Chapter 8, Problem 137P

Water at $15°$ C is to be dischaged froiti a reservoir at a rate of 18 L1s using two horizontal cast iron pipes coiuiected w series and a pump between them The first pipe is 20 w bug and lias a 6-cni diameter, while the second pipe is 35 in long and has a 3-cnt diameter. The water level in the reservoir is 30 w above the centerline of the pipe. The pipe citirance is sharp-edged. and losses associated vitli the cOiUieCtiOfl of the PiliulI) are negligible. Neglecting tite etrect of the kinetic energy correction factor. deterniùie the required plunping head and the minimum pluuping pover to maintain the indicated flow rate.

To determine

The required pumping head.

The pumping power required to maintain the flow.

Answer to Problem 137P

The required pumping head is $\text{1296}\text{.29}\text{m}$.

The pumping power required to maintain the flow is $228.69\text{kW}$.

Explanation of Solution

Given information:

The temperature of the water is $15°\text{C}$, the length of the first pipe is $20\text{m}$, the diameter of the first pipe is $6\text{cm}$, length of the second pipe is $35\text{m}$, the diameter of the second pipe is $3\text{cm}$ and the height of the water level in the reservoir is $30\text{m}$.

Write the expression for the cross-sectional flow area of the pipe 1.

  $A_1=\frac{\pi }{4}{D}_1^2$   ....... (I)

Here, the diameter of the pipe one is $D_1$.

Write the expression for the velocity of the liquid.

  $V_1=\frac{Q}{A_1}$   ....... (II)

Here, the discharge through the pipe is $Q$.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho V_1{D}_1}{\mu }$   ....... (III)

Here, the dynamic viscosity of water is $\mu$.

Write the expression for the relative roughness value.

  $R_v=\frac{\epsilon }{D}$   ...... (IV)

Here, the equivalent roughness of the pipe is $\epsilon$.

Write the expression for the head of pipe 1.

  $h_1=\frac{V_1^2}{2g}\left({\displaystyle \sum K_L+f\frac{L_1}{D_1}}\right)$   ....... (V)

Here, the friction factor is $f$, the diameter of the pipe one $D_1$ and the length of the pipe 1 $L_1$.

Write the expression for the cross-sectional flow area of the pipe second.

  $A_2=\frac{\pi }{4}{D}_2^2$   ....... (VI)

Here, the diameter of the pipe second is $D_2$.

Write the expression for the velocity of the liquid for pipe second.

  $V_2=\frac{Q}{A_2}$   ....... (VII)

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{\rho V_2{D}_2}{\mu }$   ....... (VIII)

Write the expression for the relative roughness value.

  $R_v=\frac{\epsilon }{D}$   ....... (IX)

Here, the equivalent roughness of the pipe is $\epsilon$.

Write the expression for the head of pipe 2.

  $h_2=\frac{V_2^2}{2g}\left(f\frac{L_2}{D_2}\right)$   ....... (X)

Here, the length of the pipe 1 $L_2$.

Write the expression for the total head loss.

  $h_L=h_1+h_2$   ...... (XI)

Write the expression for the pump head.

  $h_p=\frac{V_2^2}{2g}+h_1-z_1$  ......(XII)

Here, the gravitational acceleration is $g$.

Write the expression for the pumping power.

  $P=Q\rho g{h}_p$  ......(XIII)

Calculation:

Refer to table A-7 "properties of liquid" to obtain the loss coefficient for the pipe as $0.5$, density of water as $999.1\text{kg}/{\text{m}}^{\text{3}}$, dynamic coefficient of viscosity of water as $1.138\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ and roughness of the copper as $0.00026\text{m}$ at the temperature $T=15°\text{C}$.

Substitute $6\text{cm}$ for $D_1$ in Equation (I).

  $\begin{array}{c}{A}_1=\frac{\pi }{4}{\left(6\text{cm}\right)}^2\\ =0.7853\left(36{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)\\ =2.8270\times {10}^{-3}{\text{m}}^2\end{array}$

Substitute $2.8270\times {10}^{-3}{\text{m}}^2$ for $A_1$ and $18\text{L}/\text{s}$ for $Q$ in Equation (II).

  $\begin{array}{c}{V}_1=\frac{18\text{L}/\text{s}}{\left(2.8270\times {10}^{-3}{\text{m}}^2\right)}\\ =\frac{18\text{L}/\text{s}}{\left(2.8270\times {10}^{-3}{\text{m}}^2\right)}\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)\\ =6.366\text{m}/\text{s}\end{array}$

Substitute $999.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $6.366\text{m}/\text{s}$ for $V$, $6\text{cm}$ for $D_1$ and $1.138\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (III).

  $\begin{array}{c}\mathrm{Re}=\frac{\left(999.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(6.366\text{m}/\text{s}\right)\left(6\text{cm}\right)}{\left(1.138\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{\left(999.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(6.366\text{m}/\text{s}\right)\left(6\text{cm}\left(\frac{1\text{m}}{{10}^2\text{cm}}\right)\right)}{\left(1.138\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\\ =\frac{381.61}{1.138\times {10}^{-3}}\\ =335339\end{array}$

Substitute $0.00026\text{m}$ for $\epsilon$ and $6\text{cm}$ for $D_1$ in Equation (IV).

  $\begin{array}{c}{R}_v=\frac{0.00026\text{m}}{6\text{cm}}\\ =\frac{0.00026\text{m}}{6\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =0.0043\end{array}$

Refer to chart "the moody's chart" to obtain the friction factor at Reynolds number $\mathrm{Re}=335339$ and relative roughness value $R_v=0.0043$ as $0.02941$.

Substitute $6.366\text{m}/\text{s}$ for $V_1$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, 0.5 for $K_L$, $0.02941$ for $f$, $20\text{m}$ for $L_1$ and $6\text{cm}$ for $D_1$ in Equation (V).

Substitute $3\text{cm}$ for $D_2$ in Equation (VI).

  $\begin{array}{c}{A}_2=\frac{\pi }{4}{\left(3\text{cm}\right)}^2\\ =0.7853\left(9{\text{cm}}^2\left(\frac{1{\text{m}}^2}{{10}^4{\text{cm}}^2}\right)\right)\\ =7.067\times {10}^{-4}{\text{m}}^2\end{array}$

Substitute $7.067\times {10}^{-4}{\text{m}}^2$ for $A_2$ and $18\text{L}/\text{s}$ for $Q$ in Equation (VII).

  $\begin{array}{c}{V}_2=\frac{18\text{L}/\text{s}}{\left(7.067\times {10}^{-4}{\text{m}}^2\right)}\\ =\frac{18\text{L}/\text{s}}{\left(7.067\times {10}^{-4}{\text{m}}^2\right)}\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)\\ =25.46\text{m}/\text{s}\end{array}$

Substitute $999.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $25.46\text{m}/\text{s}$ for $V_2$, $3\text{cm}$ for $D_2$ and $1.138\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in Equation (VIII).

Substitute $0.00026\text{m}$ for $\epsilon$ and $3\text{cm}$ for $D_2$ in Equation (IX).

  $\begin{array}{c}{R}_v=\frac{0.00026\text{m}}{3\text{cm}}\\ =\frac{0.00026\text{m}}{3\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =0.0086\end{array}$

Refer to chart "the moody's chart" to obtain the friction factor at Reynolds number $\mathrm{Re}=670783$ and relative roughness value $R_v=0.0086$ as $0.03300$.

Substitute $25.46\text{m}/\text{s}$ for $V_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $0.03300$ for $f$, $35\text{m}$ for $L_1$ and $3\text{cm}$ for $D_2$ in Equation (X).

  $\begin{array}{c}{h}_2=\frac{{\left(25.46\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\left(0.03300\frac{\left(35\text{m}\right)}{\left(3\text{cm}\right)}\right)\\ =33.038\text{m}\left(\left(0.03300\right)\frac{\left(35\text{m}\right)}{\left(3\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)}\right)\\ =33.038\text{m}\left(38.5\right)\\ \approx 1271.9\text{m}\end{array}$

Substitute $1271.9\text{m}$ for $h_2$ and $21.3\text{m}$ for $h_1$ in Equation (XI).

  $\begin{array}{c}{h}_L=1271.9\text{m}+21.3\text{m}\\ \text{=1293}\text{.26}\text{m}\end{array}$

Substitute $\text{1293}\text{.26}\text{m}$ for $h_L$, $25.46\text{m}/\text{s}$ for $V_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $30\text{m}$ for $z_1$ in Equation (XII).

  $\begin{array}{c}{h}_p=\frac{{\left(25.46\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}+\text{1293}\text{.26}\text{m}-30\text{m}\\ =33.03\text{m}+\text{1263}\text{.26}\text{m}\\ \text{=1296}\text{.29}\text{m}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $999.1\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $18\text{L}/\text{s}$ for $Q$ and $1296.29\text{m}$ for $h_p$ in Equation (XIII).

  $\begin{array}{c}P=\left(18\text{L}/\text{s}\right)\left(999.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1296.29\text{m}\right)\\ =\left(18\text{L}/\text{s}\right)\left(\frac{{10}^{-3}{\text{m}}^{\text{3}}}{1\text{L}}\right)\left(999.1\text{kg}/{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(1296.29\text{m}\right)\left(\frac{1\text{kW}}{{\text{10}}^3\text{kg}\cdot {\text{m}}^{\text{2}}/{\text{s}}^{\text{3}}}\right)\\ =\left(0.018\right)\left(12705.15\right)\text{kW}\\ =228.69\text{kW}\end{array}$

Conclusion:

The required pumping head is $\text{1296}\text{.29}\text{m}$.

The pumping power required to maintain the flow is $228.69\text{kW}$.