1 an agitator is a device which produces turbulent mixing in a tank fo

Question

1. An agitator is a device, which produces turbulent mixing in a tank (for example, this
can be an oscillating grid without a mean flow). The following experiment was carried out
to test the mixing efficiency of an agitator in a large tank, i.e., side wall or bottom boundary
effects are negligible: 100 g of a 10% NaCl solution (25°C) was released at the water surface
and the following concentration measurements were made at a point located 2 m below
the release point:
a)
b)
Time, hrs Concentration, ppb
0
0.5
1
1.5
d)
2
3
4
2
130
175
160
127
83
67
Assuming homogeneous, isotropic and stationary turbulence determine the
diffusivity. [5 pts]
If in actuality the tank is 10 m deep, at what time would the tank bottom exert a
significant effect on the concentration value at the measuring point? It is up to you
to define "significant effects"! What is the concentration at 5, 10 and 20 hours of
the start, respectively? [5 pts]
c)
If the agitator were not operating (i.e., molecular diffusion), how long would it take
for the concentration to reach its maximum value at the distance of 2m? What is
the maximum value? The water temperature is 25°C. [5 pts]
Assuming that, for some reason, the salt solution is undergoing a first order loss
with a half time of 3.5 hours, how would the answers obtained under b) change? [5
pts]/nd)
Assuming that, for some reason, the salt solution is undergoing a first order loss
with a half time of 3.5 hours, how would the answers obtained under b) change? [5
pts]
Note: Do the problem using an analytical solution. If you also do part a and b using a 2D
rectangular numerical approach, you will receive 10 pts extra credit. This is much easier
to do in Matlab than in Excel even though Excel works great for a few time steps...
Solutions:
(a) Your solution can vary somewhat depending on your approach, but I found E=2.1
cm2/s works well.
(b) The 2 main images at z=20 m and z--20 m do not affect the solution for the 5, 10,
and 20 hrs. You would have to wait until close to 1000 hours for the images to make an
impact.
t (hours)
0
5
10
20
t (sec)
0
18000
36000
72000
(c)
Solution:
Max
concentration
Time-
(d)
t (hours) t (sec)
0
5
10
0
18000
36000
72000
M (mg)
10000
10000
10000
10000
E
(cm³/s)
2.1
2.1
M (mg)
10000
10000
10000
10000
2.1
2.1
184.03 ppb
13.00 years
E
(cm³/s)
2.1
NİNİN
Cpeak
0.00E+00
6.11E+01
2.16E+01
7.64E+00
Cpeak
0.00E+00
2.1 6.11E+01
2.1
2.16E+01
2.1 7.64E+00
C(t) ppb
0.00
46.89
18.92
7.19
C(t) ppb
0.00
17.42
2.61
0.14