Question

Ultimate BOD of a river water sample is 20 mg/L. BOD rate constant (natural log) is 0.15 day^-1. The respective values of BOD (in %) exerted and remaining after 7 days are:

1.

45 and 55

2.

55 and 45

3. 65 and 35
4. 75 and 25

Answer 1

Correct Answer - Option 3 : 65 and 35

Concept:

Biochemical oxygen demand (B.O.D):

B.O.D is defined as the amount of oxygen demand by the micro-organisms C₅H₇NO₂ (bacteria) present in it to decompose biodegredable organic matter in waste water under aerobic conditions. B.O.D is measure of strength of waste water.

The amount of organic matter removed (Y_t) is calculated by

Y_t = L_o (1 - e^-Kt)

where,

t = Time in days

L_o = Initial organic matter in waste water at time t = 0

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

K_T = K₂₀ (1.047)^{T - 20}

T = Temperature in degrees

Percentage of organic matter remaining is given by

\({\rm{\% \;exerted}} = \frac{{{\rm{B}}.{\rm{O}}.{{\rm{D}}_{\rm{t}}}}}{{{\rm{B}}.{\rm{O}}.{{\rm{D}}_{\rm{u}}}}} × 100\)

% remaining = 100 - %Exerted

Calculation:

B.O.D remaining after 7 days

B.O.D₇ = B.O.D_u (1- e^-Kt)

B.O.D₇ = 20 × (1- e^{-0.15 × 7})

B.O.D₇ = 20 × 0.65 =13 

% exerted or removed \( = \frac{{13 × 100}}{{20}} = 65\% \)

B.O.D remaining after 7 days = (100 – 65) = 35%

Mistake point:

K = B.O.D rate constant 

Normal log means log_e, K_T has following relation between the (base e) value and (base 10) value.

K_T (base e) = 2.3 × K_T (base 10)