College

2 ML/day of sewage water with a BOD of 380 mg/L is treated through five ponds with the following dimensions:

1. Pond 1: 65m x 80m
2. Pond 2: 70m x 30000cm (note: convert to meters: 70m x 300m)
3. Pond 3: 40m outer diameter with a wall thickness of 1.5m
4. Pond 4: 105m x 55m
5. Pond 5: 35m inner diameter

All ponds are 1.5m deep, except for the 35m diameter pond, which is 1m deep. Determine:

1. The average detention time for each pond.
2. The average BOD loading (g/m²·d).
3. The organic loading on the smallest pond.

Answer :

In this question, we have to use the following formulas - Detention Time = Volume of Pond / Flow Rate

BOD Loading = (Flow Rate x BOD Concentration) / Surface Area of Pond

Organic Loading = Flow Rate x BOD Concentration

Detention Time: In the context of wastewater treatment, detention time refers to the length of time that wastewater or sewage remains in a treatment process or a specific unit within a treatment system. It is a crucial parameter that determines the efficiency of various treatment processes and allows sufficient time for desired reactions to occur.

BOD Loading (Biochemical Oxygen Demand Loading): BOD loading is a measure of the amount of organic pollutants, specifically the biochemical oxygen demand, present in a given volume of wastewater over a specified time period. BOD loading is often expressed in terms of mass (e.g., kilograms or pounds) of organic matter per day or per unit volume of wastewater.

Organic Loading: Organic loading refers to the rate at which organic pollutants are added to a treatment system or process. It represents the amount of organic matter, typically expressed as chemical oxygen demand (COD) or biochemical oxygen demand (BOD), applied to a given volume or area of the treatment system per unit time.

To determine the average detention time for each pond, we need to use the formula:

Let's calculate the detention time for each pond:

1. Pond with dimensions 65m x 80m:
Volume = Length x Width x Depth = 65m x 80m x 1.5m = 7800 cubic meters
Detention Time = 7800 cubic meters / 2 ML/day = 7800 cubic meters / (2 x 10^6 liters/day) = 0.0039 days

2. Pond with dimensions 70m x 30000cm(300m):

Volume = Length x Width x Depth
= 70m x 300m x 1.5m = 31500 cubic meters
Detention Time = 31500 cubic meters / 2 ML/day
= 31500 cubic meters / (2 x 10^6 liters/day)
= 0.01575 days

3. Pond with dimensions 40m outer diameter with wall thickness of 1.5m:
Inner Diameter = Outer Diameter - 2 x Wall Thickness
Inner Diameter = 40m - 2 x 1.5m = 37m
Volume = π x (Outer Radius^2 - Inner Radius^2) x Depth
Outer Radius = Outer Diameter / 2 = 40m / 2 = 20m
Inner Radius = Inner Diameter / 2 = 37m / 2 = 18.5m
Volume = π x (20m^2 - 18.5m^2) x 1.5m = 464.5 cubic meters
Detention Time = 464.5 cubic meters / 2 ML/day
= 464.5 cubic meters / (2 x 10^6 liters/day)
= 0.00023225 days

4. Pond with dimensions 105m x 55m:
Volume = Length x Width x Depth
= 105m x 55m x 1.5m = 8651.25 cubic meters
Detention Time = 8651.25 cubic meters / 2 ML/day
= 8651.25 cubic meters / (2 x 10^6 liters/day)
= 0.004325625 days

5. Pond with dimensions 35m inner diameter:
Volume = π x (Inner Diameter^2) x Depth
= π x (35m^2) x 1m = 3848.45 cubic meters
Detention Time = 3848.45 cubic meters / 2 ML/day
= 3848.45 cubic meters / (2 x 10^6 liters/day)
= 0.001924225 days

Now, let's calculate the average BOD loading (g/m2.d):

To calculate the surface area of each pond, we'll use the following formulas:

1. Pond with dimensions 65m x 80m:
Surface Area = Length x Width = 65m x 80m = 5200 square meters

2. Pond with dimensions 70m x 30000cm(300m):

Surface Area = Length x Width = 70m x 300m = 21000 square meters

3. Pond with dimensions 40m outer diameter with wall thickness of 1.5m:
Inner Diameter = Outer Diameter - 2 x Wall Thickness = 40m - 2 x 1.5m = 37m
Surface Area = π x ((Outer Radius^2) - (Inner Radius^2))
Outer Radius = Outer Diameter / 2 = 40m / 2 = 20m
Inner Radius = Inner Diameter / 2 = 37m / 2 = 18.5m
Surface Area = π x ((20m^2) - (18.5m^2)) = 199.01 square meters

4. Pond with dimensions 105m x 55m:
Surface Area = Length x Width = 105m x 55m = 5775 square meters

5. Pond with dimensions 35m inner diameter:
Surface Area = π x ((Inner Diameter^2) / 4)
= π x ((35m^2) / 4) = 962.11 square meters

Now, let's calculate the average BOD loading for each pond:

1. Pond with dimensions 65m x 80m:
BOD Loading = (2 ML/day x 380 mg/L) / 5200 square meters
= 0.00073 g/m2.d

2. Pond with dimensions 70m x 30000cm(300m):

BOD Loading = (2 ML/day x 380 mg/L) / 21000 square meters
= 0.0003619 g/m2.d

3. Pond with dimensions 40m outer diameter with wall thickness of 1.5m:
Inner Diameter = Outer Diameter - 2 x Wall Thickness = 40m - 2 x 1.5m = 37m
BOD Loading = (2 ML/day x 380 mg/L) / 199.01 square meters
= 0.003816 g/m2.d

4. Pond with dimensions 105m x 55m:
BOD Loading = (2 ML/day x 380 mg/L) / 5775 square meters
= 0.0001307 g/m2.d

5. Pond with dimensions 35m inner diameter:
BOD Loading = (2 ML/day x 380 mg/L) / 962.11 square meters
= 0.0007891 g/m2.d

Finally, let's determine the organic loading on the smallest pond:

Organic Loading = 2 ML/day x 380 mg/L= 760 g/day

To learn more about waste water treatment: https://brainly.com/question/10022625

#SPJ11