Calculating the number of tanks in a wastewater treatment system involves several factors, including the treatment process employed, the capacity of the tanks, and the expected flow rate and retention time. Here's a general approach:

1. Determine the Treatment Process: Identify the specific treatment processes involved in your wastewater treatment system. Common processes include primary treatment (settling), secondary treatment (biological treatment), and tertiary treatment (filtration, disinfection).

2. Estimate Flow Rate: Determine the average flow rate of wastewater entering the system. This could be based on population served, industrial output, or other relevant factors.

3. Calculate Retention Time: Each treatment stage typically requires a certain retention time for effective treatment. This is the amount of time wastewater spends in a tank or treatment unit. Retention time can vary based on the treatment process and the characteristics of the wastewater. For example, in biological treatment systems, the retention time is crucial for microbial activity to break down organic matter.

4. Determine Tank Capacity: Once you have the average flow rate and desired retention time, you can calculate the required tank capacity. Tank capacity is usually calculated by multiplying the flow rate by the retention time.

   Tank Capacity = Flow Rate × Retention Time

5. Consider Redundancy and Peak Flow: It's important to account for peak flow conditions and provide redundancy in the system to handle fluctuations and maintenance downtime. This might involve adding extra tanks or designing the system to handle peak flow conditions.

6. Design for Specific Treatment Stages: Different treatment stages may require different types and sizes of tanks. For example, primary settling tanks are typically larger and shallower, while biological treatment tanks may be deeper with specific aeration systems.

7. Consult Regulations and Guidelines: Ensure that your design meets any regulatory requirements or guidelines for wastewater treatment in your region. These may specify minimum retention times, treatment efficiency targets, or other design criteria.

8. Seek Professional Assistance if Necessary: Designing a wastewater treatment system can be complex and may require input from engineers, environmental scientists, or other specialists. If you're unsure about any aspect of the design process, consider consulting with professionals experienced in wastewater treatment system design.

By following these steps and considering the specific requirements of your wastewater treatment system, you can calculate the number and size of tanks needed to effectively treat wastewater.

The collection pit will be equipped by the investor:

• Form: brick construction;
• Volume: 2 m3 ;
• Dimensions:

length x width x height = 1,000 x 1,000 x 2,000 mm.

Calculate the air conditioning tank

Calculate the volume of the conditioning tank

    V = Qh x t. In there:

• Qh: average hourly wastewater flow, Qh = 1.25 m³ /hour;
• t: average water retention time (t = 5 – 9 hours), choose t = 7 hours.

Deduce: V = 7 x 1.25 = 8.75 m³           .

Choose the good plastic tank with a volume of 10 m³ instead of the traditional reinforced concrete tank.

Calculate the air flow to the air conditioning tank

  Qkk = qkk x V. In which:

• qkk: air supply rate in the regulation tank (10 - 15 L/min.m³ ) according to the document on calculating and designing wastewater treatment works - choose qkk = 15 L/min.m³ ;
• V: volume of regulation tank.

Deduce: Qkk = 15 x 10 = 150 = L/minute = 9 m³ /hour

Calculate anoxic tank

Calculate the water retention time of anoxic tank

t = (Nv – Nr)/(P25^oCN2 x X). In there:

• Nv: input nitrogen content, Nv = 85 mg/L;
• Nr: nitrogen content after treatment, Nr = 15 mg/L;
• P 25^oCN2 : denitrification rate at 25⁰C: P 25^oCN2 = P 20^oCN2 x 1.09t – 20 x (1 – DO) = 0.1 x 1.0925– 20 x (1 – 0.15) = 0.13 (mg/mg.day)
• P 20^oCN2 : denitrification rate at 20⁰C is 0.1 (mg/mg.ng.d) o t: wastewater temperature, t = 25⁰C
• DO: dissolved oxygen in anoxic tank, DO = 0.15 mg/L o X: activated sludge concentration (2,500 – 4,000 mg VSS/L), choose X = 3,000 mg VSS/L.

In conclusion, the water retention time of the tank: t = 24 x (85 – 15)/(0.13 x 3000) = 4.3 h

    Calculate the volume of anoxic tank

   V = Qh x t. In there: Qh: average hourly wastewater flow, Qh = 1.25 m³ /hour;

• t: water retention time of anoxic tank, t = 4.3 hours.

Deduce: V = 4.3 x 1.25 = 5,375 m³

Choose an anoxic tank, the plastic tank with a volume of 10 m³ , instead of the traditional reinforced concrete tank.

Calculate the aerotank

    Calculate the volume of the aerotank

Calculation formula based on activated sludge storage time in the tank:

V = [0c x Q x (S0 – S) x Y]/[X x (1 + Kd x 0c). In there:

• 0c: sludge retention time (5- 15 days), choose 0c = 10 days;
• Q: wastewater flow, Q = 30 m³ ;
• S0: BOD₅ content of input wastewater (mg/L), S0 = 300 mg/L;
• S: BOD₅ content of wastewater after leaving the aerotank (mg/L), S = 60mg/L;
• Kd: intracellular decomposition coefficient (day-1 ), Kd = 0.06 day-1 ;
• Y: sludge load factor (0.4 – 0.8 mg VSS/mg BOD₅), choose Y = 0.5 mg VSS/mg BOD₅;
• X: activated sludge concentration (2,500 – 4,000 mg VSS/L), select X = 3,000 mg VSS/L.

Deduce: V = [10 x 30 x (300 – 60) x 0.5]/[3,000 x (1 + 0.06 x 10) = 7.5 m³

Choose the Good plastic tank with a volume of 10 m³ instead of the traditional reinforced concrete tank.

Water retention time of aerotank

t = = V/Q. In there:

• V: aerotank volume, V = 10 m³ ;
• Q: daily

Wastewater flow, Q = 30 m³                                                           .

      Deduce: t = 1/3 day = 8 hours.

Check the working criteria of the aerotank

Check the ratio of substrate mass to activated sludge mass:

F/M = S0/(tx X) = 300/(1/3 x 3,000) = 0.3 kg BOD5/kg MLSS.day

F/M ratio = 0.3 is within the allowable limit for a completely disturbed aerotank (0.2 - 0.6 kg BOD5/kg MLSS.day)

Volumetric load: La = 10-3 x (S0 x Q)/V = 10-3 x (300 x 30)/10 = 0.9 kg BOD/m³ .day

La = 0.9 is within the allowable limit for a completely disturbed aerotank (0.8 - 1.9 kg BOD/m³ .day).

Calculate the amount of activated sludge produced in 1 day

Sludge growth rate: Yb = Y/(1 + 0c x Kd). In there:

• Y: sludge load factor (0.4 – 0.8 mg VSS/mg BOD5), choose Y = 0.5 mg VSS/mg BOD₅;
• Kd: intracellular decomposition coefficient (day-1 ), Kd = 0.06

day-1 ; • 0c: sludge retention time (5- 15 days), choose 0c = 10 days.

Deduce Yb = 0.5/(1 + 10 x 0.06) = 0.3125

Amount of activated sludge produced by BOD5 reduction:

Px = Q x (S0 – S) x Yb = 30 x (300 – 60) x 0.3125 x 10-3 = 2.25 (kg/day.night)

Calculate the amount of oxygen needed

Required amount of oxygen under standard conditions (no need for nitrogen treatment):

OC0 = [Q x (S0 – S)]/[1,000 xf] – 1.42 x Px. In there:

• Q: daily wastewater flow, Q = 30 m³ ;
• S0: BOD5 content of input wastewater (mg/L), S0 = 300 mg/L;
• S: BOD5 content of wastewater after leaving the aerotank (mg/L), S = 60 mg/L;
• f: conversion constant from BOD5 to BOD₂₀, BOD5/BOD₂₀ = 0.55;
• 42: conversion factor from cells to COD;
• Px: amount of activated sludge produced by BOD5 reduction, Px = 2.25 kg/day.night. Deduce: OC0 = [30 x (300 – 60)]/[1,000 x 0.55] – 1.42 x 2.25 = 9.9 (kgO_²/day.night)

The amount of oxygen needed under real conditions:

OCt = OC0 x Cs/(Cs – C) x 1/1.024t – 20 x 1/ÿ. In there:

• Cs: saturated oxygen concentration in water at 200C, Cs ÿ 9.08 (mg/L);
• C: oxygen concentration that needs to be maintained in the tank (1.5 - 2 mg/L) according to the document Calculation and design of wastewater treatment works, choose C = 2 mg/L.
• t: wastewater temperature, t = 25⁰C
• ÿ: coefficient to adjust the amount of oxygen absorbed into wastewater due to the influence of residue and surfactant content (0.6 – 0.94), choose = 0.7.

Deduce: OCt = 9.9 x 9.08/(9.08 – 2) x 1/1.02425 – 20 x 1/0.7 = 16 (kg/day)

MBR module

MBR module is calculated and pre-assembled with the following parameters:

• Dimensions of membrane tank: length x width x height = 1,200 x 650 x 2,000 mm;
• Tank body: SS400 steel 3 mm thick;
• The exterior is anti-rust painted and covered with 2 layers of Epoxy paint;
• Inside, anti-rust paint and 5-layer Epoxy paint;
• Number of MBR membranes: 9 membranes;
• MBR membrane area: 6 m2 /membrane;
• Tank water level height: 1.5 m.

TEXAS ENVIRONMENT POWER CORPORATION (TEPCO)

Head office:         1601 Industrial Blvd  Ste. 3023 Sugar land, Texas 77478, USA

Website:              https://www.texas-environment.org

E-mail:                info@texas-environment.org

Hotline:               +1 (469) 895-5577

SCHEMMEL ALEXANDER (Sales manager)

Mobile:               +1(281) 777-5474

E-mail:                alexander@texas-environment.org