DECENTRALISED WASTEWATER TREATMENT AND REUSE Components and Designing Dr. Deblina Dwivedi Senior Research Associate-Urban Water Programme Centre for Science and Environment, New Delhi

3 factors to be considered for designing the DWWT system Population. Volume of per capita water consumption. Volume of wastewater generation. Thumb rule: 80% of the total water consumption goes out as waste

Step 1: Determine the volume of wastewater generated / day (cum) Example: Population (P) = 100, Water use = 100 litres / capita / day Volume of water consumed = 100 x 100 = 10000 litres / day or 10cum/ day Hence average volume of wastewater generated = 10000 x 0.8 = 8000 litres / day or approx 8 cum/ day.

Step 2: Calculate the peak hour wastewater production Peaking factor Harmon’s Formula: 18 + √P 4+√ P P = Population in thousands Peak hourly flow = Peaking factor x average flow of wastewater per hr Example: The average wastewater flow per day = 8 cum The average wastewater flow per hour = 0.333 cum Peaking factor = 4.24 Peak hourly flow = 1.40 cum

Step 3: Calculate the total volume of sludge generated Thumb rule: Volume of sludge produced per capita per day = 0.1 litres Example: Population = 100 Volume of sludge produced per day = 100 (P) x 0.1 = 10 litres Hence volume of sludge produced per year = 10 x 365 (days) = 3650 litres or 3.6 cum. Note: at Indian condition the volume of sludge produced in septic tank is 30 litres per capita per year Note: Sludge volume can be assumed to 0.08lpcd if desludging interval > 2 years.

System components > Modules Primary Treatment – Pretreatment and Sedimentation in Settler Secondary anaerobic treatment in Baffled reactor. Tertiary aerobic treatment in Ponds Secondary & tertiary aerobic/anaerobic treatment in Planted filter bed.

Types of Settlers 3 chambers 2 chambers

Design Specifications of settler. • Rectangular / length to breath ratio: 3 to 1 • Depth: between 1.0 to 2.5m • Two chambered: First chamber 2/3 of total length • Three chambered: First chamber ½ of total length • Manholes above each chamber • Watertight, durable and stable tank

Step 4: Calculate the dimensions of settler Thumb rule > Area required = 0.5 sq m / cum wastewater/day Volume of wastewater / day = 10 cum Then area required = 10 x 0.5 = 5 sqm. Hence the settler dimensions = L = 3.86 m B = 1.28 m 1.28 m 3.86 m

Step 5: Calculate the depth of settler The depth of the settler is based on wastewater retention time. Minimum retention time = 3 hours • Average wastewater flow per hour = 0.333 cum (8 cum/24 hr) • Hence the volume of the settler = 1 cum ( 0.333 cum x 3hrs) • Final volume of the settler = 4.00 cum (1.00 cum + 3 cum sludge) • The depth of the settler will be = 0.8 or 1 m (4.00 cum/ 5 sq m)

The final dimension of the settler will be

System components > Modules Primary Treatment – Pretreatment and Sedimentation in Settler Secondary anaerobic treatment in Baffled reactor. Tertiary aerobic treatment in Ponds Secondary & tertiary aerobic/anaerobic treatment in Planted filter bed.

Step 6: Calculate the dimensions of the ABR Thumb rule >Area required 1 sq m/cum of wastewater per day E.g. If 10 cum of wastewater is generated per day then the size of the baffled reactor will be about 10 sq m Dimensions : L = 10 m, B = 1 m, D = 1.5 to 2 m

Cross checking design parameters Retention time Upflow Velocity Organic load Sludge storage volume

Hydraulic retention time and Hydraulic load • HRT = Vol. of the reactor/ Vol. of wastewater applied per day • HL = Vol. of wastewater applied per day / Vol. of the reactor

Measuring HRT Example: 10 cum wastewater flow per day on 15 cum of reactor volume gives a Hydraulic retention time of 1.5 days I.e. more than 24 hours ( 15 cum / 10 cum) Note: 24 hours HRT is better

Hydraulic load & Hydraulic retention time 80 - 90 % of removal happens in reactor

Step 7: Calculate the area of each chamber Surface Area of each chamber (sq m) = Peak flow (cum /hr) Up flow velocity (m / hr) Up flow velocity must be kept less than 2.0m/hr 0.75 m Example: Peak flow = 1.40 cum/ hr Upflow velocity = 1.5 m /hr Surface area of each chamber = 0.93 or 1 sq m Note: The chamber length should be 50-60% of the depth. If the depth is 1.5 m, length will be 0.75 m Hence width =1 sq m / 0.75 m = 1.33 m 1.33 m

Step 8: Calculate the number of chambers Number of chambers = Total area of the ABR (sq m) / Area of each chamber (sq m) Example: Total area of chamber = 10 sq m Area of each chamber = 1.33 sq m No. of chambers = 7.52 or 8

Step 11: Calculate the dimensions of planted gravel bed Horizontal planted filter Planted horizontal gravel filters are also referred to as sub surface flow wetlands (SSF), constructed wetlands or Rootzone treatment plants. Wastewater –lead to the filter – must be pre-treated. The treatment process in horizontal ground filters is complex. Decisive parameters for the intensity of the cleaning process are physical process in horizontal ground filters is complex. Decisive parameters for the intensity of the cleaning process are physical process of filtration, the intake of oxygen as well as the infuence of plantation on the biological treatment process. The secondary treatment system consists of a planted filter body made from gravel with a bottom slope of 0-5 per cent. The flow direction in the filter body is mainly horizontal. The fitler body is permanently saturated with water. but water level is adjusted 5 cm is below filter surface. The main removal mechanisms are biological conversion, physical filtration and chemical adsorption. The mechanisms of BOD removal are aerobic, anoxic, and anerobic. As coarse particles and solids eliminated within the main treatment systems and no perforated drainage pipes are used for feeding influent into the filter, filter medium is not prone to clogging. Thumb rule >Area required 4 sq m / cum of wwpd or 0.27 sq m / user E.g. If 10 cum of wastewater is generated per day then the size of the planted filter will be about 40 sq m Dimensions: L = 20 m, B = 2 m, depth between 0.6 to 1m 20

----------------------- Dimensions of gravel bed – by adopting CPCB norms Design parameters : Expected BOD removal Volume of wastewater A = Q ( In C in – In C out)‏ ----------------------- k BOD A (m2) = Surface area of the bed Q (cum/d)= Average Wastewater flow C in = BOD at inlet (mg/l)‏ C out = BOD at outlet (mg/l)‏ KBOD = Degradation coefficient which is 0.1 m/d Example: Q = 10 cum /d BOD C in = 80 mg/l BOD C out = 30 mg /l A = 10 ( In 80 – In 30)‏ / 0.1 A = 54 sq m

Polishing Ponds Thumb rule >Area required 1.2 sq m / cum of wwpd or 0.2 sq m / user Standard depth= 1-1.5m E.g. If 10 cum of wastewater is generated per day then the size of the planted filter will be about 12 sq m 22