Hydraulic Diameter

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

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

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.

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

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

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 dimensionless quantity such as the Reynolds number, which prefers a single variable for flow analysis rather than the set of variables as listed in the table below. The Manning formula contains a quantity called the hydraulic radius. Despite what the name may suggest, the hydraulic diameter is not twice the hydraulic radius, but four times larger.

Hydraulic diameter is mainly used for calculations involving turbulent flow. Secondary flows can be observed in non-circular ducts as a result of turbulent shear stress in the turbulent flow. Hydraulic diameter is also used in calculation of heat transfer in internal-flow problems.

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:

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

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.

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 inscribed within the wetted perimeter. 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.