Waste Water Treatment System for a village in Gujarat

8/16/2019 Waste Water Treatment System for a village in Gujarat

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Total Waste Management in Rural Sector of India- proposed waste

management plan for Village Tintoda-Dist. Gandhinagar-Gujarat

Part II- System Design Chapter 5

[Type text] Page 146

WASTE WATER TREATMENT:

For the village, proposed waste water treatment system consists of following components:

Primary treatment Secondary Treatment Tertiary Treatment

1. Screens

2. Grit chamber3. Primary sedimentation tank

4. Upflow Anaerobic Sludge Blanket

5. Rotating Biological Contactor6. Secondary clarifier

7. Disinfection

[Refer Fig No.- for Treatment Train]

5.1 Flow determination

Determination of flow:

Q = w × P Where,Q= Average wastewater flow; w= per capita wastewater generation; P= Population

Hence, Q = 100 × 7000 = 700000 lit/day

Q= 700 m3/day

5.2 Characteristics of sewage

Table 22 Typical Composition of domestic wastewater29

Total Dissolved Solids Suspended Solids Settelable Solids BOD5

1 2 3 4Fixed

mg/l

Volatile

mg/l

Fixed

mg/l

Volatile

mg/l mg/l mg/l

525 325 75 275 20 400

TOC COD Total Nitrogen Total Phosphorous

5 6 7 8

mg/l mg/l

Organic

mg/l

Free ammonia

mg/l

Organic

mg/l

Inorganic-

mg/l

290 1000 35 50 5 10

Chlorides Sulphates Alkalinity Grease Total Coliform VOC59 10 11 12 13 14

mg/l mg/l mg/l mg/l CFU 100 mL-1

mg/l

100 50 200 150 10 -10 >400


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5.3 Primary Treatment

5.3.1 Design of Screens:

(i) Determination of important parameters of sewer:30

For Peak hourly flow= 3.8×Ave flow and Ratio [ Flow depth d /Sewer dia D] = 0.8

Angle of Flow Sewer Gradient Sewer Diameter

θ

rad

iminm/m

D

m

= 2 cos-

[1 – 2(d / D)]

where

d= depth of flowD= diameter of sewer

both in m

= 5.64 × 10-

q-

where

q= peak flow in m3/sec= 0.74

= n ka-

kr -

[q/(imin ) ]

where

n= roughness co-efficient= 0.0130ka= flow area co-efficient= 0.6736

kr= hydraulic radius co-efficient= 0.3042q= peak flow

imin= sewer gradient

θ= 4.42 rad imin = 6.48 × 10-

D= 0.7 m

Hydraulic Radius Flow Velocity

r m

v

m/s

= (D/4) [1 – ((sin θ) /θ)]

whereD= sewer diameter

θ= angle of flow

= (1/n) r imin ½

wheren= roughness co-efficient

r= hydraulic radius

imin= sewer gradient

r= 0.215 m v= 2.22 m/s

(ii) Determination of important parameters of screen Chamber:32

The longitudinal section of the screen chamber is divided in to four sections:

Section 1 – At sewer

Section 2 – At screen chamber u/s of bar rack

Section 3 – At d/s of bar rack

Section 4 – At u/s of the outlet of screen chamber


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1. Width of screen chamber with bar racks 2. Actual depth of flow in screen chamber

Assuming depth of flow in screen chamber

d= 0.8 D = 0.56 m

Assuming velocity of flow thru rack openings

v= 1.1 m/sec

Clear area of openings through the rack

A= q/v = 0.74/1.1 = 0.673 m2

Clear width of openings through rack

b= 0.673/0.56 = 1.2 m

Providing 48 clear spacing of 25 mm each, Number of bars = 47 of 10 mm each

Total width of screen chamber:B= 48×25 + 47×10 = 1670mm = 1.67 m

Bernoulli’s equation between sections 1 & 2,

z1 + d

1 + (v

1

2 / 2g) = z

2 + d

2 + (v

2

2 / 2g) + h

l

where

z1 & z2= datum headsd1 & d2= depth of flow at section 1 & 2

v1 & v2= velocity of flow at section 1 & 2

hl= head loss due to sudden expansion

hl= k e[v12/2g – v2

2/2g]

k e= co-efficient of expansion = 0.3

Substituting following values in above eq.z1= 0.08 m, z2= 0

v1= 2.22 m/sec, v2= q/(b×d2)

d1= 0.56 mand solving the eq,

d2= 0.85 m

v2= 0.52 m/sec

B= 1.67 m d2= 0.85 m

3. Velocity through clear openings of bar

racks

4. Head Loss through bar rack

v= flow/net area of openings through rack

v= q/Anet = 0.74/48×0.025×0.85

v= 0.725 m/sec

h= 0.0728 [v – v2

]

h= 0.0728 [(0.725)2 – (0.52)

2]

h= 0.018 m

v= 0.725 m/sec h= 0.018 m

5. Depth and flow velocity d/s of bar rack 6. Floor raising required in channel before

free fall in to sump well- critical parameters

Bernoulli’s equation between sections 2 & 3

z2 + d2 + (v22 / 2g) = z3 + d3 + (v3

2 / 2g) + h

where

z2 & z3= datum headsd2 & d3= depth of flow at section 2 & 3

v2 & v3= velocity of flow at section 2 & 3

h= head loss

d3= 0.84 m v3= 0.53m/sec

Depth of critical flow= dc= [ q /gb ] = 0.27m

Critical velocity= vc= q/B×dc = 1.64m/sec

Bernoulli’s equation between section 3 and 4,

z3 + d3 + (v32 / 2g)=z4 + zc + d4 + (v4

2 / 2g) + hn

zc= 0.45m

d3= 0.84m v3= 0.53m/sec dc= 0.27 m vc= 1.64m/sec zc= 0.45m


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5.3.2 Design of Grit chamber:33

To remove grit particles up to 0.15 mm with sp gr. = 2.65

1. Settling Velocity 2. Surface Overflow Rate

vs= [0.707 (Ss-1)d v- .

].

where,

vs= Settling velocity

Ss= Specific gravity of particles= 2.65

d= 0.15 mmv= settling rate= 1.14×10

-6 m

2/s

vs= 0.0168 m/sec

The surface overflow rate for 100 % removal

efficiency in an ideal grit chamber=

Settling velocity of the minimum size of

particle to be removed

SOR= 0.0168×24×3600

SOR= 1451.5m3/m

2/d

vs= 0.0168 m/sec SOR= 1451.5m /m /d

3. Dimensions of grit chamber 4. Hydraulic residence Time at peak flow

A= q/vs

where,

A= Plan area of grit chamberq= peak hourly flow= 0.74 m

3/s

vs= settling velocity= 0.0168m/sec

A= 0.74/0.0168= 44 m2

HRT = V/q

where,

HRT= hydraulic residence timeV= volume= (A×d)= (44×2)= 88 m

3

q= peak flow= 0.74 m3/sec

HRT= 2 minutes

A= 0.74/0.0168= 44 m HRT= 2 minutes


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5.3.3 Design of primary sedimentation tank:34

1. Required Surface area 2. Tank Length

For average flow condition,

A= Q/OR

where,

Q= Average flow= 700 m3/day

OR= Overflow rate= 30 m3/m

2.d

A= 700/30 = 23.33= 24 m2

Width of the channel as 3.0 m,

L= A/W

where,

A= Surface area

W= Width of the channel

L= 24/3 = 8m

A= 24 m2 L= 8m

3. Detention time 4. Scour Velocity

t= V/Q

where,

V= Tank Volume= 3×8×2.5= 60 m3

(let depth to be 2.5 m)

Q= Average flow= 700 m3/day

t= 2 hrs

VH= [8k(s-1)gd/f]

where,

k= cohesion constant= 0.05s= specific gravity= 1.25

g= gravitational acceleration= 9.81 m/s2

d= particle diameter= 100 µm

f= darcy-weisbach friction factor= 0.025

VH= 0.063 m/sec

t= 2 hrs VH= 0.063 m/sec

5. BOD Removal 6. TSS Removal

% BOD removed= t/(a+bt)

where,

t= detention timea= a constant= 0.018

b= a constant= 0.020

% BOD removed = 35%

% TSS removed= t/(a+bt)

where,

t= detention timea= a constant= 0.0075

b= a constant= 0.014

% TSS removed= 56 %

35% 56 %


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5.4 Secondary Treatment

5.4.1 UASB Design:35

Influent characteristics

Flow Rate: 700 m3/d COD: 1000g/m

3 sCOD: 800g/m

3 TSS: 168 g/m

3 SO4= 50g/m

3

1. Design parameters and Assumptions 2. Reactor Volume

1. γ = 0.08g VSS/g COD

k d = 0.03g VSS/g VSS.dµm= 0.25g VSS/g VSS.d

f d = 0.15g VSS cell debris/gVSS bm decay

2. CH4 production at 35 C= 0.40l CH4/gCOD

3. Reactor volume effectiveness factor= 0.85

4. Height for gas collection = 2.0 m

Vn= QSo/Lorg

where,

Q= Ave flow in m3/d= 700 m

3/d

So= Soluble COD kg/m3= 0.8 kg/m

3

Lorg= organic loading kgsCOD/m3.d= 4kg/m

3.d

Vn= 700×0.8/6= 140 m3 VL= Total reactor required Volume

= 140/0.85= 165 m3

VL= 165 m

3. Reactor Dimensions- Diameter 4. Reactor Dimensions- Height

A= Q/v

where,

A= Reactor cross sectional area

Q= Ave flow= 700 m3/day

v= Upflow velocity= 1.5 m/h

A= 700/1.5×24= 19.45 m2

A= 3.14D2/4

D= 5.0 m

HL= VL/A

where,

HL= Reactor Liquid Height

VL= Reactor Volume= 165 m3

A = Reactor Area = 19.45 m2

HL = 110.6/18.2= 8.5 mH= Total height of the reactor

= Ht of Liquid chamber + Ht of gas chamber

+ Ht of sludge chamber= 8.5 + 2.0 + 1= 10.6 m

D= 5.0 m H= 10.6 m

5. Hydraulic detention time 6. Solid Retention Time

t= VL/Qwhere,

t= hydraulic detention time

VL= Reactor Volume= 165 m3

Q= Ave daily flow= 700 m3/day

t= 165×24/700= 5.6 hrs

SRT= [µmSe/(ks + Se) - k d]-

where,

µm, k d, ks= Kinetic co-efficients

Se= effluent soluble COD concentration g/m3SRT= Solid retention time

SRT= [0.25×80/(360 + 80) – 0.03]-1

SRT= 63 days

t=5.6 hrs SRT= 63 days


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7. sCOD Removal 8. Methane production

S= k s[1+ k dSRT] / SRT(γk – k d) – 1

where,S= effluent soluble COD

SRT= Solid retention time

k s, k d, k, γ= kinetic constants

S= 360[1 + 0.03×63]/63(0.08×3.125 – 0.03) – 1

S= 72.34 mg/l

The fraction of influent sCOD in effluent=

72.34/800 = 0.09 = 9% i.e


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5.4.2 Rotating Biological Contactor:36

Influent Characteristics

Flow Rate: 700 m3/d BOD= 156 g/m

3 sBOD= 130g/m

3 TSS= 140 g/m

3

1. Number of RBC shafts 2. Number of trains & Stages

Assuming Ist stage sBOD= 12 g/m .d

sBOD loading= 120×700= 84000g/d

Disk area required= 84000/12= 7000 m2

using 9300m2/shaft (standard)

Number of shafts=1

Length of the shaft37

= 8.23 m

diameter of each disc38

=3.5 m

Assuming two trains with 2 numbers of stages

flow rate per train= 700/2= 350 m3/d

volume of tank per stage39

=45 m3

Number of shafts=1 2 trains with 2 stages

3. sBOD concentration at each stage

BOD concentration at stage 1

S1= -1+ [1+ 4(0.00974)(As/Q)So]1/2

/2(0.00974)(As/Q)

where,

S1= sBOD concentration at stage 1

As= Surface area of disc= 9300 m2

Q= Ave daily flow= 700 m3

So= sBOD in influent= 130 g/m3

S1= 28.06 g/m3

BOD concentration at stage 2

S2= -1+ [1+ 4(0.00974)(As/Q)S1]1/2

/2(0.00974)(As/Q)

S2= 1.46 g/m3

sBODe= 1.46 g/m , BODe= 27.46 g/m

4. Organic and hydraulic loading

First stage organic loading

Lorg= 700×130/9300 = 9.78 g sBOD/m2.d

Overall organic loading(BOD)

Lorg= 700×156/2×9300 = 5.87 g BOD/m2.d

Overall organic loading(COD)

Lorg= 700×272.34/2×9300 = 20.4 g COD/m2.d

Hydraulic loading

HLR= 700/2×9300= 0.037m3/m

2.d

Lorg= 5.87 g sBOD/m .d HLR= 0.037m /m .d

Effluent TSS = 0.7×140= 98 g/m

Effluent COD= 0.6×272.34= 163.4 g/m


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5.5 Effluent characteristics:

Table 23 Important parameters of effluent

Parameter Treatment

level

Unit Influent Effluent Remarks

TSS

PrimaryTreatment

PrimarySedimentation

tank

350 mg/l 168 mg/l Furthertreatment

SecondaryTreatment

UASB 168 mg/l 140 mg/l Furthertreatment

RBC 140 mg/l 98 mg/l


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On the basis of utility area of treated water, further treatment for Nitrogen and Phosphorous

removal can be selected. Presently there is no treatment is recommended for removal of Nitrogen

and Phosphorous, effluent water can be reused for irrigation purpose.

5.6 Utilization of biogas produced from UASB

Total gas generation from UASB= 221 m3/day, that can be supplied to around 200 families in the

village considering 1 m3 biogas can serve the purpose of cooking for 5 to 6 member family for

three meals.

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Waste Water Treatment System for a village in Gujarat