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Chapter:
1. 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`
2. 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`
3. 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`
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. 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. 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`
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. 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`
9. Design a sewer to serve a population of 36,000. The water supply rate is 135 lpcd of which 80 % finds its way into sewer. The slope of sewer 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. Take N = 0.012
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`
10. 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.