Chapter 11, Problem 1P

Verify the equation given in Table 11.1 for the hydraulic radius of a circular channel. Evaluate and plot the ratio R/D, for liquid depths between 0 and D.

To determine

The expression for the hydraulic radius of the circular channel and plot the ratio $R/D$ for liquid depths between 0 and D.

Explanation of Solution

Calculation:

The area of the circular channel can be molded as the summation of two triangles and the segment of the chord.

Calculate the area of the circular channel $\left(A\right)$.

  $\begin{array}{c}A=A_C+A_T+A_T\\ =A_C+2A_T\\ =\left(\frac{\frac{\alpha }{2}}{4}{D}^2\right)+2\left(\frac{1}{2}bh\right)\\ =\frac{\alpha }{8}{D}^2+bh\end{array}$

  $\begin{array}{c}=\frac{\alpha }{8}{D}^2+\left[\frac{D}{2}\sin \left(\pi -\frac{\alpha }{2}\right)\right]\left[\frac{D}{2}\cos \left(\pi -\frac{\alpha }{2}\right)\right]\\ =\frac{\alpha }{8}{D}^2+\left[\frac{D^2}{4}\sin \left(\pi -\frac{\alpha }{2}\right)\cos \left(\pi -\frac{\alpha }{2}\right)\right]\\ =\frac{\alpha }{8}{D}^2+\frac{D^2}{8}\sin \left(2\pi -\alpha \right)\end{array}$

  $\begin{array}{c}=\frac{D^2}{8}\alpha -\frac{D^2}{8}\sin \alpha \\ =\frac{D^2}{8}\left(\alpha -\sin \alpha \right)\end{array}$

Calculate the hydraulic radius of the circular channel $\left(R\right)$.

  $\begin{array}{c}R=\frac{A}{P}\\ =\frac{\left(\frac{D^2}{8}\left(\alpha -\sin \alpha \right)\right)}{\left(\frac{D\alpha }{2}\right)}\\ =\frac{1}{4}\left(1-\frac{\sin \alpha }{\alpha }\right)D\end{array}$

Thus, the hydraulic radius of the circular channel is $\frac{1}{4}\left(1-\frac{\sin \alpha }{\alpha }\right)D$.

Calculate the ratio between the hydraulic radius and the depth of the circular channel.

  $\begin{array}{l}R=\frac{1}{4}\left(1-\frac{\sin \alpha }{\alpha }\right)D\\ \frac{R}{D}=\frac{1}{4}\left(1-\frac{\sin \alpha }{\alpha }\right)\end{array}$        (I)

Calculate the relation between the height of the channel and the angle made by the channel surface.

  $\begin{array}{c}y=\frac{D}{2}+\frac{D}{2}\cos \left(\pi -\frac{\alpha }{2}\right)\\ =\frac{D}{2}+\frac{D}{2}\cos \left(\pi -\frac{\alpha }{2}\right)\end{array}$

  $\frac{y}{D}=\frac{1}{2}\left[1+\cos \left(\pi -\frac{\alpha }{2}\right)\right]$        (II)

For plotting the graph between $\frac{R}{D}$ and $\frac{y}{D}$, the term $\alpha$ has to be removed in Equation (I) and Equation (II) or else each value in terms of $\alpha$ has to be determined.

The values of $\frac{R}{D}$ and $\frac{y}{D}$ are determined for different values of $\alpha$ and the readings are tabulated as in Table (1).

$\alpha$	$\frac{R}{D}$	$\frac{y}{D}$
0.0000	0	0.0000
0.3927	0.0125	0.0096
0.7854	0.0498	0.0381
1.1781	0.1079	0.0843
1.5708	0.1817	0.1464
1.9635	0.2647	0.2222
2.3562	0.3499	0.3087
2.7489	0.4304	0.4025
3.1416	0.5000	0.5000
3.5343	0.5541	0.5975
3.9270	0.5900	0.6913
4.3167	0.6069	0.7778
4.1724	0.6061	0.8536
5.1051	0.5905	0.9157
5.4978	0.5643	0.9619
5.8905	0.5325	0.9904
6.2832	0.5000	1.0000

Plot the ratio $R/D$ for liquid depths between 0 and D as in Figure (1).