Question

A rectangular channel with bottom width 3 m, bottom slope $7\times {10}^{-4}$ carries discharge of 1.80 cumecs. The depth of flow at a certain location is 0.42 m. What is the type of GVF profile? (n=0.012 and hydraulic radius is equal to depth of flow)

• A. $S_2$
• B. $M_3$
• C. $M_2$
• D. $S_1$

Answer 1

The correct option is C $M_2$

Given, $B=3m,S=7\times {10}^{-4},Q=1.80cumecs,y=0.42m$

$Q=B{y}_n\frac{1}{n}{R}^{2/3}{S}^{1/2}$

$1.8=\frac{3y_n}{0.012}\times y_n^{2/3}(7\times {10}^{-4}{)}^{1/2}$

$y_n=0.46m$

$Criticaldepthofflow,Y_c={\left(\frac{q^2}{g}\right)}^{1/3}$

$={\left[\frac{(1.8/3{)}^2}{9.81}\right]}^{1/3}=0.332m$

$Since,y_n>y>y_c,hencetypeofprofile=M_2$