PLAIN AND CIVIL: EXAMPLE 6.4.2. HYDRAULIC RADIUS FOR PIPES FLOWING LESS THAN HALF FULL

Monday, 22 April 2019

Calculate the hydraulic radius (ft) for water flowing 6 inches deep in a 48-inch diameter storm sewer.

The diameter is 48 inches, so radius = 24 inches or 2 ft.

h= 6 inches or 0.5ft.

With only 6inches depth of water flowing in the 48-inch diameter pipe, consider the pipe less than half full. Thus hydraulic radius formula becomes:

The formulas for area and perimeter are in function of theta which is in radians.

$\theta =2\left [cos^{-1}\left ( \frac{2-0.5}{2} \right ) \right ]\left ( \frac{\pi rad}{180degrees} \right )=1.4455rad$

Then the area and wetted perimeter will be solved as functions of Radians [calculator should be in Radians mode]

$A=\frac{r^{2}\left ( \theta -sin\theta \right )}{2}$

$A=\frac{2^{2}\left ( 1.4455 -sin1.4455 \right )}{2}=0.9146ft^{2}$

$P=r\theta$

$P=2ft\left ( 1.4455 \right )=2.891ft$

Thus, the hydraulic radius becomes:

$R=\frac{A}{P}=\frac{0.9146ft^{2}}{2.891ft}$

${\color{Red} \mathbf{R=0.3164ft}}$

RELATED TOPIC:

6.4. THE HYDRAULIC RADIUS

PLAIN AND CIVIL

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Monday, 22 April 2019

EXAMPLE 6.4.2. HYDRAULIC RADIUS FOR PIPES FLOWING LESS THAN HALF FULL

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