Hydraulic jump Numerical

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written 3.0 years ago by

RakeshBhuse • 3.2k

Solution :

Given :

Discharge, $Q=7.5~m^3/s$

Wide, $b=3~m$

Velocity, $V=5~m^2/s$

Discharge per unit width

$q=\frac{Q}{b}=\frac{7.5}{3}=2.5 {~m}^{2} /{s}$

$q=\dfrac Q b =\frac{V × A}{b} =\frac{V×d_1×b}{b}$

$\implies d_1=\dfrac q V=\dfrac {2.5} 5= 0.5 ~m$

Hydraulic jump will occur if the depth of flow on upstream side is less than the critical depth on upstream side.

Critical depth ( $h_C$ )

$h_C=\left(\frac{q^{2}}{g}\right)^{\dfrac 1 3}=\left(\frac{2.5^{2}}{9.81}\right)^{\dfrac 1 3}=0.8605$

Now depth on upstream side is 0.5 m. This depth less than the critical depth and hence hydraulic jump will occur.

Height of hydraulic jump (H)

$\begin {aligned} d_{2}&=-\frac{d_{1}}{2}+\sqrt{\frac{d_{1}^{2}}{4}+\frac{2 q^{2}}{g d_{1}}}\\ &=-\frac{0.5}{2}+\sqrt{\frac{0.5^{2}}{4}+\frac{2 ×2.5^{2}}{9.81× 0.5}}\\ &=1.3658~m \end{aligned}$

$H=d_2-d_1=1.3658-0.5=0.8658 ~m$

$\frac{d_2}{d_1}=\frac{1.3658}{0.5}=2.7316$

Length of hydraulic jumps (L)

The length of hydraulic jump is experimentally found to be 5 to 7 times the height of the hydraulic jump.

It range of 4.329 m to 6.0606 m.

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