Some discarded solid chemical waste dissolves slowly in
a large drain pipe in which the water is stagnant. On a particular
day, the dissolved chemical has a concentration of 0.16 M
near the solid and is essentially 0 at a location 13.6 m further
along the pipe. The transfer rate (moles per time) of the chemical
through the water in the pipe between those two points is
a. What equation describes this kind of transfer?
b. Several days later, the chemical concentration near the
solid has decreased to 0.105 M and is essentially 0 at
a location 9.9 m away. What will be the transfer rate
between the point near the solid and the point 9.9 m
away on this later day?
c. On the next day, the heavy rains cause a current of water
to flow through the drain pipe where the dissolving solid
is located. The solid now dissolves twice as fast as on
the previous day. If the concentrations are still 0.105 M
near the solid and 0 at the more remote locations, and
if the cross-sectional area for transfer is 0.3 m2, what is
the value of the mass-transfer coefficient at this time?
In the given problem, there is diffusion of moles occurring over time. Therefore, molecular diffusion equation describes this kind of transfer.
Dissolved chemical at point 1 be represented by ‘(CA)’.
Dissolved chemical at point 2 be represented by ‘(C0)’.
Distance between two points of dissolved chemical be represented by ‘x’.
Transfer rate be represented by ‘NA’.
Mass transfer coefficient is represented by ‘kc’.
For a particular day, the data given are:
(CA)1 = 0.16 M
(C0)1 = 0 M
x1 = 13.6 m
(NA)1 = 7.3 gmol/min
From this given data, calculate the value of mass transfer coefficient, kc as:
After several days, the concentration of dissolved chemical is:
(CA)2 = 0.105M
(C0)2 = 0 M
x2 = 9.9 m
The value of mass transfer ...
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