Question

A lake with volume 5E6 m3 has a freshwater flow of 20 m3/s. Waste is dumped into lake at 50 g/s with a decay rate of 0.2/d. What is the steady state contaminant concentration in the lake in mg/L if the lake is completely mixed?

Answer 1

The input rate of waste is:
F_in = 50 g/s

Assuming well mixed behavior the output rate equals concentration of waste times volumetric flow rate out of the lake:
F_out = C ∙ Q_out

Since the volume of the lake is constant, the volumetric flow rate equals sum of the volumetric input flow rate:
Q_out = 20 m³/s

The rate of decay equals rate of reaction time volume of the lake:
F_decay = k∙C∙V
with k = 0.2 d⁻¹ = 0.2/ (24∙3600) s⁻¹ = 2..31x10⁻⁶ 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)
= 50 g/s / (20 m³/s + 2.31×10⁻⁶ s⁻¹ ∙ 5×10⁶ m³)
= 1.58 g/m³
= 1.58 mg/L

Hence the steady state contaminant concentration in the lake is 1.58 mg/L (ANS)

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