---
title: "River model depth and discharge scales"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:river-model-depth-discharge-scales"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/river-model-depth-discharge-scales
site: BCT Engineering
---

# River model depth and discharge scales

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2076 Asoj, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/river-model-depth-discharge-scales
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

A river carrying a discharge of 3500 m³/s has a depth of 2.25 m width of 1500 m. From the point of view of availability of space the horizontal scale of 1:400 is chosen. Assuming slope scale to be unity, determine the depth and discharge scales for the model.

## Short answer

Froude similarity is used. Depth scale d_r = 1/400, discharge scale Q_r = 1/(3.2×10⁶); model discharge ≈ 1.09 L/s.

## Answer

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which law is used for river flow?
   A. Reynolds law
   B. Weber law
   C. Froude law, since gravity controls it
   D. Euler law only
   Answer: C. Open channel flow with a free surface is gravity-dominated. (Key point: Model law)

2. Slope scale = depth scale / horizontal scale = 1. What is the depth scale?
   A. 1/20
   B. 1/160000
   C. 1
   D. 1/400
   Answer: D. d_r = L_r = 1/400, so the model is undistorted. (Key point: Depth scale)

3. What is the velocity scale?
   A. 1/20, sqrt(d_r)
   B. 1/400
   C. 1/8000
   D. 1/160000
   Answer: A. V_r = sqrt(d_r) = sqrt(1/400) = 1/20. (Key point: Velocity scale)

4. What is Q_r?
   A. (1/400)^2
   B. L_r d_r^(3/2) = (1/400)^(5/2) = 3.125e-7
   C. (1/400)^(1/2)
   D. (1/400)^3
   Answer: B. Q_r = A_r V_r = (L_r d_r) sqrt(d_r), giving the 5/2 power. Model Q = 3500 x 3.125e-7 = about 1.09 L/s. (Key point: Discharge scale)

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