DECENTRALISED WASTEWATER TREATMENT AND REUSE Components and Designing

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

2. 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

3. 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.

4. 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

5. 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.

6. 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.

7. Types of Settlers 3 chambers 2 chambers

8. 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

9. 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

10. 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)

11. The final dimension of the settler will be 3.86 m 1.28 m 1.0 m

12. 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.

13. 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

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

15. 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

16. 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

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

18. 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

19. 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

20. Step 11: Calculate the dimensions of planted gravel bed Horizontal planted filter 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

21. 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

22. 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