1 chapter 4 transient conduction neglect spatial variation: 4.1.1 criterion for neglecting spatial...
TRANSCRIPT
1
CHAPTER 4 TRANSIENT CONDUCTION
Neglect spatial variation:
4.1.1 Criterion for Neglecting Spatial Temperature Variation Cooling of wire by surface convection:
(4.1))(tTT
= temperature drop across radius T
4.1 Simplified Model: Lumped-Capacity Method
2
(1) radius (2) conductivity k
(3) heat transfer coefficient h
Define Biot number Bi
Factors affecting T
Neglect for smallTk
h
(4.2)k
hBi
Neglect T if
(4.3)< 0.1 k
hBi
3
• Irregular shape:
Physical significance of Bi:
Determining : • Long cylinder: or • Large plate of thickness L, convection on both sides:
2/L
sA
V= (4.4)
resistanceexternal
resistanceinternalBi
4
• Volume:V
Determine: Transient temperature
Assume:
Conservation of energy:
4.1.2 Lumped-Capacity Analysis• Surface area: sA
• Initial temperature: iT • Convection at surface: Th, • Energy generation: q
)(tTT (a)
(1.6)EEEE outgin
5
(c), (d) and (e) into (b)
EEE outg (b)
Assume that no energy is added )0( inE
Neglect radiation, is by convectionoutE
)( TThAE sout (c)
qVEg (d)
dt
dTVcTThAqV ps )(
dt
dTVcE p (e)
Energy generation gE
Rate of energy change E
6
Separating variables and rearranging
• Equation (4.5) is the lumped-capacity equation for all geometries
• Limited to:
(2) No radiation
(3) Incompressible material
Initial condition:
dtc
q
TTVqhA
dT
ps
1))(/(
(4.5)
(1) Uniform q
Bi < 0.1 (4)
iTT )0( (4.6)
7
Integrate (4.5), use (4.6)
NOTE
(1) Temperature decay is exponential
(2) Solution is independent of k
Integration of (4.5): Assume:
(5) Constant h and T
(6) Constant q
(3) Steady state temperature:set in (4.7):t
)(exp
)(1
TThA
Vqt
Vc
hA
TThA
Vq
TT
TT
isp
s
isi
(4.7)
8
Physical significance of (4.8)shA
VqTT
)( (4.8)
Special Cases:
• Case (i): No energy generation. Set in (4.7)0q
tVc
hA
TT
TT
p
s
i
exp (4.9)
Steady state: TT )(
• Case (ii): No convection. Set h=0 in (4.7)
9
Physical significance: No Steady state
tc
qTT
pi
(4.10)
• Case (iii): Initial and ambient temperatures are thesame. Set in (4.7)TTi
t
Vc
hA
hA
VqTT
p
s
s
exp1 (4.11)
Steady state: setting in (4.11) gives (4.8)t
10
• Bi > 0.1
• Governing equation: PDE
• Solution by separation of variables
• Two HBC and one initial condition
Example 4.1: Plate with Surface Convection Thickness: 2L Initial temperature:
Assume f(x) is symmetrical
L
x
L
0
Fig. 4.1
Th
Th
HBC2
4.2 Transient Conduction in Plates
)()0,( xfxTTi Heat exchange is by convection:0t
Determine ),( txT
11
(1) Observations • Temperature is symmetrical
• Convection BC is NH
(2) Origin and Coordinates
(3) Formulation (i) Assumptions.
(ii) Governing Equations. Let
One-dimensional, constant k, , Th,
TtxTtx ),(),( (a)
tx
12
2
(4.12)
12
(iii) Independent Variable with Two HBC:x-variable
(iv) Boundary and Initial Conditions
L
x
L
0
Fig. 4.1
Th
Th
HBC2 (1) 0),0(
x
t, H
IC (3) Txfx )()0,( , NH
(4) Solution (i) Assumed Product Solution
, H),(),(
tLhx
tLk
(2)
13
(b) into (4.12)
)()(),( txXtx (b)
0)(22
2
xXdx
Xdn (c)
02 ndt
d (d)
(ii) Selecting the Sign of the Terms2n
0)(22
2
xXdx
Xdn (e)
14
(iii) Solutions to the ODE
(f)02 ndt
d
02 n 000
Xyields
xBxAxX nnnnn cossin)( (g)
)exp()( 2 tCtnn
(h)
(iv) Application of Boundary and Initial Conditions
BC (1): An = 0
xBxX nnn cos)( (i)
15
where
Substituting into (b)
IC (3):
BC (2): gives n
BiLL nn tan (4.13)
khLBi / (j)
xeaTtxTn
n
t
nn
0
cos ),(2
(4.14)
0ncos)(
nn xaTxf (k)
16
(v) Orthogonality
Eq. (3.6):
in equation (k) are solutions to (e). Comparing (e) with eq. (3.5b)
xncos
021 aa and 13 a
p = w = 1 and q = 0
BC at x = 0 and x = L are homogeneous.
w(x) = 1. Multiply both sides of (k) by integrate and apply orthogonality.
Therefore, xncos are orthogonal with respect to ,cos dxx
m
))(cos(sin
cos)(2
nnn
0 nn
LLL
xTxfa
L
n
(4.15)
17
Special case: Uniform initial temperature iT
(5) Checking Dimensional check
: Limiting check Plate is initially at T Differential Equation
Boundary and initial conditions
(6) Comments
(ii) Initial temperature f(x) need not be symmetrical
))(cos(sin
))(sin(2
nnn
n
LLL
LTTa i
n
(4.16)
(i) Both separated equations have term2n
18
Example 4.2: Plate with Energy Generation,Specified Surface Temperature
Thickness = L
4.3 Non-homogeneous Equations and Boundary Conditions
Initial temperature: iTintroduce
:0t q At
Surfaces at and 1T 2T
Determine: ),( txTFig. 4.2
2T1TL
q x0
19
(1) Observations • Asymmetry
• x-variable has two NHBC
• The heat equation is non-homogeneous
(2) Origin and Coordinates
(3) Formulation (i) Assumptions.
(ii) Governing Equations: eq. (1.8)
One-dimensional, constant qk ,,
t
T
k
q
x
T
1
2
2
(4.17)
20
(iii) Independent Variable with Two HBC: x-variable
(iv) Boundary and Initial Conditions
Fig. 4.2
2T1TL
q x0
(1) 1),0( TtT
(2) 2),( TtLT
(3) iTxT )0,(
(4) Solution)(),(),( xtxtxT (a)
21
Split (b). Let
(a) into BC (1)
(a) Into eq. (4.17)
tk
q
dx
d
x
12
2
2
2
(b)
(c)tx
12
2
(d)02
2
k
q
dx
d
1)0(),0( Tt
22
0),0( t (c-1)
1)0( T (d-1)
(a) into BC (2)
IC (3)
Solution to (d)
0),( tL (c-2)
2)( TL
)()0,( xTx i (c-3)
(e)212
2)( CxCx
k
qx
23
(i) Assumed Product Solution
(f) into (c), separating variables
)()(),( txXtx (f)
022
2
nn
n Xdx
Xd (g)
(ii) Selecting the Sign of the Two Terms2n
0)(22
2
xXdx
Xdnn
n (i)
(h)02 nn
n
dt
d
24
(iii) Solutions to the Ordinary Differential Equations
(j)02 nn
n
dt
d
for ,02 n
000 X
xBxAxX nnnnn cossin)( (k)
)(exp)( 2 tCtnnn
(l)
(iv) Application of Boundary and Initial Conditions
BC (c-1) 0n
B
25
BC (c-2)
Substituting into (f) and summing
...3,2,1,,0sin nnLL nn (m)
1
2 sin )][exp(),(n
nnn xtatx (n)
Solution to :)(x
xxLk
q
L
xTTTx )(
2)()( 121
(o)
NH initial condition (c-3):
Return to BC (d-1) and (d-2) give and ),(x .2
C1
C
26
(v) Orthogonality
(p)
1
sin)(n
nni xaxT
in (p) are solutions (i). Compare (i) with eq. (3.5a): Sturm- Liouville equation with
xnsin.1)( xw
Multiply both sides of (p) by apply orthogonality, gives
,sin dxxm
integrate and
na
L
n
Lni
ndx
dxxxTa
0
2
0
sin
sin)]([
(q)
27
(o) into (q), evaluate integrals and use (m)
Complete solution
(r)
k
Lq
n
nkLq
TTn
TTn
a
n
n
i
n
n
2
3
222
121
)(
2)1)(2()2/(
)()1(2
)()1(1
2
)/sin()/exp(
)(2
)(),(
1
222
121
LxnLtna
xxLk
q
L
xTTTtxT
nn
(s)
28
(5) Checking Dimensional check
Limiting check: Temperature at steady state
(6) Comments
(i) Non-uniform initial temperature
(ii) Non-homogeneous boundary conditions
29
4.4 Transient Conduction in Cylinders
Example 4.3: Cylinder with Energy Generation
Long cylinder
Surface cooling by convectionTh,
Th,r
0q q
Fig. 4.3
2 HBC
Initial temperature i
T
Energy generation :0t q
Determine ),( trT
(1) Observations • Symmetry
30
• Heat equation is non-homogeneous
(2) Origin and Coordinates
(3) Formulation (i) Assumptions
(4) Negligible end effect
(ii) Governing Equations
makes convection BC homogeneous • , TT
(1) One-dimensional,
(2) Uniform h and ,T (3) Constant ,,k
31
TtrTtr ),(),(
Heat equation (1.11)
(iii) Independent Variable with 2 HBC: r-variable
(iv) Boundary Conditions
tk
q
rrr
11
2
2
(4.18)
or (1) 0),0(
r
t
finite),0( t Th,
Th,r
0q q
Fig. 4.3
2 HBC
32
(4) Solution
(2) ),(),(
trhr
trk o
o
IC (3) TTr i)0,(
)(),(),( rtrtr (a)
(a) into eq. (4.18)
tk
q
rd
d
rdr
d
rrr
111
2
2
2
2 (b)
Split (b)
33
(a) into BC (1)
Let
(c)trrr
11
2
2
01
2
2
k
q
rd
d
rdr
d (d)
0)0(),0(
dr
d
r
t
),0( t (c-1)0),0(
r
tor finite
34
BC (2):
Initial condition
Integrating (d)
(c-2)),(),(
trhr
trk o
o
0)0(
dr
d(d-1)
)()(
oo rh
rd
rdk
(d-2)
)()()0,( rTTr i (c-3)
(e)212 ln
4)( CrCr
k
qr
35
(f) into (c), separating variables
(i) Assumed Product Solution
(f))()(),( trRtr
01 2
2
2
kkkk R
dr
dR
rdr
Rd (g)
02 kkk
dt
d (h)
(ii) Selecting the Sign of the Terms2k
01 2
2
2
kkkk R
dr
dR
rdr
Rd (i)
36
(iii) Solutions to the ODE
(j)02 kkk
dt
d
For :02 k 000 R
)()()( 00 rYBrJArR kkkkk (k)
)exp()( 2 tCt kkk (l)
(iv) Application of Boundary and Initial Conditions
BC (c-1) and (c-2)0kB
37
(k) and (l) into (f), summing
Initial condition (c-3)
and
(m))()()( 10 okokok rJrrJBi
)()exp(),(1
02 rJtatr k
kkk
(n)
Return to solution (e) for BC (d-1) and (d-2) give
1C and 2C),(r
h
rqrr
k
qr o
o 2)(
4)( 22
(o)
38
00 )()()(
kkki rJarTT (p)
(v) Orthogonality
Eq. (3.6) gives
Functions in (p) are solutions to (i). Comparing
(i) with eq. (3.5a) shows that it is a Sturm-Liouville equation with
)(0 rJ k
and0,/1 21 ara 13 a
HBC at and therefore are orthogonal with respect to
0r ,o
rr )(0 rJ k.)( rrw
and rwp 0q
39
Multiply both sides of (p) by integrate and
apply orthogonality
,)(0 drrrJ i
(q)
o
o
r
k
r
kikk
rdrrJ
rdrrJrTTa
0
20
0 0
)(
)()()(2
(o) into (q) and evaluate the integrals
)()(
1
2
1
)(1
)()(
)())((2
0
2
22
2
22221
0
oki
o
ok
i
o
okok
okokik
rJTTk
rq
rBi
TTk
rq
rJrBi
rJrTTa
(r)
40
Complete solution
)/1()(4
),( 222
oi
o
irr
TTk
rq
TT
TtrT
)()exp()(
1
)(21
02
2
rJtaTTBiTTk
rqk
kkk
ii
o
(4.19)
(5) Checking Dimensional check
Limiting check:
(ii) Steady state
(i) and TTi 0q
41
(6) Comments
Initial temperature (i) )()0,( rfrT (ii) Two parameters: the Biot number Bi and
)(/)( 2 TTkrq io
4.5 Transient Conduction in Spheres
Example 4.4: Sphere with Surface Convection
Determine: transient temperature
Fig. 4.4
HBC2Th,
ror0
Initial temperature i
T Convection at surface :0t
42
(1) Observations
(2) Origin and Coordinates
• Convection BC is non-homogeneous
• One-dimensional, transient ),( trT
• Define TT to give r-variable 2HBC
(3) Formulation (i) Assumptions
(1) One-dimensional
(2) Constant ,k
(ii) Governing Equations
43
Eq. (1.13) gives
(iii) Independent Variable with 2 HBC: r-variable
(iv) Boundary and Initial Conditions
TtrTtr ),(),( (a)
trr
rr
11 2
2 (4.20)
or (1) 0),0(
r
t
finite),0( tFig. 4.4
HBC2Th,
ror0
44
(4) Solution (i) Assumed Product Solution
(b) into eq. (4.20), separate variables
(2) ),(),(
trhr
trk o
o
IC (3) TTr i)0,(
(b))()(),( trRtr
02 222
22 kk
kk Rrdr
dRr
dr
Rdr (c)
02 kkk
dt
d (d)
45
(iii) Solutions to the ODE
Use eqs. (2.32) and (2.33)
(ii) Selecting the Sign of the Terms2k
(e)02 222
22 kk
kk Rrdr
dRr
dr
Rdr
02 kkk
dt
d (f)
gives02 k
000 R
)()()( 2/12/12/1 rJBrJArrR kkkkk
)cossin(1
)( rBrAr
rR kkkkk (g)
46
Solution to (f)
(iv) Application of Boundary and Initial Conditions
BC (1)
BC (2)
(g) and (h) into (b)
(h))exp()( 2 tCt kkk
0kB
,tan)1(okok
rrBi k = 1, 2, … (i)
khrBi o /
47
1
2 sin)exp(),(),(
k
kkk r
rtaTtrTtr
(4.21)
IC (3):
(j)
1
sin
k
kki r
raTT
(v) Orthogonality
,/21 ra ,02 a 13 a
,2rp ,0q 2rw
The characteristic functions in (j) are
it is a Sturm-Liouville problem with
)/1( r rksin solutions to (e). Comparing (e) with eq. (3.5a) shows that
48
are orthogonal with respect to
Multiply (j) by
orthogonality
)/1( r rksin .)( 2rrw ,))(sin/1( 2drrrr i integrate and invoke
(5) Checking
Dimensional check
(k)
]cossin[
cossin)(2
sin
sin)(
0
2
0
okokokk
okokoki
r
k
r
ki
k
rrr
rrrTT
drr
drrrTTa
o
o
49
(ii) Steady state
(6) Comments
Limiting check:
(i) Initial temperature = T
(i) Non-uniform initial temperature, )()0,( rfrT (ii) Solution is expressed in terms of a single
parameter: Biot number Bi
50
Examples:
• Solar heating• Reentry aerodynamic heating
• Reciprocating surface friction
• Periodic oscillation of temperature or flux
• Limitation: Linear equations
4.6 Temperature Dependent Boundary Conditions: Duhamel’s Superposition Integral
51
Duhamel’s superposition integralUse solution of an auxiliary problem with constantboundary condition to construct the solution to the
same problem with time dependent condition
4.6.1 Formulation of Duhamel’s Integral
Initial temperature: 0iT:0
it boundary is at 01T
:i
t boundary is at 02T01T
02TT
1
0102 TT
t11.4.Fig
52
SolutionDecompose into two problems: One starts and the
second at Each problem has a constant BC
0t.
1t
),( txT Solution to auxiliary problem, constant surfaceTemperature of magnitude unity.Thus
(b)),(),( 011 txTTtxT
),(),(),( 21 txTtxTtxT (a)
Second problem starts at with surface temperature
Solution to 1
).(0102
TT :),(2
txT
53
Adding (b) and (c)
• Generalize to arbitrarily boundary condition
(c)),()(),( 101022 txTTTtxT
),()(),(),( 1010201 txTTTtxTTtxT (d)
= surface temperature, or)(tF
= ambient temperature, or
= heat flux
to many problems, each havinga small step change, in B.C.
• Solution: superimpose solutions
F
t
)(tF
F
tFig. 4.12
54
solution
• BC for first problem: F(0) starts at time t = 0,
t
)(tF
F
tFig. 4.12
),()0(),(0 txTFtxT (e)
• Contribution of the ith problem,
BC is )( iF
),()(),( iii txTFtxT
Adding all solutions
(f)
n
iii txTFtxTFtxT
1
),()(),()0(),(
55
or
i
n
ii
i
i txTF
txTFtxT
1
),()(
),()0(),(
,n , di
,)()(
d
dFF
i
i
Integration by parts with 0)0,( xT
(4.29)
t
dt
txTFtxT
0
),()(),(
t
dtxTdt
dFtxTFtxT
0),(
)(),()0(),(
(4.28)
56
NOTE:
• Duhamel’s method applies to linear equations
• Equations (4.28) and (4.29) apply to any coordinate system
• 0iT
Define • If :0i
T iTT
equal to unity • Auxiliary solution: ),( txT is based on a constant BC
• Method applies to:
(1) )(tq (2) Lumped-capacity models )(tT
• Integrals in (4.28) and (4.29) are with respect to the dummy variable
57
4.6.2 Extension to Discontinuous Boundary Conditions
and tat
)(1 tF
)(2 tF)(tF
Fig. 4.13
If is discontinuous, modify Duhamel’s integral. )(tF
atttFtF 0)()( 1
atttFtF )()( 2
• :a
tt use )()(1
tFtF in (4.28) & (4.29)
• :a
tt modify (4.28) and (4.29)
and with discontinuity)(1 tF )(2
tF at :at
58
Result:
For (4.28) add another solution, :),( txTa
),()]()([),( 12 aaaa ttxTtFtFtxT
Integration by parts with gives0)0,( xT
dtxTd
dFdtxT
d
dF
ttxTtFtF
txTFtxT
t
at
at
aaa
),()(
),()(
),()]()([
),()0(),(
2
0
1
12
1
(4.30)
59
dt
txTF
dttxT
F
ttxTtFtFtxT
t
t
taaa
a
a
),()(
),()(
),()()(),(
2
01
12
(4.31)
• Extension to several discontinuities
60
4.6.3 ApplicationsExample 4.5: Plate with Time Dependent
Surface TemperatureBC and IC:
(1) 00 ),( tT
(2) AttftLT )(),(
(3) )0,(xT 0iT
Determine ),( txT
(1) Observations
0 x
xAT 0T 0)0,( xT
L
Fig. 4.14
• One BC depends on time
0iT •
61
(2) Origin and Coordinates
(3) Formulation (i) Assumptions
(1) 1-D
(ii) Governing equation
(iii) Boundary and Initial conditions
t
T
x
T
1
2
2 (a)
(1) 0),0( tT
(2) constant ,,k
62
0 x
xAT 0T 0)0,( xT
L
Fig. 4.14
(2) tAtLT ),(
(3) 0)0,( xT
(4) Solution
Eq.(a) is linear, apply Duhamel’s integral, eq .(4.28)
tAtF )(
),( txT solution to problem with BC (2) replaced by1),( tLT
t
dtxTdt
dFtxTFtxT
0),(
)(),()0(),(
(4.28)
63
Use result of Example 4.2: set
)/sin(])/(exp[)1(2
),(
1
2 LxntLnnL
x
txT
n
n
(b)
where
nn n
a )1(2 (c)
Find and )0(F ddF /)(
,)( tAF 0)0(, FAd
dF
(d)
0,01
i
TTq 12
Tand
64
Substituting (b) and (d) into eq. (4.28)
dtLnLxnn
A
dL
xAtxT
n
tn
t
)]()/(exp[)/sin()1(2
),(
10
2
0
(e)
Evaluating the integrals in (e)
)]/exp(1[)/sin()1(2
),(
222
133
2LtnLxn
n
LA
tL
xAtxT
n
n
(f)
65
(5) Checking
(i) A = 0(ii) t
Dimensional check
Limiting check:
Example 4.6: Lumped-Capacity Method with Time Dependent Ambient Temperature
Metal foil, 1.0Bi
Initial temperature iT
Oven temperature starts at and changes with time asiT
66
(1) Observations
io TtTtT )]exp(1[)( (a)
Determine: )(tT
• Oven temperature is time dependent
• Use Duhamel’s method if lumped-capacity equation is linear
appears in DE • T
• 0iT
(2) Origin and Coordinates
(3) Formulation
67
(ii) Governing Equation
Define
(i) Assumptions
(1) Bi < 0.1
(2) constant hck p ,,,
Set in (4.5)0q
(a)dtVc
hA
tTT
dT
p
s
)(
iTtTt )()( (b)
Oven temperature
iTtTt )()( (c)
68
Substitute into (a)
Initial condition
(4) Solution
Apply Duhamel’s integral
dtVc
hA
tt
d
p
s
)()(
(d)
0)0( (e)
t
dtxTdt
dFtxTFtxT
0),(
)(),()0(),(
(4.28)
69
Constant oven temperature equal to unity:
Rewrite (a)
= solution to auxiliary problem of constantoven temperature equal to unity and zeroinitial temperature
)(t
(f))]exp(1[)()()( tTTtTttF oi
1 iTT (g)
dtVc
hAd
p
s
1
(h)
= time dependent oven temperature)(tF
70
Initial condition
0)0( (i)
Integrate and use IC
tVc
hAt
p
s
exp1)( (j)
Determine ddF /)(
(k))exp(
oTd
dF
Substituting (j) and (k) into eq. (4.28), use 0)0( F
71
(l)
dtVc
hATt
t
p
so
0)(exp1)exp()(
Performing the integration
(5) Checking
Dimensional check
)/exp()exp()/(
)exp()(
VcthAtVchA
T
tTTTtT
psps
o
ooi
(m)
Limiting check:
t
72
Differential equation check
Initial condition check: 0t
4.7 Conduction in Semi-infinite Regions: The Similarity Method
• Limitations of separation of variables: 2 HBC separated by a finite distance
• Method fails for semi-infinite regions
• Alternate approach: Similarity method
Combine two independent variables into one andtransform PDE to ODE
• Basic idea:
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(i) Semi-infinite region
• Limitations:
(ii) Limitations on BC and IC
Fig. 4.15
oT
0 x
iT
Example: Semi-infinite plate, uniform initial Surface is suddenly temperature .
iT
maintained at temperature .o
T
t
T
x
T
1
2
2
(a)
BC (1) oTtT ),0(
(2) iTtT ),(
IC (3) iTxT )0,(
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Define:
io
i
TT
TT
(b)
Substitute into (a)
BC and IC become
tx
12
2
(c)
Fig. 4.15
oT0 x
iT
(1) 1),0( t
(2) 0),( t(3) 0)0,( x
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Assume that x and t can be combined into a singlevariable ),( tx
)(),( tx (d)
Try:
Conditions on Must transform PDE, BC and IC:),( txin terms of and eliminate x and t.
t
x
4 (4.32)
Using (4.32), construct the derivatives in (c)
td
d
xd
d
x
4
1
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2
2
2
2
4
1
4
1
xd
d
txtd
d
d
d
x
dd
tdd
ttx
tx
dd
tdd
t
21
42
422/3
(c) becomes
022
2
d
d
d
d (4.33)
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NOTE:
(1) x and t are eliminated and replaced by(2) Solution to PDE (c) must satisfy 3 conditions.
(3) Solution to ODE (4.33) can only satisfy two conditions.
Transformation of BC and IC:
Using t
x
4
ttx ,0 transforms to 0ttx , transforms to 0, txx transforms to
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BC and IC transform to:
NOTE: One BC and IC coalesce. Fig. 4.15
oT0 x
iT
(1) 1)0( (2) 0)( (3) 0)(
Solution to (4.33): Separating variables
ddd
ddd2
/
)/(
Integrating
Ad
dlnln 2
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or
Integrating and using first BC
dAd e2
dAd e
2
01
dA e
2
01 (e)
Define
de
2
0
2erf (f)
Where00erf
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and
Equation (e) becomes
BC (2) gives B = -1.
(g) becomes
1erf
erf1)( B (g)
erf1)( (4.33a)
tx
tx
4
erf1),( (4.33b)
Derivative of :erf 22
)(erf
ed
d (4.34)
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