Air flow through openings numerical (0.36 m, 0.4 m)

Given

• Inlet and outlet openings: 0.36 m × 0.36 m and 0.4 m × 0.4 m, so $A_1 = 0.1296\ \text{m}^2$ and $A_2 = 0.1600\ \text{m}^2$
• Wind speed $v = 5$ km/h $= 5/3.6 = 1.3889$ m/s, at 45° to the opening
• Coefficient of effectiveness $E = 0.32$ (already accounts for the wind direction; the room size does not enter the formula)

Formula

For wind-driven ventilation through two openings in series, the effective area is

$$A = \frac{A_1 A_2}{\sqrt{A_1^2 + A_2^2}}, \qquad Q = E\, A\, V$$

Calculation

$$\begin{aligned} A &= \frac{0.1296 \times 0.1600}{\sqrt{0.1296^2 + 0.1600^2}} = 0.10071\ \text{m}^2 \\ Q &= 0.32 \times 0.10071 \times 1.3889 = 0.04476\ \text{m}^3/\text{s} \end{aligned}$$

Answer: $Q = 0.0448$ m³/s $= 161.1$ m³/h.