Chapter 8, Problem 96P

Water to a residential area is transported at a rate of 1.5 m³/s via 70-cm-internal-diameter concrete pipes with a surface roughness of 3 mm and a total length of 1500 m. In order to reduce pumping power requirements, it is proposed to line the interior surfaces of the concrete pipe with 2-cm-thick petroleum-based lining that has a surface roughness thickness of 004 mm, this is a concern that the reduction of pipe diameter to 66 cm and the increase in average velocity may offset any gains. Taking p = 1000 kg/m³ and v = 1 × 10-6 m²/s for water, determine the percent increase or decrease in the pumping power requirements due to pipe frictional losses as a result of lining time concrete pipes.

To determine

The percentage change in pumping power requirement due to frictional losses.

Answer to Problem 96P

The percentage decrease in pumping power requirement due to frictional losses is $54.157\%$.

Explanation of Solution

Given information:

The density of the water is $1000\text{kg}/{\text{m}}^{\text{3}}$, kinematic viscosity of the water is $1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}$, diameter of the pipe without line interior surface is $70\text{cm}$, diameter of the pipe with line interior surface is $66\text{cm}$, roughness of the line interior concrete pipe surfaces is $0.04\text{mm}$, roughness of the concrete pipe surfaces is $3\text{mm}$, volume flow rate of water is $1.5{\text{m}}^{\text{3}}/\text{s},$ and length of the pipe is $1500\text{m}$.

Write the expression for the Reynolds number.

  $\mathrm{Re}=\frac{vD}{\upsilon }$........... (I)

Here, Reynolds number is $\mathrm{Re}$, velocity of the water is $v$, diameter of the pipe is $D,$ and kinematic viscosity of the water is $\upsilon$.

Write the expression for the volume flow rate of the water.

  $\dot{Q}=\frac{\pi }{4}{D}^{2}v$........... (II)

Here, the volume flow rate of the water is $\dot{Q}$.

Write the expression for the friction factor for turbulent flow.

  $\frac{1}{\sqrt{f}}=-2\log \left(\frac{\frac{\epsilon }{D}}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$........... (III)

Here, friction factor for turbulent flow is $f$ and roughness of the plastic pipe surfaces is $\epsilon$.

Write the expression for the head loss through the pipe.

  $h_f=\frac{fL{v}^2}{2gD}$........... (IV)

Here, length of the pipe is $L$, gravitational acceleration is $g,$ and the head loss through the pipe is $h_f$.

Write the expression for the pressure drop through the pipe.

  $\Delta P=\rho g{h}_f$........... (V)

Here, the pressure drop through the pipe is $\Delta P$ and density of the water is $\rho$.

Write the expression for the pumping power requirement to maintain the flow rate of water.

  ${\dot{W}}_{\text{pump}}=\dot{Q}\Delta P$........... (VI)

Here, pumping power requirement to maintain the flow rate of water is ${\dot{W}}_{\text{pump}}$.

Substitute $\rho g{h}_f$ for $\Delta P$ and $\frac{\pi }{4}{D}^{2}v$ for $\dot{Q}$ in Equation (VI).

  $\begin{array}{c}{\dot{W}}_{\text{pump}}=\left(\frac{\pi }{4}{D}^{2}v\right)\left(\rho g{h}_f\right)\\ =\frac{\pi }{4}{D}^{2}v\rho g{h}_f\end{array}$........... (VII)

Write the expression for the pumping power for without line interior surface pipe.

  ${\left({\dot{W}}_{\text{pump}}\right)}_{\text{without}\text{line}}=\frac{\pi }{4}\rho g{\left(v{h}_f{D}^2\right)}_{\text{without}\text{line}}$........... (VIII)

Here, subscript for without line interior pipe is without line.

Write the expression for the pumping power for with line interior surface pipe.

  ${\left({\dot{W}}_{\text{pump}}\right)}_{\text{with}\text{line}}=\frac{\pi }{4}\rho g{\left(v{h}_f{D}^2\right)}_{\text{with}\text{line}}$........... (IX)

Here, subscript for with line interior pipe is with line.

Write the expression for the percentage change in power due to frictional losses.

  $m=\left(\frac{{\left({\dot{W}}_{\text{pump}}\right)}_{\text{with}\text{line}}}{{\left({\dot{W}}_{\text{pump}}\right)}_{\text{without}\text{line}}}\right)\times 100\%$........... (X)

Here percentage change in power due to frictional losses is $m$.

Substitute $\frac{\pi }{4}\rho g{\left(v{h}_f{D}^2\right)}_{\text{without}\text{line}}$ for ${\left({\dot{W}}_{\text{pump}}\right)}_{\text{without}\text{line}}$ and $\frac{\pi }{4}\rho g{\left(v{h}_f{D}^2\right)}_{\text{with}\text{line}}$ for ${\left({\dot{W}}_{\text{pump}}\right)}_{\text{with}\text{line}}$ in Equation (X).

  $\begin{array}{c}m=\left(\frac{\frac{\pi }{4}\rho g{\left(v{h}_f{D}^2\right)}_{\text{with}\text{line}}}{\frac{\pi }{4}\rho g{\left(v{h}_f{D}^2\right)}_{\text{without}\text{line}}}\right)\times 100\%\\ =\left(\frac{{\left(v{h}_f{D}^2\right)}_{\text{with}\text{line}}}{{\left(v{h}_f{D}^2\right)}_{\text{without}\text{line}}}\right)\times 100\%\end{array}$........... (XI)

Calculation:

Substitute $1.5{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $70\text{cm}$ for $D$ and $v_{\text{without}\text{line}}$ for $v$ in Equation (II).

  $\begin{array}{l}\left(1.5{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(70\text{cm}\right)}^2{\left(v\right)}_{\text{without}\text{line}}\\ \left(1.5{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(\left(70\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]\right)}^2{\left(v\right)}_{\text{without}\text{line}}\\ {\left(v\right)}_{\text{without}\text{line}}=3.897\text{m}/\text{s}\end{array}$

Substitute $3.897\text{m}/\text{s}$ for $v$, $1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}$ for $\upsilon$, $70\text{cm}$ for $D$ and ${\mathrm{Re}}_{\text{without}\text{line}}$ for $\mathrm{Re}$ in Equation (I).

  $\begin{array}{c}{\mathrm{Re}}_{\text{without}\text{line}}=\frac{\left(3.897\text{m}/\text{s}\right)\left(70\text{cm}\right)}{\left(1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =\frac{\left(3.897\text{m}/\text{s}\right)\left(70\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}{\left(1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =2.728\times {10}^6\end{array}$

Substitute $2.728\times {10}^6$ for $\mathrm{Re}$, $3\text{mm}$ for $\epsilon$, $70\text{cm}$ for $D$ and $f_{\text{without}\text{line}}$ for $f$ in Equation (III).

  $\begin{array}{l}\frac{1}{\sqrt{f_{\text{without}\text{line}}}}=-2\log \left(\frac{\frac{\left(3\text{mm}\right)}{\left(70\text{cm}\right)}}{3.7}+\frac{2.51}{\left(2.728\times {10}^6\right)\sqrt{f_{\text{without}\text{line}}}}\right)\\ \frac{1}{\sqrt{f_{\text{without}\text{line}}}}=-2\log \left(\frac{\frac{\left(3\text{mm}\right)\left[\frac{1\text{cm}}{10\text{mm}}\right]}{\left(70\text{cm}\right)}}{3.7}+\frac{2.51}{\left(2.728\times {10}^6\right)\sqrt{f_{\text{without}\text{line}}}}\right)\\ f_{\text{without}\text{line}}=0.029\end{array}$

Substitute $0.029$ for $f$, $1500\text{m}$ for $L$, $3.897\text{m}/\text{s}$ for $v$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $70\text{cm}$ for $D$ and ${\left(h_f\right)}_{\text{without}\text{line}}$ for $h_f$ in Equation (IV).

  $\begin{array}{c}{\left(h_f\right)}_{\text{without}\text{line}}=\frac{\left(0.029\right)\left(1500\text{m}\right){\left(3.897\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(70\text{cm}\right)}\\ =\frac{\left(0.029\right)\left(1500\text{m}\right){\left(3.897\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(70\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}\\ =48.1\text{m}\end{array}$

Substitute $1.5{\text{m}}^{\text{3}}/\text{s}$ for $\dot{Q}$, $66\text{cm}$ for $D$ and $v_{\text{with}\text{line}}$ for $v$ in Equation (II).

  $\begin{array}{l}\left(1.5{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(66\text{cm}\right)}^2{\left(v\right)}_{\text{with}\text{line}}\\ \left(1.5{\text{m}}^{\text{3}}/\text{s}\right)=\frac{\pi }{4}{\left(\left(66\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]\right)}^2{\left(v\right)}_{\text{with}\text{line}}\\ {\left(v\right)}_{\text{with}\text{line}}=4.384\text{m}/\text{s}\end{array}$

Substitute $4.384\text{m}/\text{s}$ for $v$, $1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}$ for $\upsilon$, $66\text{cm}$ for $D$ and ${\mathrm{Re}}_{\text{with}\text{line}}$ for $\mathrm{Re}$ in Equation (I).

  $\begin{array}{c}{\mathrm{Re}}_{\text{with}\text{line}}=\frac{\left(4.384\text{m}/\text{s}\right)\left(66\text{cm}\right)}{\left(1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =\frac{\left(4.384\text{m}/\text{s}\right)\left(66\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}{\left(1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}\right)}\\ =2.893\times {10}^6\end{array}$

Substitute $2.893\times {10}^6$ for $\mathrm{Re}$, $0.04\text{mm}$ for $\epsilon$, $66\text{cm}$ for $D$ and $f_{\text{with}\text{line}}$ for $f$ in Equation (III).

  $\begin{array}{l}\frac{1}{\sqrt{f_{\text{with}\text{line}}}}=-2\log \left(\frac{\frac{\left(0.04\text{mm}\right)}{\left(66\text{cm}\right)}}{3.7}+\frac{2.51}{\left(2.893\times {10}^6\right)\sqrt{f_{\text{with}\text{line}}}}\right)\\ \frac{1}{\sqrt{f_{\text{with}\text{line}}}}=-2\log \left(\frac{\frac{\left(0.04\text{mm}\right)\left[\frac{1\text{cm}}{10\text{mm}}\right]}{\left(66\text{cm}\right)}}{3.7}+\frac{2.51}{\left(2.893\times {10}^6\right)\sqrt{f_{\text{with}\text{line}}}}\right)\\ f_{\text{with}\text{line}}=0.0117\end{array}$

Substitute $0.0117$ for $f$, $1500\text{m}$ for $L$, $4.384\text{m}/\text{s}$ for $v$, $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$, $66\text{cm}$ for $D$ and ${\left(h_f\right)}_{\text{with}\text{line}}$ for $h_f$ in Equation (IV).

  $\begin{array}{c}{\left(h_f\right)}_{\text{with}\text{line}}=\frac{\left(0.0117\right)\left(1500\text{m}\right){\left(4.384\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(66\text{cm}\right)}\\ =\frac{\left(0.0117\right)\left(1500\text{m}\right){\left(4.384\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(66\text{cm}\right)\left[\frac{1\text{m}}{100\text{cm}}\right]}\\ =26.048\text{m}\end{array}$

Substitute $3.897\text{m}/\text{s}$ for $v_{\text{without}\text{line}}$, $70\text{cm}$ for $D_{\text{without}\text{line}}$, $48.1\text{m}$ for ${\left(h_f\right)}_{\text{without}\text{line}}$, $4.384\text{m}/\text{s}$ for $v_{\text{with}\text{line}}$, $66\text{cm}$ for $D_{\text{with}\text{line}}$ and $26.048\text{m}$ for ${\left(h_f\right)}_{\text{with}\text{line}}$ in Equation (XI).

  $\begin{array}{c}m=\left(\frac{\left(4.384\text{m}/\text{s}\right)\left(26.048\text{m}\right){\left(66\text{cm}\right)}^2}{\left(3.897\text{m}/\text{s}\right)\left(48.1\text{m}\right){\left(70\text{cm}\right)}^2}\right)\times 100\%\\ =\left(\frac{\left(4.384\text{m}/\text{s}\right)\left(26.048\text{m}\right){\left(66\text{cm}\right)}^2}{\left(3.897\text{m}/\text{s}\right)\left(48.1\text{m}\right){\left(70\text{cm}\right)}^2}\right)\times 100\%\\ =54.157\%\end{array}$

Conclusion:

The percentage decrease in pumping power requirement due to frictional losses is $54.157\%$.