Chapter 14, Problem 77P

To determine

The discharge rate of water.

Explanation of Solution

Given:

Temperature of water is $15°\text{C}$.

Diameter of the first pipe is $10\text{}\text{cm}$.

Length of the first pipe is $20\text{ m}$.

Diameter of the second pipe is $4\text{cm}$.

Length of the second pipe is $\text{35 m}$.

Level of water in the reservoir is 18 m.

Calculation:

Refer Table A-15 “Properties of saturated water” from Appendix 1 to obtain the following properties of water:

  $\begin{array}{l}\text{Density,}\rho =999.1\text{kg}/{\text{m}}^3\\ \text{Dynamic viscosity,}\mu =1.138\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\end{array}$

Applying energy equation at points 1 at the free surface of water in the tank and point 2 at the exit of the pipe.

  $\begin{array}{l}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1{}^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2{}^2}{2g}+z_2+h_{turbine}+h_L\\ z_1={\alpha }_2\frac{V_2{}^2}{2g}+h_L\\ 18\text{m}=\frac{V_2{}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}+h_L\end{array}$        (I)

From the continuity equation,

  $\begin{array}{l}{V}_1{A}_1=V_2{A}_2\\ V_1=\frac{A_2}{A_1}{V}_2\\ V_1={\left(\frac{D_2}{D_1}\right)}^2{V}_2\\ V_1={\left(\frac{4\text{ cm}}{10\text{ cm}}\right)}^2{V}_2\\ V_1=0.16V_2\end{array}$        (II)

The head loss is,

  $\begin{array}{l}{h}_L=\left(f_1\frac{L_1}{D_1}+K_{L,entrance}\right)\frac{V_1{}^2}{2g}+\left(f_2\frac{L_2}{D_2}+K_{L,contraction}\right)\frac{V_2{}^2}{2g}\\ h_L=\left(f_1\frac{20\text{ m}}{0.10\text{ m}}+0.5\right)\frac{V_1{}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}+\left(f_2\frac{\text{35 m}}{0.04\text{ m}}+0.46\right)\frac{V_2{}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\end{array}$        (III)

The flow rate is,

  $\begin{array}{c}\dot{V}=V_2{A}_2\\ =\frac{\pi }{4}\left(0.04{\text{m}}^2\right)V_2\end{array}$        (IV)

The Reynolds number in the first pipe is,

  $\begin{array}{c}{\mathrm{Re}}_1=\frac{\rho V_1{D}_1}{\mu }\\ =\frac{\left(999.1\text{kg}/{\text{m}}^3\right)V_1\left(0.10\text{ m}\right)}{\left(1.38\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\end{array}$        (V)

The Reynolds number in the second pipe is,

  $\begin{array}{c}{\mathrm{Re}}_2=\frac{\rho V_2{D}_2}{\mu }\\ =\frac{\left(999.1\text{kg}/{\text{m}}^3\right)V_2\left(0.04\text{ m}\right)}{\left(1.38\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\end{array}$        (VI)

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

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

  $\frac{1}{\sqrt{f_2}}=-2.0\log \left(\frac{2.51}{{\mathrm{Re}}_2\sqrt{f_2}}\right)$        (VII)

We have 8 equations with 8 unknowns. Solving the system of simultaneous equations, we get;

  $\begin{array}{l}\dot{V}=0.00595{\text{m}}^3/\text{s}\\ V_1=0.757\text{m}/\text{s}\\ V_2=4.73\text{m}/\text{s}\\ h_L=16.86\text{ m}\\ {\text{Re}}_1=66500\\ {\mathrm{Re}}_2=166200\\ f_1=0.0196\\ f_2=0.0162\end{array}$

Thus, the discharge rate of water is $0.00595{\text{m}}^3/\text{s}$.