x

Integrating the first term of equation (4) by parts and rearranging,

)5(dx

dTANKdxNThPdxNQAdxNThPdx

dx

dT

dx

dNAk

2

1

2

1

2

2

2

1

2

1

x

x

x

x

x

x

ixi

x

xii

x

x

ix


Integration by parts:

dx

dTKAvanddx

dx

dTKA

dx

ddv

,dxdx

dNdu,NuWhere

vduuvudv

xxxx

ii


Substituting for T , we get

)6(dx

dTANKdxNThPdxNQA

dxtNtNNhPdxtdx

dNt

dx

dN

dx

dNAk

2

1

2

2

2

1

2

1

2

1

x

x

x

x

ixi

x

xi

x

x2211i2

21

1

x

x

ix

Or

(7)dx

dTNAKdxNThPdxNQA

dxtNNhPdxtdx

dN

dx

dNAk

2

1

2

2

2

1

2

1

2

1

x

x

x

x

T

x

Tx

x

T

x

x

T

Tx

xx


(8)ffftk e

h

e

q

e

Q

e

The above equations are of the form,

)9(kk

dxNNhPdxdx

dN

dx

dNAkk

e

h

e

c

x

x

T

Tx

xx

e2

1

2

1

And

Let x1= 0 and x2 = L, then, N I T CALICUT 17/08/2014

If one substitute for N1 & N2 in equation (9) and solve, then,

(10)kk

21

12

6

hPL

11

11

L

kAk

e

h

e

c

xe

The forcing function vectors on the right hand side of the equation (7) are given by


1

1

2

QALAf

02

01

e

Q L

L

dxQN

dxQN

2

1

Lx

0x

L

0

x

e

q q-

qA

q-

qA

dx

dT

dx

dT

Akf

)11(1

1

2

PLhTPhTf

02

01

h

L

L

dxN

dxN


Two Dimensional Finite Element Formulation:

1-d elements are lines2-d elements are either triangles,

quadrilaterals, or a mixture as shown

Label the nodes so that the difference between two nodes on any element is minimized.


Three noded triangular element:

Assume (Choose) a Temperature Function:

etemperatur nodalt

t

t

t

NNNT

tNtNtNT

m

j

i

mji

mmjjii


Define Temperature Gradient Relationships

m

j

i

mji

mji

t

t

t

y

N

y

N

y

Nx

N

x

N

x

N

y

Tx

T

g

ijmjimjijim

mijimjimimj

jmimjimjmji

mmmm

jjjj

iiii

xxγ,yyβ,xyyxα

xxγ,yyβ,xyyxα

xxγ,yyβ,xyyxαand

yγxβα2A

1N

yγxβα2A

1N

yγxβα2A

1NWhere,

Analogous to strain matrix: {g}=[B]{t}


yy

xx

yy

xx

y

x

K0

0 KDWhere

gDgK0

0 K

q

q

mji

mji

A2

1N

xB

[B] is derivative of [N]:

Heat flux/ temperature gradient relationship is:


Derive the element conduction matrix and equations:

hS C

TT

VKKdSNNhdVBDBK

3

Where,

dVK0

0K

A2

1dVBDBK

mji

mji

yy

xx

mm

jj

ii

V 2

T

VC

If thickness is assumed constant and all terms of integrand constant, then the conduction portion of the total stiffness matrix is,

BDBtAdVBDBKTT

VC


Now the convection portion of the total stiffness matrix is,

dS

NNNNNN

NNNNNN

NNNNNN

h

dSNNhK

3

3

S

mmjmim

mjjjij

mijiii

S

T

h

Consider the side between nodes i and j of the element subjected to convection, then, Nm = 0 along side i-j


000

021

012

6

tLhK ji

h

We Obtain

Where Li-j is the length of side i-j

Force Matrices:

1

1

1

3

QV

dVNQdVQNfT

V

T

VQ


imsideon

1

1

0

2

tL*qmjsideon

1

1

0

2

tL*q

jisideon

0

1

1

2

tL*qdS

N

N

N

*q

dSN*qf

i-mm-j

j-i

m

j

i

S

T

Sq

2

2

The integral

dSNThfT

Sh3

Can be found in a same manner by simply replacing

T h with*qN I T CALICUT 17/08/2014

imsideon

1

1

0

2

tLTh

mjsideon

1

1

0

2

tLTh

jisideon

0

1

1

2

tLThdS

N

N

N

Th

dSNThf

i-m

m-j

j-i

m

j

i

S

T

Sh

2

3


Thank You


HEAT TRANSFER FINITE ELEMENT FORMULATION VIJAYAVITHAL BONGALE DEPARTMENT OF MECHANICAL ENGINEERING MALNAD COLLEGE OF ENGINEERING HASSAN - 573 202. Mobile.

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