Question

A hydraulic jump occurs in a rectangular, horizontal, frictionless channel. What would be the pre-jump depth if the discharge per unit width is $2m^{3}s/m$ and the energy loss is 1 m?

• A. 0.9 m
• B. 0.3 m
• C. 0.2 m
• D. 0.8 m

Answer 1

The correct option is B 0.3 m

$\frac{2q^2}{g}=y_1{y}_2(y_1+y_2)$ ....(i)
$H_L=\frac{(Y_2-Y_1{)}^3}{4Y_1{Y}_2}$ ....(ii)

Given,

$H_L=1m$

$(Y_2-Y_1{)}^3=4Y_1{Y}_2$

$Put,Y_2=m{y}_1$ in equation (ii),

$Y_1=\frac{4m}{(m-1{)}^3}$

Similarly, put
$y_2=m{y}_1$ in equation (i),

we get

$m(m+1)y_1^3=\frac{2\times 2^2}{9.81}=0.8155$

$m(m+1){\left\{\frac{4m}{(m-1{)}^3}\right\}}^3=0.8155$

$\frac{m(m+1)\left(m^3\right)}{(m-1{)}^9}=\frac{0.8155}{4^3}$

by trial and error, m = 5.09

$y_1=0.3m$

$y_2=1.527m$