Sewer Line Design

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Quiz No. 3
Wastewater Collection
SOLUTION
1. The present population of a certain colony in a city is 10000.? The present (Year
2003) per capita average water supply to the colony is 180 lpcd (liters per capita
per day).? The population of the colony is expected to be 45000 at the end of the
design period of 20 years (Year 2023).? Per capita average water supply at the end
of the design period is expected to be 240 lpcd. The entire wastewater generated
in the colony is collected and discharged through a connecting sewer to one of the
trunk sewers of the city.? Determine an acceptable slope and diameter of this
sewer.? Make appropriate assumptions wherever required and state them
explicitly.
d/D

v/V

q/Q

1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1

1.000
1.124
1.140
1.120
1.072
1.000
0.902
0.776
0.615
0.401

1.000
1.066
0.968
0.838
0.671
0.500
0.337
0.196
0.088
0.021

Manning?s Formula
V

1 2 3 12
.R .S
n

V = Velocity in m/s
R = Hydraulic Radius, m
S = Slope in m/1000m
Assume, n = 0.013
Assume circular sewer

Solution:
Present Population (in Year 2003) = 10,000
Present Average Water Supply (in Year 2003) = 10,000.(180) = 1.8 million liters per day
Assuming that 80 percent water becomes wastewater,
Present Average Wastewater Generation = 1.8.(0.8) = 1.44 million liters per day
Assuming a Peaking Factor of 3,
Present Peak Wastewater Generation = 3.(1.44) = 4.32 million liters per day
106
1
4.32. 3 .
 0.05m3 / s
10 86400
Present Peak Wastewater Generation =
?
Design Population (in Year 2023) = 45,000
Design Average Water Supply (in Year 2023) = 45,000.(240) = 10.8 million liters/day
Assuming 80 percent water becomes wastewater,
Design Average Wastewater Generation = 10.8.(0.8) = 8.64 million liters per day
Assuming a Peaking Factor of 3,
Design Peak Wastewater Generation = 3.(8.64) = 25.92 million liters per day

Design Peak Wastewater Generation =
For a pipe flowing full:
1  .D 2 
 .

n  4..D 

Q Full

2
3

25.92.

106
1
.
 0.30m3 / s
3
86400
10

1
 .D 2  1 2 1
8
.
 .S  .(0.3117 ). D  3 .S 2
n
 4 

At q = 0.3 m3/s, the pipe is flowing 0.8 full

When,

Therefore,

d
q
 0.8,
 0.968
D
Q full
.

Q full 

q
0 .3

 0.3099 m 3 / s
0.968 0.968
8

Assuming,

n = 0.013,
Solving,

1

1
0.3099 
.(0.3117 ).(D) 3 .(0.002) 2
0.013
D = 0.628 m, ??say, 0.700 m or 700 mm

Corresponding to D = 700 mm,
8

1

1
Q full 
.(0.3117 ).(0.7) 3 .(0.002) 2  0.4142m 3 / s
0.013
q
0 .3

 0.724
Q
0
.
4142
full
Therefore,the new value of
v
d
 1.100
 0.65
V
D
full
Corresponding
?, and
2

1

1  0.7  3
Vfull 
.
 .(0.002) 2  1.076m / s
0.013  4 
;
Therefore, vDesign = Velocity at design peak flow = (1.076).(1.1) = 1.184 m/s
(which is within the recommended range of 0.8 m/s to 3 m/s)
q
0.05
v

 0.121
 0.65
At present peak flow, ?? Qfull 0.4142
; corresponding Vfull
Therefore, vpresent = velocity at present peak flow = (0.65).(1.076) = 0.699 m/s

(which is greater than the recommended minimum value of 0.6 m/s)
Hence use 700 mm circular concrete pipe at a slope 0f 2 m per 1000 m for
accommodating the above flow.

2. Consider the sewer system shown in the figure below.? A main sewer (MS5) with manholes, M5.4 (starting manhole) to M 5.1
discharges into manhole M5 of the trunk sewer.? Branch sewers discharge into manhole M5.1 ? M5.4 of the main sewer (MS5)
as shown in the figure.? Invert levels of the branch sewers discharging into manhole M5.4 ? M5.1 are 97.223, 98.775, 98.311,
and 96.334 m respectively.? Invert level of the trunk sewer leaving manhole M5 is 93.700 m. Based on this information
explain how values in the last three columns of the design table for main sewer (MS5) have been calculated. All branch sewers
have 400 mm diameter.? The ground level in the area under question is between 101 and 102 m.
TRUNK SEWER
M5
M5.2
M5.4

Table 1.
1

2

Manhole
From

To

M5.4
M5.3
M5.2
M5.1

M5.3
M5.2
M5.1
M5

M5.1

M5.3

?????Tabular Calculations for Design of main Sewer M5
3

5
Peak
Flow,
(2023)
m3/s

6
Diameter,
mm

7
Slope

Length,
m

4
Peak?
Flow,
(2003)
m3/s

88
77
101
122

0.05
0.09
0.12
0.15

0.15
0.28
0.33
0.45

500
700
700
800

0.002
0.002
0.002
0.002

8

9

Discharge,
(2023)
?Litre/s
Full Actual
0.169
0.414
0.414
0.591

0.15
0.28
0.33
0.45

10

11

Velocity,
(2023)
m/s
Full Actual
0.860
1.076
1.076
1.176

0.972
1.152
1.184
1.235

12
Present
Peak
Flow,
Year
2020
Litre/s
0.05
0.09
0.12
0.15

13

14

15

16

SelfClensing
Velocity
(2000)
m/s
0.688
0.829
0.861
0.929

Total
Fall,
m

Invert
Elevation,
m
Upper
Lower

0.176
0.154
0.202
0.244

96.773
96.397
96.213
95.734

96.597
96.243
96.011
95.490

Solution:
Elevation of the centerline of branch sewer at manhole M5.4 = 97.223 + 0.2 = 97.423
Elevation of centerline of the upper end of the main sewer from M5.4 to M5.3
= 97.423 ?0.4 = 97.023
Invert level of the upper end of the main sewer for M5.4 to M5.3
= 97.023 ?0.25 = 96.773
Fall from M5.4 to M5.3 = 88.(0.002) = 0.176
Invert level of the lower end of the main sewer for M5.4 to M5.3
= 96.773 ? 0.176 = 96.597
Invert level of the upper end of the main sewer for M5.3 to M5.2
= 96.597 ? 0.2 = 96.397
Considering the above invert level, minimum invert level of the branch sewer at M5.3 should be 96.397 + 0.35 + 0.4 ? 0.2 = 96.957,
while the actual invert level is 98.775.?
Hence okay.
Fall from M5.3 to M5.2 = 77.(0.002) = 0.154
Invert level of the lower end of the main sewer for M5.3 to M5.2
= 96.397 ? 0.154 = 96.243
Invert level of the upper end of the main sewer for M5.2 to M5.1
= 96.243 ? 0.03 = 96.213
Considering the above invert level, minimum invert level of the branch sewer at M5.2 should be 96.213 + 0.35 + 0.4 ? 0.2 = 96.763,
while the actual invert level is 98.311.?
Hence okay.
Fall from M5.2 to M5.1 = 101.(0.002) = 0.202
Invert level of the lower end of the main sewer for M5.2 to M5.1
= 96.213 ? 0.202 = 96.011
Invert level of the upper end of the main sewer for M5.1 to M5
= 96.011 ? 0.1 = 95.911

Considering the above invert level, minimum invert level of the branch sewer at M5.1 should be 95.911 + 0.4 + 0.4 ? 0.2 = 96.511,
while the actual invert level is 96.334.?
Hence not okay.
Hence, actual invert level of the upper end of the main sewer for M5.1 to M5
= 96.334 + 0.2 ? 0.4 ? 0.4 = 95.734
Fall from M5.1 to M5 = 122.(0.002) = 0.244
Invert level of the lower end of the main sewer for M5.1 to M5
= 95.734 ? 0.244 = 95.490

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