Chapter 10, Problem 10.91P

(a)

To determine

The oblique angle, $\beta$.

Answer to Problem 10.91P

The value of wave angle $\beta$ is, $\beta =56.65$

Explanation of Solution

Given information:

Average velocity, V = 0.94m/s

Depth of flat-water channel, y₁=1cm

Wedge half angle,θ = 20⁰

Calculation:

$\begin{array}{l}Fr1=\frac{V}{\sqrt{g{y}_1}}\mathrm{...............}(1)\\ F{r}_1=\frac{0.94}{\sqrt{9.81\times 1\times {10}^{-2}}}\\ F{r}_1=\frac{0.94}{\sqrt{0.0981}}\\ F{r}_1=\frac{0.94}{0.3132}\\ F{r}_1=3.0013\end{array}$

As we know the following relation between y₂ and y₁

$\begin{array}{l}\frac{y_2}{y_1}=\frac{1}{2}\sqrt{1+8F{r}_1}-1\\ \frac{y_2}{1\times {10}^{-2}}=\frac{1}{2}\sqrt{1+8\times 3.0013}-1\\ \frac{y_2}{1\times {10}^{-2}}=\frac{1}{2}.\sqrt{25.0104}-1\\ \frac{y_2}{1\times {10}^{-2}}=\frac{1}{2}\times (5.001)-1\\ \frac{y_2}{1\times {10}^{-2}}=1.5005\\ y_2=1.5005\times {10}^{-2}\\ y_2=0.015\text{}m\\ \end{array}$

As per the continuity equation we have following relation

$\begin{array}{l}{V}_1{y}_1=V_2{y}_2\\ \frac{V_1}{V_2}=\frac{y_2}{y_1}\\ V_2=\frac{y_2}{y_1}.V_1\\ V_2=\frac{0.015}{0.01}\times 0.94\\ V_2=1.41m/s\end{array}$

As we know the equation for oblique angle

$\begin{array}{l}\frac{\tan \left(\beta -\theta \right)}{\tan \theta }=\frac{V_2}{V_1} \\ \beta ={\tan }^{-1}\left(\frac{V_2}{V_1}\times \tan \theta \right)+\theta \\ \\ \beta ={\tan }^{-1}(\frac{1.41}{0.94})+\text{0}\text{.349}\\ \beta ={\tan }^{-1}(1.5)+\text{0}\text{.349}\\ \beta =56.301+\text{0}\text{.349}\\ \beta =56.65\\ \end{array}$.

To determine

(b)

The Froud number Fr₂ is, $F{r}_2=0.343$.

Answer to Problem 10.91P

]

The Froud number Fr2

Explanation of Solution

Given information:

Average velocity, V = 0.94m/s

Depth of flat-water channel, y₁=1cm

Wedge half angle,θ = 20⁰

Solution:

$\begin{array}{l}F{r}_1=\frac{V}{\sqrt{g{y}_1}}\mathrm{...............}(1)\\ F{r}_1=\frac{0.94}{\sqrt{9.81\times 1\times {10}^{-2}}}\\ F{r}_1=\frac{0.94}{\sqrt{0.0981}}\\ F{r}_1=\frac{0.94}{0.3132}\\ F{r}_1=3.0013\end{array}$

$\begin{array}{l}F{r}_2=\frac{F{r}_1\sqrt{8}}{{\left(\sqrt{1+8F{r}_1{}^2-1}\right)}^{\frac{3}{2}}}\\ F{r}_2=\frac{3.0013\sqrt{8}}{{\left(\sqrt{1+8{(3.0013)}^2-1}\right)}^{\frac{3}{2}}}\\ F{r}_2=\frac{8.489}{{\left(\sqrt{1+8\times 9.0078-1}\right)}^{\frac{3}{2}}}\\ F{r}_2=\frac{8.489}{{\left(\sqrt{72.062}\right)}^{\frac{3}{2}}}\\ F{r}_2=\frac{8.489}{{\left(8.489\right)}^{\frac{3}{2}}}=\\ F{r}_2={(}^{8.489}\\ F{r}_2=\frac{1}{\sqrt{8.489}}\\ F{r}_2=\frac{1}{2.913}\\ F{r}_2=0.343(Asthisisa\sup ercriticalflow,F{r}_2<1)\end{array}$.

c

To determine

The value y₂.

Answer to Problem 10.91P

The value y₂ is, $y_2=0.015\text{}m$

Explanation of Solution

Given information:

Average velocity, V = 0.94m/s

Depth of flat-water channel, y₁=1cm

Wedge half angle,θ = 20⁰

Solution:

As we know the following relation between y₂ and y₁

$\begin{array}{l}\frac{y_2}{y_1}=\frac{1}{2}\sqrt{1+8F{r}_1}-1\\ \frac{y_2}{1\times {10}^{-2}}=\frac{1}{2}\sqrt{1+8\times 3.0013}-1\\ \frac{y_2}{1\times {10}^{-2}}=\frac{1}{2}.\sqrt{25.0104}-1\\ \frac{y_2}{1\times {10}^{-2}}=\frac{1}{2}\times (5.001)-1\\ \frac{y_2}{1\times {10}^{-2}}=1.5005\\ y_2=1.5005\times {10}^{-2}\\ y_2=0.015\text{}m\\ \end{array}$

$\begin{array}{l}\\ \end{array}$.