06 finite difference-6
DESCRIPTION
Heat transfer (finite difference method)TRANSCRIPT
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