Question

A hydraulic jump occurs in a $1.0$ m wide horizontal, frictionless, rectangular channel, with a pre-jump depth of $0.2$ m and a post-jump depth of $1.0$ m. Take $g=10\ \text{m/s^2$. The values of the specific force at the pre-jump and post-jump sections are the same and are equal to (in m$^3$, rounded off to two decimal places) \underline{\hspace{2cm}}.}

Hint:

For a hydraulic jump in a rectangular channel, the conserved specific force can be written directly as \( M=\dfrac{y_1^2+y_1y_2+y_2^2}{2}\;(\text{per unit width}), \) once the sequent depths \(y_1,y_2\) are known.

Answer 1

For a rectangular channel of width $b$ the specific force (momentum function) is
\[ M \;=\; \frac{Q^2}{gA} \;+\; \frac{A^2}{2T} \;=\; \frac{Q^2}{g\,b\,y} \;+\; \frac{b\,y^2}{2}, \] where $A=by$, $T=b$, $y=$ depth. Here $b=1$ m.
Let $y_1=0.2$ m (pre-jump) and $y_2=1.0$ m (post-jump). Equating $M_1=M_2$ gives
\[ \frac{Q^2}{g\,y_1}+\frac{y_1^2}{2} \;=\; \frac{Q^2}{g\,y_2}+\frac{y_2^2}{2} \;\Rightarrow\; \frac{Q^2}{g} = \frac{y_1y_2(y_1+y_2)}{2}. \] Hence the common specific force (use either side) is
\[ M = \frac{Q^2}{g\,y_1}+\frac{y_1^2}{2} = \frac{y_2(y_1+y_2)}{2}+\frac{y_1^2}{2} = \frac{y_1^2+y_1y_2+y_2^2}{2}. \] Substituting $y_1=0.2$, $y_2=1.0$:
\[ M=\frac{0.2^2+0.2\times1.0+1.0^2}{2} =\frac{0.04+0.20+1.00}{2} =\frac{1.24}{2} =0.62\ \text{m}^3. \] \[ \boxed{M \approx 0.62\ \text{m}^3} \]