Chapter 14, Problem 107RQ

To determine

The flow rate through the pipes.

Explanation of Solution

Given:

Temperature of water is $100°\text{C}$.

Length of the pipe A is $500\text{ m}$.

Length of pipe B is $800\text{ m}$.

Diameter of the pipe A is $30\text{}\text{cm}$.

Diameter of the pipe B is $45\text{cm}$.

Flow rate is $3{\text{m}}^3/\text{s}$.

Calculation:

Refer Table 15 “Properties of saturated water” from Appendix 1 and obtain the following properties of water at $100°\text{C}$:

  $\begin{array}{l}\text{Density,}\rho =957.9\text{kg}/{\text{m}}^3\\ \text{Dynamic viscosity,}\mu =0.282\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\end{array}$

From the rate of flow through the pipes,

  $\begin{array}{l}{\dot{V}}_1=V_1{A}_1\\ {\dot{V}}_1=V_1\left[\frac{\pi }{4}{\left(0.3\text{ m}\right)}^2\right]\end{array}$        (I)

  $\begin{array}{l}{\dot{V}}_2=V_2{A}_2\\ {\dot{V}}_2=V_2\left[\frac{\pi }{4}{\left(0.45\text{ m}\right)}^2\right]\end{array}$        (II)

The Reynolds number in the first pipe is,

  $\begin{array}{c}{\mathrm{Re}}_1=\frac{\rho V_1{D}_1}{\mu }\\ =\frac{\left(957.9\text{kg}/{\text{m}}^3\right)V_1\left(0.3\text{ m}\right)}{\left(0.282\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\end{array}$        (III)

The Reynolds number in the second pipe is,

  $\begin{array}{c}{\mathrm{Re}}_2=\frac{\rho V_2{D}_2}{\mu }\\ =\frac{\left(957.9\text{kg}/{\text{m}}^3\right)V_2\left(0.45\text{ m}\right)}{\left(0.282\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\end{array}$        (IV)

The friction factor is determined from the Moody chart as follows:

  $\begin{array}{l}\frac{1}{\sqrt{f_1}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f_1}}\right)\\ \frac{1}{\sqrt{f_1}}=-2.0\log \left(\frac{0.000045\text{m}/0.3\text{m}}{3.7}+\frac{2.51}{{\mathrm{Re}}_1\sqrt{f_1}}\right)\end{array}$        (V)

  $\frac{1}{\sqrt{f_2}}=-2.0\log \left(\frac{0.000045\text{m}/0.45\text{m}}{3.7}+\frac{2.51}{{\mathrm{Re}}_2\sqrt{f_2}}\right)$        (VI)

The head loss is,

  $h_L=h_{L,1}=h_{L,2}$        (VII)

The head loss in pipe 1 is,

  $\begin{array}{l}{h}_{L,1}=f_1\frac{L_1}{D_1}\frac{V_1{}^2}{2g}\\ h_{L,1}=f_1\left(\frac{\text{500 m}}{0.3\text{ m}}\right)\left[\frac{V_1{}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\right]\end{array}$        (VIII)

The head loss in pipe 2 is,

  $\begin{array}{l}{h}_{L,2}=f_2\frac{L_2}{D_2}\frac{V_2{}^2}{2g}\\ h_{L,2}=f_2\left(\frac{\text{800 m}}{0.45\text{ m}}\right)\left[\frac{V_2{}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\right]\end{array}$        (IX)

The total flow rater between the reservoirs is,

  $\dot{V}={\dot{V}}_1+{\dot{V}}_2$        (X)

We have 10 equations with 10 unknowns. Solving the system of simultaneous equations, we get;

  $\begin{array}{l}{\dot{V}}_1=0.919{\text{m}}^3/\text{s}\\ {\dot{V}}_2=2.081{\text{m}}^3/\text{s}\\ V_1=13.0\text{m}/\text{s}\\ V_2=13.1\text{m}/\text{s}\\ h_L=187\text{ m}\\ {\text{Re}}_1=1.324\times {10}^7\\ {\mathrm{Re}}_2=2.00\times {10}^7\\ f_1=0.0131\\ f_2=0.0121\end{array}$

Thus, the flow rate through pipe A and pipe B is $0.919{\text{m}}^3/\text{s}$ and $2.081{\text{m}}^3/\text{s}$ respectively.