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CHAPTER 3
TWO-DIMENSIONAL STEADY STATECONDUCTION
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
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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
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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 withvariable coefficients
3.2 Method of Solution and Limitations
Method: Separation of variables
Basic approach: Replace PDE with sets of ODE
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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.
variabledependenttheiflinearisequationAn unitypowertoraisedappearsderivativeitsoroccurnotdoproductstheirifand
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3.3 Homogeneous Differential Equations and
Boundary Conditions
Example: Eq. (3.1): Replace T bycTand divide
through byc
02
2
2
2
ck
q
x
T
k
Uc
y
T
x
T p NH (a)
notisitifshomogeneouisconditionbuondaryorequationAnconstantabymultipliedisvariabledependentthewhenaltered
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Example: Boundary condition
TThxTk (b)
Replace T bycT
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
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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)
)()()(3
2
212
2xaxa
dx
dxa
dx
dn
nn 0n
(3.5b))()()( 2 xwxqdx
dxp
dx
dn
n 0n
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where
,exp)( 1dxaxp ),()( 2 xpaxq )()( 3 xpaxw
(3.6)
NOTE:
w(x) = weighting function, plays a special role
Equation (3.5) represents a set ofn equations
corresponding ton values of n
The solutionsn
are known as thecharacteristic
functions
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Important property of the Sturm-Liouville problem:
orthogonality
Two functions,n
,m
and are said to be orthogonal
in the intervala andb 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:
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(1)p(x) , q(x) and w(x) are real, and
(2) BC atx = a andx = b are homogeneous of the
form
0n
(3.8a)
(3.8b)0dx
dn
(3.8c)0dx
d nn
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Special Case: Ifp(x) = 0 atx = a orx = b , these
conditions can be extended to include
)()( bann
(3.9a)
and
dx
bd
dx
adnn
)()((3.9b)
(3.9a) and (3.9b) areperiodic boundary conditions.
Physical meaning of (3.8) and (3.9)
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Relationship between the Sturm-Liouville problem,
orthogonality, and the separation of variablesmethod:
PDE + separation of variables2 ODE
One of the 2 ODE =Sturm-Liouville problem
Solution to this ODE = n= orthogonal functions
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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
4BC are needed
3 BC are homogeneous
2 HBC
x
y
1.3Fig.
T = f(x)
T = 0 T = 0
T = 0
00
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(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) Constantk
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(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 H
T = 02 HBC
x
y
1.3Fig.
T = f(x)
T = 0
T = 00
0(3) ,0)0,(xT H
(2) ,0),( yLT H
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(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
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2
2
2
2
)(
1
)(
1
dy
Yd
yYdx
Xd
xX(c)
)()( yGxF
where 2n
is known as theseparation constant
2
2
2
2
2
)(
1
)(
1n
dy
Yd
yYdx
Xd
xX(d)
0)()(2
2
2
2
dy
YdxX
dx
XdyY
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NOTE:
The separation constant is squared
The constant is positive or negative
The subscriptn. Many values:n
,...,,321
are
known as eigenvalues orcharacteristic values
Special case: constant = zero must be considered
Equation (d) represents two sets of equations
022
2
nnn X
dx
Xd (e)
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022
2
nnn Y
dy
Yd(f)
The functionsn
Xn
Yand
or characteristic values
are known as eigenfunctions
(ii) Selecting the sign of then
terms
Select the plus sign in the equation representing the variable
with 2 HBC. Second equation takes the minus sign
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022
2
nnn X
dx
Xd(g)
02
2
2
nnn Y
dy
Yd(h)
Important case: ,0n
equations (g) and (h) become
02
0
2
dx
Xd (i)
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020
2
dy
Yd(j)
(iii) Solutions to the ODE
xBxAxX nnnnn cossin)( (k)
yDyCyYnnnnn
coshsinh)( (l)
xBAxX 000 )( (m)
yDCyY000
)( (n)
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NOTE: Each product is a solution
)()(),(000
yYxXyxT (p)
)()(),( yYxXyxT nnn (o)
The complete solution becomes
100
)()()()(),(n nn
yYxXyYxXyxT (q)
(iv) Applying Boundary Conditions
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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
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Therefore
0n
B
Similarly, (r) and (m) give
00
B
B.C. (2)
0)()(),( yYLXyLT nnn
Therefore
0)(LXn
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Applying (k)
0sin LAnn
Therefore
0sin Ln
(t)
Thus
,L
nn
...3,2,1n (u)
Similarly, BC 2 and (m) give
0)(000
BLALX
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Therefore
00
A
With ,000
BA the solution corresponding to
0n
vanishes.
BC (3)
0)0()()0,(nnn
YxXxT
0)0(nY
00cosh0sinh)0(nnn
DCY
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Which gives
0n
D
Results so far: ,000 nn
DBBA therefore
and
Temperature solution:
1
))(sinh(sin),(n
nnnyxayxT (3.10)
xAxXnnn
sin)(
yCyYnnn
sinh)(
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Remaining constant
nnn CAaB.C. (4)
xWaxfWxT
n nnn1
sin)sinh()(),((3.11)
To determinen
a from (3.11) we applyorthogonality.
(v) Orthogonality
Use eq. (7) to determinen
a
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The function xn
sin in eq. (3.11) is thesolution 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
nnn X
dx
Xd(g)
Compare with eq. (3.5a)
0)()()(3
2
212
2
nnnn xaxa
dx
dxa
dx
d
(3.5a)
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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
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xdxxwxf
L
m0
sin)()(
xdxxwxWa
L
mn
nnn0 1
sin)(sin)sinh(
(3.12)
Interchange the integration and summation, and use
1)(xw
Introduce orthogonality (3.7)
0)()()( dxxwxx
b
amn
mn (3.7)
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Apply (3.7) to (3.12)
xdxWaxdxxfn
L
nn
L
n0
2
0
sinsinhsin)(
Solve for na
(3.13)xdxxf
WL
a
L
n
n
n
0
sin)(
sinh
2
(5) Checking
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The solution is
1))(sinh(sin),(
nnnn yxayxT (3.10)
Dimensional check
Limiting checkBoundary conditions
Differential equations
(6) Comments
Role of the NHBC
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3.6 Cartesian Coordinates: Examples
Example 3.3: Moving Plate with SurfaceConvection
Outside furnace the plate
exchanges heat by convection.
Determine the temperature distribution in the plate.
y
xU
0
Th,
L
L
3
.
3
Fig.
HBC2
Th,
oT
Semi-infinite plate leaves a
furnace atoT
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Solution
(1) Observations
Symmetry
NHBC
(2) Origin and Coordinates
(3) Formulation
(i) Assumptions
(1) 2-D
(2) Steady state
At 0x plate is atoT
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(4) Uniform velocity
,k(3) Constantp
c and ,
(ii) Governing Equations
Define TT
022
2
2
2
xyx
(a)
k
Ucp
2(b)
(5) Uniform andh T
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(iii) Independent variable with 2 HBC: y-
variable
(iv) Boundary Conditions
(1) 0)0,(
y
x
(2) ),(),(
Lxhy
Lxk
(3) 0),( y
(4) TTy0
),0(
y
x
U
0
Th,
L
L
Fig. 3.3
HBC2
Th,oT
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(4) Solution
(i) Assumed Product Solution)()( yYxX
02 22
2
nnnn X
dxdX
dxXd (c)
(d)0
2
2
2
nn
n
Ydy
Yd
(ii) Selecting the sign of the 2n terms
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022
2
2
nnnn X
dx
dX
dx
Xd(e)
(f)02
2
2
nnn Y
dy
Yd
For :0n
(g)02 02
0
2
dx
dX
dx
Xd
(h)02
0
2
dy
Yd
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(iii) Solutions to the ODE
)(xXn
2222 expexp)exp( nnnn xBxAx
(i)
yDyCyYnnnnn
cossin)( (j)
(k)000
2exp)( BxAxX
And
(l)yDCyY 000 )(
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100 )()()()(),(
nnn yYxXyYxXyx (m)
Complete solution:
(iv) Application of Boundary Conditions
BC (1)
00
DCn
BC (2)
BiLLnn
tan (n)
00
C
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0n
A
BC (3)
With ,000
DC
000
YX
(3.17)
yxayxn
nnn1
22 cosexp),(
BC (4)
10
cosn
nnyaTT (o)
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(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-
and021 aa .13aLiouville equation with
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L
n
L
n
nydy
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
sin2 0 (q)
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(5) Checking
Dimensional check
Limiting check: If00
),(,0 TyxTqTT
(6) Comments
.0UFor a stationary plate,
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3.7 Cylindrical Coordinates: Examples
Example 3.5: Radial and Axial Conduction ina 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,
L
0
2 HBC
Fig. 3.5
L
Th, Th,
Th,
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Find the interface temperature.
Solution
(1) Observations
Symmetry or
r
z
Th,
L
0
2 HBC
Fig. 3.5
LTh,
Th,
Th,
Interface frictional heat
= tangential forcex velocity
Transformation of B.C., TT
(2) Origin and Coordinates
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(3) Formulation
(i) Assumptions
(1) Steady
(2) 2-D
(4) Uniform interface pressure
(3) Constant ,k and
(6) Radius of rod holding cylinders is smallcompared to cylinder radius
(ii) Governing Equations
(5) Uniformh and T
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01
2
2
zr
r
rr
(3.17)
(iii) Independent variable with 2 HBC: r-
variable
(iv) Boundary conditions
(1) ,0),0(
r
z or ),0( z finite
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(2) ),(),(
00 zrh
r
zrk
(3) ),(),(
Lrhz
Lrk
(4),
2
0
( 0)( )
r Fk r f r
z r
(4) Solution
(i) Assumed Product Solution
orr
z
Th,
L0
2 HBC
Fig. 3.5
LTh,
Th,
Th,
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)()(),( zZrRzr (a)
01 2R
dr
dRr
dr
d
r kk (b)
022
2
kkk Z
dz
Zd (c)
(ii) Selecting the sign of the 2k
r-variable has 2 HBC
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0222
22
kkkk Rr
dr
dRr
dr
Rdr (d)
022
2
kkk Z
dz
Zd(e)
0020
2
dr
dR
dr
Rdr (f)
For :02k
02
0
2
dz
Zd(g)
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(iii) Solutions to the ODE
(d) is a Bessel equation:
,0BA 0n,1C ,k
D
Since 0n andD
is real
rYBrJArRkkkkk 00
)( (h)
Solutions to (e), (f) and (g):
zDzCzZkkkkk
coshsinh)( (i)
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000BlnrAR (j)
000 DxCZ (k)
Complete solution
)()()()(),(1
00 zZrRzZrRzr kk
k(l)
(iv) Application of Boundary Conditions
BC (1)
)0(0
Y and 0lnNote:
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00
ABk
BC (2)
)()(00,0
0
rhJrJdr
dk
kzrk
(m)BirJ
rJr
k
kk )(
)(
00
010
00
B
khrBi 0 (n)
BC (2)
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Since ,000
BA 0n
LLBiL
LLBiLCD
kkk
kkkkk cosh)(sinh
sinh)(cosh
1cosh)(sinh
sinh)(cosh[sinh),(
k kkk
kkkkk LLBiL
LLBiLzazr
)](cosh0
rJzkk
(3.20)
BC (3)
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BC (4)
)()( 01
rJakrf kkk
k (o)
(v) Orthogonality
)(0
rJ k is a solution to (d) with 2HBC in .r
and
Comparing (d) with eq. (3.5a) shows that it is a Sturm-
,/11
ra .13
aLiouville equation with ,02
a
Eq. (3.6):
rwp and 0q (p)
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)(0
rJk
and )(0
rJi
are orthogonal with respect to
ka.)( rrw Applying orthogonality, eq. (3.7), gives
)(])()[
)()(2
)(
)()(
0
2
0
2
0
2
0
0
0
0
2
0
0
0
0
0
0
rJkhrrk
drrrJrf
drrrJk
drrrJrf
a
kk
r
kk
r
kk
r
k
k
(q)
Interface temperature: set 0z in eq. (3.20)
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)(cosh)(sinh
sinh)(cosh)0,(
0
1
rJLLBiL
LLBiLar
k
k kkk
kkkk
(r)
(5) Checking
Dimensional check: Units of ka
Limiting check: If 0 or 0 then TzrT ),(
(6) Comments
)(xw is not always equal to unity.
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3.8 Integrals of Bessel Functions
0
0
2 )(
r
knndrrrJN (3.21)
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Table 3.1 Normalizing integrals for solid cylinders [3]
Boundary Condition at
(3.8a):
(3.8b):
(3.8c):
0
0
2 )(
r
knn drrrJN0r
0)(0
rJkn
0)(
0
dr
rdJkn
)()(
00 rhJ
dr
rdJk
knkn
2
0
2
2
0)(
2dr
rdJrkn
k
)()(2
10
222
02rJnr
knk
k
)()()(2
10
222
0
2
2rJnrBi
knk
k
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3.9 Non-homogeneous Differential Equations
Example 3.6: Cylinder with EnergyGeneration
L
aT
oT
r
or q
z0 q0
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 .aTFind the steady state temperature distribution.
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Solution
(1) Observations Energy generation leads to
NHDE
Use cylindrical coordinates
L
aT
oT
r
or qz0q
0
Fig. 3.6 Definea
TT to make
BC at surface homogeneous
(2) Origin and Coordinates
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(3) Formulation
(i) Assumptions(1) Steady state
(2) 2-D
k(3) Constant and
(ii) Governing Equations
Definea
TT
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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
TTr0
)0,(
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(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)
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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)
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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
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Thus
0)(
dz
Ld(d-1)
BC (4)
aTTr
0)0()0,(
Let
(c-4)0)0,(r
Thus
aTT0)0( (d-2)
Solution to (d)
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FEzzk
qz 2
2)( (e)
(i) Assumed Product Solution
)()(),( zZrRzr (f)
(f) into (c), separating variables
(g)022
2
kkk Z
dz
Zd
0222
22
kkkk Rr
dr
Rdr
dr
Rdr (h)
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(ii) Selecting the sign of the 2k
terms
(i)022
2
kkk Z
dz
Zd
0222
22
kkkk Rr
dr
Rdr
dr
Rdr (j)
For,0k
(k)02
0
2
dz
Zd
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00
2
0
22
dr
Rdr
dr
Rdr (l)
(iii) Solutions to the ODE
(m)kkkkk
BAzZ cossin)(
(j) is a Bessel equation with ,0nBA ,1C
.iD k
(n))()()( 0 rKDrICzR kkkokk
S l i (k) d (l)
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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)
00
CDk
BC ( 4) t ( ) d ( )
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BC (c-4) to (m) and (o)
00
BBk
BC (c-3) to (m) and (o)
00
A
Equation for k
,0cos Lk
or2
)12( kL
k,...2,1k (r)
With 000 BA0
00ZR
The solutions to ),( zr become
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zrIazrk
kkk
10
sin)(),( (s)
BC (d-1) and (d-2)
TTF0
kLqE / and
)()/()/(22)( 02
2
TTLzLzk
Lqz (t)
BC (c-2)
)()/()/(22 0
2
2
TTLzLzkLq
zrIak
k
kk
1
00sin)(
(u)
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(v) Orthogonality
zksin are solutions to equation (i).
and
Comparing (i) with eq. (3.5a) shows that it is a Sturm-
.13
aLiouville equation with ,021
aa
Eq. (3.6) gives
0z and ,Lz2 HBC at zk
sin are orthogonal
with respect to 1w
rwp and 0q
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dzTTLzLz
k
Lqk
L
sin)()/()/(2
200
22
L
kkkdzrIa
0
2
00sin)(
Evaluating the integrals and solving fork
a
20
/)()()(
2kao
okkk kqTTrILa (v)
Solution to )(
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Solution to ),( zr
(5) Checking
Limiting check: 0q and .0
TTa
Dimensional check: Units of kLq /2 and of ka insolution (s) are in C
(w)a
TzrTzr ),(),(
22
0)/()/(2
2)( LzLz
k
LqTT
a
zrIa kk
kk1
0 sin)(
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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
Solution:
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Solution:
),(),(),(),(),(4321
yxTyxTyxTyxTyxT
)(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
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