(a) Explain why distorted scale models of rivers are commonly used.
(b) A river model is constructed to a vertical scale of 1:50 and a horizontal scale of 1:200.
The model is to be used to investigate a flood alleviation scheme. At the design flood of 450 m³/s, the average width and depth of flow are 60 and 4.2 m, respectively. Determine the corresponding discharge in the model and check the Reynolds number of the model flow.

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Accepted Answer

(a) The size of a river model is determined by the laboratory space available (although in some cases special buildings are constructed to house a particular model, notably the Eastern Schelde model in the Delft Hydraulics Laboratory, the Netherlands). In the case of a long river reach, a natural model scale may result in such small flow depths that depth and water elevations cannot be measured with sufficient accuracy, the flow in the model may become laminar, surface tension effects may become significant and sediment studies may be precluded because of the low tractive force.
To avoid these problems, geometrical distortion wherein the vertical scale [latex]\lambda_{y}[/latex] is larger than the horizontal scale [latex]\lambda_{x}[/latex] is used, typical vertical distortions being in the range of 5–100 with a vertical scale not smaller than 1:100.

(b) Velocity in prototype = [latex]\frac{450}{60 imes 4.2}[/latex] = 1.786 m/s
The Froude scaling law is based on the vertical scale ratio

[latex]V_{m} = V_{p}\sqrt{ \lambda_{y} } = 1.786 \sqrt{\frac{1}{50}}[/latex] = 0.25 m/s

[latex]\frac{Q_{m} }{Q_{p}} = \frac{V_{m}}{V_{p}} \frac{(xy)_{m}}{(by)} = \lambda ^{3∕2}_{y} \lambda_{x}[/latex]

[latex]=\left(\frac{1}{50} ight)^{3∕2} imes \frac{1}{200} = 1.414 imes 10^{−5}[/latex]

[latex]\Rightarrow Q_{m} = 450 imes 1.414 imes 10^{-5}[/latex] m³ ∕s
= 6.36 L/s

Average dimensions of model channel:

Width = [latex]\frac{60}{200} [/latex] = 0.30 m

Depth = [latex]\frac{4.2}{50} [/latex] = 0.084 m

Hydraulic radius [latex]\simeq \frac{0.3 × 0.084}{0.468} [/latex] = 0.0538

Reynolds number = [latex]\frac{VR}{ u } =\frac{0.25 imes 0.0538}{1 imes 10^{−6} } [/latex] = 13 450

The flow therefore will be turbulent.

Question 9.8: (a) Explain why distorted scale models of rivers are commonl...

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