Chapter 10, Problem 36P

To determine

To Find: The speed of the air flow in the duct

Explanation of Solution

Given information:

The radius of air duct, $r=15\mathrm{cm}$

Room’s dimension $=9.2m\times 5m\times 4.5m$

Formula used:

  $A_{\text{duct}}\times v_{\text{duct}}=\pi r^2{v}_{\text{duct}}=\frac{v_{\text{room}}}{t_{t\text{o}\text{fill}\text{room}}}$

Here the equation of continuity at constant density is used which states that:

Flow rate out of duct becomes equal to the Flow rate into room

Calculations:

By using the equation of continuity at constant density, then the result will be

  $\begin{array}{l}{A}_{\text{duct}}\times v_{\text{duct}}=\pi r^2{v}_{\text{duct}}=\frac{v_{\text{room}}}{t_{t\text{o}\text{fill}\text{room}}}\\ v_{\text{duct}}=\frac{v_{\text{room}}}{t_{t\text{o}\text{fill}\text{room}}\times \pi r^2}\\ =\frac{9.2m\times 5m\times 4.5m}{p{\left(15cm\right)}^2\left(16minute\right)\left(\frac{60s}{1minute}\right)}=3.1m/s\end{array}$

Conclusion:

The speed of the flow of the air in the duct is $3.1\text{m/s}$ .