Chapter 14, Problem 41P

Suppose that the free surface of the inlet reservoir in Prob. 13-39 is 3.0 m higher in elevation, such that Z₂- z₁= 4.85 m. All the constants and parameters are identical to those of Prob. 14-39 except for the elevation difference. Calculate the volume flow rate for this case and compare with the result of Prob. 14-39. Discuss.

To determine

The volume flow rate by the pump.

Answer to Problem 41P

The volume flow rate by the pump is $14.58755\text{Lpm}$.

Explanation of Solution

Given information:

The diameter of pipe is $2.03\text{cm}$, elevation difference is $4.85\text{m}$, 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 $0.25\text{mm}$, 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).

  $\begin{array}{c}{h}_{\text{L}}=\left(\frac{f\times 176.5\text{m}}{2.03\text{cm}}+23.65\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(\frac{f\times 176.5\text{m}}{2.03\text{cm}\times \frac{1\text{m}}{100\text{cm}}}+23.65\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(\frac{f\times 176.5\text{m}}{0.0203\text{m}}+23.65\right)\frac{V^2}{2\times 9.81\text{m}/{\text{s}}^{\text{2}}}\\ =\left(8694.58f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\end{array}$

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 $z_2-z_1$, $\left(8694.58f+23.65\right)\frac{V^2}{19.62}$for $h_{\text{L}}$in Equation (I).

  $\begin{array}{c}{H}_{\text{required}}=\frac{0-0}{\rho g}+\frac{0-0}{2g}+\left(4.85\text{m}\right)+\left(8694.58f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\\ =\left(4.85\text{m}\right)+\left(8694.58f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\end{array}$

Substitute $2.03\text{cm}$for $D$in Equation (VII).

  $\begin{array}{c}\dot{V}=\frac{\pi }{4}{\left(2.03\text{cm}\right)}^{2}V\\ =\frac{\pi }{4}{\left(2.03\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^{2}V\\ =\frac{\pi }{4}{\left(0.0203\text{m}\right)}^{2}V\\ =3.2365\times {10}^{-4}{\text{m}}^2\times V\end{array}$

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 $\left(4.85\text{m}\right)+\left(8694.58f+23.65\right)\frac{V^2}{19.62}$for $H_{\text{available}}$in Equation (IV).

  $\begin{array}{l}\left[\begin{array}{l}\left(4.85\text{m}\right)+\\ \left(8694.58f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\end{array}\right]=\left[\begin{array}{l}24.4\text{m}-\\ 0.0678\text{m}/{\text{Lpm}}^{\text{2}}{\left(3.2365\times {10}^{-4}{\text{m}}^{\text{2}}\times V\right)}^2\end{array}\right]\\ \left[\left(\begin{array}{l}8694.58f\\ +23.65\end{array}\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\right]=\left[\begin{array}{l}19.55\text{m}-\left(\begin{array}{l}0.0678\text{m}/{\text{Lpm}}^{\text{2}}\times \\ \frac{3.6\times {10}^8{\text{s}}^{\text{2}}/{\text{m}}^{\text{5}}}{1\text{m}/{\text{Lpm}}^{\text{2}}}\end{array}\right)\\ {\left(3.2365\times {10}^{-4}{\text{m}}^{\text{2}}\times V\right)}^2\end{array}\right]\\ \left[\left(8694.58f+23.65\right)\frac{V^2}{19.62\text{m}/{\text{s}}^{\text{2}}}\right]=\left[\begin{array}{l}19.55\text{m}-\\ \left(244080000{\text{s}}^{\text{2}}/{\text{m}}^{\text{5}}\right)\\ {\left(3.2365\times {10}^{-4}{\text{m}}^{\text{2}}\times V\right)}^2\end{array}\right]\\ \left(443.148{\text{s}}^{\text{2}}/\text{m}f+1.205{\text{s}}^{\text{2}}/\text{m}\right)V^2=19.55\text{m}-25.56{\text{s}}^{\text{2}}/\text{m}\times V^2\end{array}$

  $f=\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041$

Substitute $998\text{kg}/{\text{m}}^{\text{3}}$for $\rho$, $2.03\text{cm}$for $D$and $1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}$for $\mu$in Equation (V).

  $\begin{array}{c}\mathrm{Re}=\frac{998\text{kg}/{\text{m}}^{\text{3}}\times 2.03\text{cm}\times V}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =\frac{998\text{kg}/{\text{m}}^{\text{3}}\times 2.03\text{cm}\times \frac{1\text{m}}{100\text{cm}}\times V}{1.002\times {10}^{-3}\text{kg}/\text{m}\cdot \text{s}}\\ =20218.96\text{s}/\text{m}\times V\end{array}$

Substitute $0.25\text{mm}$for $\epsilon$, $2.03\text{cm}$for $D$, $20218.96\text{s}/\text{m}\times V$for $\mathrm{Re}$and $\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041$for $f$in Equation (VI).

  $\begin{array}{l}\left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041}}\\ =-2.0\log \left(\begin{array}{l}\frac{0.25\text{mm}/2.03\text{cm}}{3.7}\\ +\frac{2.51}{20218.96\text{s}/\text{m}\times V\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041}}\end{array}\right)\end{array}\right]\\ \left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041}}\\ =-2.0\log \left(\begin{array}{l}\frac{0.25\text{mm}/2.03\text{cm}\times \frac{10\text{mm}}{1\text{cm}}}{3.7}\\ +\frac{2.51}{20218.96\text{s}/\text{m}\times V\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041}}\end{array}\right)\end{array}\right]\\ \left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041}}\\ =-2.0\log \left(\begin{array}{l}0.00332432\\ +\frac{2.51}{20218.96\text{s}/\text{m}\times V\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{V^2}-0.06041}}\end{array}\right)\end{array}\right]\end{array}$  ....(VIII)

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

Trial-(1)

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

  $\begin{array}{l}\left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{{\left(1\text{m}/\text{s}\right)}^2}-0.06041}}\\ =-2.0\log \left(\begin{array}{l}0.00332432\\ +\frac{2.51}{20218.96\text{s}/\text{m}\times 1\text{m}/\text{s}\sqrt{\frac{0.044116{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{{\left(1\text{m}/\text{s}\right)}^2}-0.06041}}\end{array}\right)\end{array}\right]\\ \frac{1}{\sqrt{-0.016294}}=-2.0\log \left(\begin{array}{l}0.00332432\\ +\frac{2.51}{20218.96\sqrt{-0.016294}}\end{array}\right)\end{array}$

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

Trial-(2)

Substitute $0.7512\text{m}/\text{s}$for $V$in Equation (VIII).

  $\begin{array}{l}\left[\begin{array}{l}\frac{1}{\sqrt{\frac{0.03734{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{{\left(0.7512\text{m}/\text{s}\right)}^2}-0.06041}}\\ =-2.0\log \left(\begin{array}{l}0.00332432\\ +\frac{2.51}{20218.96\text{s}/\text{m}\times 0.7512\text{m}/\text{s}\sqrt{\frac{0.03734{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}{0.7512\text{m}/\text{s}}-0.06041}}\end{array}\right)\end{array}\right]\\ \frac{1}{\sqrt{0.0057074}}=-2.0\log \left(\begin{array}{l}0.00332432\\ +\frac{2.51}{20218.96\text{s}/\text{m}\times 0.7512\text{m}/\text{s}\sqrt{0.0057074}}\end{array}\right)\\ 13.2366=13.2363\end{array}$

Here, the values on both sides are equal, so velocity of flow is $0.7512\text{m}/\text{s}$

Substitute $0.7512\text{m}/\text{s}$for $V$in Equation (VII).

  $\begin{array}{c}\dot{V}=3.2365\times {10}^{-4}{\text{m}}^2\times 0.7512\text{m}/\text{s}\\ =2.43125\times {10}^{-4}{\text{m}}^{\text{3}}/\text{s}\times \frac{60000\text{Lpm}}{1{\text{m}}^{\text{3}}/\text{s}}\\ =14.58755\text{Lpm}\end{array}$

Here, the volume flow rate for elevation difference of $3\text{m}$is $14.58755\text{Lpm}$.

Since the volume flow rate for the for the elevation difference of $7\text{m}$is $11.576\text{Lpm}$

Hence on decreasing elevation difference between the reservoir the volume flow rate increases.

Conclusion:

The volume capacity delivered by the pump is $14.58755\text{Lpm}$.