3-Dispersion in Rivers and Lakes_S13

8/13/2019 3-Dispersion in Rivers and Lakes_S13

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3-Dispersion Coefficients in Rivers and Lakes_S13

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Diffusion and Dispersion

Molecular Diffusion

The atoms in any phases (solid, liquid, or gas) are always in motion (rotational, vibrational, andtranslational) at temperatures above absolute zero. The amount of the movement increases with

increasing temperature because the kinetic energy (molecular velocity) is associated with thermalenergy of the molecules. In a gas or liquid phase, the individual atoms bounce at random as theycollide with other atoms. The bouncing of individual atoms causes the random motion of atoms,molecules, and microscopic particles. The bouncing motion is called Brownian motion (C. Sato).If there is an unbalance in the number of particles (atoms, molecules) from one region to another, the

particles migrate from the region of higher concentration to the region of lower concentrationdiminishing the concentration gradient.

Diffusion refers to the transport process by which particles (atoms, molecules) intermingle dueto the random motion and a concentration gradient (C. Sato). If sufficient time is given, the particleswill be distributed uniformly throughout the system driven by the concentration gradient and Brownianmotion.

Grahams Law

When two gases exist at the same temperature, their molecules must have the same average kineticenergy.

K.E. 1 = K.E. 2 (x.1.1)

where K.E. = the kinetic energy of gas; subscripts 1 and 2 denote different types of gases 1 and 2,respectively.

The kinetic energy is expressed as

K E m v. . 12

2 (x.1.2)

where m = molecular mass of gasv = molecular velocity of gas

Thus, Eq. (x.1.1) is written as

2 21 1 2 2

1 1

2 2m v m v

and

m v m v1 12

2 22 (x.1.3)

where m 1 = molecular mass of gas 1m2 = molecular mass of gas 2v1 = molecular velocity of gas 1v2 = molecular velocity of gas 2


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The ratio of v 12 to v 22 gives

v

v

mm

12

22

2

1 (x.1.4)

The relative rates of two gases that exist under the same temperature and pressure is

vv

mm

1

2

2

1 (x.1.5)

Eq. (x.1.5) shows that two gases travel (diffuse) with the velocities that are inversely proportional tothe square root of their molecular masses. At given temperature and pressure, less massive gasesdiffuse more rapidly than more massive gases.

The diffusion rate is expressed as ( Reference )

Diffusion Rate K T

m (x.1.6)

where K = the constant that depends on geometric factors including the area across which the diffusionis occurring, m = molecular mass, and T = temperature.

The relative diffusion rate for two different molecules is given by ( Reference )

Diffusion Rate of molecule Diffusion Rate of molecule

K T m

K T m

mm

12

1

2

2

1 (x.1.7)

Eq. x.1.7 indicates that the diffusion rates of two gases are inversely proportional to the square root oftheir molecular masses. Grahams law of diffusion states that the rate at which gases diffuse isinversely proportional to the square root of their densities ( gas) under given temperature and pressure(Reference ):

Diffusion Rate gas

1

Since the density of a gas ( gas) is directly proportional to its molecular mass (m),

Diffusion Ratem

1 (x.1.8)


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When gases are introduced into a liquid, the relative rate of diffusion of a given gas is proportional to its solubility in the liquid, and inversely proportional to the square root of its molecularmass. Therefore, the relative rate of diffusion of gas 2 to gas 1 is given by

Diffusion rate of molecule gas

Diffusion rate of molecule gas

solubility of gas

solubility of gas

MW of gas

MW of gas

2

1

2

1

1

2

(x.1.9)

where MW = molecular weight.

Example x.1.1 (Sato) What is the relative rate of diffusion of SO 2 to O 2 assuming that SO 2 is 1,250times more soluble in water than O 2?

(Solution)

MW of O 2 = 32 and MW of SO 2 = 64, and SO 2 is 1,250 times more soluble in water than O 2.According to Grahams law of diffusion (Eq. x.1.9), the relative rate of diffusion is

2

2

1250 32884

1 64 Diffusion Rate of molecule SO Diffusion Rate of molecule O

The diffusion rate of SO 2 into water is 884 times faster than O 2.

Example x. 1.2 (Sato) What is the relative rate of diffusion of H 2S to O 2 if H 2S is 85 times moresoluble in water than O 2?

(Solution)

MW of O 2 = 32 and MW of H 2S = 34, and H 2S is 85 times more soluble in water than O 2. Accordingto Grahams law of diffusion (Eq. x.1.9 ), the relative rate of diffusion is

Diffusion Rate of molecule H S Diffusion Rate of molecule O

2

285

3234

82

Diffusion of H 2S is 82 times faster than O 2.

Example x.1.3 (Sato) What is the ratio of the diffusion velocity of oxygen atoms (O 2) to that ofhydrogen sulfide (H 2S) atoms when both gases are at the same temperature and pressure?

(Solution)


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Molecular mass of oxygen (O 2) = 32 and molecular mass of hydrogen sulfide (H 2S) = 34,

vv

mm

1

2

2

1 (1.1.5)

where v 1 = diffusion velocity of O 2 v2 = diffusion velocity of H 2Sm1 = mass of O 2 m2 = mass of H 2S

2

2

341.0

32O

SO

v

v

The velocity of oxygen atoms (O 2) and the velocity of hydrogen sulfide (H 2S) atoms are approximatelythe same.

Example x.1.4 (Sato) What is the ratio of the velocity of oxygen atoms (O 2) to the velocity of sulfurdioxide (SO 2) atoms when both gases are at the same temperature and pressure?

(Solution)

Molecular mass of oxygen (O 2) = 32 and molecular mass of sulfur dioxide (SO 2) = 64.Using Eq. (1.1.5),

vv

mm

1

2

2

1

v

vO

SO

2

2

6432

14 .

The ratio of the velocity of oxygen atoms (O 2) to the velocity of sulfur dioxide (SO 2) atoms is 1.4.

Example (Web site) What is the relative rate of diffusion of CO2 to O

2 if CO

2 is 22 times more soluble

in water than O 2?

Solution

MW of O 2 = 32 and MW of CO 2 = 44, and CO 2 has 22 times the solubility of O 2. According toGrahams law of diffusion, the relative rate of diffusion is


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Diffusion Rate of molecule CO Diffusion Rate of molecule O

2

222

3244

19

Diffusion of CO 2 is 19 times faster than O 2.

Example (Web site) What is the ratio of the velocity of oxygen atoms to the velocity of radon atomswhen both gases are at the same temperature and pressure?

Solution

Molecular mass of oxygen = 32 and molecular mass of radon = 222,vv

mm

1

2

2

1

vv

O

Rn

2 22232

2 6 .

Modeling of Diffusion (Diffusive Transport)

In 1855, Adolf Fick introduced two differential equations to quantify the diffusive transport in water.These equations are called Ficks first law and Ficks second law . Modeled on Graham's experiments

with gas, but Fick measured the fluxes of salt migrating from one solution reservoir to the otherthrough tubes (A. Fick, Poggendorff's Annel. Physik. (1855), 94 , 59; A. Fick, Phil. Mag. (1855), 10 ,30. (in English) (http://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion, Accessed 1/20/11).

Ficks first law of diffusion Ficks first law of diffusion describes diffusive transport of a solute across an interface layer (e.g.,membrane) of unit area. The law can be developed based on experimental observations. Consider twocontainers filled with solutions of two different concentrations of a chemical. The containers areconnected by a pipe but separated by a valve ( Fig. x.1.1 ). The molecules in the containers are inrandom motion making numerous collisions with the walls of the containers, tube, and partition of thevalve. If the valve is opened, the solution will mix because of the random motion of the molecules. Intime a uniform concentration of the chemical will be formed in the containers. If the concentration ofthe chemical (such as a tracer) in the reservoir 1 (C 1) is greater than the concentration in the reservoir 2(C2), we observe that the chemical in the container 1 diffuses into the container 2 through an area A ofthe tube, across which diffusive transport is occurring.
http://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusionhttp://en.wikipedia.org/wiki/Fick%27s_laws_of_diffusion


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Figure x.1.1 An experimental setup for diffusive transport: A = cross-sectional area of the pipex2 x1 = x = total length of the pipes between the two containers.

The experiment leads to the following observations:

W (C 1 C2) (a)W A (b)W 1/ (x 2 x1 ) (c)

where W = the rate of the mass transport.

Consideration of the observations (a), (b), and (c) yields

1 2 2 1

2 1 2 1

C C C C C W A A A

x x x x x

(x.01)

By introducing a proportionality constant, D, the mass transport due to diffusion (W d) may beexpressed by

d C

W DA x

(x.02)

where D = the molecular diffusion coefficient, L2

/T.

as x 0

d

dC W DA

dx Unit: M T

LT

L M

L L

22

31 (x.0.3)

The diffusive mass flux, J diffusion , is given by

(1) (2)

A

x1 x2


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d diffusion

W dC J D

A dx

(x.0.4)

Eq. (x.0.4) describes a one- dimensional flux, and is known as the Ficks first law of diffusion.

The law states that a magnitude of the diffusive flux is proportional to the concentration differentialacross that interface. For a three-dimensional flux, the Ficks first law of diffusion is given as

diffusionW

J D C A

where J = the flux of a solute due to diffusion (M T -1 L-2),W = the rate of mass transport (MT -1),A = the interfacial area (L 2) ,C = the concentration of a solute (ML -3),D = the proportionality constant known as the diffusion coefficient (L 2T-1 generally cm 2/sec).

= the differential vector operator, and C is given by

C i C

x j

C y

k C

z

where x, y, and z = the distances in x, y, and z directions, respectively.

Molecular diffusion coefficient

The diffusion coefficient, D, is always a positive quantity, commonly expressed using a unit of cm 2s-1.

The magnitude of diffusion coefficients in a fluid depends on types of solute (molecules), temperature,and viscosity of the fluid. The diffusion coefficients for gases vary with types of diffusing molecules,temperature, and pressure (Thibodeaux, 1979). According to Einstein,

D k T

f k T B i B (x.02)

where k B = the Boltzmanns constantT = the absolute temperature f = the frictional coefficient of the molecule (species) i = mobility of species i = 1/ f = v d /F

vd = the particles terminal drift velocity F = an applied force

Diffusion is slower for a molecule with larger friction coefficient (e.g., larger, irregularly shapedmolecule) and faster at higher temperature. For a spherical particle, Stokes law gives:

f r 6 (x.03)


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where r = the radius of a spherical particle = the viscosity of the medium in which the spherical particle is diffusing

By combining Eq. x.02 and Eq. x.03, the diffusion coefficient is expressed by

D k T

r

B

6 (x.04)

Eq. (x.04) is often referred to as the Einstein-Stokes equation for diffusion of spherical particlesthrough fluid with a low Reynolds number(http://en.wikipedia.org/wiki/Einstein_relation_%28kinetic_theory%29 ; Accessed 1-24-2011). Themolecular diffusion coefficient, D, is extremely small with the order of 10 -5 cm 2/s, and is dependent on

properties of molecule (e.g., geometry) and fluid properties (e.g., viscosity) ( Schnoor, 1996).

Turbulent (Eddy) Diffusion and Dispersion

Turbulence is irregular fluctuations (eddy fluctuations) of fluid properties occurring in macro-scale

fluid motions. Turbulence can occur in a broad range of spatial and temporal scale, and cause mixingof fluid properties (e.g., temperature, dissolved and suspended particles). At the micro-scale level,

particles are transported advectively with a glob of surrounding fluid in turbulent shear flow.Turbulent diffusion is a process of mixing due to the micro-scale turbulent motions of fluid, caused

by shear force within the body of fluid.

The term dispersion is generally used for the mixing process caused by the macro-scaleturbulent motions of fluid. Dispersion, which comprises interactions of eddy diffusion, occurs resultingfrom velocity gradients caused by shear force at the boundaries of water body such as air-waterinterface (wind share), river bed-water interface (bed share), and river bank-water interface. Thevelocity gradients also occur at the changes of stream morphology such as meandering, bends, and

backwater in rivers. The examples of fluid mixing in a large-scale include the spring and fall overturnsin lakes due to density instability, tidal oscillation in estuaries, and catastrophic earth-movement andstorms. The mixing is also introduced by man-made objects such as ships, barges, cascades, and dams.

Although molecular motion does not greatly resemble turbulence, we often justify the use ofthe Ficks expression by appealing to an analogy among molecular diffusion, turbulent diffusion, anddispersion. Eqs (1.1.10) and (1.1.11) are the expressions for turbulent diffusion and dispersion(Schnoor, 1996).

eddydC

J dx

dispersiondC

J E dx

where J e = the mass flux rate due to turbulent (eddy) diffusion, ML -2T-1 = turbulent diffusion coefficient or eddy mass diffusivity, L 2T-1 Jd = the mass flux rate due to dispersion, MT -1 E = dispersion coefficient, L 2T-1
http://en.wikipedia.org/wiki/Reynolds_numberhttp://en.wikipedia.org/wiki/Einstein_relation_%28kinetic_theory%29http://en.wikipedia.org/wiki/Einstein_relation_%28kinetic_theory%29http://en.wikipedia.org/wiki/Einstein_relation_%28kinetic_theory%29http://en.wikipedia.org/wiki/Einstein_relation_%28kinetic_theory%29http://en.wikipedia.org/wiki/Reynolds_number


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The coefficients D, , and E are measure of mixing. Both the turbulent dispersion coefficient anddispersion coefficient E are dependent on the flow regime. The molecular diffusion coefficients D aresmallest, the eddy diffusion coefficients , are considerably larger than the molecular diffusioncoefficients, the dispersion coefficients, E, are considerably larger than the eddy diffusion coefficients,

. D


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Table . Dispersion Coefficients of Radioactive Waste Forms (invalidated, unknown source)

Example 2.2 (Schnoor, p. 44) Molecular Diffusion of a Chemical in Water

Calculate the mass flux rate (in mg/d) for a chemical diffusing between two beakers. Assume the

chemical is diffusing through a 10-cm distance with a concentration gradient of -1 mg L -1 cm -1. D =10 -5 cm 2 s-1, A = 3.14 cm 2.

Solution:

J DAdC dx

J cm s cm mg L cm

L

cm

sd

mg d

10 314 1

10

86 4000 00271

5 22

3 3.

,. /

Radioactive Waste Form Dispersion Coefficientscm 2s -1

Salt Grout 10 -6 - 10 -8

Salt Grout with Diffusion Barrier 10 -10

Mineral Grout 10 -7 - 10 -10

Mineral Grout with Diffusion Barrier 10 -10

Salt/polyethylene 10 -10 - 10 -11

Glass Canisters 10 -13 - 10 -15

ISV Monoliths 10 -13 - 10 -15

Ceramic Clinkers 10 -13 - 10 -15

Grout Matrix 10 -7 - 10 -8

Glass Cullet 10 -13 - 10 -15

Sulfur Matrix 10 -10 -10 -14


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Example 2.3 (Schnoor, p. 44) Molecular diffusion through a thin film

The molecular diffusivity of caffeine (C 9H8O) in water is 0.63 x 10 -5 cm 2 s-1. For a 1.0 mg L -1 solution,calculate the mass flux in mg s -1 through an intestinal membrane (0.1-m 2 area) with a liquid film

approximately 60 m thick. How long would it take 1 mg of caffeine to move through 0.1 m2

ofintestine, assuming the above flux rate? (Note: We assume transport through the film is the rate-limiting step in transport and metabolism.)

Solutions:

Given: D = 0.63 x 10 -5 cm 2 s-1 A = 0.1-m 2

C x

mg L

x m

( . ) /0 10

60 10 6

Using Ficks first law J DA dC dx

J x cm s m mg L

x m

cm

m

L

cm

m

cm

x mg s

065 10

01 0 10

60 10

10

1

1

10 10

105 10

5 22

6

4 2

2 3 3 2

3

..

. /

. /

time mass

flux

mass

mass time

mg

x mg s s

/ . /

min. min

1

105 10 60

1593

Example 2.1 (Sato): Experiments were conducted using a dual-chamber microbial fuel cell (MFC)with a phosphate buffer as catholite. The anode and cathode chambers are separated by Nafionmembrane. The cross-sectional area and thickness of the membrane was 81 cm 2 and 0.051 mm,respectively. pH of the medium in the anode chamber was 6.7 and pH of the phosphate buffer in thecathode camber was 7.2. The molecular diffusion coefficient of proton (H +) is 9.3 x 10 -5 cm 2/s. UsingFicks first law of diffusion, determine the diffusion rate (mass flux rate) of protons from the anodechamber to the cathode chamber through the Nafion membrane.

(Solution)

In the anode chamber (pH = 6.7), [H +] = 10 -6.7 = 1.995 x 10 -7 In the cathode chamber (pH = 7.2), [H +] = 10 -7.2 = 6.310 x 10 -8

d dC C

J DA DAdx x


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where D = 9.3 x 10 -5 cm 2/sA = 81 cm 2

-7-8 -7

-7 -10

3 3 3

-1.364x10 moles(6.310 x 10 ) - (1.995 x 10 ) =

L-1.364x10 moles L -1.364x10 moles

= =L 10 cm cm

C

x = 0.051 mm = 0.0051 cm

10 3

-5 2 2

10 10 7

4

1.364 10 /9.3 x 10 cm /s 81 cm

0.0051 cm2.015 10 2.015 10 3600 7.25 10

7.25 10

d

C x moles cm J DA

x x moles x moles s x moles

s s hr hr x mmol

hr


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Ficks Second Law of Diffusion

Since the concentration gradient of substances in a finite volume of media decreases with time, Ficksfirst law is often inadequate in describing the diffusion process. The second law describes theconcentration with respect to time at any location.

According to the first law, the mass flux per unit area in a one-dimensional flow can also be described by

J D

C x t

x A

, x.01

where J A = the mass flux per unit area = J/A

In this expression, the concentration C is a function of distance x and time t. From the conservation ofmass,

C t

J x A x.02

M M------ = -----------L3 T T L 2 L

Combining Eq (1) and Eq (2) yields

C

t D

C

x

2

2 x.03

Eq. x.0 3 is known as Ficks second law of diffusion.

The second law can also be derived from the first law as follows (Schnoor, 1996):

J DA C x

J V C

t DA

C

x

M L 3 M L 2 L2 M---- = --------- = -----------------T L 3 T T L 3 L

Since V = A x


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C t

D A A x

C x

D C x x

lim t 0,

C t D

C

x

2

2

The following steps illustrate the derivation of Ficks second law based on the conservation of mass.

EA C x

V A x

EA C

xC x

x

Accumulation of Mass = Mass Inputs - Mass Outputs

V C

t EA

C x

EA C

xC x

(1)

Noting that

C x

x xC x

or

C x x

C x

x

(2)

Substitute (2) into (1)

V C

t EA

C x

EA C

x xC x

x

A


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V C

t EA

C x

EA C

x EA

xC x

x

V C

t EA

xC x

x

divided by V = A x

C t

E A A x x

C x

x

C t

E x

C x

C t

E C x

2

2 (3)

This is Ficks Second Law of diffusion.

To solve Ficks Second Law, one initial condition and two boundary conditions are required.

Solution of Eq. x.03 is given by

C M

E t e

x E t

2

2

4


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Solving the Partial Differential Equation (P. D. E.)

Solve the following partial differential equation (3) for C by separation of variables (transformation) ortrial and error

C t

E C x

2

2 (3)

The general solution with arbitrary constant A is

C A

t e

x E t

1 24

2

/ (2)

Derivative of C,

C

t must equal E

C

x

2

2

From integral tables for derivative of a product

d dx

uv u d vd x

v d ud x

.. chain rule

Similarly

C t

d d t

A

t e

A

t e e

A

t

x E t

x E t

x E t

1 24

1 24 4

1 2

2 2 2

/ /

'

/

'

From integral table

d dx e e du

dxu u

Similarly

C t

A

t e

x E t

e A

t

x E t

x E t

1 24

24

1 2

2 2

4/

'

/

'

Since

d dt t d dt

t t t

11

11 22

d dt t d dt

t t 1 1

21 21 2 3 2

// /


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C t

A

t e

x

E t e

A

t

x E t

x E t

1 24

2

24

3 2

2 2

4

12/ /

C t

A x

E t e

A

t e

x E t

x E t

2

5 24

3 24

4 2

2 2

/ /

C t

A x

E t

A

t e

x E t

2

5 2 3 24

4 2

2

/ / (3)

Right side of Eq. (1) :

E C

x E

xC x

2

2

(4)

Since C A

t e

x E t

1 24

2

/

C t

At

e At

e

x

E t

x

E t

1 24

1 24

2 2

/

'

/

'

C t

A

t e

x E t

x E t

1 2

4

2

24/

C t Ax E t e

x

E t

2 3 2 4

2

/ (5)

Substitute (5) into (4)


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E C

x E

x Ax

E t e

x Et

2

2 3 24

2

2

/

E C x

E Ax E t

e Ax E t

e x

Et

x

Et

2

2 3 24

3 24

2 2

2 2

/

'

/

'

E C

x E

Ax

E t e

x Et

A

E t e

x Et

x Et

2

2 3 24

3 24

2

24 2

2 2

/

'

/

E C x

E Ax E t

e A E t

e

x

Et

x

Et

2

2

2

5 2 4 3 2 44 2

2 2

/ /

E C

x E

Ax

E t

A

E t e

x Et

2

2

2

5 2 3 24

4 2

2

/ /

Since Eq. (3) = Eq. (6),

C t

E C x2

2

C A

t e

x E t

1 24

2

/

A can be any constant.

Determination of constant A

For a given concentration profile (given time, t),

Mass diffused at any time,

M C dx


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where

C A

t e

x E t

1 24

2

/

Thus,

M A

t e dx

x Et

1 24

2

/

M A

t e dx

x Et

2

1 24

0

2

/

Since A is constant and t is fixed.

M At

e dx x Et 2

1 24

0

2

/ (7)

From Integral table,

where

at E t

1

4

1

2

e dx

E t

E t E t x

14

0

2

1

2 1

2

Substitute into (7)

M A

t e dx

A

t E t A E

x Et 2 2 2

1 24

0

2

/

Solve for A

e dxa

aa x 2 2

0

12

0


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A M

E 2

Since C A

t

e

x E t

1 24

2

/,

C

M

E

t e

x E t

2

2

4

C M

E t e

x E t

2

2

4


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x.x Dispersion Coefficients in Rivers and Streams (Schnoor, p.55)

In streams and rivers, an extent of dispersion of constituents varies considerably with thestream characteristics such as flow velocity, width, and depth. The dispersion coefficients E x, E y, E z,in the longitudinal (x), lateral (y), and vertical (z) directions, respectively, vary orders of magnitude

from small creeks to large rivers. Formulations to estimate the values of the dispersion coefficients arefound in various literatures (Elder, 1959; Fisher, 1966; Sayre, 1973; Bowie et al., 1985).

x.01 Longitudinal Dispersion Coefficient, E x

Liu (1977) used the work of Fischer (1957) to develop an expression for the longitudinal dispersioncoefficient in rivers and streams:

22 3

3* *

x B x

B E

U A U DQu (1)

where xu = mean stream velocity, [L T -1] (m s -1)U* = bed shear velocity, [L T

-1] (m s -1)B = mean width, [L] (m)A = cross sectional area, [L 2] (m 2)QB = river flow (discharge), [L 3 T -1] (m 3 /s)D = mean depth, [L] (m)

*0.5 x

U u

(2)

does not depend on stream morphometry, but on the dimensionless bottom roughness. The bed shear

velocity is given as

*oU

dimensions:2 2

3 2

//

M LT L L M L T T

where = density of water, [M L -3]o = bed shear stress, [M L -1 T -2

The bed shear stress o is given as

2

8o x f

u

where f = friction factor = Darcy-Weisbach friction factor; 0.02 for natural, fully turbulent flowdepends on Raynolds number .

The bed shear velocity U * is related empirically to the bed friction factor and the mean stream velocityas


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2

* 8 8 x x

f f U u u

If we use f = 0.02,

*0.02

0.05 0.18 8

x x x x f

U u u u u

Example (Sato): Prove the following relationship

2 3 2

3* *

x Bu B Qu A u D

where xu = mean stream velocity, [L T-1]

u* = bed shear velocity, {L T -1]B = mean width, [L]A = cross sectional area, [L 2]QB = ricer flow (discharge), [L

3 T -1]D = mean depth, [L]

(Solution)

Since Q B = u x A and D = A / B

2 22 2 3

33* *

* 3

B x xQ u A u B Au D u A

u B

To estimate the longitudinal dispersion coefficient (E x), the following equation can be usedover a wide range of river conditions (Sayer, 1973; Fisher et al., 1979):

Ex = (u 2 B2) / (30 d U *) ... (5)

where E x = longitudinal dispersion coefficient [L 2 T -1], (m 2/s)d = river depth [L], (m)B = river width [L], (m)u = river flow velocity [LT -1], (m/s)U* = shear velocity [LT

-1], (m/s)

The bed shear velocity can also determined using the relationship:


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*U g h S

where g = gravityh = depth of the river (= D)S = slope of the river

As previously shown, the bed shear velocity, U *, can be estimated by (Fischer et al., 1979):

U* = 0.1 u (4)

Substituting Eq. (4) into Eq. (5) yields:

Ex = (u B 2) / (3 h) (6)

For a simplified conservative risk assessment, the 30-year low annual flow conditions may be used; i.e.,h = the river depth (m), B = the river width (m) and u = the river flow velocity (m/s), all for the 30-yearlow annual flow.

Brown and Barnwell (1987) used the following relationship in the Stream Water Quality Model:

5/63.82 x E K n u d

where E x = longitudinal dispersion coefficient, [L2 T -1], ( ft 2/s);

K = dispersion constant, [-], (dimensionless) = 5.93 (Brown and Barnwell, 1987)n = Mannings roughness coefficient, [ -], (dimensionless)u = mean stream velocity, [L T -1], (ft/s)d = mean river depth, [L], (ft)

The Manning roughness coefficient n values are given for different channel (stream) types in TableII-1 (Brown and Barnwell, 1987). Elder (1959) used a value of 5.93 for K (cited by Brown andBarnwell, 1987).


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x.02 Lateral (Transverse) Dispersion Coefficient, E y

Elder (1959) proposed the equation for predicting the lateral dispersion coefficient, E y:

Ey = d U *

where E y = lateral dispersion coefficient, [L 2 T -1], ( m 2/s)d = mean depth, [L], (m)U* = bed shear velocity, [LT -1], (m/s) = proportionality coefficient, [-], (dimensionless)

The value of = 0.23 was obtained by experiment in long, wide laboratory flumes.

Many authors have investigated the value of in both laboratory flumes and natural streams: Sayre and Chang (1968) reported = 0.17 in a straight laboratory flume. Yotsukura and Cobb (1972) report values of for natural streams and irrigation canals varying

from 0.22 to 0.65, with most values being near 0.3. Other reported values of a range from 0.17 to 0.72. The effect of bends in the channel on E y is significant. Yotsukura and Sayre (1976) reported

that: varies from 0.1 to 0.2 for straight channels, ranging in size from laboratory flumes to medium

size irrigation channels; varies from 0.6 to 10 in the Missouri River varies from 0.5 to 2.5 in curved laboratory flumes.

Fischer (1968) reports that higher values of are also found near the banks of rivers. The highervalues for are all for very fast rivers.

Okoye (1970) refined the determination of somewhat by use of the aspect ratio, = d/B, theratio of the stream depth to stream width. It was found that decreased from 0.24 to 0.033 as increased from 0.015 to 0.200. The value of varies with the aspect ratio, , ranging from 0.1 to 0.2for small laboratory flumes and medium-sized irrigation canals to 0.6 to 2.0 for the Missouri andMacKenzie Rivers (Sayre, 1973; Fisher et al., 1979; Lau and Krishnappen, 1981).

The form of Eq. 6 can be used to estimate the lateral dispersion coefficient (E y) over a widerange of river conditions (Sayer, 1973; Fisher et al., 1979). But it should be noted that may vary, andapplication of Fickian theory to lateral dispersion can be justified as long as there are no appreciablelateral currents in the stream.

For a simplified assessment, the value of 0.6 may be used if the data is not readily available(NCRP, 1996). Substitution of Eq. (5) into Eqs. (7) and using = 0.6 yields:

Ey = 0.6 d U * = 0.6 d (0.1 u) = 0.06 d u (8)

Note that the higher values for are reported for very fast rivers, and found near the banks of rivers(Fisher, 1968). For the 30-year low annual flow conditions, d is the river depth (m) for the 30-year lowannual flow rate.


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Table xx.xx. Ranges of for lateral dispersion coefficient, E y

Condition Reference0.23 Long, wide lab flumes Elder (1959)0.71 Straight lab flumes Sayre & Chang (1968)0.22 - 0.65 (~0.3) Natural stream, Irrigation canals Yotsukura & Cobb (1972)0.1 - 0.2 Straight channels, Lab flumes

Medium size irrigation channels Yotsukura & Sayre (1976)0.6 - 10 Missouri River Yotsukura & Sayre (1976)0.5 - 2.5 Curved laboratory flumes Yotsukura & Sayre (1976)0.17 - 0.72 Other reported values


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x.03 Vertical Dispersion Coefficient, E z

A fewer experimental work has been done on the vertical dispersion coefficient, E z. For a logarithmicvertical velocity distribution, Jobson and Sayre (1970) reported Eq. (7) for marked fluid particles:

* 1 Z z E U z d

(7)

where z = the vertical depth dimension, d = the mean depth, and = the von Karman coefficient,which is shown, experimentally, to be approximately = 0.4 (Tennekes and Lumley, 1972). Eq. (7)agrees with experimental data fairly closely.

The vertical dispersion coefficient, E z, in a river can also be estimated by (Sayre, 1973):

Ez = 0.067 (U *) d ... (9)

where E z = vertical dispersion coefficient [L2 T

-1], ( m

2/s)

U* = shear velocity [LT-1], (m/s)

When the approximation ( U * = 0.1 u ) is used,

Ez = 0.0067 u d ... (10)

For the 30-year low annual flow conditions, u and d refer to the river flow velocity (m/s) and the riverdepth (m), respectively, for the 30-year low annual flow.

Brown and Barnwell (1987) summarized typical values for the dispersion coefficients (E x)

along with values of the dispersion constant in streams (Table II-2). Schnoor et al. (1987) also gatheredthe available dispersion coefficients (E x) in Table 2.01.


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x.01 Vertical Eddy Diffusivity in Lakes

Vertical mixing in lakes is mechanistically different from that in rivers. The term "eddy diffusivity" isoften used to describe the turbulent diffusion coefficient for dissolved substances in lakes. Chemicaland thermal stratification serve to limit vertical mixing in lakes, and the eddy diffusivity is usuallyobserved to be a minimum at the thermocline.

Many authors have correlated the vertical eddy diffusivity in stratified lakes to mean depth,hypolimnion depth, and to the stability frequency. Mortimer (1942) correlated the vertical diffusioncoefficient with the mean depth of the lake. He found the following relationship:

1.49 z = 0.0142 d E

where E z = vertical eddy diffusivity, m2/day

d = mean depth, m

Vertical eddy diffusivities can be calculated from vertical temperature data by solving the verticalheat balance equation or by the simplified estimations of McEwen (1929) and Edinger and Geyer(1965). Schnoor (1975) has demonstrated that the mineralization and release of dissolved substancesfrom anaerobic sediment can be used to calculate average hypolimnetic eddy diffusivities. Thisapproach avoids the problem of assuming that heat (temperature) and mass (dissolved substances) willmix with the same rate constant, i.e., that the eddy diffusivity must equal the eddy conductivity.


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3-Dispersion in Rivers and Lakes_S13

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