Chapter 3, Problem 3.135P

To determine

The nozzle diameter $D$ for which cavitation began to occur.

Whether the diameter of the nozzle increases or decrease to avoid cavitation.

Answer to Problem 3.135P

The nozzle diameter $D$ for which cavitation began to occur is $0.132\text{ft}$.

To avoid cavitation. in the nozzle, the diameter $D$ should be less than $0.132\text{ft}$.

Explanation of Solution

Given information:

The temperature of the water flowing inside the nozzle is $35°\text{C}$, The diameter at section (1) is $1\text{in}$ . The diameter at section (2) is $3\text{in}$ . The atmospheric pressure is $14.3\text{lbf}/{\text{in}}^{\text{2}}$ . The height of the water level in the left side above the throat is $6\text{ft}$.

Write the expression for the Bernoulli’s equation between the section (1) and section (3) of the given system.

$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_3}{\rho g}+\frac{V_3^2}{2g}+z_3$        …… (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 (3) is $p_3$, the velocity at section (3) is $V_3$, the height of the section (3) from the datum is $z_3$

Since the fluid at section (1) and section (3) is at the same level

$z_1=z_3=0$

Substitute $0$ for $z_1$ and $z_3$ in Equation (I).

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}=\frac{p_3}{\rho g}+\frac{V_3^2}{2g}\\ \frac{V_1^2}{2g}=\left(\frac{p_3-p_1}{\rho g}\right)+\frac{V_3^2}{2g}\end{array}$        …… (II)

Substitute $p_v$ for $p_1$ to avoid cavitation in the nozzle in Equation (II).

$\frac{V_1^2}{2g}=\left(\frac{p_3-p_v}{\rho g}\right)+\frac{V_3^2}{2g}$ ... (III)

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

$A_1=\frac{\pi }{4}{D}_1^2$

Here, the diameter at section (1) is $D_1$.

Write the expression for the area at section (3).

$A_3=\frac{\pi }{4}{D}_3^2$

Here, the diameter at section (3) is $D_3$.

Write the expression for the continuity equation between section (1) and at section (3).

$A_1{V}_1=A_3{V}_3$        …… (IV)

Substitute $\frac{\pi }{4}{D}_1^2$ for $A_1$ and $\frac{\pi }{4}{D}_3^2$ for $A_3$ in Equation (IV).

$\begin{array}{l}\frac{\pi }{4}{D}_1^2{V}_1=\frac{\pi }{4}{D}_3^2{V}_3\\ D_1^2{V}_1=D_3^2{V}_3\end{array}$        …… (V)

Write the expression for Torricelli equation for the velocity at section (3).

$V_3=\sqrt{2gh}$        …… (VI)

Here the acceleration due to gravity is $g$ and height of the tank above the throat is $h$

Calculation:

Substitute $6\text{ft}$ for $h$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (VI).

$\begin{array}{c}{V}_3=\sqrt{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}\times 6\text{ft}}\\ V_3=\sqrt{386}\text{ft}/\text{s}\\ V_3=19.65\text{ft}/\text{s}\end{array}$

Convert the pressure at section (3) from $\frac{\text{lbf}}{{\text{in}}^{\text{2}}}$ to $\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}$.

$\begin{array}{c}{p}_3=\left(14.3\frac{\text{lbf}}{{\text{in}}^{\text{2}}}\right)\\ =\left(14.3\frac{\text{lbf}}{{\text{in}}^{\text{2}}}\right)\left(\frac{144{\text{in}}^{\text{2}}}{1{\text{ft}}^{\text{2}}}\right)\\ =2059.2\text{lbf}/{\text{ft}}^{\text{2}}\end{array}$

Refer to the property table to obtain the vapour pressure of water at $35°\text{C}$ as $5.8085\text{kPa}$.

Convert the vapour pressure from $\text{kPa}$ to $\frac{\text{lbf}}{{\text{ft}}^{\text{2}}}$.

$\begin{array}{c}{p}_v=5.8085\text{kPa}\\ =\left(\text{5}.8085\text{kPa}\right)\left(\frac{20.88\text{lb}/{\text{ft}}^{\text{2}}}{1\text{kPa}}\right)\\ \approx 121.28\text{lbf}/{\text{ft}}^{\text{2}}\end{array}$

Substitute $2059.2\text{lbf}/{\text{ft}}^{\text{2}}$ for $p_3$, $121.28\text{lbf}/{\text{ft}}^{\text{2}}$ for $p_v$, $1.93\text{lb}/{\text{ft}}^{\text{3}}$ for $\rho$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$, $19.65\text{ft}/\text{s}$ for $V_3$ in Equation (III).

$\begin{array}{l}\frac{V_1^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)}=\left(\frac{2059.2\text{lbf}/{\text{ft}}^{\text{2}}-121.28\text{lbf}/{\text{ft}}^{\text{2}}}{\left(1.93\text{lb}/{\text{ft}}^{\text{3}}\right)\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)}\right)+\frac{{\left(19.65\text{ft}/\text{s}\right)}^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)}\\ \frac{V_1^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)}=\left(\frac{1937.92\text{lbf}/{\text{ft}}^{\text{2}}}{62.146\text{lb}/{\text{ft}}^{\text{2}}{\text{s}}^{\text{2}}}\right)+5.99\text{ft}\\ \frac{V_1^2}{2\left(32.2\text{ft}/{\text{s}}^{\text{2}}\right)}=37.17\text{ft}\\ V_1^2=2393.748{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}\end{array}$

$\begin{array}{l}{V}_1^2=\sqrt{2393.748{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ V_1=48.92\text{ft}/\text{s}\end{array}$

Substitute $48.92\text{ft}/\text{s}$ for $V_1$, $19.65\text{ft}/\text{s}$ for $V_3$ in Equation (V).

$\begin{array}{l}{\left(1\text{in}\right)}^2\left(48.92\text{ft}/\text{s}\right)=D_3^2\left(19.65\text{ft}/\text{s}\right)\\ {\left(\left(1\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2\left(48.92\text{ft}/\text{s}\right)=D_3^2\left(19.65\text{ft}/\text{s}\right)\\ \frac{0.339{\text{ft}}^{\text{3}}/\text{s}}{\left(19.65\text{ft}/\text{s}\right)}=D_3^2\\ 0.001725{\text{ft}}^{\text{2}}=D_3^2\end{array}$

$\begin{array}{l}\sqrt{0.001725{\text{ft}}^{\text{2}}}=D_3\\ D_3=0.132\text{ft}\end{array}$

Conclusion:

The nozzle diameter $D$ for which cavitation began to occur is $0.132\text{ft}$.

To avoid cavitation in the nozzle diameter $D$ should be less than $0.132\text{ft}$ ..