Chapter 10, Problem 10.127P

To determine

i.

The normal depth of the river.

Answer to Problem 10.127P

The normal depth of the river, y_n = 2.21 ft.

Explanation of Solution

Given:

S₀=0.0025

Q = 600 ft³ / s

b = 50 ft

increase in water depth = 8 ft

Concept:

$Q=\frac{\alpha }{n}A{R}_h{}^{2/3}{S}_o{}^{1/2}$

$\begin{array}{l}A=b{y}_n=50y_n\\ R_h=\frac{b{y}_n}{b+2y_n}=\frac{50y_n}{50+2y_n}\\ Forcleanearth,n=0.022\\ Q=\frac{\alpha }{n}A{R}_h{}^{2/3}{S}_o{}^{1/2}\\ 600=\frac{1.486}{0.024}(50y_n)*{\left(\frac{50y_n}{50+2y_n}\right)}^{2/3}{\left(0.0025\right)}^{1/2}\\ y_n=2.21ft\end{array}$

Conclusion:

The normal depth of the river is 2.21 ft.

To determine

ii.

The new water depth half mile upstream.

Answer to Problem 10.127P

The new water depth half mile upstream is 1.4 ft.

Explanation of Solution

Given:

S₀=0.0025

Q = 600 ft³ / s

b = 50 ft

y₁ = 10 ft

Concept Used:

$\begin{array}{l}Foruniformflow,\\ q=y_1{V}_1=y_2{V}_2\\ E_1=y_1+\frac{V_1{}^2}{2g}\\ E_2=y_2+\frac{V_2{}^2}{2g}\\ S_{avg}=\frac{n^2{V}_{avg}{}^2}{{\alpha }^2{R}_{h,avg}{}^2}\\ S_o-S_{avg}=\frac{E_2-E_1}{\Delta x}\end{array}$

Calculation:

$\begin{array}{l}Foruniformflow,\\ Q=b{y}_1{V}_1\\ V_1=\frac{Q}{b{y}_1}\\ V_1=\frac{600}{50*10}\\ V_1=1.2ft/s\\ q=y_1{V}_1=y_2{V}_2\\ V_2=\frac{y_1{V}_1}{y_2}\\ V_2=\frac{0.24}{y_2}\\ E_1=y_1+\frac{V_1{}^2}{2g}\\ E_1=10+\frac{{1.2}^2}{2*32.2}\\ E_1=10.02ft\\ E_2=y_2+\frac{V_2{}^2}{2g}\\ E_2=y_2+\frac{{\left(\frac{0.24}{y_2}\right)}^2}{2*32.2}\\ E_2=y_2+\frac{0.00089}{y_2{}^2}\\ V_{avg}=\frac{V_1+V_2}{2}\\ V_{avg}=\frac{1.2+\frac{0.24}{y_2}}{2}\\ R_1=\frac{b{y}_1}{b+2y_1}\\ R_1=\frac{50*10}{50+2*10}\\ R_1=7.14ft\\ R_2=\frac{b{y}_2}{b+2y_2}\\ R_2=\frac{50y_2}{50+2y_2}\end{array}$

$\begin{array}{l}\\ R_{h,avg}=\frac{R_1+R_2}{2}\\ R_{h,avg}=\frac{7.143+\frac{50y_2}{50+2y_2}}{2}\\ \\ S_{avg}=\frac{n^2{V}_{avg}{}^2}{{\alpha }^2{R}_{h,avg}{}^2}\\ n=0.022\\ S_{avg}=\frac{{0.022}^2{\left(\frac{1.2+\frac{0.24}{y_2}}{2}\right)}^2}{{1.468}^2{\left(\frac{7.143+\frac{50y_2}{50+2y_2}}{2}\right)}^2}\\ S_{avg}=0.000219\left(\frac{0.6y_2+0.12}{y_2}\right)\left(\frac{50+2y_2}{178.5+32.14y_2}\right)\\ S_o-S_{avg}=\frac{E_2-E_1}{\Delta x}\\ 0.0025-\left[0.000219\left(\frac{0.6y_2+0.12}{y_2}\right)\left(\frac{50+2y_2}{178.5+32.14y_2}\right)\right]=\frac{y_2+\frac{0.00089}{y_2{}^2}-10.02}{0.5mi*5280\frac{ft}{mi}}\\ y_2=3.6ft\\ Thenewwaterdepth,\Delta y=y_2-2.2\\ \Delta y=3.6-2.2\\ \Delta y=1.4ft\end{array}$.

Conclusion:

The new water depth half mile upstream is 1.4 ft.