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 = litres / day or 10cum/ day
Hence average volume of wastewater generated = 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: √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 = 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 = cum (8 cum/24 hr)
• Hence the volume of the settler = 1 cum ( 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

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
% 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
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

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 22