---
title: "Froude model of a spillway"
subject: "Fluid Mechanics"
chapter: 5
key: "semester_4/fluid_mechanics_mechanical:spillway-model-froude-scaling"
source: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/spillway-model-froude-scaling
site: BCT Engineering
---

# Froude model of a spillway

Fluid Mechanics · Chapter 5: Dimensional Analysis and Dynamic Similitude

- Asked in: year not given, 6 marks
- Video: [Dynamic Similitude between Prototype and Model using Froude's Model Law|Solved Problem](https://www.youtube.com/watch?v=OHBU9kLX-t8) — Master Civil Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/fluid_mechanics_mechanical/spillway-model-froude-scaling
- Whole chapter: https://www.bctengineer.com/md/semester_4/fluid_mechanics_mechanical/chapter-5

## Question

A 1:25 scale model of a dam spillway is built and tested using water. The prototype discharge is 1200 m$^3$/s. (a) Find the discharge in the model. (b) Find the velocity ratio and time ratio. (c) If the measured force on a model gate is 40 N, find the force on the prototype gate. (d) If a surge in the model takes 2 minutes to pass, find the time for the prototype. State the criterion of similarity used.

## Short answer

Using Froude scaling, model discharge is 0.384 m³/s, velocity and time ratios are both 5, prototype gate force is 625 kN, and prototype surge time is 10 minutes.

## Answer

#### Criterion of similarity
Flow over a spillway is a free-surface flow dominated by gravity, so the **Froude number** is the same in the model and prototype. The scale ratio $L_r = L_p/L_m = 25$. The same fluid (water) and the same $g$ are used.

#### Ratios from Froude law
$$
\begin{aligned}
V_r &= \sqrt{L_r} = \sqrt{25} = 5 \\
T_r &= \frac{L_r}{V_r} = \sqrt{L_r} = 5 \\
Q_r &= V_r L_r^2 = L_r^{5/2} = 25^{2.5} = 3125 \\
F_r &= \frac{\rho_p}{\rho_m} L_r^3 = L_r^3 = 15\,625
\end{aligned}
$$

#### (a) Model discharge
$$
Q_m = \frac{Q_p}{Q_r} = \frac{1200}{3125} = 0.384\ \text{m}^3/\text{s}
$$

#### (b) Velocity and time ratios
$V_p/V_m = 5$ and $T_p/T_m = 5$.

#### (c) Force on prototype gate
$$
F_p = 40 \times 15\,625 = 625\,000\ \text{N} = 625\ \text{kN}
$$

#### (d) Prototype time
$$
T_p = 5 \times 2 = 10\ \text{minutes}
$$

#### Comment
The Reynolds number is not matched in the model ($Re_r = V_r L_r = 125$), so viscous effects and surface tension are not scaled correctly. For a large model with a turbulent flow this error is small. This is incomplete similarity.

**Answer: $Q_m = 0.384\ \text{m}^3/\text{s}$; $V_r = T_r = 5$; $F_p = 625\ \text{kN}$; $T_p = 10\ \text{min}$ (Froude law)**

## Check yourself

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

1. Which criterion is used for a spillway model and why?
   A. Reynolds, because viscosity dominates
   B. Mach, because of compressibility
   C. Weber, because of surface tension
   D. Froude, because the free-surface flow is dominated by gravity
   Answer: D. Spillway flow is free-surface and gravity-driven, so the Froude number is equated. (Key point: Similarity criterion)

2. With L_r = 25 under Froude law, what is Q_r and the model discharge for Q_p = 1200 m^3/s?
   A. Q_r = 25, Q_m = 48 m^3/s
   B. Q_r = 625, Q_m = 1.92 m^3/s
   C. Q_r = 15625, Q_m = 0.0768 m^3/s
   D. Q_r = 3125, Q_m = 0.384 m^3/s
   Answer: D. Q_r = L_r^(5/2) = 25^2.5 = 3125, so Q_m = 1200/3125 = 0.384 m^3/s. (Key point: Model discharge)

3. The model gate force is 40 N. What is the prototype force and why?
   A. 625 kN, since F_r = L_r^3 = 15625 for the same fluid
   B. 1000 N, since F_r = L_r
   C. 25 kN, since F_r = L_r^2
   D. 200 N, since F_r = sqrt(L_r)
   Answer: A. With same density and g, F_r = L_r^3 = 15625, so F_p = 40 x 15625 = 625 000 N. (Key point: Force scale)

4. A surge takes 2 min in the model. How long in the prototype?
   A. 50 min
   B. 2 min
   C. 10 min
   D. 0.4 min
   Answer: C. T_r = sqrt(L_r) = 5, so T_p = 5 x 2 = 10 min. (Key point: Time scale)

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