Chapter 8, Problem 99P

To determine

The required power input for the recirculating pump.

Answer to Problem 99P

The required power input for the recirculating pump is $0.111\text{kW}$.

Explanation of Solution

Given information:

The length of the pipe of the circulating loop is $40\text{m}$, the diameter of the pipe of the circulating loop is $1.2\text{cm}$, the number of bends in the loop is $6$, the angle of the threads of the bends is $90°$, the average flow velocity through the loop is $2\text{m}/\text{s}$, the average water temperature is $60°C$, the efficiency of the pump is $70\%$, the density of the water is $983.3\text{kg}/{\text{m}}^{\text{3}}$ and the dynamic viscosity of water is $0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$.

The loss coefficient of $90°$ threaded smooth bend from the turbulent flow tables is $0.9$, the loss coefficient of fully open gate valve from the turbulent flow tables is $0.2$ and the surface roughness of the cast iron pipe from the new commercial pipe table is $0.00026\text{m}\text{.}$.

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

$A_C=\frac{\pi }{4}{D}^2$ …… (I)

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

Write the expression for the volume flow rate.

$\dot{\vee }=AV$ …… (II)

Substitute $\frac{\pi D^2}{4}$ for area in the Equation (II).

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

Here, the average flow velocity of water is $V$

Write the expression for Reynolds number.

$R_e=\frac{\rho VD}{\mu }$ …… (IV)

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

Write the expression for relative roughness of the pipe.

$\frac{\epsilon }{D}$ …… (V)

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

Write the expression for the friction factor by the Colebrook equation.

$\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{R_e\sqrt{f}}\right)$ …… (VI)

Here, the relative roughness of the pipe is $\frac{\epsilon }{D}$, the Reynolds number is $R_e$ and the friction factor is $f$.

Write the expression for total head loss.

$\begin{array}{c}{h}_L=\left(f\frac{L}{D}+{\displaystyle \sum K_L}\right)\frac{V^2}{2g}\\ =\left(f\frac{L}{D}+6K_{L,bend}+2K_{L,valve}\right)\frac{V^2}{2g}\end{array}$ …… (VII)

Here, the length of the pipe is $L$, the acceleration due to gravity is $g$, The loss coefficient of threaded smooth bend from the turbulent flow tables is $K_{L,bend}$ and the loss coefficient of fully open gate valve from the turbulent flow tables is $K_{L,valve}$.

Write the expression for the pressure drop.

$\Delta P=\rho g{h}_L$ …… (VIII)

Here, the density of the water is $\rho$ and the total head loss is $h_L$.

Write the expression for the pumping power.

${\dot{W}}_{elect}=\frac{{\dot{W}}_{pump}}{{\eta }_{pump}}$ …… (IX)

Here, the work done by the pump is ${\dot{W}}_{pump}$, the work done by the motor is ${\dot{W}}_{elect}$ and the efficiency of the pump is ${\eta }_{pump}$.

Write the expression for the work done by the pump.

${\dot{W}}_{pump}=\dot{\vee }\Delta P$

Here, the pressure drop is $\Delta P$ and the volume flow rate is $\dot{\vee }$ ,

Substitute $\dot{\vee }\Delta P$ for ${\dot{W}}_{pump}$ in the Equation (IX).

${\dot{W}}_{elect}=\frac{\dot{\vee }\Delta P}{{\eta }_{pump}}$ …… (X)

Calculation:

Substitute $1.2\text{cm}$ for $D$ in the Equation (I).

$\begin{array}{c}{A}_C=\frac{\pi }{4}{\left(1.2\text{cm}\right)}^2\\ =\frac{\pi }{4}{\left(12\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}^2\\ =\frac{\pi }{4}{\left(0.012\text{m}\right)}^2\\ =1.13096\times {10}^{-4}{\text{m}}^{\text{2}}\end{array}$

Substitute $1.13096\times {10}^{-4}{\text{m}}^{\text{2}}$ for $A$ and $2\text{m}/\text{s}$ for $V$ in the Equation (III).

$\begin{array}{c}\dot{\vee }=\left(1.13096\times {10}^{-4}{\text{m}}^{\text{2}}\right)\left(2\text{m}/\text{s}\right)\\ =2.26194\times {10}^{\text{-4}}{\text{m}}^{\text{3}}/\text{s}\\ =0.0002261{\text{m}}^{\text{3}}/\text{s}\end{array}$

Substitute $2\text{m}/\text{s}$ for $V$, $1.2\text{cm}$ for $D$, $983.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$ for $\mu$ in the Equation (III).

$\begin{array}{c}{R}_e=\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(2\text{m}/\text{s}\right)\left(1.2\text{cm}\right)}{0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{\left(983.3\text{kg}/{\text{m}}^{\text{3}}\right)\left(2\text{m}/\text{s}\right)\left(1.2\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\right)}{0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{1966.6\text{kg}/{\text{m}}^{\text{2}}\cdot \text{s}\left(0.012\text{m}\right)}{0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{23.5992\text{kg}/\text{m}\cdot \text{s}}{0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\end{array}$

$R_e=50533.61$

Substitute $0.00026\text{m}$ for $\epsilon$ and $1.2\text{cm}$ for $D$ in the Equation (V).

$\begin{array}{c}\frac{\epsilon }{D}=\frac{0.00026\text{m}}{1.2\text{cm}}\\ =\frac{0.00026\text{m}}{1.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}\\ =\frac{0.00026\text{m}}{0.12\text{m}}\\ =0.0217\end{array}$

Substitute $0.0217$ for $\frac{\epsilon }{D}$ and $50533.61$ for $R_e$ in the Equation (VI).

$\begin{array}{l}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{0.0217}{3.7}+\frac{2.51}{50533.61\sqrt{f}}\right)\\ \frac{1}{\sqrt{f}}=-2.0\log \left(5.8648\times {10}^{-3}+\frac{4.9669\times {10}^{-5}}{\sqrt{f}}\right)\\ f=0.05075\end{array}$

Substitute $0.05075$ for $f$, $0.9$ for $K_{L,bend}$, 0.2 for $K_{L,valve}$, $40\text{m}$ for $L$, $1.2\text{cm}$ for $D$, $2\text{m}/\text{s}$ for $V$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the Equation (VII).

$\begin{array}{c}{h}_L=\left(0.05075\frac{40\text{m}}{1.2\text{cm}}+6\left(0.9\right)+2\left(0.2\right)\right)\frac{{\left(2\text{m}/\text{s}\right)}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(0.05075\frac{40\text{m}}{1.2\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)}+6\left(0.9\right)+2\left(0.2\right)\right)\frac{{\left(2\text{m}/\text{s}\right)}^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(0.05075\left(\frac{40\text{m}}{0.012}\right)+5.8\right)0.203873\text{m}\\ =35\text{m}\end{array}$

Substitute $983.3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $35\text{m}$ for $h_L$ in the Equation (VIII).

$$

   $\begin{array}{c}\Delta P=983.3\text{kg}/{\text{m}}^{\text{3}}\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 35\text{m}\\ \text{=337616}\text{.055}\text{kg}/\text{m}\cdot {\text{s}}^{\text{2}}\left(\frac{1\text{m}}{1\text{m}}\right)\\ =\text{37616}\text{.055}\text{kg}\cdot \text{m}/{\text{m}}^2\cdot {\text{s}}^{\text{2}}\\ =\text{37616}\text{.055}\left(\text{kg}\cdot \text{m}/{\text{s}}^2\right)\left(1/{\text{m}}^2\right)\end{array}$

$\begin{array}{c}\text{=37616}\text{.055}\left(\text{kg}\cdot \text{m}/{\text{s}}^2\frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^2}\right)\left(1/{\text{m}}^2\right)\\ =\text{37616}\text{.055}\text{N}/{\text{m}}^{\text{2}}\left(\frac{1000\text{kN}}{1\text{N}}\right)\\ =337.616\text{kN}/{\text{m}}^{\text{2}}\end{array}$

Substitute $337.616\text{kN}/{\text{m}}^{\text{2}}$ for $\Delta P$, $70\%$ for ${\eta }_{pump}$ and $0.0002261{\text{m}}^{\text{3}}/\text{s}$ for $\dot{\vee }$ in the Equation (X).

$\begin{array}{c}{\dot{W}}_{elect}=\frac{0.0002261{\text{m}}^{\text{3}}/\text{s}\times 337.616\text{kN}/{\text{m}}^{\text{2}}}{70\%}\\ =\frac{0.076334\text{kN}\cdot \text{m}/\text{s}}{70\left(\frac{1}{100}\right)}\\ =\frac{0.076334\text{kN}\cdot \text{m}/\text{s}\left(\frac{1\text{kW}}{\text{kN}\cdot \text{m}/\text{s}}\right)}{0.7}\\ =0.111\text{kW}\end{array}$

Conclusion:

The required power input for the recirculating pump is $0.111\text{kW}$.