Solution to Problem 1.3
|
t (day) |
t (sec) |
Cmax |
0.6 Cmax |
|
|
1 |
86400 |
13.5 |
8.1 |
3.0 |
|
4 |
345600 |
6.8 |
4.1 |
6.0 |
Assuming you released the dye in a very
thin horizontal layer, such that initially
= 0, then:
Thus, our two estimates of D are:
D1 = (3.0 cm)2 / 2(86400 s) = 5.2 x 10-5 cm2s-1.
and
D2 = (6.0 cm)2 / 2(345600 s) = 5.2 x 10-5 cm2s-1.
Therefore, the estimated diffusion coefficient of the dye is 5.2 x 10-5 cm2s-1.
We
must consider the boundaries of the tank when the cloud width (4
) is 40 cm.
That is, when:
4(2Dt)1/2 = 40 cm
(i.e. when Dt = 50 cm2)
t = 50 cm2/
5.2 x 10-5 cm2s-1
= 9.6 x 105 s = 11.1 days.