If the BOD3 of a waste water sample is 75 mg/l and reaction

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If the BOD₃ of a waste water sample is 75 mg/l and reaction rate constant k (base e) is 0.345 per day, the amount of BOD remaining in the given sample after 10 days is

1. 3.21 mg/l
2. 3.45 mg/l
3. 3.69 mg/l
4. 3.92 mg/l

Correct Option: C

L_t = L_o e^–kt
k = 0.345/d
BOD = L_o – L_t = L_o (1 – e^–kt)
BOD₃ = 75 = L_o(1 – e^–0.345×3)
∴ L_o = 116.28 mg/l
∴ BOD₁₀
Remaining BOD,
L_t = L_o e^–kt
= 116.28 × e^{–10 × 0.345} = 3.69 mg/l.