Civil Engineering - Waste Water Engineering - Discussion
Discussion Forum : Waste Water Engineering - Section 6 (Q.No. 37)
37.
If the discharge of a sewer running half is 628 1.p.s., i = 0.001, and n = 0.010, the diameter of the sewer, is
Discussion:
10 comments Page 1 of 1.
Yorum Guna said:
3 years ago
Since running half full then, d/D=.5.
q/Q=.5 ( hence, Q = 1.256 m3/s)
Use, Manning's equation.
Q= 1/n * A*(D/4)power 2/3 * (s)power 1/2
Here, S = i = 0.0001 not 0.001.
Solve it, the answer comes approx to 1.66 m.
q/Q=.5 ( hence, Q = 1.256 m3/s)
Use, Manning's equation.
Q= 1/n * A*(D/4)power 2/3 * (s)power 1/2
Here, S = i = 0.0001 not 0.001.
Solve it, the answer comes approx to 1.66 m.
(1)
Hridoy Hasan said:
3 years ago
Q/A=1/n*(R^(2/3))*(i^(1/2)).
Q/((π*d^2)/8) = 1/n*((d/4)^(2/3))*(i^(1/2).
Here,
Q = 0.628 m^3/s;
n=0.01; i= 0.001.
By solving.
d=1.0952m.
Q/((π*d^2)/8) = 1/n*((d/4)^(2/3))*(i^(1/2).
Here,
Q = 0.628 m^3/s;
n=0.01; i= 0.001.
By solving.
d=1.0952m.
(3)
Anirban Roy said:
5 years ago
Q=av; use a=π*d^2÷8 get the results.
SHAHZAD IQBAL KAZMI said:
5 years ago
Q = (0.3116/n) * (dia)^8/3 * (i)^1/2.
dia is taken as d/2 because running half.
dia is taken as d/2 because running half.
Raj said:
7 years ago
Thank you @Orai.
Orai said:
7 years ago
Solve by Manning's equation:
V=1/n*R*(2/3)*i*(1/2),
Q/A=1/n*R*(2/3)*i*(1/2),
Q/(&radicD*2/4)=1/n*(D/4)*(2/3)*i*(1/2).
Sewer running half;
So put D=d/2,
and put other given data,
Finally d=1.688 m.
V=1/n*R*(2/3)*i*(1/2),
Q/A=1/n*R*(2/3)*i*(1/2),
Q/(&radicD*2/4)=1/n*(D/4)*(2/3)*i*(1/2).
Sewer running half;
So put D=d/2,
and put other given data,
Finally d=1.688 m.
(1)
Anii said:
7 years ago
It's 628 litres per sec (lps).
Zahir said:
8 years ago
Please anyone explain the unit of Discharge as above.
Susant said:
8 years ago
Anyone can explain it?
Gayatri said:
8 years ago
I didn't get it. Please explain anyone.
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