Introduction & Context

The hydraulic diameter is a representative length used to characterize flow in ducts whose cross‑section is not circular. It allows engineers to apply correlations and design criteria—such as Reynolds number, pressure drop, and heat‑transfer relationships—that were originally derived for circular pipes. In process engineering, the hydraulic diameter is essential for sizing ventilation ducts, heat‑exchanger channels, reactor manifolds, and any non‑circular conduit where fluid flow performance must be predicted.

Methodology & Formulas

The calculation proceeds by converting practical units to SI, evaluating geometric properties, and then applying fundamental fluid‑mechanics relationships.

Symbol	Description	Units
w	duct width	m
h	duct height	m
Q	volumetric flow rate	m³·s⁻¹
μ	dynamic viscosity	Pa·s
ρ	fluid density	kg·m⁻³
A	cross‑sectional area	m²
P_w	wetted perimeter	m
D_h	hydraulic diameter	m
V̅	average flow velocity	m·s⁻¹
ν	kinematic viscosity	m²·s⁻¹
Re	Reynolds number (dimensionless)	–

Geometric calculations

\[ A = w \times h \] \[ P_w = 2\,(w + h) \]

Hydraulic diameter

\[ D_h = \frac{4\,A}{P_w} \]

Average velocity

\[ \overline{V} = \frac{Q}{A} \]

Kinematic viscosity

\[ \nu = \frac{\mu}{\rho} \]

Reynolds number

\[ Re = \frac{\overline{V}\,D_h}{\nu} \]

Flow‑Regime Classification (based on Reynolds number)

Regime	Re range
Laminar	Re < 2 300
Transitional	2 300 ≤ Re ≤ 4 000
Turbulent	Re > 4 000

By following the steps above, the hydraulic diameter, average velocity, kinematic viscosity, and Reynolds number can be obtained for any rectangular duct. These results feed directly into pressure‑drop calculations, heat‑transfer coefficient estimations, and equipment‑sizing decisions in process‑industry applications.

The hydraulic diameter (D_h) is a characteristic length used to relate non‑circular flow passages to an equivalent circular pipe. It is defined as:

D_h = 4·A / P

where A is the cross‑sectional flow area and P is the wetted perimeter (the perimeter in contact with the fluid). This definition allows engineers to apply correlations developed for circular pipes to ducts of arbitrary shape.

For a rectangle:

A = w·h
P = 2·(w + h)
D_h = 4·A / P = 4·w·h / [2·(w + h)] = 2·w·h / (w + h)

This expression reduces to the height for a very wide duct (w » h) and to the width for a very tall duct (h » w).

For an annulus with inner diameter D_i and outer diameter D_o:

A = (π/4)·(D_o² – D_i²)
P = π·(D_o + D_i)
D_h = 4·A / P = (D_o² – D_i²) / (D_o + D_i) = D_o – D_i

Thus the hydraulic diameter of an annular gap equals the radial clearance between the two pipes.

Hydraulic diameter is appropriate when:

The cross‑section is constant along the flow direction.
The flow is fully developed and the duct walls are smooth enough that roughness effects are captured by standard correlations.
The shape does not introduce strong secondary flows (e.g., sharp corners) that would invalidate circular‑pipe assumptions.

If any of these conditions are violated, consider using shape‑specific correlations or CFD for more accurate results.

Worked Example – Hydraulic Diameter for a Rectangular Duct

A process engineer at a chemical plant must size a rectangular ventilation duct that carries a coolant flow. The engineer needs the hydraulic diameter to assess the Reynolds number and determine whether the flow will be laminar or turbulent.

Knowns

Width of duct, W = 150 mm = 0.15 m
Height of duct, H = 30 mm = 0.03 m
Volumetric flow rate, Q = 12 L s⁻¹ = 0.012 m³ s⁻¹
Dynamic viscosity, μ = 150 cP = 0.15 Pa·s
Fluid density, ρ = 1200 kg m⁻³

Step‑by‑Step Calculation

Calculate the cross‑sectional area: \[ A = W \times H = 0.15 \times 0.03 = 0.0045\ \text{m}^2 \]
Determine the wetted perimeter: \[ P = 2\,(W + H) = 2\,(0.15 + 0.03) = 0.36\ \text{m} \]
Compute the hydraulic diameter: \[ D_h = \frac{4A}{P} = \frac{4 \times 0.0045}{0.36} = 0.050\ \text{m} \;(= 50\ \text{mm}) \]
Find the average fluid velocity: \[ V_{\text{avg}} = \frac{Q}{A} = \frac{0.012}{0.0045} = 2.667\ \text{m s}^{-1} \]
Calculate the kinematic viscosity: \[ \nu = \frac{\mu}{\rho} = \frac{0.15}{1200} = 0.000125\ \text{m}^2\text{s}^{-1} \]
Evaluate the Reynolds number: \[ Re = \frac{V_{\text{avg}}\,D_h}{\nu} = \frac{2.667 \times 0.050}{0.000125} = 1{,}066.667 \]

Final Answer

The hydraulic diameter of the rectangular duct is 0.050 m (50 mm). Using this value, the Reynolds number for the given flow conditions is ≈ 1.07 × 10³, indicating laminar flow (Re < 2 300).

"Un projet n'est jamais trop grand s'il est bien conçu." — André Citroën

"La difficulté attire l'homme de caractère, car c'est en l'étreignant qu'il se réalise." — Charles de Gaulle