Chapter 8, Problem 27P

To determine

Express the volume flow rate of the duct in $\frac{{\text{ft}}^3}{\text{s}}$, $\frac{{\text{ft}}^3}{\text{min}}$, $\frac{{\text{m}}^3}{\text{s}}$ and find the average speed of air in the duct.

Answer to Problem 27P

The volume flow rate of the duct in $\frac{{\text{ft}}^3}{\text{s}}$, $\frac{{\text{ft}}^3}{\text{min}}$ and $\frac{{\text{m}}^3}{\text{s}}$ are $\underset{\_}{16.7\frac{{\text{ft}}^3}{\text{s}}\text{, }2004\frac{{\text{ft}}^3}{\text{min}},0.95\frac{{\text{m}}^3}{\text{s}}}$ respectively and the average speed of air is $\underset{\_}{19.1\frac{\text{ft}}{\text{s}}}$.

Explanation of Solution

Given data:

Refer to the Figure problem 8.27 in the textbook for the duct system.

Formula used:

Write the expression for the volume flow rate.

$Q=V_{\text{average}}{A}_c$ (1)

Here,

$A_c$ is cross-sectional area, and

$V_{\text{average}}$ is average velocity.

Calculation:

Substitute $30\frac{\text{ft}}{\text{s}}$ for $V_{\text{average}}$, $8\text{in}\text{.}\times 10\text{in}\text{.}$ for $A_c$ in equation (1) to find the volume flow rate in each of the $8\text{in}\text{.}\times 10\text{in}\text{.}$ duct.

$\begin{array}{c}Q=\left(30\frac{\text{ft}}{\text{s}}\right)\left(8\text{in}\text{.}\times 10\text{in}\text{.}\right)\\ =\left(30\frac{\text{ft}}{\text{s}}\right)\left(8\times 0.083333\text{ft}\times 10\times 0.083333\text{ft}\right)\left\{\because 1\text{in}=0.083333\text{ft}\right\}\\ =16.7\frac{{\text{ft}}^3}{\text{s}}\end{array}$

Give the volume flow rate in $18\text{in}\text{.}\times 14\text{in}\text{.}$ duct as below.

$Q_{\text{total}}=16.7\frac{{\text{ft}}^3}{\text{s}}+16.7\frac{{\text{ft}}^3}{\text{s}}$

$Q_{\text{total}}=33.4\frac{{\text{ft}}^3}{\text{s}}$ (2)

Convert the unit volume flow rate in equation (2) from $\frac{{\text{ft}}^3}{\text{s}}$ to $\frac{{\text{ft}}^3}{\text{min}}$.

$\begin{array}{c}{Q}_{\text{total}}=33.4\frac{{\text{ft}}^3}{\text{s}}\left(\frac{60\text{s}}{1\min }\right)\left\{\because 1\min =60\text{s}\right\}\\ =2004\frac{{\text{ft}}^3}{\text{min}}\end{array}$

Convert the unit volume flow rate in equation (2) from $\frac{{\text{ft}}^3}{\text{s}}$ to $\frac{{\text{m}}^3}{\text{s}}$.

$\begin{array}{c}{Q}_{\text{total}}=33.4\frac{{\text{ft}}^3}{\text{s}}{\left(\frac{1\text{m}}{3.28\text{ft}}\right)}^3\left\{\because 1\mathrm{m}=3.28\text{ft}\right\}\\ =0.95\frac{{\text{m}}^3}{\text{s}}\end{array}$

Find the average speed of air in the $18\text{in}\text{.}\times 14\text{in}\text{.}$ duct.

Substitute $33.4\frac{{\text{ft}}^3}{\text{s}}$ for $Q_{\text{total}}$, $18\text{in}\text{.}\times 14\text{in}\text{.}$ for $A_c$ in equation (1).

$\begin{array}{l}33.4\frac{{\text{ft}}^3}{\text{s}}=V_{\text{average}}\left(18\text{in}\text{.}\times 14\text{in}\text{.}\right)\\ V_{\text{average}}=\frac{33.4\frac{{\text{ft}}^3}{\text{s}}}{\left(18\text{in}\text{.}\times 14\text{in}\text{.}\right)}\\ V_{\text{average}}=\frac{33.4\frac{{\text{ft}}^3}{\text{s}}}{\left(18\times 0.083333\text{ft}\times 14\times 0.083333\text{ft}\right)}\left\{\because 1\text{in}=0.083333\text{ft}\right\}\\ V_{\text{average}}=19.1\frac{\text{ft}}{\text{s}}\end{array}$

Conclusion:

Hence, the volume flow rate of the duct in $\frac{{\text{ft}}^3}{\text{s}}$, $\frac{{\text{ft}}^3}{\text{min}}$ and $\frac{{\text{m}}^3}{\text{s}}$ are $\underset{\_}{16.7\frac{{\text{ft}}^3}{\text{s}}\text{, }2004\frac{{\text{ft}}^3}{\text{min}},\text{and}0.95\frac{{\text{m}}^3}{\text{s}}}$ respectively and the average speed of air is $\underset{\_}{19.1\frac{\text{ft}}{\text{s}}}$.