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## Q. 1.3

Water is flowing at a rate of 800 cfs in a 6 ft wide rectangular channel at a constant depth of 4 ft. The bottom of the channel changes slope by means of a circular arc of radius 50 ft. The depth of flow through this arc remains constant at 4 ft. Assuming that the streamlines of this flow are concentric circles (have the same center of curvature), determine the pressure along a radial line at the beginning of the arc where it connects to the straight upstream channel that has a bottom slope $S_{o}$ = 0.15, and also determine the pressure distribution along a vertical radial line.

## Verified Solution

Using the upstream bottom slope θ = arctan(0.15) = 8.531°. The velocity in the channel equals 33.333 fps and therefore,

$\frac{dp}{dr}=\rho \left\{g\cos(8.531^{\circ})+\frac{(33.333)^{2}}{r}\right\}$

or after integrating p from 0 to p as r is integrated between 46 and r, this results in

$p=\rho\left\{31.844r+1111.11 Ln\left\lgroup\frac{r}{46}\right\rgroup-1464.81\right\}$

to give the pressure distribution as a function of r. On the bottom r = 50, and the pressure here is p = 426.75 psf = 2.964 psi, at the channel bottom, where r = 50 ft. Ignoring the added pressure due to the radius of curvature, the pressure at the bottom of the channel is p = 4γ/144 = 1.73 psi. This latter amount is 1.234 psi too small.
Flip-bucket spillways cause high-velocity water to have a considerable normal component of acceleration, as illustrated in this example. To find the added forces on such structures caused by changing the direction of the fluid can be handled easier through the use of the momentum principle that will be discussed in detail in Chapter 3.