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

 

Manning’s Formula

1.0

1.000

1.000

 

V = Velocity in m/s

R = Hydraulic Radius, m

S = Slope in m/1000m

 

Assume, n = 0.013

Assume circular sewer

0.9

1.124

1.066

0.8

1.140

0.968

0.7

1.120

0.838

0.6

1.072

0.671

0.5

1.000

0.500

0.4

0.902

0.337

0.3

0.776

0.196

0.2

0.615

0.088

0.1

0.401

0.021

 

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

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:

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

When,              .  Therefore,

Assuming,         n = 0.013,       

                        Solving,            D = 0.628 m,   say, 0.700 m or 700 mm

 

Corresponding to D = 700 mm,           

Therefore,        the new value of

Corresponding  , and

;

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)

 

At present peak flow,   ; corresponding

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.

 

 

 

 

 

 

 

 

 


Table 1.      Tabular Calculations for Design of main Sewer M5

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

 

 

Manhole

 

 

Length,

m

Peak 

Flow,

(2003)

m3/s

Peak

Flow,

(2023)

m3/s

Diameter,

mm

Slope

 

Discharge,

(2023)

 Litre/s

 

Velocity,

(2023)

m/s

Present

Peak Flow,

Year

2020

Litre/s

 

Self-Clensing

Velocity

(2000)

m/s

 

Total

Fall,

m

 

Invert

Elevation,

m

From

To

Full

Actual

Full

Actual

Upper

Lower

M5.4

M5.3

88

0.05

0.15

500

0.002

0.169

0.15

0.860

0.972

0.05

0.688

0.176

96.773

96.597

M5.3

M5.2

77

0.09

0.28

700

0.002

0.414

0.28

1.076

1.152

0.09

0.829

0.154

96.397

96.243

M5.2

M5.1

101

0.12

0.33

700

0.002

0.414

0.33

1.076

1.184

0.12

0.861

0.202

96.213

96.011

M5.1

M5

122

0.15

0.45

800

0.002

0.591

0.45

1.176

1.235

0.15

0.929

0.244

95.734

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