A pond with a volume of 1200 m3 is at equilibrium with inlet and outlet streams;

Question

A pond with a volume of 1200 m3 is at equilibrium with inlet and outlet streams; the concentration of a non-reactive (conservative) substance is 8 mg/L in all locations. If the inlet concentration suddenly increased to 40 mg/L, what would be the concentration in the pond (and outlet stream) after 2 days? Inlet and outlet flows are both 280 m3 per day.

Explanation / Answer

1. The input rate of pollutant is: F_in = 100 mg/L · 5 m³/s = 100 mg/L · 5000 L/s = 500 g/s Assuming well mixed behavior the output rate equals concentration of pollutant times volumetric flow rate out of the pond: F_out = C · Q_out Since the volume of the lake is constant, the volumetric flow rate equals sum of the volumetric input flow rates: Q_out = 5m³/s + 50 m³/s = 55 m³/s The rate of decay equals rate of reaction time volume of the lake: F_decay = k·C·V with k = 0.25 d?¹ = 0.25 / (24·3600) s?¹ = 2.89×10?6 s?¹ Hence, F_in = F_out + F_decay F_in = C·Q_out + k·C·V Solve for C C = F_in / (Q_out + k·V) = 500 g/s / (55 m³/s + 2.89×10?6 s?¹ · 10×106 m³) = 5.96 g/m³ = 5.96 mg/L 2. 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 => ? 1/(100 - C) dC = ? (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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