written 2.7 years ago by |
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
$$\begin {aligned} d_{2} &=-\frac{d_{1}}{2}+\frac{d_{1}}{2} \sqrt{1+8\left(F_{e}\right)_{1}^{2}}\\ &=\frac{d_{1}}{2} \left(\sqrt {1+8 \left(F_ {e}\right)_{1}^{2}}-1\right) \end {aligned}$$