ALERT! ALERT! CORRECTION TO LAST LECTURE
Suppose that source enters the up gradient end of a column
At a continuous concentration of Co=1000mg/l
K = 0.1 cm/sec
dh = 10 cm
dl = 100 cm
φ = 0.2
02
Dispersivity αx = 5 cm
What will the concentration be at 50 cm after 1000sec?
cm
sec 10cm = 0.05 cm
average linear velocity
0.2 100cm
sec
cm
distance traveled in 1000sec? d = v t = 0.05
1000 sec = 50cm
sec
Kdh
v=
=
φdl
0 .1
By inspection we expect the concentration should be 0.5*Co=500mg/l
But let’s carry out the calculation
ALERT! ALERT! CORRECTION TO LAST LECTURE – EXCEL WAS CORRECT
cm
cm
m2 10000cm2
cm2
v = 0.05
Dx = v α + D* = 0.05 5cm + 1x10−10
= 0.25
2
sec
sec
sec 1m
sec
⎞⎞
⎛
⎛ x − vt ⎞
C ⎛
⎟ + exp⎛⎜ v x ⎞⎟erfc⎜ x + vt ⎟ ⎟
Often this term is ignored
C = o ⎜ erfc⎜
⎟
⎜
⎜2 D t ⎟
2 ⎜⎝
because it is typically small, but
⎝ Dx ⎠ ⎜⎝ 2 Dxt ⎟⎠ ⎟⎠
x ⎠
⎝
we carry it⎞through here.
⎛
⎛
⎞
⎜
⎟
⎜ 50cm − 0.05 cm 1000sec ⎟
⎜
⎟ including the
In this case
⎜
⎟
sec
erfc
⎜
⎟ provides a more
⎜
⎟
2
second
term
cm
mg ⎜
⎟
⎜
⎟
2
0
.
25
1000
sec
⎜
⎟
1000
accurate result
⎜
⎟
sec
⎝
⎠
l
C=
⎜
⎟
2
⎞⎟
C is greater than
⎜
⎛
⎞ ⎛⎜
cm
cm
⎟
⎜ 0.05 50cm ⎟
50cm + 0.05 1000sec ⎟
⎜
500 at the
⎟
sec
sec
⎟erfc⎜
⎜ + exp⎜
⎟
2
⎜
⎟
2
average travel
⎜
cm ⎟
cm
⎜
⎟
⎟ ⎜⎜ 2 0.25
⎜ 0.25
1000sec ⎟⎟ ⎟
time distance
⎜
sec ⎠ ⎝
⎝
sec
⎠⎠
⎝
because
⎞ dispersion
⎛
⎛
cm2 ⎞
⎟
⎜
⎟
⎜ 2.5
mg ⎜
⎛ 50cm − 50cm ⎞
sec ⎟erfc⎛⎜ 50cm + 50cm ⎞⎟ ⎟ dominates
C = 500
erfc⎜
⎟ + exp⎜
⎜
cm2 ⎟ ⎝ 31.62cm ⎠ ⎟ velocity near the
l ⎜
⎝ 31.62cm ⎠
⎟
⎜ 0.25
⎟
⎜
sec ⎠
⎝
⎠ source
⎝
mg
(erfc(0.0) + exp(10)erfc(3.1623)) = 500 mg (1.0 + 22026.46(7.743x10−7 ) = 585.3 mg
C = 500
l
l
l
1
http://inside.mines.edu/~epoeter/_GW/22ContamTrans/C1d.xls
1000
900
800
700
600
alpha = 0.5 cm
alpha = 5 cm
alpha = 50 cm
500
400
300
Dispersion dominates
advection when
200
V X i.e.
ie
100
D
0
0
1000
2000
VX
< 10
Vα + D*
3000
CONCLUSION DOES NOT CHANGE
The second term is important for calculating C @ early times near the source.
source
plume front
Normally
distributed
spreading
degree of spreading
early time
later
later
Early time is not normally distributed because the
mechanical dispersion does not move contaminants
backwards so the contaminants must travel some distance
from the source before the normal distribution is realized.
NOTE: In fact in heterogeneous environments (which
means everywhere although more pronounced some
places than others) contaminants often lag behind in low K
materials biasing the spreading and causing long late tails
2
Suppose a source continuously enters that uniform flow field
With an initial concentration of Co=1,000mg/l
pause to consider relationship of mass and concentration
Mass = Conc * Volume
Mass/Time = Conc * Velocity * Area = Conc * Q (Q is discharge)
Mass = Conc * Q * Time
Envision the source is submerged and
emanates from a 0.5cm high x 1cm wide zone
pause to consider the character of the source geometry
v = 0.05 cm/sec
dispersivity
d
spe s ty αx = 5 c
cm
dispersivity αy = 1/5 αx
dispersivity αz = 1/10 αx
What will the concentration be at 50 cm
directly down gradient after 1000sec?
pause to consider the coordinate system
v = 0.05
cm
sec
cm
m2 10000cm2
cm2
=
5cm + 1x10−10
0
.
25
sec
sec 1m2
sec
2
2
cm
m 10000cm
cm2
1
1
=
Dy = v α x + D* = 0.05
5cm + 1x10−10
0
.
05
5
sec
5
sec 1m2
sec
2
2
cm
m 10000cm
cm2
1
1
= 0.025
Dz = v α x + D* = 0.05
5cm + 1x10−10
2
10
sec 1m
sec
10
sec
Dx = v α x + D* = 0.05
⎛ x − vx t
C ⎛
C ( x, y, z, t ) = o ⎜ erfc⎜
⎜2 D t
8 ⎜⎝
x
⎝
⎛ ⎛
Y
⎜
⎞ ⎞⎜ ⎜⎜ y + 2
⎟
⎟ erf
⎟ ⎟⎜ ⎜
⎠ ⎠⎜ ⎜ 2 D x
y
⎜
v
⎝ ⎝
2 Dxt = 2 0.25
2 Dy
2 Dz
⎛
⎞
⎜ y−Y
⎟
⎟ − erf ⎜
2
⎜
⎟
x
⎜ 2 Dy
⎟
v
⎝
⎠
⎞ ⎞⎛ ⎛
⎟ ⎟⎜ ⎜ z + Z
⎟ ⎟⎜ erf ⎜
2
⎟ ⎟⎜ ⎜
x
⎟ ⎟⎟⎜⎜ ⎜ 2 Dz
v
⎠ ⎠⎝ ⎝
⎛
⎞
⎜ z−Z
⎟
⎟ − erf ⎜
2
⎜
⎟
x
⎜ 2 Dz
⎟
v
⎝
⎠
⎞⎞
⎟⎟
⎟⎟
⎟⎟
⎟ ⎟⎟
⎠⎠
cm2
1000 sec = 31.62cm
sec
x
cm2 50cm
= 2 0.05
= 14.14cm
sec 0.05 cm
v
sec
x
cm2 50cm
= 2 0.025
= 10cm
sec 0.05 cm
v
sec
3
⎛ x − vxt
C ⎛
C ( x, y, z, t ) = o ⎜ erfc⎜
⎜2 D t
8 ⎜
x
⎝
⎝
2 Dxt = 2 0.25
2 Dy
2 Dz
cm2
1000 sec = 31.62cm
sec
x
cm2 50cm
= 2 0.05
= 14.14cm
sec 0.05 cm
v
sec
x
cm2 50cm
= 2 0.025
= 10cm
v
sec 0.05 cm
sec
⎛ ⎛
Y
⎜
⎞ ⎞⎜ ⎜⎜ y + 2
⎟ ⎟ erf
⎜
⎟⎟
⎜
⎠ ⎠⎜ ⎜ 2 D x
y
⎜
v
⎝ ⎝
⎛
⎞
⎜ y−Y
⎟
⎟ − erf ⎜
2
⎜
⎟
x
⎜ 2 Dy
⎟
v
⎝
⎠
⎞ ⎞⎛ ⎛
⎟ ⎟⎜ ⎜ z + Z
⎟ ⎟⎜ erf ⎜
2
⎟ ⎟⎜ ⎜
x
⎟ ⎟⎟⎜⎜ ⎜ 2 Dz
v
⎠ ⎠⎝ ⎝
⎛
⎞
⎜ z−Z
⎟
⎟ − erf ⎜
2
⎜
⎟
x
⎜ 2 Dz
⎟
v
⎝
⎠
⎞⎞
⎟⎟
⎟⎟
⎟⎟
⎟ ⎟⎟
⎠⎠
cm
1000 sec = 0cm
sec
Y
1cm
1cm
= 0+
= 0.5cm y − = 0 −
= −0.5cm
2
2
2
0.5cm
Z
0.5cm
= 0+
= 0.25cm z − = 0 −
= −0.25cm
2
2
2
x − vxt = 50cm − 0.05
Y
2
Z
z+
2
y+
mg
0
⎛ − 0.25cm ⎞ ⎞
⎛ − 0.5cm ⎞ ⎞⎛ ⎛ 0.25cm ⎞
⎞ ⎞⎛ ⎛ 0.5cm ⎞
l ⎛⎜ erfc⎛⎜
C=
⎟ − erf ⎜
⎟ ⎟⎟
⎟ − erf ⎜
⎟ ⎟⎟⎜⎜ erf ⎜
⎟ ⎟⎟⎜⎜ erf ⎜
⎜
8
31
.
62
cm
14
.
14
cm
14
.
14
cm
10
cm
⎠
⎝ 10cm ⎠ ⎠
⎠
⎝
⎠ ⎠⎝ ⎝
⎝
⎠ ⎠⎝ ⎝
⎝
mg
(erfc(0))(erf (0.0354) − erf (− 0.0354))(erf (0.025) − erf (− 0.025))
C = 125
l
1000
C = 125
mg
(1)(0.0399 − (− 0.0399))(0.0282 − (− 0.0282))
l
C = 0.56
mg
l
What do you make of the concentration relative to the C we obtained for the slug source?
How much mass enters the system in 1000sec?
M = CQT = CAVDT
How would you go about developing a contour map of the plume?
If you did not know the dispersivities, how could you use this equation to estimate them?
How might you set up the problem if 8g/d arrived at the water table over a 1m2 area in an
aquifer with the properties and conditions used for the example?
What do you make of the concentration relative to the C we obtained for the slug source?
The slug source concentration at 50cm and 1000sec is 2 orders of magnitude higher. This is intuitively
reasonable. The slug added 1000mg at time zero.
How much mass enters the system in 1000sec? ? M = CQT = CAVDT
With the continuous source we have 1000mg/l entering an area of 0.5cm2 for 1000sec at a Darcy
velocity of 0.01cm/sec so the mass is lower and mass that entered later has not had the chance to
travel far.
M = Conc Area VelocityDarcy Time
Total Mass = 1000mg * 1liter
* 0.5cm2 * 0.01cm * 1000sec = 5mg
sec
liter
1000cm3
On the other hand all the mass for the slug case has been dispersing for the full 1000sec.
How would you go about developing a contour map of the plume?
For a given point in time calculate C at many x,y,z values then map then and contour them
If you did not know the dispersivities, how could you use this equation to estimate them?
Collect concentration data by sampling water in the field then create a contour map using
different values of dispersivity until a good match is obtained. There are some automated
techniques for calibration that we will not go into here.
How might you set up the problem if 8g/d arrived at the water table over a 1m2 area in an
aquifer with the properties and conditions used for the example?
Use formula for 2D downward spreading with a small Z, perhaps 0.0001 cm, and Y=100cm.
The formula assumes the Q is the area of the source * DarcyV = 0.01cm2 * 0.01cm/sec
So to get 8g/d Co needs to be = M / (Area VelocityDarcy Time) =
Concentration = 8000mg / 0.01cm2 * 0.01cm * 86400sec
sec
= 926mg 1liter
= 9.26mg
cm3 1000cm3
liter
4
For a material with a half-life of 12 yrs, how much is
left after 40 yrs? (Hint figure it as a % of initial mass)
N = N o e ( − λt )
λ=
0.693
T1
0.693 0.693
λ=
=
= 0.05775yrs −1
T1
12yrs
2
2
N = N o e (− λt )
N = 1 * e (− 0.05775 yrs
−1
*40 yrs
) = 0.099
or about 10%
It is often said that material is essentially gone after
7 half-lives. How much is left then?
N = N o e ( − λt )
0.693 0.693
λ=
=
= 0.693units−1
T1
1unit
λ=
0.693
T1
2
2
N = Noe(−λt )
N = 1* e(−0.693units
−1
) = 0.0078~ 0.008
*7units
or less than1%
5
What is the Retardation Coefficient for a site with
⎞
⎛ ρ
Vwater
Kd = 0.01ml
R=
= ⎜⎜1 + b K d ⎟⎟
Vcontaminant ⎝ φe
mg
⎠
effective porosity of 0.3
particle density of 2.65 g/cc
ρ b = (1 − 0.3) * 2.65
g 1000mg 1cc
mg
= 1855
cc
g
1ml
ml
⎛
⎞
ρ
= ⎜⎜ 1 + b K d ⎟⎟
φe
Vcontaminant ⎝
⎠
mg
⎛
⎞
⎜ 1855
ml ⎟
ml
⎟ = 62.8 ~ 63
0.01
R = ⎜1 +
0. 3
mg ⎟
⎜
⎜
⎟
⎝
⎠
R=
Vwater
What is the Retardation Coefficient for a site with
Ground water velocity = 0.05 cm/sec
Contaminant velocity = 0.0009 cm/sec
R=
R=
R=
Vwater
Vcontaminant
Vwater
Vcontaminant
⎞
⎛ ρ
Vwater
= ⎜1 + b K d ⎟⎟
Vcontaminant ⎜⎝ φe
⎠
⎛
⎞
ρ
= ⎜⎜ 1 + b K d ⎟⎟
φe
⎝
⎠
cm
0.05
sec = 55.56 ~ 56
=
cm
0.0009
sec
6