Chapter 14, Problem 57EP

Repeat Prob. 14—55E. Ignoring all minor losses. How important arc the minor losses in this problem? Discuss.

To determine

The importance of the minor losses.

Answer to Problem 57EP

The volume flow rate when the minor losses that are considered at the entrance, elbow and damper are $9.96{\text{ft}}^{\text{3}}/\text{s}$ and the volume flow rate of the system by neglecting all the minor losses are $5.2022{\text{ft}}^{\text{3}}/\text{s}$. The minor losses could not be neglected.

Explanation of Solution

Write the expression for steady energy equation.

   $\frac{P_1}{\rho g}+\frac{{\alpha }_1{V}_1^2}{2g}+z_1+H_{required}=\frac{P_2}{\rho g}+\frac{{\alpha }_2{V}_2^2}{2g}+z_2+h_{L,Total}$

...... (I)

Here, the required head of the fan is $H_{required}$, the velocity if the water at section 1 is $V_1$, the pressure at section 1is $P_1$, and the height at section 1 is $z_1$, the velocity if the water at section 2 is $V_2$, the pressure at section 2 is $P_2$, the constant is ${\alpha }_1,{\alpha }_2$ and the height at section 2 is $z_2$, the density of water is $\rho$, the gravitational constant is $g$ and the total irreversible head loss is $h_{L,Total}$.

Write the expression for irreversible head loss.

   $h_{L,Total}=\frac{fl{V}^2}{2gD}+{\displaystyle \sum K_L}\frac{V^2}{2g}$ ...... (II)

Here, the Darcy’s friction factor is $f$, the length of the pipe is $l$, the diameter of the pipe is $D$ ,the velocity of the air is $V$ and the total head loss is $K_L$.

Write the expression for Reynolds number.

   $\mathrm{Re}=\frac{VD}{\upsilon }$ ...... (III)

Here, the Reynolds number is $\mathrm{Re}$ and kinematic viscosity is $\upsilon$.

Write the expression for relative roughness.

   $\frac{\epsilon }{D}$ ...... (IV)

Here, $\epsilon$ is average roughness.

Write the expression for volume flow rate.

   $\dot{V}=AV$ ...... (V)

Here, the area of pipe is $A$.

Write the expression for head of air in water column from inches of water column to inches of air column.

   $H_{0,air}=\frac{H_{0,water}\times {\rho }_{water}}{{\rho }_{air}}$ ...... (VI)

Here, head of water in water column is $H_{o,water}$, the density of water is ${\rho }_{water}$ and density of air is ${\rho }_{air}$.

Write the expression for height of air in water column from inches of water column to inches of air column.

   $a_{air}=\frac{a_{water}\times {\rho }_{water}}{{\rho }_{air}}$ ...... (VII)

Here, the height of water in water column is $a_{water}$.

Write the expression for available net head.

   $H_{available}=H_{0,air}-\left(a_{air}\right){\left(\dot{V}\right)}^2$ ...... (VIII)

Write the expression for irreversible head loss.

   $h_{L,Total}=\frac{fl{V}^2}{2gD}$ ...... (IX)

Here, the Darcy’s friction factor is $f$, the length of the pipe is $l$, the diameter of pipe is $D$ ,the velocity of air is $V$.

Write the expression for loss.

   ${\displaystyle \sum K_L}=K_{L,entrance}+3K_{L,elbow}+K_{L,damp}$ ...... (X)

Here, head loss coefficient at entrance is $K_{L,entrance}$, head loss coefficient at elbow is $K_{L,elbow}$ and head loss coefficient at damper is $K_{L,damp}$.

Write the expression for Darcy’s friction factor.

   $\frac{1}{\sqrt{f}}=-2{\log }_{10}\left(\frac{\epsilon /D}{3.7}+\frac{2.51}{\mathrm{Re}\sqrt{f}}\right)$ ...... (XI)

Write the expression for ideal head condition.

   $H_{required}=H_{available}$ ...... (XII)

Write the expression for interpolation to calculate the density of air at temperature $T$.

   ${\rho }_{air}={\rho }_1-\left(\frac{{\rho }_1-{\rho }_2}{T_H-T_L}\right)\left(T-T_L\right)$ ...... (XIII)

Here, the density of air at higher temperature is ${\rho }_1$, the density of air at lower temperature is ${\rho }_2$, the higher temperature is $T_H$ and the lower temperature is $T_L$ and the temperature at which the density is to be calculated is $T$.

Write the expression for interpolation to calculate the kinematic viscosity of air at temperature $T$.

   ${\upsilon }_{air}={\upsilon }_2-\left(\frac{{\upsilon }_1-{\upsilon }_2}{T_H-T_L}\right)\left(T-T_L\right)$ ...... (XIV)

Here, the kinematic viscosity of air at higher temperature is ${\rho }_1$, the kinematic viscosity of air at lower temperature is ${\rho }_2$, the higher temperature is $T_H$ and the lower temperature is $T_L$ and the temperature at which the kinematic viscosity is to be calculated is $T$.

Write the expression for interpolation to calculate the dynamic viscosity of air at temperature $T$.

   ${\mu }_{air}={\mu }_2-\left(\frac{{\mu }_1-{\mu }_2}{T_H-T_L}\right)\left(T-T_L\right)$ ...... (XV)

Here, the dynamic viscosity of air at higher temperature is ${\mu }_1$, the kinematic viscosity of air at lower temperature is ${\mu }_2$, the higher temperature is $T_H$ and the lower temperature is $T_L$ and the temperature at which the dynamic viscosity is to be calculated is $T$.

Calculation:

Refer to Table A-9E, “Properties of air at $1$ atm pressure” to obtain the values of ${\rho }_1$ as $0.07489\text{lbm}/{\text{ft}}^{\text{3}}$, ${\rho }_2$ as $0.0735\text{lbm}/{\text{ft}}^{\text{3}}$, $T_H$ as $80\text{°F}$, $T_L$ as $70\text{°F}$ and $T$ as $77\text{°F}$.

The various value of temperature and density of air are given in Table-(1).

Table-(1)

Temperature,    $\text{°F}$	density of the air,    $\text{lbm}/{\text{ft}}^{\text{3}}$
$70$	$0.0735$
$77$	?
$80$	$0.07489$

Substitute $0.07489\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_1$, $0.0735\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_2$, $80\text{°F}$ for $T_H$, $70\text{°F}$ for $T_L$ and $77\text{°F}$ for $T$ in Equation (XIII).

   $\begin{array}{c}{\rho }_{air}=0.07489\text{lbm}/{\text{ft}}^{\text{3}}-\left(\frac{0.07489\text{lbm}/{\text{ft}}^{\text{3}}-0.0735\text{lbm}/{\text{ft}}^{\text{3}}}{80\text{°F}-70\text{°F}}\right)\left(77\text{°F}-70\text{°F}\right)\\ =0.07489\text{lbm}/{\text{ft}}^{\text{3}}-\left(\frac{1.39\text{lbm}/{\text{ft}}^{\text{3}}}{10\text{°F}}\right)\\ =0.07392\text{lbm}/{\text{ft}}^{\text{3}}\end{array}$

Refer to Table A-9E, “Properties of air at $1$ atm pressure” to obtain the values of ${\upsilon }_1$ as $1.643\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$, ${\upsilon }_2$ as $1.643\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$, $T_H$ as $80\text{°F}$, $T_L$ as $70\text{°F}$ and $T$ as $77\text{°F}$.

The various value of temperature and density of air are given in Table-(2).

Table-(2)

Temperature,    $\text{°F}$	Kinematic viscosity,    ${\text{ft}}^{\text{2}}/\text{s}$
$70$	$1.643\times {10}^{-4}$
$77$	?
$80$	$1.697\times {10}^{-4}$

Substitute $1.643\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$ for ${\upsilon }_1$, $1.643\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$ for ${\upsilon }_2$, $T_H$ for $80\text{°F}$, $70\text{°F}$ for $T_L$ and

   $77\text{°F}$ for $T$ in Equation (XIV)

   $\begin{array}{c}{\upsilon }_{air}=1.643\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}+\left(\frac{1.697\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}-1.643\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}{80\text{°F}-70\text{°F}}\right)\left(77\text{°F}-70\text{°F}\right)\\ =1.643\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}+\left(\frac{0.054\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}{10\text{°F}}\right)\left(7\text{°F}\right)\\ =1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}\end{array}$

Refer to Table A-9E, “Properties of air at $1$ atm pressure” to obtain the values of ${\mu }_1$ as $1.230\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}$, ${\upsilon }_2$ as $1.230\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}$, $T_H$ as $80\text{°F}$, $T_L$ as $70\text{°F}$ and $T$ as $77\text{°F}$.

The various value of temperature and density of air are given in Table-(3).

Table-(3)

Temperature,    $\text{°F}$	Dynamic viscosity,    $\text{lbm}/\text{ft}\cdot \text{s}$
$70$	$1.230\times {10}^{-5}$
$77$	?
$80$	$1.247\times {10}^{-5}$

Substitute $1.247\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}$ for ${\mu }_1$ and $1.230\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}$ for ${\mu }_2$, $80\text{°F}$ as $T_H$, $70\text{°F}$ as $T_L$ and $77\text{°F}$ as $T$ in Equation (XV).

   $\begin{array}{c}{\mu }_{air}=\left[1.230\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}+\left(\frac{\begin{array}{l}1.247\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}-\\ 1.230\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}\end{array}}{80\text{°F}-70\text{°F}}\right)\left(77\text{°F}-70\text{°F}\right)\right]\\ =1.230\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}+\left(\frac{0.017\times {10}^{-5}\text{lbm}/\text{ft}\cdot \text{s}}{10\text{°F}}\right)\left(7\text{°F}\right)\\ =1.242\times {10}^{-3}\text{lbm}/\text{ft}\cdot \text{s}\end{array}$

Refer Table A-3E, “Properties of saturated water” to obtain the value of ${\rho }_{water}$ as $62.24\text{lbm}/{\text{ft}}^{\text{3}}$.

Substitute $0$ for $V_1$, $P_{atm}$ for $P_1$, $P_{atm}$ for $P_2$ and $0$ for $\left(z_1-z_2\right)$ in Equation (I).

   $\begin{array}{l}{H}_{required}=\frac{0}{\rho g}+\frac{{\alpha }_2{V}_2^2-0}{2g}+\left(0\right)+h_{L,Total}\\ H_{required}=\frac{{\alpha }_2{V}_2^2}{2g}+h_{L,Total}\end{array}$ ...... (XVI)

Substitute $\frac{fl{V}^2}{2gD}$ for $h_{L,Total}$ in Equation (XVI).

   $\begin{array}{c}{H}_{required}=\frac{{\alpha }_2{V}_2^2}{2g}+\frac{fl{V}^2}{2gD}\\ =\left({\alpha }_2+\frac{fl}{D}\right)\frac{V^2}{2g}\end{array}$ ...... (XVII)

Assume $\alpha =1.05$

Substitute $1.05$ for ${\alpha }_2$, $34\text{ft}$ for $l$, $9.06\text{in}$ for $D$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ in Equation (XVII).

   $\begin{array}{c}{H}_{available}=\left(1.05+\frac{f\times 34\text{ft}}{9.06\text{in}}\right)\left(\frac{V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}\right)\\ =\left(1.05+\frac{f\times 34\text{ft}}{\left(9.06\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)}\right)\left(\frac{V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}\right)\\ =\left(1.05+\frac{f\times 34\text{ft}}{0.755\text{ft}}\right)\left(\frac{V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}\right)\end{array}$

   $=\left(1.05+45.033f\right)0.0155V^2\text{ft}$

Substitute $9.06\text{in}$ for $D$ and $1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$ for $\upsilon$ in Equation (III).

   $\begin{array}{c}\mathrm{Re}=\frac{V\times 9.06\text{in}}{1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =\frac{V\times \left(9.06\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)}{1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =4491.37\left(V\text{ft}/\text{s}\right)\end{array}$

Substitute $0.0059\text{in}$ for $\epsilon$ and $9.06\text{in}$ for $D$ in Equation (IV).

   $\begin{array}{c}\frac{\epsilon }{D}=\frac{0.0059\text{in}}{9.06\text{in}}\\ =6.51\times {10}^{-3}\end{array}$

Substitute $9.06\text{in}$ for $D$ in equation (V).

   $\begin{array}{c}\dot{V}=\frac{\pi }{4}{\left(9.06\text{in}\right)}^{2}V\\ =\frac{\pi }{4}{\left(9.06\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)}^{2}V\\ =0.4477\left(V\text{ft}/\text{s}\right)\end{array}$

Substitute $23\text{in}$ for $H_{0,water}$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{water}$, $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{air}$ in Equation (VI).

   $\begin{array}{c}{H}_{0,air}=\frac{\left(2.3\text{in}\right)\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{\left(2.3\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =161.38\text{ft}\end{array}$

Substitute $8.50\times {10}^{-6}\text{in}$ for $a_{water}$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{water}$, $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{air}$ in Equation (VII).

   $\begin{array}{c}{a}_{air}=\frac{8.50\times {10}^{-6}\text{in}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{8.50\times {10}^{-6}\text{in}\times \frac{1\text{ft}}{12\text{in}}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =5.96\times {10}^{-4}\text{ft}\end{array}$

Substitute $161.38\text{ft}$ for $H_{0,air}$, $5.96\times {10}^{-4}$ for $a_{air}$, and $0.4477$ for $\dot{V}$ in Equation (VIII).

   $\begin{array}{c}{H}_{available}=161.38\text{ft}-5.96\times {10}^{-4}\text{ft}{\left(0.4477V\right)}^2\\ =161.38\text{ft}-1.1946\times {10}^{-4}{V}^2\text{ft}\end{array}$

Substitute $161.38-1.1946\times {10}^{-4}{V}^2$ for $H_{available}$ and $\left(1.05+45.033f\right)0.0155V^2$ for $H_{required}$ in Equation (XII).

   $\begin{array}{l}\left(1.05+45.033f\right)0.0155V^2=161.38-1.1946\times {10}^{-4}{V}^2\\ \left(1.05+45.033\times 1.688\right)0.0155V^2+1.1945\times {10}^{-4}{V}^2=161.38\\ V=\sqrt{\frac{161.38}{1.1946}}\end{array}$

   $V=11.62\text{ft}/\text{s}$

Substitute $11.62\text{ft}/\text{s}$ for $V$ and $9.06\text{in}$ for $D$ in Equation (V).

   $\begin{array}{c}\dot{V}=\frac{\pi }{4}\left(9.06\text{in}\right)\left(11.62\text{ft}/\text{s}\right)\\ =\frac{\pi }{4}\left(9.06\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)\left(11.62\text{ft}/\text{s}\right)\\ =5.2022{\text{ft}}^{\text{3}}/\text{s}\end{array}$

Substitute $4.6$ for $K_{L,enterance}$ and $0.21$ for $K_{L,elbow}$ and $1.8$ for $K_{L,damp}$ in Equation (X).

   $\begin{array}{c}{\displaystyle \sum K_L}=4.6+3\left(0.21\right)+1.8\\ =4.6+0.63+1.8\\ =7.03\end{array}$

Substitute $1.05$ for ${\alpha }_2$, $0.025$ for $f$, $34\text{ft}$ for $l$, $9.06\text{in}$ for $D$, $5.23+K_{L,damp}$ for $K_L$, $5.608\text{ft}/\text{s}$ for $V$, and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for $g$ .in Equation (VIII).

   $\begin{array}{c}{H}_{available}=\left(1.05+\frac{f\times 34\text{ft}}{9.06\text{in}}+7.03\right)\left(\frac{V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}\right)\\ =\left(1.05+\frac{f\times 34\text{ft}}{9.06\text{in}\times \frac{1\text{ft}}{12\text{in}}}+7.03\right)\left(\frac{V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}\right)\\ =\left(8.08+45.033f\right)0.0155V^2\text{ft}\end{array}$

Substitute $9.06\text{in}$ for $D$ and $1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$ for $\upsilon$ in Equation (III).

   $\begin{array}{c}\mathrm{Re}=\frac{V\times 9.06\text{in}}{1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =\frac{V\times 9.06\text{in}\times \frac{1\text{ft}}{12\text{in}}}{1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}}\\ =4491.37\left(V\text{ft}/\text{s}\right)\end{array}$

Substitute $0.0059$ for $\epsilon$ and $9.06$ for $D$ in Equation (IV).

   $\begin{array}{c}\frac{\epsilon }{D}=\frac{0.0059}{9.06}\\ =6.51\times {10}^{-3}\end{array}$

Substitute $6.51\times {10}^{-4}$ for $\frac{\epsilon }{D}$ and $4491.37V$ for $\mathrm{Re}$ in Equation (XI).

   $\frac{1}{\sqrt{f}}=-2{\log }_{10}\left(\frac{6.51\times {10}^{-4}}{3.7}+\frac{2.51}{4491.37V\sqrt{f}}\right)$ ...... (XVIII)

Substitute $9.06\text{in}$ for $D$ in Equation (V).

   $\begin{array}{c}\dot{V}=\frac{\pi }{4}{\left(9.06\text{in}\right)}^{2}V\text{ft}/\text{s}\\ =\frac{\pi }{4}{\left(9.06\text{in}\times \frac{1\text{ft}}{12\text{in}}\right)}^{2}V\text{ft}/\text{s}\\ =0.4477V{\text{ft}}^3/\text{s}\end{array}$

Substitute $2.3\text{in}$ for $H_{0,water}$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{water}$, $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{air}$ in Equation (VI).

   $\begin{array}{c}{H}_{0,air}=\frac{2.3\text{in}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{2.3\text{in}\times \frac{1\text{ft}}{12\text{in}}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =161.38\text{ft}\end{array}$

Substitute $8.50\times {10}^{-6}\text{ft}$ for $a_{water}$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{water}$, $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_{air}$ in Equation (VII).

   $\begin{array}{c}{a}_{air}=\frac{8.50\times {10}^{-6}\text{in}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{8.50\times {10}^{-6}\text{in}\times \frac{1\text{ft}}{12\text{in}}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =5.96\times {10}^{-4}\text{ft}\end{array}$

Substitute $161.38\text{ft}$ for $H_{0,air}$, $5.96\times {10}^{-4}\text{ft}$ for $a_{air}$, and $0.4477V$ for $\dot{V}$ in Equation (VIII).

   $\begin{array}{c}{H}_{available}=161.38\text{ft}-5.96\times {10}^{-4}\text{ft}{\left(0.4477V\right)}^2\\ =161.38\text{ft}-1.1945\times {10}^{-4}{V}^2\text{ft}\end{array}$

Substitute $161.38\text{ft}-1.1945\times {10}^{-4}{V}^2\text{ft}$ in for $H_{available}$ and $\left(8.08+45.033f\right)0.0155V^2\text{ft}$ for $H_{required}$ Equation (XII).

   $\begin{array}{l}161.38\text{ft}-1.1945\times {10}^{-4}{V}^2\text{ft}=\left(8.08+45.033f\right)0.0155V^2\text{ft}\times 0.0155V^2\\ 45.033f=\left(\frac{161.38\text{ft}-1.1945\times {10}^{-4}{V}^2\text{ft}}{0.0155V^2}\right)-8.08\\ f=\frac{\left(\frac{161.38\text{ft}-1.1945\times {10}^{-4}{V}^2\text{ft}}{0.0155V^2{\text{ft}}^{\text{2}}/\text{s}}\right)-8.08}{45.033}\end{array}$

Substitute $\frac{\left(\frac{161.38\text{ft}-1.1945\times {10}^{-4}{V}^2\text{ft}}{0.0155V^2{\text{ft}}^{\text{2}}/\text{s}}\right)-8.08}{45.033}$ for $f$ in Equation (XVIII).

   $\left[\frac{1}{\sqrt{\frac{\begin{array}{l}\left(\frac{\begin{array}{l}161.38\text{ft}\\ -1.1945\times {10}^{-4}{V}^2\text{ft}\end{array}}{0.0155V^2{\text{ft}}^{\text{2}}/\text{s}}\right)\\ -8.08\end{array}}{45.033}}}\right]=-2{\log }_{10}\left(\begin{array}{l}\frac{6.51\times {10}^{-4}}{3.7}\\ +\frac{2.51}{4491.37V\sqrt{\frac{\left[\begin{array}{l}\left(\frac{\begin{array}{l}161.38\text{ft}\\ -1.1945\times {10}^{-4}{V}^2\text{ft}\end{array}}{0.0155V^2{\text{ft}}^{\text{2}}/\text{s}}\right)\\ -8.08\end{array}\right]}{45.033}}}\end{array}\right)$

   $\left[\frac{6.71}{\begin{array}{l}\left(\frac{\begin{array}{l}12.703\text{ft}\\ -\text{0}\text{.0109}V\text{ft}\end{array}}{0.1244V{\text{ft}}^{\text{2}}/\text{s}}\right)\\ -2.84\end{array}}\right]=-2{\log }_{10}\left(\begin{array}{l}\left(1.7594\times {10}^{-4}\right)\\ +\left(\frac{1}{\left[\begin{array}{l}669.354V\left(\frac{12.703\text{ft}-0.00109V\text{ft}}{0.1244V{\text{ft}}^{\text{2}}/\text{s}}\right)\\ -8.08\end{array}\right]}\right)\end{array}\right)$

   $\left(\frac{0.1244V{\text{ft}}^{\text{2}}/\text{s}}{\left[\begin{array}{l}12.703\text{ft}-\text{0}\text{.0109}V\text{ft}\\ -\text{0}\text{.3532}V{\text{ft}}^{\text{2}}/\text{s}\end{array}\right]}\right)=-2{\log }_{10}\left(\begin{array}{l}\left(1.7594\times {10}^{-4}\right)\\ +\left(\frac{0.1224V{\text{ft}}^{\text{2}}/\text{s}}{\left[\begin{array}{l}8502.803\text{ft}-0.72959V\text{ft}-\\ 1.005152V{\text{ft}}^{\text{2}}/\text{s}\end{array}\right]}\right)\end{array}\right)$

   $V=16.8\text{ft}/\text{s}$

Substitute $\text{9}\text{.06}\text{in}$ for $D$ and $16.8\text{ft}/\text{s}$ for $V$ in Equation (V).

   $\begin{array}{c}\dot{V}=\frac{\pi }{4}\left(9.06\text{in}\right)\left(16.8\text{ft}/\text{s}\right)\\ =\frac{\pi }{4}\left(9.06\text{in}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\left(16.8\text{ft}/\text{s}\right)\\ =\frac{\pi }{4}\left(0.755\text{ft}\right)\left(16.8\text{ft}/\text{s}\right)\\ =9.96{\text{ft}}^{\text{3}}/\text{s}\end{array}$

The volume flow rate when the minor losses that are considered at the entrance, elbow and damper are $9.96{\text{ft}}^{\text{3}}/\text{s}$ and the volume flow rate of the system by neglecting all the minor losses are $5.2022{\text{ft}}^{\text{3}}/\text{s}$. Therefore, minor losses cannot be neglected. This is because if these minor losses are not neglected then there will be loss in power when the system when the system will be under working condition. Most of the losses are due such minor and major losses in the system and to overcome these losses the manufacturers and the designers will have to design the system in such a way that would still give the desired output even though there will be losses in the system.

Conclusion:

The volume flow rate when the minor losses that are considered at the entrance, elbow and damper are $9.96{\text{ft}}^{\text{3}}/\text{s}$ and the volume flow rate of the system by neglecting all the minor losses are $5.2022{\text{ft}}^{\text{3}}/\text{s}$ .The minor losses cannot be neglected.