flow rate

[logistic_guy]

In climates with low night-time temperatures, an energy-efficient way of cooling a house is to install a fan in the ceiling that draws air from the interior of the house and discharges it to a ventilated attic space. Consider a house whose interior air volume is \(\displaystyle 720 \ \text{m}^3\). If air in the house is to be exchanged once every \(\displaystyle 20\) minutes, determine \(\displaystyle \bold{(a)}\) the required flow rate of the fan and \(\displaystyle \bold{(b)}\) the average discharge speed of air if the fan diameter is \(\displaystyle 0.5 \ \text{m}\).

 

[khansaheb]

In climates with low night-time temperatures, an energy-efficient way of cooling a house is to install a fan in the ceiling that draws air from the interior of the house and discharges it to a ventilated attic space. Consider a house whose interior air volume is \(\displaystyle 720 \ \text{m}^3\). If air in the house is to be exchanged once every \(\displaystyle 20\) minutes, determine \(\displaystyle \bold{(a)}\) the required flow rate of the fan and \(\displaystyle \bold{(b)}\) the average discharge speed of air if the fan diameter is \(\displaystyle 0.5 \ \text{m}\).

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[logistic_guy]

determine \(\displaystyle \bold{(a)}\) the required flow rate of the fan

\(\displaystyle \bold{flow \ rate} = \frac{V}{t} = \frac{720}{20 \times 60} = \textcolor{blue}{0.6 \ \text{m}^3\text{/s}}\)

 

[logistic_guy]

determine \(\displaystyle \bold{(b)}\) the average discharge speed of air if the fan diameter is \(\displaystyle 0.5 \ \text{m}\).

\(\displaystyle \bold{average \ discharge \ speed} = \frac{\bold{flow rate}}{A} = \frac{0.6}{\pi\left(\frac{0.5}{2}\right)^2} = \textcolor{blue}{3.056 \ \text{m/s}}\)