Answer:
The following are the answer to this question:
Explanation:
In point a, Calculating the are of flow:
[tex]\bold{Area =B \times D_f}[/tex]
[tex]=6\times 5\\\\=30 \ ft^2[/tex]
In point b, Calculating the wetter perimeter.
[tex]\bold{P_w =B+2\times D_f}[/tex]
[tex]= 6 +2\times (5)\\\\= 6 +10 \\\\=16 \ ft[/tex]
In point c, Calculating the hydraulic radius:
[tex]\bold{R=\frac{A}{P_w}}[/tex]
[tex]=\frac{30}{16}\\\\= 1.875 \ ft[/tex]
In point d, Calculating the value of Reynolds's number.
[tex]\bold{Re =\frac{4VR}{v}}[/tex]
[tex]=\frac{4V \times 1.875}{1 \times 10^{-5} \frac{ft^2}{s}}\\\\[/tex]
[tex]=750,000 V[/tex]
Calculating the velocity:
[tex]V= \sqrt{\frac{8gRS}{f}}[/tex]
[tex]= \sqrt{\frac{8\times 32.2 \times 1.875 \times 0.02}{f}}\\\\=\frac{3.108}{\sqrt{f}}\\\\[/tex]
[tex]\sqrt{f}=\frac{3.108}{V}\\\\[/tex]
calculating the Cole-brook-White value:
[tex]\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}})\\[/tex]
[tex]\frac{V}{3.108} =-2\log(8.88 \times 10^{-5} + \frac{3.346 \times 10^{-6}}{750,000(3.108)})[/tex]
After calculating the value of V it will give:
[tex]V= 25.18 \ \frac{ft}{s^2}\\[/tex]
In point a, Calculating the value of Froude:
[tex]F= \frac{V}{\sqrt{gD}}[/tex]
[tex]= \frac{V}{\sqrt{g\frac{A}{\text{Width flow}}}}\\[/tex]
[tex]= \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[/tex]
The flow is supercritical because the amount of Froude is greater than 1.
Calculating the channel flow rate.
[tex]Q= AV[/tex]
[tex]=30x 25.18\\\\= 755.4 \ \frac{ft^3}{s}\\[/tex]
