Hydraulic diameter

(Difference between revisions)

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:

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

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.

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

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

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

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

@@ Line 21: / Line 21: @@
 ==Coaxial circular tube==
-For a coaxial circular tube with an inner diameter of <math>d_i</math> and an outer diameter of <math>d_o</math> the hydraulic diameter is:
+For a coaxial circular tube with an inner diameter <math>d_i</math> and an outer diameter <math>d_o</math> the hydraulic diameter is:
 :<math>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</math>