Chapter 10, Problem 10.111P

To determine

(a)

The water depth one mile upstream for initial depth of $7.7\text{ft}$.

Answer to Problem 10.111P

The water depth one mile upstream of the dam is $7.325\text{ft}$.

Explanation of Solution

Given information:

The average flow rate is $900{\text{ft}}^{\text{3}}/\text{s}$, upstream width is $150\text{ft}$, slope is $10\text{ft}/\text{mile}$ and water depth just upstream of the dam is $7.7\text{ft}$.

Write the expression for the flow:

$Fr=\frac{V}{\sqrt{gy}}$     …… (I)

Here, Froude number is $Fr$, velocity of flow is $V$, acceleration due to gravity is $g$ and water depth is $y$.

Write the expression for flow rate in open channel.

$\dot{q}=w\times y\times V$     …… (II)

Here, flow rate in open channel is $\dot{q}$, width of channel is $w$ and velocity of flow in open channel is $V$.

Write the expression for width upstream.

$w_m=w+m\times s$     …… (III)

Here, width $s\text{mile}$ upstream is $w_m$, width at flash board is $w$, slope of width is $m$ and distance from flashboard is $s$.

Calculation:

Substitute $900{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{q}$, $150\text{ft}$ for $w$, and $7.7\text{ft}$ for $y$ .in Equation (II)

$\begin{array}{l}900{\text{ft}}^{\text{3}}/\text{s}=150\text{ft}\times \text{7}\text{.7}\text{ft}\times \text{V}\\ 900{\text{ft}}^{\text{3}}/\text{s}=1155{\text{ft}}^2\times \text{V}\end{array}$

$\begin{array}{c}V=\frac{900{\text{ft}}^{\text{3}}/\text{s}}{1155{\text{ft}}^{\text{2}}}\\ =0.78\text{ft}/\text{s}\end{array}$

Substitute $0.78\text{ft}/\text{s}$ for $V$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ and $7.7\text{ft}$ for $y$ in Equation (I).

$\begin{array}{c}Fr=\frac{0.78\text{ft}/\text{s}}{\sqrt{\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\times \left(7.7\text{ft}\right)}}\\ =\frac{0.78\text{ft}/\text{s}}{\sqrt{247.94{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}}\\ =0.05\end{array}$

Substitute $150\text{ft}$ for $w$, $10\text{ft}/\text{mile}$ for $m$ and $1\text{mile}$ for $s$.

$\begin{array}{c}{w}_m=150\text{ft}+\left(10\text{ft}/\text{mile}\right)\times \left(1\text{mile}\right)\\ =150\text{ft}+10\text{ft}\\ \text{=160}\text{ft}\end{array}$

Substitute $900{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{q}$, $160\text{ft}$ for $w$, $V$ for velocity and $y$ for depth in Equation (II).

$\begin{array}{l}900{\text{ft}}^{\text{3}}/\text{s}=160\text{ft}\times \text{y}\times \text{V}\\ V=\frac{900{\text{ft}}^{\text{3}}/\text{s}}{160\text{ft}\times \text{y}}\end{array}$

Substitute $\frac{900{\text{ft}}^{\text{3}}/\text{s}}{160\text{ft}\times \text{y}}$ for $V$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $0.05$ for $Fr$ and $y$ for depth in Equation (I).

$\begin{array}{c}0.05=\frac{\frac{900{\text{ft}}^{\text{3}}/\text{s}}{160\text{ft}\times \text{y}}}{\sqrt{32.2\text{ft}/{\text{s}}^{\text{2}}\times y}}\\ =\frac{0.9912{\text{ft}}^{\text{3}/\text{2}}}{y^{3/2}}\end{array}$

$y^{3/2}=\frac{0.9912{\text{ft}}^{\text{3}/\text{2}}}{0.05}$

$\begin{array}{c}y={\left(\frac{0.9912{\text{ft}}^{\text{3}/\text{2}}}{0.05}\right)}^{2/3}\\ ={\left(19.824{\text{ft}}^{\text{3}/\text{2}}\right)}^{2/3}\\ =7.325\text{ft}\end{array}$

Conclusion:

The water depth one mile upstream of the dam is calculated by the flow expression: $Fr=\frac{V}{\sqrt{gy}}$.

To determine

(b)

The water depth one mile upstream for initial depth of $10.7\text{ft}$.

Answer to Problem 10.111P

The water depth one mile upstream of the dam is $10.3\text{ft}$.

Explanation of Solution

Given information:

The average flow rate is $900{\text{ft}}^{\text{3}}/\text{s}$, upstream width is $150\text{ft}$, slope is $10\text{ft}/\text{mile}$ and water depth just upstream of the dam is $10.7\text{ft}$.

Write the expression for the

$Fr=\frac{V}{\sqrt{gy}}$     …… (I)

Here, Froude number is $Fr$, velocity of flow is $V$, acceleration due to gravity is $g$ and water depth is $y$.

Write the expression for flow rate in open channel.

$\dot{q}=w\times y\times V$     …… (II)

Here, flow rate in open channel is $\dot{q}$, width of channel is $w$ and velocity of flow in open channel is $V$.

Write the expression for width upstream.

$w_m=w+m\times s$     …… (III)

Here, width $s\text{mile}$ upstream is $w_m$, width at flash board is $w$, slope of width is $m$ and distance from flashboard is $s$.

Calculation:

Substitute $900{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{q}$, $150\text{ft}$ for $w$, and $10.7\text{ft}$ for $y$ .in Equation (II)

$\begin{array}{l}900{\text{ft}}^{\text{3}}/\text{s}=150\text{ft}\times 10.\text{7}\text{ft}\times \text{V}\\ 900{\text{ft}}^{\text{3}}/\text{s}=1605{\text{ft}}^2\times \text{V}\end{array}$

$\begin{array}{c}V=\frac{900{\text{ft}}^{\text{3}}/\text{s}}{1605{\text{ft}}^{\text{2}}}\\ =0.56\text{ft}/\text{s}\end{array}$

Substitute $0.78\text{ft}/\text{s}$ for $V$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ and $7.7\text{ft}$ for $y$ in Equation (I).

$\begin{array}{c}Fr=\frac{0.56\text{ft}/\text{s}}{\sqrt{\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)\times \left(10.7\text{ft}\right)}}\\ =\frac{0.56\text{ft}/\text{s}}{\sqrt{344.54{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}}\\ =0.03\end{array}$

Substitute $150\text{ft}$ for $w$, $10\text{ft}/\text{mile}$ for $m$ and $1\text{mile}$ for $s$.

$\begin{array}{c}{w}_m=150\text{ft}+\left(10\text{ft}/\text{mile}\right)\times \left(1\text{mile}\right)\\ =150\text{ft}+10\text{ft}\\ \text{=160}\text{ft}\end{array}$

Substitute $900{\text{ft}}^{\text{3}}/\text{s}$ for $\dot{q}$, $160\text{ft}$ for $w$, $V$ for velocity and $y$ for depth in Equation (II).

$\begin{array}{l}900{\text{ft}}^{\text{3}}/\text{s}=160\text{ft}\times \text{y}\times \text{V}\\ V=\frac{900{\text{ft}}^{\text{3}}/\text{s}}{160\text{ft}\times \text{y}}\end{array}$

Substitute $\frac{900{\text{ft}}^{\text{3}}/\text{s}}{160\text{ft}\times \text{y}}$ for $V$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $0.05$ for $Fr$ and $y$ for depth in Equation (I).

$\begin{array}{c}0.03=\frac{\frac{900{\text{ft}}^{\text{3}}/\text{s}}{160\text{ft}\times \text{y}}}{\sqrt{32.2\text{ft}/{\text{s}}^{\text{2}}\times y}}\\ =\frac{0.9912{\text{ft}}^{\text{3}/\text{2}}}{y^{3/2}}\end{array}$

$y^{3/2}=\frac{0.9912{\text{ft}}^{\text{3}/\text{2}}}{0.03}$

$\begin{array}{c}y={\left(\frac{0.9912{\text{ft}}^{\text{3}/\text{2}}}{0.03}\right)}^{2/3}\\ ={\left(33.04{\text{ft}}^{\text{3}/\text{2}}\right)}^{2/3}\\ =10.3\text{ft}\end{array}$

Conclusion:

The water depth one mile upstream of the dam is calculated by flow expression in open channel : $\dot{q}=w\times y\times V$.