River model depth and discharge scales

River flow is controlled by gravity, so the Froude law is used. The slope scale is $S_r = \dfrac{\text{vertical scale}}{\text{horizontal scale}}$, and here it is to be unity.

Given

Prototype: $Q_p = 3500\ \text{m}^3/\text{s}$, depth $2.25$ m, width $1500$ m. Horizontal scale $L_r = 1/400$. $S_r = 1$.

Depth scale

$$S_r = \frac{d_r}{L_r} = 1 \;\Rightarrow\; d_r = L_r = \frac1{400}$$

So the model is geometrically undistorted. Model depth $= 2.25/400 = 5.625$ mm; model width $= 1500/400 = 3.75$ m.

Velocity and discharge scale (Froude law)

$$V_r = \sqrt{d_r} = \sqrt{1/400} = \frac1{20}$$ $$Q_r = A_rV_r = (L_rd_r)\sqrt{d_r} = L_r\,d_r^{3/2} = \left(\frac1{400}\right)^{5/2} = \frac1{3.2\times10^{6}} = 3.125\times10^{-7}$$

Model discharge

$$Q_m = 3500 \times 3.125\times10^{-7} = 1.094\times10^{-3}\ \text{m}^3/\text{s} \approx 1.09\ \text{L/s}$$

Answer: depth scale $d_r = 1/400$; discharge scale $Q_r = 1/(3.2\times10^6)$; model discharge $\approx 1.09$ L/s. Such a shallow model (5.6 mm deep) would suffer scale effects (surface tension, laminar flow), which is why real river models are usually distorted.