Chapter 14, Problem 62P

Repeat Prob. 14-60, but with the pipe diameter increased by a factor of 2 (all else being equal). Does the volume flow rate at which cavitation occurs in the pump increase or decrease with the larger pipe? Discuss.

To determine

Whether the volume flow rate at which cavitations occurs in the pump increase or decrease of the larger pipe.

Answer to Problem 62P

The diameter of the pipe is doubled then the flow rate increase by $7.69\%$. But the average speed is decreased by the four.

Explanation of Solution

Given information:

The pipe diameter is increased by the factor two.

Write the expression for the net positive suction head.

  $NPSH=\left(\frac{P_{\text{atm}}-P_v}{\rho g}\right)+z_1-z_2-h_L$   ....... (I)

Here, the net positive suction head is $NPSH$, the atmospheric pressure is $P_{\text{atm}}$, the vapor pressure is $P_v$, the acceleration due to gravity is $g$, the density of flowing fluids is $\rho$, the datum head difference between point 1 and point 2 is $z_1-z_2$ and the total head loss is $h_L$.

Write the expression for the total head loss.

  $h_L=\left(f\frac{L}{D}+K_{L1}+K_{L2}\right)\frac{V^2}{2g}$   ....... (II)

Here, the friction factor is $f$, the minor loss due to sharp edged reentrant inlet is $K_{L1}$, the minor loss due to flanged smooth at right angle regular elbow, the velocity is $V$, the length of the pipe is $L$ and the diameter of the pipe is $D$.

Write the expression for the average speed.

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

Here, the diameter of the pipe is $D$ and the volume flow rate is $\dot{V}$.

Write the expression for the Reynolds number.

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

Here, the Reynolds number is $\mathrm{Re}$ and the kinematic viscosity is $\mu$.

Write the expression for the friction factor.

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

Substitute $\left(f\frac{L}{D}+K_{L1}+K_{L2}\right)\frac{V^2}{2g}$ for $h_L$ in Equation (I).

  $NPSH=\left(\frac{P_{\text{atm}}-P_v}{\rho g}\right)+z_1-z_2-\left(f\frac{L}{D}+K_{L1}+K_{L2}\right)\frac{V^2}{2g}$   ....... (VI)

Write the expression for the required net positive suction head.

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

Here, the required net positive head is $NPS{H}_{\text{required}}$.

Write the expression for the percentage increase.

  $E=\left(\frac{{\left(V\right)}_{60}-{\left(V\right)}_{20}}{{\left(V\right)}_{60}}\right)\times 100$   ......... (VIII)

Here, the percentage increase is $E$.

Calculation:

Refer the Table B-1, "The physical properties of saturated liquid" to obtain the value of the density is $997\text{kg}/{\text{m}}^3$ corresponding to the temperature $25°\text{C}$.

Refer the Table B-1, "The physical properties of saturated liquid" to obtain the value of the kinematic viscosity is $8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}$ corresponding to the temperature $25°\text{C}$.

Refer the Table B-1, "The physical properties of saturated liquid" to obtain the value of the vapor pressure is $3.169\text{kPa}$ corresponding to the temperature $25°\text{C}$.

Substitute $997\text{kg}/{\text{m}}^3$ for $\rho$, $101.3\text{kPa}$ for $P_{\text{atm}}$, $3.169\text{kPa}$ for $P_v$, $2.8\text{m}$ for $L$, $24\text{mm}$ for $D$, $9.81\text{m}/{\text{s}}^2$ for $g$, $0.85$ for $K_{L1}$, $0.3$ for $K_{L2}$ and $2.2\text{m}$ for $\left(z_2-z_1\right)$ in Equation (VI).

  $\begin{array}{c}NPSH=\left[\begin{array}{l}\left(\frac{101.3\text{kPa}-3.169\text{kPa}}{997\text{kg}/{\text{m}}^3\times 9.81\text{m}/{\text{s}}^2}\right)-2.2\text{m}\\ -\left(f\frac{2.8\text{m}}{24\text{mm}}+0.85+0.3\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^2}\end{array}\right]\\ NPSH=\left[\begin{array}{l}\left(\frac{98.131\text{kPa}\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)}{9780.57\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\left(\frac{1\text{Pa}}{1\text{kg}/{\text{m}}^1\cdot {\text{s}}^2}\right)}\right)-2.2\text{m}\\ -\left(f\frac{2.8\text{m}}{24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}+1.15\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^2}\end{array}\right]\\ NPSH=7.833\text{m}-5.946f{V}^2-0.0586V^2\text{m}/{\text{s}}^2\end{array}$.......(IX)

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

  $\begin{array}{c}V=\frac{4\times 20\text{Lpm}}{\pi {\left(24\text{mm}\right)}^2}\\ =\frac{80\text{Lpm}}{\pi {\left(24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}\\ =\frac{80\text{Lpm}\left(\frac{1{\text{m}}^3/\text{s}}{60000\text{Lpm}}\right)}{0.001809{\text{m}}^2}\\ =0.736828\text{m}/\text{s}\end{array}$

Substitute $0.736828\text{m}/\text{s}$ for $V$, $997\text{kg}/{\text{m}}^3$ for $\rho$, $24\text{mm}$ for $D$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (IV).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(997\text{kg}/{\text{m}}^3\right)\left(0.736828\text{m}/\text{s}\right)\left(24\text{mm}\right)}{8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{\left(997\text{kg}/{\text{m}}^3\right)\left(0.736828\text{m}/\text{s}\right)\left(24\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{17.63082\text{kg}/\text{m}\cdot \text{s}}{8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\\ =19787.6772\end{array}$

Substitute $19787.6772$ for $\mathrm{Re}$ in Equation (V).

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-1.8\log \left[\frac{6.9}{19787.6772}\right]\\ \frac{1}{\sqrt{f}}=-1.8\log \left(0.000348\right)\\ \sqrt{f}=0.160679\\ f=0.02581\end{array}$

Substitute $0.02581$ for $f$ and $0.736828\text{m}/\text{s}$ for $V$ in Equation (IX).

  $\begin{array}{c}NPSH=\left[\begin{array}{c}7.833-5.946\left(0.02581\right){\left(0.736828\text{m}/\text{s}\right)}^2\text{m}/{\text{s}}^2\\ -0.0586{\left(0.736828\text{m}/\text{s}\right)}^2\text{m}/{\text{s}}^2\end{array}\right]\\ =7.833\text{m}-0.0837\text{m}-0.03181\text{m}\\ =\text{7}\text{.7174}\text{m}\end{array}$

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

  $\begin{array}{c}NPS{H}_{\text{required}}=2.2\text{m}+\left[0.0013\text{m}/{\text{Lpm}}^2\right]{\left(20\text{Lpm}\right)}^2\\ =2.2\text{m}+\left[0.0013\text{m}/{\text{Lpm}}^2\right]\left(400{\text{Lpm}}^2\right)\\ =2.2\text{m}+0.52\text{m}\\ =2.72\text{m}\end{array}$

The following table represents the net positive suction head, the required net positive suction head, and discharge.

$\dot{V}$	$NPSH$   $\left(\text{m}\right)$	$NPS{H}_{\text{required}}$   $\left(\text{m}\right)$
$0$	$7.833$	$2.2$
$10$	$7.75$	$2.33$
$20$	$\text{7}\text{.7174}$	$2.72$
$50$	$\text{7}\text{.1666}$	$5.46$
$60$	$\text{7}\text{.8244}\text{m}$	$6.88$

Plot the graph by using tabulated values.

  

Figure-(1)

From Figure-(1), the cavitations occurs at the flow rate $60\text{Lpm}$.

Substitute $997\text{kg}/{\text{m}}^3$ for $\rho$, $101.3\text{kPa}$ for $P_{\text{atm}}$, $3.169\text{kPa}$ for $P_v$, $2.8\text{m}$ for $L$, $24\text{mm}$ for $D$, $9.81\text{m}/{\text{s}}^2$ for $g$, $0.85$ for $K_{L1}$, $0.3$ for $K_{L2}$ and $2.2\text{m}$ for $\left(z_2-z_1\right)$ in Equation (VI).

  $\begin{array}{c}NPSH=\left[\begin{array}{l}\left(\frac{101.3\text{kPa}-3.169\text{kPa}}{997\text{kg}/{\text{m}}^3\times 9.81\text{m}/{\text{s}}^2}\right)-2.2\text{m}\\ -\left(f\frac{2.8\text{m}}{48\text{mm}}+0.85+0.3\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^2}\end{array}\right]\\ NPSH=\left[\begin{array}{l}\left(\frac{98.131\text{kPa}\left(\frac{1000\text{Pa}}{1\text{kPa}}\right)}{9780.57\text{kg}/{\text{m}}^2\cdot {\text{s}}^2\left(\frac{1\text{Pa}}{1\text{kg}/{\text{m}}^1\cdot {\text{s}}^2}\right)}\right)-2.2\text{m}\\ -\left(f\frac{2.8\text{m}}{48\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)}+1.15\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^2}\end{array}\right]\\ NPSH=7.833\text{m}-2.973f{V}^2-0.0586V^2\text{m}/{\text{s}}^2\end{array}$   ....... (X)

Substitute $60\text{Lpm}$ for $\dot{V}$ and $48\text{mm}$ for $D$ in Equation (III).

  $\begin{array}{c}V=\frac{4\times 20\text{Lpm}}{\pi {\left(24\text{mm}\right)}^2}\\ =\frac{80\text{Lpm}}{\pi {\left(48\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}^2}\\ =\frac{80\text{Lpm}\left(\frac{1{\text{m}}^3/\text{s}}{60000\text{Lpm}}\right)}{0.007238{\text{m}}^2}\\ =0.18420\text{m}/\text{s}\end{array}$

Substitute $0.18420\text{m}/\text{s}$ for $V$, $997\text{kg}/{\text{m}}^3$ for $\rho$, $48\text{mm}$ for $D$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (IV).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(997\text{kg}/{\text{m}}^3\right)\left(0.18420\text{m}/\text{s}\right)\left(48\text{mm}\right)}{8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{\left(997\text{kg}/{\text{m}}^3\right)\left(0.18420\text{m}/\text{s}\right)\left(48\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\right)}{8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{17.63082\text{kg}/\text{m}\cdot \text{s}}{8.91\times {10}^{-4}\text{kg}/\text{m}\cdot \text{s}}\\ =9893.46\end{array}$

Substitute $9893.46$ for $\mathrm{Re}$ in Equation (V).

  $\begin{array}{l}\frac{1}{\sqrt{f}}=-1.8\log \left[\frac{6.9}{9893.46}\right]\\ \frac{1}{\sqrt{f}}=-1.8\log \left(0.000697\right)\\ \sqrt{f}=0.1750\\ f=0.03097\end{array}$

Substitute $0.03097$ for $f$ and $0.18420\text{m}/\text{s}$ for $V$ in Equation (X).

  $\begin{array}{c}NPSH=\left[\begin{array}{c}7.833-5.946\left(0.03097\right){\left(0.18420\text{m}/\text{s}\right)}^2\text{m}/{\text{s}}^2\\ -0.0586{\left(0.18420\text{m}/\text{s}\right)}^2\text{m}/{\text{s}}^2\end{array}\right]\\ =7.833\text{m}-0.0062339\text{m}-\mathrm{0.00.1988}\text{m}\\ =\text{7}\text{.8244}\text{m}\end{array}$

Substitute $60\text{Lpm}$ for $\dot{V}$ in Equation (VII).

  $\begin{array}{c}NPS{H}_{\text{required}}=2.2\text{m}+\left[0.0013\text{m}/{\text{Lpm}}^2\right]{\left(60\text{Lpm}\right)}^2\\ =2.2\text{m}+\left[0.0013\text{m}/{\text{Lpm}}^2\right]\left(3600{\text{Lpm}}^2\right)\\ =2.2\text{m}+4.68\text{m}\\ =6.88\text{m}\end{array}$

The following table represents the net positive suction head, the required net positive suction head, and discharge.

$\dot{V}$	$NPSH$   $\left(\text{m}\right)$	$NPS{H}_{\text{required}}$   $\left(\text{m}\right)$
$0$	$7.833$	$2.2$
$10$	$7.75$	$2.33$
$20$	$\text{7}\text{.7174}$	$2.72$
$50$	$\text{7}\text{.1666}$	$5.46$
$60$	$\text{7}\text{.8244}\text{m}$	$6.88$

Plot the graph by using tabulated values.

  

Figure-(2)

From Figure-(2), the cavitations occurs at the flow rate $65\text{Lpm}$.

Substitute $65\text{Lpm}$ for ${\left(V\right)}_{60}$ and $20\text{Lpm}$ for ${\left(V\right)}_{20}$ in Equation (VIII).

  $\begin{array}{c}E=\left(\frac{65\text{Lpm}-60\text{Lpm}}{65\text{Lpm}}\right)\times 100\\ =\left(\frac{5\text{Lpm}}{65\text{Lpm}}\right)\times 100\\ =0.07692\times 100\\ =7.69\%\end{array}$

The diameter of the pipe is doubled then the flow rate increase by $7.69\%$. But the average speed is decrease by the four.

Conclusion:

The diameter of the pipe is doubled then the flow rate increase by $7.69\%$. But the average speed is decrease by the four.