Chapter 11, Problem 11.4CP

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To determine

(a)

To select:

The pump that runs at the given speed performing the task with efficiency being maximum.

Answer to Problem 11.4CP

The pump that runs at the given speed performing the task with efficiency being maximum is determined below.

Explanation of Solution

Given Information:

$$

The given figure,

The system delivers 20-degree Celsius water from the sea level reservoir to another iron pipe with diameter 38 cm. $$

Minor loss before pump entrance $\sum K1=0.5$

Minor loss after pump entrance $\sum K2=7.2$

Concept Used:

Velocity of the flow:

$V=\frac{Q}{A}$

$\begin{array}{l}A=area\\ Q=disch\arg e\end{array}$

$V=\frac{Q}{\frac{\pi }{4}{d}^2}$

Calculation:

According to the figure:

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f-H_{pump}\\ 0+0+0=0+0+z_2+h_f-H_{pump}\\ 0=0+z_2+h_f-H_{pump}\end{array}$

$z_2+h_f=H_{pump}$

$\to (1)$

$V=\frac{Q}{\frac{\pi }{4}{d}^2}\Rightarrow \frac{1.39}{\frac{3.14}{4}*{0.38}^2}$

$\begin{array}{l}=\frac{1.39}{0.785*0.144}\Rightarrow \frac{1.39}{0.11304}\\ =12.24m/s\end{array}$

Reynold’s number:

${\mathrm{Re}}_d=\frac{\rho Vd}{\mu }$

Density $\rho$ = 998 kg/m³

Dynamic Viscosity $\mu$ = 0.001kg/m.s

Diameter=380mm

Volume=12.24m/s

Substituting the values:

$\begin{array}{l}=\frac{\rho Vd}{\mu }\\ =\frac{998*12.24*0.380}{0.001}\\ =\frac{4641.8976}{0.001}\Rightarrow 4641897.6=4.64*{10}^6\\ \end{array}$

The flow is turbulent.

Friction factor is given as:

$f^{-1/2}=-2.0{\log }_{10}\left(\frac{\epsilon /d}{3.7}+\frac{2.51}{{\mathrm{Re}}_d\sqrt{f}}\right)$

$[\because \epsilon =0.26mm,aspertherecommendedductcommercialvalue]$

$\begin{array}{l}=-2.0{\log }_{10}\left(\frac{0.000684}{3.7}+\frac{2.51}{4.64*{10}^6\sqrt{f}}\right)\\ =-2.0{\log }_{10}\left(\frac{0.000684}{3.7}+\frac{2.51}{4.64*{10}^6\sqrt{f}}\right)\Rightarrow 0.0180\end{array}$

Head loss using the friction factor is calculated.

$h_f=\frac{V^2}{2g}\left[f\frac{L}{d}+{\displaystyle \sum K}\right]$

$h_f=\frac{{12.24}^2}{2*9.81}\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]$

$\begin{array}{l}=\frac{{12.24}^2}{2*9.81}\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]\\ =\frac{149.82}{2*9.81}\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]\\ =7.64\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]\\ =7.64\left[1.326+(0.5+7.2)\right]\\ =68.9m\\ =226ft\end{array}$

Using equation (1)

$\begin{array}{l}{z}_2+h_f=H_{pump}\\ 11+68.9=H_{pump}\Rightarrow 79.9m\end{array}$

$=262ft$

Frictional heal loss is calculated using flow rates having different values:

Q(gal/min)	4000	8000	12000	16000	20000	22000	24000
$h_f(ft)$	44	66	103	156	223	262	305

From the table and the reference figure at the flow rate of 20,000 gal/min, 88% efficiency. Hence the pump is given as 38 in.

Conclusion:

Thus, the pump that runs at the given speed performing the task with efficiency being maximum is determined.

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To determine

(b)

To calculate:

The flow rate.

Answer to Problem 11.4CP

The flow rate is given as 20,000 gal/min

Explanation of Solution

Given Information:

$$

The given figure:

The system delivers 20-degree Celsius water from the sea level reservoir to another iron pipe with diameter 38 cm. $$

Minor loss before pump entrance $\sum K1=0.5$

Minor loss after pump entrance $\sum K2=7.2$

Concept Used:

Velocity of the flow:

$V=\frac{Q}{A}$

$\begin{array}{l}A=area\\ Q=disch\arg e\end{array}$

$V=\frac{Q}{\frac{\pi }{4}{d}^2}$

Calculation:

$\begin{array}{l}\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f-H_{pump}\\ 0+0+0=0+0+z_2+h_f-H_{pump}\\ 0=0+z_2+h_f-H_{pump}\end{array}$

$z_2+h_f=H_{pump}$

$\to (1)$

$V=\frac{Q}{\frac{\pi }{4}{d}^2}\Rightarrow \frac{1.39}{\frac{3.14}{4}*{0.38}^2}$

$\begin{array}{l}=\frac{1.39}{0.785*0.144}\Rightarrow \frac{1.39}{0.11304}\\ =12.24m/s\end{array}$

Reynold’s number

${\mathrm{Re}}_d=\frac{\rho Vd}{\mu }$

Density $\rho$ = 998 kg/m³

Dynamic Viscosity $\mu$ = 0.001kg/m.s

Diameter=380mm

Volume=12.24m/s

Substituting the values:

$\begin{array}{l}=\frac{\rho Vd}{\mu }\\ =\frac{998*12.24*0.380}{0.001}\\ =\frac{4641.8976}{0.001}\Rightarrow 4641897.6=4.64*{10}^6\\ \end{array}$

The flow is turbulent.

Friction factor is given as:

$f^{-1/2}=-2.0{\log }_{10}\left(\frac{\epsilon /d}{3.7}+\frac{2.51}{{\mathrm{Re}}_d\sqrt{f}}\right)$

$[\because \epsilon =0.26mm,aspertherecommendedductcommercialvalue]$ :

$\begin{array}{l}=-2.0{\log }_{10}\left(\frac{0.000684}{3.7}+\frac{2.51}{4.64*{10}^6\sqrt{f}}\right)\\ =-2.0{\log }_{10}\left(\frac{0.000684}{3.7}+\frac{2.51}{4.64*{10}^6\sqrt{f}}\right)\Rightarrow 0.0180\\ \end{array}$

Head loss using the friction factor is calculated:

$h_f=\frac{V^2}{2g}\left[f\frac{L}{d}+{\displaystyle \sum K}\right]$

$h_f=\frac{{12.24}^2}{2*9.81}\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]$

$\begin{array}{l}=\frac{{12.24}^2}{2*9.81}\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]\\ =\frac{149.82}{2*9.81}\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]\\ =7.64\left[0.018\frac{28}{0.38}+(0.5+7.2)\right]\\ =7.64\left[1.326+(0.5+7.2)\right]\\ =68.9m\\ =226ft\end{array}$

Using equation (1)

$\begin{array}{l}{z}_2+h_f=H_{pump}\\ 11+68.9=H_{pump}\Rightarrow 79.9m\end{array}$

$=262ft$

Frictional heal loss is calculated using flow rates having different values:

Q(gal/min)	4000	8000	12000	16000	20000	22000	24000
$h_f(ft)$	44	66	103	156	223	262	305

Conclusion:

Thus, the flow rate is determined.

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To determine

(c)

Brake horse power.

Answer to Problem 11.4CP

Brake horse power=1250hp

Explanation of Solution

Given Information:

$$

The given figure,

The system delivers 20-degree Celsius water from the sea level reservoir to another iron pipe with diameter 38 cm. $$

Minor loss before pump entrance $\sum K1=0.5$

Minor loss after pump entrance $\sum K2=7.2$

Concept Used:

$BHP=\rho gQ{h}_f$

$\rho =density$

$Q=flowrate$

$h_f=frictionfactor$

Calculation:

Substituting we have:

$BHP=\rho gQ{h}_f$

$\begin{array}{l}=998*1000*1.39*68.9\\ =95.58*{10}^{6}watts\end{array}$

$=\frac{95.58*{10}^6}{745.6}$

$=1250hp$

Conclusion:

Thus, the brake horse power is calculated.

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To determine

(d)

The given pump is safe from cavitation.

Answer to Problem 11.4CP

The pump replacement is needed as the pump will cavitate.

Explanation of Solution

Given Information:

$$

The given figure:

The system delivers 20-degree Celsius water from the sea level reservoir to another iron pipe with diameter 38 cm. $$

Minor loss before pump entrance $\sum K1=0.5$

Minor loss after pump entrance $\sum K2=7.2$

Concept Used:

$NPSH=\frac{p_a}{\gamma }-z_h-h_{fi}-\frac{p_v}{\gamma }$

$\begin{array}{l}{p}_a=totalpressure\\ p_v=vaporpressure\\ h_{fi}=head\text{ loss}\\ z_h=elevation\end{array}$

Calculation:

Substituting we have:

$NPSH=\frac{p_a}{\gamma }-z_h-h_{fi}-\frac{p_v}{\gamma }$

$=\frac{101350}{9790}-1-\frac{11.13}{2(9.81)}(1+0.5)-\frac{2337}{9790}$

$=10.352-1-0.567-0.239$

$=-0.36m$

For the flow rate Q=20000gal/min, the NPSH=16m(-0.36m).

So, the pump cavitates.

Conclusion:

Thus, the given pump is safe from cavitation is determined.