Waste Water Plant Design

1
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/user
Hotels
400 l/bed
Tuition classes
12 l/student
Loading rates per serviced property (KS)

Sewage flows are estimated based on water consumption figures
subject 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 liters
per capita per day.
Average number of persons per household is approximately 5. The
average number of employees in commercial shops is approximately
6 per 100 m2 of floor space.
Non-residential water consumption is estimated on the basis of
actual water production and consumption data.
Design Parameters (KS)

Daily Variations
The peak hourly flow is equal to 2.5 x the average daily flow
Infiltration and Inflow

An additional 10% is added to the peak hourly flow estimates for
groundwater infiltration and wet weather inflow/infiltration into
sewers.
Minimum Grades For Self-Cleansing
Self cleansing velocity of grit and debris shall be based on achieving a
wetted cross section average velocity of at least 0.8m/sec a peak dry
weather flow.
4

Predicted population to design year
Population growth rate is p% and number of years is n (to design year)
Population of current year = P
Population 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

Total average flow rate
Total average flow rate (Qave) = domestic WW + Urban and Commercial
WW
Consider the Infiltration flow
Assume infiltration is i% of average flow rate
Infiltration flow rate = i% x Qave
Peak flow rate
Consider peak flow factor is 2.5 times average flow

Peak flow rate = 2.5 x Qave
Design flow rate
Design flow rate= QD = i% x Qave + 2.5 x Qave
6
Screen Design
Bar width = w cm
Bar space = s cm
Assume channel width is a
Angle of inclination is to the horizontal
Number of unit is 2
Flow for each screen=Q= (QD )/2
Assume horizontal velocity (VH)
Effective area of the screen (AE)
S+w
AE = Q/ VHs
=
+
Screen Design
Inclined net area of the screen
Height of the channel
Height of the steel bar (hb)
= Anet / sin
= Anet/ a
= 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
U=V
+
8
Screen Design
Head loss
1
2 2
=
x
< 15
0.7
2
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 peak

flow, min
2-5
Air supply,
m3/min .m of length
0.15-0.45
0.3
Grit and scum

quantities:
Grit, m3/103 m3
0.004-0.200
0.015
Source: Metcalf & Eddy, Inc. [5-36]
12
Grit chamber Design

Number of unit is 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)

Design Length (LD)
= t x VH
= 1.5 L
13
Grit chamber Design

Settle down velocity (VS)
=
( 0.6 1.2
)
18
< ( )

Assume free board is f and depth of grit collection is d

DD= Dtotal= D+f+d
Minimum settle down time
= Dtotal /Vs t
14
Grit chamber Design

Check for the detention time

= =
= HRT
(45-90 s)
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
0.015
1000
15
Primary Sedimentation Tank Design

Average 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 m3/m2/day
L:W=4:1
L
Weir load 200 m3/m/day
W
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
16

Design of weir
Weir length Lw = Qa+i per tank / Weir load
Assumption
Number of weir is n
d:w = 1:4
w
d
n x 2 (d+w) = Lw
d=
w=
17
Design criteria for primary sedimentation tank

Parameter
Value
Range
Typical
1.5-2.5
2.0
Average flow
30-50
40
Peak flow
80-120
100
Length
15-90
24-40 (usually 30)
Width
3-24
10
Depth
3-4.9
Depth
3-4.9
Diameter
3.0-60
12-45
1/16-1/6
1/12
Detention time, h
Surface loading rate / Overflow rate,
m3/m2.d
Dimensions, m
Rectangular
Circular
Bottom slope, mm/mm
18

Design of weir
Number of weir width per one layer (c) = W/w
Number of layers
= n/c
19
Aeration Tank Design

OLR calculation
Using BOD data, you can calculate the influent BOD5 loading rate
BOD5
= OLR kg-BOD5/day
Total average flow
= Qavg
BOD5 concentration = OLR/Qavg
20
X0, Q0, S0
Xe Qe S
Influent
Effluent
V, X, S
Xr Qr
Return Activated sludge
Xr Qw
Waste Activated sludge
21

=

()
=

coefficient =

=
=
= 0.8
22

(0 )
=
;
1 +
0 ,
=
1 +
(1 + )
=
1
(0 )
= 0
= 0 (0 )
1 +
5 =
5 =
(0 )
0
(0 )
0
23

0
( ) =

=
0
=
= 20 (1.056)20
@ = @20 (1.056)20
24

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
25

= 20 (1.056)20
@ = @20 (1.056)20
Flow for each tank=Q0= (Qavg + Qinfiltration)/2

S0= (OLR/Qavg) x 0.75
Xr = 10,000 x 0.8 = 8,000 mg/L
Stotal =Ssoluble + Ssuspended
Biodegradable solid
= Stotal x 0.65
Biodegradable solid
Ultimate BODL
= Stotal x 0.65
= Stotal x 0.65 x 1.42
26

BOD5
BOD5
S
(Ssuspended)
5
5
= 0.65 x BODL
= (Stotal x 0.65 x 1.42) x 0.65
= Stotal - (Stotal x 0.65 x 1.42) x 0.65
(0 )
=
0
(0 )
=
0
Calculate the reactor volume (V)

=
(0 )
1 +
=
0
V=
27

Calculate the HRT ()
=
0
Acceptable level = 3 10 hr
Calculate the Sludge return rate (Qr)

0
=

=
0
Acceptable level = 0.25 1.0
28
X0, Q0, S0
Xe Qe S
Influent
Effluent
V, X, S
Xr Qr
Xr Qw
Mass balance equation

Q0= Qe+ Qw
Qe = Q0 - Qw
29
=
+
=
= 20
0.8
+ (0 )
Qw
Calculate the F/m ratio
0
=
Acceptable level = 0.2 0.5 /day
Calculate the volumetric loading rate

=
0 0
Acceptable level = 0.8 2.5

kg/m3/day/BOD5
30

Calculate the biomass production rate
= 0
(0 )
1 +
Calculate the O2 required per day

Calculate the O2 required per day = BODL 1.42 Qw Xr
= Q0 (S0-S)/0.68 1.42 Qw Xr
31
Secondary Sedimentation Tank Design
X0, Q0, S0
Xe Qe S
Influent
Effluent
V, X, S
Xr Qr
Xr Qw
32

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)]
Flow for each tank=Q0= (Qavg + Qinfiltration)/2
Q0+ Qr
Qe
Mass balance equation

Qr + Q0= Qr + Qe+ Qw
Qe = Q0 - Qw
Qr + Qw
33

Calculate the Surface area of the tank
=
0
. .
2
=
4
Calculate the solid loading rate

0 +
=
0
= 0.777
= /0.8
Calculate the weir loading rate

. . =
Acceptable level = 100-200 m2/day

34

Calculate the HRT
=
=
+ + 0 +
Acceptable level = 2-4 days
35

Waste Water Plant Design

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1

Calculation of wastewater flow rate

Loading rates per serviced property (KS)

Design Parameters (KS)

Infiltration and Inflow

Calculation of wastewater flow rate

Calculation of wastewater flow rate

Consider peak flow factor is 2.5 times average flow

Number of bars (n)

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

Design parameters for aerated grit chamber

Width- Depth ratio

Detention time at peak

Grit and scum

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

Grit chamber Design

Length of the chamber (L)

Grit chamber Design

Assume free board is f and depth of grit collection is d

Grit chamber Design

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

= Air rate X L m3/min

Assume 0.015 m3 : 1000 m3 is the average grit removed

Primary Sedimentation Tank Design

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

Primary Sedimentation Tank Design

Design criteria for primary sedimentation tank

24-40 (usually 30)

Bottom slope, mm/mm

Primary Sedimentation Tank Design

Aeration Tank Design

Aeration Tank Design

Aeration Tank Design

Aeration Tank Design

Aeration Tank Design

Aeration Tank Design

Aeration Tank Design

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

Aeration Tank Design

Calculate the reactor volume (V)

Aeration Tank Design

Calculate the Sludge return rate (Qr)

Acceptable level = 0.25 1.0

Aeration Tank Design

Mass balance equation

Aeration Tank Design

Calculate the F/m ratio

Acceptable level = 0.2 0.5 /day

Calculate the volumetric loading rate

Acceptable level = 0.8 2.5

Aeration Tank Design

Calculate the O2 required per day

Secondary Sedimentation Tank Design

Secondary Sedimentation Tank Design

Mass balance equation

Secondary Sedimentation Tank Design

Calculate the solid loading rate

Calculate the weir loading rate

Acceptable level = 100-200 m2/day