Environmental Engineering Test 3

Variations in the sewerage flow:

(i) Maximum Daily discharge = 2 × Average Daily Discharge

(ii) Maximum Hourly discharge = 3 × Average Daily Discharge

(iii) Minimum Daily discharge = \(\frac{2}{3}\) × Average Daily Discharge

Concept:

Biochemical Oxygen demand (BOD): It is the total amount of oxygen required to oxidize the biodegradable organic matter present in wastewater through microbial utilization of organics.

Chemical Oxygen demand (COD):- It is the total amount of oxygen required to oxidize both the biodegradable and non-biodegradable organic matter present in wastewater.

COD - BOD = Non biodegradable organic matter

Calculation:

Sample 1

\(\frac{{{\rm{COD}}}}{{{\rm{BO}}{{\rm{D}}_5}}} = 1.5\) ⇒ COD = 1.5 BOD₅

Sample 2

\(\frac{{{\rm{COD}}}}{{{\rm{BO}}{{\rm{D}}_{\rm{s}}}}} = 3\) ⇒ COD = 3 BOD_s

⇒ The lower value of COD means less non-biodegradable organic matter and a higher value of COD means more amount of non-biodegradable organic matter.

Hence, the lower value of \(\frac{{{\rm{COD}}}}{{{\rm{BO}}{{\rm{D}}_{\rm{s}}}}}\) means, higher percentage of biodegradable organic matter.

Important Point:

Industrial sources produce wastewater having higher value of non-biodegradable organic matter compared to that in a domestic sewage. Also industrial sources contains inorganic matter in waste water.

Hence COD/BOD value is generally higher for an industrial source.

Concept:

Population Equivalent:

The number of person which produce the amount of BOD at the rate average standard BOD of domestic sewage per person per day equal to that produced by industrial sewage is called the population equivalent of industrial sewage.

Calculation:

Average standard BOD of domestic sewage is 80 gms per person per day.

Total BOD produced by industry = 240 × 10³ × 1500 mg = 360 × 10⁶ mg = 360000 gm

∴ Population equivalent \(= \frac{{Total\;BOD\;produced}}{{Average\;standard\;BOD}} = \frac{{360000}}{{80}} = 4500\)

Concept:

Dry weather Flow: It is the sanitary sewage flown into sewer and it excludes the discharge cause due to precipitation. It is calculated as:

Dry weather flow = Total population × Contribution by per capita

Wet Weather Flow: It includes both sanitary sewage as well as storm water.

It is calculated as:

Wet weather flow = Dry weather flow + Storm discharge

To calculate storm discharge, rational formula is used. According rational formula,

\({\rm{Q}} = \frac{1}{{36}} \times {\rm{k}} \times {\rm{A}} \times {\rm{i}}\)

Where,

k = runoff coefficient or impermeability factor, A = catchment area in hectares, and i = intensity of rainfall in cm/hr.

If time of concentration is given and

If 5 min ≤ T_c ≤ 20 min, then \({\rm{i}} = \frac{{75}}{{10 + {{\rm{T}}_{\rm{c}}}}}\)

If T_c ≥ 20 min, then \({\rm{i}} = \frac{{100}}{{20 + {{\rm{T}}_{\rm{c}}}}}\)

Where T_c is in minute and i is in cm/hr.

Combined sewer is always designed for wet weather flow.

Calculation:

Given: Total population = 50000, Water supply rate = 200 lpcd, and Wastewater generation rate = 75%

Design dry weather flow = 3 × 50000 × 200 × 0.75 = 22.5 × 10⁶ l/day = \(\frac{{22.5 \times {{10}^6} \times {{10}^{ - 3}}}}{{86400}} = 0.26{\rm{\;}}{{\rm{m}}^3}/{\rm{s}}\)

Given: Time of concentration = 25 min, k = 0.6, area (A) = 100 hectares

So, rainfall intensity \(\left( {\rm{i}} \right) = \frac{{100}}{{20 + {{\rm{T}}_{\rm{c}}}}} = \frac{{100}}{{20 + 25}} = 2.22{\rm{cm}}/{\rm{hr}}\)

Storm discharge \(= \frac{1}{{36}} \times {\rm{k}} \times {\rm{A}} \times {\rm{i}} = \frac{1}{{36}} \times 0.6 \times 100 \times 2.22 = 3.7{\rm{\;}}{{\rm{m}}^3}/{\rm{s}}\)

Concept:

Theoretical oxygen Demand: If the chemical formulae and the quantity of all organic matter present in the wastewater is known, the exact amount of oxygen required to oxidise them can be calculated stoichiometrically. This is called theoretical oxygen demand.

Calculation:

Balanced Reaction of C₈H₁₆O₈ is

C₈ H₁₆ O₈ + 8 O₂ → 8 CO₂ + 8 H₂O

∴ For 1 mole C₈H₁₆O₈, 8 moles of O₂ is required

i.e 240 gm of C₈H₁₆O₈ required 256 gm of O₂

∴ 240 mg/L of C₈H₁₆O₈ requires \(\frac{{256}}{{240}} \times 240 = 256\;mg\) of O₂

Hence the Theoretical oxygen demand is 256 mg/L

Important point:

1) Number of moles \(= \frac{{{\rm{Weight\;in\;grams}}}}{{{\rm{Molecular\;Weight}}}}\)

2) Biochemical Oxygen Demand (BOD) → Biochemical Oxygen Demand is the quantity of oxygen required for oxidation of biodegradable organic matter present in the water sample by aerobic biochemical action.

3) Chemical Oxygen Demand (COD) → It is the quantity of oxygen required to oxidize both biodegradable and non-biodegradable organic matter present in wastewater.

4) Total Organic Carbon (TOC) → Total organic carbon is the method of expressing organic matter in terms of carbon content.

5) ThOD > COD > BOD > TOC.

Concept:

Seeded water is used for the dilution in the BOD test. The seeded water is the water with seeding of mixed bacterial culture.

For a seeded sample:

\(BO{D_5} = \frac{{\left( {{D_1} - {D_2}} \right) - \left( {{B_1} - {B_2}} \right)\left( {1 - P} \right)}}{P}\)

Where,

D₁ = DO of the diluted sample immediately after dilution (mg/l)

D₂ = DO of diluted sample after 5 days (mg/l)

B₁ = DO of seeded control sample before incubation (mg/l)

B₂ = DO of seeded control sample after 5 days incubation (mg/l)

P = Decimal volumetric fraction of sample used = Volume of undiluted sample/Volume of the diluted sample.

Calculation:

D₁ = 9 mg/L,, D₂ = 3.6 mg/ℓ, B₁ – B₂ = 2 mg/L

\(P = \frac{{Volume\;of\;undiluted\;sample}}{{Volume\;of\;diluted\;sample}} = \frac{{30}}{{300}} = \frac{1}{{10}}\)

\(\therefore BO{D_5} = \frac{{\left( {{D_1} - {D_2}} \right)-\left( {{B_1} - {B_2}} \right)\left( {1 - P} \right)}}{P}\)