Chapter 14, Problem 91P

To determine

The flow rate between the reservoirs and through each pipe.

Explanation of Solution

Given:

Temperature of water is $20°\text{C}$.

The elevation of reservoir A is 2 m.

The elevation of reservoir B is 9 m.

Length of the pipes is $25\text{ m}$.

Diameter of the first pipe is $3\text{}\text{cm}$.

Diameter of the second pipe is $5\text{cm}$.

Efficiency of the pump is 68%.

Power of the pump is 7 kW.

Calculation:

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

  $\begin{array}{l}\text{Density,}\rho =998\text{kg}/{\text{m}}^3\\ \text{Dynamic viscosity,}\mu =1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\end{array}$

The power required by the pump is,

  $\begin{array}{c}{\dot{W}}_{pump}=\frac{\rho \dot{V}g{h}_{pump}}{{\eta }_{pump}}\\ 7000\text{ W}=\frac{\left(998\text{kg}/{\text{m}}^3\right)\dot{V}\left(9.81\text{m}/{\text{s}}^2\right)h_{pump}}{0.68}\end{array}$        (I)

Applying energy equation at points A and B at the free surface of water in reservoir A and B.

  $\begin{array}{l}\frac{P_A}{\rho g}+{\alpha }_A\frac{V_A{}^2}{2g}+z_A+h_{pump}=\frac{P_B}{\rho g}+{\alpha }_B\frac{V_B{}^2}{2g}+z_B+h_{turbine}+h_L\\ h_{pump}=\left(z_B-z_A\right)+h_L\end{array}$        (II)

From the rate of flow through the pipes,

  $\begin{array}{l}{\dot{V}}_1=V_1{A}_1\\ {\dot{V}}_1=V_1\left[\frac{\pi }{4}{\left(0.03\text{ m}\right)}^2\right]\end{array}$        (III)

  $\begin{array}{l}{\dot{V}}_2=V_2{A}_2\\ {\dot{V}}_2=V_2\left[\frac{\pi }{4}{\left(0.05\text{ m}\right)}^2\right]\end{array}$        (IV)

The Reynolds number in the first pipe is,

  $\begin{array}{c}{\mathrm{Re}}_1=\frac{\rho V_1{D}_1}{\mu }\\ =\frac{\left(998\text{kg}/{\text{m}}^3\right)V_1\left(0.03\text{ m}\right)}{\left(1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\end{array}$        (V)

The Reynolds number in the second pipe is,

  $\begin{array}{c}{\mathrm{Re}}_2=\frac{\rho V_2{D}_2}{\mu }\\ =\frac{\left(998\text{kg}/{\text{m}}^3\right)V_2\left(0.05\text{ m}\right)}{\left(1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}\right)}\end{array}$        (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)

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

The head loss is,

  $h_L=h_{L,1}$        (IX)

  $h_L=h_{L,2}$        (X)

The head loss in pipe 1 is,

  $\begin{array}{l}{h}_{L,1}=f_1\frac{L_1}{D_1}\frac{V_1{}^2}{2g}\\ h_{L,1}=f_1\left(\frac{25\text{ m}}{0.03\text{ m}}\right)\left[\frac{V_1{}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\right]\end{array}$        (XI)

The head loss in pipe 2 is,

  $\begin{array}{l}{h}_{L,2}=f_2\frac{L_2}{D_2}\frac{V_2{}^2}{2g}\\ h_{L,2}=f_2\left(\frac{25\text{ m}}{0.05\text{ m}}\right)\left[\frac{V_2{}^2}{2\left(9.81\text{m}/{\text{s}}^2\right)}\right]\end{array}$        (XII)

The total flow rater between the reservoirs is,

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

We have 13 equations with 13 unknowns. Solving the system of simultaneous equations, we get;

  $\begin{array}{l}\dot{V}=0.0183{\text{m}}^3/\text{s}\\ {\dot{V}}_1=0.0037{\text{m}}^3/\text{s}\\ {\dot{V}}_2=0.0146{\text{m}}^3/\text{s}\\ V_1=5.30\text{m}/\text{s}\\ V_2=7.42\text{m}/\text{s}\\ h_L=19.5\text{ m}\\ h_{pump}=26.5\text{ m}\\ {\text{Re}}_1=158300\\ {\mathrm{Re}}_2=369700\\ f_1=0.0164\\ f_2=0.0139\end{array}$

Thus, the flow rate between the reservoirs, flow rate through pipe 1 and pipe 2 are $0.0183{\text{m}}^3/\text{s}$, $0.0037{\text{m}}^3/\text{s}$ and $0.0146{\text{m}}^3/\text{s}$ respectively.