# Hydraulic diameter

(Difference between revisions)
 Revision as of 05:22, 17 December 2008 (view source)m (darc4tchiri)← Older edit Latest revision as of 09:58, 17 December 2008 (view source)Peter (Talk | contribs) m (Reverted edits by NorolRelor (Talk) to last version by Jola) Line 1: Line 1: - olorollali The hydraulic diameter, $d_h$, is commonly used when dealing with non-circular pipes, holes or ducts. The hydraulic diameter, $d_h$, is commonly used when dealing with non-circular pipes, holes or ducts.

## Latest revision as of 09:58, 17 December 2008

The hydraulic diameter, $d_h$, is commonly used when dealing with non-circular pipes, holes or ducts.

The definition of the hydraulic diamater is:

$d_h \equiv 4 \; \frac{\mbox{cross-sectional-area of duct}}{\mbox{wetted perimeter of duct}}$

## Use of hydraulic diameter

### Estimating the turbulent length-scale

For fully-developed flow in non-circular ducts the turbulent length scale can be estimated as $0.07 \, d_h$. This is as usefull estimation for setting turbulence boundary conditions for inlets that have fully developed flow.

### Computing Reynolds number

The hydraulic diamater is often used when computing the dimensionless Reynolds number for non-circular ducts.

## Hydraulic diameters for different duct-geometries

Using the definition above the hydraulic diamater can easily be computed for any type of duct-geometry. Below follows a few examples.

### Circular pipe

For a circular pipe or hole the hydraulic diamater is:

$d_h = 4 \; \frac{\frac{\pi d^2}{4}}{\pi d} = d$

Where d is the real diameter of the pipe. Hence, for circular pipes the hydraulic diameter is the same as the real diameter of the pipe.

### Rectangular tube

For a rectangular tube or hole with the width $a$ and the height $b$ the hydraulic diamter is:

$d_h = 4 \; \frac{a b}{2 a + 2 b} = 2 \; \frac{a b}{a + b}$

### Coaxial circular tube

For a coaxial circular tube with an inner diameter $d_i$ and an outer diameter $d_o$ the hydraulic diameter is:

$d_h = 4 \; \frac{\frac{\pi d_o^2}{4} - \frac{\pi d_i^2}{4}}{\pi d_o + \pi d_i} = d_o - d_i$