Chapter 14, Problem 104RQ

To determine

The optimum diameter for the pipe.

Explanation of Solution

Given:

The mass flow rate is $10000\text{kg}/\text{s}$.

The length of the pipe is $10\text{ km}$.

The friction factor is 0.015.

Total life is 30 years.

Electricity cost is $\$0.06/\text{kWh}$.

Overall pump efficiency is 80%.

Initial cost is $\${10}^6{D}^2$.

Calculation:

Apply energy equation,

  $\begin{array}{l}\frac{P_1}{\rho g}+{\alpha }_1\frac{V_1{}^2}{2g}+z_1+h_{pump}=\frac{P_2}{\rho g}+{\alpha }_2\frac{V_2{}^2}{2g}+z_2+h_{turbine}+h_L\\ h_{pump}=h_L\end{array}$

The volume flow rate is,

  $\begin{array}{c}\dot{V}=\frac{\dot{m}}{\rho }\\ =\frac{10000\text{kg}/\text{s}}{1000\text{kg}/{\text{m}}^3}\\ =10{\text{m}}^3/\text{s}\end{array}$

The head loss is,

  $\begin{array}{c}\Delta P=f\frac{L}{D}\frac{\rho V^2}{2}\\ =f\frac{L}{D}\frac{\rho }{2}{\left(\frac{\dot{V}}{\pi D^2/4}\right)}^2\\ =f\frac{16L}{D}\frac{\rho {\dot{V}}^2}{2{\pi }^2{D}^4}\\ =\frac{8fL\rho {\dot{V}}^2}{{\pi }^2{D}^5}\end{array}$

The power required by the pump is,

  $\begin{array}{c}{\dot{W}}_{pump}=\frac{\dot{V}\Delta P}{{\eta }_{pump}}\\ =\frac{\dot{V}}{{\eta }_{pump}}\times \frac{8fL\rho {\dot{V}}^2}{{\pi }^2{D}^5}\\ =\frac{8fL\rho {\dot{V}}^3}{{\eta }_{pump}{\pi }^2{D}^5}\\ =\frac{8\left(0.015\right)\left(10000\text{ m}\right)\left(1000\text{kg}/{\text{m}}^3\right){\left(10{\text{m}}^3/\text{s}\right)}^3}{\left(0.80\right){\pi }^2{D}^5}\\ =\frac{1.52\times {10}^5}{D^5}\text{kW}\end{array}$

The electric power used for one year is,

  $\begin{array}{c}{E}_{pump}={\dot{W}}_{pump}\Delta t\\ =\frac{1.52\times {10}^5}{D^5}\left(24\times 365\right)\\ =\frac{1.332\times {10}^9}{D^5}\text{kWh}/\text{yr}\end{array}$

The energy cost for one year is,

  $\begin{array}{c}Energy\cos t=E_{pump}\times \text{ unit cost}\\ \text{=}\left(\frac{1.332\times {10}^9}{D^5}\text{kWh}/\text{yr}\right)\left(\$0.06/\text{kWh}\right)\\ =\frac{7.99\times {10}^7}{D^5}\$/\text{yr}\end{array}$

The installation cost per year is,

  $\begin{array}{c}System\cos t=\frac{Total\cos t}{Lifetime}\\ =\frac{\${10}^6{D}^2}{30\text{ yr}}\\ =3.33\times {10}^4{D}^2\$/\text{yr}\end{array}$

The total annual cost is,

  $Total\cos t=\left(\frac{7.99\times {10}^7}{D^5}+3.33\times {10}^4{D}^2\right)\$/\text{yr}$

For the optimum cost,

  $\begin{array}{l}\frac{\partial \left(Total\cos t\right)}{\partial D}=0\\ \frac{\partial }{\partial D}\left(\frac{7.99\times {10}^7}{D^5}+3.33\times {10}^4{D}^2\right)=0\\ -5\frac{7.99\times {10}^7}{D^6}+6.66\times {10}^{4}D=0\\ 6.66\times {10}^4{D}^7=5\left(7.99\times {10}^7\right)\\ D^7=5998\\ D=3.5\text{ m}\end{array}$

Thus, the optimum diameter of the pipe is $3.5\text{ m}$.