1 chapter 3 two-dimensional steady state conduction 3.1 the heat conduction equation assume: steady...
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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
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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 with
variable 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.
variabledependent the iflinear is equation Anunitypower to raisedappear sderivative itsor
occurnot do productstheir if and
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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
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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
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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) )()()(
32
212
2
xaxadx
dxa
dx
dn
nn
0n
(3.5b) )()()( 2 xwxqdx
dxp
dx
dn
n
0
n
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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
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• 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:
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(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
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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)
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• 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
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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
• 4 BC are needed
• 3 BC are homogeneous
2 HBC
x
y
1.3Fig.
T = f(x)
T = 0 T = 0
T = 00
0
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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) Constant k
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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 HT = 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
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
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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)
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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
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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)
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020
2
dy
Yd(j)
(iii) Solutions to the ODE
xBxAxXnnnnn cossin)( (k)
yDyCyYnnnnn coshsinh)( (l)
xBAxX000
)( (m)
yDCyY000
)( (n)
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NOTE: Each product is a solution
)()(),(000
yYxXyxT (p)
)()(),( yYxXyxTnnn
(o)
The complete solution becomes
100
)()()()(),(n
nnyYxXyYxXyxT (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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Therefore0
nB
Similarly, (r) and (m) give
00B
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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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
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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)
xAxXnnnsin)(
yCyYnnnsinh)(
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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
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• 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)
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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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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)
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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
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The solution is
1
))(sinh(sin),(n
nnnyxayxT (3.10)
Dimensional check
Limiting check
Boundary conditions
Differential equations
(6) Comments
Role of the NHBC
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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
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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 at oT
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(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
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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) TTy
0),0(
y
xU
0
Th,
L
L
Fig. 3.3
HBC2
Th,oT
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(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
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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
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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) yDCyY000
)(
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100
)()()()(),(n
nnyYxXyYxXyx (m)
Complete solution:
(iv) Application of Boundary Conditions
BC (1)0
0DC
n
BC (2)
BiLLnn tan (n)
00C
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0n
ABC (3)
With
,000DC
000YX
(3.17)
yxayxn
nnn
1
22 cosexp),(
BC (4)
10
cosn
nnyaTT (o)
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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
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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)
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(5) Checking
Dimensional check
Limiting check: If 00
),(,0 TyxTqTT
(6) Comments
.0UFor a stationary plate,
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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,
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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
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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 small compared to cylinder radius
(ii) Governing Equations
(5) Uniform
h and T
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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
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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,
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)()(),( 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
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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)
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(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)
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000Bln rAR (j)
000DxCZ (k)
Complete solution
)()()()(),(1
00zZrRzZrRzr
kk
k
(l)
(iv) Application of Boundary Conditions
BC (1)
)0(0
Y and 0lnNote:
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00AB
k
BC (2)
)()(00,0
0
rhJrJdr
dk
kzrk
(m) BirJ
rJr
k
kk
)(
)(
00
010
00B
khrBi0
(n)
BC (2)
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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)
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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)
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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)
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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.
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60
3.8 Integrals of Bessel Functions
0
0
2 )(
r
knndrrrJN (3.21)
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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
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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.
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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
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64
(3) Formulation
(i) Assumptions
(1) Steady state
(2) 2-D
k(3) Constant and
(ii) Governing Equations
Define
aTT
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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,(
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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)
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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)
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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
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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)
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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)
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(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
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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
![Page 73: 1 CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation Assume: Steady state, isotropic, stationary, k = = constant (3.1)](https://reader035.vdocuments.us/reader035/viewer/2022062301/56649ea05503460f94ba32ce/html5/thumbnails/73.jpg)
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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)
00CD
k
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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
![Page 75: 1 CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation Assume: Steady state, isotropic, stationary, k = = constant (3.1)](https://reader035.vdocuments.us/reader035/viewer/2022062301/56649ea05503460f94ba32ce/html5/thumbnails/75.jpg)
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)
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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
![Page 77: 1 CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation Assume: Steady state, isotropic, stationary, k = = constant (3.1)](https://reader035.vdocuments.us/reader035/viewer/2022062301/56649ea05503460f94ba32ce/html5/thumbnails/77.jpg)
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)
![Page 78: 1 CHAPTER 3 TWO-DIMENSIONAL STEADY STATE CONDUCTION 3.1 The Heat Conduction Equation Assume: Steady state, isotropic, stationary, k = = constant (3.1)](https://reader035.vdocuments.us/reader035/viewer/2022062301/56649ea05503460f94ba32ce/html5/thumbnails/78.jpg)
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)(
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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
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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
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81