Question 9.1: The flow over a spillway is 140 m³/s. The model flow is limi......
The flow over a spillway is 140 m³/s. The model flow is limited to 1.25 m³/s, based on the capacity of the laboratory pumps. What should be the model scale to maintain dynamic similarity? If the force on a certain area of the model is measured as 4 N, what would be the force on the corresponding area of the prototype?
Given, Q_{p} = 140 m³/s; Q_{m} = 1.25 m³/s
This being a gravity driven flow, Froude number should be the same in the model and the prototype.
Thus F_{r}=1\;\;\text{or}\;\left\lgroup{\frac{V}{\sqrt{g L}}}\right\rgroup_{r}=1
Therefore, considering g_{r} = 1, the velocity ratio V_{r}={\sqrt{L_{r}}}
Discharge ratio, Q_{r}=A_{r}\,V_{r}=L_{r}^{2}\,L_{r}^{\frac{1}{2}}=L_{r}^{\frac{5}{2}}
Hence from the given flow rates,
{\frac{1.25}{140}}={L_{r}^{\frac{5}{2}}}\;\;\mathrm{or}\;\;L_{r}=\left\lgroup{\frac{1.25}{140}}\right\rgroup^{2/5}=0.151\;\mathrm{i.e.,}\;1:6.6Equal Froude number in the model and prototype leads to equal Euler number in the two.
Therefore \left\lgroup\frac{\Delta p}{\rho V^{2}}\right\rgroup_{r}=1;\text{ or }\Delta p_{r}=\rho_{r}V_{r}^{2}
Force ratio (F)_{r}=A_{r}\,\Delta p_{r}=L_{r}^{2}\,\rho_{r}\,V_{r}^{2}=L_{r}^{3}\,\rho_{r}
Assuming \rho_r = 1 ,the force ratio = L_r^3 = 0.151³ = 0.00344
Therefore, force in prototype = {\frac{4}{0.00344}}=1163\ \mathrm{N}