PLAIN AND CIVIL: EXAMPLE 6.4.1. HYDRAULIC RADIUS FOR PIPES FLOWING IN FULL AND HALF FULL

Sunday, 21 April 2019

a. Calculate the hydraulic radius (ft) for water flowing full in a 48-inch storm sewer.

b. What is the hydraulic radius (ft) if water flows only half the storm sewer?

From diameter = 48 inches, radius = 24 inches or 2 ft.

A. WATER FLOWING FULL

$A=\pi r^{2}=\pi \left ( 2ft \right )^{2}=12.5664ft^{2}$

$P=2\pi r=2\pi \left ( 2ft \right )=12.5664ft$

Thus, for hydraulic radius:

$R=\frac{A}{P}=\frac{12.5664ft^{2}}{12.5664ft}$

${\color{Red} \mathbf{R=1ft}}$

Alternately, from the formula:

$R=\frac{D}{A}=\frac{4ft}{4}=1ft$

B. PIPE IS HALF-FULL

$A=\frac{\pi R^{2}}{2}=\frac{\pi \left ( 2ft \right )^{2}}{2}=2\pi ft^{2}$

$P=\frac{2\pi r}{2}=2\pi ft$

So, the hydraulic radius will become:

$R=\frac{A}{P}=\frac{2\pi ft^{2}}{2\pi ft }$

${\color{Red} \mathbf{R=1ft}}$

Same formula can be applied for pipes half full - $R=\frac{D}{4}$ .

RELATED TOPIC:

6.4. THE HYDRAULIC RADIUS

PLAIN AND CIVIL

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Sunday, 21 April 2019

EXAMPLE 6.4.1. HYDRAULIC RADIUS FOR PIPES FLOWING IN FULL AND HALF FULL

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