---
title: "Model liquid viscosity, Froude and Reynolds"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:model-liquid-viscosity-froude-reynolds"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/model-liquid-viscosity-froude-reynolds
site: BCT Engineering
---

# Model liquid viscosity, Froude and Reynolds

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2078 Bhadra, 4 marks
- Video: [Model Studies and Similitude - 4 Example Problems using Reynolds Number](https://www.youtube.com/watch?v=9YUzt53kBsE) — Brian Bernard
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/model-liquid-viscosity-froude-reynolds
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

Oil of kinematic viscosity $4.645\times10^{-5}$ m²/s is to be used in a prototype in which both viscous and gravity force dominate. A model scale of 1:5 is also desired. What viscosity of model liquid is necessary to make both the Froude number and the Reynolds number same in model and prototype?

## Short answer

For equal Froude and Reynolds numbers, Vr = √Lr, so νr = Lr3/2 = (1/5)1.5 = 0.0894. The required model kinematic viscosity is νm = 4.645×10−5 × 0.0894 ≈ 4.16×10−6 m²/s.

## Answer

If both gravity and viscous forces are to be modelled, the Froude and Reynolds numbers must both be equal in model and prototype. The liquid in the model must then have a different viscosity.

#### Froude number equal

$$
\frac{V_m}{\sqrt{g L_m}} = \frac{V_p}{\sqrt{g L_p}}
\;\Rightarrow\; V_r = \sqrt{L_r}
$$

(the same $g$ for both).

#### Reynolds number equal

$$
\frac{V_mL_m}{\nu_m} = \frac{V_pL_p}{\nu_p}
\;\Rightarrow\; \nu_r = V_rL_r
$$

#### Combine

$$
\nu_r = L_r^{1/2}L_r = L_r^{3/2} = \left(\frac15\right)^{1.5} = 0.0894
$$

$$
\nu_m = \nu_p\times 0.0894 = 4.645\times10^{-5}\times0.0894 = 4.155\times10^{-6}\ \text{m}^2/\text{s}
$$

**Answer: the model liquid must have a kinematic viscosity of $\nu_m \approx 4.15\times10^{-6}$ m²/s** (about 0.0415 stokes). The dynamic viscosity then follows as $\mu_m = \rho_m\nu_m$ for the chosen liquid; if its density equals that of the oil, $\mu_m = 0.0894\,\mu_p$.

## Check yourself

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

1. Equal Froude number with the same g gives which velocity ratio?
   A. V_r = L_r
   B. V_r = 1/L_r
   C. V_r = sqrt(L_r)
   D. V_r = L_r^2
   Answer: C. V/sqrt(gL) equal and same g gives V_r = sqrt(L_r). (Key point: Froude condition)

2. Equal Reynolds number gives which viscosity ratio?
   A. nu_r = V_r/L_r
   B. nu_r = L_r/V_r
   C. nu_r = V_r^2
   D. nu_r = V_r L_r
   Answer: D. VL/nu equal gives nu_r = V_r L_r. (Key point: Reynolds condition)

3. Combining both conditions gives nu_r equal to what?
   A. L_r^(3/2)
   B. L_r^(1/2)
   C. L_r^2
   D. L_r^(5/2)
   Answer: A. nu_r = L_r^(1/2) x L_r = L_r^(3/2). (Key point: Combined result)

4. With L_r = 1/5 and nu_p = 4.645e-5 m^2/s, what is nu_m?
   A. About 9.3e-6 m^2/s
   B. About 4.15e-6 m^2/s
   C. About 4.15e-4 m^2/s
   D. About 1.04e-5 m^2/s
   Answer: B. (1/5)^1.5 = 0.0894, and 4.645e-5 x 0.0894 = 4.155e-6 m^2/s. (Key point: Model viscosity)

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