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


\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

Return to Exercise 14

Return to where text linked to Exercise 14


Introduction to Karst Aquifers Copyright © 2022 by Eve L. Kuniansky, Charles J. Taylor, and Frederick Paillet. All Rights Reserved.