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© meg/aol ‘02 Transform Methods where Kernel of the transform Laplace kernel Independent variable (time). Laplace transform converts an object function, F(t), to its image function, F(p). ~
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© meg/aol ‘02
Solutions To The Linear Diffusion Equation
Martin Eden GlicksmanAfina Lupulescu
Rensselaer Polytechnic InstituteTroy, NY, 12180
USA
© meg/aol ‘02
Outline• Transform methods
• Linear diffusion into semi-infinite medium– Boundary conditions
– Laplace transforms
• Behavior of the concentration field
• Instantaneous planar diffusion source in an infinite medium
• Conservation of mass for a planar source– Error function and its complement
– Estimation of erf(x) and erfc(x)
• Thin-film configuration
© meg/aol ‘02
Transform Methods
whereK t, p Kernel of the transform
e pt Laplace kernel
t Independent variable (time).
L F t F t K t, p d t
a
b
˜ F p
L F t Laplace transform converts an object function, F(t), to its image function, F(p). ~
© meg/aol ‘02
Linear diffusion into a semi-infinite
medium
time
Initial state
Final state
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
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Boundary Conditions
1) Initial state:C = 0, for x > 0, t = 0.
2) Left-hand boundary: At x = 0, C0 is maintained for all t > 0.
Ct
D 2Cx2
• The diffusion equation is a 2nd-order PDE and requires two boundary or initial conditions to obtain a unique solution.
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Laplace Transform of the Diffusion Equation
object functionC x, t
e ptLaplace kernel
e pt
0
2 Cx2 dt
2
x2 C e pt
0
dt d2 ˜ C d x2
˜ C x, p C x,t e pt d t0
2 Cx 2
Linear Diffusion Equation
˜ C image function
Laplace transformof C(x,t)
˜ C x, p
1D
Ct
0
e pt
0
2 Cx2 dt
Laplace transform ofthe spatial derivative.
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Laplace Transforms
The Laplace transform of Fick’s second law when C(x,0)=0
d2 ˜ C d x2
pD
˜ C 0
1D
e pt
0
Ct
dt
1D
Ce pt0
p Ce pt
0
dt
pD
˜ C
integration by parts
Ce pt0
0 C t 0 0 boundary
condition
1D
e pt
0
Ct
dt
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Laplace Transforms
˜ C 0 C0e pt
0
dt C0
pe pt
0
˜ C 0 C0
p
Transform of the boundary condition:
˜ C C0
pe
pD
xGeneral transform solution:
˜ C x C0
pe
pD
x
The particular transform solution for the image function arises from the negative root, because the positive root leads to non-physical behavior, . ˜ C (x)
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Laplace Transforms
C x, t C0 erfc x2 Dt
The concentration field associated with the image field is found by inverting the transform either by formal means, a look-up table,or using a computer-based mathematics package.
erfc z 1 erf z 1 2
e 2
d0
z
The error function, erf (z) , and its complement, erfc (z) , are defined
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Estimation of the Error Function
erf(z) erf( z) 2
z z3
31 !
z5
52 !
z7
73 ! ...
, ( z 1)
• For small arguments:
erf(z) 1 e z 2
z 1
12z2 ...
, (z )
• For very large arguments:
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Estimation of the Error Function
erf(z)
1.12838z, (0 z 0.15) 0.0198 z 1.2911 0.4262z , (0.15 z 1.5)0.8814 0.0584z, (1.5 z 2)1, (2 z)
• Piecewise approximations for restricted ranges of the argument:
erf z 1 10.278393z 0.230389z2
0.000972z3 0.078108z4
4
z 5 10 5
• Rational approximation for positive arguments, z > 0:
Useful for spreadsheet calculations.
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Error Function
-1
-0.5
0
0.5
1
-3 -2 -1 0 1 2 3
Err
or F
unct
ion,
erf
(z)
zz
Erf(
z)
Antisymmetric: erf(z)=-erf(-z)
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Complementary Error Function
0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
erfc
(z)
zz
Erfc
(z)
Non-antisymmetric: erfc(z) -erfc(-z)
© meg/aol ‘02
Linear diffusion into a semi-infinite
medium
time
Initial state
Final state
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
t
t =4
t =16
C0
t =1
t =0
x
© meg/aol ‘02
Concentration versus the similarity variable
0
0.2
0.4
0.6
0.8
1
0 0.5 1 1.5 2 2.5 3
Rel
ativ
e co
ncen
tratio
n, C
/C0
Space-time similarity variable, x/2(Dt)1/2
C x,t C0 erfc x2 Dt
Rel
ativ
e C
onc.
C/C
0
Similarity Variable, x/2(Dt)1/2
© meg/aol ‘02
Concentration field versus distance
= 2(Dt)1/2 is the “time tag”
(Note: has the units of distance!)
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10Rela
tive
Conc
entra
tion,
C(x
)/C0
Distance, x [ in units of =1]
0.20.5
1
23
510
20=100
Rel
ativ
e C
onc.
C/C
0
Distance, x [units of =1]
time
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Diffusion Penetration X*
0
0.2
0.4
0.6
0.8
1
0 2 4 6 8 10Rela
tive
Conc
entra
tion,
C(x
)/C0
Distance, x [ in units of =1]
0.20.5
1
23
510
20=100
X* = K t1/2
Rel
ativ
e C
onc.
C(x
)/C0
Rel
ativ
e C
onc.
C/C
0
Distance, x [units of =1]
© meg/aol ‘02
Penetration versus square-root of time
0
0.2
0.4
0.6
0.8
1
0 1 2 3 4 5
X*(0.8)X*(0.6)X*(0.4)X*(0.9)
Pen
etra
tion
Dis
tanc
e, X
* (C/C
0)
Time tag, 2(Dt)1/2
Pen
etra
tion
Dis
tanc
e, X
*(C
/C0)
0.40.6
0.8
0.9
© meg/aol ‘02
Instantaneous planar diffusion source in an
infinite medium
These diffusion problems concern placing a finite amount of diffusant that spreads into the adjacent semi-infinite solid.
Initial state
Final state
Tim
e
0 x
M
t = 0
-
t
t =1
t =10
© meg/aol ‘02
0 x
M
t = 0
-
t
t =1
t =10
0 x
M
t = 0
-
t
t =1
t =10
0 x
M
t = 0
-
t
t =1
t =10
0 x
M
t = 0
-
t
t =1
t =10
Instantaneous planar diffusion source in an
infinite medium
These diffusion problems concern placing a finite amount of diffusant that spreads into the adjacent semi-infinite solid. Ti
me
© meg/aol ‘02
Instantaneous planar diffusion source
pD
˜ C x, p d2 ˜ C x, p
d x2 C x, 0
D
C x,t
dx M
Application of the Laplace transform to Fick’s second law gives:
The diffusion process is subject to the mass constraint for a unit area:
t = 0, C (x, 0), for all x 0
Initial condition
C (∞, t) = 0
Boundary condition
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Instantaneous planar source
pD
˜ C x, p d2 ˜ C x, p
d x2 0
˜ C Ae p D x Be p D x
Reduction of the Laplace transform:
The general solution for which is:
˜ C Ae p Dx , x 0, B 0
˜ C Be p D x, x 0 , A 0
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Instantaneous planar source
or e pt
0
C x,t dx dt0
M20
e ptdt
or Bp
De
pD
x
0
M2 p
and so B M
2 pD
C x,t 0
dx M2
Mass constraint for the field:
L C x,t d x
0
L M
2
Laplace transform the mass constraint:
˜ C x, p d x0
M2p
The integral constraint for the image function is:
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Instantaneous planar source solution
˜ C x, p M
2 De
xD
p
p
Laplace transforms table shows, L-1 e a p
p
1 t
e a24t
The transform solution
where a = x / (D)1/2.
L-1 ˜ C C x,t = M2 D
L-1 p 12e a p
Inverting the transform solution
C x,t = M
2 Dte
x2
4 Dt .
Diffusion solution
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0
0.5
1
1.5
2
-3 -2 -1 0 1 2 3
C(x
)/M [
leng
th]-1
x [distance]
.075
.05
Dt=0
0.025
0.25
13
C (x
) M [l
engt
h] -1
x [distance]
Normalized plot of the planar source solution
time
© meg/aol ‘02
Conservation of mass for a planar source
x
dx 1C x,t M
(x), thus
C x,t M
d x
1
1
e u2
du
1Gauss’s integral
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Thin-film configuration
Thin-film diffusion configuration is used in many experimental studies for determining tracer diffusion coefficients. It is mathematically similar to the instantaneous planar source solution.
C x,t = M
2 Dte
x2
4 Dt .C x, t Mthin film
Dte x 2
4 Dt
where Mthin-film represents the instantaneous thin-film source “strength.”
2Mthin-film=M
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Procedure for Analysis of Thin-Film Data
C x,t Mthin film
Dte x 2
4 DtTake logs of both sides of
ln C x, t lnMthin film
Dt
x 2
4Dt
A plot of lnC versus x2 yields a slope=-1/4Dt
© meg/aol ‘02
Thin-Film Experiment
0
1 105
2 105
3 105
4 105
5 105
0 10 20 30 40 50
Counts 100s
Cou
nts
in 1
00 s
Distance, [microns]
Geiger counter data after microtoning 25 slices from the thin-film specimen.
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Log Concentration versus x2
0.3
0.4
0.5
0.6
0.7
0.8
0 0.5 1 1.5 2 2.5 3 3.5 4
log e R
adio
activ
ty, l
nA*
x2, [cm2 25104]
Slope=-1/4Dt
© meg/aol ‘02
Exercise1. Show by formal integration of the concentration distribution, C(x,t), given by eq.(3.31), that the initial surface mass, M, redistributed by the diffusive flow is conserved at all times, t>0.
C(x,t)d x
M
2 Dte
x 2
4 Dt d x
The mass conservation integral is given by:
C(x, t)d x
2 M2 Dt
e
x2
4Dt d x0
Symmetry of diffusion flow allows:
© meg/aol ‘02
Exercise
Introduce the variable substitution u= x /2 (Dt )1/2 and obtain:
C(x,t)d x
MDt
e u2
2 Dt d u0
Simplification yields:
C(x, t)d x
M 2
e u2
du0
The diffusant mass is conserved according to:
C(x,t)d x
M erf M
© meg/aol ‘02
Exercise2a) Two instantaneous planar diffusion sources, each of “strength” M, are symmetrically placed about the origin (x=0) at locations =1, respectively, and released at time t=0. Using the linearity of the diffusion solution develop an expression for the concentration, C(x,t), developed at any arbitrary point in the material at a fixed time t>0. Plot the concentration field as a function of x for several fixed values of the parameter Dt to expose its temporal behavior.
2b) Find the peak concentration at x=0 and determine the time, t*, at which it develops, if D=10-11 cm2/sec, M=25 g/cm2, and the sources are both located 1m to either side of the origin.
2c) Plot the concentration at the plane x=0 against time.
© meg/aol ‘02
Exercise
2a) C x, t M
2 Dte x 1 2 4 Dt e x1 2 4 Dt
0
0.5
1
1.5
2
-4 -3 -2 -1 0 1 2 3 4
C(x
,t)/M
x
0.5
0.05
2
0.025
Dt=0
0.1
0.25
© meg/aol ‘02
Exercise
C 0,t;ˆ x MDt
e ˆ x 2 4 Dt
2b)For two sources of strength M the concentration is:
C 0,t;ˆ x t
MDt
e ˆ x 24 Dt ˆ x 2
4Dt 2 12t
Differentiate with respect to t :
Dt ˆ x 2
2
The concentration reaches its maximum at t *:
Cmax 0,t M
ˆ x e / 2 0.484 M
ˆ x
The maximum concentration reached at x = 0
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Exercise
2c)
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0 1 2 3 4 5 6 7
C(0
,t) [g
/cm
3 ]
Time [103 sec]
D=10-11 cm/sM=5.10-4 g/cm2
© meg/aol ‘02
Key Points• Solutions to the linear diffusion equations require two initial or boundary conditions. Examples of
problems with constant composition and constant diffusing mass are demonstrated.
• Laplace transform methods were employed to obtain the desired solutions.
• Solutions are in the form of fields, C(r, t). Exposing the behavior of such fields requires careful parametric description and plotting.
• Similarity variables and time tags are used, because they capture special space-time relationships that hold in diffusion.
• Diffusion solutions in infinite, or semi-infinite, domains often contain error and complementary error functions. These functions can be “called” as built-in subroutines in standard math packages, like Maple® or Mathematica® or programmed for use in spreadsheets.
• The theory for the classical “thin-film” method of measuring diffusion coefficients was derived using the concept of an instantaneous planar source in linear flow.
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