PLAIN AND CIVIL: EXAMPLE 6.4.4. HYDRAULIC RADIUS FOR PIPES FLOWING MORE THAN HALF FULL

Monday, 22 April 2019

Calculate the hydraulic radius for water flowing 3.4 ft deep in a 48- inch diameter storm sewer.

   $\theta =2cos^{-1}\left ( \frac{r-h}{r} \right )$

The angle formula remains the same, but this time, h is the available height of air, not water. The value for h is:

$h=4ft-3.4ft =0.60ft$

So the angle can be computed as [calculator in radians mode]:

   $\theta =2cos^{-1}\left ( \frac{2-0.60}{2} \right )=1.5908$

So the area and wetted perimeter can be computed as follows:

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

   $A=\pi 2^{2}-\frac{2^{2}\left ( 1.5908 -sin\left (1.5908 \right ) \right )}{2}=11.3844ft^{2}$

$P=2\pi r-r\theta$

$P=2\pi \left ( 2 \right )-2\left ( 1.5908 \right )=9.3848ft$

The hydraulic radius is then computed as:

   $R=\frac{A}{P}=\frac{11.3844ft^{2}}{9.3848ft}$

${\color{Red} \mathbf{R=1.2131ft}}$

RELATED TOPIC:

6.4. THE HYDRAULIC RADIUS

PLAIN AND CIVIL

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Monday, 22 April 2019

EXAMPLE 6.4.4. HYDRAULIC RADIUS FOR PIPES FLOWING MORE THAN HALF FULL

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