McKay, Donald. "Front matter" Multimedia Environmental Models ...

or
or
When t is zero, C is zero, thus,
and
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or
©2001 CRC Press LLC
Input – Output = d(contents)/dt
2 – 2.78C – 10C = d(10 6 C)/dt
dC/(2 – 12.78C) = 10 –6 dt
ln(2 – 12.78C)/(–12.78) = 10 –6 t + IC
IC = –ln(2)/12.78
ln[(2 – 12.78C)/2] = –12.78 ¥ 10 –6 t
(2 – 12.78 C) = 2 exp(–12.78 ¥ 10 –6 t)
C = (2/12.78)[1 – exp(–12.78 ¥ 10 –6 t)]
Note that when t is zero, exp(0) is unity and C is zero, as dictated by the initial
condition. When t is very large, the exponential group becomes zero, and C
approaches (2/12.78) or 0.157 mol/m 3 . At such times, the input of 2 mol/s is equal
to the total of the output by flow of 10 ¥ 0.157 or 1.57 mol/s plus the output by
reaction of 2.78 ¥ 0.157 or 0.44 mol/s. This is the steady-state solution, which the
lake eventually approaches after a long period of time.
When t is 1 day or 86400s, C will be 0.105 mol/m 3 or 67% of the way to its
final value. C will be halfway to its final value when 12.78 ¥ 10 –6 t is 0.693 or t is
54200 s or 15 h. This time is largely controlled by the residence time of the water
in the lake, which is
Worked Example 2.10
(10 6 m 3 )/(10 m 3 /s) or 10 5 s or 27.8 h
A well mixed lake of 10 5 m 3 is initially contaminated with chemical at a concentration
of 1 mol/m 3 . The chemical leaves by the outflow of 0.5 m 3 /s, and it reacts
with a rate constant of 10 –2 h –1 . What will be the chemical concentration after 1 and
10 days, and when will 90% of the chemical have left the lake?