Chapter 3, Problem 3.114P

To determine

(a)

The velocity at section (1).

The velocity at section (2).

Answer to Problem 3.114P

The velocity at section (1) is $0.896\text{m}/\text{s}$.

The velocity at section (2) is $9.96\text{m}/\text{s}$.

Explanation of Solution

Given information:

The force required to hold the plate steady is $70\text{N}$, diameter of the nozzle at section (1) is $10\text{cm}$ and diameter of nozzle at section (2) is $3\text{cm}$.

Concept Used:

The below figure shows the two sections of the nozzle as section (1) and section (2).

Figure (1)

Write the expression for the force with which the jet strikes a plate.

$\begin{array}{l}F={\rho }_W{A}_2{V}_2{}^2\\ V_2{}^2=\frac{F}{{\rho }_W{A}_2}\\ V_2=\sqrt{\frac{F}{{\rho }_W{A}_2}}\end{array}$ .... (I)

Here, the jet force on the plate is $F$, density of water is ${\rho }_W$, cross section area at point (2) is $A_2$, and the velocity of jet at section (2) is $V_2$.

Write the expression for cross section area in terms of nozzle diameter for section (1).

$A_1=\frac{\pi }{4}{d}_1{}^2$ .... (II)

Write the expression for cross section area in terms of nozzle diameter for section (2).

$A_2=\frac{\pi }{4}{d}_2{}^2$ .... (III)

Here, the diameter of the nozzle at section (1) is $d_1$, area of the nozzle at section (1) is $A_1$, and the diameter of the nozzle at section (2) is $d_2$.

Write the expression for momentum equation at section (1) and section (2).

$\begin{array}{l}{A}_1{V}_1=A_2{V}_2\\ V_1=\frac{A_2{V}_2}{A_1}\end{array}$

$$ .... (IV)

Here, the cross-section area of the nozzle at section (1) is $A_1$, and velocity of water at section (1) is $V_1$.

Calculation:

Substitute $10\text{cm}$ for $d_1$ in Equation (II).

$\begin{array}{c}{A}_1=\frac{\pi }{4}\times {\left(10\text{cm}\right)}^2\\ =\left(\frac{3.1415\times 100{\text{cm}}^{\text{2}}}{4}\right)\left(\frac{1{\text{m}}^{\text{2}}}{10000{\text{cm}}^2}\right)\\ =\frac{314.15{\text{m}}^{\text{2}}}{40000}\\ =7.853\times {10}^{-3}{\text{m}}^2\end{array}$

Substitute $3\text{cm}$ for $d_2$ in Equation (III).

$\begin{array}{c}{A}_2=\frac{\pi }{4}\times {\left(3\text{cm}\right)}^2\\ =\left(\frac{3.1415\times 9{\text{cm}}^{\text{2}}}{4}\right)\left(\frac{1{\text{m}}^{\text{2}}}{10000{\text{cm}}^2}\right)\\ =\frac{28.2735{\text{m}}^{\text{2}}}{40000}\\ =7.068\times {10}^{-4}{\text{m}}^2\end{array}$

Substitute $7.068\times {10}^{-4}{\text{m}}^2$ for $A_2$, $70\text{N}$ for $F$ and $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_W$ in Equation (I).

$\begin{array}{c}{V}_2=\sqrt{\frac{70\text{N}}{\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(7.068\times {10}^{-4}{\text{m}}^{\text{2}}\right)}}\\ =\sqrt{\left(\frac{70\text{N}}{0.7053\text{kg}/\text{m}}\right)\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)}\\ =\sqrt{99.25{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ \approx 9.96\text{m}/\text{s}\end{array}$

Thus, the velocity at section (2) is $9.96\text{m}/\text{s}$.

Substitute $7.068\times {10}^{-4}{\text{m}}^2$ for $A_2$, $7.853\times {10}^{-3}{\text{m}}^2$ for $A_1$ and $9.96\text{m}/\text{s}$ for $V_2$ Equation (III).

$\begin{array}{c}{V}_1=\frac{\left(7.068\times {10}^{-4}{\text{m}}^2\right)\left(9.96\text{m}/\text{s}\right)}{7.853\times {10}^{-3}{\text{m}}^2}\\ =\frac{70.4{\text{m}}^{\text{3}}/\text{s}}{78.53{\text{m}}^2}\\ =0.896\text{m}/\text{s}\end{array}$

Conclusion:

Thus, the velocity at section (1) is $0.896\text{m}/\text{s}$.

Thus, the velocity at section (2) is $9.96\text{m}/\text{s}$.

To determine

(b)

The mercury manometer reading.

Answer to Problem 3.114P

The mercury manometer reading is $0.398\text{m}$.

Explanation of Solution

Concept Used:

Write the expression for Bernoulli equation at section (1) and section (2).

$\begin{array}{l}{P}_1+\frac{{\rho }_W{V}_1{}^2}{2}=P_2+\frac{{\rho }_W{V}_2{}^2}{2}\\ P_1-P_2=\frac{{\rho }_W{V}_2{}^2}{2}-\frac{{\rho }_W{V}_1{}^2}{2}\\ P_1-P_2=\frac{{\rho }_W}{2}\left(V_2{}^2-V_1{}^2\right)\end{array}$ .... (IV)

Here, pressure at section (1) is $P_1$, and the pressure at section (2) is $P_2$.

Write the expression for continuity equation.

$\begin{array}{l}{P}_1+{\rho }_{W}g{h}_1=P_2+{\rho }_{W}g{h}_2+{\rho }_{Hg}g\left(h_1-h_2\right)\\ \left(P_1-P_2\right)+{\rho }_{W}g\left(h_1-h_2\right)={\rho }_{Hg}g\left(h_1-h_2\right)\\ \left(P_1-P_2\right)=g\left(h_1-h_2\right)\left({\rho }_{Hg}-{\rho }_W\right)\\ \left(h_1-h_2\right)=\frac{\left(P_1-P_2\right)}{\left({\rho }_{Hg}-{\rho }_W\right)g}\end{array}$ .... (V)

Here, height of lower mercury level from the axis of the nozzle is $h_1$, the height of the upper mercury level from the axis of the nozzle is $h_2$, density of water is ${\rho }_W$ and density of mercury is ${\rho }_{Hg}$.

Calculation:

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_W$, $9.96\text{m}/\text{s}$ for $V_2$ and $0.896\text{m}/\text{s}$ for $V_1$ in Equation (IV).

$\begin{array}{c}{P}_1-P_2=\frac{998\text{kg}/{\text{m}}^{\text{3}}}{2}\times \left[{\left(9.96\text{m}/\text{s}\right)}^2-{\left(0.896\text{m}/\text{s}\right)}^2\right]\\ =499\text{kg}/{\text{m}}^{\text{3}}\times 98.398{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}\\ \approx 49101\text{Pa}\end{array}$

Substitute $49101\text{Pa}$ for $P_1-P_2$, $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_W$ and $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{Hg}$ in Equation(V).

$\begin{array}{c}{h}_1-h_2=\frac{49101\text{Pa}}{\left(13550\text{kg}/{\text{m}}^{\text{3}}-998\text{kg}/{\text{m}}^{\text{3}}\right)\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(\frac{49101\text{Pa}}{123135\text{kg}/{\text{m}}^{\text{2}}\cdot {\text{s}}^{\text{2}}}\right)\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)\left(\frac{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{1\text{N}}\right)\\ =0.398\text{m}\end{array}$

Conclusion:

Thus, the mercury manometer reading is $0.398\text{m}$.