Chapter 3, Problem 3.139P

To determine

(a)

The velocity $V_2$.

Answer to Problem 3.139P

The velocity $V_2$ is $9.28\text{m}/\text{s}$.

Explanation of Solution

Given information:

The flow is uniform and hydrostatic. The height of the water level at section (1) is $5\text{m}$ and the jet height at section (2) is $0.7\text{m}$ . The losses are neglected.

The Figure-(1) shown below defines the section (1) and section (2) in the given system.

Figure-(1)

Write the expression for the Bernoulli’s equation between the section (1) and section (2) of the given system.

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ \left(\frac{p_1-p_2}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=0\end{array}$ …… (I)

Here, the pressure at section (1) is $p_1$, the velocity at section (1) is $V_1$, the height of the section (1) from the datum is $z_1$, the pressure at section (2) is $p_2$, the velocity at section (2) is $V_2$, the height of the section (2) from the datum is $z_2$, the density of water is $\rho$, and the acceleration due to gravity is $g$.

Since the pressure at the section (1) and section (2) is atmospheric, so the difference between the pressures is zero.

$p_1-p_2=0$

Substitute $0$ for $p_1-p_2$ in Equation (I).

$\begin{array}{l}\left(\frac{0}{\rho g}\right)+\left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=0\\ \left(\frac{V_1^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=0\end{array}$ …… (II)

Since the flow is uniform at both the sections, so the flow remains conserved.

Write the expression for the conservation of flow.

$\begin{array}{l}{V}_1{h}_1=V_2{h}_2\\ V_1=\frac{h_2}{h_1}{V}_2\end{array}$

Substitute $\frac{h_2}{h_1}{V}_2$ for $V_1$ in Equation (II).

$\begin{array}{l}\left(\frac{{\left(\frac{h_2}{h_1}{V}_2\right)}^2}{2g}-\frac{V_2^2}{2g}\right)+\left(z_1-z_2\right)=0\\ \frac{V_2^2}{2g}\left\{{\left(\frac{h_2}{h_1}\right)}^2-1\right\}+\left(z_1-z_2\right)=0\end{array}$ …… (III)

Calculation:

Substitute $5\text{m}$ for $h_1$, $0.7\text{m}$ for $h_2$, $5\text{m}$ for $z_1$, $0.7\text{m}$ for $z_2$ and $9.81\text{m}/{\text{s}}^2$ for $g$ in Equation (III).

$\begin{array}{l}\frac{V_2^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\left\{{\left(\frac{0.7\text{m}}{5\text{m}}\right)}^2-1\right\}+\left\{\left(5\text{m}\right)-\left(0.7\text{m}\right)\right\}=0\\ -\left(0.0499{\text{s}}^{\text{2}}/\text{m}\right)V_2^2+\left(4.3\text{m}\right)=0\\ V_2=\sqrt{\frac{\left(4.3\text{m}\right)}{\left(0.0499{\text{s}}^{\text{2}}/\text{m}\right)}}\\ V_2=9.28\text{m}/\text{s}\end{array}$

Conclusion:

The velocity $V_2$ is $9.28\text{m}/\text{s}$.

To determine

(b)

The force per unit width of the water.

Answer to Problem 3.139P

The force per unit width of the water is $68248.19\text{N}/\text{m}$.

Explanation of Solution

Given information:

The flow is uniform and hydrostatic. The height of the water level at section (1) is $5\text{m}$ and the jet height at section (2) is $0.7\text{m}$ . The losses are neglected.

Write the expression for the force momentum equation.

$F-\frac{1}{2}\left(p_2{A}_2-p_1{A}_1\right)=\dot{m}\left(V_1-V_2\right)$ …… (IV)

Here, the force per unit width is $F$, the mass flow rate is $\dot{m}$, the area at section (1) is $A_1$ and the area at section (2) is $A_2$.

Write the expression for the area at section (1) for unit width.

$A_1=h_1$

Write the expression for the area at section (2) for unit width.

$A_2=h_2$

Write the expression for the hydrostatic pressure at section (1).

$p_1=\rho g{h}_1$

Write the expression for the hydrostatic pressure at section (2).

$p_2=\rho g{h}_2$

Write the expression for the mass flow rate.

$\dot{m}=\rho A_1{V}_1$ …… (V)

Substitute $h_1$ for $A_1$ in Equation (V).

$\dot{m}=\rho h_1{V}_1$

Substitute $h_1$ for $A_1$, $h_2$ for $A_2$, $\rho g{h}_1$ for $p_1$, $\rho g{h}_2$ for $p_2$ and $\rho h_1{V}_1$ for $\dot{m}$ in Equation (I).

$\begin{array}{l}F-\frac{1}{2}\left\{\left(\rho g{h}_2\right)h_2-\left(\rho g{h}_1\right)h_1\right\}=\left(\rho h_1{V}_1\right)\left(V_1-V_2\right)\\ F=\left(\rho h_1{V}_1^2-\rho h_1{V}_1{V}_2\right)+\frac{1}{2}\left\{\left(\rho g{h}_2^2\right)-\left(\rho g{h}_1^2\right)\right\}\\ F=\left\{\rho h_1\left(V_1^2-V_1{V}_2\right)\right\}+\frac{1}{2}\left\{\rho g\left(h_2^2-h_1^2\right)\right\}\end{array}$ …… (VI)

Write the expression for the conservation of flow.

$\begin{array}{l}{V}_1{h}_1=V_2{h}_2\\ V_1=\frac{h_2}{h_1}{V}_2\end{array}$ …… (VII)

Calculation:

Substitute $9.28\text{m}/\text{s}$ for $V_2$, $5\text{m}$ for $h_1$ and $0.7\text{m}$ for $h_2$ in Equation (VII).

$\begin{array}{c}{V}_1=\left(\frac{0.7\text{m}}{5\text{m}}\right)\left(9.28\text{m}/\text{s}\right)\\ =0.14\times \left(9.28\text{m}/\text{s}\right)\\ =1.299\text{m}/\text{s}\end{array}$

Substitute $9.28\text{m}/\text{s}$ for $V_2$, $1.299\text{m}/\text{s}$ for $V_1$, $5\text{m}$ for $h_1$, $0.7\text{m}$ for $h_2$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $998\text{kg}/{\text{m}}^3$ for $\rho$ in Equation (VI).

$\begin{array}{c}F=\left[\begin{array}{l}\left[\left(998\text{kg}/{\text{m}}^3\right)\left(5\text{m}\right)\left\{{\left(1.299\text{m}/\text{s}\right)}^2-\left(1.299\text{m}/\text{s}\right)\left(9.28\text{m}/\text{s}\right)\right\}\right]\\ +\frac{1}{2}\times \left[\left(998\text{kg}/{\text{m}}^3\right)\left(9.81\text{m}/{\text{s}}^2\right)\left\{{\left(5\text{m}\right)}^2-{\left(0.7\text{m}\right)}^2\right\}\right]\end{array}\right]\\ =\left(-51732.92\text{N}/\text{m}\right)+\left(119981.11\text{N}/\text{m}\right)\\ =68248.19\text{N}/\text{m}\end{array}$

Conclusion:

The force per unit width of the water is $68248.19\text{N}/\text{m}$.