A lagoon with volume 1,200 m3 has been receiving a steady flow of a conservative

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-state conditions 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.

Explanation / Answer

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)

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