Question

A triangular duct flow partly as shown in the figure. If the critical depth is 0.6 m, the critical flow rate is $m^3/sec.$

• A. 1.388

Answer 1

The correct option is A 1.388
Let the top width of flow be T.

$\frac{T}{1}=\frac{\frac{\sqrt{3}}{2}-0.6}{\left(\sqrt{3}/2\right)}$

T = 0.307 m

Area of flow,

$A=\frac{1}{2}(0.307+1)\times 0.6=0.3921m^2$

For critical condition ,

$\frac{Q^{2}T}{g{A}^3}=1$

$\frac{Q^2\times 0.307}{9.81\times (0.3932{)}^3}=1$

$Q=1.388m^3/sec$