If the energy dissipated in the model is$\frac{1}{10}kW$, What corresponding value in prototype?

Solution:

Given:

1) Liner scale ratio $L_r=20$

2) Height of hydraulic jump in mode = $h_m=0.20 m$

3) Energy dissipated in model , $P_m=\frac{1}{10}kW$

Step No (1)

$\frac{h_p}{h_m}=L_r$

=20

$h_p=h_m$ $\times$20

=0.20$\times$20

[$h_p=4m$]

Step No (2) Energy dissipated in prototype = $F_p$

$\frac{P_p}{P_m}=L_r^{3.5}$

=$20^{3.5}$

=35777.088

$P_p=P_m\times$335777.088

$\frac{1}{10}\times $ 35777.088

[$P_p=3577.708Kw$]

In a 1 in 75 model of a styling basin the height of the hydraulic jump in the model is observed to be 0.28 meters. What is the height of the hydraulic jump in the prototype. If the energy dissipated in the model is 0.068 kW what is the corresponding value in the prototype. Use Froud’s model law.