FDM_lecturenotes-4

8/22/2019 FDM_lecturenotes-4

1/34

Finite Difference Method

MEL 807

Computational Heat Transfer (2-0-4)

Dr. Prabal TalukdarAssistant Professor

Department of Mechanical Engineering

IIT Delhi


2/34

Discretization Methods Required to convert the general transport

equation to set of algebraic equations Finite difference method

Finite volume method

Finite element method


3/34

Introduction to Finite Difference

+++

+

+=+

!n)x(

xf

2)x(

xfx

xf)x(f)xx(f

n

n

n2

2

2

Taylor series expansion

gradient curvature


4/34

Discretization

+

+

+= + 6

)x(

x2

)(

xxx

3

j,i3

32

j,i2

2

j,ij,ij,1i

x


5/34

Representation of a Derivative

)x(xx

j,ij,1i

j,i

+

=

+

O

=

+

6

)x(

x2

)x(

xxx

2

j,i

3

3

j,i

2

2j,ij,1i

j,i

Finite difference

representation

+

+

+= +

6

)x(

x2

)x(

x

x

x

3

j,i

3

32

j,i

2

2

j,i

j,ij,1i

Forward difference


6/34

Backward Difference

+

+

+=

6

)x(

x2

)x(

x

)x(

x

3

j,i

3

32

j,i

2

2

j,i

j,ij,1i

+

=

6

)x(

x2

)x(

x

x

x

3

j,i

3

32

j,i

2

2

j,i

j,ij,1i

)x(xx

j,1ij,i

j,i+

=

OBackward difference


7/34

Central difference

+

+

+= +

6

)x(

x2

)x(

x

x

x

3

j,i

3

32

j,i

2

2

j,i

j,ij,1i

+

= 6

)x(

x2

)x(

xx

x

3

j,i

3

32

j,i

2

2

j,i

j,ij,1i

2j,1ij,1i

j,i

)x(x2x

+

=

+

OSubtracting,

Central difference


8/34

Forward, backward and central

difference stencil

)x(Oxx

j,ij,1i

j,i

+

=

+

)x(xx

j,1ij,i

j,i

+

=

O

2j,1ij,1i

j,i

)x(x2x

+

=

+

O


9/34

Stencil in y direction

)y(yy

j,i1j,i

j,i

+

=

+ O

)y(yy

1j,ij,i

j,i

+

=

O

2j,1ij,1i

j,i

)y(y2y

+

=

+

O


10/34

2nd

Order and Mixed Derivative

))x()x(

2

x2

2j,1ij,ij,1i

j,i

2

2

+

+=

+ 0

))y()y(

2

y

2

2

1j,ij,i1j,i

j,i

2

2

+

+=

+

0

])y(,)x[(

yx4yx

22

1j,1i1j,1i1j,1i1j,1i

j,i

2

+

+=

++++

O


11/34

Boundary Consideration What kind of differencing scheme is possible at the

boundaries?

Central difference approximationis not possible as point 2 is

beneath the boundary

How we can get a second orderaccurate scheme?

Possibility: Polynomial approach

)y(yy

12

1

+

=

O


12/34

Polynomial Approach Assume

At grid point 1, 1 = a

At grid point 2 where y = y,

At grid point 3 where y = 2y,

2cybya ++=

2

2 )y(cyba ++=

2

2 )y2(cy2ba ++=


13/34

Polynomial Approach (contd) Solving these equations,

Differentiation of

gives

At point 1 (boundary),

What is the order of

approximation??

y2

43b 321

+=

by

cy2b

y

cybya

1

2

=

+=

++=

y2

43

y

321

1

+=


14/34

Order of Approximation Taylor series gives,

Comparing with the polynomial expression ,

we can say that our polynomial is of O(

y)

3

1, 2, 3 can all be expressed in terms of the polynomial

Represents one-sided difference of 2nd order accuracy

+

+

+

+=

6y

y2y

yy

y)y(

3

1

3

32

1

2

2

1

1

23

321

1

)y(

y

)y(

y2

43

y

=

=

+=

O

O

2cybya ++=


15/34

FDM for 1D diffusion Uses truncated Taylor series expansion to approximate the

derivative of the DE

Consider 1-D diffusion equation

Expand in Taylor series about

point 2

0Sdx

d2

2

=+

Subtracting these equations yields


16/34

FDM (contd)

))x()x(

2

dx

d 22

231

2

2

2

+

+=

0

2

231

2

2

2

)x(

2

dx

d

+=

Second ordertruncation error

Dropping the truncated terms

Adding the equations

2321222S

)x()x()x(

2+

+

=

The final discretized equation

S2 = S(2)


17/34

FDM (contd)

We can write one such equation for each gridpoint

Boundary conditions gives us boundary value

Second order accurate Need to find a way to solve the couple

algebraic equation set

2321222 S)x()x()x(

2

+

+

=


18/34

1D Steady State Conduction Consider the steady state heat conduction in a slab of thickness L,

in which energy is generated at a constant rate of S W/m3. The

boundary surface at x = 0 is maintained at a constant temperatureTo, while the boundary surface at x = 0 dissipates heat by

convection with a heat transfer coefficient h into an ambient at

temperature T .

Compute the temperature inside the slab for h = 200W/(m2.C), k = 18 W/(m.C), L = 0.01 m, T = 100C, To = 50C, andS = 7.2 x 107.

To = 50 C

0.01 m

T = 100 C

x

x

0 1 32 54


19/34

Solution

0.01mLat xhThT

dx

)x(dTk

0at xT)x(T

0Sdx

Tdk

o

2

2

===+

==

=+

B.C.

16Sk

)5/L(S

k

)x(TT2T

ST)x(

kT

)x(

k2T

)x(

k

ST)x(

kT

)x(

kT

)x(

k2

i

2

i

2

1ii1i

i1i2i21i2

i1i21i2i2

==

=+

=+

+

+

=

+

+

+


20/34

Treatment of Boundary Condition

16TT2T 1ii1i =+ +

To = 50 C

0.01 m

x

x

01 32 N=54

Applicable to node 1-4

x/2

044.416T044.2T2

0Tk

xh2

k

S)x(T

k

xh22T2

02

xS)TT(h

x

TTk

54

N

2

N1N

N1NN

=++

=

+

+

+

=

+

T

cond conv

Applying it to

node 5


21/34

Treatment of Boundary Condition

(Another Way)

16TT2T 654 =+

456

546

T)TT(k

xh2T

0)TT(hx2

TTk

+

=

=

16TT2T 1ii1i =+ +

To = 50 C

0.01 m

x

x

01 32 N=54

Apply to node 5

T

B.C. gives

44.416T044.2T2

Tk

xh216T

k

xh22T2

54

54

+=

+=

+


22/34

Algebraic Equations16TT2T 210 =+

16TT2T 321 =+

16TT2T 432 =+16TT2T 543 =+

44.416T044.2T2 54 =

5016TT2 21 =+

Can be solved by Thomas algorithm

Matrix inversion as shown in next slide


23/34

Matrix Form

=

44.20

16

1616

66

T

T

TT

T

044.22000

12100

0121000121

00012

5

4

3

2

1

[ ][ ] [ ][ ] [ ] [ ]BAx

BxA1=

=


24/34

Prescribed Heat Flux

X=0

x

0 1 32 NN-1

x/2

X=L

qoqN

Energy balance gives,

0xS2

1

x

TTkq

0xS2

1

x

TTkq

NN1N

N

001

0

=++

=+

+


25/34

FDM representation

Ni0k

xq2

k

S)x(T2T2

0i0kxq2

kS)x(T2T2

NN

2

N1N

00

2

01

==

+

+

==++

for

for

Ni0k

S)x(

T2T2

0i0k

S)x(T2T2

N

2

N1N

0

2

01

==

+

==

+

for

for

For insulated or symmetry boundary,

Flux boundary,


26/34

Unsteady Heat Conduction with

FDM Unsteady heat conduction in 1-

D with constant thermal

conductivity

Expand the individual

terms with Taylor series,

2

2

x

T

t

T

=

+

=

+

2

t

t

T

t

TT

t

Tn

i

2

2n

i

1n

i

n

i

+

+=

+12

)x(xT

)x(TT2T

xT

2n

i

4

4

2

n

1i

n

i

n

1i

n

i

2

2


27/34

Unsteady Heat Conduction (contd)

2

n

1i

n

i

n

1i

n

i

1n

i

)x(

)TT2T(

t

TT

+=

++


28/34

Explicit Solutions

)TT2T()x(

tTT

)x(

)TT2T(

t

TT

n1i

ni

n1i2

ni

1ni

2

n1i

ni

n1i

ni

1ni

++

++

+

+=

+=

cm105

50x == s/cm10x17 22

=

Tw = 30CTw = 100C

50 cm

Find 1-D unsteady temperature distribution till steady state

Initial temp Tin = 30C, t = 10 sec

2

1

)x(

t2

will talk later


29/34

Results

0

20

40

60

80

100

120

0 1 2 3 4 5 6 7 8

20sec

90sec

260sec

360

T(C)

Nodes


30/34

2D Steady State Heat Conduction

0k

)y,x(S

y

T

x

T2

2

2

2

=+

+

2

j,1ij,ij,1i

j,i

2

2

)x(TT2T

xT

+=

+

2D steady state with heat generation

21j,ij,i1j,i

j,i

2

2

)y(TT2T

yT

+=

+

lyx

0klST4TTTT

2

j,ij,i1j,i1j,ij,1ij,1i

==

=++++ ++

where


31/34

Flux Boundary Condition

Ti,,j+1

Ti,,j-1

Ti+1,,j

Ti,,j

l

qoW/m2

l/2

Nodes (i,j) on a prescribed heat flux boundary

0k

lq2

k

SlT4TT2T

0Sl2

1

l

TT

2

lk

l

TTkl

l

TT

2

lklq

oj,i2

j,i1j,ij,1i1j,i

j,i

2j,i1j,ij,ij,1ij,i1j,i

0

=++++

=+

+

+

+

++

++

After rearrangement

l d C d


32/34

Flux Boundary Condition

(another way)

0k

SlT4TTTT

j,i2

j,i1j,i1j,ij,1ij,1i =++++ ++

j,1io

j,1i

j,1ij,1i

o

Tklq2T

l2

TTk

x

Tkq

+

+

+=

=

=

Ti,,j+1

Ti,,j-1

Ti+1,,j

Ti,,j

l

qoW/m2

l/2

Applying the finite difference equation at the

boundary node (i, j)

B.C.

0

k

lq2

k

SlT4TT2T o

j,i2

j,i1j,ij,1i1j,i =++++ ++


33/34

Convective Boundary Condition

0Sl

4

3)TT(hl

l

TT

2

lk

l

TTkl

l

TTkl

l

TT

2

lk j,i

2j,i

j,ij,1ij,i1j,ij,ij,1ij,i1j,i =++

+

+

+

++

After rearrangement,

0T

k

hl2S

k

l

2

3T

k

hl26TT2T2T j,i

2

j,ij,1i1j,ij,1i1j,i =++

++++ ++

Ti,,j+1

Ti,,j

l

l/2

Ti-1,,jTi+1,j

Convection h,T

Ti,,j-1

l

Energy balance gives,


34/34

Insulated Boundary

xb6

cos100)x(T

=

0 3b

02b

T1

T2

T5T3

T4 T6

100 86.66 50

Insulated

Maintained at

zero temperature

Insulated

Node 1: 2T2+2T3-4T1= 0

Node 2: T1+2T4+100-4T2 = 0Node 3: T1+2T4+T5-4T3 = 0

Node 4: T2+T3+T6+86.66-4T4 = 0

Node 5: T3+2T6-4T5 = 0

Node 6: T4+T5+50-4T6 = 0

=

50

0

66.86

0

100

0

T

T

T

T

T

T

411000

240100

104110

012401

002041

000224

6

5

4

3

2

1

Matrix Form

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