Chapter 14, Problem 55EP

A local ventilation system (a hood and duct system) is used to remove air and contaminants produced by a welding operation (Fig. P 14-55E). The inner diameter (ID) of the duct is D = 9.06 in, its average roughness is 0.0059 in, and its total length is L = 34.0 ft. There are three elbows along the duct, each with a minor loss coefficient of 0.21. Literature from the hood manufacturer lists the hood entry loss coefficient as 4.6 based on duct velocity. When the damper is fully open, its loss coefficient is 1.8. A squirrel cage centrifugal fan with a 9.0-in inlet is available. Its performance data fit a parabolic curve of the form $H_{available}=H_0-a{\stackrel{\dot{}}{V}}^2$ , where shutoff head $H_0=2.30$ inches of water column, coefficient $a=8.50\times {10}^{-6}$ inches of water column per (SCFM)2, available head $H_{available}$is in units of inches of water column, and capacity is in units of standard cubic feet per minute (SCFM, at 77°F). Estimate the volume flow rate in SCFM through this ventilation system.

To determine

The volume flow rate.

Answer to Problem 55EP

The volume flow rate is $\text{451}\text{.277}\text{SCFM}$.

Explanation of Solution

Given Information:

The inner diameter of the duct is $9.06\text{in}$, the average roughness of the duct is $0.0059\text{in}$, the total length of the duct is $34.0\text{ft}$, the number of elbows is $3$ each having minor loss coefficient of $0.21$, the hood entry loss coefficient is $4.6$, the loss coefficient is $1.8$, the given equation for available head is $H_{avialable}=H_o-a{\dot{V}}^2$, the shutoff head is $2.30\text{in}$ of water column and the 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

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

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 $n$.

Expression for Reynold'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 $a$ from inches of water column to inches of air column.

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

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).

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

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 (II).

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

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 (X).

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

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).

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

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.

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

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

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

Solve Equation (XII) and Equation (XIV) 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).

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

Conclusion:

The volume flow rate is $\text{451}\text{.277}\text{SCFM}$.