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Chapter:
1. The population of a town is 80,000 persons with a water supply rate of 145 lpcd. Assuming 80% of water supply contributes for sewage flow, taking Manning's N as 0.013, average slope as 1:400 and peak factor as 3. Determine minimum diameter of sewer required to carry maximum discharge if it runs at 0.75 depths?
Solution:
Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by
`Q_(DWF)=(0.8**80000**0.145)/(86400)`
`=0.1074\ m^3/s`
The peak sanitary discharge is, `=3**0.1074=0.3222\ m^3/s`
Given,Â
`d/D=1/3`
or, `1/2(1-cos (theta/2))=0.75`
or, `theta=240^0`
Now, we know,
`q/Q=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)=0.9118`
Solving this equation, we get,
`0.3222/Q=0.9118\ m^3/s`
or, `Q=0.3533\ m^3/s`
or, `(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/400)^(1/2)=0.4559`
or, `d=0.632\ m`
2. Calculate diameter and velocity of circular sewer at a slope of 1 in 400 when running full at a discharge of 1.0 m 3 /sec. What will be discharge and velocity when flowing half full? Take n =0.012
Solution:
Case I: When Running Full
We know,
`Q=A/n R^(2/3) S^(1/2)`
Where, `R=A/P=((pi d^2)/4)/(pi d)=d/4`
or, `1=1/0.012**(pid^2)/4**(d/4)^(2/3)**(1/400)^(1/2)`
or, `d=0.906\ m`
Thus, the velocity is, `V=Q/A=1/((pi**0.906^2)/4)=1.55\ m/s`
Now, when the sewer is running half full,
`V=1/0.012 (0.906/4)^(2/3)(1/400)^(1/2)`
`=1.548\ m/s`
and the discharge is
`Q=AV=1/2(pid^2)/4**1.548`
`=0.499\ m^3/s`
3. Determine diameter and velocity of a circular sewer at a slope of 1 in 500 when it is running just full at a discharge of 1m3/sec. The value of n in Manning's formula is 0.012. What will be discharge and velocity when flowing half full for same slope?
Solution:
When running full,
`Q=1=(pi d^2)/4**(1/0.012)**(d/4)^(3/2)**(1/500)^(1/2)`
Solving this, we get,
`d=0.95\ m`
 and the velocity is, `v=Q/A`
 `=1/((pi**0.95^2)/4)`
 `=1.411\ m^3/s`
 Again, when the sewer is running half full,
 `Q=(pi d^2)/8**(1/0.012)**(d/4)^(3/2)**(1/500)^(1/2)`
or, `Q=0.5\ m^3/s`
and the velocity is,
`v=0.5/(pi*0.95^2/8)`
`=1.41\ m/s`
4. A 60 cm diameter sewer is to discharge `0.07` `m^3`/sec at a velocity as self cleansing as sewer flowing full at 0.85 m/sec. Find depth and velocity of flow and required slope. Take uniform value of manningâ??s coefficients (n) = 0.015
Solution:
When sewer is running full,
`d=0.6\ m`
`v=0.85=1/0.015(d/4)^(2/3)**(S)^(1/2)`
or, `S=1/490.269` and the discharge is,
`Q=pi**0.6^2/4**0.85=0.24\ m^3/s`
For partial depth self cleansing flow
`q_s=0.07\ m^3/s`
or, `q_s/Q=0.07/0.24=0.2912`
or, `0.2912=(theta/360-(sin theta)/(2pi))(1-(360 sin theta)/(2 pi theta))^(1/6)`
or, `theta=143.63^0`
Also, we have,
`d/D=1/2(1-cos theta/2)=0.3439`
or, `d=0.2063\ m`
Also,
`r/R=(1-(260 sin theta)/(2 pi theta))`
`=0.763`
Now,
`V_s/V=(r/R)^(1/6)`
or, `V_s=0.956**0.85=0.8162\ m/s`
Also,
`s=(R/r)S=1/0.763**2.093**10^-3`
`=1/374.20`
5. Calculate velocity and discharge through rectangular sewer of `2.5\ m` width and `1.5\ m` depth and laid at a gradient of `1:200` and running full. Take Chezy's coefficient ,`C =58.8`.
Solution:
From Chezy formula,
`V=C root()(RS)`
`=58.8 root()(RS)`
Where, `R=A/P=(2.5**1.5)/(2(2.5+1.5))=0.4687\ m`
Thus,
`V=58.8 root()(0.4687**1/200)`
`=2.846\ m/s`
And the discharge is,
`Q=AV=(1.5**2.5)**2.846`
`=10.67\ m^3/s`
6. A circular sewer and a rectangular sewer are hydraulically equivalent. Find relation between depth of rectangular sewer and diameter of circular sewer if width of rectangular sewer is 2.5 times depth. Assume that only three sides of rectangular sewer are wetted.
Solution:
Let `b` and `d` be the width and depth of rectangular sewer and `D` be the diameter of circular sewer.
Here, width of rectangular sewer is, `b=2.5 d`
For hydraulically equivalent sewers, discharge in both are same,
`Q_R=Q_C`
or, `bd**1/n **((bd)/(b+2d))^(2/3)**(s)^(1/2)=(piD^2)/4**(d/4)^(2/3)**s^(1/2)`
Solving this equation for d, we get,
`d=0.531 D`
 Thus, the depth of rectangular sewer is 0.531 times the diameter of circular sewer.
7. A main combined sewer is to be designed to serve an area of 12 sq.km with a population density of 250 persons/ha. The average rate of sewage flow is 250 lpcd. The maximum flow is 100% in excess of average toger with rainfall equivalent of 15 mm in 24 hours, all of which are runoff. Determine capacity of sewer. Taking maximum velocity of flow as 3 m/sec. Determine size of sewer.
Solution:
The dry weather flow is given by,
`Q_(DWF)=((250**1200)**(250)*1**1.5)/(1000**86400)`
`=1.302\ m^3/s`
Here, `i=15/24=0.625 (m m)/(h r)`
The storm flow is given by,
`Q_(W W F)=(CiA)/(360)`
`=1**0.625**1200)/360`
`=2.083\ m^3/s`
Thus, the total discharge of the combined sewer is,
`Q=1.302+2.083`
`=3.385\ m^3/s`
or, `3.385=(pi D^2)/4**3`
or, `D=1.2\ m`
8. A town has a population of 80000 persons with a per capita water supply of 180 liter per day. Design a sewer running 0.7 times full at maximum discharge. Take a constant value of n=0.013 at all depths of flow. The sewer is to be laid at a slope of 1 in 500.Take peak factor of 3.
Solution:
Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by
`Q_(DWF)=(0.8**80000**0.18)/(86400)`
`=0.133\ m^3/s`
The peak sanitary discharge is, `=3**0.133=0.4`
By question,
`d/D=0.7`
or, `1/2(1-cos (theta/2))=0.7`
or, `theta=227.15^0`
Now, `q/Q=theta/360(1-(360 sin theta)/(2 pi theta))^(5/3)=0.8372`
 or, `0.4/Q=0.8372\ m^3/s`
or, `Q=0.4778\ m^3/s`
or, `(pi d^2)/4**(1/0.013)**(d/4)^(2/3)**(1/500)^(1/2)=0.4559`
or, `d=0.738\ m`
Adopt commercially available size, `d=0.75\ m`
`v=1/0.013**0.1875^(2/3)**(1/500)^(1/2)`
or, `v=1.127\ 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**1.127=1.01`
Here 0.60 < 1.01 < 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**1.127=0.83`
Here 0.60 < 0.83 < 3 m/s. Okay
Hence the design is okay.
9. Design a section of combined circular sewer from following data :
Area to be served = 250 ha
- Population = 60,000
- Maximum permissible velocity = 3.2 m/sec
- Time of entry = 5 minutes
- Time of flow =20 minutes
- Rate of water supply = 280 lpcd
- Impermeability factor =0.5
- Maximum discharge =1.85 * DWF.
Solution:
Assuming `80%` of the supplied water reaches the sewer, the average sanitary discharge is given by
`Q_(DWF)=(0.8**60000**0.280)/(86400)`
`=0.1556\ m^3/s`
The peak sanitary discharge is, `=1.85**0.1556=0.28778\ 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`
`=(0.5**22.67**250)/360`
`=7.8715\ m^3/s`
Therefore, the combined discharge is,
`Q=0.28778+7.8715=8.1593\ m^3/s`
Now, 8.1593=(pi d^2)/4**3.2`
or, `d=1.801\ m`
Adopting commercially available size `d=1.85\ m`
`A=(pid^2)/4=2.688\ m^2`
`V=Q/A=3.035/ m/s` which is less than `3.2\ m/s`.
Check for self Cleansing velocity during dry weather flow:
`q/Q=0.1556/8.1593=0.0191`
or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]^(5/3)`
Solving this, we get,
`theta=72.142^0`
and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.391`
or, `v=0.391**3.035=1.18`
Here 0.60 < 1.18 < 3.2 m/s. Okay
Check for self cleansing velocity during minimum flow.
Let, `Q_(m i n)=1/2 Q_(DWF)`
Thus,
`q/Q=0.00955`
or, `q/Q=(theta)/360[1-(360 sin theta)/(2 pi theta)]`
Solving this, we get,
`theta=60.94^0`
and `v/V=(1-(360 sin theta)/(2 pi theta))^(2/3)=0.32`
or, `v=0.32**3.035=0.97`
Here 0.60 < 0.97 < 3.2 m/s. Okay
Hence the design is okay.
10. Calculate velocity of flow and discharge in a circular sewer having diameter `1` m and laid at a gradient of `1:600` and - running full and
- running half full.
Take N = 0.012.
Solution:
From Mannings formula, we have, the velocity of flow is,
`v=1/n R^(2/3)S^(1/2)`
Where `R=A/P` is the hydraulic radius and `S` is the gradient
Case I: When the sewer is Running Full:
`A=(Pid^2)/4=(pi**1^2)/4=0.785\ m^2`
and `R=A/P=((Pid^2)/4)/(pi d)`
`=d/4=0.25\ m`
Thus, `v=1/0.012**0.25^(2/3)**(1/600)^(1/2)`
`=1.35\ m/s`
And the discharge is,
`Q=Av=0.785**1.35=1.059\ m^3/s`
Again for the sewers running half full,
`A=1/2(Pid^2)/4=1/2(pi**1^2)/4=0.392\ m^2`
and `R=A/P=((Pid^2)/8)/((pi d)/2)`
`=d/4=0.25\ m`
Thus, `v=1/0.012**0.25^(2/3)**(1/600)^(1/2)`
`=1.35\ m/s`
And the discharge is,
`Q=Av=0.392**1.35=0.529\ m^3/s`
