Chapter 5, Problem 58P

Water at 20°C is siphoned from a reservoir as shown in Fig. P5-53. For d = 8 cm and D = 16 cm, determine (a) the minimum flow rate that can be achieved without cavitation occurring in the piping system and (b) the maximum elevation of the highest point of the piping system to avoid cavitation. (c) Also, discuss the ways of increasing the maximum elevation of the highest point of the piping system to avoid cavitation.

To determine

(a)

The minimum flow rate in the piping system.

Answer to Problem 58P

The minimum flow rate in the piping system is $0.08{\text{m}}^{\text{3}}/\text{s}$.

Explanation of Solution

Given information:

The diameter of smaller the pipe is $8\text{cm}$, the diameter of the larger pipe is $16\text{cm,}$ and the temperature on which water is introduced from reservoir is $20°\text{C}$.

The figure below shows the different sections in the piping system.

  

Figure-(1)

Write the expression for Bernoulli's equation between section $1$ and $4$.

  $\begin{array}{l}\frac{P_{atm}}{\rho g}+\frac{V_1{}^2}{2g}+Z_1=\frac{P_{atm}}{\rho g}+\frac{V_4{}^2}{2g}+Z_4\\ \frac{V_1{}^2}{2g}+Z_1=\frac{V_4{}^2}{2g}+Z_4\end{array}$   .......(I)

Here, the atmospheric pressure is $P_{atm}$, the density of flowing fluid is $\rho$, acceleration due to gravity is $g$, the velocity at section $1$ is $V_1$, the velocity at section $4$ is $V_4$, the datum head at section $1$ is $Z_1,$ and the datum head at section $4$ is $Z_4$.

Write the expression for discharge of fluid from the pipe.

  $Q=A_4{V}_4$   .......(II)

Here, the area of the section $4$ is $A_4$ and the velocity of the fluid at section $4$ is $V_4$.

Write the expression for the area of the section $4$.

  $A_4=\frac{\pi }{4}{D}^2$  .......(III)

Here, the diameter of the larger pipe is $D$.

Write the expression for velocity of the fluid at section $2$.

  $V_2=\frac{4Q}{\pi {\left(d\right)}^2}$  .......(IV)

Here, the diameter of smaller the pipe is $d$.

Write the expression for Bernoulli's equation between section $1$ and $2$.

  $\frac{P_{atm}}{\rho g}+\frac{V_1{}^2}{2g}+Z_1=\frac{P_2}{\rho g}+\frac{V_2{}^2}{2g}+Z_2$   .......(V)

Here, the datum head at section $2$ is $Z_2$.

Write the expression for minimum flow rate in the pipe.

  $\dot{Q}=A_2{V}_2$   .......(VI)

Here, the area of the section $2$ is $A_2$ and the velocity of the fluid at section $2$ is $V_2$.

Write the expression for area of the section $2$.

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

Calculation:

The velocity at section $1$ is $0\text{m}/\text{s}$.

The line $X-X^{\prime }$ is taken as datum line.

Substitute $0\text{m}/\text{s}$ for $V_1$, $5\text{m}$ for $Z_1$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $0\text{m}$ for $Z_4$ in Equation (I).

  $\begin{array}{l}\frac{0\text{m}/\text{s}}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+5\text{m}=\frac{V_4{}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+0\text{m}\\ 0+5\text{m}=\frac{V_4{}^2}{19.62\text{m}/{\text{s}}^{\text{2}}}+0\text{m}\\ 5\text{m}\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)=V_4{}^2\\ V_4=9.904\text{m}/\text{s}\end{array}$

Substitute $16\text{cm}$ for $D$ in Equation (III).

  $\begin{array}{c}{A}_4=\frac{\pi }{4}{\left(16\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(16\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =0.785{\left(0.16\text{m}\right)}^2\\ =0.02{\text{m}}^2\end{array}$

Substitute $0.02{\text{m}}^2$ for $A_4$

  $9.904\text{m}/\text{s}$ for $V_4$ in Equation (II).

  $\begin{array}{c}Q=0.02{\text{m}}^2\left(9.904\text{m}/\text{s}\right)\\ =0.199{\text{m}}^3/\text{s}\end{array}$

Substitute $8\text{cm}$ for $d$

  $0.199{\text{m}}^3/\text{s}$ for $Q$ in Equation (IV).

  $\begin{array}{c}{V}_2=\frac{4\left(0.199{\text{m}}^3/\text{s}\right)}{\pi {\left(8\text{cm}\right)}^2}\\ =\frac{4\left(0.199{\text{m}}^3/\text{s}\right)}{\pi {\left(8\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =\frac{0.796{\text{m}}^{\text{3}}/\text{s}}{2.01{\text{m}}^2}\\ =25.33\text{m}/\text{s}\end{array}$

Refer to the Table A-3 "Properties of saturated water" to obtain the value of absolute pressure as $2.339\text{KPa}$ at the temperature of $20°\text{C}$.

The density of the water is $998\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $1\text{atm}$ for $P_{\text{atm}}$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0\text{m}/\text{s}$ for $V_1$, $5\text{m}$ for $Z_1$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $2\text{m}$ for $Z_2$ in Equation (V).

  $\begin{array}{l}\left[\begin{array}{l}\frac{1\text{atm}}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{0\text{m}/\text{s}}{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+5\text{m}\end{array}\right]=\left[\begin{array}{l}\frac{P_2}{998\text{kg}/{\text{m}}^{\text{3}}g\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{{\left(25.33{\text{m}/\text{s}}^2\right)}^2}{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+2\text{m}\end{array}\right]\\ \frac{1\text{atm}\left(\frac{101325\text{Pa}}{1\text{atm}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+3\text{m}=\frac{P_2}{998\text{kg}/{\text{m}}^{\text{3}}g\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+32.735\text{m}/{\text{s}}^2\\ \frac{101325\text{Pa}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)}{97804\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}+3\text{m}=\frac{P_2}{97804\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}+32.735\text{m}/{\text{s}}^2\\ P_2=-28641.6\text{N}/{\text{m}}^2\end{array}$

  $\begin{array}{c}{P}_2=-28641.6\text{N}/{\text{m}}^2\left(\frac{1\text{Pa}}{1\text{N}/{\text{m}}^2}\right)\\ =-28641.6\text{Pa}\end{array}$

The value of pressure at section $2$ is less than absolute pressure hence cavitation occurs in the pipe.

Substitute $1\text{atm}$ for $P_{\text{atm}}$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0\text{m}/\text{s}$ for $V_1$, $5\text{m}$ for $Z_1$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.339\text{KPa}$ for $P_2$ and $2\text{m}$ for $Z_2$ in Equation (V).

  $\left[\begin{array}{l}\frac{1\text{atm}}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{0\text{m}/\text{s}}{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+5\text{m}\end{array}\right]=\left[\begin{array}{l}\frac{2.339\text{KPa}\left(\frac{{10}^3\text{Pa}}{1\text{KPa}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}g\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{{\left(V_2\right)}^2}{2\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+2\text{m}\end{array}\right]$

  $\frac{1\text{atm}\left(\frac{101325\text{Pa}}{1\text{atm}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+3\text{m}=\frac{2.339\text{KPa}\left(\frac{{10}^3\text{Pa}}{1\text{KPa}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)}+\frac{{\left(V_2\right)}^2}{19.6\text{m}/{\text{s}}^{\text{2}}}$

  $\frac{101325\text{Pa}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)}{97804\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}+3\text{m}=\frac{2.339\text{KPa}\left(\frac{{10}^3\text{Pa}}{1\text{KPa}}\right)\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)}{97804\text{kg}/{\text{m}}^{\text{2}}{\text{s}}^{\text{2}}}+\frac{{\left(V_2\right)}^2}{19.6\text{m}/{\text{s}}^{\text{2}}}$

  $V_2=16.038\text{m}/\text{s}$

Substitute $8\text{cm}$ for $d$ in Equation (VII).

  $\begin{array}{c}{A}_2=\frac{\pi }{4}{\left(8\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(8\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2\\ =5.024\times {10}^{-3}{\text{m}}^2\end{array}$

Substitute $5.024\times {10}^{-3}{\text{m}}^2$ for $A_2$ and $16.038\text{m}/\text{s}$ for $V_2$ in Equation (VI).

  $\begin{array}{c}\dot{Q}=5.024\times {10}^{-3}{\text{m}}^2\left(16.038\text{m}/\text{s}\right)\\ =0.08{\text{m}}^{\text{3}}/\text{s}\end{array}$

Conclusion:

The minimum flow rate in the piping system is $0.08{\text{m}}^{\text{3}}/\text{s}$.

To determine

(b)

The maximum elevation of the highest point of the piping system.

Answer to Problem 58P

The maximum elevation of the highest point of the piping system is $14.30\text{m}$.

Explanation of Solution

Write the expression for the velocity at section $3$.

  $V_3=\frac{4\dot{Q}}{\pi {\left(D\right)}^2}$   .......(VIII)

Here, the diameter of the larger pipe is $D$.

Write the expression for Bernoulli's equation between section $1$ and $3$.

  $\frac{P_{atm}}{\rho g}+\frac{V_1{}^2}{2g}+Z_1=\frac{P_3}{\rho g}+\frac{V_3{}^2}{2g}+Z_3$   .......(IX)

Here, the datum head at section $3$ is $Z_3$.

Calculation Substitute $16\text{cm}$ for $D$ and $0.08{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$ in Equation (VIII).

  $\begin{array}{c}{V}_3=\frac{4\left(0.08{\text{m}}^{\text{3}}/\text{s}\right)}{\pi {\left(16\text{cm}\right)}^2}\\ =\frac{4\left(0.08{\text{m}}^{\text{3}}/\text{s}\right)}{\pi {\left(16\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}\\ =\frac{0.32{\text{m}}^{\text{3}}/\text{s}}{0.08{\text{m}}^{\text{2}}}\\ =3.97\text{m}/\text{s}\end{array}$

Substitute $1\text{atm}$ for $P_{\text{atm}}$, $998\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0\text{m}/\text{s}$ for $V_1$, $3.97\text{m}/\text{s}$ for $V_3$, $5\text{m}$ for $Z_1$ and $2.339\text{KPa}$ for $P_3$ in Equation (IX).

  $\left[\begin{array}{l}\frac{1\text{atm}}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{0}{2g}+5\text{m}\end{array}\right]=\left[\begin{array}{l}\frac{2.339\text{KPa}}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{{\left(3.97\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+Z_3\end{array}\right]$

  $\left[\frac{1\text{atm}\left(\frac{101325\text{Pa}}{1\text{atm}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+5\text{m}\right]=\left[\begin{array}{l}\frac{2.339\text{KPa}\left(\frac{{10}^3\text{Pa}}{1\text{KPa}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +\frac{{\left(3.97\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+Z_3\end{array}\right]$

  $\left[\frac{101325\text{Pa}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+5\text{m}\right]=\left[\begin{array}{l}\frac{2.339\times {10}^3\text{Pa}\left(\frac{1\text{N}/{\text{m}}^{\text{2}}}{1\text{Pa}}\right)}{998\text{kg}/{\text{m}}^{\text{3}}\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ +0.8033\text{m}/\text{s}+Z_3\end{array}\right]$

  $Z_3=14.30\text{m}$

Conclusion:

The maximum elevation of the highest point of the piping system is $14.30\text{m}$.

To determine

(c)

The way of increasing the maximum elevation of the highest point of the piping system to avoid cavitation.

Explanation of Solution

By increasing the diameter of the siphoned pipe, the elevation of the highest point of the piping system can be increased because the value of pressure at section is greater than absolute pressure, hence cavitation can be avoided in the pipe.