Chapter 14, Problem 61P

Repeat Prob. 14-60. but at a water temperature of 80°C. Repeat for 90°C. Discuss.

To determine

The maximum volume flow rate at which cavitation occur.

Answer to Problem 61P

The maximum volume flow rate at which cavitation occur is $28\text{Lpm}$.

Explanation of Solution

Given information:

The temperature of the water is $25°\text{C}$, the elevation difference is $2.2\text{m}$, the length of the pipe is $2.8\text{m}$, the internal diameter of the pipe is $24\text{mm}$, the required net positive suction head is $NPS{H}_{required}=2.2\text{m}+\left[0.0013{\text{m}/\left(\text{Lpm}\right)}^2\right]{\dot{V}}^2$, the minor loss coefficient due to sharp edge is $0.85$ and the minor loss coefficient due to the smooth edge is $0.3$.

Write the expression for the required head using the energy balance equation.

  $\frac{P_1}{\rho g}+\frac{{\alpha }_1{V}_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{{\alpha }_2{V}_2^2}{2g}+z_2+h_{L,total}$....... (I)

Here, the initial pressure is $P_1$, the final pressure is $P_2$, the initial velocity is $V_1$, the final velocity is $V_2$, the density of the water is $\rho$, the acceleration due to gravity is $g$, the potential head of water at section 1 is $z_1$, the potential head of the water at section 2 is $z_2$, the total loss is $h_{Ltotal}$.

Write expression for the available net positive suction head at the pump inlet.

  $NPS{H}_{available}=\left(\frac{P_{atm}-P_s}{\rho g}-\left(z_2-z_1\right)-\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\right)$....... (II)

Here, the saturation pressure is $P_s$, the friction factor is $f$ and the length of the pipe is $L$.

Write the expression for the minor losses.

  ${\displaystyle \sum K_L=K_{L,1}+K_{L,2}}$..... (III)

Here, the minor loss coefficient due to sharp edge is $K_{L,1}$ and the minor loss coefficient due to the smooth edge is $K_{L,2}$.

Write the expression for the velocity of the water.

  $V=\frac{4\dot{V}}{\pi D^2}$....... (IV)

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho VD}{\mu }$....... (V)

Here, the density of the water is $\rho$ and the dynamitic viscosity is $\mu$.

Write the expression for the minor losses.

Write the expression for the required net positive suction head.

  $NPS{H}_{required}=2.2\text{m}+\left[0.0013{\text{m}/\left(\text{Lpm}\right)}^2\right]{\dot{V}}^2$....... (VI)

Write the expression for the friction factor.

  $\frac{1}{\sqrt{f}}=-1.8\log \left[\frac{6.9}{\mathrm{Re}}\right]$....... (VII)

Here, the friction factor is $f$.

Calculation:

Refer to the Table-A-7E, "Properties of saturated liquid" to obtain the density of the water as $971.8\text{kg}/{\text{m}}^{\text{3}}$, the dynamic viscosity of water as $0.355\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$, the vapor pressure as $47.39\text{kPa}$ at temperature $T=80°\text{C}$.

Substitute $0.85$ for $K_{L,1}$ and $0.3$ for $K_{L,2}$ in Equation (III).

  $\begin{array}{c}{\displaystyle \sum K_L}=0.85+0.3\\ =1.15\end{array}$

Substitute $971.8\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $101.3\text{kPa}$ for $P_{atm}$, $2.8\text{m}$ for $L$, $24\text{mm}$ for $D$, $47.39\text{kPa}$ for $P_s$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.2\text{m}$ for $\left(z_2-z_1\right)$ and $1.15$ for ${\displaystyle \sum K_L}$ in Equation (II).

  $\begin{array}{c}NPS{H}_{available}=\left[\left(\begin{array}{l}\frac{101.3\text{kPa}-47.39\text{kPa}}{\left(971.8\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}-\left(2.2\text{m}\right)-\\ \left(f\frac{2.8\text{m}}{24\text{mm}}+1.15\right)\frac{V^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\end{array}\right)\right]\\ =\left[\left(\begin{array}{l}\frac{53.91\text{kPa}}{\left(9533.358\text{kg}/{\text{m}}^2{\text{s}}^2\right)}-\left(2.2\text{m}\right)-\\ \left(f\frac{2.8\text{m}}{24\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}+1.15\right)\frac{V^2}{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)}\end{array}\right)\right]\\ =\left[\left(\begin{array}{l}\frac{53.91\text{kPa}}{\left(9533.358\text{kg}/{\text{m}}^2{\text{s}}^2\right)}-\left(2.2\text{m}\right)-\\ \left(f\left(1166.66\right)+1.15\right)\frac{V^2}{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)}\end{array}\right)\right]\\ =3.454\text{m}-\left(5.946f+0.0586\right){\text{s}}^{\text{2}}/\text{m}\times V^2\end{array}$....... (VIII)

Use hit and trial method to obtain the volume flow rate.

Iteration 1 consider the volume flow rate as $20\text{Lpm}$.

Substitute $20\text{Lpm}$ for $\dot{V}$ and $24\text{mm}$ for $D$ in Equation (IV).

  $\begin{array}{c}V=\frac{4\left(20\text{Lpm}\right)}{\pi {\left(24\text{mm}\right)}^2}\\ =\frac{4\times 20\text{Lpm}\left(\frac{\text{1}\text{L}}{1\min }\times \frac{1{\text{m}}^3}{1000\text{L}}\times \frac{1\min }{60\sec }\right)}{\pi {\left(24\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2}\\ =\frac{1.33\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}}{1.8095\times {10}^{-3}{\text{m}}^2}\\ =0.7368\text{m}/\text{s}\end{array}$

Substitute $0.7368\text{m}/\text{s}$ for $V$, $0.355\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ and $24\text{mm}$ for $D$ in Equation (V).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(0.7368\text{m}/\text{s}\right)\left(971.8\text{kg}/{\text{m}}^3\right)\left(24\text{mm}\right)}{0.355\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{\left(0.7368\text{m}/\text{s}\right)\left(971.8\text{kg}/{\text{m}}^3\right)\left(24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{0.355\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{17.6301}{0.355\times {10}^{-3}}\\ =48407.137\end{array}$

Substitute $48407.137$ for $\mathrm{Re}$ in Equation (VII).

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-1.8\log \left[\frac{6.9}{48407.137}\right]\\ \frac{1}{\sqrt{f}}=-1.8\log \left[0.0001425\right]\\ \sqrt{f}=0.14445\\ f=0.02086\end{array}$

Substitute $0.02086$ for $f$ and $0.7368\text{m}/\text{s}$ for $V$ in Equation (VIII).

  $\begin{array}{c}NPS{H}_{available}=3.454\text{m}-\left(5.946\left(0.02086\right)+0.0586\right){\text{s}}^{\text{2}}/\text{m}\times {\left(0.7368\text{m}/\text{s}\right)}^2\\ =3.454\text{m}-\left(0.09138\text{m}+0.043176\text{m}\right)\\ =3.454\text{m}-0.13455648\text{m}\\ =3.31944\text{m}\end{array}$

Substitute $20\text{Lpm}$ for $\dot{V}$ in Equation (VI).

  $\begin{array}{c}NPS{H}_{required}=2.2\text{m}+\left[0.0013{\text{m}/\left(\text{Lpm}\right)}^2\right]{\left(20\text{Lpm}\right)}^2\\ =2.2\text{m}+\left[0.0013{\text{m}/\left(\text{Lpm}\right)}^2\right]\left(400{\left(\text{Lpm}\right)}^2\right)\\ =2.2\text{m}+0.52\text{m}\\ =\text{2}\text{.72}\text{m}\end{array}$

Since, the available net positive suction head is greater than the required net positive suction head hence, cavitation does not occur at this flow rate.

The different iteration are shown in the below Table.

S.No	$\dot{V}\left(\text{Lpm}\right)$	$NPS{H}_{available}$	$NPS{H}_{required}$
1	$20$	$3.454$	$2.72$
2	$30$	$3.24$	$3.37$
3	$40$	$3.09$	$4.28$
4	$50$	$2.9094$	$5.45$
5	$60$	$2.687$	$6.88$
6	$70$	$2.43$	$8.57$
7	$80$	$2.139$	$10.52$

Draw the plot between the flow rate and available net positive suction head and the required net positive suction head.

Figure-(1)

Figure-(1) shows that the cavitation occur at volume flow rates above the $28\text{Lpm}$.

Refer to the Table-A-7E, "Properties of saturated liquid" to obtain the density of the water as $965.3\text{kg}/{\text{m}}^{\text{3}}$, the dynamic viscosity of water as $0.315\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$, the vapor pressure as $70.14\text{kPa}$ at temperature $T=90°\text{C}$.

Substitute $0.85$ for $K_{L,1}$ and $0.3$ for $K_{L,2}$ in Equation (III).

  $\begin{array}{c}{\displaystyle \sum K_L}=0.85+0.3\\ =1.15\end{array}$

Substitute $965.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $101.3\text{kPa}$ for $P_{atm}$, $2.8\text{m}$ for $L$, $24\text{mm}$ for $D$, $70.14\text{kPa}$ for $P_s$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $2.2\text{m}$ for $\left(z_2-z_1\right)$ and $1.15$ for ${\displaystyle \sum K_L}$ in Equation (II).

  $\begin{array}{c}NPS{H}_{available}=\left[\left(\begin{array}{l}\frac{101.3\text{kPa}-70.14\text{kPa}}{\left(965.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}-\left(2.2\text{m}\right)-\\ \left(f\frac{2.8\text{m}}{24\text{mm}}+1.15\right)\frac{V^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\end{array}\right)\right]\\ =\left[\left(\begin{array}{l}\frac{31.16\text{kPa}}{\left(9469.593\text{kg}/{\text{m}}^2{\text{s}}^2\right)}-\left(2.2\text{m}\right)-\\ \left(f\frac{2.8\text{m}}{24\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)}+1.15\right)\frac{V^2}{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)}\end{array}\right)\right]\\ =\left[\left(\begin{array}{l}\frac{31.16\text{kPa}}{\left(9469.593\text{kg}/{\text{m}}^2{\text{s}}^2\right)}-\left(2.2\text{m}\right)-\\ \left(f\left(1166.66\right)+1.15\right)\frac{V^2}{\left(19.62\text{m}/{\text{s}}^{\text{2}}\right)}\end{array}\right)\right]\\ =1.0905\text{m}-\left(5.946f+0.0586\right){\text{s}}^{\text{2}}/\text{m}\times V^2\end{array}$....... (VIII)

Use hit and trial method to obtain the volume flow rate.

Iteration 1 consider the volume flow rate as $20\text{Lpm}$.

Substitute $20\text{Lpm}$ for $\dot{V}$ and $24\text{mm}$ for $D$ in Equation (IV).

  $\begin{array}{c}V=\frac{4\left(20\text{Lpm}\right)}{\pi {\left(24\text{mm}\right)}^2}\\ =\frac{4\times 20\text{Lpm}\left(\frac{\text{1}\text{L}}{1\min }\times \frac{1{\text{m}}^3}{1000\text{L}}\times \frac{1\min }{60\sec }\right)}{\pi {\left(24\text{mm}\left(\frac{1\text{m}}{{10}^3\text{mm}}\right)\right)}^2}\\ =\frac{1.33\times {10}^{-3}{\text{m}}^{\text{3}}/\text{s}}{1.8095\times {10}^{-3}{\text{m}}^2}\\ =0.7368\text{m}/\text{s}\end{array}$

Substitute $965.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0.7368\text{m}/\text{s}$ for $V$, $0.315\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ and $24\text{mm}$ for $D$ in Equation (V).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(0.7368\text{m}/\text{s}\right)\left(965.3\text{kg}/{\text{m}}^3\right)\left(24\text{mm}\right)}{0.315\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{\left(0.7368\text{m}/\text{s}\right)\left(965.3\text{kg}/{\text{m}}^3\right)\left(24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{0.315\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{17.0695}{0.315\times {10}^{-3}}\\ =54189.184\end{array}$

Substitute $54189.184$ for $\mathrm{Re}$ in Equation (VII).

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-1.8\log \left[\frac{6.9}{54189.184}\right]\\ \frac{1}{\sqrt{f}}=-1.8\log \left[0.00012733\right]\\ \sqrt{f}=0.14263\\ f=0.0203435\end{array}$

Substitute $0.0203435$ for $f$ and $0.7368\text{m}/\text{s}$ for $V$ in Equation (VIII).

  $\begin{array}{c}NPS{H}_{available}=1.0905\text{m}-\left(5.946\left(0.0203435\right)+0.0586\right){\text{s}}^{\text{2}}/\text{m}\times {\left(0.7368\text{m}/\text{s}\right)}^2\\ =1.0905\text{m}-\left(0.089125\text{m}+0.043176\text{m}\right)\\ =1.0905\text{m}-0.13455648\text{m}\\ =0.9581988\text{m}\end{array}$

Substitute $20\text{Lpm}$ for $\dot{V}$ in Equation (VI).

  $\begin{array}{c}NPS{H}_{required}=2.2\text{m}+\left[0.0013{\text{m}/\left(\text{Lpm}\right)}^2\right]{\left(20\text{Lpm}\right)}^2\\ =2.2\text{m}+\left[0.0013{\text{m}/\left(\text{Lpm}\right)}^2\right]\left(400{\left(\text{Lpm}\right)}^2\right)\\ =2.2\text{m}+0.52\text{m}\\ =\text{2}\text{.72}\text{m}\end{array}$

Since, the available net positive suction head is greater than the required net positive suction head hence, cavitation does not occur at this flow rate.

The different iteration are shown in the below Table.

S.No	$\dot{V}\left(\text{Lpm}\right)$	$NPS{H}_{available}$	$NPS{H}_{required}$
1	$20$	$0.9581988$	$2.72$
2	$30$	$0.883661$	$3.37$
3	$40$	$0.736891$	$4.28$
4	$50$	$0.5537$	$5.45$
5	$60$	$0.335078$	$6.88$
6	$70$	$0.081399$	$8.57$

Draw the plot between the flow rate and available net positive suction head and the required net positive suction head.

Figure-(2)

From Figure-(2), available net positive suction head and the required net positive suction head curves are not crossing each other as the temperature of water is near to boiling temperature. The pump cavitates at any flow rate.

Conclusion:

The maximum volume flow rate at which cavitation occurs is $28\text{Lpm}$.