Chapter 14, Problem 47P

Calculate the volume flow rate between the reservoirs of Prob. 14-39 for the case in which the pipe diameter is doubled, all else remaining the same. Discuss.

To determine

The volume capacity delivered by the pump.

Answer to Problem 47P

The volume capacity delivered by the pump is $15.3798\text{Lpm}$.

Explanation of Solution

Given information:

The diameter of pipe is $4.06\text{cm}$, elevation difference is $7.85\text{m}$, total length of pipe is $176.5\text{m}$, shutoff head is $24.4\text{m}$, coefficient $a$is $0.0678\text{m}/{\text{Lpm}}^{\text{2}}$, pipe roughness is $0.25\text{mm}$, coefficient of entrance loss is $0.5$, coefficient of valve loss is $17.5$, coefficient of elbow loss is $0.92$and coefficient of exit loss is $1.05$.

Write the expression for head required.

  $H_{\text{required}}=\frac{P_2-P_1}{\rho g}+\frac{V_2^2-V_1^2}{2g}+\left(z_2-z_1\right)+h_{\text{L}}$   ....(I)

Here, pressure at exit is $P_2$, pressure at inlet is $P_1$, velocity at exit is $V_2$, velocity inlet is $V_1$, elevation difference is $z_2-z_1$, head loss is $h_{\text{L}}$.

Write the expression for coefficient of total minor head loss.

  $K=K_1+K_2+K_3+K_4$  ....(II)

Here, coefficient of entrance loss is $K_1$, coefficient ofvalve loss is $K_2$, coefficient of elbow loss is $K_3$, coefficient of exit loss is $K_4$.

Write the expression for head loss.

  $h_{\text{L}}=\left(\frac{fL}{D}+K\right)\frac{V^2}{2g}$   ....(III)

Here, length of pipe is $L$, diameter of pipe is $D$, coefficient of friction loss is $f$, velocity of flow is $V$and acceleration due to gravity is $g$.

Write the expression for $H_{\text{available}}$.

  $H_{\text{available}}=H_0-a{\dot{V}}^2$  ....(IV)

Here, shutoff head is $H_0$, coefficient is $a$and volume flow rate is $\dot{V}$.

Write the expression for Reynolds number.

  $\mathrm{Re}=\frac{\rho VD}{\mu }$.  ....(V)

Here, density of water is $\rho$and dynamic viscosity is $\mu$.

Write the expression for Colebrook equation.

  $\frac{1}{\sqrt{f}}=-2.0\log \left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$  ....(VI)

Here, coefficient of friction is $f$and pipe roughness factor is $\epsilon$.

Write the expression for volume flow rate.

  $\dot{V}=\frac{\pi }{4}{D}^{2}V$   ....(VII)

Calculation:

Substitute $0.5$for $K_1$, $17.5$for $K_2$, $5\times 0.92$for $K_3$, $1.05$for $K_4$in Equation (II).

  $\begin{array}{c}K=0.5+17.5+\left(5\times 0.92\right)+1.05\\ =0.5+17.5+4.6+1.05\\ =23.65\end{array}$

Substitute $23.65$for $K$, $176.5\text{m}$for $L$, $2.03\text{cm}$for $D$and $9.81\text{m}/{\text{s}}^{\text{2}}$for $g$in Equation (III).

  $\begin{array}{c}{h}_{\text{L}}=\left(\frac{f\times 176.5\text{m}}{4.06\text{cm}}+23.65\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(\frac{f\times 176.5\text{m}}{4.06\text{cm}\times \frac{1\text{m}}{100\text{cm}}}+23.65\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(\frac{f\times 176.5\text{m}}{0.0406\text{m}}+23.65\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(4347.29f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\end{array}$

Substitute $0$for $P_1$, $0$for $P_2$, $0$for $V_1$, $0$for $V_2$, $9.81\text{m}/{\text{s}}^{\text{2}}$for $g$, $7.85\text{m}$for $z_2-z_1$, $\left(4347.29f+23.65\right)\frac{V^2}{19.62}$for $h_{\text{L}}$in Equation (I).

  $\begin{array}{c}{H}_{\text{required}}=\frac{0-0}{\rho g}+\frac{0-0}{2g}+\left(7.85\text{m}\right)+\left(4347.29f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ =\left(7.85\text{m}\right)+\left(4347.29f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\end{array}$

Substitute $4.06\text{cm}$for $D$in Equation (VII).

  $\begin{array}{c}\dot{V}=\frac{\pi }{4}{\left(4.06\text{cm}\right)}^{2}V\\ =\frac{\pi }{4}{\left(4.06\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^{2}V\\ =\frac{\pi }{4}{\left(0.0406\text{m}\right)}^{2}V\\ =1.2946\times {10}^{-3}{\text{m}}^2\times V\end{array}$

Refer Table-A-3 "Properties of saturated water" at $20°\text{C}$to obtain density of water as $998\text{kg}/{\text{m}}^{\text{3}}$and dynamic viscosity as $1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$

Substitute $3.2365\times {10}^{-4}{\text{m}}^{\text{2}}\times V$for $\dot{V}$, $24.4\text{m}$for $H_0$, $0.0678\text{m}/{\text{Lpm}}^{\text{2}}$for $a$and $\left(7.85\text{m}\right)+\left(4347.29f+23.65\right)\frac{V^2}{19.62}$for $H_{\text{available}}$in Equation (IV).

  $\begin{array}{l}\left[\begin{array}{l}\left(7.85\text{m}\right)+\left(4347.29f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ =24.4\text{m}-0.0678\text{m}/{\text{Lpm}}^{\text{2}}{\left(3.2365\times {10}^{-4}{\text{m}}^{\text{2}}\times V\right)}^2\end{array}\right]\\ \left[\begin{array}{l}\left(4347.29f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ =16.55\text{m}-\left(0.0678\text{m}/{\text{Lpm}}^{\text{2}}\times \frac{3.6\times {10}^8{\text{s}}^{\text{2}}/{\text{m}}^{\text{5}}}{1\text{m}/{\text{Lpm}}^{\text{2}}}\right){\left(3.2365\times {10}^{-4}{\text{m}}^{\text{2}}\times V\right)}^2\end{array}\right]\\ \left[\begin{array}{l}\left(4347.29f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ =16.55\text{m}-\left(244080000{\text{s}}^{\text{2}}/{\text{m}}^{\text{5}}\right){\left(3.2365\times {10}^{-4}{\text{m}}^{\text{2}}\times V\right)}^2\end{array}\right]\\ \left(221.5744{\text{s}}^{\text{2}}/\text{m}f+1.205{\text{s}}^{\text{2}}/\text{m}\right)V^2=16.55\text{m}-25.56{\text{s}}^{\text{2}}/\text{m}\times V^2\end{array}$

  $f=\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079$

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$for $\rho$, $4.06\text{cm}$for $D$and $1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$for $\mu$in Equation (V).

  $\begin{array}{c}\mathrm{Re}=\frac{998\text{kg}/{\text{m}}^{\text{3}}\times 4.06\text{cm}\times V}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{998\text{kg}/{\text{m}}^{\text{3}}\times 4.06\text{cm}\times \frac{1\text{m}}{100\text{cm}}\times V}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =40437.92\text{s}/\text{m}\times V\end{array}$

Substitute $0.25\text{mm}$for $\epsilon$, $4.06\text{cm}$for $D$, $40437.92\text{s}/\text{m}\times V$for $\mathrm{Re}$and $\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079$for $f$in Equation (VI).

  $\begin{array}{l}\left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\\ =-2.0\log \left(\begin{array}{l}\frac{0.25\text{mm}/4.06\text{cm}}{3.7}\\ +\frac{2.51}{40437.92\text{s}/\text{m}\times V\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\end{array}\right)\end{array}\right]\\ \left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\\ =-2.0\log \left(\begin{array}{l}\frac{0.25\text{mm}/4.06\text{cm}\times \frac{10\text{mm}}{1\text{cm}}}{3.7}\\ +\frac{2.51}{40437.92\text{s}/\text{m}\times V\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\end{array}\right)\end{array}\right]\\ \left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\\ =-2.0\log \left(\begin{array}{l}0.001664225\\ +\frac{2.51}{40437.92\text{s}/\text{m}\times V\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\end{array}\right)\end{array}\right]\end{array}$  ....(VIII)

Here, apply hit and trial method to obtain value of velocity of flow.

Trial-(1)

Substitute $1\text{m}/\text{s}$for $V$in Equation (VIII).

  $\begin{array}{l}\left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\\ =-2.0\log \left(\begin{array}{l}0.00166216\\ +\frac{2.51}{40437.92\text{s}/\text{m}\times 1\text{m}/\text{s}\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.12079}}\end{array}\right)\end{array}\right]\\ \frac{1}{\sqrt{-0.046098}}=-2.0\log \left(\begin{array}{l}0.00166216\\ +\frac{2.51}{40437.92\sqrt{-0.046098}}\end{array}\right)\end{array}$

Since the value on both sides are not equal, so assumption of $V=1\text{m}/\text{s}$is not correct.

Trial-(2)

Substitute $0.198\text{m}/\text{s}$for $V$in Equation (VIII)

  $\begin{array}{l}\left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{{\left(0.198\text{m}/\text{s}\right)}^2}-0.12079}}\\ =-2.0\log \left(\begin{array}{l}0.00166216\\ +\frac{2.51}{40437.92\text{s}/\text{m}\times 0.198\text{m}/\text{s}\sqrt{\frac{0.074692{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{{\left(0.198\text{m}/\text{s}\right)}^2}-0.12079}}\end{array}\right)\end{array}\right]\\ \frac{1}{\sqrt{1.784423}}=-2.0\log \left(\begin{array}{l}0.00166216\\ +\frac{2.51}{40437.92\sqrt{1.784423}}\end{array}\right)\\ 0.7486020=0.7486021\end{array}$

Here, the values on both sides are equal, so velocity of flow is $0.198\text{m}/\text{s}$

Substitute $0.198\text{m}/\text{s}$for $V$in Equation (VII).

  $\begin{array}{c}\dot{V}=1.2946\times {10}^{-3}{\text{m}}^2\times 0.198\text{m}/\text{s}\\ =2.563308\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\times \frac{60000\text{Lpm}}{1{\text{m}}^{\text{3}}/\text{s}}\\ =15.3798\text{Lpm}\end{array}$

Conclusion:

The volume capacity delivered by the pump is $15.3798\text{Lpm}$.