Post on 16-Jan-2016
1
CHAPTER 3
TWO-DIMENSIONAL STEADY STATE CONDUCTION
3.1 The Heat Conduction Equation
• Assume: Steady state, isotropic, stationary, k = = constant
(3.1) 02
2
2
2
k
q
x
T
k
Uc
y
T
x
T p
2
• Special case: Stationary material, no energy generation
• Cylindrical coordinates:
02
2
2
2
y
T
x
T (3.2)
(3.3) 01
2
2
z
T
r
Tr
rr
Eq. (3.1), (3.2) and (3.3) are special cases of
3
0)()(
)()()()(
01
2
2
2012
2
2
Tygy
Tyg
y
TygTxf
x
Txf
x
Txf
(3.4)
Eq. (3.4) is homogenous, second order PDE with
variable coefficients
3.2 Method of Solution and Limitations
• Method: Separation of variables
• Basic approach: Replace PDE with sets of ODE
4
• The number of sets depends on the number of independent variables
• Limitations:
(1) Linear equations
• Examples: Eqs. (3.1)-(3.4) are linear
(2) The geometry is described by an orthogonal coordinate system. Examples: Rectangles, cylinders, hemispheres, etc.
variabledependent the iflinear is equation Anunitypower to raisedappear sderivative itsor
occurnot do productstheir if and
5
3.3 Homogeneous Differential Equations and Boundary Conditions
• Example: Eq. (3.1): Replace T by cT and divide
through by c
02
2
2
2
ck
q
x
T
k
Uc
y
T
x
T p
NH (a)
not isit if shomogeneou is condition buondaryor equation Anconstant a by multiplied is variabledependent the when altered
6
Example: Boundary condition
TThx
Tk (b)
Replace T by cT
c
TTh
x
Tk NH (c)
• Simplest 2-D problem: HDE with 3 HBC and 1 NHBC
• Separation of variables method applies to NHDE and NHBC
7
3.4 Sturm-Liouville Boundary Value Problem: Orthogonality
• PDE is split into sets of ODE. One such sets is known as Sturm-Liouville equation if it is of the form
Rewrite as
(3.5a) )()()(
32
212
2
xaxadx
dxa
dx
dn
nn
0n
(3.5b) )()()( 2 xwxqdx
dxp
dx
dn
n
0
n
8
where
,exp)(1dxaxp ),()(
2xpaxq )()(
3xpaxw
(3.6)
NOTE:
• w(x) = weighting function, plays a special role
• Equation (3.5) represents a set of n equations
corresponding to n values of n
• The solutions n are known as the characteristic
functions
9
• Important property of the Sturm-Liouville problem: orthogonality
• Two functions, n ,
mand are said to be orthogonal
in the interval a and b with respect to a weighting
function w(x), if
b
a mndxxwxx 0)()()( (3.7)
• The characteristic functions of the Sturm-Liouville problem are orthogonal if:
10
(1) p(x) , q(x) and w(x) are real, and
(2) BC at x = a and x = b are homogeneous of the form
0n (3.8a)
(3.8b) 0dx
dn
(3.8c) 0dx
d nn
11
Special Case: If p(x) = 0 at x = a or x = b , these conditions can be extended to include
)()( bann (3.9a)
and
dx
bd
dx
adnn
)()( (3.9b)
(3.9a) and (3.9b) are periodic boundary conditions.
• Physical meaning of (3.8) and (3.9)
12
• Relationship between the Sturm-Liouville problem, orthogonality, and the separation of variables method:
PDE + separation of variables 2 ODEOne of the 2 ODE = Sturm-Liouville problem
Solution to this ODE = n = orthogonal functions
13
3.5 Procedure for the Application of Separation of Variables Method
Example 3.1: Steady state 2-D conduction in a rectangular plate
Find T (x,y)
(1) Observations
• 2-D steady state
• 4 BC are needed
• 3 BC are homogeneous
2 HBC
x
y
1.3Fig.
T = f(x)
T = 0 T = 0
T = 00
0
14
(2) Origin and Coordinates
• Origin at the intersection of the two simplest BC
• Coordinates are parallel to the boundaries
(3) Formulation
(i) Assumptions
(1) 2-D
(2) Steady
(3) Isotropic
(4) No energy generation
(5) Constant k
15
(ii) Governing Equations
02
2
2
2
y
T
x
T(3.2)
(iii) Independent Variable with 2 HBC: x- variable
(iv) Boundary Conditions
(1) ,0),0( yT HT = 02 HBC
x
y
1.3Fig.
T = f(x)
T = 0
T = 00
0 (3) ,0)0,( xT H
(2) ,0),( yLT H
16
(4) Solution
(i) Assumed Product Solution
)()(),( yYxXyxT (a)
(a) into eq. (3.2)
0
)()()()(2
2
2
2
y
yYxX
x
yYxX (b)
(4) ),(),( xfWxT non-homogeneous
17
2
2
2
2
)(1
)(1
dy
YdyYdx
XdxX
(c)
)()( yGxF
where 2n is known as the separation constant
22
2
2
2
)(
1
)(
1ndy
Yd
yYdx
Xd
xX (d)
0)()(2
2
2
2
dy
YdxX
dx
XdyY
18
NOTE:
• The separation constant is squared
• The constant is positive or negative
• The subscript n. Many values: n ,...,,
321are
known as eigenvalues or characteristic values
• Special case: constant = zero must be considered
• Equation (d) represents two sets of equations
022
2
nn
n Xdx
Xd (e)
19
022
2
nn
n Ydy
Yd (f)
• The functionsn
Xn
Yand
or characteristic values
are known as eigenfunctions
(ii) Selecting the sign of the n terms
Select the plus sign in the equation representing the variable
with 2 HBC. Second equation takes the minus sign
20
022
2
nn
n Xdx
Xd (g)
022
2
nn
n Ydy
Yd (h)
• Important case: ,0n equations (g) and (h) become
02
02
dx
Xd (i)
21
020
2
dy
Yd(j)
(iii) Solutions to the ODE
xBxAxXnnnnn cossin)( (k)
yDyCyYnnnnn coshsinh)( (l)
xBAxX000
)( (m)
yDCyY000
)( (n)
22
NOTE: Each product is a solution
)()(),(000
yYxXyxT (p)
)()(),( yYxXyxTnnn
(o)
The complete solution becomes
100
)()()()(),(n
nnyYxXyYxXyxT (q)
(iv) Applying Boundary Conditions
23
NOTE: Each product solution must satisfy the BC
BC (1))()0(),0( yYXyT
nnn
)()0(),0(000
yYXyT
Therefore
0)0( n
X (r)
0)0(0
X (s)
Applying (r) to solution (k)
00cos0sin)0( nnn
BAX
24
Therefore0
nB
Similarly, (r) and (m) give
00B
B.C. (2)
0)()(),( yYLXyLT nnn
Therefore
0)( LXn
25
Applying (k)
0sin LAnn
Therefore
0sin Ln (t)
Thus
,L
nn
...3,2,1n (u)
Similarly, BC 2 and (m) give
0)(000 BLALX
26
Therefore0
0A
With ,000BA the solution corresponding to
0n vanishes.
BC (3)
0)0()()0,( nnn
YxXxT
0)0( n
Y
00cosh0sinh)0( nnn
DCY
27
Which gives
0n
D
Results so far: ,000
nn
DBBA therefore
and
Temperature solution:
1
))(sinh(sin),(n
nnnyxayxT (3.10)
xAxXnnnsin)(
yCyYnnnsinh)(
28
Remaining constant
nnnCAa
B.C. (4)
xWaxfWxTn
nnn
1
sin)sinh()(),( (3.11)
To determinen
a from (3.11) we apply orthogonality.
(v) Orthogonality
• Use eq. (7) to determinen
a
29
• The function xnsin in eq. (3.11) is the
solution to (g)
• If (g) is a Sturm-Liouville equation with 2 HBC, orthogonality can be applied to eq. (3.11)
Return to (g)
022
2
nn
n Xdx
Xd (g)
Compare with eq. (3.5a)
0)()()(3
2212
2
nn
nn xaxadx
dxa
dx
d
(3.5a)
30
0)()(21
xaxa 1)(3
xaand
Eq. (3.6) gives
1)()( xwxp and 0)( xq
• BC (1) and (2) are homogeneous of type (3.8a).
• The characteristic functions xxXxnnn sin)()(
are orthogonal with respect to 1)( xw
0x to
Multiply eq. (3.11) by ,sin)( xdxxwm integrate from
Lx
31
xdxxwxfL
m0
sin)()(
xdxxwxWaL
mn
nnn
0 1
sin)(sin)sinh(
(3.12)
Interchange the integration and summation, and use
1)( xw
Introduce orthogonality (3.7)
0)()()( dxxwxxb
amn mn (3.7)
32
Apply (3.7) to (3.12)
xdxWaxdxxfn
L
nn
L
n
0
2
0
sinsinhsin)(
Solve for n
a
(3.13) xdxxfWL
aL
nn
n 0
sin)(sinh
2
(5) Checking
33
The solution is
1
))(sinh(sin),(n
nnnyxayxT (3.10)
Dimensional check
Limiting check
Boundary conditions
Differential equations
(6) Comments
Role of the NHBC
34
3.6 Cartesian Coordinates: Examples
Example 3.3: Moving Plate with Surface Convection
Outside furnace the plate exchanges heat by convection.
Determine the temperature distribution in the plate.
y
xU
0
Th,
L
L
3.3Fig.
HBC2
Th,
oT
Semi-infinite plate leaves a furnace at oT
35
Solution
(1) Observations
• Symmetry
• NHBC
(2) Origin and Coordinates
(3) Formulation (i) Assumptions
(1) 2-D
(2) Steady state
• At 0x plate is at oT
36
(4) Uniform velocity
,k(3) Constantp
c and ,
(ii) Governing EquationsDefine TT
022
2
2
2
xyx
(a)
k
Ucp
2
(b)
(5) Uniform andhT
37
(iii) Independent variable with 2 HBC: y- variable
(iv) Boundary Conditions
(1) 0)0,(
y
x
(2) ),(),(
Lxhy
Lxk
(3) 0),( y
(4) TTy
0),0(
y
xU
0
Th,
L
L
Fig. 3.3
HBC2
Th,oT
38
(4) Solution
(i) Assumed Product Solution
)()( yYxX
02 22
2 nn
nn Xdx
dX
dx
Xd (c)
(d) 022
2
nnn Y
dy
Yd
(ii) Selecting the sign of the 2n terms
39
02 22
2 nn
nn Xdx
dX
dx
Xd (e)
(f) 022
2 nn
n Ydy
Yd
For :0n
(g) 02 02
02
dx
dX
dx
Xd
(h) 020
2
dy
Yd
40
(iii) Solutions to the ODE
)(xXn
2222 expexp)exp( nnnn xBxAx
(i)
yDyCyYnnnnn cossin)( (j)
(k) 000
2exp)( BxAxX
And
(l) yDCyY000
)(
41
100
)()()()(),(n
nnyYxXyYxXyx (m)
Complete solution:
(iv) Application of Boundary Conditions
BC (1)0
0DC
n
BC (2)
BiLLnn tan (n)
00C
42
0n
ABC (3)
With
,000DC
000YX
(3.17)
yxayxn
nnn
1
22 cosexp),(
BC (4)
10
cosn
nnyaTT (o)
43
(v) Orthogonality
Characteristic Functions:
ynn cos are solutions to equation (f).
Thus eq. (3.6) gives
1wp and 0q (p)
2 HBC at 0y and Ly
ynn cos are orthogonal
Comparing (f) with eq. (3.5a) shows that it is a Sturm-and0
21aa .1
3aLiouville equation with
44
L
n
L
n
n
ydy
ydyTT
a
0
2
00
cos
cos
to
Multiply both sides of (o) by ,cos xdxm integrate
from 0x Lx and apply orthogonality
LLL
LTTa
nnn
nn
cossin2
sin20
(q)
45
(5) Checking
Dimensional check
Limiting check: If 00
),(,0 TyxTqTT
(6) Comments
.0UFor a stationary plate,
46
3.7 Cylindrical Coordinates: Examples
Example 3.5: Radial and Axial Conduction in a Cylinder
Two solid cylinders are pressed co-
axially with a force F and rotated in opposite directions.
Coefficient of friction is .
Convection at the outer surfaces.
or
r
z
Th,
L0
2 HBC
Fig. 3.5
LTh, Th,
Th,
47
Find the interface temperature.
Solution
(1) Observations
• Symmetryor
r
z
Th,
L0
2 HBC
Fig. 3.5
LTh, Th,
Th,
• Interface frictional heat
= tangential force x velocity
• Transformation of B.C., TT
(2) Origin and Coordinates
48
(3) Formulation
(i) Assumptions
(1) Steady
(2) 2-D
(4) Uniform interface pressure
(3) Constant ,k and
(6) Radius of rod holding cylinders is small compared to cylinder radius
(ii) Governing Equations
(5) Uniform
h and T
49
01
2
2
zrr
rr
(3.17)
(iii) Independent variable with 2 HBC: r- variable
(iv) Boundary
conditions (1) ,0),0(
r
zor ),0( z finite
50
(2) ),(),(
00 zrhr
zrk
(3) ),(),(
Lrhz
Lrk
(4),
2
0
( 0)( )
r Fk r f r
z r
(4) Solution
(i) Assumed Product Solution
or
r
z
Th,
L0
2 HBC
Fig. 3.5
LTh, Th,
Th,
51
)()(),( zZrRzr (a)
01 2
R
dr
dRr
dr
d
r kk (b)
022
2
kk
k Zdz
Zd (c)
(ii) Selecting the sign of the 2k
r-variable has 2 HBC
52
0222
22
kkkk Rr
dr
dRr
dr
Rdr (d)
022
2
kk
k Zdz
Zd (e)
0020
2
dr
dR
dr
Rdr (f)
For :02 k
02
02
dz
Zd(g)
53
(iii) Solutions to the ODE
(d) is a Bessel equation:
,0BA 0n,1C ,k
D
Since 0n and D is real
rYBrJArRkkkkk
00)( (h)
Solutions to (e), (f) and (g):
zDzCzZkkkkk coshsinh)( (i)
54
000Bln rAR (j)
000DxCZ (k)
Complete solution
)()()()(),(1
00zZrRzZrRzr
kk
k
(l)
(iv) Application of Boundary Conditions
BC (1)
)0(0
Y and 0lnNote:
55
00AB
k
BC (2)
)()(00,0
0
rhJrJdr
dk
kzrk
(m) BirJ
rJr
k
kk
)(
)(
00
010
00B
khrBi0
(n)
BC (2)
56
Since ,000BA 0n
LLBiL
LLBiLCD
kkk
kkkkk
cosh)(sinh
sinh)(cosh
1 cosh)(sinh
sinh)(cosh[sinh),(
k kkk
kkkkk LLBiL
LLBiLzazr
)](cosh0
rJzkk (3.20)
BC (3)
57
BC (4)
)()(0
1
rJakrfkk
kk
(o)
(v) Orthogonality
)(0
rJk is a solution to (d) with 2HBC in .r
andComparing (d) with eq. (3.5a) shows that it is a Sturm-
,/11
ra .13aLiouville equation with ,0
2a
Eq. (3.6):
rwp and 0q (p)
58
)(0
rJk and )(
0rJ
i are orthogonal with respect to
ka.)( rrw Applying orthogonality, eq. (3.7), gives
)(])()[
)()(2
)(
)()(
020
20
20
00
0
20
00
0
0
0
rJkhrrk
drrrJrf
drrrJk
drrrJrf
akk
r
kk
r
kk
r
k
k
(q)
Interface temperature: set 0z in eq. (3.20)
59
)(cosh)(sinh
sinh)(cosh)0,(
01
rJLLBiL
LLBiLar
kk kkk
kkkk
(r)
(5) Checking
Dimensional check: Units ofk
a
Limiting check: If 0 or 0 then TzrT ),(
(6) Comments
)(xw is not always equal to unity.
60
3.8 Integrals of Bessel Functions
0
0
2 )(
r
knndrrrJN (3.21)
61
Table 3.1 Normalizing integrals for solid cylinders [3]
Boundary Condition at
(3.8a):
(3.8b):
(3.8c):
0
0
2 )(
r
knndrrrJN 0
r
0)(0rJ
kn
0)(
0 dr
rdJkn
)()(
00 rhJ
dr
rdJk
knkn
2
02
20
)(
2
dr
rdJrkn
k
)()(2
10
22202
rJnrknk
k
)()()(2
10
2220
22
rJnrBiknk
k
62
3.9 Non-homogeneous Differential Equations
Example 3.6: Cylinder with Energy Generation
L
aT
oT
r
or q z0 q
0
Fig. 3.6
Solid cylinder generates
heat at a rate One end is at.q
0T while the other is insulated.
Cylindrical surface is at .a
T
Find the steady state temperature distribution.
63
Solution
(1) Observations
• Energy generation leads to
NHDE
• Use cylindrical coordinates
L
aT
oT
ror q
z0q0
Fig. 3.6 • Definea
TT to make BC at surface homogeneous
(2) Origin and Coordinates
64
(3) Formulation
(i) Assumptions
(1) Steady state
(2) 2-D
k(3) Constant and
(ii) Governing Equations
Define
aTT
65
01
2
2
k
q
zrr
rr
(3.22)
(iii) Independent variable with 2 HBC
(iv) Boundary
conditions (1) ),0( z finite
(2) 0),(0
zr L
aT
oT
ror q
z0q0
Fig. 3.6
(3) 0),(
z
Lr
(4)a
TTr 0
)0,(
66
(4) Solution
)(),(),( zzrzr (a)
Substitute into eq. (3.22)
Split (b): Let
(c) 01
2
2
zrr
rr
Let:
01
2
2
2
2
k
q
zd
d
zrr
rr
(b)
67
Therefore
NOTE:
• (d) is a NHODE for the )(z
• Guideline for splitting PDE and BC:
),( zr• (c) is a HPDE for
should be governed by HPDE and three HBC.Let take care of the NH terms in PDE and BC)(z
),( zr
02
2
k
q
zd
d (d)
68
BC (1)
),0( z finite (c-1)
BC (2)
)(),(0
zzr (c-2)
BC (3)
0)(),(
dz
Ld
z
Lr
Let
(c-3) 0),(
z
Lr
69
Thus
0)(
dz
Ld(d-1)
BC (4)
aTTr
0)0()0,(
Let(c-4) 0)0,( r
Thus
aTT
0)0( (d-2)
Solution to (d)
70
FEzzk
qz
2
2)( (e)
(i) Assumed Product Solution
)()(),( zZrRzr (f)
(f) into (c), separating variables
(g) 022
2
kk
k Zdz
Zd
0222
22
kkkk Rr
dr
Rdr
dr
Rdr (h)
71
(ii) Selecting the sign of the 2k terms
(i) 022
2
kk
k Zdz
Zd
0222
22
kkkk Rr
dr
Rdr
dr
Rdr (j)
For ,0k
(k) 02
02
dz
Zd
72
0020
22
dr
Rdr
dr
Rdr (l)
(iii) Solutions to the ODE
(m) kkkkkBAzZ cossin)(
(j) is a Bessel equation with ,0 nBA ,1C
.iDk
(n) )()()(0
rKDrICzRkkkokk
73
Solution to (k) and (l)
(o) 000)( BzAzZ
(p) 000
ln)( DrCzR
Complete Solution:
100
)()()()()(),(k
kkzZrRzZrRzzr (q)
(iv) Application of Boundary Conditions
BC (c-1) to (n) and (p)
00CD
k
74
BC (c-4) to (m) and (o)
00BB
k
BC (c-3) to (m) and (o)
00A
Equation fork
,0cos Lk or
2
)12(
kL
k,...2,1k (r)
With 000BA
000ZR
The solutions to ),( zr become
75
zrIazrk
kkk
10
sin)(),( (s)
BC (d-1) and (d-2)
TTF0
kLqE / and
)()/()/(22
)(0
22
TTLzLzk
Lqz (t)
BC (c-2)
)()/()/(22 0
22
TTLzLzk
Lq
zrIak
kkk
100
sin)(
(u)
76
(v) Orthogonality
zksin are solutions to equation (i).
andComparing (i) with eq. (3.5a) shows that it is a Sturm-
.13aLiouville equation with ,0
21aa
Eq. (3.6) gives
0z and ,Lz 2 HBC at zksin are orthogonal
with respect to 1w
rwp and 0q
77
dzTTLzLzk
Lqk
L
sin)()/()/(22
00
22
L
kkkdzrIa
0
200
sin)(
Evaluating the integrals and solving for k
a
2
0/)(
)()(
2kao
okkk kqTT
rILa
(v)
78
Solution to ),( zr
(5) Checking
Limiting check: 0q and .0
TTa
Dimensional check: Units of kLq /2 and ofk
a in
solution (s) are in °C
(w) a
TzrTzr ),(),(
22
0)/()/(2
2)( LzLz
k
LqTT
a
zrIak
kkk
10
sin)(
79
3.10 Non-homogeneous Boundary Conditions
Method of Superposition:
• Decompose problem
Example:
)(yg
)(xfx
y
L
Woq
Th,
0
3.7Fig.
Problem with 4 NHBC is decomposed
into 4 problems each having one NHBC
DE
02
2
2
2
y
T
x
T
80
Solution:),(),(),(),(),(
4321yxTyxTyxTyxTyxT
)( yg
)(xfx
y
L
Woq
Th,
0
x
y
L
Woq
0
0,h
0
0
x
y
L
W
0 0
0
Th,
4T
)(xf x
y
L
W
0
0,h
0
3T)( yg
x
y
L
W
0
0,h
0
2T
81