Numerical MethodFinite Differencing

x = 1

1 2 3 i-1 i i+1 N-2 N-1 N

x = 0 h

Consider uniform interval h is constant

f(x) is assigned at each node, f1, f2, fi-1, fi, fi+1, fN-2, fN-1, fN

nodeeachatxf

xf

eApproximat 2

2

&

)(Accuracy2

)(21

)(21

)12(NodesInternalFor:ExpansionSeriesTaylor

sideeachatnodeoneuse:Example

311

322

2

1

322

2

1

hOhff

xf

xf

forSolve

hOhxf

hxf

ff

hOhxf

hxf

ff

Ni

ii

i

i

iiii

iiii

Discretization

)(Accuracy2

)(21

)(21

)12(NodesInternalFor:ExpansionSeriesTaylor

sideeachatnodeoneuse:Example

42

112

2

2

2

322

2

1

322

2

1

hOh

fffxf

xf

forSolve

hOhxf

hxf

ff

hOhxf

hxf

ff

Ni

iii

i

i

iiii

iiii

Discretization

)(Accuracy234

Eliminate

)(4212

)(21

0orknownis,specifiedisGradient)(

knownis:boundarytheatspecifiedFunction)(

3

1132

12

2

32

12

2

113

32

12

2

112

111

1

hOhxf

fff

xf

hOhxf

hxf

ff

hOhxf

hxf

ff

Bfxf

Axf

b

fa

For Nodes at the Boundary Nodes (i=1, i=N)

Example: Numerical Solution of the Fin Problem

0)1()1(&1)0(

)()(

1000

2

H

specifiedareXfandXawhere

XfAd

a

x = 1

1 2 3 i-1 i i+1 N-2 N-1 N

x = 0 h

For Interior Nodes (2 ≤ i ≤ N-1)

12

02

2

2&2

eapproximatweside,eachatnodeoneUsing

0

211

211

21111

Nifor

fAd

ha

ha

hh

iiii

iiii

i

iiii

iii

x = 1

1 2 3 i-1 i i+1 N-2 N-1 N

x = 0 h

Boundary Nodes (i = 1 and i = N)

NNN

N

NNNNNN

N

NNNN

NNNN

Hh

Hh

hOhh

hOhh

NiForiFor

234obtain,Eliminate

0since234Eliminate

)(4212

)(21

1,1

21

21

322

321

1

Algebraic Equation

matrixt coefficien

0...,0,0,2/22/2

vector known :,1...,3,2

vector unknown ::formmatrixtheIn

:11

xNNA

Thahab

b

TNNx

xbAx

Algebraic Equation

hHAAA

hiahiaA

AifdhiaA

hiahiaA

hahaA

AfdhaA

NNNNNN

ii

ii

ii

2341

2/2/

0/22/2

2/2/

2/22/2

0/222/22

:formmatrix Aoft coefficienThe

1,12,13,1

1,

,

1,

2,1

1,1

Solution

solution Iterative - MethodIndirect

1matrixInvertMethodDirect

:solveTo

bAx

A

ASSIGNMENT 2 – due Feb 16 Fin

w0Tb

T0

k=k11

k=k22

Streaming fluid

r1 r2r3

h=h1h=h2

r1

r2

r3

Top view

h=0

Cross-section

ASSIGNMENT 2

For two fin whose cross - sections are shown calculate the temperature and the heat flux at any cross -section. Run the program for a perfect interface.

Find the temperature at the cross-section where the fin changes shape. Plot the temperature as a function of r in r1 ≤r ≤ r3.

.15,0.9,0.7,5,0.3,/60,/40

,/150,/100,20,150

3210

21

22

210

cmrcmrcmrcmwCmWkCmWk

CmWhCmWhCTCTb

ASSIGNMENT 2

Use the fact that the temperature and the heat flux are continuous at the cross-section for a perfect interface.

xTk

xTk

tzyxTtzyxT II

22

11

21 ),,,(),,,(

Conduction-Conditions at the Interface

Perfect Interface

xTk

xTk

tzyxTtzyxT II

22

11

21 ),,,(),,,(

Imperfect Interface

x = xI

Material 2k2 T2(x,y,z)

Material 1k1 T1(x,y,z)

Interface

xTk

xTk

xTkTThc

22

11

1121Rc is the contact resistance

Rc ≈ 10-6 – 10-3 m2K/W

hc ≈ 102 – 106 W/m2K

RkhrRfR

AARa

RTTTTRfkSh

wrArwArrR

b

1

111

0

11

1001

1111

010011

4)( and)(

).()()(

2,2,/:1Region

Introduce Normalized Variables:

10

2112

11211

1

1222

2 where0)(

:)(for Equation where1 :1Region

kwrhRR

RrrRRR

Introduce Normalized Variables:

0

1

2

2122

0

22

2002

2222

2021

13332

tan2with

4)( and1)(

).()()(

tan2,2,/ where:2Region

wrc

RkhrRfRRcR

AARa

RTTTTRfkSh

rrwwrwArrRrrRRRR

b

0

122

20

2122

222222

2

32

tan2 where1)(

2 where0)(

:)(for Equation :2Region

wrcRRcRRa

kwrhRa

RRRR

0)( tipat the

whereinterface at the continuousflux heat and

)continuous re(temperatuinterface, at the

1)1(1base at the

323

212221

2221

2

1

RRR

kkRR

RRRR

R

Boundary Conditions – with the Perfect Interface

0)(tip theat

whereinterface theat continuousflux heat and

where0interface, theat

1)1(1base theat

323

212221

11c211

2

1

RRR

kkRR

krhHHRR

R

cc

Boundary Conditions – with the imperfect Interface

212221

2221

321

20

2122

22222

32

10

2112

11211

2

where

interfaceperfect aFor 0)(,1)1(

2 where0)(

For

2 where0)(

1 For

kkRRRR

RkwrhRR

RRRkwrhRR

RR

Special Case = 0

:conditionsboundary theapplyingbydeterminedbecan),,,(Constants

)(and

)(kind second andfirst zero,order of

funtions Bessel modifiedin expressed becan )( and )(for solution The

20202

10101

21

DCBARDKRCIR

RBKRAIR

RR

Solution

2212212

2112111

220220

210210

321321

1010

and,

,0,1

:are for Equations

RDKRCIRBKRAI

RDKRCIRBKRAI

RDKRCIBKAI(A,B,C,D)

hiRkwrh

RRcRaN-iI

hiRkwrh

hiRa-Ii

Rh

ah

a

i

iii

i

ii

iiii

iiii

i

21with2

,1 11for

11with2

11 12for

02

2nodesinterior for ng,differenci centralUsing

20

2122

22

10

2112

1

211211

R = R3

1 2 3 I1-1 (I1,I2) I2+1 N-2 N-1 N

R = 1 h R = R2

Finite Difference Discretization – for each regionUse two nodes at the interface one for region 1 (I1) and one for region 2 (I2)

Boundary condition: Nodes i = 1 and i = N

034 insulated is Tip

1,11 base at the re temperatuSpecified

21

1

NNNNiForNi

iFori

Condition at the interface: Interface Nodes, i = I1 ,I2

(I1 ,I2 are at the same point)

hh

h

h

IIIIII

III

III

II

243

243

243

243

)continuousheatflux (

)continuous eTemperatur(:interfaceperfect aFor

1212

12I

12I

II

222111

222

2

111

1

21

21

Condition at the interface: Interface Nodes (i = I1 ,I2)

hh

Hh

H

IIIIII

IIcIII

IIc

243

243

)continuousflux heat (

02

43

)resistancecontact (0:interfaceimperfect an For

1212

II

12

I

222111

21

21

111

211

22312

232321

11

1IIRegion For IRegion For

and11

dd

RRdRd

dd

RdRd

RRRRRR

Another Approach: Multi-region Problems - Transformation

R

1 R2 R3

0 01

ξ1

ξ2

variablesdependent areandst variableindependen areand

10 For

0110 For

0111

1

21

21

2

22323222222

23

1

121211112

2

RRRaRR

RaR

Multi-region Problems – Transformation

,0)0(,1)0(

01

0111

110 For

21

223322222

23

1221112

2

with

RRRaRR

RaR

Multi-region Problems – Final

)1(1)1(1

1

0)1()1()1(1

1:interfaceimperfect an For

)1(1)1(1

1)1()1(

:interfaceperfect aFor

223

12

2112

223

12

21

RRR

HR

RRR

c

Multi-region Problems – Final