Question

Consider the following statements regarding biochemical oxygen demand (BOD) of river water:

1. The BOD rate constant varies with river water temperature

2. The BOD rate constant does not depend on the BOD of the river water

3. The BOD rate constant is often different for different river waters

4. The BOD rate constant cannot be determined in a laboratory

Which of the above statements are correct?
1. 1 and 4
2. 1 and 3
3. 2 and 3
4. 2 and 4

Answer 1

Correct Answer - Option 2 : 1 and 3

Concept:

Biological Oxygen Demand (BOD): It can be defined as the amount of oxygen required by aerobic organisms to break down organic material present in a given sample of water at a certain temperature over a specific time period.

The BOD value is commonly expressed in “mg of oxygen consumed per litre of the sample during 5 days of incubation at 20℃” i.e. (mg/l) and is often used as a surrogate of the degree of organic pollution of water.

Dilution Method:

• This method is based upon the determination of dissolved oxygen originally present in undiluted sample of sewage and the dissolved oxygen present in diluted sample of sewage after it is subjected to incubation at a constant temperature of 20℃ for a period of 5days.
• The dilution water may be prepared by adding 1.0 ml each of phosphate buffer solution, magnesium sulphate solution, calcium chloride solution and ferric chloride solution in 1.0 litre of distilled water.
• The phosphate buffer is added to maintain pH value between ( 7 and 7.6) which is the optimum pH value for biological activity and the other salts are added to provided nutrients necessary for biological activity.
• For BOD test 300ml size bottles are used.
• Mathematically,

\(BO{D_5} = \left( {D{O_i} - D{O_f}} \right) \times Dilution\;Factor\)

Here, DO_i = Dissolved oxygen initial (mg/l) , DO_f = Dissolved Oxygen final (mg/l) and 

\(Dilution\;factor = \frac{{Final\;volume}}{{initial\;volume}} = \frac{{pure\;water\;volume + sewage\;volume}}{{sewage\;volume}}\)

Direct Method:

\({\rm{BO}}{{\rm{D}}_5} = {L_o}\left( {1 - {e^{ - kt}}} \right) = {L_o}\left( {1 - {{10}^{ - {k_D}t}}} \right)\)

Here, L_o = Oxygen equivalent of organic matter initially (mg/l), k_D and k = Deoxygenation constant at base (10 and e) (day^-1) at 20℃  and t – incubation period (days)

BOD rate constant at any temperature T° C is given by,

K_DT = K_D20°C [1.047]^{T - 20}

There are several other methods of determining k and L_o from a series of BOD measurements which include:

a. Method of least squares

b. Method of moments

c. Daily difference method

d. Rapid ratio method

e. Thomas method

Explaination for statements:

From above discussion about BOD, it is clear that:

1.BOD rate constant depends on the temperature of the river water as it is given by –

KDT = KD20°C [1.047]T - 20 

2. BOD rate constant value depends on the type of water

Water Type	KD value per day
Tap waters	< 0.05
Surface waters	0.05 – 0.1
Municipal wastewaters	0.1 – 0.15
Treated sewage effluents	0.05 – 0.1

3. It can be determined in a laboratory