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1

UNIVERSITY OF KARLSRUHE

Institute for Hydromechanics

Mixing, Transport, and Transformation

Lecture 2:

Instantaneous Point-source Solution

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Table of Contents 2

Instantaneous Point-source Solution

Table of Contents

1. Review of the advective diffusion equation

1.1. Fickian diffusion

1.2. Advection

1.3. Advective-diffusion equation

2. Moving coordinate system

3. Similarity solution to the one-dimensional diffusion equa-

tion

3.1. Dimensional analysis

3.2. Coordinate transformation

3.3. Solution

4. Interpretation of the similarity solution

5. Point-source solution with advection

A. References


Section 1: Review of the advective diffusion equation 3

1. Review of the advective diffusion equation

Diffusion has two primary properties: it is random in nature, andtransport is from high concentration to low concentration regions,with an equilibrium state of uniform concentration.

1.1. Fickian diffusion

A static model of one-dimensional diffusion is given in Figure 1. Thismodel can also be viewed as an animation.

This motion is described by Fick’s law, given by

~q = −D

(

∂C

∂x,∂C

∂y,∂C

∂z

)

= −D∇C

= −D∂C

∂xi

. (1)

Processes that obey this relationship are called Fickian diffusion pro-cesses. In water, D is of order 2·10−9 m2/s (See Table 1).


http://www.ifh.uni-karlsruhe.de/ifh/stud/StoTra/ws/images/animations/diff_anim.gif


1.2. Advection

In a current, the molecules experience two types of motion:

• Due to diffusion, each molecule in time δt will move either onestep to the left or one step to the right (i.e. ±δx).

• Due to advection, each molecule will also move uδt in the cross-flow direction.

These processes are clearly additive and independent: we can usesuperposition.

Thus, the advective-diffusive flux vector becomes

~J = ~uC + ~q

= uiC − D∂C

∂xi

. (2)



1.3. Advective-diffusion equation

To derive the governing equation we use the control volume shown inFigure 2. From the conservation of mass, the net flux through thecontrol volume is

∂M

∂t=

∑

Jin −∑

Jout, (3)

and for the x-direction, we have

∆Jx =

(

uC − D∂C

∂x

)∣

∣

∣

∣

1

δyδz −(

uC − D∂C

∂x

)∣

∣

∣

∣

2

δyδz. (4)

After using Taylor-series expansion we obtain

∂C

∂t+

∂(uiC)

∂xi

= D∂2C

∂x2

i

, (5)

the governing Advective-Diffusion (A-D) equation.


Section 2: Moving coordinate system 6

2. Moving coordinate system

For one-dimensional diffusion in a steady flow, we can simplify (5)using a moving coordinate transformation. The new coordinates are

ξ = x − (x0 + ut) (6)

τ = t, (7)

The one-dimensional advective-diffusion equation with a steady cur-rent ~u = (u, 0, 0) is

∂C

∂t+ u

∂C

∂x= D

∂2C

∂x2. (8)

To perform the coordinate transformation, we must substitute thenew coordinates into this equation. To do this we use the chain rule,giving

∂C

∂t=

∂C

∂τ

∂τ

∂t+

∂C

∂ξ

∂ξ

∂t

∂C

∂x=

∂C

∂τ

∂τ

∂x+

∂C

∂ξ

∂ξ

∂x



From the definitions of the coordinate transformations we have

∂ξ

∂t= −u

∂τ

∂t= 1

and

∂ξ

∂x= 1

∂τ

∂x= 0.

Substituting into the governing equation (8) gives

∂C

∂τ

∂τ

∂t+

∂C

∂ξ

∂ξ

∂t+ u

[

∂C

∂ξ

∂ξ

∂x+

∂C

∂τ

∂τ

∂x

]

=

D

(

∂

∂ξ

∂ξ

∂x+

∂

∂τ

∂τ

∂x

)(

∂C

∂ξ

∂ξ

∂x+

∂C

∂τ

∂τ

∂x

)

(9)



which reduces to∂C

∂τ= D

∂2C

∂ξ2. (10)

This is just the one-dimensional pure diffusion equation in the coor-dinates τ and ξ.

Hence, we only need to find the point-source solution for purediffusion.


Section 3: Similarity solution to the one-dimensional diffusion equation 9

3. Similarity solution to the one-dimensional dif-

fusion equation

Consider the one-dimensional inviscid problem of a narrow, infinitepipe (radius a) as depicted in Figure 3. A mass of tracer, M , isinjected uniformly across the cross-section of area A = πa2 at thepoint x = 0 at time t = 0. We seek a solution for the spread of tracerin time due to pure diffusion.

The governing equation is

∂C

∂t= D

∂2C

∂x2(11)

which requires two boundary conditions and an initial condition:

• As boundary conditions, we impose that the concentration at±∞ remain zero:

C(±∞, t) = 0. (12)



• The initial condition is that the dye tracer is injected instanta-neously and uniformly across the cross-section over an infinites-imally small width in the x-direction:

C(x, 0) = (M/A)δ(x) (13)

where δ(x) is zero everywhere accept at x = 0, where it is infi-nite, but the integral of the delta function from −∞ to ∞ is 1.

Thus, the total injected mass is given by

M =

∫

V

C(x, t)dV (14)

=

∫

∞

−∞

∫

a

0

(M/A)δ(x)2πrdrdx. (15)

To find the solution we use the similarity method.



3.1. Dimensional analysis

To use dimensional analysis, we must consider all the parameters thatcontrol the solution. Table 4 summarizes the dependent and indepen-dent variables for our problem. There are m = 5 parameters andn = 3 dimensions; thus, we can form two dimensionless groups:

π1 =C

M/(A√

Dt)(16)

π2 =x√Dt

(17)

From dimensional analysis we have that π1 = f(π2), which impliesfor the solution of C:

C =M

A√

Dtf

(

x√Dt

)

(18)

where f is a yet-unknown function with arguement π2. (18) is calleda similarity solution because C has the same shape in x at all times t.



3.2. Coordinate transformation

The similarity solution represented by (18) is really just a coordi-nate transformation. The new coordinate η = x/

√Dt is called the

similarity variable. For the coordinate transformation we have:

η =x√Dt

(19)

∂η

∂t= − η

2t(20)

∂η

∂x=

1√Dt

. (21)

Thus, we must

1. Substitute C = M/(A√

Dt)f(x/√

Dt).

2. Substitute η = x/√

Dt.



Using the chain rule to compute ∂C/∂t we have:

∂C

∂t=

∂

∂t

[

M

A√

Dtf(η)

]

=M

A√

Dt

(

−1

2

)

1

tf(η) +

M

A√

Dt

∂f

∂η

∂η

∂t

= − M

2At√

Dt

(

f + η∂f

∂η

)

. (22)

Using the chain rule to compute ∂2C/∂x2 we have:

∂2C

∂x2=

∂

∂x

[

∂

∂x

(

M

A√

Dtf(η)

)]

=∂

∂x

[

∂η

∂x

∂f

∂η

M

A√

Dt

]

=M

ADt√

Dt

∂2f

∂η2. (23)



After substituting these two results into the diffusion equation, weobtain the ordinary differential equation in η

d2f

dη2+

1

2

(

f + ηdf

dη

)

= 0. (24)

3.3. Solution

To solve (24) we first make use of the identity

d(fη)

dη= f + η

df

dη. (25)

Substituting gives us

d

dη

[

df

dη+

1

2fη

]

= 0. (26)

To solve this equation requires two integrations.



1. Integrating once leaves us with

df

dη+

1

2fη = C0. (27)

It can be shown that choosing C0 = 0 satisfies both boundaryconditions and the initial condition (see Appendix B in the classnotes).

2. The second integration requires a little algebra. First, move thesecond term to the right-hand-side.

df

dη= −1

2fη. (28)

Next, like terms:df

f= −1

2ηdη. (29)

Finally, integrate both sides:

f = C1 exp

(

−η2

4

)

. (30)



To find C1 we must use the conservation of mass:

M =

∫

V

C(x, t)dV

=

∫

∞

−∞

∫

a

0

C(η)2πrdr√

Dtdη

= M

∫

∞

−∞

f(η)dη

which gives the constraint∫

∞

−∞

C1 exp(

−η

4

)

dη = 1. (31)

To solve this integral we need to make one more change of variablesto remove the 1/4 from the exponential. Thus, we introduce ζ suchthat

ζ2 =1

4η2 (32)

2dζ = dη. (33)



Solving for C1 leaves

C1 =1

2∫

∞

−∞exp(−ζ2)dζ

, (34)

and after looking up the integral in a table, we obtain C1 = 1/(2√

π).

Thus, f = exp(η2/4)/(2√

π), and our similarity solution is

C(x, t) =M

A√

4πDtexp

(

− x2

4Dt

)

. (35)


Section 4: Interpretation of the similarity solution 18

4. Interpretation of the similarity solution

Figure 4 shows the solution (35) in non-dimensional space. Theanimation also plots this solution, confirming that our solution is validfor a Fickian diffusion process.

Comparing (35) with the Gaussian probability distribution revealsthat (35) is the normal bell-shaped curve with a standard deviation,σ, of width

σ2 = 2Dt. (36)

The concept of self similarity is now also evident: the concentra-tion profile shape is always Gaussian. By plotting in non-dimensionalspace, the profiles also collapse into a single profile; thus, profiles forall times t > 0 are given by the result in the figure.


http://www.ifh.uni-karlsruhe.de/ifh/stud/StoTra/ws/images/animations/diff_anim.gif

Section 5: Point-source solution with advection 19

5. Point-source solution with advection

The final step is to substitute the coordinate transformation for themoving reference frame into the point source solution (35). In themoving coordinate system, this solution is

C(ξ, τ) =M

A√

4πDτexp

(

− ξ2

4Dτ

)

. (37)

Our coordinate transformation was ξ = x − (x0 + ut) and τ = t.Substituting into (37) gives:

C(x, t) =M

A√

4πDtexp

(

− (x − (x0 + ut)2

4Dt

)

. (38)

Figure 5 shows the schematic behavior of this solution for three dif-ferent times, t1, t2, and t3. A second animation shows this solutionin action.


http://www.ifh.uni-karlsruhe.de/ifh/stud/StoTra/ws/images/animations/AD_anim.gif

Figures and tables 20

Figure 1: Schematic of the one-dimensional molecular (Brownian) mo-tion of a group of molecules illustrating the Fickian diffusion model.The upper part of the figure shows the particles themselves; the lowerpart of the figure gives the corresponding histogram of particle loca-tion, which is analogous to concentration.

(a.) Initial (b.) Randommotionsdistribution

(c.) Finaldistribution

n

x

n

x

n

x0 0 0



Table 1: Molecular diffusion coefficients for typical solutes in waterat standard pressure and temperature (20◦C).a

Solute name Chemical symbol Diffusion coefficientb

(10−4 cm2/s)

hydrogen ion H+ 0.85

hydroxide ion OH− 0.48

oxygen O2 0.20

carbon dioxide CO2 0.17

bicarbonate HCO−

3 0.11

carbonate CO2−

3 0.08

methane CH4 0.16

ammonium NH+

4 0.18

ammonia NH3 0.20



Table 2: Molecular diffusion coefficients (continued).


(10−4 cm2/s)

nitrate NO−

3 0.17

phosphoric acid H3PO4 0.08

dihydrogen phosphate H2PO−

4 0.08

hydrogen phosphate HPO2−

4 0.07

phosphate PO3−

4 0.05

hydrogen sulfide H2S 0.17

hydrogen sulfide ion HS− 0.16

sulfate SO2−

4 0.10

silica H4SiO4 0.10



Table 3: Molecular diffusion coefficients (continued).


(10−4 cm2/s)

calcium ion Ca2+ 0.07

magnesium ion Mg2+ 0.06

iron ion Fe2+ 0.06

manganese ion Mn2+ 0.06

a Taken from diffusion coefficients web pageb for water at 20◦C with salinity of 0.5 ppt.c for water at 10◦C with salinity of 0.5 ppt.


http://www.talknet.de/~alke.spreckelsen/roger/thermo/difcoef.html


Figure 2: Schematic of a control volume with crossflow.

Jx,in Jx,out

x

-y

z

δxδy

δz

u



Figure 3: Definitions sketch for one-dimensional pure diffusion in aninfinite pipe.

A M

-x x



Table 4: Dimensional variables for one-dimensional pipe diffusion.

Variable Dimensions

dependent variable C M/L3

independent variables M/A M/L2

D L2/T

x L

t T



Figure 4: Self-similarity solution for one-dimensional diffusion of aninstantaneous point source in an infinite domain.

−4 −2 0 2 40

0.2

0.4

0.6

0.8

1Point source solution

η = x / (4Dt)1/2

C A

(4π

D t)

1/2 /

M



Figure 5: Schematic solution of the advective-diffusion equation inone dimension. The dotted line plots the maximum concentration asthe cloud moves downstream.

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5Solution of the advective−diffusion equation

Position

Con

cent

ratio

n

t1

t2 t

3

Cmax


References

Fischer, H. B., List, E. G., Koh, R. C. Y., Imberger, J. & Brooks,N. H. (1979), Mixing in Inland and Coastal Waters, AcademicPress, New York, NY.

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