Chapter 10, Problem 10.3CP

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To determine

(a)

To compute:

The water depth 10 m downstream at section (3) for the $5°$ unfinished concrete slope with use gradually varied theory.

Answer to Problem 10.3CP

The water depth 10 m downstream at section (3) for the $5°$ unfinished concrete slope with use gradually varied theory is $\left(y_n\right)=0.0611m$ At $x=10m$ nearly critical.

Explanation of Solution

Figure:

Given:

For unfinished concrete $n\approx 0.014$, $y\left(0\right)=0.0462m$, $q=0.2m^3/s$, $g=32.2ft/s^2$

$S_o=\sin 5°$

Concept Used:

Annuity problem requires the use of the formula as follows:

$q=V{y}_n$

$\Rightarrow$

$V=\frac{1}{n}{\left(y_n\right)}^{2/3}{S}_o{}^{1/2}$

$\Rightarrow$

$q=\frac{1}{n}\left(y_n\right){\left(y_n\right)}^{2/3}{S}_o{}^{1/2}$

Here,

Constant $=q$, Normal depth $=\left(y_n\right)$, Slope $=S_o$

Calculation:

As per the given problem

$n\approx 0.014$, $q=0.2m^3/s$, $S_o=\sin 5°$

Annuity problem requires the use of this formula

$q=\frac{1}{n}\left(y_n\right){\left(y_n\right)}^{2/3}{S}_o{}^{1/2}$

Substitute these values in the formula

$0.2=\frac{1}{0.014}\left(y_n\right){\left(y_n\right)}^{2/3}{\left(\sin 5°\right)}^{1/2}$

$\Rightarrow$

$\left(y_n\right)\approx 0.0611m$

$\left(y_n\right)\approx 0.0611m$ At $x=10m$

For the steep (S-3) slope yields $\left(y_n\right)=0.0611m$ At $x=10m$ nearly critical as shown in figure

Conclusion:

The water depth 10 m downstream at section (3) for the $5°$ unfinished concrete slope with use gradually varied theory $\left(y_n\right)=0.0611m$ At $x=10m$ nearly critical.

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To determine

(b)

To compute:

The water depth 10 m upward (adverse) slope of $5°$ unfinished concrete slope with use gradually varied theory.

Answer to Problem 10.3CP

Integration for the adverse (- $5°$ ) slope goes to critical at point about $x=4.5m$

$\to$ The theory fails for $x>4.5m$.

Explanation of Solution

Given:

For unfinished concrete $n\approx 0.014$, $y\left(0\right)=0.0462m$, $q=0.2m^3/s$, $g=9.81m/s^2$

$S_o=\sin 5°$

Concept Used:

Annuity problem requires the use of the formula as follows:

$\frac{dy}{dx}=\frac{S_o-n^2{q}^2/y^{10/3}}{1-q^2/(g{y}^3)}$ The channel is “wide “

Here,

Constant $=q$, depth $=\left(y\right)$, Slope $=S_o$

Take smaller adverse slope of $-1°$, we obtain $y=0.160m$ at $x=9.8m$ as shown in figure.

Annuity problem requires the use of this formula.

$n\approx 0.014$, $y\left(0\right)=0.0462m$, $q=0.2m^3/s$, $g=9.81m/s^2$

$S_o=-\sin 5°.$

$\frac{dy}{dx}=\frac{S_o-n^2{q}^2/y^{10/3}}{1-q^2/(g{y}^3)}$ The channel is “wide “

Substitute these values in the formula:

$\frac{dy}{dx}=\frac{\sin -5°-{\left(0.014\right)}^2{\left(0.2\right)}^2/y^{10/3}}{1-{\left(0.2\right)}^2/\left(9.81\right)(y^3)}$

Integration for the adverse (- $5°$ )slope goes to critical at point about $x=4.5m$

$\to$ The theory fails for $x>4.5m$ as shown in figure:

Conclusion:

Integration for the adverse (- $5°$ ) slope goes to critical at point about $x=4.5m$

$\to$ The theory fails for $x>4.5m$.

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To determine

(c)

Find that part and take smaller (adverse) slope of $(-1)°$.

Answer to Problem 10.3CP

That part and take smaller (adverse) slope of $(-1)°$ is $y=0.160m$ at $x=9.8m$

Explanation of Solution

Given:

For unfinished concrete $n\approx 0.014$, $q=0.2m^3/s$, $g=9.81m/s^2$

$S_o=-\sin (-1)°$

Concept Used:

Annuity problem requires the use of the formula as follows

$y_c={\left(\frac{q^2}{g}\right)}^{1/3}$

Here,

Critical depth $=y_c$

Calculation:

As per the given problem

$n\approx 0.014$, $q=0.2m^3/s$, $g=9.81m/s^2$

$S_o=-\sin (-1)°$

Annuity problem requires the use of this formula

$y_c={\left(\frac{q^2}{g}\right)}^{1/3}$

Substitute these values in the formula

$y_c={\left(\frac{{0.2}^2}{9.81}\right)}^{1/3}=0.160m$

$y_c=0.160m$

Annuity problem requires the use of this formula

$n\approx 0.014$, $y\left(0\right)=0.0462m$, $q=0.2m^3/s$, $g=9.81m/s^2$

$S_o=-\sin -1°$

$\frac{dy}{dx}=\frac{S_o-n^2{q}^2/y^{10/3}}{1-q^2/(g{y}^3)}$ The channel is “wide “

Substitute these values in the formula

$\frac{dy}{dx}=\frac{\sin (-1°)-{\left(0.014\right)}^2{\left(0.2\right)}^2/y^{10/3}}{1-{\left(0.2\right)}^2/\left(9.81\right)(y^3)}$

$x=9.8m$

take smaller adverse slope of $-1°$, we obtain $y=0.160m$ at $x=9.8m$, The flow goes critical near x=L ,as shown in graph

The flow goes critical near x=L ,as shown in graph

Conclusion:

That part and take smaller (adverse) slope of $(-1)°$ is $y=0.160m$ at $x=9.8m$.