written 6.3 years ago by
teamques10
★ 68k
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modified 6.3 years ago
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Given:
$n=0.013$
$d=30cm$
$s=\frac{1}{200}$
To find:
1. Velocity self cleansing:-
$Q=\frac{1}{n}AR^\frac{2}{3}s\frac{1}{2}=\frac{1}{0.013}*(\frac{π }{4}*(0.3)^2)*(\frac{1}{200})^\frac{1}{2}$
$=76.92*(0.071)*0.176*0.071=0.067cumsec $
Velocity=$\frac{Q}{A}=\frac{0.067}{\frac{π }{4}*(0.3)^2}=\frac{0.067}{0.071}=0.94m/sec$
$assume \frac{q}{Q}=0.333$
From given table:-
$\frac{v}{V}=0.898$
$v=0.898*v$
$=0.898*0.94$
$=0.844m/sec \gt 0.45 m/sec$
This is greater than self cleansing of 0.45m/sec
2. Velocity and discharge when the flawing with 0.2 and 0.8 of its full length:-
for, $\frac{d}{D}=0.2$
$\frac{V}{v}=0.615$
$V=0.615*0.94=0.5781m/sec=0.58m/sec$
similarly, for $\frac{d}{D}=0.2$
$\frac{q}{Q}=0.088$
$q=0.088*0.067$
$q=0.0059 cumecs$
$\frac{d}{D}=0.8$
$\frac{v}{V}=1.140$
$V=1.140*0.94$
$=1.0716m/sec$
$V=1.072m/sec$
$\frac{q}{Q}=0.9781$
$q=0.9781*0.067$
$q=0.066 cumecs$