Quality and Characteristics of Sewage MCQ Quiz - Objective Question with Answer for Quality and Characteristics of Sewage - Download Free PDF

Concept:

Population equivalent:

• The number of individuals releasing the BOD at the rate of average standard BOD of domestic sewage equivalent to that released by industries is termed as population equivalent.

Industrial BOD = average standard BOD × population equivalent ( PE )

PE = Industrial BOD / Average standard BOD

• The average standard BOD of domestic sewage is 80 gm/c/day.
• Population equivalent represents the strength of the industrial sewage, hence help in finding the charge to be born by industries for the treatment of industrial sewage 

Calculation:

Given,

Industrial BOD = 200 kg/day = 200 × 10³ gm/day

Assume, average standard BOD = 80 gm/c/day

PE = Industrial BOD / Average standard BOD

PE = (200 × 10³)/80 = 2500

Concept:

BOD5 = (DO5 - DOi) × Dilution factor

\({{\rm{Y}}_{\rm{t}}} = {{\rm{L}}_{\rm{o}}}\left[ {1 - {{10}^{ - {{\rm{k}}_{\rm{D}}} × {\rm{t}}}}} \right]\)

Yt = BOD at ‘t’ days, Lo = ultimate BOD, kD = deoxygenation constant and t = time

Calculation:

Y₅ = 180 ppm, kD = 0.8 per day

\({{\rm{Y}}_{\rm{5}}} = {{\rm{L}}_{\rm{o}}}\left[ {1 - {{10}^{ - {{\rm{k}}_{\rm{D}}} × {\rm{t}}}}} \right]\)

180 = L_o [1 - 10^{(- 0.8 × 5)}]

L_o = 180. 018 ppm 

Biochemical Oxygen Demand (BOD) of Sewage

Biochemical Oxygen Demand (BOD) is a measure of the quantity of oxygen used by microorganisms in the aerobic decomposition of organic matter in water. It is an important indicator of water quality. The following points clarify the options given:

• BOD during 5 days at 27°C: This is generally taken as the standard measure and represents approximately 88% of the total BOD.

• 10-day BOD: This typically accounts for about 90% of the total BOD.

• Ultimate First Stage BOD (Yu): This is the total amount of oxygen required to stabilize the biodegradable organic matter in the sewage. It is a fixed quantity for a given sewage but does depend on temperature because the rate of biological activity changes with temperature.

Analyzing the Given Options

1. Statement 1 is incorrect, while statements 2 and 3 are correct.

   • Statement 1 is incorrect because the BOD of water during 5 days at 27°C is generally about 68% of the total demand, not 88%.

   • Statement 2 is correct because a 10-day BOD is indeed about 90% of the total BOD.

   • Statement 3 is correct because Yu is a fixed quantity for a given sewage but does depend on temperature.

Explanation:

Anaerobic decomposition of glucose occurs in the absence of oxygen, leading to the production of:

• Carbon dioxide (CO₂)

• Methane (CH₄)

• Water (H₂O) (as a byproduct in some cases)

However, Hydrogen sulfide (H₂S) is not a direct product of anaerobic glucose decomposition. H₂S is typically produced by sulfate-reducing bacteria when they break down sulfur-containing compounds, which is a different biochemical process.

Concept:

Biochemical oxygen demand (B.O.D.): It is the amount of dissolved oxygen needed by aerobic biological organisms in a body of water or wastewater sample to break down organic material present in a given water or wastewater sample at certain temperature (20˚C) over a specific time period (5 days).

The carbonaceous stage (First stage): It represents that portion of oxygen demand involved in the conversion of organic carbon to carbon dioxide.

The nitrogenous stage (Second stage): It represents a combined carbonaceous plus nitrogenous demand, when organic nitrogen, ammonia, and nitrite are converted to nitrate.

Nitrogenous oxygen demand generally begins after 5th day.

COD provides a measure of the amount of organic compounds in water.

The difference is that COD is less specific since it measures everything that can be chemically oxidized rather than just levels of bio-degradable organic matter.

COD also is different in that it reflects the oxidation based on a specific chemical oxidant (dichromate).COD value is always greater than BOD for every wastewater sample.

Explanation:

COD/TOC ratio:

(i) The total carbonaceous orgaincs present in given wastewater can be ascertained computing TOC of the wastewater, by converting the carbonaceous organics to CO_2, which is measured by infra-red analysis, and converted instrumentally to the organic carbon content.

(ii) COD/TOC ratio is considered to be an important factor in monitoring wastewater treatments.

(iii) TOC is related to COD through a carbon-oxygen balanced, such as in the oxidation of glucose.

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

We will have, \(\frac{{COD}}{{TOC}} = \frac{{6\ mole\ of\ {O_2}}}{{6\ mol\ of\ C}} = \frac{{6 \times 32}}{{6 \times 12}} = 2.66\)

Important Points

The TOC is rapid and accurate, but has not become popular in our country, because it needs a costly instrument and sufficient skill in using the same. 

Concept:

BOD is defined as the amount of oxygen demanded by microorganisms to decompose biodegradable organic matter present in wastewater under anaerobic conditions.

BOD is the measure of the strength of wastewater.

Within 5 days 60 – 70 % of organic content is oxidized.

Therefore 5 day BOD is taken as the standard reference

\(BOD = \left( {Initial\;BOD\; - Final\;BOD} \right) \times Dilution\;Factor\)

Calculation:

Given Initial DO = 12 mg/l

Final DO = 4 mg/l 

2 % Dilution sample, ∴ Dilution factor = 100/2 

\(BO{D_5} = \left( {12 - 4} \right) \times \frac{{100}}{2}\)

∴ BOD5 = 400 mg/l

Explanation:

1. F/M (Food to microorganisms) is the ratio of organic substances which can be decomposed by micro-organisms available in wastewater to microorganisms present in wastewater.

Mathematically, 

F/M = Daily BOD load applied to the aerator system / Total microbial mass in the system

2. Sludge age signifies that the time of biomass remains in the aeration.

Mathematically, Sludge age = Mass of suspended solids in the system / Mass of the solids leaving the system per day.

Explanation:

Biochemical oxygen demand (B.O.D.): It is the amount of dissolved oxygen needed by aerobic biological organisms in a body of water or wastewater sample to break down organic material present in a given water or wastewater sample at certain temperature (20˚C) over a specific time period (5 days).

The carbonaceous stage (First stage): It represents that portion of oxygen demand involved in the conversion of organic carbon to carbon dioxide.

The nitrogenous stage (Second stage):It represents a combined carbonaceous plus nitrogenous demand, when organic nitrogen, ammonia, and nitrite are converted to nitrate.

Nitrogenous oxygen demand generally begins after 5th day.

Explanation:

Chemical Oxygen Demand (COD) is the total measurement of all chemicals in the water that can be oxidized. A COD test measures all organic carbon with the exception of certain aromatics (benzene, toluene, phenol, etc.) which are not completely oxidized in the reaction.

Biochemical Oxygen Demand (BOD) is supposed to measure the amount of food (or organic carbons) that bacteria can oxidize.

Note:

• A high COD to BOD ratio means that the organic pollutant has very less organic matter which can be oxidized by the pollutant.
• This ratio is generally greater than 1 but can be equal to 1 if the sample has only biodegradable organics.
• For fresh water, as both BOD and COD should be negligible, hence the ratio can be equal to 1.

The COD/BOD ratio for the different forms of wastewater are as follows:

COD/BOD < 2 ⇒ Readily biodegradable effluent

2 < COD/BOD < 4 ⇒ Moderately biodegradable effluent

COD/BOD > 4: hardly biodegradable effluent

Explanation

Population equivalent

The total nos. of individuals releasing the BOD at the rate of average standard BOD of the domestic sewage to that released by the industry is called population equivalent.

Population equivalent = \(\frac{Industrial BOD}{Average Standard BOD}\)

Calculation:

Average savage from city = 95 × 10⁶ l\day

Average 5 day of BOD = 300 mg/l

Domestic sewage quantity i.e Avg = 0.08 Kg person/day

Now Industrial BOD = 95 × 10⁶ l/day × 300 mg/l

\(= 95\times10^6\times\frac{l}{day}\)\(\times300\times10^{-6}\times\frac{kg}{l}\)

= 28500 \(\frac{kg}{day}\)

Now population equivalent

 \(=\frac{2850\ kg/day}{0.08\ kg\ person\ /day}\)

= 356250

Explanation:

(i) The COD (Chemical oxygen demand) test is used to measure the content of organic matter of waste water, both biodegradable and non-biodegradble. So, we can say that COD determines the strength of sewage

(ii) The COD test uses potassium dichromate (K2Cr2O7) in presence of concentrated sulfuric acid (H2SO4) solution that oxidizes both organic (predominate) and inorganic substances in a waste water sample.

(iii) The reagents for COD test are followings:

• Standard potassium dichromate solution (0.25N)
• Concentrated sulfuric acid
• Standard ferrous ammonium sulphate titrant (0.1N)
• Ferroin indicator solution.

Characteristics of fresh and septic sewage are:

1. Colour: Fresh sewage is light brownish grey or light brown colour. The old sewage is converted to black or dark Brown due to anaerobic activities, known as stale or septic colour.

2. Temperature: The normal temperature of sewage is commonly higher than water supply due to Biological activity in it. The average temperatures of sewage in india 20⁰C, which is near about ideal temperature for the biological activities.

3. Odour: odours in waste water usually are caused by gases produced by the decomposition of organic matter. The most characteristics odour of stale or septic waste water is that of hydrogen sulphide, which is produced by anaerobic microorganisms that reduce sulphates to sulphide.

4. Turbidity: The turbidity of sewage is very high as compared to raw water.

5. Alkalinity: The alkalinity of fresh waste water sample is alkaline but as time passes it becomes acidic, because of the bacterial action in anaerobic or nitrification processes.

Explanation:

(i) The COD (Chemical oxygen demand) test is widely used as an alternative to BOD test to estimate the strength of domestic and industrial wastes as COD test can give results in few hours as compared to BOD test which takes normally 5 days.

(ii) The COD test uses potassium dichromate (K2Cr2O7) in presence of concentrated sulfuric acid (H2SO4) solution that oxidizes both organic (predominate) and inorganic substances in a waste water sample.

(iii) The reagents for COD test are followings:

• Standard potassium dichromate solution (0.25N)
• Concentrated sulfuric acid
• Standard ferrous ammonium sulphate titrant (0.1N)
• Ferroin indicator solution.

Concept:

• The negative logarithm of hydrogen ion concentration is known as pH.
• pH = - log10[H+]

If any solution is acidic then pH is less than 7 and if the solution is alkaline pH is more than 7.

• The fresh sewage is generally alkaline in nature which means pH of fresh sewage is generally more than 7.
• The exact value of pH for fresh sewage entirely depends upon the composition and percentage of sewage constituents and normally ranges from 7.5 to 9.5.
• However as the time passes, its pH tends to fall due to production of acids by bacterial action and fresh sewage turns into septic condition which is acidic in nature.

Confusion Points

pH of fresh sewage is more than 7 (average range is 7.5-9.5) and fresh sewage is alkaline in nature.

pH of septic sewage is less than 7 (average range is 6.5-4.5) and septic sewage is acidic in nature.