Guest
Chapter:
1. 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.
2. 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.
3. 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`
4. Design a sanitary sewer to carry dry wear flow of `0.26\ m^3` /sec laid at a gradient of 1 in 500.Assume manningâ??s `n=0.012` and sewer runs at `3/5` th depth during peak flow condition. Take peak factor =3.
Solution:
`q=a/n (r)^(2/3) s^(1/2)`
where, `q=0.26\ m^3/s`
`s=1/500`
`n=0.012`
`d/D=3/5`
or, `1/2 (1-cos (theta/2))=3/5`
or, `theta=203.07^0`
Thus, the area is,
`a/A=theta/360-(sin theta)/(2 pi)`
`=0.6265`
`r/R=[1-(360 sin theta)/(2 pi theta)]`
`=1.1106`
Thus,Â
`q/Q=(a/A)(r/R)^(2/3)`
`=0.6265**(1.1106)^(2/3)`
`=0.6718`
or, `(0.26**3)/Q=0.6719`
or, `Q=1.1609`
or, `1/0.012**(pi D^2)/4**(D/4)^(2/3)**(1/500)^(1/2)=1.1609`
or, `D=1\ m`
5. Calculate diameter and discharge of a circular sewer laid at a slope of 1 in 500 when running half full with velocity of 2 m/sec. Take n =0.012
Solution:
We have, from Mannings formula,
`v=1/n R^(2/3) S^(1/2)`
or, `2=1/0.012 (d/4)^(2/3)(1/500)^(1/2)`
or, `d=1.572\ m`
And the discharge is,
`Q=AV`
`=1/2(pi**1.572^2)/4**2`
`=1.942\ m^3/s`
6. A one meter diameter sewer is required to flow at 60% depth on a grade ensuring a degree of self cleansing equivalent to that obtained at full depth at a velocity of 0.9 m/s. Find required grade, associated velocities and rates of discharged at full depth and 0.6 depth. Take a uniform value of n=0.012 at all depth of flow.
Solution:
For the sewer running at full depth,
`v=0.9=1/0.012 **(1/4)^(2/3)**(S)^(1/2)`
Solving this, we get,
`S=1/1350`
and the discharge is, `Q=(pi **1^2)/4**0.9=0.706\ m^3/s`
Again for the sewer running at partial depth,
`d/D=0.6`
or, `1/2(1-cos (theta/2))=0.6`
or, `theta=203.07^0`
Now, `v/0.9=(1-(360 sin theta)/(2 pi theta))^(2/3)=1.072`
or, `v=0.965\ m/s`
 and the discharge is,
 `q/0.706=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)`
 or, `q=0.474\ m^3/s`
 For the sewer at 60% depth to ensure equivalent self cleansing velocity to that obtained at full depth, it should have gradient `S_s`
 `S_s=(R/r)S=1/1499.3`
 Now,Â
 `V_s/V=n/n_d(r/R)^(1/6)=1.017`
 or, `V_s=0.9305\ m/s`
 Again,
 `q_s/Q=n/n_d**a_s/A**(r/R)^(1/6)=0.718`
 or, `q_s=0.5069\ m^3/s`
7. Design a sewer to serve a population of 36000; daily per capita water supply allowance being 135 liters, of which 80% finds its way into sewer. The slope available for sewer to be laid is 1 in 625 and sewer should be designed to carry four times dry wear flow when running full. What would be velocity of flow in sewer when running full? Assume n=0.012 in manningâ??s formula.
Solution:
Here, the design discharge is,
`Q=(0.8**0.135**36000**4)/86400`
`=0.18\ m^3/s`
From Mannings Formula,
`0.18=1/0.012 **(pi D^2)/4(D/4)^(2/3)**(1/625)^(1/2)`
or, `D=0.518\ m`
Now, the velocity is,
`V=0.18/((pi**0.518^2)/4)`
`=0.85\ m/s`
8. 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.
9. 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,Â
`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.
10. 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.
