Chapter 12, Problem 31P

(a)

To determine

The minimum flow rate required for not occurring the cavitation in the given piping system.

Explanation of Solution

Given:

The temperature (T) of the water in the siphon is $20°\text{C}$.

The diameter (d) of the small pipe is 10 cm.

The diameter (D) of the large pipe is 16 cm.

Calculation:

The vapor pressure $\left(P_V\right)$ of water at $T=20°C$ is 2.338 kPa.

Consider the atmospheric pressure as 101.325 kPa.

Consider the density $\left(\rho \right)$ of the water as $1,000{\text{kg/m}}^3$.

Consider the acceleration due to gravity (g) as $9.81{\text{m/s}}^2$.

Write the Bernoulli’s equation between the points 1 and 4 as follows;

  $\begin{array}{l}\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_4}{\rho g}+\frac{V_4^2}{2g}+z_4\\ \frac{P_{atm}}{\rho g}+\frac{0^2}{2g}+z_1=\frac{P_{atm}}{\rho g}+\frac{V_4^2}{2g}+0\\ z_1=\frac{V_4^2}{2g}\\ V_4=\sqrt{2g{z}_1}\end{array}$

Substitute $g=9.81{\text{m/s}}^2$ and $z_1=\left(1+4\right)\text{m}$ in the above equation.

  $\begin{array}{c}{V}_4=\sqrt{2\times 9.81\times \left(1+4\right)}\\ =9.905\text{m/s}\\ V_D=V_4=9.905\text{m/s}\end{array}$

Write the continuity equation for the small and large pipes as follows;

  $\begin{array}{l}{A}_d{V}_d=A_D{V}_D\\ \frac{\pi d^2}{4}\times V_d=\frac{\pi D^2}{4}\times V_D\\ d^2{V}_d=D^2{V}_D\end{array}$        (I)

Substitute $d=10\text{cm}$, $D=16\text{cm}$, and $V_D=9.905\text{m/s}$ in the equation (I).

  $\begin{array}{l}{10}^2\times V_d={16}^2\times 9.905\\ V_d=25.357\text{m/s}\\ V_2=V_d=25.357\text{m/s}\end{array}$

Check the possibility of the velocity 25.357 m/s.

Write the Bernoulli’s equation between the points 1 and 2 as follows;

  $\begin{array}{l}\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ \frac{101.325\text{kPa}\times \frac{\text{1,000}\text{Pa}}{\text{1}\text{kPa}}}{1,000\times 9.81}+\frac{0^2}{2\times 9.81}+\left(1+4\right)=\frac{P_2}{\rho g}+\frac{{25.357}^2}{2\times 9.81}+2\\ 10.329+5=\frac{P_2}{\rho g}+32.772+2\\ \frac{P_2}{\rho g}=-19.443<0\end{array}$

Therefore, the velocity $25.357\text{m/s}$ cannot be maximum.

Use the Bernoulli’s equation for the points 1 and 2 to find the maximum velocity. Consider the vapor pressure at point 2.

Write the Bernoulli’s equation between the points 1 and 2 as follows;

  $\begin{array}{l}\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+z_2\\ \frac{101.325\text{kPa}\times \frac{\text{1,000}\text{Pa}}{\text{1}\text{kPa}}}{1,000\times 9.81}+\frac{0^2}{2\times 9.81}+\left(1+4\right)=\frac{2.338\text{kPa}\times \frac{\text{1,000}\text{Pa}}{\text{1}\text{kPa}}}{1,000\times 9.81}+\frac{V_{\max }^2}{2\times 9.81}+2\\ 10.329+5=0.238+\frac{V_{\max }^2}{2\times 9.81}+2\\ V_d=V_{\max }=16\text{m/s}\end{array}$

Find the maximum flowrate $\left(\dot{V}\right)$  using the relation;

  $\begin{array}{c}\dot{V}=A_d{V}_d\\ =\frac{\pi d^2}{4}{V}_d\\ =\frac{\pi \times {\left(10\text{cm}\times \frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}^2}{4}\times 16\\ =0.125{\text{m}}^{\text{3}}\text{/s}\end{array}$

Thus, the minimum flow rate required for not occurring the cavitation in the given piping system is $\underset{\_}{0.125{\text{m}}^{\text{3}}\text{/s}}$.

(b)

To determine

The maximum elevation of the highest point to avoid cavitation.

Explanation of Solution

Find the velocity in the large pipe using the Equation (I).

Substitute $d=10\text{cm}$, $D=16\text{cm}$, and $V_d=16\text{m/s}$ in the equation (I).

  $\begin{array}{l}{10}^2\times 16={16}^2\times V_D\\ V_D=6.25\text{m/s}\\ V_3=V_D=6.25\text{m/s}\end{array}$

The maximum pressure occurs at point 3, and the atmospheric pressure at point 3 is $P_3=2.338\text{kPa}$.

Write the Bernoulli’s equation between the points 1 and 3 as follows;

  $\begin{array}{l}\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{P_3}{\rho g}+\frac{V_3^2}{2g}+z_3\\ \frac{101.325\text{kPa}\times \frac{\text{1,000}\text{Pa}}{\text{1}\text{kPa}}}{1,000\times 9.81}+\frac{0^2}{2\times 9.81}+\left(1+4\right)=\frac{2.338\text{kPa}\times \frac{\text{1,000}\text{Pa}}{\text{1}\text{kPa}}}{1,000\times 9.81}+\frac{{6.25}^2}{2\times 9.81}+z_3\\ 10.329+5=0.238+2+z_3\\ z_{\max }=z_3=13\text{m}\end{array}$

Thus, the maximum elevation of the highest point to avoid cavitation is $\underset{\_}{13\text{m}}$.