ch5_advection_v2.pdf

Chapter 5

The Advection-Diffusion Equation

Now that we have a strong foundation in understanding and solving the diffusion equation(Chapter 4), let’s consider the effects of advection on both analytical and numericalsolutions.

5.1 Analytical Solutions of the ADE

We can find some analytical solutions for the advection-diffusion equation (ADE) if wehave incompressible flow, Fickian diffusion, and uniform diffusivities. Under these con-ditions, the ADE simplifies to

∂C

∂t+ u

∂C

∂x+ v

∂C

∂y+ w

∂C

∂z= Dx

∂2C

∂x2+Dy

∂2C

∂y2+Dz

∂2C

∂z2(5.1)

Eqn. 5.1 is equivalent to Eqn. 2.30 without a source term – see Chapter 2 for its derivation.

5.1.1 Instantaneous Point Source

Moving Coordinates

Let’s consider the 1D case first. If the coordinate system is chosen such that cross-streamand vertical velocities are zero (v = w = 0), and if mixing in the y and z directions isfast compared to longitudinal advection and diffusion, then Eqn. 5.1 simplifies to

∂C

∂t+ u

∂C

∂x= Dx

∂2C

∂x2(5.2)

If we release scalar mass per unit area M/A at location x = x0 and time t = t0, itspreads due to diffusion. The only effect of advection is to carry the cloud downstream

112

Advection and Diffusion 113

as it diffuses. In fact, if we float along in a boat moving at velocity u (the same velocityas the fluid), the cloud will appear to evolve just like it would have without any advectionat all because the center of mass will not move relative to us. To describe the evolutionof the cloud from this perspective, let’s define the moving coordinate, ξ, to be in thesame direction as x but with its origin in the moving boat as illustrated below:

If the boat starts at time t = 0 and location x = 0, its location is given by x = ut, andthe moving coordinate ξ must be related to the stationary coordinate x by the equation

ξ = x− ut (5.3)

to ensure that ξ = 0 in the boat at all times. From the perspective of the boat, theconcentration distribution simply follows the diffusion equation (with zero advection),for which the solution is

C(ξ, t) =M/A√

4πDx(t− t0)exp

[− (ξ − ξ0)2

4Dx(t− t0)

](5.4)

where ξ0 = x0 − ut0 is the location of the point source from the perspective of the boat.Eqn. 5.4 is equivalent to Eqn. 4.11, but with ξ taking the place of x.

We found Eqn. 5.4 from our intuitive understanding of advection as a process that simplycarries a diffusing scalar cloud downstream. The more rigorous way to show that Eqn. 5.4is the solution of Eqn. 5.2 is to transform Eqn. 5.2 to moving coordinates and show thatwe obtain the diffusion equation, whose solution is given by Eqn. 5.4. We can transformthe Eqn. 5.2 from stationary coordinates (x, t) to moving coordinates (ξ, t) using thechain rule as follows

∂C(x, t)

∂t=∂C(ξ(x, t), t)

∂ξ

∂ξ

∂t−u

+∂C(ξ(x, t), t)

∂t

∂t

∂t1

= −u∂C(ξ, t)

∂ξ+∂C(ξ, t)

∂t

∂C(x, t)

∂x=∂C(ξ(x, t), t)

∂ξ

∂ξ

∂x1

=∂C(ξ, t)

∂ξ

∂2C(x, t)

∂x2=

∂

∂ξ

∂C(ξ(x, t), t)

∂ξ

∂ξ

∂x1

∂ξ

∂x1

=∂2C(ξ, t)

∂ξ2

114

Substituting into Eqn. 5.2, we find

��−u∂C∂ξ

+∂C

∂t+��

u∂C

∂ξ= Dx

∂2C

∂ξ2⇒

∂C

∂t= Dx

∂2C

∂ξ2

which is, indeed, the diffusion equation, which for an instantaneous point source in anunbounded domain has the solution given by Eqn. 5.4 (recall, we found this solutionusing the similarity approach in Section 4.1.1). Note that by defining similar movingcoordinates in the y and z directions, we may tranform the 2D and 3D ADE to thediffusion equation as well.

1D Solution

Transforming back to stationary coordinates, i.e., plugging in ξ = x − ut, we find thefollowing solution to the 1D ADE for an instantaneous point source of mass M/A releasedat location x0 and time t0

C(x, t) =M/A√

4πDx(t− t0)exp

[−(x− x0 − u(t− t0))2

4Dx(t− t0)

](5.5)

2D Solution

The solution for 2D advection and diffusion of an instantaneous point source of mass perunit length M/L released at location (x0, y0) and time t0 is

C(x, y, t) =M/L

4π(t− t0)√DxDy

×

exp

[−(x− x0 − u(t− t0))2

4Dx(t− t0)− (y − y0 − v(t− t0))2

4Dy(t− t0)

] (5.6)

3D Solution

The solution for 3D advection and diffusion of an instantaneous point source of mass Mreleased at location (x0, y0, z0) and time t0 is

C(x, y, t) =M

[4π(t− t0)]3/2√DxDyDz

×

exp

[−(x− x0 − u(t− t0))2

4Dx(t− t0)− (y − y0 − v(t− t0))2

4Dy(t− t0)− (z − z0 − w(t− t0))2

4Dz(t− t0)

] (5.7)


Spatial and Temporal Records for a 1D Instantaneous Point Source

In this section, we will examine spatial and temporal distributions of concentration for1D advection and diffusion of an instantaneous point source released at location x0 andtime t0. The spatial distribution of concentration is fairly intuitve by now – a Gaussianscalar cloud spreads via diffusion such that the standard deviation is given by

σx =√

2Dx(t− t0)

The cloud is carried downstream by advection such that the center of mass is given by

xc = x0 + u(t− t0)

This spatial distribution of concentration is plotted below at times t− t0 = βL2/Dx forβ = 10, 30, and 50 at Peclet numbers of 0.1 (diffusion dominating) and 10 (advectiondominating):

0 100 200 300 400 5000

0.02

0.04

0.06

0.08

(x!x0)/L

C(x)

L1

/2

34 5 0.134 5 10

Figure 5.1: Some spatial concentration distributions for 1D advection and diffusion.

The peak concentration over all space at a given time for a 1D point source under theinfluence of both advection and diffusion is

Cmax(t) =M/A√

4πDx(t− t0)(5.8)

This is the same maximum as for diffusion alone (see p. 75). Let us define xmax, thelocation of maximum concentration, such that Cmax(t) ≡ C(xmax, t). When there is noadvection, the maximum concentration occurs at the location of the release, xmax = x0.However, when there is advection, the maximum concentration occurs at location

xmax = x0 + u(t− t0)

116

which moves downstream over time. Since the concentration distribution is symmetricabout xmax at any given time, the maximum concentration over all space coincides withthe center of mass of the cloud.

Using aerial photography or various laboratory visualization techniques, we can measurethe spatial distribution of concentration C(x) at a given time, and for an instantaneouspoint source, we will obtain nice Gaussian concentration distributions like the ones inFigure 5.1. However, in the field, it is more common to obtain a time history of con-centration C(t) from a stationary instrument. Therefore, we need to understand thetemporal evolution of concentration at a fixed point in space as well as the more intuitivespatial distribution of concentration at a given point in time.

The temporal record of concentration at a point in space depends on the Peclet numberin a more complicated way than the spatial record of concentraiton at a point in time, solet’s work with the dimensionless form of Eqn. 5.5. Recall from Section 3.4 that there aretwo ways to scale the 1D ADE: using the diffusive time scale T = L2/Dx (appropriatewhen diffusion is important, at low to moderate Peclet numbers) and using the advectivetime scale T = L/u (appropriate when advection is important, at moderate to high Pecletnumbers). Let’s define dimensionless variables so that the coordinate system is centeredat the point of the release as follows:

x∗ ≡ x− x0

L, t∗ ≡ t− t0

T, C∗ ≡ CLA

M

The Peclet number is defined by

Pe =uL

Dx

Using the diffusive time scale, T = L2/Dx, we find the nondimensional solution

C(x∗, t∗) =1√4πt∗

exp

[−(x∗ − Pe t∗)2

4t∗

](5.9)

and alternatively, using the advective time scale T = L/u, we find the nondimensionalsolution

C(x∗, t∗) =

√Pe

4πt∗exp

[−Pe(x∗ − t∗)2

4t∗

](5.10)

Time histories of concentration at location x = x0 + L are plotted on the next pagefor the diffusive scaling (Figure 5.2) and the advective scaling (Figure 5.3) for somedifferent Peclet numbers. To fit all of these curves on the same figure, we have normalizedconcentration by Cmax(x0 + L), the maximum concentration measured over all time atx = x0 + L.


0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

Dx(t!t0) /L2

C(x

0+L,t)

/ C m

ax(x

0+L)

Figure 5.2: Concentration time histories C(t) measured at location x = x0 + L for somedifferent Peclet numbers Pe = uL/Dx. Time is normalized by the diffusive time scale.

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

u(t!t0) /L

C(x

0+L,t)

/ C m

ax(x

0+L)

Figure 5.3: Concentration time histories C(t) measured at location x = x0 + L for somedifferent Peclet numbers Pe = uL/Dx. Time is normalized by the advective time scale.

Setting the time derivative of Eqn. 5.5 to zero, we find that Cmax(x), the maximumconcentration measured over all time at location x, occurs at time

tmax = t0 −Dx

u2+

√(Dx

u2

)2

+(xu

)2

= t0 +x2

Dx

(−1 +

√1 + Pe2x

Pe2x

)

= t0 +x

u

(−1 +

√1 + Pe2x

Pex

) (5.11)

118

where we have defined a new Peclet number based on L = x− x0

Pex ≡u(x− x0)

Dx

We may find the following two limiting solutions using Taylor series expansion aboutPex = 0 and 1/Pex = 0, respectively:

tmax = t0 +

x2

2Dx

for Pex . 0.1

x

ufor Pex & 1000

(5.12)

These limiting solutions are clearly visible in Figures 5.2 and 5.3 with L = x− x0.

Plugging tmax into the solution of the 1D ADE for an instantaneous point source (Eqn. 5.5),we find Cmax(x), the maximum concentration measured over all time at location x. Theresult is a rather ugly function of Pex. Since this function ugly, we will not write it out,but it is plotted below vs. Pex and also vs.

√Pex.

5 10 15

0.2

0.4

0.6

0.8

1

Pex

C max

(x) L

A/M

1 2 3 4

0.2

0.4

0.6

0.8

1

Pex1/2

general solutionlimiting solution for small Pexlimiting solution for large Pex

Figure 5.4: Maximum concentration over all time at a given point in space, as a functionof Peclet number and its square root. Limiting cases for small and large Peclet numberare also plotted.

Also included in this plot are the limiting solutions for small and large values of Pex.These are given by

Cmax(x) ≈

M

AL

√1

2πe≈ 0.242

M

ALfor Pex . 0.1

M

AL

√Pex

4π≈ 0.282

M

AL

√Pex for Pex & 1000

(5.13)


5.1.2 Advection and Diffusion of a Concentration Front in 1D

Now consider 1D advection and diffusion of an initially sharp concentration front. Forexample, consider fluid with concentration C0 displacing fluid with zero concentration ina pipe where the velocity is u. The governing equation is

∂C

∂t+ u

∂C

∂x= Dx

∂2C

∂x2

At time t = 0, there is a sharp front so that

C(x, 0) =

{C0 for x ≤ 0

0 for x > 0

The center of the front will travel at velocity u as the front is smoothed out by diffusionas illustrated below.

Again, it is useful to imagine riding the front as it moves with the water velocity. Fromthis perspective, advection is zero, and the problem looks a lot like one we solved inChapter 4. Transforming to moving coordinates, ξ = x−ut and t, the governing equationand initial conditions become

∂C

∂t= Dx

∂2C

∂ξ2

C(ξ, 0) =

{C0 for ξ ≤ 0

0 for ξ > 0

This is the mirror image of the problem we solved in Section 4.1.2, for which the solutionis given by Eqn. 4.22. Adjusting for the mirror image effect1, and plugging in the movingcoordinates, we find

C(x, t) =C0

2erfc

(x− ut√

4Dxt

)(5.14)

1To adjust for the mirror image effect, we may transform to C ′ = C − C0 to obtain a problem thatlooks exactly like the one solved by Eqn. 4.1.2. We also make use of the equality 1 + erf(u) = −erfc(u)to find the solution.

120

5.1.3 Lateral Mixing of Two Streams

Now consider the lateral mixing of two streams with different concentrations of somechemical, as might occur when two rivers converge. This problem is illustrated below:

For now, let’s neglect the effects of boundaries in the lateral direction (you can add themlater using the method of images). Assume that concentration is fully mixed across theriver depth, so the problem is 2D. Also assume the solution has reached steady state.Consider the result far downstream of the convergence point so that the Peclet number islarge. Under all these assumptions, the governing equation may be simplified as follows:

��7

0∂C

∂t+ u

∂C

∂x=��>

0

Dx∂2C

∂x2+Dy

∂2C

∂y2+��>

0

Dz∂2C

∂z2

and we are left with

u∂C

∂x= Dy

∂2C

∂y2

We may convert this equation into the 1D diffusion equation

∂C

∂τ= Dy

∂2C

∂y2

using the following coordinate transformation:

τ =x

u

The “initial conditions” are given by

C(y, τ = 0) =

{C0 for y ≤ 0

0 for y > 0


The solution is apparent from Section 5.1.2 if we recognize the equivalence of ξ and yand of t and τ

C(x, y) =C0

2erfc

(y√

4Dxx/u

)(5.15)

5.1.4 Concentration Specified at a Fixed Point in Space

For the problems we have considered thus far, it was possible to transform to somecoordinate system in which the advection-diffuison problem became simply a diffusionproblem. In the case of concentration specified at a fixed point in space, this is notpossible. Consider 1D advection and diffusion

∂C

∂t+ u

∂C

∂x= Dx

∂2C

∂x2

where the initial concentration at time t = 0 is zero everywhere, i.e.

C(x, 0) = 0

and after time t = 0, the concentration at x = 0 is set to a constant concentration C0

C(0, t) = C0

For x > 0, the solution is given by

C(x, t) =C0

2

[erfc

(x− ut√

4Dxt

)+ erfc

(x+ ut√

4Dxt

)exp

(ux

Dx

)](5.16)

This solution is important for soil column studies, as illustrated below:

Note that for small Peclet number (common in soils), the second term is small.

122

5.1.5 Steady, Continuous Point Source

It is common for pollutants to be released steadily from a pipe into a river. Considereffluent released into a river at rate M and at location (x0, y0, z0). As you saw in Home-work 2, an effluent plume in steady state will travel through three zones. In zone 1, theplume will spread in 3D until it is fully mixed over the depth of the river. Then, in zone2, it will spread in 2D until it is mixed over the width as well. Finally, in zone 3, theplume is fully mixed over the river cross-section, and the concentration is constant in xdue to mass conservation. These three zones are illustrated below:


Steady, Continuous Point Source in 3D

In 3D, with coordinates chosen so that advection is in the x direction, the steady-stateadvection-diffusion equation is

��7

0∂C

∂t+ u

∂C

∂x=��>

0

Dx∂2C

∂x2+Dy

∂2C

∂y2+Dz

∂2C

∂z2

As you saw in Homework 2, the Peclet number is quite large in rivers, except very nearto the source, so the longitudinal diffusion term is small, and our governing equation canbe simplified to

u∂C

∂x= Dy

∂2C

∂y2+Dz

∂2C

∂z2

A sheet of water having width ∆x passes the continuous point source within time ∆t =∆x/u. Within this small time, it picks up mass ∆M = M∆t = M∆x/u. Since thereis very little diffusion in the x direction, mass within a sheet of water having width ∆xspreads like an instantaneous point source in 2D (in y and z) as it is carried downstream.

The mass per unit width in a given sheet is ∆M/∆x = M/u. Plugging this mass per unitwidth into the solution for a point source diffusing in 2D (Eqn. 4.16), and noting thatthe mass within a sheet located at x has been spreading for time t− t0 = (x− x0)/u, wefind the following solution for a 3D, steady, continuous point source located at (x0, y0, z0)

C(x, y, z) =M

4π(x− x0)√DyDz

exp

[− u(y − y0)

2

4Dy(x− x0)− u(z − z0)

2

4Dz(x− x0)

](5.17)

This is the solution for an unbounded domain. Once the plume has traveled far enoughdownstream that it reaches the water surface and the bed, image sources are necessaryto meet the vertical boundary conditions. Eventually, concentration becomes well-mixedacross the depth and the plume enters zone 2.

124


In zone 2, where the plume has become fully mixed across the depth of the river, it willcontinue to spread in the y direction as it travels downstream. Since Peclet number islarge (even bigger in zone 2 than in zone 1), the steady-state advection-diffusion equationin 2D becomes

u∂C

∂x= Dy

∂2C

∂y2

Sheets of water having width ∆x pick up mass M∆x/u as they pass the source. Thesesheets have cross-sectional area Lz∆x perpendicular to the direction of spreading (y)where Lz is the depth of the stream, so the mass per unit area within a single sheet isM/(uLz). The mass within a single sheet spreads like an instantaneous point source in1D (spreading in y), resulting in the following solution for a 2D, steady, continuous pointsource located at (x0, y0)

C(x, y) =M/Lz

u√

4πDy(x− x0)/uexp

[− u(y − y0)

2

4Dy(x− x0)

](5.18)

This is the solution for a domain unbounded in y. Image sources are required to satisfythe boundary conditions once the plume spreads to the banks of the river.


Once the plume has mixed fully across the depth and the width of the river, concentrationbecomes uniform. Using a control volume approach, we may find the following solutionfor the steady state concentration far downstream from a continuous point source

C =M

uA=M

Q(5.19)

where A is the cross-sectional area of the river and Q is the flow rate.


5.1.6 Unsteady, Continuous Point Source in 1D

If a 1D continuous point source is turned on at time t = t0, so that

M(t)/A =

{0 for t < t0

M/A for t ≥ t0

slabs of fluid having thickness ∆x that pass the point source at times t > t0 will pick upmass M∆x/u.

These slabs have volume ∆xA, and in the 1D case, we can assume that mass instantlymixes across the slabs. Thus, the slabs pick up concentration C0 = M/(uA) as they passthe point source. Since the slabs that passed the point source before it was turned onhave zero concentration, there is a sharp front located at x = x0 +u(t− t0). The solutionfor the evolution of such a sharp front under the influence of advection and diffusion isgiven by Eqn. 5.14 for the special case of x0 = t0 = 0. Generalizing this solution for x0

and t0, and plugging in C0 = M/(uA), we find the following solution for a continuouspoint source located at x = x0 that releases mass at rate M beginning at time t = t0

C(x, t) =M

2uAerfc

(x− x0 − u(t− t0)√

4Dx(t− t0)

)(5.20)

If M(t)/A is a more complicated function of time, we may follow the method outlined inSection 4.1.6 to find the solution

C(x, t) =1

2uA

∫ t

−∞

∂M

∂τerfc

(x− x0 − u(t− τ)√

4Dx(t− τ)

)dτ (5.21)

126

5.1.7 Unsteady Velocity

Note that the solutions developed in this chapter using moving coordinates may be easilyadapted for the the case of unsteady velocity by replacing u(t− t0) with the more general∫ t

t0

u(τ)dτ

For example, this is appropriate in the case of an instantaneous point source or anunsteady continuous point source. Moving coordinates in the y and z directions may begeneralized in a similar manner.

ch5_advection_v2.pdf

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