Question

Asked Dec 4, 2019

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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

7.3 gmol/min.

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?

Step 1

(a)

In the given problem, there is diffusion of moles occurring over time. Therefore, molecular diffusion equation describes this kind of transfer.

Step 2

(b)

Let,

Dissolved chemical at point 1 be represented by ‘(C_{A})’.

Dissolved chemical at point 2 be represented by ‘(C_{0})’.

Distance between two points of dissolved chemical be represented by ‘x’.

Transfer rate be represented by ‘N_{A}’.

Mass transfer coefficient is represented by ‘k_{c}’.

For a particular day, the data given are:

(C_{A})_{1} = 0.16 M

(C_{0})_{1} = 0 M

x_{1} = 13.6 m

(N_{A})_{1} = 7.3 gmol/min

From this given data, calculate the value of mass transfer coefficient, k_{c} as:

Step 3

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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