Chapter 8, Problem 82P

Water to a residential area is transported at a rate of 1.5 ${\text{m}}^3\text{/s}$ 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 0.04 mm. There is a concern that the reduction of pipe diameter to 66 cm and the increase in average velocity may offset any gains. Taking $\rho =100{\text{kg/m}}^3$ and $v=1\times {10}^{-6}{\text{m}}^2\text{/s}$ for water, determine the percent increase or decrease in the pumping power requirements due to pipe frictional losses as a result of lining the concrete pipes.

To determine

The increase or decrease of percent in the pumping power requirements due to pipe frictional losses as a result of lining concrete pipes.

Answer to Problem 82P

The percent decrease in the pumping power requirements is $54.12\%$ .

Explanation of Solution

Given information:

The volume flow rate is $1.5{\text{m}}^{\text{3}}/\text{s}$, the internal diameter of the pipe is $70\text{cm}$, the surface roughness thickness of the pipe is $3\text{mm}$, the total length of the pipe is $1500\text{m}$, the thickness of petroleum based lining in the pipe is $2\text{cm}$, the surface roughness thickness of the petroleum based lining is $0.04\text{mm}$, the reduction of pipe diameter is $66\text{cm}$, the density of water is $1000\text{kg}/{\text{m}}^{\text{3}}$, and the kinematic viscosity of water is $1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}$.

Write the expression for the cross-sectional area of pipe without coating.

   $A_C=\frac{\pi }{4}{D}^2$ ...... (I)

Here, the internal diameter of the pipe is $D$.

Write the expression for the average velocity of flow in the pipe.

   $V=\frac{\dot{\vee }}{A_C}$ ...... (II)

Here, the volume flow rate of water is $\dot{\vee }$, and the cross-sectional area of pipe is $A_C$.

Write the expression for finding the Reynolds number.

   $R_e=\frac{VD}{v}$ ...... (III)

Here, the kinematic viscosity is $v$, average velocity of flow in the pipe is $V$, and the internal diameter of the pipe is $D$.

Write the expression for the Colebrook relation of the friction factor for turbulent flow.

   $\frac{1}{\sqrt{f_1}}=-0.2\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{R_e\sqrt{f_1}}\right)$ ...... (IV)

Here, the friction factor is $f$, the Reynolds number is $R_e$ ,and the surface roughness of the pipe is $\epsilon$.

Write the expression for the head loss in the pipe1.

   $h_L=f_1\frac{L}{D}\frac{V^2}{2g}$ ...... (V)

Here, the length of the pipe is $L$, the acceleration due to gravity is $g$ ,and the average velocity of flow in the pipe is $V$.

Write the expression for the new diameter of the pipe with the lining.

   $D_r=D-2t$ ...... (VI)

Here, the thickness of the pipe is $t$, and the internal diameter of the pipe is $D$.

Write the expression for the new cross-sectional area of the pipe.

   ${A^{\prime }}_C=\frac{\pi }{4}{D}_r$ ....... (VII)

Here, the new diameter of the pipe with the lining is $D_r$.

Write the expression for the new average velocity.

   $V_r=\frac{\dot{\vee }}{{A^{\prime }}_C}$ ...... (VIII)

Here, the new cross-sectional area of the pipe is ${A^{\prime }}_C$.

Write the expression for finding the Reynolds number.

   $R_e=\frac{V_r{D}_r}{v}$ ...... (IX)

Here, the new average velocity is $V_r$.

Write the expression for the new friction function from the Colebrook relation.

   $\frac{1}{\sqrt{f_2}}=-0.2\log \left(\frac{{\epsilon }_2/D_r}{3.7}+\frac{2.51}{R_e\sqrt{f_2}}\right)$ ...... (X)

Here, the Reynolds number is $R_e$, the surface roughness of the lining is ${\epsilon }_2$.

Write the expression for finding the new head loss.

   ${h^{\prime }}_L=f_2\frac{L}{D_r}\frac{V_r^2}{2g}$ ...... (XI)

Write the expression for the percentage reduction.

   $\text{\%reduction}=\frac{{h^{\prime }}_L}{h_L}\times 100$ ...... (XII)

Here, the new head loss is ${h^{\prime }}_L$, and the head loss is $h_L$.

Calculation:

Substitute $70\text{cm}$ for $D$ in the Equation (I).

   $\begin{array}{c}{A}_C=\frac{\pi }{4}{\left(70\text{cm}\left(1\text{m}/100\text{cm}\right)\right)}^2\\ =\frac{\pi }{4}{\left(0.7\text{m}\right)}^2\\ =0.3848{\text{m}}^{\text{2}}\end{array}$

Substitute $1.5{\text{m}}^{\text{3}}/\text{s}$ for $\dot{\vee }$, and $0.3848{\text{m}}^{\text{2}}$ for $A_C$ in the Equation (II).

   $\begin{array}{c}V=\frac{1.5{\text{m}}^{\text{3}}/\text{s}}{0.3848{\text{m}}^{\text{2}}}\\ =3.8976\text{m}/\text{s}\end{array}$

Substitute $1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}$ for $v$, $70\text{cm}$ for $D$, and $3.8976\text{m}/\text{s}$ for $V$ in the equation (III).

   $\begin{array}{c}{R}_e=\frac{\left(3.8976\text{m}/\text{s}\right)\left(70\text{cm}\right)}{1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{\left(3.8976\text{m}/\text{s}\right)\left(70\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{\left(3.8976\text{m}/\text{s}\right)\left(0.7\text{m}\right)}{1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}}\\ =2.7283\times {10}^6\end{array}$

Substitute $3\text{mm}$ for $\epsilon$, $70\text{cm}$ for $D$, and $2.7283\times {10}^6$ for $R_e$ in the Equation (IV).

   $\begin{array}{c}\frac{1}{\sqrt{f_1}}=-0.2\log \left(\begin{array}{l}\frac{3\text{mm}/70\text{cm}}{3.7}\\ +\frac{2.51}{2.7283\times {10}^6\sqrt{f_1}}\end{array}\right)\\ \frac{1}{\sqrt{f_1}}=-0.2\log \left(\begin{array}{l}\frac{3\text{mm}\left(\frac{\text{1}\text{m}}{\text{1000}\text{mm}}\right)/70\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}{3.7}\\ +\frac{2.51}{2.7283\times {10}^6\sqrt{f_1}}\end{array}\right)\\ \frac{1}{\sqrt{f_1}}=-0.2\log \left(1.1583\times {10}^{-3}+\frac{9.1996\times {10}^{-7}}{\sqrt{f_1}}\right)\\ f_1=0.029\end{array}$

Substitute $0.029$ for $f_1$, $1500\text{m}$ for $L$, $70\text{cm}$ for $D$, $3.8976\text{m}/\text{s}$ for $V$, and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the Equation (V).

   $\begin{array}{c}{h}_L=0.029\left(\frac{1500\text{m}}{70\text{cm}}\right)\left(\frac{{\left(3.8976\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)\\ =0.029\left(\frac{1500\text{m}}{70\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)}\right)\left(\frac{{\left(3.8976\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)\\ =0.029\left(\frac{1500\text{m}}{0.7\text{m}}\right)\left(0.77427\text{m}\right)\\ =48.11\text{m}\end{array}$

Substitute $70\text{cm}$ for $D$, and $2\text{cm}$ for $t$ in the Equation (VI).

   $\begin{array}{c}{D}_r=70\text{cm}-2\left(2\text{cm}\right)\\ =70\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)-2\left(2\text{cm}\right)\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right)\\ =0.7\text{m}-\left(2\times 0.02\text{m}\right)\\ =0.66\text{m}\end{array}$

Substitute $0.66\text{m}$ for $D_r$ in the Equation (VII).

   $\begin{array}{c}{A^{\prime }}_C=\frac{\pi }{4}{\left(0.66\text{m}\right)}^2\\ =\frac{\pi }{4}\left(0.4356{\text{m}}^{\text{2}}\right)\\ =0.3421{\text{m}}^2\end{array}$

Substitute $1.5{\text{m}}^{\text{3}}/\text{s}$ for $\dot{\vee }$, and $0.3421{\text{m}}^2$ for ${A^{\prime }}_C$ in the Equation (VIII).

   $\begin{array}{c}{V}_r=\frac{1.5{\text{m}}^{\text{3}}/\text{s}}{0.3421{\text{m}}^{\text{2}}}\\ =4.384\text{m}/\text{s}\end{array}$

Substitute $1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}$ for $v$, $0.66\text{m}$ for $D_r$, and $4.384\text{m}/\text{s}$ for $V_r$ in the equation (IX).

   $\begin{array}{c}{R}_e=\frac{\left(4.384\text{m}/\text{s}\right)\left(0.66\text{m}\right)}{1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}}\\ =\frac{2.89344{\text{m}}^{\text{2}}/\text{s}}{1\times {10}^{-6}{\text{m}}^{\text{2}}/\text{s}}\\ =2.894\times {10}^6\end{array}$

Substitute $0.04\text{mm}$ for $\epsilon$, $0.66\text{m}$ for $D_r$, and $2.894\times {10}^6$ for $R_e$ in the Equation (X).

   $\begin{array}{c}\frac{1}{\sqrt{f_2}}=-0.2\log \left(\frac{0.04\text{mm}/0.66\text{m}}{3.7}+\frac{2.51}{2.894\times {10}^6\sqrt{f_2}}\right)\\ \frac{1}{\sqrt{f_2}}=-0.2\log \left(\begin{array}{l}\frac{0.04\text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)/0.66\text{m}}{3.7}\\ +\frac{2.51}{2.894\times {10}^6\sqrt{f_2}}\end{array}\right)\\ \frac{1}{\sqrt{f_2}}=-0.2\log \left(\frac{4\times {10}^{-5}\text{m}}{3.7}+\frac{2.51}{2.894\times {10}^6\sqrt{f_2}}\right)\\ \frac{1}{\sqrt{f_2}}=-0.2\log \left(1.08108\times {10}^{-5}+\frac{8.6731\times {10}^{-7}}{\sqrt{f_2}}\right)\end{array}$

   $f_2=0.0117$

Substitute $0.0117$ for $f_2$, y $1500\text{m}$ for 13 $L$, $0.66\text{m}$ for $D_r$,

   $4.384\text{m}/\text{s}$ for $V_r$, and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the Equation (XI).

   $\begin{array}{c}{h^{\prime }}_L=0.0117\left(\frac{1500\text{m}}{0.66\text{m}}\right)\left(\frac{{\left(4.384\text{m}/\text{s}\right)}^2}{2\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\right)\\ =0.0117\left(2272.72\text{m}\right)\left(0.2234\text{m}\right)\\ =26.04\text{m}\end{array}$

Substitute $26.04\text{m}$ for ${h^{\prime }}_L$, and $48.11\text{m}$ for $h_L$ in the Equation (XII).

   $\begin{array}{c}\text{\%reduction}=\frac{26.04\text{m}}{48.11\text{m}}\times 100\\ =0.54125\times 100\\ =54.12\%\end{array}$

Conclusion:

The percent decrease in the pumping power requirements is $54.12\%$ .