Sewerage RETICULATION Design

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Sewerage RETICULATION Design

Ts. Dr ROSLINDA SESWOYA

AREA 1 AREA 2 AREA 3 AREA 4

STP

To manufr’s detail

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Sewerage System Plan MH 1

MH 2 MH

3

MH 4 MH

5 MH

6 MH

7

MH 9

MH

8

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Hydraulic Profile and Processes and

Instrumentation of Sewage Treatment System

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Example 1

A sewerage system plan as below is proposed for a development of the petrol station with 6 unit toilets. Design the sewerage reticulation system using new vitrified clay sewer.

Consider:

Sewer diameter = 225 mm Ground level = 24.200 m Slope gradient of 1: 150 Upper invert level = 23.088 m

MH 1

MH 2

MH

3

SEPTIC TANK

Population : 30 Population : 60

7.5 m 8.1 m

2 toilets 4 toilets

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Solution MH 1 to MH 2

1. Population on line (PE) = 30

2. Dry weather flow (DWF) = 30 x 225 L/cap.day = 6750 L/d = 0.08 L/s 3. Peak flow factor = 4.7 ( PE/1000)

^-0.11

= 6.91

4. Peak DWF = 6.91 x 0.08 L/s =0.54 L/s 5. Diameter of sewer = 0.225 m

Condition : full pipe flow 5a. Area, A =

^𝜋𝐷₄²

= 0.04 m

²

5b. Wetted perimeter, P = πD =0.71 m 5c. Hydraulic radius, Rh = A/P = 0.06 m 5d. Gradient, S = 1:150 = 0.007

5e. Velocity, v = (R

^2/3

S

^1/2

)/n = 1.19 m/s (within the range of 0.8 -4.0 OK!, and bigger than sediment cleaning velocity, 0.8 m/s  OK!)

5f. Maximum flowrate, Q = AV = 47.18 L/s ( > peak DWF  OK!)

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Solution

MH 1 to MH 2 (continued)

6. Length of sewer = 7.5 m

6a. Fall = length of sewer /gradient = 7.5/150 = 0.05 m 6b. Ground level, GL ( as observed by surveyor) = 24.200 m 6c. Upper invert level ( as fixed by consultant) = 23.088 m 6d. Average depth = GL - upper invert level = 1.112 m

6e. Lower invert level = Upper invert level – fall = 23.038 m GL = 24.200m

Upper IL = 23.088m Lower IL = 23.038m

Ave depth

1.112 m

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U P P E R M H

L O W E R M H

P O P U L A T I O N

C U M M U L A T I V E

P O P U L A T I O N

D W F

P E A K D W F

D I A M E T E R S E W E R

V E L O C I T Y - F U L L

P I P E

F L O W

M A X I M U M F L O W R A T E

L E N G T H O F S E W E R

F A L L

U P P E R I N V E R T

L E V E L

L O W E R I N V E R T

L E V E L

A V E R A G E D E P T H

G R O U N D L E V E L

- - PE PE L/s L/s m m/s L/s m m m m m m

MH1 MH2 30 30 0.08 0.54 0.225 1.19 47.18 7.5 0.050 23.088 23.038 1.112 24.200 MH2 MH3 60 90 0.23 1.44 0.225 1.19 47.18 8.1 0.054 23.088 23.034 1.112 24.200

SEWERAGE RETICULATION DESIGN

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U P P E R M H

L O W E R M H

P O P U L A T I O N

D W F

P E A K D W F

P I P E

F L O W

F A L L

L E V E L

- - PE PE L/s L/s m m/s L/s m m m m m m

MH1 MH2 30 30 0.08 0.54 0.225 1.19 47.18 7.5 0.050 23.088 23.038 1.112 24.200 MH2 MH3 60 90 0.23 1.44 0.225 1.19 47.18 8.1 0.054 23.088 23.034 1.112 24.200

SEWERAGE RETICULATION DESIGN

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Example 2

Design the sewerage reticulation system using new ductile iron sewer (225 mm diameter) for the following sewerage system plan. Given

Ground level = 24.200 m Upper invert level = 23.088 m

MH 1

MH 2

MH 3

SEWAGE TREATMENT PLANT

Commercial area = 2000 m

²

Residential area = 200 houses

10 m S= 1: 150

10 m S = 1: 80

market = Wet (15 stalls) Dry ( 20 stalls)

MH

10 m

4

S = 1:60

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U P P E R M H

L O W E R M H

P O P U L A T I O N

D W F

P E A K D W F

G R A D I E N T

P I P E

F L O W

F A L L

L E V E L

- - PE PE L/s L/s m - m/s L/s m m m m m m

MH1 MH2 60 60 0.16 1.00 0.225 0.0067 0.99 39.32 10 0.067 23.088 23.021 1.112 24.200 MH2 MH3 1000 1060 2.76 12.89 0.225 0.0125 1.35 53.84 10 0.125 23.088 22.963 1.112 24.200 MH3 MH4 65 1125 2.93 13.59 0.225 0.0167 1.56 62.17 10 0.067 23.088 22.921 1.112 24.200

Answer (Hint)

?

? ?

?

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Sewage is valueable resource, but it is seen as a

burden to disposed of.

This perception

shall be changed.

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Tutorial

Refer to Figure 1 and considering the full pipe flow, recommend the sewerage reticulation for MH3 – MH4 (only) using sewer of 225 mm diameter. Use the followings:

Ground level: 45.000 m Upper invert level: 43.500 m Slope: 1 to 80

n: 0.010

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Sewerage RETICULATION Design