Chapter 13, Problem 128P

To determine

The velocity before the jump.

The velocity after the jump.

The mechanical power dissipated.

Answer to Problem 128P

The velocity before the jump is $14.132\text{m}/\text{s}$.

The velocity after the jump is $1.978\text{m}/\text{s}$.

The mechanical power dissipated is $559.5787\text{kW}$.

Explanation of Solution

Given information:

The flow depth before the jump is $0.7\text{m}$, the flow depth after the jump is $5\text{m}$, the flow is steady and uniform and the end effect is neglected.

Write the expression for flow depth.

  $y_2=0.5y_1\left(-1+\sqrt{1+8{\text{Fr}}_1^2}\right)$....... (I)

Here, Froude number is ${\text{Fr}}_{\text{1}}$, height before jump is $y_1$, height after the jump is $y_2$.

Write the expression for Froude number.

  $\text{Fr}=\frac{V_1}{\sqrt{g{y}_1}}$...... (II)

Here, acceleration due to gravity is $g$ and velocity before jump is $V_1$.

Write the expression for conservation of mass.

  $y_1{V}_1=y_2{V}_2$...... (III)

Here, velocity after jump is $V_2$.

Write the expression for volume flow rate.

  $\dot{V}=V_{1}b{y}_1$...... (IV)

Write the expression for head loss during the jump.

  $h_{\text{L}}=y_1-y_2+\frac{V_1^2-V_2^2}{2g}$...... (V)

Write the expression for mass flow rate.

  $\dot{m}=\rho \dot{V}$...... (VI)

Here, density of water is $\rho$.

Write the expression for energy dissipated.

  ${\dot{E}}_{\text{dissipated}}=\dot{m}g{h}_{\text{L}}$...... (VII)

Calculation:

Substitute $0.7\text{m}$ for $y_1$, $5\text{m}$ for $y_2$ in Equation (I).

  $\begin{array}{l}5\text{m}=0.5\left(0.7\text{m}\right)\left(-1+\sqrt{1+8{\text{Fr}}_1^2}\right)\\ \left(-1+\sqrt{1+8{\text{Fr}}_1^2}\right)=\frac{10\text{m}}{0.7\text{m}}\\ \sqrt{1+8{\text{Fr}}_1^2}=15.286\\ 1+8{\text{Fr}}_1^2=233.662\\ {\text{Fr}}_1^2=29.083\end{array}$

  $\text{Fr}=5.393$

Substitute $0.7\text{m}$ for $y_1$, $5.393$ for $\text{Fr}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (II).

  $\begin{array}{l}5.393=\frac{V_1}{\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\times 0.7\text{m}}}\\ V_1=\left(5.393\right)\sqrt{9.81\text{m}/{\text{s}}^{\text{2}}\times 0.7\text{m}}\\ V_1=14.132\text{m}/\text{s}\end{array}$

Substitute $0.7\text{m}$ for $y_1$, $5\text{m}$ for $y_2$ and $14.132\text{m}/\text{s}$ for $V_1$ in Equation (III).

  $\begin{array}{l}\left(0.7\text{m}\right)\left(14.132\text{m}/\text{s}\right)=\left(5\text{m}\right)V_2\\ V_2=\frac{0.7\text{m}}{5\text{m}}\times 14.132\text{m}/\text{s}\\ V_2=1.978\text{m}/\text{s}\end{array}$

Substitute $0.7\text{m}$ for $y_1$, $14.132\text{m}/\text{s}$ for $V_1$ and $1\text{m}$ for $b$ in Equation (IV).

  $\begin{array}{c}\dot{V}=\left(14.132\text{m}/\text{s}\right)\left(1\text{m}\right)\left(0.7\text{m}\right)\\ =14.132\text{m}/\text{s}\times 0.7{\text{m}}^2\\ =9.892{\text{m}}^3/\text{s}\end{array}$

Substitute $0.7\text{m}$ for $y_1$, $5\text{m}$ for $y_2$ and $14.132\text{m}/\text{s}$ for $V_1$, $1.978\text{m}/\text{s}$ for $V_2$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (V).

  $\begin{array}{c}{h}_{\text{L}}=\left(0.7\text{m}\right)-\left(5\text{m}\right)+\frac{{\left(14.132\text{m}/\text{s}\right)}^2-{\left(1.978\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =-4.3\text{m}+\frac{{\left(14.132\text{m}/\text{s}\right)}^2-{\left(1.978\text{m}/\text{s}\right)}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ =5.778\text{m}\end{array}$

Refer to table "Liquid water at $20°\text{C}$ " to obtain density of water as $998\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $9.892{\text{m}}^3/\text{s}$ for $\dot{V}$ in Equation (VI).

  $\begin{array}{c}\dot{m}=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.892{\text{m}}^3/\text{s}\right)\\ =9872.216\text{kg}/\text{s}\end{array}$

Substitute $9872.216\text{kg}/s$ for $\dot{m}$, $5.68\text{m}$ for $h_{\text{L}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VII).

  $\begin{array}{c}{\dot{E}}_{\text{dissipated}}=\left(9872.216\text{kg}/s\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(5.778\text{m}\right)\\ =559.5787\times {10}^3{\text{kg}\cdot {\text{m}}^2/\text{s}}^3\times \frac{1\text{W}}{1{\text{kg}\cdot {\text{m}}^2/s}^3}\\ =559.5787\times {10}^3\text{W}\times \frac{1\text{kW}}{1000\text{W}}\\ =559.5787\text{kW}\end{array}$

Conclusion:

The velocity before the jump is $14.132\text{m}/\text{s}$.

The velocity after the jump is $1.978\text{m}/\text{s}$.

The mechanical power dissipated is $559.5787\text{kW}$.