Chapter 14, Problem 57EP

Repeat Prob. 14-55E, ignoring all minor losses. How important are the minor losses in this problem? Discuss.

To determine

Whether the minor losses are important or not.

Answer to Problem 57EP

The minor losses are important because the volume flow rate is increased by $11\%$ when neglecting the minor losses.

Explanation of Solution

Given Information:

Inner diameter of the duct is $9.06\text{in}$, the average roughness of the duct is $0.0059\text{in}$, total length of the duct is $34.0\text{ft}$, number of elbows is $3$ each having minor loss coefficient of $0.21$, hood entry loss coefficient is $4.6$, loss coefficient is $1.8$, given equation for available head is $H_{avialable}=H_o-a{\dot{V}}^2$, shutoff head is $2.30\text{in}$ of water column, volume flow rate is $226\text{SCFM}$, and constant $a$ is $8.5\times {10}^{-6}\text{in}$ of water column.

Expression for steady energy equation from point 1 in the stagnant air region to point 2 at the duct outlet.

  $\frac{P_1}{\rho g}+\frac{{\alpha }_1{V}_1^2}{2g}+z_1+H_{required}=\frac{P_2}{\rho g}+\frac{{\alpha }_2{V}_2^2}{2g}+z_2+h_L$...... (I)

Here, the required head for the fan is $H_{required}$, the velocity at the section 1 is $V_1$, the pressure at section 1 is $P_1$, the velocity of water at section 2 is $V_2$, the pressure at section 2 is $P_2$, the density of the water is $\rho$, the elevation at section 1 is $z_1$, the elevation at section 2 is $z_2$, the acceleration due to gravity is $g$ and the total head loss is $h_L$.

Expression for the total head loss.

  $h_L=\frac{fL{V}^2}{2gD}$...... (II)

Here, the velocity of the air is $V$, the friction factor is $f$, the diameter of the pipe is $D$ and the length of the pipe is $L$.

Expression for Reynolds's number.

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

Here, the kinematic viscosity is $v$.

Expression for relative roughness.

  $R=\frac{\epsilon }{D}$....... (IV)

Here, the roughness of the pipe is $\epsilon$.

Expression for the friction factor.

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

Expression for the volume flow rate.

  $\dot{V}=AV$...... (VI)

Here, the area of the pipe is $A$.

Expression for the area of the pipe.

  $A=\frac{\pi D^2}{4}$

Substitute $\frac{\pi D^2}{4}$ for $A$ in Equation (V).

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

Expression to convert the shutoff head from inches of water column to inches of air column.

  $H_{o,air}=\frac{H_{o.water}\times {\rho }_w}{{\rho }_a}$...... (VIII)

Here, the density of the water is ${\rho }_w$, the shutoff head is $H_o$ and the density of the air is ${\rho }_a$.

Expression to convert the $a$ from inches of water column to inches of air column.

  $a_{air}=\frac{a_{water}\times {\rho }_w}{{\rho }_a}$...... (IX)

Expression for the total head loss.

  $h_L=\frac{fL{V}^2}{2gD}+\left(K_e+n{K}_{elbow}+K_{damper}\right)\frac{V^2}{2g}$...... (X)

Here, the velocity of the air is $V$, the friction factor is $f$, the diameter of the pipe is $D$, the length of the pipe is $l$, the loss coefficient at the entrance is $K_e$, the loss coefficient for the elbow is $K_{elbow}$, the loss coefficient for the damper is $K_{damper}$ and the number of elbow is $n$.

Calculation:

Refer to the Table-A-9E, "Properties of air at 1 atm pressure" to obtain the density of the air as $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$, the dynamic viscosity of air as $1.242\times {10}^{-3}\text{lbm}/\text{ft}\cdot \text{s}$, the value of kinematic viscosity as $1.681\times {10}^{-4}{\text{ft}}^{\text{2}}/\text{s}$.

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

  $\begin{array}{l}\frac{P_{atm}}{\rho g}+\frac{{\alpha }_1\times 0}{2g}+z+H_{required}=\frac{P_{atm}}{\rho g}+\frac{{\alpha }_2{V}_2{}^2}{2g}+z+h_L\\ H_{required}=\frac{{\alpha }_2{V}_2^2}{2g}+h_L+0\\ H_{required}=\frac{{\alpha }_2{V}_2^2}{2g}+h_L\end{array}$...... (XI)

Substitute $9.06\text{in}$ for $D$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ fro $g$ and $34\text{ft}$ for $L$ in Equation (II).

  $\begin{array}{c}{h}_L=\frac{f\left(34\text{ft}\right)V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}\times 9.06\text{in}}+\\ =\frac{f\left(34\text{ft}\right)V^2}{64.4\text{ft}/{\text{s}}^{\text{2}}\times 9.06\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}\\ =\left(0.6992\right)f{V}^2{\text{s}}^{\text{2}}/\text{ft}\end{array}$

Substitute $\left(0.6992\right)f{V}^2{\text{s}}^{\text{2}}/\text{ft}$ for $h_L$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ fro $g$, $1.05$ for ${\alpha }_2$ and $V$ For $V_2$ in Equation (X).

  $\begin{array}{c}{H}_{required}=\frac{1.05V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}+\left(\left(0.6992\right)f{V}^2{\text{s}}^{\text{2}}/\text{ft}\right)\\ =\frac{1.05V^2}{64.4\text{ft}/{\text{s}}^{\text{2}}}+\left(\left(0.6992\right)f{V}^2{\text{s}}^{\text{2}}/\text{ft}\right)\\ =\left(0.0163+0.6992f\right)V^2{\text{s}}^{\text{2}}/\text{ft}\end{array}$...... (XII)

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

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

Substitute $\mathrm{0..0059}\text{in}$ for $\epsilon$ and $9.06\text{in}$ for $D$ in Equation (IV).

  $\begin{array}{c}R=\frac{0.0059\text{in}}{9.06\text{in}}\\ =6.51\times {10}^{-4}\end{array}$

Substitute $6.51\times {10}^{-4}$ for $R$, $4491.37V$ for $\mathrm{Re}$ in Equation (V).

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

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

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

Substitute $2.3\text{ft}$ for $H_{o,water}$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_w$ and $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ in Equation (VIII).

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

Substitute $8.5\times {10}^{-6}{\text{in}/\text{SCFM}}^2$ for $a$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_w$ and $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ in Equation (IX).

  $\begin{array}{c}{a}_{air}=\frac{\left(8.5\times {10}^{-6}{\text{in}/\text{SCFM}}^2\right)\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{\left(8.5\times {10}^{-6}{\text{in}/\text{SCFM}}^2\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =5.96\times {10}^{-4}\text{ft}/{\text{SCFM}}^{\text{2}}\end{array}$

Substitute $161.38\text{ft}$ for $H_o$, $5.96\times {10}^{-4}\text{ft}/{\text{SCFM}}^{\text{2}}$ for $a_{air}$ and $0.4477V{\text{ft}}^{\text{2}}$ for $\dot{V}$ in the given available head Equation.

  $\begin{array}{c}{H}_{avialable}=161.38\text{ft}-\left(5.96\times {10}^{-4}\text{ft}/{\text{SCFM}}^{\text{2}}\right){\left(0.4477V{\text{ft}}^{\text{2}}\right)}^2\\ =161.38\text{ft}-\left(5.96\times {10}^{-4}\text{ft}\right)\left(0.2004V^2{\text{ft}}^5/{\text{SCFM}}^{\text{2}}\right)\\ =\left[161.38\text{ft}-\left(1.1946V^2\times {10}^{-4}{\text{ft}}^5/{\text{SCFM}}^{\text{2}}\right)\right]\end{array}$...... (XIV)

Since at the operating point the available head and the required head are equal, therefore equate Equation (XIII) and (XIV).

  $\begin{array}{l}\left[161.38\text{ft}-\left(1.1946V^2\times {10}^{-4}{\text{ft}}^5/{\text{SCFM}}^{\text{2}}\right)\right]=\left[\left(0.0163+0.6992f\right)V^2{\text{s}}^{\text{2}}/\text{ft}\right]\\ \left[\begin{array}{l}161.38\text{ft}-\\ \left(1.1946V^2\times {10}^{-4}{\text{ft}}^5/{\text{SCFM}}^{\text{2}}\right)\\ \left(\frac{{\text{SCFM}}^{\text{2}}}{0.000289{\text{ft}}^6/{\text{s}}^2}\right)\end{array}\right]=\left(0.0163+0.6992f\right)V^2{\text{s}}^{\text{2}}/\text{ft}\\ \left[161.38\text{ft}-\left(0.413V^2{\text{s}}^{\text{2}}/\text{ft}\right)\right]=\left(0.0163+0.6992f\right)V^2{\text{s}}^{\text{2}}/\text{ft}\end{array}$.(XV)

Solve Equation (XIII) and Equation (XV) to obtain the value of velocity as $18.72\text{ft}/\text{s}$.

Substitute $18.72\text{ft}/\text{s}$ for $V$ and $9.06\text{in}$ for $D$ in Equation (VII).

  $\begin{array}{c}\dot{V}=\left(\frac{\pi {\left(9.06\text{in}\right)}^2}{4}\right)\left(18.72\text{ft}/\text{s}\right)\\ =\left(\frac{\pi {\left(9.06\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2}{4}\right)\times 18.72\text{ft}/\text{s}\\ =8.3808{\text{ft}}^{\text{3}}/\text{s}\left(\frac{1\text{SCFM}}{0.017{\text{ft}}^{\text{3}}/\text{s}}\right)\\ =492.99\text{SCFM}\end{array}$

Substitute $3$ for $n$, $0.21$ for $K_{elbow}$, $4.6$ for $K_{entrance}$, $1.8$ for $K_{damper}$, $9.06\text{in}$ for $D$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ fro $g$ and $34\text{ft}$ for $l$ in Equation (X).

  $\begin{array}{c}{h}_L=\frac{f\left(34\text{ft}\right)V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}\times 9.06\text{in}}+\left(4.6+3\times 0.21+1.8\right)\frac{V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}\\ =\frac{f\left(34\text{ft}\right)V^2}{64.4\text{ft}/{\text{s}}^{\text{2}}\times 9.06\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)}+\left(7.03\right)\frac{V^2}{64.4\text{ft}/{\text{s}}^{\text{2}}}\\ =\left(0.699f+0.109\right)V^2{\text{s}}^{\text{2}}/\text{ft}\end{array}$

Substitute $\left(0.699f+0.109\right)V^2{\text{s}}^{\text{2}}/\text{ft}$ for $h_L$, $32.2\text{ft}/{\text{s}}^{\text{2}}$ fro $g$, $1.05$ for ${\alpha }_2$ and $V$ For $V_2$ in Equation (XI).

  $\begin{array}{c}{H}_{required}=\frac{1.05V^2}{2\times 32.2\text{ft}/{\text{s}}^{\text{2}}}+\left(0.699f+0.109\right)V^2{\text{s}}^{\text{2}}/\text{ft}\\ =\frac{1.05V^2}{64.4\text{ft}/{\text{s}}^{\text{2}}}+\left(0.699f+0.109\right)V^2{\text{s}}^{\text{2}}/\text{ft}\\ =\left(8.08+45.033f\right)\times 0.0155V^2{\text{s}}^{\text{2}}/\text{ft}\end{array}$...... (XVI)

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

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

Substitute $\mathrm{0..0059}\text{in}$ for $\epsilon$ and $9.06\text{in}$ for $D$ in Equation (IV).

  $\begin{array}{c}R=\frac{0.0059\text{in}}{9.06\text{in}}\\ =6.51\times {10}^{-4}\end{array}$

Substitute $6.51\times {10}^{-4}$ for $R$, $4491.37V$ for $\mathrm{Re}$ in Equation (V).

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

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

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

Substitute $2.3\text{ft}$ for $H_{o,water}$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_w$ and $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ in Equation (VIII).

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

Substitute $8.5\times {10}^{-6}\text{in}/{\text{SCFM}}^{\text{2}}$ for $a$, $62.24\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_w$ and $0.07392\text{lbm}/{\text{ft}}^{\text{3}}$ for ${\rho }_a$ in Equation (IX).

  $\begin{array}{c}{a}_{air}=\frac{8.5\times {10}^{-6}\text{in}/{\text{SCFM}}^{\text{2}}\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =\frac{\left(8.5\times {10}^{-6}\text{in}/{\text{SCFM}}^{\text{2}}\right)\left(\frac{1\text{ft}}{12\text{in}}\right)\times 62.24\text{lbm}/{\text{ft}}^{\text{3}}}{0.07392\text{lbm}/{\text{ft}}^{\text{3}}}\\ =8.5\times {10}^{-6}\text{in}/{\text{SCFM}}^{\text{2}}\end{array}$

Substitute $161.38\text{ft}$ for $H_o$, $5.96\times {10}^{-4}\text{ft}/{\text{SCFM}}^{\text{2}}$ for $a$ and $0.4477V{\text{ft}}^{\text{2}}$ for $\dot{V}$ in the given available head equation.

  $\begin{array}{c}{H}_{avialable}=161.38\text{ft}-\left(5.96\times {10}^{-4}\text{ft}/{\text{SCFM}}^{\text{2}}\right){\left(0.4477V{\text{ft}}^{\text{2}}\right)}^2\\ =161.38\text{ft}-\left(5.96\times {10}^{-4}\text{ft}/{\text{SCFM}}^{\text{2}}\right)\left(0.2004V^2{\text{ft}}^{\text{4}}\right)\\ =161.38\text{ft}-\left(1.1946V^2\times {10}^{-4}{\text{ft}}^5/{\text{SCFM}}^{\text{2}}\right)\end{array}$...... (XVIII)

Since at the operating point the available head and the required head are equal, therefore equate equation (XVI) and (XVIII).

  $\begin{array}{l}\left[\begin{array}{l}161.38\text{ft}-\\ \left(1.1946V^2\times {10}^{-4}{\text{ft}}^5/{\text{SCFM}}^{\text{2}}\right)\end{array}\right]=\left[\begin{array}{l}\left(8.08+45.033f\right)\times \\ 0.0155V^2{\text{s}}^{\text{2}}/\text{ft}\end{array}\right]\\ \left[\begin{array}{l}161.38\text{ft}-\\ \left(1.1946V^2\times {10}^{-4}{\text{ft}}^5/{\text{SCFM}}^{\text{2}}\right)\\ \left(\frac{{\text{SCFM}}^{\text{2}}}{0.000289{\text{ft}}^6/{\text{s}}^2}\right)\end{array}\right]=\left[\begin{array}{l}\left(8.08+45.033f\right)\times \\ 0.0155V^2{\text{s}}^{\text{2}}/\text{ft}\end{array}\right]\\ \left[\begin{array}{l}161.38\text{ft}-\\ \left(0.413V^2{\text{s}}^{\text{2}}/\text{ft}\right)\end{array}\right]=\left[\begin{array}{l}\left(8.08+45.033f\right)\times \\ 0.0155V^2{\text{s}}^{\text{2}}/\text{ft}\end{array}\right]\end{array}$...... (XIX)

Solve Equation (XVII) and Equation (XIX) to obtain the value of velocity as $16.8\text{ft}/\text{s}$.

Substitute $16.8\text{ft}/\text{s}$ for $V$ and $9.06\text{in}$ for $D$ in Equation (VII).

  $\begin{array}{c}\dot{V}=\left(\frac{\pi {\left(9.06\text{in}\right)}^2}{4}\right)\left(16.8\text{ft}/\text{s}\right)\\ =\left(\frac{\pi {\left(9.06\text{in}\left(\frac{1\text{ft}}{12\text{in}}\right)\right)}^2}{4}\right)\times 16.8\text{ft}/\text{s}\\ =7.5213{\text{ft}}^{\text{3}}/\text{s}\left(\frac{1\text{SCFM}}{0.017{\text{ft}}^{\text{3}}/\text{s}}\right)\\ =451.277\text{SCFM}\end{array}$

Therefore, the volume flow rate when considering the minor losses is $451.277\text{SCFM}$ and the volume flow rate when the minor losses are neglected is $492.99\text{SCFM}$. Hence, the volume flow rate is increased by approximately by $11\%$ while neglecting the minor losses. So the minor losses are not negligible.

Conclusion:

The minor losses are important because the volume flow rate is increased by $11\%$ when neglecting the minor losses.