Question 12.7: A rectangular channel carries 5.66 m^3/s. Find the critical ...

(i) Critical depth is defined as the depth at which the flow velocity is given by its critical value as

Again,        V_{c}=Q / b h_{c}=5.66 / 3.66 h_{c} 

Therefore, 5.66 / 3.66 h_{c}=\left(g h_{c}\right)^{1 / 2}

Or     h_{c}=\left[\frac{5.66 \times 5.66}{3.66 \times 3.66 \times 9.81}\right]^{1 / 3}=0.625 \mathrm{~m} 

Now, V_{c}=(9.81 \times 0.625)^{1 / 2}=2.48 \mathrm{~m} / \mathrm{s}

(ii) When the width is 2.74 m

And, V_{c}=(9.81 \times 0.758)^{1 / 2}=2.73 \mathrm{~m} / \mathrm{s}

(iii) Applying Eq. (12.9), we can write

V=(1 / n) R_{h}^{2 / 3} S^{1 / 2}                     (12.9)

where R_{c}  is the hydraulic radius at the critical flow and is given by

Hence,       2.48=\frac{(0.466)^{2 / 3}}{0.02} S^{1 / 2} 

which gives                 S = 0.0068