# Hydraulic diameter

(Difference between revisions)
 Revision as of 13:51, 24 March 2006 (view source)Jola (Talk | contribs)← Older edit Revision as of 13:52, 24 March 2006 (view source)Jola (Talk | contribs) mNewer edit → Line 21: Line 21: ==Coaxial circular tube== ==Coaxial circular tube== - For a coaxial circular tube with an inner diameter of $d_i$ and an outer diameter of $d_o$ the hydraulic diameter is: + For a coaxial circular tube with an inner diameter $d_i$ and an outer diameter $d_o$ the hydraulic diameter is: :$d_h = 4 \; \frac{\frac{\pi d_o^2}{4} - \frac{\pi d_i^2}{4}}{\pi d_o + \pi d_i} = d_o - d_i$ :$d_h = 4 \; \frac{\frac{\pi d_o^2}{4} - \frac{\pi d_i^2}{4}}{\pi d_o + \pi d_i} = d_o - d_i$

## Revision as of 13:52, 24 March 2006

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

The definition of the hydraulic diamater is:

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

## 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:

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