Chapter 14, Problem 95RQ

To determine

The pressure drop across a single tube and power required by the pump and the reduction in flow rate due to scaling.

Explanation of Solution

Given:

Diameter of the tube is 1cm.

Length of the tube is 1.5m.

Temperature of water is $60°\text{C}$.

Rate of flow is $15\text{L}/\text{s}$.

Number of brass tubes is 80.

Calculation:

Refer Table A-15 “Properties of saturated water” from Appendix 1 and obtain the following properties of ammonia.

  $\begin{array}{l}\text{Density},\rho =983.3\text{kg}/{\text{m}}^3\\ \text{Dynamic viscosity, }\mu =0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\end{array}$

Calculate the velocity of the flow.

  $\begin{array}{c}V=\frac{\dot{V}}{A_c}\\ =\frac{15\text{L}/\text{s}\times \frac{1{\text{ m}}^3}{1000\text{ L}}}{80\times \frac{\pi }{4}\times {\left(1\times {10}^{-2}\text{m}\right)}^2}\\ =2.387\text{m}/\text{s}\end{array}$

The Reynolds number is

  $\begin{array}{c}\mathrm{Re}=\frac{\rho VD}{\mu }\\ =\frac{\left(983.3\text{kg}/{\text{m}}^3\right)\left(2.387\text{m}/\text{s}\right)\left(\text{0}\text{.01 m}\right)}{0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =50270\end{array}$

Since the Reynolds number is greater than 2300, the flow is turbulent. Hence, the friction factor is obtained from Moody chart as follows:

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /\text{D}}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ =-2.0\log \left[\frac{1.5\times {10}^{-6}\text{m}/0.01\text{ m}}{3.7}+\frac{2.51}{50270\sqrt{f}}\right]\end{array}$

On solving the above equation, we get $f=0.0214$.

Calculate the pressure drop.

  $\begin{array}{c}\Delta P=f\frac{L}{D}\frac{\rho V^2}{2}\\ =0.0214\times \frac{\text{1}\text{.5 m}}{0.01\text{ m}}\times \frac{\left(983.3\text{kg}/{\text{m}}^3\right){\left(2.387\text{m}/\text{s}\right)}^2}{2}\\ =8992\text{ Pa}\\ =8.992\text{ kPa}\end{array}$

Thus, the drop in pressure across a single tube is $\text{8}\text{.99 kPa}$.

Calculate the power required by the pump.

  $\begin{array}{c}{\dot{W}}_{pump}=\dot{V}\Delta P\\ =\left(15\text{L}/\text{s}\times \frac{1{\text{ m}}^3}{1000\text{ L}}\right)\left(\text{8}\text{.99 kPa}\right)\\ =0.135\text{ kW}\end{array}$

Thus, the power required by the pump is $0.135\text{ kW}$.

After the buildup of scale:

The velocity of the flow is,

  $\begin{array}{c}{V}_{new}=\frac{{\dot{V}}_{new}}{A_c}\\ =\frac{{\dot{V}}_{new}}{80\times \frac{\pi }{4}\times {\left(8\times {10}^{-3}\text{m}\right)}^2}\end{array}$        (I)

The Reynolds number is

  $\begin{array}{c}\mathrm{Re}=\frac{\rho V_{new}D}{\mu }\\ =\frac{\left(983.3\text{kg}/{\text{m}}^3\right)V_{new}\left(\text{0}\text{.008 m}\right)}{0.467\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\end{array}$        (II)

The friction factor is obtained from Moody chart.

  $\begin{array}{c}\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /\text{D}}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)\\ =-2.0\log \left[\frac{\text{0}\text{.0004m}/0.008\text{ m}}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right]\end{array}$        (III)

The pressure drop is,

  $\begin{array}{c}\Delta P=f\frac{L}{D}\frac{\rho V_{new}{}^2}{2}\\ =f\frac{\text{1}\text{.5 m}}{0.008\text{ m}}\times \frac{\left(983.3\text{kg}/{\text{m}}^3\right)V_{new}{}^2}{2}\end{array}$        (IV)

The power required by the pump is,

  $\begin{array}{c}{\dot{W}}_{pump}=\dot{V}\Delta P\\ 0.135\text{ kW}=\dot{V}\Delta P\end{array}$        (V)

We have 5 equations and 5 variables. Solving the system of simultaneous equations, we get:

  $\begin{array}{l}{V}_{new}=1.714\text{m}/\text{s}\\ \mathrm{Re}=28870\\ f=0.0723\\ \Delta P=19.6\text{ kPa}\\ {\dot{V}}_{new}=6.89\text{L}/\text{s}\end{array}$

Calculate the reduction in the flow rate.

  $\begin{array}{c}\text{Reduction}=\frac{\dot{V}-{\dot{V}}_{new}}{\dot{V}}\\ =\frac{\left(15-6.89\right)\text{L}/\text{s}}{15\text{L}/\text{s}}\\ =0.54\\ =54\%\end{array}$

Thus, the reduction in the flow rate after scaling is $54\%$.