Derive the expression for the depth of Hydraulic Jump in case of Upstream Froudes number.

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written 3.0 years ago by

RakeshBhuse • 3.2k

Let

$V_1=$ Velocity of flow on the upstream side

$d=$ Depth of flow on upstream side

Then Froude Number $\left(F_{e}\right)_{1}$ on the upstream side of the jump is given by

$\left(F_{e}\right)_{1}=\frac{V_{1}}{\sqrt{g d_{1}}} .....(i)$

Depth of flow after the hydraulic jump is $d_{2}$ given by,

$\begin{aligned} d_{2} &=-\frac{d_{1}}{2}+\sqrt{\frac{d_{1}^{2}}{4}+\frac{2 V_{1}^{2} d_{1}}{g}}\\ &=-\frac{d_{1}}{2}+\sqrt{\frac{d_{1}^{2}}{4}\left(1+\frac{8 V_{1}^{2}}{g d_{1}}\right)} \\ &=-\frac{d_{1}}{2}+\frac{d_{1}}{2} \sqrt{1+\frac{8 V_{1}^{2}}{g d_{1}}} .....(ii) \end{aligned}$

But from equation (i),

$\left(F_{e}\right)_{1}=\frac{V_{1}}{\sqrt{g d_{1}}}\\ \left(F_{e}\right)_{1}^{2}=\frac{V_{1}^{2}}{g d_{1}}$

Substituting this value in equation (ii), we get

$\begin {aligned} d_{2} &=-\frac{d_{1}}{2}+\frac{d_{1}}{2} \sqrt{1+8\left(F_{e}\right)_{1}^{2}}\\ &=\frac{d_{1}}{2} \left(\sqrt {1+8 \left(F_ {e}\right)_{1}^{2}}-1\right) \end {aligned}$

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