Hydraulic diameter

The hydraulic diameter, D_H, is a commonly used term when handling flow in non-circular tubes and channels. Using this term, one can calculate many things in the same way as for a round tube. When the cross-section is uniform along the tube or channel length, it is defined as^[1][2]

${\displaystyle D_{\text{H}}={\frac {4A}{P}},}$

where

A is the cross-sectional area of the flow,

P is the wetted perimeter of the cross-section.

More intuitively, the hydraulic diameter can be understood as a function of the hydraulic radius R_H, which is defined as the cross-sectional area of the channel divided by the wetted perimeter. Here, the wetted perimeter includes all surfaces acted upon by shear stress from the fluid.^[3]

${\displaystyle R_{\text{H}}={\frac {A}{P}},}$

${\displaystyle D_{\text{H}}=4R_{\text{H}},}$

Note that for the case of a circular pipe,

${\displaystyle D_{\text{H}}={\frac {4\pi R^{2}}{2\pi R}}=2R}$

The need for the hydraulic diameter arises due to the use of a single dimension in the case of a

Hydraulic diameter is mainly used for calculations involving

In the more general case, channels with non-uniform non-circular cross-sectional area, such as the Tesla valve, the hydraulic diameter is defined as:^[5]

${\displaystyle D_{\text{H}}={\frac {4V}{S}},}$

where

V is the total wetted volume of the channel,

S is the total wetted surface area.

This definition is reduced to ${\displaystyle {\frac {4A}{P}}}$ for uniform non-circular cross-section channels, and ${\displaystyle 2R}$ for circular pipes.

List of hydraulic diameters

Geometry	Hydraulic diameter	Comment
Circular tube	${\displaystyle D_{\text{H}}={\frac {4\cdot {\frac {\pi D^{2}}{4}}}{\pi D}}=D}$	For a circular tube the hydraulic diameter is simply the diameter of the tube.
Annulus	${\displaystyle D_{\text{H}}={\frac {4\cdot {\frac {\pi (D_{\text{out}}^{2}-D_{\text{in}}^{2})}{4}}}{\pi (D_{\text{out}}+D_{\text{in}})}}=D_{\text{out}}-D_{\text{in}}}$
Square duct	${\displaystyle D_{\text{H}}={\frac {4a^{2}}{4a}}=a}$	here a represents the length of a side, not the cross sectional area
Rectangular duct (fully filled). The duct is closed so that the wetted perimeter consists of the 4 sides of the duct.	${\displaystyle D_{\text{H}}={\frac {4ab}{2(a+b)}}={\frac {2ab}{a+b}}}$	For the limiting case of a very wide duct, i.e. a slot of width b, where b ≫ a, then DH = 2a.
Channel of water or partially filled rectangular duct. Open from top by definition so that the wetted perimeter consists of the 3 sides of the duct (2 on the side and the base).	${\displaystyle D_{\text{H}}={\frac {4ab}{2a+b}}}$	For the limiting case of a very wide duct, i.e. a slot of width b, where b ≫ a, and a is the water depth, then DH = 4a.

For a fully filled duct or pipe whose cross-section is a regular polygon, the hydraulic diameter is equivalent to the diameter ${\displaystyle D}$ of a circle

. This can be seen as follows: The ${\displaystyle N}$-sided regular polygon is a union of ${\displaystyle N}$ triangles, each of height ${\displaystyle D/2}$ and base ${\displaystyle B=D\tan(\pi /N)}$. Each such triangle contributes ${\displaystyle BD/4}$ to the total area and ${\displaystyle B}$ to the total perimeter, giving

${\displaystyle D_{\text{H}}=4{\frac {NBD/4}{NB}}=D}$

for the hydraulic diameter.

References