UNIVERSITI TUN HUSSEIN ONN MALAYSIA

SESSION I 2019/2020

BFB 40403

DESIGN OF WASTE WATER

SECTION 1

GROUP PROJECT
LECTURE NAME: DR. NURADILA BINTI ABD AZIZ

PREPARED BY:

STUDENT NAME MATRIX NO.

ARIFF HAIQAL BIN JOHARI DF160065

MUHAMMAD ASMAWI BIN MOHD AYOB DF160067

AHMAD YAMIN BIN ABD RAHMAN DF160093

MUHAMMAD NAIM BIN MOHD YUSOF DF160073

MUHAMMAD ILHAM BIN ZULKIPLLY DF160015

1.0 INTRODUCTION

The purpose of this report is to propose a wastewater treatment facility for a

new military camp will be construct in the year 2020. The number of
population equivalent for the new military camp is 1464.6 to cover the whole
of resident at camp site. To cater the huge number of residents for military
camp and the production of wastewater, a well design of wastewater
treatment facility is important for the whole project. Therefore, our company
are required to design the waste water treatment facilities, which will cater for
the wastewater expected to be produced by the year 2020 in military camp.

The object of this project is to design the wastewater treatment facilities

for new military camp. The criteria of the design project are as follows:

1) Bar screen
2) Balancing tank
3) Grit removal facilities
4) Primary sedimentation tank
5) Secondary treatment (Activated sludge reactor/ trickling filter/rotating
biological contactor)

Premise Unit / area m2 PE Total PE

Residential 200 5 1000
Office 300 1 300
Canteen 400 0.2 80
Weapon room 360 0.2 72
Guard office 63 0.2 12.6
Total 1464.6

Average daily flow = 225 x 1464.6

= 329.54 x 103 Lpcd x = 0.33 MLD

=329.54 m3pcd
PFF = 4.7x 1.46 -0.11
= 4.5
Peak flow rate = 4.5 x 0.33x103
=1485 m3pcd

1485 m3/d x = 0.017 m3/s

Group Profile

NAME PHOTO SPECIFIC TASKS

AHMAD YAMIN 1) Designed balancing

BIN ABDUL tank
RAHMAN

MUHAMMAD 1) Designed grit

ASMAWI BIN Chamber
MOHD AYOB 2) Design of secondary
treatment facility
MUHAMMAD 1) Get the site plan
NAIM BIN MOHD 2) Designed screen
YUSOF chamber

ARIFF HAIQAL 1) Designed Primary

BIN JOHARI sedimentation tank

MUHAMMAD 1) Designed secondary

ILHAM BIN treatment
ZULKIPLLY 2) Edit the video
2.0 Design of Screen Chamber – Bar Rack

Q max = 0.017 m3/s = 1485 m3/d

Assume:
Flow through velocity, V h = 0.05 m/s
Depth: Width, D: B = 1:2
Detention time, t =3 minutes

= = = 0.34 m2

0.34 m/s = = 2D2

D = 0.41 m 0.5 m

B==1m

Volume = Q max t = 0.017 m3/s 180 s = 3.06 m3 3.5 m3

L= = = 10.3 m 10.5m

Design summary for Screen Chamber:

B = 1.0 m
L = 10.5 m
D = 0.5 m
Freeboard = 0.3 m
Compute the number of bars

Provide bars of 10 mm x 50 mm with a clear opening of 25 mm. Let be the

number of bars, then
Opening (n + 1) size of bars (n) = B (total width)
0.025 (n+ 1) + 0.01 n = 1.0 m
n=28

Assume data:
Q =0.017 m3/s
B = 1.0 m
D = 0.5 m
n= 28

Compute flow velocity through the screen bars

The effective width of channel, Be

Be = total width - width of 16 bars
Be =1.0 - (0.01 x n)
Be = 0.72 m

Therefore, the effective cross-sectional area of the screen is A effective = Be× D

Then, the velocity of flow through screen bars, V h, at screen bars is given by

, V h, at screen bars = = 0.05 m/s

=
Compute head loss through the rack

Available data:
Q =0.017 m3/s
V h, at screen bars = 0.05 m/s
V h, before the bar screen = 0.025 m/s
n= 28

h L = 0.0729 (

= 0.0729 (
= 0.0002 m

Compute the quantity of screening

Since the Q max = 0.017 m3/s = 1.5 MLD

Quantity of screening produced = 0.0015 (m3 /ML) ×1.5 MLD = 0.00025 MLD
= 0.00025 MLD = 0.3 L/d

Design of perforated plate

Provided the length of the plate = width of the chamber = 1.0 m, width of the
plate is half of its length, and the depth of the pocket equal to 0.15 m for
collecting screenings, the capacity of the screening pocket, C s,

Cs =L×B×D = 1.0×0.5×10.5 = 5.25 m3

Cleaning needs to be done manually every day for an approximate quantity of
screenings of 5.25 m3/d
Compute the inclined length of the bars

Inclined length of bars, =1m

=

So, provide bars of total length of 1 m

Compute the length of the screen chamber

Horizontal projected length is 1 × Cos 45° = 0.7 m

The length of the inlet zone = Length of the perforated plate + 0.2 m
= 0.6 m + 0.2 m
= 0.8 m

The length of outlet zone = Width of perforated plate + 0.2 m

= 0.3 m + 0.2 m
= 0.5 m.

The total length of the screen channel = 0.8 m + 0.7 m + 0.5 m = 2.0 m
3.0 DESIGN OF EQUALISATION TANK

Design a horizontal equalization tank flow (2 units required for duty and standby
during cleaning). The average design flow rate is 330 day with peak factor 2.5.
Solution:

Average design flow, Q = 330 day

Detention time, t = 2 minute

Maximum design flow, Q max = 1485 day

Volume max, V max =

=2

Assuming the depth of the liquid in a tank is 1.0m and 0.5m as free board. Hence,
the total depth, D = 1.5m

The total surface area of a tank is 1.33 m2 1.5 m2

By providing 2 unit of equalization tank,

(Ratio of rectangular tank size, L: B = 2: 1)
1.5 = 2 L2
L = 0.9 m 1.0 m

Design Summary:
No. of unit = 2
Depth of tank = 1.5 m
Freeboard in each unit = 0.5 m
Length, L = 1.0 m
Width, B = 2.0 m
4.0 DESIGN OF GRIT CHAMBER

Design a horizontal grit chamber flow (2 units required for duty and standby
during cleaning). The average design flow rate is 329 day with peak factor
2.5.
Solution:
Average quantity of sewage, considering sewage generation 80% of water
supply, is

= (329 x 0.8) / 2 = 131.6 day = 0.0015 /sec

Maximum flow = 2.5 x average flow
= 0.0015 x 2.5 = 0.0038 /sec
Keeping the horizontal velocity at Qp as 0.2 m/sec and detention time period
as 3 minutes.
Volume of the grit chamber = Discharge x detention time
= 0.0038 x (3x60) = 0.68
With ratio depth: width, 1:2, Provide width of the chamber = 1.5 m, hence
depth = 0.8 m 1.0 m
Length of the grit chamber = volume/ (width x depth)
=0.68/ (1.5 x 0.8) = 0.56m
Cross section area of flow ‘A’ = Volume / Length = 0.68/0.56 = 1.2
Provide 20% additional length to accommodate inlet and outlet zones.
Hence, the length of the grit chamber = 0.56 + 0.24 = 0.8 m 1.0 m
and width = 1.5 m
check for volume
V=LxBxD
= 1.0m x 1.5m x 1.0m = 1.5
Design summary:
The number of channels, n = 2
Total length of channel, L = 1.0 m
Total depth of channel, D = 1.0 m
Width of channel, B = 1.5 m
Detention time, t = 180 s
5.0 Primary Sedimentation Tank (Rectangular Clarifiers)
Given:
Q avg = 330 m3/d Estimated overflow rate = 45 m3/m2.d
Q peak = 1485 m3/d Width: Depth, B: D = 2:1

Surface area for average flow condition

A avg = = = 7.3 m2

A avg = 2D2
7.3 m2 = 2D2
D= 1.9 m 2.0 m

W = = 3.65 m 4.0 m

Length of the tank for average flow condition

L= = 0.91 m 1.0 m
=

Design Summary for Primary Sedimentation tank:

Width, W = 4.0 m
Depth, D = 2.0 m
Length, L = 1.0 m

Volume of the tank for average flow condition,

Vol. No. of tank (D L W) = 2 (2 1 4) =16 m3

 Overflow rate for peak flow condition

SLR peak
= = 372 m3/m2.d
=

Detention time for peak flow condition, t

= 0.011day = 15.84 min 16 min

t= =

6.0 DESIGN OF SECONDARY TREATMENT FACILITY

It is assumed that the hydraulic residence time is 8 hours. The BOD 5 is

20mg/L

The volume of activated sludge reactor, V

8hours = 0.33day

V=Qxθ

Q = 1485 m3/day

V = 1485 m3/day x 0.33 day = 490m3

Amount of food (BOD5) fed to the reactor, F in kg/day.

S0 = 20mg/L x 1g/1000mg x 1000L/m3 = 20g/m3

F = S0 x Q

F = 20g/m3 x 1kg/1000g x 7800m3/day = 156kg kg/day

7.0 CONCLUSION

From the military camp drawing, there are five block of residential which each
block consists 5 floor. Then each floor have 8 house at there. Total unit of
residential is 200 units. Besides that, the military camp have a office, a
canteen, a weapon room and a guard office. The size of the office is 20m x
15m. The size for the canteen is 20 m x 20 m. For the weapon room, the size
is 20 m x 18 m and for the guard office the size is 7 m x 9 m. From that, the
total population equivalent (PE) can be determine. Total PE for this military
camp is 1464.6. Based on the total PE, the Q max = 0.017 m3/s or 1485 m3/d
and Q average = 330 m3/d

From the information above, the screen chamber, balancing tank, grit
chamber, and primary sedimentation tank can be designed. The function of
the screen chamber is as protection against clogging and damage upstream
of pumps and mechanical equipment. Figure below shows the design of
screen chamber.

Figure 7.1: 3D view for screen chamber

 Figure 7.2: Plan view for screen chamber

Total length of screen chamber, L= 10.5 m

Total depth of screen chamber, D = 0.5 m
Total width of screen chamber, B = 1.0 m
Freeboard = 0.3 m
Number of bar, n = 28
Size of bar = 10 mm x 50 mm with 25 mm opening
Velocity through screen bar = 0.05 m/s
Quantity of screening = 0.3 L/d or 0.0003 MLD

A balancing tank has sufficient volume to permit a non-uniform flow of

wastewater to be collected, mixed and pumped forward to a treatment system
at a more uniform rate. Pumping is controlled by level sensors and its rate
varies according to the depth of liquid in the balancing tank. For this military
camp, 2 units of balancing tank is provided for the maintenance and cleaning
purposes. Below is show the dimension of one balancing tank.
 Figure 7.3: 3D view for balancing tank

Figure 7.4: Plan view for balancing tank

Total length of balancing tank, L = 1.0 m

Total depth of balancing tank, D = 1.5 m
Total width of balancing tank, B = 2.0 m
Freeboard = 0.5 m

Grit chambers are used to remove grit present in the wastewater. Their
functions are to protect mechanical equipments used in the wastewater
treatment plant from abrasion and to prevent heavy deposits in pipelines and
channels. For this military camp, 2 units of grit chamber are provided for the
maintenance and cleaning purposes. Figure below show the dimension of
one grit chamber.

Figure 7.5: 3D view for grit chamber

Figure 7.6: Plan view for grit chamber

Total length of grit chamber, L = 1.0 m
Total depth of grit chamber, D = 1.0 m
Total width of grit chamber, B = 1.5 m
Detention time, t = 180 s
The primary sedimentation tank are designed to reduce the velocity of the
wastewater flow for organic solids or called raw sludge to settle. In this
Military Camp, 2 unit of primary sedimentation tank was provided. Figure
below shows the dimension of 1 unit primary sedimentation tank.

Figure 7.7: 3D view for primary sedimentation tank

Figure 7.8: Plan view for primary sedimentation tank

Total length of primary sedimentation tank, L = 1.0 m

Total depth of primary sedimentation tank, D = 2.0 m
Total width of primary sedimentation tank, B = 4.0 m
Detention time for peak flow, t = 16 min