Design of sewers Tutorial

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1. Design section of combined sewer from following data: Area to be served =150 ha, population of locality = 50000, maximum permissible velocity =3.2 m/sec, time of entry = 5 minutes, time of flow = 20 minutes, rate of water supply =270 lpcd, impermeability factor =0.45, maximum discharge =1.8 times DWF. Assume any or suitable data if needed.

Solution:

Assuming `100%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(1**50000**0.270)/(86400)`

`=0.15625\ m^3/s`

The peak sanitary discharge is, `=1.8**0.15625=0.28125\ m^3/s`

The sewer is designed for maximum discharge.

Now,

`T_c=T_e+T_f=5+20=25\ m i n`

Where, `T_c` is the time of concentration

`T_e` is the time of entry.

`T_f` is the time of flow.

The quantity of storm water will be maximum when storm duration is equal to the time of concentration.

`Thus, t=T_c=25 mi n`

Now, the intensity of rainfall is,

`i=1020/(t+20)=22.67\ (m m)/(h r)`

Again, the storm discharge is given by,

`Q_(W W F)=(CiA)/360=(0.45**22.67**150)/360=4.35\ m^3/s`

Thus, the discharge for the combined sewer is,

`Q=4.35+0.281`

`=4.531\ m^3/s`

Now, `Q=AV`

or, `4.531=(pi d^2)/4**3.2`

or, `d=1.34\ m`

Adopting commercially available size, `d=1.40\ m`

`A=(pid^2)/4=1.54\ m^2`

`V=Q/A=0.4.531/1.54=2.94\ m/s`

Here, `0.6<2.94<3`. Hence okay.

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.15625/4.531=0.034`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=83.25^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.464`

or, `v=0.464**2.94=1.36`

Here 0.60 < 1.36 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.017`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=70.11^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.377`

or, `v=0.377**2.94=1.11><0.60`

Hence the design is okay.

2. Design combined circular sewer from following available data and draw neat sketch of sewer.

Population of locality `= 75000`

Rate of water supply =275 lpcd

Area to be served = 175 hectares

Maximum permissible velocity = 3 m/sec

Time of entry = 5 minutes

Time of flow = 20 minutes

Average permeability factor =0.45

Solution:

Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(0.8**75000**0.275)/(86400)`

`=0.1909\ m^3/s`

The peak sanitary discharge is, `=2**0.1909=0.3819\ m^3/s`

Now,

`T_c=T_e+T_f=5+20=25\ m i n`

Where, `T_c` is the time of concentration

`T_e` is the time of entry.

`T_f` is the time of flow.

The quantity of storm water will be maximum when storm duration is equal to the time of concentration.

Thus, t=T_c=25\ mi n`

Now, the intensty of rainfall is,

`i=1020/(t+20)`

`=1020/(45)=22.67\ ( m m)/(hr)`

Again, the storm discharge is given by,

`Q_(W W F)=(CiA)/360`

`=(0.45**22.67**175)/360`

`=4.959\ m^3/s`

Therefore, the combined discharge is,

`Q=0.3819+4.959=5.3409\ m^3/s`

Now, 5.3409=(pi d^2)/4**3`

or, `d=1.505\ m`

Adopting commercially available size `d=1.6\ m`

`A=(pid^2)/4=2.0106\ m^2`

`V=Q/A=2.66\ m/s` which is less than `3\ m/s`.

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.1909/5.3409=0.03574`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=84.303^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.471`

or, `v=0.471**2.66=1.25`

Here 0.60 < 1.25 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.01787`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]`

Solving this, we get,

`theta=70.97^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.38`

or, `v=0.38**2.66=1.01`

Here 0.60 < 1.01 < 3 m/s. Okay

Hence the design is okay.

3. A town has population of 1 lakh with per capita water supply of 200 lpcd. Design a sewer taking n=0.013; slope =1 in 600 and peak factor =2.25.Assume 80% of water supply is converted to sewage.

Solution:

Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(0.8**100000**0.2)/(86400)`

`=0.1852\ m^3/s`

The peak sanitary discharge is, `=2.25**0.1852=0.4167\ m^3/s`

The sewer is designed for maximum discharge.

Now,

`Q=(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/600)^(1/2)`

Solving this, we get,

`d=0.73\ m`

Adopting commercially available size, `d=0.75\ m`

`A=(pid^2)/4=0.442\ m^2`

`V=Q/A=0.4167/0.442=0.94\ m/s`

Here, `0.6<0.94<3`. Hence okay.

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.1852/0.4167=0.444`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=169.9^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.96`

or, `v=0.96**0.94=0.90`

Here 0.60 < 0.90 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.222`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=137.84^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.804`

or, `v=0.804**0.94=0.755 >0.60`

Hence the design is okay.

4. Determine size of circular sewer for a discharge of `500` lps running half full. The gradient is 1 in 1000 and `n=0.015`.

Solution:

Here, `Q=500 lps=0.5 m^3/s`

We have from Mannings formula,

`Q=A**1/n R^(2/3) S^(1/2)`

or, `0.5=1/0.015**1/2**pid^2/4**(d/4)^(2/3)**(1/1000)^(1/2)`

or, `d=1.17\ m`

5. Design a combined sewer section for a 45 hector residential area having runoff coefficient `0.40, 0 70. 0.25, 0.80, 0.10` for area of `15,20,25,10` and `30%` respectively with altoger `1500` population? Average rainfall duration is `21` min. Self-cleansing velocity is `0.88` m/s. Assume water supply rate = 100 lpcd and time of concentration =20 min.

Solution:

Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(0.8**1500**0.10)/(86400)`

`=1.388**10^-3\ m^3/s`

The peak sanitary discharge is, `=2**1.388**10^-3=2.78**10^-3`

Now,

Time of concentration, `T_c=20\ m i n`

And the storm duration is 21 min.

Since the storm duration is greater than the time of concentration, whole of the area (45 hectares) will contribute for the runoff.

Now, the intensity of rainfall is,

`i=1020/(t+20)`

`=1020/(21+20)`

`=24.88\ (m m)/(hr)`

Now, average impermeability coefficient is given by,

`C=(C_1A_1+C_2A_2+C_3A_3+C_4A_4+C_5A_5)/(A_1+A_2+A_3+A_4+A_5)`

`=0.3725`

Now, The storm discharge is,

`Q_( W W F)=(CiA)/360`

`=(0.375**24.87**45)/360`

`=1.1583\ m^3/s`

Thus, the combined discharge is,

`Q=2.77**10^-3+1.1583=1.16115\ m^3/s`

Given, Self cleansing velocity is `0.88\ m/s`

Thus,

`A=Q/V=1.16115/0.88=1.319`

Thus, `D=1.3\ m`

6. Calculate diameter and velocity of a circular sewer at a slope of 1 in 400 when it is running just full at a discharger of `2` `m^3`/sec. The Manning's coefficient n = 0.013. What will be discharge and velocity when flowing one third full?

Solution:

When running just full,

We know,

`Q=A/nR^(2/3)S^(1/2)`

or, `2=(pid^2)/4**(1/0.013)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=1.2115\ m`

And the velocity is, `V=Q/A=2/((pi**1.2115^2)/4)`

or, `V=1.735\ m/s`

When flowing at a one third depth,

`d/D=1/3`

or, `1/2(1-cos (theta/2))=1/3`

or, `theta=141.057^0`

Now, we know,

`q/Q=q/2=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`q=0.4794\ m^3/s`

And the velocity is,

`v/1.735=(1-(360 sin theta)/(2pi theta))^(2/3)`

Solving this equation, we get,

`v=1.425\ m/s`

7. Calculate diameter and velocity of circular sewer at a slope of 1 in 400 when it is running just full at a discharge of `1\ m^3`/sec. The value of n in manning's coefficient is 0.012. Will self cleansing velocity be maintained in sewer when flow drops to `0.6\ m^3`/sec?

Solution:

When running just full,

We know,

`Q=A/nR^(2/3)S^(1/2)`

or, `1=(pid^2)/4**(1/0.012)(d/4)^(2/3)**(1/400)^(1/2)`

Solving this, we get,

`d=0.9066\ m`

A=pi d^2/4=0.65\ m^2`

And the velocity is, `V=Q/A=1/((pi**0.9066^2)/4)`

or, `V=1.5489\ m/s`

Now, when flows drops to `0.6\ m^3/s`,

`q/Q=0.6/1=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`

Solving this equation, we get,

`theta=193.38^0`

Also,

`v/1.5489=(1-(360 sin theta)/(2pi theta))^(2/3)`

or, `v=1.618\ m/s`

Since `1.618 > 0.6`, Self cleansing velocity will be maintained in the sewer.

8. Design separate sewer system for a town having population of 1 lakh with water supply of 180 lpcd. The permissible sewer slope is 1 in 1000 and n=0.012. Assume DWF is` 1/3` rd of maximum discharge. Also check velocity of flow.

Solution:

The design discharge is,

`Q=(3**0.8**100000**0.18)/86400`

`=0.5\ m^3/s`

Now, Considering the circular sewer running full,

`Q=(pi D^2)/4**1/0.012**(D/4)^(2/3)**(1/1000)^(1/2)`

or, `D=0.83\ m`

Thus, `V=0.5/((pi**0.83^2)/4)=0.924\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.333`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=156.31^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.899`

or, `v=0.899**0.924=0.831`

Here 0.60 < 0.831 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.1665`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]`

Solving this, we get,

`theta=126.77^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.74`

or, `v=0.74**0.924=0.683>0.6`

Hence the design is okay.

9. Calculate quantity of wastewater to be carried by separate system with following data and design a half flowing sanitary sewage with a slope of 1 in 600.Check for self cleaning velocity and limiting velocity for concrete sewer pipe. Area to be served = 500 ha, population density = 100 persons/ha, waters supply rate = 20 lpcd, peak factor =3 and 80% of water supply is converted to sewage. Assume any data if required.

Solution:

Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(0.8**100**500**0.02)/(86400)`

`=0.00926\ m^3/s`

The peak sanitary discharge is, `=3**0.00926=0.0278\ m^3/s`

Now, For half full circular sewer,

`q=0.0278=(pid^2)/8**(1/0.014)**(d/4)^(2/3)**(1/600)^(1/2)`

or, `d=0.351\ m`

Adopting commercially available size, `d=0.35\ m`

`A=(pid^2)/8=0.048\ m^2`

`V=Q/A=0.1874/0.2513=0.57\ m/s`

Here, `v=0.57 <0.6\ m/s`.

For cement concrete pipes the limiting value i.e, maximum value is about 3.0 and the minimum self cleansing velocity is 0.6m/s

Since the velocity is less than self cleansing velocity, flow does not maintain the self cleansing velocity of `0.6\ m/s`

10. Design a sewer for a population of 100,000 persons with water supply per capita of `120` l/d. It is expected that `80%` of water is converted into sewage. The DWF estimated will be `1/3` rd of maximum discharge in this separate sewer. The permissible slope is `1:1000` and rugosity coefficient is taken as 0.012. For self-cleaning purpose at least `0.75` m/sec velocity need to be developed in drain.

Solution:

Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by

`Q_(DWF)=(0.8**100000**0.12)/(86400)`

`=0.11\ m^3/s`

The peak sanitary discharge is, `=3**0.11=0.33\ m^3/s`

Now,

`Q=(pi d^2)/4**(1/0.012)**(d/4)^(2/3)**(1/1000)^(1/2)`

or, `d=0.713\ m`

Adopting commercially available size, `d=0.75\ m`

Now, velocity is,

`v=0.33/((pi**0.713^2)/4)`

`=0.8346\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.11/0.33=0.333`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=156.31^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.899`

or, `v=0.899**0.8346=0.7503`

Here 0.75 < 0.7503 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.1665`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]`

Solving this, we get,

`theta=126.77^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.74`

or, `v=0.74**0.8346=0.61<0.75`

Hence the design is not okay.

let, `d=0.6\ m`

Now, velocity is,

`v=0.33/((pi**0.6^2)/4)`

`=1.17\ m/s`

Check for self Cleansing velocity during dry weather flow:

`q/Q=0.11/0.33=0.333`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`

Solving this, we get,

`theta=156.31^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.899`

or, `v=0.899**1.17=1.05`

Here 0.75 < 1.05 < 3 m/s. Okay

Check for self cleansing velocity during minimum flow.

Let, `Q_(m i n)=1/2 Q_(DWF)`

Thus,

`q/Q=0.1665`

or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]`

Solving this, we get,

`theta=126.77^0`

and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.74`

or, `v=0.74**1.17=0.86 > 0.75`

Hence the design is okay.

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Design of sewers Tutorial

Design section of combined sewer from following data: Area to be served =150 ha, population of locality = 50000, maximum permissible velocity =3.2 m/sec, time of entry = 5 minutes, time of flow = 20 minutes, rate of water supply =270 lpcd, impermeability factor =0.45, maximum discharge =1.8 times DWF. Assume any or suitable data if needed.

;

Design combined circular sewer from following available data and draw neat sketch of sewer.

Population of locality `= 75000`

Rate of water supply =275 lpcd

Area to be served = 175 hectares

Maximum permissible velocity = 3 m/sec

Time of entry = 5 minutes

Time of flow = 20 minutes

Average permeability factor =0.45

;

A town has population of 1 lakh with per capita water supply of 200 lpcd. Design a sewer taking n=0.013; slope =1 in 600 and peak factor =2.25.Assume 80% of water supply is converted to sewage.

;

Determine size of circular sewer for a discharge of `500` lps running half full. The gradient is 1 in 1000 and `n=0.015`.

;

Design a combined sewer section for a 45 hector residential area having runoff coefficient `0.40, 0 70. 0.25, 0.80, 0.10` for area of `15,20,25,10` and `30%` respectively with altoger `1500` population? Average rainfall duration is `21` min. Self-cleansing velocity is `0.88` m/s. Assume water supply rate = 100 lpcd and time of concentration =20 min.

;

Calculate diameter and velocity of a circular sewer at a slope of 1 in 400 when it is running just full at a discharger of `2` `m^3`/sec. The Manning's coefficient n = 0.013. What will be discharge and velocity when flowing one third full?

;

Calculate diameter and velocity of circular sewer at a slope of 1 in 400 when it is running just full at a discharge of `1\ m^3`/sec. The value of n in manning's coefficient is 0.012. Will self cleansing velocity be maintained in sewer when flow drops to `0.6\ m^3`/sec?

;

Design separate sewer system for a town having population of 1 lakh with water supply of 180 lpcd. The permissible sewer slope is 1 in 1000 and n=0.012. Assume DWF is` 1/3` rd of maximum discharge. Also check velocity of flow.

;

Calculate quantity of wastewater to be carried by separate system with following data and design a half flowing sanitary sewage with a slope of 1 in 600.Check for self cleaning velocity and limiting velocity for concrete sewer pipe. Area to be served = 500 ha, population density = 100 persons/ha, waters supply rate = 20 lpcd, peak factor =3 and 80% of water supply is converted to sewage. Assume any data if required.

;

Design a sewer for a population of 100,000 persons with water supply per capita of `120` l/d. It is expected that `80%` of water is converted into sewage. The DWF estimated will be `1/3` rd of maximum discharge in this separate sewer. The permissible slope is `1:1000` and rugosity coefficient is taken as 0.012. For self-cleaning purpose at least `0.75` m/sec velocity need to be developed in drain.

;

Design of sewers Tutorial

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Design of sewers Tutorial

A town has population of 1 lakh with per capita water supply of 200 lpcd. Design a sewer taking n=0.013; slope =1 in 600 and peak factor =2.25.Assume 80% of water supply is converted to sewage.

Determine size of circular sewer for a discharge of `500` lps running half full. The gradient is 1 in 1000 and `n=0.015`.

Calculate diameter and velocity of a circular sewer at a slope of 1 in 400 when it is running just full at a discharger of `2` `m^3`/sec. The Manning's coefficient n = 0.013. What will be discharge and velocity when flowing one third full?

Calculate diameter and velocity of circular sewer at a slope of 1 in 400 when it is running just full at a discharge of `1\ m^3`/sec. The value of n in manning's coefficient is 0.012. Will self cleansing velocity be maintained in sewer when flow drops to `0.6\ m^3`/sec?

Design separate sewer system for a town having population of 1 lakh with water supply of 180 lpcd. The permissible sewer slope is 1 in 1000 and n=0.012. Assume DWF is` 1/3` rd of maximum discharge. Also check velocity of flow.

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