Question

A hydraulic jump takes place in a triangular channel of vertex angle ${90}^o$, as shown in figure. The discharge is 1 $m^3/s$ and the pre-jump depth is 0.5 m. What will be the post-jump depth?(Take g=9.81 m/s${}^2$

• A. 1.57 m
• B. 0.91 m
• C. 1.02 m
• D. 0.57 m

Answer 1

The correct option is C 1.02 m

The specific force will remain constant on both sides of hydraulic jump.

$\therefore \frac{Q^2}{g{A}_1}+A_1\overline{Z_1}=\frac{Q^2}{g{A}_2}+A_2\overline{Z_2}$

$A_1=y_1^2;,A_2=y_2^2$

${\overline{Z}}_1=\left(\frac{y_1}{3}\right);{\overline{Z}}_2=\left(\frac{y_2}{3}\right)$

$GivenQ=1m^3/s;y_1=0.5m$

$\therefore \frac{(1{)}^2}{9.81\times (0.5{)}^2}+(0.5{)}^2\times \left(\frac{0.5}{3}\right)$

$=\frac{(1{)}^2}{9.81\times y_2^2}+(y_2{)}^2\times \left(\frac{y_2}{3}\right)$

$\Rightarrow 0.4494=\frac{0.1019}{y_2^2}+\frac{y_2^3}{3}$

By hit and trial, we get $y_2=1.02m$