design of sewers example_2

8/10/2019 Design of Sewers Example_2

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Wastewater Engineering

Dr. Sudipta Sarkar

Design of Sewers: Examples


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D

d

/2 /2

]2

cos1[21

Dd

]

2

sin3601[

R

r

3/2

2sin360

1

V v

3/2

2

sin3601

2

sin

360

Q

q

In all the above expressions, is the only variable, all other

parameters are constant. Thus at different values of

, the aboveproportional elements can be easily calculated


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d/D a/A v/V q/Q

1.00 1.00 1.00 1.00

0.9 0.949 1.124 1.0660.8 0.858 1.140 0.988

0.7 0.748 1.120 0.838

0.5 0.5 1.000 0.500

0.4 0.373 0.902 0.337

Capital Letters denote the situationwhen the sewers run full

Maximum velocity is achievedwhen the sewers are designedto run at 80% of the full depth.


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Designing Sewer Systems

Sewers are designed taking consideration of 30 years.

Population in the initial years of the design period are low compared to thedesign population at the end of design period

Peak flow rate in the initial years is low compared to the designed peakflow rate

Sizing should be such that it will attain the self-cleansing velocity at theaverage flow rate or at least at the maximum flow rate at the beginning of the

design period


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s

1000

2/13/21 sr n

v

]2

sin3601[

4 D

pa

r

3/2

2sin360

1

V vVelocity at partially full flow

Velocity at full flow

For Partially-full flow v is not influenced by the diameter of the

pipe, rather is much influenced by the slope of the channel


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After finding the minimum slope required, the pipe size is decided on the basisof ultimate design peak flow rate and the permissible depth of flow. Adoptionof the above slopes would ensure minimum flow velocity of 0.6 m/s

Minimum size for a public sewer is 150 mm diameter

Minimum size for a public sewer in hilly terrain is 100 mm diameter


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What will be the diameter of the sewer designed with the followingconditions:a) Population to be served is 100,000;

b) Water consumption= 180 lpcdc) 80 % of supplied water appears as wastewaterd) Self-cleansing velocity to maintained in the sewer = 0.6 m/s;e) Maximum velocity in the sewer 3 m/s;f) Minimum size of the sewer = 150 mm;g) Peak factor = 2.5h) n=0.015

d/D a/A v/V q/Q

1.00 1.00 1.00 1.00

0.9 0.949 1.124 1.066

0.8 0.858 1.140 0.988

0.7 0.748 1.120 0.838

0.5 0.5 1.000 0.500

0.4 0.373 0.902 0.337


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Average Discharge or flowrate = 130,000 people * 130 lpcd * 0.8

= 0.156 cum/s

Peak Flowrate = 2.5* 0.156 cum/s = 0.391 cum/s = 391 L/s

Slope to be provided = s=1 in 1000 = 1/1000 = 0.001

Find out D considering that there will be full flow.

Q = A.V

STEP 1. Find out the average flowrate and maximum flow rate

STEP 2. Find out the optimum slope to be provided

STEP 3. Find out the size based on the maximum flowrate.

2/13/21 s Rn

V 2

4

D

A

44

2

D D

D

P A

R

2/13/22

*41

*4. s D

n D

V AQ


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Q=0.391 m 3/s S= 0.001 n =0.015

391.0)001.0(*

4

*

015.0

1*

4

2/13/22

D D

595.03/8 D

823.0)595.0( 8/3 D m

sm AQ

V /778.0

4)8.0(*

391.02

Say, D = 800 mm

2/13/22

*4*1

*4. s D

n D

V AQ

>0.6 m/s (OK)


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design of sewers example_2

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