Question

For a hydraulic jump formed in a rectangular horizontal channel, the sequent depth ratio is 2. The Froude number of supercritical stream is

• A. \(\sqrt{3}\)
• B. \(\sqrt{5}\)
• C. \(\sqrt{6}\)
• D. \(\sqrt{8}\)

Hint:

Use the Belanger sequent-depth equation \(y_2/y_1=\frac{1}{2}(\sqrt{1+8Fr_1^2}-1)\) and substitute the given ratio of 2 to solve for \(Fr_1\).

Answer 1

The Correct Option is A

Step 1: Start from the momentum (specific force) balance across the jump, instead of quoting the ratio formula directly.
For a short hydraulic jump in a rectangular channel, friction on the bed is neglected, and the specific force per unit width, $M(y) = \frac{q^2}{gy} + \frac{y^2}{2}$, is the same just before and just after the jump, where $q$ is the discharge per unit width.

Step 2: Write $q$ in terms of the Froude number.
The Froude number is $Fr_1 = \frac{V_1}{\sqrt{gy_1}}$, and $V_1 = q/y_1$, so $Fr_1^2 = \frac{q^2}{gy_1^3}$, giving $q^2 = Fr_1^2\, g\, y_1^3$.

Step 3: Substitute into the specific force balance.
Setting $M(y_1) = M(y_2)$:
\[ \frac{Fr_1^2 g y_1^3}{gy_1} + \frac{y_1^2}{2} = \frac{Fr_1^2 g y_1^3}{gy_2} + \frac{y_2^2}{2} \]
Let $r = y_2/y_1$ (the sequent depth ratio) and substitute $y_2 = ry_1$, then divide every term by $y_1^2$:
\[ Fr_1^2 + \frac{1}{2} = \frac{Fr_1^2}{r} + \frac{r^2}{2} \]

Step 4: Solve this for $Fr_1^2$ in terms of $r$.
Rearrange: $Fr_1^2\left(1-\frac{1}{r}\right) = \frac{r^2-1}{2}$, so $Fr_1^2\cdot\frac{r-1}{r} = \frac{(r-1)(r+1)}{2}$.
Dividing both sides by $(r-1)$ (valid since $r\ne1$ for a real jump):
\[ Fr_1^2 = \frac{r(r+1)}{2} \]
This is an equivalent, directly usable form of the same sequent-depth relation, derived here from the momentum principle instead of just being recalled as a memorized ratio formula.

Step 5: Substitute the given ratio $r=2$.
\[ Fr_1^2 = \frac{2(2+1)}{2} = \frac{6}{2} = 3 \]
\[ Fr_1 = \sqrt3 \]
This matches option (A), and confirms the same value found by the ratio-formula method, but by deriving it directly from momentum conservation across the jump.

Final Answer:
\[ Fr_1 = \sqrt3 \]