Question

The cross-section of a rough, rectangular, concrete() channel measures . The channel slope is 0.02ft/ft. Using the Darcy-Weisbach friction method, determine the maximum allowable flow rate through the channel to maintain one foot of free board(freeboard is the vertical distance form the water surface to the overtopping level of the channel). For these conditions, find the following characteristics(note that FlowMaster may not directly report all of these):
a) Flow area
b) Wetted perimeter
c) Hydraulic radius(A/P) :
d) Velocity
e) Froude number

Answer 1

Answer:

The following are the answer to this question:

Explanation:

In point a, Calculating the are of flow:

$\bold{Area =B \times D_f}$

         $=6\times 5\\\\=30 \ ft^2$

In point b, Calculating the wetter perimeter.

$\bold{P_w =B+2\times D_f}$

      $= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft$

In point c, Calculating the hydraulic radius:

$\bold{R=\frac{A}{P_w}}$

   $=\frac{30}{16}\\\\= 1.875 \ ft$

In point d, Calculating the value of Reynolds's number.

$\bold{Re =\frac{4VR}{v}}$

     $=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}$

     $=750,000 V$

Calculating the velocity:

$V= \sqrt{\frac{8gRS}{f}}$

   $= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}$

$\sqrt{f}=\frac{3.108}{V}$

calculating the Cole-brook-White value:

$\frac{1}{\sqrt{f}}= -2 \log (\frac{K}{12 R} +\frac{2.51}{R_e \sqrt{f}})\\\\ \frac{1}{\frac{3.108}{V}}= -2 \log (\frac{2 \times 10^{-2}}{12 \times 1.875} +\frac{2.51}{750,000V\sqrt{f}})$

$\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})$

After calculating the value of V it will give:

$V= 25.18 \ \frac{ft}{s^2}$

In point a, Calculating the value of Froude:

$F= \frac{V}{\sqrt{gD}}$

$= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}$

$= \frac{25.18}{\sqrt{32.2\frac{30}{6}}}\\\\= \frac{25.18}{\sqrt{32.2 \times 5}}\\\\= \frac{25.18}{\sqrt{161}}\\\\= \frac{25.18}{12.68}\\\\= 1.98$

The flow is supercritical because the amount of Froude is greater than 1.  

Calculating the channel flow rate.

$Q= AV$

   $=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}$