PPT Sewerage System

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Sewerage System

Ts. Dr Roslinda Seswoya

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Contents

Design of Force Mains Flowrate Estimations

Sewer Cleansing Velocities

Requirements and Limitations for Use of Certain Pipe Material Design of Gravity Sewer

Sewer Types

Sewer Design – general requirements

Hydraulic element for circle sewer

RS 2023

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Sewer Design –General Requirement

The sewerage system shall be suitably designed to carry all sewage flows including sullage to the approved disposal point. Unauthorised

connections of surface waters or excessive infiltration to the sewerage system are not permitted.

Unless otherwise agreed by the Commission, all sewers shall be sited in public road reserve so that access can be gained for maintenance

purposes. Under special circumstances where the sewer cannot be sited in public road reserve then vehicular access for the sewerline

of at least 3 m in width and road bearing capacity of not less 5 tonne shall be provided.

RS 2023

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Sewer pipes should not be

constructed on slope or within slope failure envelope. In the event where it is unavoidable, the said structures must be designed not to encounter settlement or the sorts and at any time at risk of collapse during its operating lifespan.

An overflow pipe shall be provided at the last manhole before network pump station and/or sewage

treatment plant. Otherwise it should be located at the manhole sited at the lowest ground level.

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Sewer Types

I. Separate sewer systems One pipe carries stormwater(rain water) from storm drains to waterbodies without treatment.

Another pipe carries sewage to the

wastewater/sewage treatment plant

Widely practices in Malaysia

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II.Combined Sewer System Combined sewer systems only have one pipe.

Sewage from domestic area (residential/commercial/

school etc) drains to the pipe.

When it rains,

stormwater flows into the same pipe and mixes with raw sewage.

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to cater for peak flow.

to ensure that there will be a sufficient velocity

during each day to sufficiently cleanse the sewer

of slime and sediment.

to limit the velocity to avoid scouring of sewers Few principal considerations when selecting the diameter and gradient of a sewer are:

Flowrate Estimations

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I) Average Flow

The volume of sewage that needs to be treated per day is based on an assumed contribution per population equivalent of 225 litres (225 L/cap.day)from various types of premises where the contribution from each premise type is defined in terms of a PE

II) Peak Flow

The flow used to determine the diameter and gradient of the pipeline is the peak flow. Peak flow is the most severe flow that could occur on any day when considering daily flow fluctuations and infiltrations.

The peak flow is derived from the average flow by applying a peak factor for daily flow fluctuations. The peak factor shall be estimated from the following formula:

Peak Factor = 4.7 (PE/1000)^-^0.11

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Example 1

By considering dry weather flow of 225 L/cap.day, calculate peak flow rate for the sewerage reticulation of 500 PE.

Solution

Hydraulic flow = 500 cap x 225 L/cap/day = 112500 L/day = 0.0013 m³/s Peak Factor = 4.7 (PE/1000)^-0.11

= 4.7 ( 500/1000) ^-0.11

= 5.07

Peak flow = Peak factor x hydraulic flow

= 5.07 x 0.0013 m³/s

= 0.0066 m³/s

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III Infiltration

Infiltration is the amount of groundwater that enters sewers through damage in the network such as cracked pipes, leaked joint seals and manhole walls, etc.

There are many variables affecting infiltration such as quality of workmanship, joint types, pipe

materials, height of water table above pipeline, soil type, etc.

The peak factor above has included the contribution of infiltrations.

The maximum allowable infiltration rate shall be 50 litre/ (mm diameter.km of sewer length.day).

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Sewer Cleansing Velocities The principal accumulants in sewers are slimes and sediments. The hydraulic

requirements for cleansing the

sediments of sewer differ from those required for

cleansing the slimes

of sewer.

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I) Sediment Cleansing

For the removal of sediments, the traditional design

approach has been to set a minimum velocity to be achieved at least once daily

Minimum velocity values at full bore of 0.8 m/s are

commonly specified. However, it has been found that larger pipe diameters require higher velocity to cleanse the

sediment. This is mainly due to higher sediment depths in

large diameter pipes

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II) Slime Cleansing

Removal of large portion of slimes requires high sewage velocities.

It has been found that 85% or more of the sulphide producing slimes are removed when the grade of the sewer is 2.5 times of that for sediment cleansing.

In many instances, it may not be practical to design a sewer to achieve such velocities due on the excessive cost of constructing such a deep and steep sewer. Although increasing the velocity up to the critical velocity will increase the amount of slime being

sloughed off, the rate of sulphide production remains substantially unaffected by the thinner slime layer.

Therefore, the selection of steep gradient to achieve velocities for full slime stripping is not a design requirement

.

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Requirements and Limitations for Use of Certain Pipe Material

Unless the exemption is granted by the Commission (SPAN), the following Limitations:

I) Gravity Sewer

a) Vetrified Clay (VC)

i) Only size 150 mm or above shall be used.

ii) The minimum size for public sewer shall be at least 225 mm.

iii) Pipe shall not be used in unstable ground.

iv) Flexible joints are recommended.

b) Reinforced Concrete (RC)

i) Pipe protection linings are required.

ii) Only sizes 600 mm or above are allowed in compliance to the policy.

iii) Flexible joints are recommended.

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c) Glass reinforced plastic (GRP)

i) Pipe shall not be used in ground contaminated with high concentration of chemicals such as solvent that can degrade the pipe.

ii) Pipe shall not accept any industrial or other aggressive discharges that may affect the pipe integrity.

iii) Pipe shall be used only when no fittings are required.

iv) Only sizes 600 mm or above are allowed.

d) Ductile iron (DI)

i) The use is only allowed for applications needed high pipe strength.

ii) Pipe protection linings and coatings are required.

iii) Polyethylene sleeving is required for all buried applications.

e) High-density polyethylene (HDPE)

iii) Only pipe with profile wall is permitted.

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II) Force Mains a) DI

i) Pipe shall not be used in unstable ground.

iii) Polyethylene sleeving is required for all buried applications.

iv) Flexible joints are recommended.

b) GRP

iii) Fittings shall be made of ductile iron.

iv) Only sizes 600 mm or above are allowed.

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c) Acrylonitrile butadiene styrene (ABS) i) Where VC or RC pipes are not suitable.

ii) Only for nominated projects or as permitted by the relevant authority.

d) HDPE

e) Steel

i) Pipe is allowed only for sizes 700 mm or above.

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Convey sewage from discharge side of a pump to sewage treatment plant.

The key elements of force main are pipe, valves, pressure surge control devices and force main cleaning system

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Design of Gravity Sewer

Unless special arrangements have been agreed for the

structural protection of pipes, the minimum depth of soil cover over the sewer shall be 1.2 m. Sewers are not to be constructed under buildings.

The minimum size of public gravity sewers shall be 225 mm in diameter. The minimum size of domestic connections to the public sewer shall be 150 mm in diameter.

The maximum design velocity at peak flow shall not be more than 4.0 m/s.

The design shall be based on the worst case scenario.

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The selection of the gravity sewer diameter and gradient to cope with the peak flow shall be based on the following equations:

.

Typical ks values for various types

of sewer pipes are presented in Table 2.2 :

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.

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.

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Design of Force Mains

The minimum diameter of force mains (also known as rising mains) shall be 100 mm diameter. There shall be no reduction in force main diameter with distance downstream.

All bends on force mains shall be securely anchored to resist lateral thrusts and subsequent joint movements.

Air release valves and washouts shall be provided at appropriate locations along the longitudinal profile.

For long and undulating force mains, hydraulic pressure transient analyses may be required to ensure that the force main can cope with water hammer pressures.

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Retention times in force mains must not exceed 2 hours without special precautions to mitigate septicity.

All force main shall be designed to withstand at least 1.5 times the working pressure. Approval from the Commission is required if any force main is to be designed to withstand pressure less than the pressure stated above.

Where retention times in the force mains exceed two hours and where concrete pipe are laid downstream of the force mains, an induct vent shall be provided at manholes receiving pumping discharges.

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Force mains shall be designed to handle the full range of flows from present minimum to future peak.

The design velocity shall fall within the range of 0.8 to 3.0m/sec over the full range of design flows.

The hydraulic resistance of force main fittings and bends shall be includedin the hydraulic design.

Friction losses are normally calculated using either Darcy- Weisbach (Colebrook-White) Equation or Hazen-Williams

Equations.

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This equation with the Moody Diagram are used to

determine the coefficient of friction, f.

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Hydraulic element for circle sewer

For this case of less than half full flow, h is

simply equal to the depth of flow y

where Radius of the pipe = r

Cross sectional area of flow= A Wetted perimeter = P

Hydraulic radius, = Rh

RS 2023

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For more than half full pipe

flow, the parameter h is 2r – y, instead of simply being equal to y as for less than half full pipe flow.

where Radius of the pipe = r

Cross sectional area of flow= A Wetted perimeter = P

Hydraulic radius, = Rh

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For full pipe the hydraulic radius of a full pipe is simply half of its radius.

Cross sectional area of flow= A = ^{𝜋 𝐷}² Wetted perimeter = P = 𝜋𝐷 4

Hydraulic radius = Rh = ^𝐴

𝑃 = ^𝐷

4 = ^𝑟

2

where r = Radius of the pipe

r

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Example 2

(a) Calculate the hydraulic radius (m) for sewage flowing 20 mm deep in a pipe of 100 mm diameter.

(b) Use answer from (a),design a concrete circular gravity sewer with gradient of 0.05

Solution:

(a)

r = D/2 = 50 mm = 0.050 m;

h = y = 20 mm = 0.020 m;

θ = 2 arccos [ (0.050 –

0.020)/0.050) ] = 1.85 radians A = [ 0.05

²

(1.85 – sin (1.85)) ] / 2 = 0.00111 m

²

P = (0.05)(1.85) = 0.0925 m Rh = 0.00111/0.0925 = 0.0120 m

RS 2023

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(b)

Knowing that

Manning coefficient for concrete pipe = 0.012 ( Table 2.3) Velocity =

^𝑅

2 3𝑆

1 2

𝑛 = ^0.012

2 3(0.05)

1 2

0.012 = 0.962 m/s ok! ( 0.8 – 4.0 m/s)

Assume the flow is uniform, the capacity of sewage flowing 20 mm deep in a pipe of 100 mm diameter is

Q = AV = 0.00111 m

²

x 0.962 m/s

= 0.0011 m

³

/s

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Example 3

Design a circular gravity sewer which running half full. Does the sediment in the pipe will be removed? Give your comment.

Given

Pipe diameter = 225 mm

Type of pipe = Ductile iron (D1) Pipe gradient = 1:80

Solution:

Wetted perimeter = P =¹

2𝜋𝐷 = ¹

2𝜋 225 = 353 .25 𝑚𝑚 Hydraulic radius = Rh = ^𝐴

𝑃 = ^{19870.31𝑚𝑚}²

353.25 𝑚𝑚 = 56.25 mm Cross sectional area of flow= A = ¹

2 𝜋 𝐷²

4 = ¹

2

𝜋 225²

4 = 19870.31 mm²

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Knowing that

Manning coefficient for ductile iron (DI) pipe = 0.012 ( Table 2.3) Velocity = ^𝑅

2 3𝑆

1 2

𝑛 = ^(0.056)

2 3 1

80 1 2

0.012 = 1.36 m/s ok! ( 0.8 – 4.0 m/s)

1.36 m/s velocity is bigger then 0.8 m/s sediment cleansing velocity.

Therefore, the sediment in the pipe will be removed

Assume the flow is uniform, the capacity of flow when this pipe is flowing half full is

Q = AV = 19870.31 mm² x 1.36 m/s

= 0.02 m² x 1.36 m/s

= 0.27 m³/s

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Example 4

Recommend a circular gravity sewer that appropriate to operate at peak flow rate for population equivalent of 3000. Assume the sewer is running full.

Given

Pipe diameter = 225 mm

Type of pipe = Ductile iron (D1) Pipe gradient = 1:80

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Solution

Dry weather flow per capita = 225L/cap/day

1 Population 3000

2Dry weather flow ( DWF) =

3000 X 225 L/cap.day = 675000L/day

7.81L/s 0.0078m3/s

3Peak flow factor = 4.16

4.7 (PE/1000)^-0.11

4Peak DWF = 2811372L/day

Peak flow factor x DWF 32.54 L/S

0.0325m3/s

RS 2023

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Solution

5Diameter of sewer, D (given ) = 0.225m

flow in full condition

5aArea, A = (πD2)/ 4 = 0.039740625m2

5bWet parameter, WP = πD = 0.7065m

5cThen hydraulic radius, R =A/WP = 0.05625m

5dGradient = 1: 80 ( given) 0.0125 1: 80

Knowing that,

R^2/3 = 0.145

S^1/2 = 0.112

n (Ductile iron) 0.012

5eVelocity = ( R^{2 /3}S^1/2)/n = 1.35m/s OK! ( 0.8 m/s - 4.0 m/s) RS 2023

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Solution

5fMaximum flowrate , Q = AV= 0.05m3/s

(running half full flow) 53.84L/s > peak DWF, OK!

Ductile iron sewer with diamter of 225 mm is appropriate to handle sewage from 3000 PE because the the velocity ( 1.35m/s) is within the recommended velocity ( 0.8 to 4.0 m/s) and the maximum flowrate is bigger than peak DWF. Sediment cleansing is possible

because the velocity of 1.68 m/s is bigger than 0.8 m/s

RS 2023

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Tutorial

Q1. Calculate the velocity of flow in a sewer of diameter 1.36 m. The gradient of sewer line is 1 to 420. If Manning coefficient, n is 0.012, calculate the discharge when running one- half full

(Answer : V = 1.9 m/s Q = 1.44 m

³

/s )

Q2. Calculate the discharge for a sewer running full. The diameter of sewer is 200 mm and it is laid at the slope of 1 to 72 and Manning coefficient, n is 0.013

(Answer : V = 1.23 m/s Q = 0.044 m

³

/s )

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Tutorial

Q3. Refer to Example 2. Assume the sewage flows in

1) full pipe and 2) less than half full (h=70 mm).

Recommend the sewer pipe for each condition and give comments on the changes

Answer: 1)Q = 0.05 m

³

/s 2) Q = 0.012 m

³

/s

Q4. Refer to Example. Comment on the changes if the sewage is running half full.

Q5. Recommend a circular gravity sewer that appropriate to operate at peak flow rate for population equivalent of 5000 that comply with the requirement as stated in

MSIG.

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Thank You