For a wastewater sample, the three-day biochemical oxygen demand at incubation temperature of 20°C (BOD_{3day, 20°C}) is estimated as 200 mg/L. Taking the value of the first order BOD reaction rate constant as 0.22 day^{-1}, the five-day BOD (expressed in mg/L) of the wastewater at incubation temperature of 20°C (BOD_{5day, 20°c}) would be _________

This question was previously asked in

GATE CE 2016 Official Paper: Shift 2

CT 1: Ratio and Proportion

2672

10 Questions
16 Marks
30 Mins

__Concept:__

B.O.D at any temperature is given by,

**\({\rm{B}}.{\rm{O}}.{\rm{D}} = {{\rm{L}}_{\rm{o}}} \times \left[ {1 - {{\rm{e}}^{ - {\rm{KT}}}}} \right]\)**

Where,

L_{o} = initial organic matter in wastewater. i.e at time t = 0

K = B.O.D rate constant depends on temperature.

T = number of days

**Calculation:**

K = 0.22 day^{-1}

T_{1} = 5 days

T_{2} = 3 days

B.O.D_{3} = 200 mg/lit

\(BO{D_5} = {L_0}\left( {1 - {e^{ - 0.22 \times 5}}} \right)\) ………(i)

\(BO{D_3} = {L_0}\left( {1 - {e^{ - 0.22 \times 3}}} \right)\) ………(ii)

From equation (i) and (ii), we get

\(\begin{array}{l} \frac{{BO{D_5}}}{{BO{D_3}}} = \frac{{\left( {1 - {e^{ - 0.22 \times 5}}} \right)}}{{\left( {1 - {e^{ - 0.22 \times 3}}} \right)}}\\ BO{D_5} = 200 \times \frac{{0.667}}{{0.483}} \end{array}\)

B.O.D_{5 }= 200 × 1.38