If the depth of partial flow in a sewer of diameter 200 cm is 1/4^th of running full then its wetted perimeter is

Solution:

For partial flow in circular sewer the relationship between depth of flow and diameter of sewer is given by \(\frac{d}{D}=\frac{1}{2}\left ( 1-cos\frac{\alpha }{2} \right ) \) 

Given depth of flow = 1/4th of running full ⇒ d = D/4 ⇒\(\frac{1}{4}=\frac{1}{2}\left ( 1-cos\frac{\alpha }{2} \right ) \)

⇒ \(\alpha = 120\)° 

Therefore wetted perimeter \(p= \frac{\pi D}{360^{\circ}}\times 120^{\circ} = \frac{\pi\times D}{3}\)

Additional Information

\(\frac{{\rm{q}}}{{\rm{Q}}} = \left( {\frac{ \propto }{{360^\circ }} - \frac{{\sin \propto }}{{2{\rm{\pi }}}}} \right){\left( {\frac{{\frac{ \propto }{{360^\circ }} - \frac{{\sin \propto }}{{2{\rm{\pi }}}}}}{{ \propto /{\rm{\;}}360^\circ }}} \right)^{2/3}}\)\({\rm{q}} = \frac{1}{{\rm{n}}} \times {\rm{a\;}}{{\rm{r}}^{2/3}}{\rm{\;}}{{\rm{s}}^{1/2}}{\rm{\;and\;r}} = \frac{{\rm{a}}}{{\rm{p}}}\)