Exercise 14 Solution

Eve L. Kuniansky; Charles J. Taylor; John H. Williams; Frederick Paillet

Exercise 14 Solution

a. The hydraulic radius for image a) is

$\displaystyle \frac{\textrm{Area}}{\textrm{Circumference}}=\frac{\pi r^{2}}{2\pi r}=\frac{r}{2}=\frac{2}{2}=1\;\textrm{meter}$

for image b) the hydraulic radius is

$\displaystyle \frac{\textrm{Area}}{\textrm{Wetted perimeter}}=\frac{(3.6)(3.8)+(0.2)(3.8)}{3.6+2\sqrt{0.2^{2}+3.8^{2}}}$

= $\displaystyle \frac{13.68+0.76}{3.6+2\sqrt{14.48}}=\frac{14.44}{3.6+7.61}=1.29\;\textrm{meter}$

b. For image a) (0.005 m³/s)/(12.57 m²) = 0.000398 m/s

and for image b) (0.005 m³/s)/(14.44 m²) = 0.000346 m/s

c.

$\displaystyle Re=\frac{VD\rho }{\mu }=\frac{VD}{\nu }$

For the pipe (a) D is equal to the pipe diameter of 4 m, but for the open channel (b), we would use the equation for a pipe diameter that has the same hydraulic radius

r = 2 (1.29) = 2.58, then D = 2r = 5.16 m

for image a) Re= (0.000398 m/s)(4 m)/(10⁻⁶ m²/s) = 1600

and for image b) Re = (0.000346 m/s)(5.16 m)/(10⁻⁶ m²/s) = 1800

d. for image a) Most likely laminar as Re< 2100

for image b) Most likely turbulent as Re > 500

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