Chapter 10, Problem 10.99P

To determine

i.

If the solution curve is mild, critical or steep slope when the local water depth is 1 m.

Answer to Problem 10.99P

The required curve is of critical slope and the local depth is lesser than y_cand y_n, it lies on C-3 curve.

Explanation of Solution

Given:

Q = 1.98 m³ /s

Slope = 0.33°

Local depth = 1m

Concept Used:

$\begin{array}{l}Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ y_c={\left(\frac{2Q^2}{g{\left(\cot 60°\right)}^2}\right)}^{1/5}\end{array}$

Calculation:

$\begin{array}{l}Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ Forclaytile,n=0.025\\ ForV-shapedchannel,\\ A=y_n{}^2\cot (60)\\ where{y}_{n}isdepth\\ R_h=\frac{y_n}{2}\cos 60°\\ Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ 1.98=\frac{1}{0.014}*y_n{}^2\cot (60)*{\left(\frac{y_n}{2}\cos 60°\right)}^{2/3}*{\left(\sin (0.33°)\right)}^{1/2}\\ 0.63=y_n{}^2*{\left(\frac{y_n}{2}\cos 60°\right)}^{2/3}\\ 1.59=y_n{}^{8/3}\\ y_n=1.19m\\ y_c={\left(\frac{2*Q^2}{g{\left(\cot 60°\right)}^2}\right)}^{1/5}\\ y_c={\left(\frac{2*{1.98}^2}{g{\left(\cot 60°\right)}^2}\right)}^{1/5}\\ y_c=1.19m\\ y_n=y_c\\ Theslopeiscritical\\ y=1<y_n=y_c\\ \mathrm{Sin}cethegivenlocaldepthislessserthan{y}_{n}and{y}_c.\\ ThecurveliesonC-3type\\ \end{array}$

Conclusion:

The required curve is of critical slope and the local depth is lesser than y_cand y_n, it lies on C-3 curve.

To determine

ii.

If the solution curve is mild, critical or steep slope when the local water depth is 2 m.

Answer to Problem 10.99P

The required curve is of critical slope and the local depth is greater than y_cand y_n, it lies on C-1 curve.

Explanation of Solution

Given:

Q = 1.98m³ /s

Slope = 0.33°

Local depth = 2m

Concept Used:

$\begin{array}{l}Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ y_c={\left(\frac{q^2}{g}\right)}^{1/3}\end{array}$

Calculation:

$\begin{array}{l}Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ Forclaytile,n=0.025\\ ForV-shapedchannel,\\ A=y_n{}^2\cot (60)\\ where{y}_{n}isdepth\\ R_h=\frac{y_n}{2}\cos 60°\\ Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ 1.98=\frac{1}{0.014}*y_n{}^2\cot (60)*{\left(\frac{y_n}{2}\cos 60°\right)}^{2/3}*{\left(\sin (0.33°)\right)}^{1/2}\\ 0.63=y_n{}^2*{\left(\frac{y_n}{2}\cos 60°\right)}^{2/3}\\ 1.59=y_n{}^{8/3}\\ y_n=1.19m\\ y_c={\left(\frac{2*Q^2}{g{\left(\cot 60°\right)}^2}\right)}^{1/5}\\ y_c={\left(\frac{2*{1.98}^2}{g{\left(\cot 60°\right)}^2}\right)}^{1/5}\\ y_c=1.19m\\ y_n=y_c\\ Theslopeiscritical.\\ y_n=y_c<y=2\\ \mathrm{Sin}ce.thegivenlocaldepthislessserthan{y}_{n}and{y}_c.\\ ThecurveliesonC-1type.\\ \end{array}$

Conclusion:

The required curve is of critical slope and the local depth is greater than y_cand y_n, it lies on C-1 curve.

To determine

iii.

If the solution curve is mild, critical or steep slope when the local water depth is 3 m.

Answer to Problem 10.99P

The required curve is of critical slope and the local depth is greater than y_cand y_n, it lies on C-1 curve.

Explanation of Solution

Given:

Q = 1.98 m³ /s

Slope = 0.33°

Local depth = 3m

Concept Used:

$\begin{array}{l}Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ y_c={\left(\frac{q^2}{g}\right)}^{1/3}\end{array}$

Calculation:

$\begin{array}{l}Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ Forclaytile,n=0.025\\ ForV-shapedchannel,\\ A=y_n{}^2\cot (60)\\ where{y}_{n}isdepth\\ R_h=\frac{y_n}{2}\cos 60°\\ Q=\frac{\alpha }{n}*A*R^{2/3}*Slop{e}^{1/2}\\ 1.98=\frac{1}{0.014}*y_n{}^2\cot (60)*{\left(\frac{y_n}{2}\cos 60°\right)}^{2/3}*{\left(\sin (0.33°)\right)}^{1/2}\\ 0.63=y_n{}^2*{\left(\frac{y_n}{2}\cos 60°\right)}^{2/3}\\ 1.59=y_n{}^{8/3}\\ y_n=1.19m\\ y_c={\left(\frac{2*Q^2}{g{\left(\cot 60°\right)}^2}\right)}^{1/5}\\ y_c={\left(\frac{2*{1.98}^2}{g{\left(\cot 60°\right)}^2}\right)}^{1/5}\\ y_c=1.19m\\ y_n=y_c\\ Theslopeiscritical.\\ y_n=y_c<y=3\\ \mathrm{Sin}cethegivenlocaldepthislessserthan{y}_{n}and{y}_c.\\ ThecurveliesonC-1type.\\ \end{array}$

Conclusion:

The required curve is of critical slope and the local depth is greater than y_cand y_n, it lies on C-1 curve.