Contents.docx

Contents: 2: Design of Preliminary & Primary Treatment Units 2.1 Design of Equalization Tank (ET) 2.2 Design of Grit Chamber (GC) with Bar Racks 2.3 Design of Primary Sedimentation Tank (PST)

2.1

Design of Equalization Tank (ET):

Given Data: Average wastewater flow rate =90 MLD = 90000m3/d =1.04167 m3/s

Design Data: Table showing the variations in wastewater flow with time of day.

Time

Avg. inflow,

Inflow

Cum. Inflow

Interval

MLD

volume, ML

Volume ML

𝒂𝒗𝒈 𝒊𝒏𝒇𝒍𝒐𝒘 [ ) 𝟐𝟒 12 - 1 am 1- 2 am 2-3 am 3-4 am 4-5 am 5 -6 am 6-7 am 7-8 am 8-9 am 9-10 am 10-11 am 11-12 am 12-1 pm 1-2 pm 2-3 pm 3-4 pm 4-5 pm 5-6 pm 6-7 pm 7-8 pm 8-9 pm 9-10 pm 10-11 pm 11-12 pm

50 40 40 40 70 80 120 200 190 190 110 80 80 65 65 80 100 100 110 95 95 70 50 40

2.08 1.67 1.67 1.67 2.92 3.33 5 8.33 7.92 7.92 4.58 3.33 3.33 2.71 2.71 3.33 4.16 4.16 4.58 3.96 3.96 2.92 2.08 1.67

2.08 3.75 5.42 7.09 10.01 13.34 18.34 26.67 34.59 42.51 47.09 50.42 53.75 56.46 59.17 62.5 66.66 70.82 75.4 79.36 83.32 86.24 88.32 90

Cumulative outflow

Cumulative storage (CIV-COV)

ML

ML

3.75 7.5 11.25 15 18.75 22.5 26.25 30 33.75 37.5 41.25 45 48.75 52.5 56.25 60 63.75 67.5 71.25 75 78.75 82.5 86.25 90

-1.67 -3.75 -5.83 -7.91 -8.74 -9.16 -7.91 -3.33 0.84 5.01 5.84 5.42 5.0 3.96 2.92 2.5 2.91 3.33 4.15 4.36 4.57 3.74 2.070 0

Calculations: Required volume of the equalization tank (from Mass Flow Curve) V=Net(deficit+surplus)=(9.16+5.84)ML=15000m3 Flow rate to the equalization tank (given)= 90MLD =90000 m3/d = 1.04167m^3/s

Assuming depth of the tank, d=6m Area required by the tank, A=V/d=

15000 6

= 2500𝑚2

Providing 4 no. of equalization tanks and area of each tank =

Diameter of the tank, D=√

4×625 𝜋

2500 4

= 625 m2

= 28.20𝑚

Total head, H= (6-0.5)=5.5m 𝜌𝑄ℎ

1000×1.04167×5.5

= =76.39HP 75 75 Justification- As sewage contains very less amount of solids and rest is water So its density can be taken equal to that of water. Pumping requirement, HP =

Provide 4 pumps of 20 HP and 2 standby pumps of 20 HP.

Item No.

Parameter

Value

15000m3

1.

Remark

Mass Flow Curve

Volume

2.

Depth

6m

3.

Diameter

28.20m

4.

Base level

-

Assumed

5.

Design Specification

Water level

-

2.2

Design of Grit Chamber (GC) with Bar Racks:

Design Data:

Items

Value

Average wastewater flow

90MLD =90000 m3/d = 1.04167m3/s

Rate

Remark

Given 0.10 x 10-3 to 0.20 x 10-3 m [Clause: 5.6.2.7.1.1,

Particle Size (d)

0. 15 × 10−3m

Manual on Sewerage & Sewage Treatment, CPHEEO, 2013] 2.4 – 2.65

Specific gravity of particle (SS)

[Clause: 5.6.2.7.1.1, 2.64

Manual on Sewerage & Sewage Treatment, CPHEEO, 2013]

Settling velocity

.01578m/s

Calculated Calculated; Should not exceed 60 s

Retention time

53S

[Clause: 5.6.2.7.1.3, Manual on Sewerage & Sewage Treatment, CPHEEO, 2013]

Ground level

-

Calculations:

1. Computation of settling velocity: In winter generally the temperature goes upto 10℃ so at that time flow is not laminar,it is in transient condition.so the formula for settling velocity for this condition is given by; 𝑉𝑠 = [0.707(𝐺𝑠 − 1) × 𝑑1.6 × 𝜗 −.6 ].714 𝑚/𝑠 = [0.707(2.64 − 1) × (. 15 × 10−3 )1.6 × (1.3065 × 10−6 )−.6 ].71 = .01578𝑚/𝑠 2.Computation of surface overflow rate, SOR: The surface overflow rate for 100% removal efficiency in an ideal grit chamber =

Settling

velocity of the minimum size of particle to be removed =.01578 m/s =1363.4 m3/m2/d However, due to turbulence and short circuiting due to several factors as eddy, wind and density currents, the actual value to be adopted has to be reduced taking into account the performance of the basin and the desired efficiency of the particles removal. To determine the actual overflow rate, the following formula may be used – ƞ = 1 − [1 + (𝑛 × 𝑉𝑠)/((𝑄/𝐴) )]^((−1)/𝑛)

Assumed= efficiency of removal of desired particles = 75% n = measure of settling basin performance =

1 8

for very good performancee A=99.91= 100𝑚2

2. Determination of the dimensions of grit chamber: Plan area of grit chamber[Q /(Q / A)]= 100m2

Provide 3 channels of 3.5 m wide and 10.5 m long and one additional channel as a stand-by.

4. Check for scouring velocity (VC) and determination of the depth (d) of grit chamber 𝑉𝐶 = 𝐾𝐶 [(𝐺𝑆 − 1)𝑔𝑑]0.5 = 4[(2.64 − 1) × 9.81 × 0.15 × 10−3 ]0.5=0.1965m/s Horizontal flow velocity 𝑉ℎ < 𝑉𝐶 𝑄 𝑊×𝑑

< 𝑉𝐶

1.04167

< 0.1965 0.1693<0.1965 4.1×1.5

Assuming d=1.5m

5. Estimate hydraulic retention time (HRT)=V/Q=

1.5×3.5×10.5 1.04167

= 53𝑠

Provide 3 channels of grit chamber, each 10.5 m × 3.5 m × 1.5 m and one additional channel as a stand-by. Total depth D= 1.5+0.25(free board)+0.25(grit storage space)=2m 6.

Design of Proportional Flow Weir (Velocity Control Device):

Flow (Q) through a proportional weir is given by – 𝑎 Q=Cb√(2𝑎𝑔) × [(𝐻 − )] 3 Where, C = a coefficient of weir = 0.61 for symmetrical sharpedged weir; a = dimension of weir (m) usually between 25 – 50 mm (40 mm adopted); h = depth of flow (m); b = base width of weir (m) 1.04167 3

0.04

= 0.61 × 3.5 × √(2× 0.04 × 9.81) × [ℎ − (

3

)]

ℎ = 0.19691𝑚 To determine the coordinates [X(m), Y(m)] of the curve forming the edge of the weir, the following equation can be used: 𝑏

2

𝑌

2

𝜋

𝑎

𝑋 = [1 − tan−1 √( − 1)]

Y a=0.04 10a=0.4 20a=0.8 30a=1.2 40a=1.6 49a=1.96

X 1.75 0.36 0.251 0.204 0.177 0.16

Proportional flow wier profile 2.5

2

1.5

Y

Y 0.4 0.8 1.2 1.6 1.96

1

0.5

0 -2

-1

0 X

1

2

Design of bar racks: Width of each chamber =3.5 m Let the bar screens (coarse) be of 30 mm × 10 mm with 50 mm openings and let numbers of bars required = N Therefore, B=[(N× 𝑤) + (𝑁 + 1) × 𝑠𝑝𝑎𝑐𝑖𝑛𝑔] 3.5 = [(𝑁 × 0.01) + (𝑁 + 1) × 0.05] N=58 Similarly, for bar screen (medium) (30 mm openings) and bar screen (fine) (15 mm openings), the numbers of bars required are 87 and 140, respectively.

Design Specifications:

Item No.

Parameter

Value

Remark

Collection chamber

Number

1

Retention time

30s

Capacity

31.25m3

Length

10.20m

Width

2.04m

Base Level

m

Water Depth

1.5m

Water Level

m

Free Board

0.5 m

1.

Usually 30 s Q*t

Same as in GC

Transition zone Length 2.

Width (initial)

0.73m 2.04m

Width (final)

3.5m

Water depth

1.5m

Same as in GC

Item No. Parameter

Result

Remark

Grit Chamber One additional unit for

Number

3

165.375m3

Capacity

3.

stand-by

Area required

100m2

Water depth

1.5m

Length

10.5m

Individual unit

Width

3.5m

Individual unit

Excluding free board & grit storage space

Free board + grit storage Space

0.5 m

4.

Base level

-

5.

Water level

-

6.

Bar Screen (Coarse) Bar specifications Number Opening Bar Screen (Medium) Bar specifications

7.

8.

30mm ×10 mm 58 50 mm 30mm ×10 mm

Number Opening Bar Screen (Fine) Bar specifications

87 30 mm

Number Opening

140 15 mm

30mm ×10 mm

Same as Bar Screen (Coarse)

Same as Bar Screen (Coarse)

2.3

Design of Primary Sedimentation Tank (PST):

Given Data: Average wastewater flow rate =90MLD=90000𝑚3 /𝑑=1.04167𝑚3 /𝑠 Assumed Data:

Parameter

Remark

Adopted Value

35 – 50 m3/m2/d [ Table 5.8, Manual on Sewerage & Surface loading rate

3

2

40𝑚 /𝑚 /𝑑

Sewage Treatment, CPHEEO, 2013] 2 – 2.5 h

Hydraulic retention

Range: 1.65 – 4.0 h

time

[Clause 5.7.4.2.4, 2h

Manual on Sewerage & Sewage Treatment, CPHEEO, 2013] Raw Sewage Characteristics

TSS concentration

590mg/l

[Table 5.1, Manual on Sewerage & Sewage Treatment, CPHEEO, 2013] Weir loading should not be greater than 125 m3/d/m

Weir loading

125 m3/d/m

[Table 5.8, Manual on Sewerage & Sewage Treatment, CPHEEO, 2013]

Assuming 60% Underflow solid

removal of VSS in 530× 0.6 = 318mg/l

concentration

PST [Clause 5.7.5, Manual on Sewerage & Sewage Treatment, CPHEEO, 2013]

Ground level

414 m 8.0136𝑚3

Assuming 2%

Sludge volume

Consistency [Clause 5.7.4.2.6.1, 8

Manual on Sewerage & Sewage Treatment, CPHEEO, 2013]

Calculations: Flow Rate =90 𝑚3

Inflow discharge (Q)=90MLD=90000 𝑑 = 1.04167𝑚3 /𝑠 Let the overflow rate=40 m3 /m2/d Let the detention period (t) of tank=2h Q/A = 40m3 /m2/d

A =2250, 𝑚2 Volume of the tank, V = t × Q =2×

90000 24

= 7500𝑚3

Depth of the tank

H = V/A =3.33 m

Provide 0.25 m freeboard & 0.25 m sludge zone Overall depth of tank=3.33+0.5=3.83m Provide overall depth of tank=4m Provide 3 units Area of each tank=2250/3=750𝑚2 4𝐴 Diameter of each tank d=√ 𝜋 Hence d=31m

𝑄

Weir loading=𝜋𝑑 =

308.04𝑚3 𝑑

/𝑚

Weir loading should not be greater than 125 m3/d/m. but it is so, provide effluent launder with weir on both sides of the launder.

Influent/Inlet Arrangement: Circular baffle diameter=3.1m (Range: 10 – 20 % of tank diameter; Clause 5.7.4.2.7, Manual on Sewerage & Sewage Treatment, CPHEEO, 2013) Circular baffle depth

=1 m below water surface (Range: 1 – 2 m; Clause 5.7.4.2.7, Manual on Sewerage &

Sewage Treatment, CPHEEO, 2013)

Outlet Arrangement: 1. Effluent weir Length (L) of effluent weir plate on each side of launder = π(d – 1)=94.25m Provide 90° 𝑉 notches@20cm c/c Total number of notches=L/0.2=472

30×1000

Average discharge per notch (qʹ) at average design flow =

8

472

𝜃

=63.56𝑚3 /𝑑

m3/s

5

The discharge through a V-notch,𝑞 = 15 × 𝐶𝑑 × √2𝑔 × tan 2 × 𝐻 2

m3/s

For peak flow per notch, q = 2.25qʹ m3/s = For Cd = 0.584, θ = 900 Head over V-notch at peak flow = H =

m

Provide safe allowance of 2.5 – 3.0 cm in addition to head over V-notch at peak flow. Therefore, Provide [numbers] of 900 V-notches [total head] cm deep at 20 cm c/c distance. 2. Effluent Launder Let width of launder (b)

= 0.6 m

Critical depth at the end of effluent launders 1

Y2

Q

23

b2 g m

= Depth of water at upper end of channel

Y

Y2

1

2

2Q 2 2

g b =

Q Q / 2; Half of the flow divides

Y2

on each side of the launder

m Hence, provide effluent launder of depth with safe allowance =

m

Design specifications:

Item No.

1.

Parameter

Result

No. of Units

5

Remark

Capacity Internal diameter (d) 2.

m

Depth + Free board + sludge zone

m

Base level

m

Water level

m

Influent Zone Influent pipe 3.

diameter

0.4 m 10 – 20% of

Circular baffle diameter Circular baffle depth

m m

Effluent Channel

4.

Width

0.6 m

Depth

0.3 m

Diameter at inner periphery

25.5m

Diameter at outer periphery

27.5m

diameter 1–2m

Design Specifications:

Item No.

Parameter

Result

Remark

Hopper Bottom 7.5 – 10.0% from periphery to centre [Clause Slope

4.50

5.7.4.2.9, Manual on Sewerage

4.

Sewage Treatment, CPHEEO, 2013] Diameter of sludge withdrawal pipe

0.4 m

Depth

1.0 m

Base level

413 m

&