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Design-of-sewers-Tutorial
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`
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