If we only consider advection and
start with a "point" of material with Co=1000mg/l
A point has no volume so can it have a concentration? So why do we say a “point”?
K = 0.1 cm/sec
dh = 10 cm
dl = 100 cm
φ = 0.2
How long will it take for the material to move 50cm?
What will the concentration be at that location at that time?
in the down gradient direction
cm
0 .1
Kdh
sec 10 cm = 0 .05 cm
v=
=
φ dl
0 .2 100 cm
sec
d = vt
d
50 cm
t= =
= 1000 sec = 0 .28 hr
v 0 .05 cm
sec
mg
C = C o = 1000
l
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 know that the concentration should be 0.5*Co=500mg/l
But let’s carry out the calculation
1
ALERT! ALERT! CORRECTION IN SUBSEQUENT LECTURE
Experiment with the spreadsheet
http://inside.mines.edu/~epoeter/_GW/22ContamTrans/C1d.xls
http://inside.mines.edu/~epoeter/_GW/22ContamTrans/C1d.xls
Note the values of C using only the first term and then both terms at
times and locations where y
your intutition allows y
you to know the
concentration.
When is use of the second term important? When does excel cause it
to be in error?
Try
x = 50, 49, 51, 0, 100 then 10, 30, 200
Consider other times
TRY 10000sec
x = 500, 490, 510, 0, 1000
then x = 100, 200, 300, 400, 450, compare 450 to 550 ?symmetrical?
Where and when can you know the correct C?
The second term is important for calculating C @ early times near the source.
v = 0.05
cm
sec
x = 0.05
cm
1000sec = 50cm so X = Y = Z = 0 and we want Cmax
sec
cm
m2 10000cm2
cm2
5cm + 1x10−10
0
.
25
=
sec
sec 1m2
sec
2
2
1
cm
1
m 10000cm
cm2
Dy = v α x + D* = 0.05
5cm + 1x10−10
=
0
.
05
sec
5
sec
5
sec 1m2
2
2
1
1
cm
m 10000cm
cm2
5cm + 1x10−10
Dz = v α x + D* = 0.05
=
0
.
025
10
sec
10
sec 1m2
sec
Dx = v α x + D* = 0.05
C=
M
8(πt )
3
2
Dx Dy Dz
1000mg
C=
8(π 1000sec)
C = 0.0402
3
2
0.25
cm2
cm2
cm2
0.05
0.025
sec
sec
sec
mg 1000cm3
mg
mg
= 40.2
~ 40
3
cm
l
l
l
2