Question

A lagoon with volume 1,200 m3 has been receiving a steady flow of a conservative waste at a rate of 100 m3/day for a long enough time to assume that steady-stateconditions apply. The waste entering the lagoon has a concentration of 10 mg/L. assuming completely mixed conditions,
(a) What should be the concentration of pollutant in the effluent leaving the lagoon?
(b) If the input waste concentration suddenly increased to 100 mg/L, what the concentration in the effluent be 7 days later?
C)A Nonconservative of pullutant with rate constant k=0.20/d.

Answer 1

a)
In steady flow regime entering volumetric flow rate and effluent flow rate are equal

So rate of change of pollutant is given by:
d(C·V) dt = C_in·Q - C·Q
(C pollutant concentration, V volume of the lagoon, Q volumetric flow rate)
Since V is constant
V·(dC/dt) = Q·(C_in - C)
<=>
(dC/dt) = (Q/V)·(C_in - C)

At steady state conditions
dC/dt = 0
Hence
C = C_in = 10 mg/L

b)
For this part you need to solve the differential equation from part a)
In difference to part a)) input concentration is 100 mg/l
(dC/dt) = (100 m³/d / 1200 m³)·(100 mg/L - C)
Or simply (with C in mg/L and t in days)
(dC/dt) = (1/12)·(100 - C)

To solve this 1st order differential equation separate variables and integrate
1/(100 - C) dC = (1/12) dt
=>
1/(100 - C) dC = (1/12) dt
=>
integral 1/(100 - C) dC =integral (1/12) dt
=>
- ln(100 - C) = (1/12)·t + a
(a is the constant of integration)
Apply initial condition to find it:
C(t=0) = 10
<=>
- ln(100 - 10) = (1/12)·0 + a
a = - ln(90)

Hence,
- ln(100 - C) = (1/12)·t - ln(90)
<=>
100 - C = e^(ln(90) - (1/12)·t) = e^(ln(90)) · e^(-(1/12)·t)
<=>
C = 100 - 90·e^(-(1/12)·t)

After 7 days:
C = 100 - 90·e^(-(1/12)·7) = 49.8 (mg/L)