Question

A river model is constructed to a horizontal scale of 1 : 1000 and a vertical scale of 1 : 100. A model discharge of 0.1 $m^3$/s would correspond to a discharge in the prototype, of what magnitude?

• A. ${10}^2{m}^3/s$
• B. ${10}^3{m}^3/s$
• C. ${10}^4{m}^3/s$
• D. ${10}^5{m}^3/s$

Answer 1

The correct option is D ${10}^5{m}^3/s$

$Horizontalscale,L_r=\frac{L_m}{L_p}=\frac{1}{1000}$
$Verticalscale,h_r=\frac{y_m}{L_p}=\frac{1}{100}$
By Froude model law, we have
$\frac{V_m}{\sqrt{g{y}_m}}=\frac{V_p}{\sqrt{g{y}_p}}\Rightarrow \frac{V_m}{V_p}=\sqrt{\frac{y_m}{y_p}}\Rightarrow \frac{V_m}{V_p}=\sqrt{h_r}\Rightarrow V_r=\sqrt{h_r}\therefore Dischargeratio,Q_r=\frac{Q_m}{Q_p}=A_r\times V_r\Rightarrow \frac{Q_m}{Q_p}=L_r{h}_r\times \sqrt{h_r}\Rightarrow \frac{Q_m}{Q_p}=L_r{h}_r^{3/2}\Rightarrow \frac{0.1}{Q_p}=\frac{1}{1000}\times {\left(\frac{1}{100}\right)}^{3/2}\Rightarrow Q_p=0.1\times 1000\times 1000\Rightarrow Q_p={10}^5{m}^3/s$