waste water plant design

7/24/2019 Waste Water Plant Design

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Calculation of wastewater flow rate

Category Water consumption

Households 120 l/c/d

School 12 l/student

Boarding school 120 / boarder

Hospital 400 l/bed

Cinemas 8 l/seat

Office 24 l/employee

Shops 24 l/employee

Public toilets 12 l/userHotels 400 l/bed

Tuition classes 12 l/student


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Loading rates per serviced property (KS)

Sewage flows are estimated based on water consumption figuressubject to a wastewater return factor of 0.8.

Residential water consumption is taken as 120 liters per capita per

day and estimated to correspond to a wastewater flow of 96 litersper capita per day.

Average number of persons per household is approximately 5. Theaverage number of employees in commercial shops is approximately

6 per 100 m2 of floor space.

Non-residential water consumption is estimated on the basis ofactual water production and consumption data.


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Design Parameters (KS)

Daily Variations

The peak hourly flow is equal to 2.5 x the average daily flow

Infiltration and InflowAn additional 10% is added to the peak hourly flow estimates forgroundwater infiltration and wet weather inflow/infiltration intosewers.

Minimum Grades For Self-Cleansing

Self cleansing velocity of grit and debris shall be based on achieving awetted cross section average velocity of at least 0.8m/sec a peak dryweather flow.


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Predicted population to design year

Population growth rate is p% and number of years is n (to design year)Population of current year = PPopulation in design year = PD= P(1+p/100)

n


Wastewater Flow rate for domestic

Domestic wastewater flow rate = PD x 120 l/c/d x 0.8

Wastewater Flow rate for Urban and Commercial population


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Total average flow rate

Total average flow rate (Qave) = domestic WW + Urban and CommercialWW


Consider the Infiltration flowAssume infiltration is i% of average flow rateInfiltration flow rate = i% x Qave

Peak flow rate

Consider peak flow factor is 2.5 times average flowPeak flow rate = 2.5 x Qave

Design flow rate

Design flow rate= QD = i% x Qave + 2.5 x Qave


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Inclined net area of the screen = Anet / sin Height of the channel = Anet/ aHeight of the steel bar (hb) = Anet/ (a x sin )

Number of bars (n) =

Check the total width of screen = [ (n+1) w + nx s] a

V- Horizontal velocity (flow through velocity)

U- Approach velocity

Screen Design

U = V

+


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Screen Design

=1

0.7 x

2 < 15

Head loss


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Design parameters for aerated grit chamber

Item Value

Range Typical

Dimensions:

Depth, m 2:5

Length, m 7.5:20

Width, m 2.5:7.0

Width- Depth ratio 1:1 5:1 1.5:1

Length-width ratio 3:1 5:1 4:1

Detention time at peakflow, min

2-5 3

Air supply,m3/min .m of length

0.15-0.45 0.3

Grit and scumquantities:Grit, m3/103 m3

0.004-0.200 0.015

12Source: Metcalf & Eddy, Inc. [5-36]


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Grit chamber Design

Number of unit is 2

Flow for each screen=Q= (QD )/2

Assumptions: Horizontal velocity (VH) (0.3 -0.45 m/s) Detention time is t (45-90 s)

=

=

2

1

Length of the chamber (L) = t x VHDesign Length (LD) = 1.5 L


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Grit chamber Design

Settle down velocity (VS)

=

18 ( 0.6 1.2

)

=

< ( )

Assume free board is f and depth of grit collection is dDD= Dtotal= D+f+d

Minimum settle down time = Dtotal /Vs t


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Grit chamber Design

=

=

= HRT

Check for the detention time

Assume supply of air rate is 0.15 0.45 m3/min/m of length

Air requirement = Air rate X L m3/min

Assume 0.015 m3 : 1000 m3 is the average grit removed

The total quantity of grit to be handled per day in one unit

=.

3

(45-90 s)


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Flow for each screen=Q= (QD )/2Average flow rate with infiltration flow = Qa+i

Assumption Number of unit is 2 Detention time t (1.5-2.5h) Surface loading rate @ peak flow (Qsur) is 100 m

3/m2/day L:W=4:1

Weir load 200 m3/m/day

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Primary Sedimentation Tank Design

Floor area of each tank (A) = Q/ Qsur

Detention time = (A x D)/Q

Depth of tank D = (3- 5.5 m)Design depth DD = D + free board

D

LW


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Weir length Lw = Qa+i per tank / Weir load

Assumption Number of weir is n d:w = 1:4

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Design of weir

d

w

w

n x 2 (d+w) = Lwd=w=


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Parameter Value

Range Typical

Detention time, h 1.5-2.5 2.0Surface loading rate / Overflow rate,m3/m2.d

Average flow 30-50 40

Peak flow 80-120 100

Dimensions, m

Rectangular

Length 15-90 24-40 (usually 30)

Width 3-24 10

Depth 3-4.9 4

Circular

Depth 3-4.9 4

Diameter 3.0-60 12-45

Bottom slope, mm/mm 1/16-1/6 1/12

Design criteria for primary sedimentation tank

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Number of weir width per one layer (c) = W/wNumber of layers = n/c

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Design of weir


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Aeration Tank Design

OLR calculation

Using BOD data, you can calculate the influent BOD5 loading rateBOD5 = OLR kg-BOD5/day

Total average flow = QavgBOD5 concentration = OLR/Qavg


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Influent Effluent

Waste Activated sludge

Return Activated sludge

X0, Q0, S0

V, X, S

Xe Qe S

Xr Qw

Xr Qr


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=

()

coefficient =

= 0.8 =

=

=

+


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=

( )

1 + ; ,

=

1 +

=( 1 + )

1

= ( )1 +

= ( )

=( )

= ( )


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() =

=

=

= (1.056)

@ = @(1.056)


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Assuming Primary sedimentation tank removes 25% of BOD5 BOD5 = 0.68 BODL X0= 0

MLSS in return sludge = 10,000 mg/L Aerator basin MLVSS (X) = 3,500 mg/L Effluent biological solution concentration = 20 mg/L 65% of it is biodegradable Y 20= 0.4

kd 20 = 0.025 /day Number of units 2 Room temperature is 28 C c = 10 days


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= (1.056)

@ = @(1.056)

Flow for each tank=Q0= (Qavg + Qinfiltration)/2

S0= (OLR/Qavg) x 0.75Xr = 10,000 x 0.8 = 8,000 mg/L

Stotal =Ssoluble + Ssuspended

Biodegradable solid = Stotal x 0.65Biodegradable solid = Stotal x 0.65Ultimate BODL = Stotal x 0.65 x 1.42


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BOD5 = 0.65 x BODLBOD5 (Ssuspended) = (Stotal x 0.65 x 1.42) x 0.65S = Stotal - (Stotal x 0.65 x 1.42) x 0.65

=(

)

=( )

Calculate the reactor volume (V)

=

( )

1 +

=

V=


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Acceptable level = 0.25 1.0

Calculate the HRT ()

=

Calculate the Sludge return rate (Qr)

=

=

Acceptable level = 3 10 hr


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Influent Effluent



X0, Q0, S0

V, X, S

Xe Qe S

Xr Qw

Xr Qr

Mass balance equationQ0= Qe+ Qw

Qe = Q0 - Qw


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=

+

=

+ () Qw

Calculate the F/m ratio

=

Acceptable level = 0.2 0.5 /day

Calculate the volumetric loading rate

=

Acceptable level = 0.8 2.5

kg/m3/day/BOD5

= 20

0.8


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Calculate the biomass production rate

Calculate the O2 required per day = BODL 1.42 Qw Xr

= Q0 (S0-S)/0.68 1.42 Qw Xr

= ( )

1 +

Calculate the O2 required per day


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Secondary Sedimentation Tank Design

Influent Effluent



X0, Q0, S0

V, X, S

Xe Qe S

Xr Qw

Xr Qr


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Assuming Number of units is 2 Average surface loading rate (30-50 m3/m2/day) Free board is 0.4 m

Depth of tank is D [3 - 4.9 m (base of the diameter)]

Q0+ Qr Qe

Qr + Qw

Mass balance equationQr + Q0= Qr + Qe+ Qw

Qe = Q0 - Qw

Flow for each tank=Q0= (Qavg + Qinfiltration)/2


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Calculate the Surface area of the tank

=

. .

=

4

=+

Calculate the solid loading rate

Calculate the weir loading rate

. . =

Acceptable level = 100-200 m2/day

= 0.777

= /0.8


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Calculate the HRT

=

Acceptable level = 2-4 days

=

+ + =

+

waste water plant design

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