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

# Rayleigh's method for orifice discharge

Fluid Mechanics · Chapter 10: Similitude and physical modeling

- Asked in: 2075 Chaitra, 6 marks
- Video: [Rayleigh's Method Problem Example](https://www.youtube.com/watch?v=k_Tsm3IaJ-I) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/orifice-discharge-rayleigh-method
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-10

## Question

In a flow through a small orifice discharging freely into atmosphere under a constant head (H), the flow discharge (Q) depends on diameter of pipe (d), constant head, dynamic viscosity ($\mu$), density of fluid ($\rho$) and acceleration due to gravity (g). Using Rayleigh's methods develop the relation in terms of non-dimensional terms.

## Short answer

Q = √g d⁵/² φ(H/d, μ/(ρ√g d³/²)). The dimensionless groups are head ratio and a Reynolds-type number; standard form is Q = C_d A√(2gH).

## Answer

#### Relation

$Q$ depends on $d, H, \mu, \rho, g$. By Rayleigh's method:

$$
Q = K\,d^{a}H^{b}\mu^{c}\rho^{e}g^{f}
$$

#### Dimensions

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

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

There are 5 unknowns and 3 equations. Express in terms of $b$ and $c$:

$$
e = -c,\qquad f = \frac{1-c}{2},\qquad a = \frac52 - \frac{3c}{2} - b
$$

#### Substitute

$$
Q = K\,d^{\,\frac52-\frac{3c}{2}-b}\,H^{b}\,\mu^{c}\,\rho^{-c}\,g^{\frac{1-c}{2}}
$$

Collect terms with the same powers:

$$
Q = K\,\sqrt g\,d^{5/2}\left(\frac Hd\right)^{b}\left(\frac{\mu}{\rho\,g^{1/2}d^{3/2}}\right)^{c}
$$

**Answer:**

$$
Q = \sqrt{g}\,d^{5/2}\ \phi\left[\frac Hd,\ \frac{\mu}{\rho\sqrt g\,d^{3/2}}\right]
$$

Equivalently, $Q = d^2\sqrt{gH}\ \phi_1\!\left(H/d,\ Re\right)$. The dimensionless groups are the head ratio $H/d$ and a Reynolds-type number. For a given orifice, experiments give $Q = C_dA\sqrt{2gH}$.

## Check yourself

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

1. How is the relation set up?
   A. Q = pi terms with repeating variables
   B. Q = Cd A sqrt(2gH) directly
   C. Q = K d^a H^b mu^c rho^e g^f
   D. Q = K + d + H + mu
   Answer: C. Rayleigh's method uses a product of powers. (Key point: Rayleigh form)

2. Why are powers expressed in terms of b and c?
   A. K is unknown
   B. H and d are equal
   C. Q has no dimensions
   D. 5 unknowns but only 3 equations
   Answer: D. M, L and T give three equations for five unknown powers. (Key point: Why free powers)

3. What is the final non-dimensional relation?
   A. Q = sqrt(g) d^(5/2) phi[H/d, mu/(rho sqrt(g) d^(3/2))]
   B. Q = sqrt(g) d phi[H, mu]
   C. Q = d^2 phi[rho g]
   D. Q = sqrt(H) mu phi[d]
   Answer: A. The groups are the head ratio H/d and a Reynolds-type term. (Key point: Result)

4. What does dimension M give?
   A. 0 = c - e
   B. 0 = c + e
   C. 1 = c + e
   D. 0 = c + f
   Answer: B. Only mu and rho contain M, so c + e = 0. (Key point: Mass equation)

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