The Eastern Treatment Plant, which is a WWTP, in Melbourne in Australia treats 40% of Melbourne’s sewage, servicing approximately 1.2 million people living in the southern and eastern suburbs. The WWTP consists of primary, secondary and tertiary treatment steps. The effluent is treated with ozone, UV and chlorine in order to re-use a part of the water for irrigation of gardens and in agriculture.


The waste water flows after the grid and the primary sedimentation tanks to the aeration tanks, from which it flows to the secondary sedimentation tanks (clarifiers). The sludge is being recycled to the aeration tanks.

The WWTP has the following data:
Q = 380000 m3/day

Qr = 80000 m3/day

VA = 110000 m3

VS = 150000 m3

Xi ≈ 0

X = 2 kg/m3

Xr = 8.0 kg/m3

Xe = 0.02 kg/m3

BODinf = 240 mg/L

BODeff = 6 mg/L

Calculate the BOD loading (in kg/m3·day) and the F/M ratio (sludge loading) (in kg BOD/kg sludge·day) in the aeration tank.

To solve the problem, we need to calculate two things:

1. **BOD Loading (in kg/m³·day):**  
   BOD loading represents the amount of Biochemical Oxygen Demand (BOD) applied per unit volume of the aeration tank per day.

2. **F/M Ratio (Sludge Loading) (in kg BOD/kg sludge·day):**  
   The F/M ratio is the ratio of the influent BOD to the amount of mixed liquor suspended solids (MLSS) in the aeration tank. It measures the food (BOD) to microorganism (MLSS) ratio.

Let's begin with the calculations:

### 1. **BOD Loading (in kg/m³·day)**

The BOD loading can be calculated using the formula:

\[
\text{BOD Loading} = \frac{Q \times \text{BOD}_{\text{inf}}}{V_A}
\]

Where:
- \( Q \) is the influent flow rate = 380,000 m³/day
- \( \text{BOD}_{\text{inf}} \) is the influent BOD concentration = 240 mg/L = 0.240 kg/m³
- \( V_A \) is the volume of the aeration tank = 110,000 m³

Substituting the values:

\[
\text{BOD Loading} = \frac{380,000 \, \text{m}^3/\text{day} \times 0.240 \, \text{kg/m}^3}{110,000 \, \text{m}^3}
\]

\[
\text{BOD Loading} = \frac{91,200 \, \text{kg/day}}{110,000 \, \text{m}^3}
\]

\[
\text{BOD Loading} = 0.829 \, \text{kg/m}^3\cdot\text{day}
\]

### 2. **F/M Ratio (Sludge Loading) (in kg BOD/kg sludge·day)**

The F/M ratio is calculated using the formula:

\[
\text{F/M Ratio} = \frac{Q \times \text{BOD}_{\text{inf}}}{V_A \times X}
\]

Where:
- \( Q \) is the influent flow rate = 380,000 m³/day
- \( \text{BOD}_{\text{inf}} \) is the influent BOD concentration = 240 mg/L = 0.240 kg/m³
- \( V_A \) is the volume of the aeration tank = 110,000 m³
- \( X \) is the concentration of mixed liquor suspended solids (MLSS) in the aeration tank = 2 kg/m³

Substituting the values:

\[
\text{F/M Ratio} = \frac{380,000 \, \text{m}^3/\text{day} \times 0.240 \, \text{kg/m}^3}{110,000 \, \text{m}^3 \times 2 \, \text{kg/m}^3}
\]

\[
\text{F/M Ratio} = \frac{91,200 \, \text{kg/day}}{220,000 \, \text{kg}}
\]

\[
\text{F/M Ratio} = 0.415 \, \text{kg BOD/kg sludge}\cdot\text{day}
\]

### Final Results:
- **BOD Loading:** \( 0.829 \, \text{kg/m}^3\cdot\text{day} \)
- **F/M Ratio:** \( 0.415 \, \text{kg BOD/kg sludge}\cdot\text{day} \)

Would you like further details on any of these calculations, or do you have any other questions?

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Here are some related questions you might find interesting:

1. How does the F/M ratio influence the operation of a wastewater treatment plant?
2. What are the implications of a high or low BOD loading in an aeration tank?
3. What role does the sludge recycle flow rate \( Q_r \) play in the overall treatment process?
4. How would changes in influent BOD concentration affect the treatment process?
5. What are the potential effects of temperature on the aeration tank's performance?

**Tip:** Always monitor the F/M ratio in real-time to ensure that your wastewater treatment plant operates efficiently and meets effluent quality standards.

The Eastern Treatment Plant, which is a WWTP, in Melbourne in Australia treats 40% of Melbourne’s sewage, servicing approximately 1.2 million people living in the southern and eastern suburbs. The WWTP consists of primary, secondary and tertiary treatment steps. The effluent is treated with ozone, UV and chlorine in order to re-use a part of the water for irrigation of gardens and in agriculture.

The waste water flows after the grid and the primary sedimentation tanks to the aeration tanks, from which it flows to the secondary sedimentation tanks (clarifiers). The sludge is being recycled to the aeration tanks.

The WWTP has the following 
Q = 380000 m3/day
Qr = 80000 m3/day
VA = 110000 m3
VS = 150000 m3
Xi ≈ 0
X = 2 kg/m3
Xr = 8.0 kg/m3
Xe = 0.02 kg/m3
BODinf = 240 mg/L
BODeff = 6 mg/L

Calculate the BOD loading (in kg/m3·day) and the F/M ratio (sludge loading) (in kg BOD/kg sludge·day) in the aeration tank.

Calculate the BOD loading and F/M ratio for the Eastern Treatment Plant in Melbourne, treating 380,000 m³/day of sewage with a BOD concentration of 240 mg/L. This problem covers key concepts in wastewater treatment, including BOD loading and sludge loading (F/M ratio), using environmental engineering principles.

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