---
title: "Rayleigh's method, sphere terminal velocity"
subject: "Fluid Mechanics"
chapter: 10
key: "semester_3/fluid_mechanics:rayleigh-sphere-terminal-velocity"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/rayleigh-sphere-terminal-velocity
site: BCT Engineering
---

# Rayleigh's method, sphere terminal velocity

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2079 Baisakh, 4 marks
- Video: [Fluid Mechanics: Dimensional Analysis: Rayleigh Method](https://www.youtube.com/watch?v=Ch6Iqog9gm8) — Study from Home with Us
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/rayleigh-sphere-terminal-velocity
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

Sphere of diameter d and density $\rho_s$ settles at a terminal velocity V in a liquid of density $\rho_l$ and dynamic viscosity $\mu$. Determine an expression of velocity in which velocity also depends on acceleration due to gravity g. Use Rayleigh's Method.

## Short answer

Using Rayleigh's method, V = K da ρsf ρlb μc ge. Dimensional analysis yields V/√(gd) = φ( μ/(ρl d3/2 g1/2), ρs/ρl ), where the function φ is found from experiments.

## Answer

Rayleigh's method writes the dependent variable as a product of the independent variables, each raised to an unknown power. The powers are found by equating the dimensions on both sides.

#### Functional relation

$$
V = K\, d^{a}\,\rho_s^{\,f}\,\rho_l^{\,b}\,\mu^{c}\,g^{e}
$$

#### Dimensions

$V = LT^{-1}$, $d = L$, $\rho = ML^{-3}$, $\mu = ML^{-1}T^{-1}$, $g = LT^{-2}$.

| Dimension | Equation |
|---|---|
| M | $0 = f + b + c$ |
| L | $1 = a - 3f - 3b - c + e$ |
| T | $-1 = -c - 2e$ |

There are 5 unknown powers and only 3 equations, so express three of them in terms of $c$ and $f$:

$$
e = \frac{1-c}{2},\qquad b = -c - f,\qquad a = \frac12 - \frac{3c}{2}
$$

#### Substitute

$$
V = K\, d^{\,\frac12-\frac{3c}{2}}\,\rho_s^{\,f}\,\rho_l^{\,-c-f}\,\mu^{c}\,g^{\,\frac{1-c}{2}}
$$

Group terms with the same power:

$$
V = K\sqrt{gd}\left[\frac{\mu}{\rho_l\, d^{3/2}g^{1/2}}\right]^{c}\left[\frac{\rho_s}{\rho_l}\right]^{f}
$$

**Answer:**

$$
\frac{V}{\sqrt{gd}} = \phi\left[\frac{\mu}{\rho_l\,d^{3/2}g^{1/2}},\ \frac{\rho_s}{\rho_l}\right]
$$

This says the dimensionless velocity (Froude number) depends on a viscous term (a form of inverse Reynolds number) and on the density ratio. The functions of these groups are found by experiment.

## Check yourself

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

1. What does Rayleigh's method do?
   A. Forms n - m pi terms using repeating variables
   B. Uses the Froude number only
   C. Writes the dependent variable as a product of the independent variables raised to unknown powers and equates dimensions
   D. Takes ratios of forces
   Answer: C. Rayleigh's method uses a power-product form and equates exponents of M, L and T. (Key point: Rayleigh's method)

2. Why are the powers left in terms of c and f?
   A. The constant K is unknown
   B. The viscosity is not measurable
   C. Dimension T gives no equation
   D. There are 5 unknown powers but only 3 dimensional equations
   Answer: D. Only M, L and T give equations, so two powers stay free. (Key point: Why powers left unknown)

3. Which equation comes from dimension T?
   A. -1 = -c - 2e
   B. 0 = f + b + c
   C. 1 = a - 3f - 3b - c + e
   D. -1 = -c + 2e
   Answer: A. mu has T^-1 and g has T^-2, V has T^-1, giving -1 = -c - 2e. (Key point: Time equation)

4. What is the final dimensionless form?
   A. V/sqrt(gd) = phi[rho_s d V/mu]
   B. V/sqrt(gd) = phi[mu/(rho_l d^(3/2) g^(1/2)), rho_s/rho_l]
   C. V = K d g only
   D. V/(g d) = phi[rho_s x rho_l]
   Answer: B. The groups are the Froude-type velocity, a viscous term and the density ratio. (Key point: Final result)

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