DECENTRALISED WASTEWATER TREATMENT AND REUSE Components and Designing Dr. Deblina Dwivedi Senior...

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

Transcript of DECENTRALISED WASTEWATER TREATMENT AND REUSE Components and Designing Dr. Deblina Dwivedi Senior...

DECENTRALISED WASTEWATER TREATMENT AND REUSE

Components and Designing

Dr. Deblina DwivediSenior Research Associate-Urban Water

ProgrammeCentre 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 factorHarmon’s Formula: 18 + √P

4+√ PP = 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 litresHence 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 inSettler

Secondary anaerobic treatment in Baffled reactor.

Secondary & tertiary aerobic/anaerobic treatment in Planted filter bed.

Tertiary aerobic treatment in Ponds

Types of Settlers

2 chambers

3 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 cumThen area required = 10 x 0.5 = 5 sqm.Hence the settler dimensions =L = 3.86 mB = 1.28 m

1.28 m

3.86 m

Step 5: Calculate the depth of settlerThe 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 cumsludge)• The depth of the settler will be = 0.8 or 1 m (4.00 cum/ 5 sq m)

1.28 m1.0 m3.

86 m

The final dimension of the settler will be

System components > Modules

Primary Treatment – Pretreatment and Sedimentation inSettler

Secondary anaerobic treatment in Baffled reactor.

Secondary & tertiary aerobic/anaerobic treatment in Planted filter bed.

Tertiary aerobic treatment in Ponds

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 thenthe size of the baffled reactor will be about 10 sq mDimensions :L = 10 m,B = 1 m,D = 1.5 to 2 m

Cross checking design parameters

UpflowVelocity

Retentiontime

Organicload

Sludgestoragevolume

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 reactorvolume gives a Hydraulic retention time of 1.5 days I.e. morethan 24 hours ( 15 cum / 10 cum)

Note: 24 hours HRT is better

80 - 90 % of removal happens in reactor

Hydraulic load & Hydraulic retention time

Step 7: Calculate the area of each chamber

Surface Area of each chamber (sq m) = Peak flow (cum /hr)

Up flow velocity (m / hr)

Example: Peak flow = 1.40 cum/ hrUpflow 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

Up flow velocity must be kept less than 2.0m/hr

0.75 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 mArea 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

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

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