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A wastewater sample has a 5-day BOD of 190 mg/L with a first-order deoxygenation rate constant k = 0.01 h^-1. What is the ultimate BOD (oxygen demand), in mg/L, of the sample?

Difficulty: Medium

Correct Answer: 271

Explanation:


Introduction / Context:
Ultimate BOD (L0) represents the total oxygen demand exerted as biodegradable organics are oxidised to completion. The 5-day BOD (at 20°C) measures only a portion of this demand, depending on the reaction rate constant k. Converting BOD5 to L0 is routine in wastewater engineering for mass balance and treatment design.


Given Data / Assumptions:

  • BOD5 = 190 mg/L at 20°C.
  • First-order kinetics with k = 0.01 h^-1.
  • Time for BOD5 test t = 5 days = 120 h.


Concept / Approach:
For first-order BOD decay, the BOD exerted over time t is BOD_t = L0 * (1 − e^(−k * t)). Rearranging gives L0 = BOD_t / (1 − e^(−k * t)). This uses the classic Thomas/first-order model without nitrification adjustment.


Step-by-Step Solution:

Compute k * t = 0.01 * 120 = 1.2.Evaluate exponential term: e^(−1.2) ≈ 0.301.Compute fraction exerted: 1 − 0.301 = 0.699.Calculate L0 = 190 / 0.699 ≈ 271.8 mg/L ≈ 271 mg/L.


Verification / Alternative check:

Back-calculate expected BOD5 using L0 = 271 mg/L: 271 * (1 − e^(−1.2)) ≈ 190 mg/L, confirming consistency.


Why Other Options Are Wrong:

190 mg/L ignores unexerted demand; 475 mg/L or 3800 mg/L are inconsistent with first-order kinetics and the given k; 225 mg/L underestimates L0.


Common Pitfalls:

Forgetting to convert 5 days to hours for k in h^-1; using k for a different temperature without adjustment.


Final Answer:

271
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