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Air flow through openings numerical (0.3 m, 0.4 m)

Short answer

For wind-driven ventilation through two openings in series, effective area A = (A1×A2)/√(A1²+A2²) = 0.07844 m²; airflow Q = E×A×V = 0.68 × 0.07844 × 1.111 = 0.0593 m³/s = 213.4 m³/h.

Air is flowing through a two square opening of size 0.3 m and 0.4 m in a 5m × 5m × 5m room, with a wind of 4 km/hr blowing at 90° to the opening. Take coefficient of effectiveness = 0.68. Calculate the rate of air flow.

Given

  • Inlet and outlet openings: 0.3 m × 0.3 m and 0.4 m × 0.4 m, so A1=0.0900 m2A_1 = 0.0900\ \text{m}^2 and A2=0.1600 m2A_2 = 0.1600\ \text{m}^2
  • Wind speed v=4v = 4 km/h =4/3.6=1.1111= 4/3.6 = 1.1111 m/s, at 90° to the opening
  • Coefficient of effectiveness E=0.68E = 0.68 (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=A1A2A12+A22,Q=E A VA = \frac{A_1 A_2}{\sqrt{A_1^2 + A_2^2}}, \qquad Q = E\, A\, V

Calculation

A=0.0900×0.16000.09002+0.16002=0.07844 m2Q=0.68×0.07844×1.1111=0.05927 m3/s\begin{aligned} A &= \frac{0.0900 \times 0.1600}{\sqrt{0.0900^2 + 0.1600^2}} = 0.07844\ \text{m}^2 \\ Q &= 0.68 \times 0.07844 \times 1.1111 = 0.05927\ \text{m}^3/\text{s} \end{aligned}

Answer: Q=0.0593Q = 0.0593 m³/s =213.4= 213.4 m³/h.

✓ Answer checked by Claude Opus 5.5 (high effort)

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