---
title: "Air flow through openings numerical (0.36 m, 0.4 m)"
subject: "Building Technology"
chapter: 1
key: "semester_6/building_technology:airflow-078p"
source: https://www.bctengineer.com/q/semester_6/building_technology/airflow-078p
site: BCT Engineering
---

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

Building Technology · Chapter 1: Functional Requirements of Buildings

- Asked in: 2078 Poush, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/building_technology/airflow-078p
- Whole chapter: https://www.bctengineer.com/md/semester_6/building_technology/chapter-1

## Question

Calculate the rate of air flow through two square openings of size 0.36 m and 0.4 m in a 4m × 4m × 4m room, with a wind of 5km/hr blowing inclined at 45° to the opening. Take coefficient of effectiveness = 0.32.

## Short answer

Effective area A = (A1*A2)/√(A1²+A2²) = 0.1007 m². Airflow Q = E*A*V = 0.32*0.1007*1.389 = 0.0448 m³/s (161.1 m³/h).

## Answer

#### 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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Before using the formula, the wind speed 5 km/h must be converted to:
   A. 5 x 3.6 = 18 m/s
   B. 5/3.6 = 1.389 m/s
   C. 5/60 = 0.083 m/s
   D. 5 x 1000 = 5000 m/s
   Answer: B. Divide km/h by 3.6 to get m/s, giving 1.389 m/s. (Key point: Unit conversion)

2. What does the coefficient of effectiveness 0.32 account for here, and does room size matter?
   A. Wind direction effect; room size does not enter the formula
   B. Room volume; wind direction is ignored
   C. Opening shape; room size matters
   D. Air temperature; room size matters
   Answer: A. E already includes the 45 degree wind direction, and the room volume is not used in Q = E A V. (Key point: Role of E)

3. Squares of 0.36 m and 0.4 m give areas of:
   A. 0.72 m^2 and 0.8 m^2
   B. 0.1296 m^2 and 0.16 m^2
   C. 0.36 m^2 and 0.4 m^2
   D. 0.13 m^2 and 0.4 m^2
   Answer: B. Area is side squared: 0.36^2 = 0.1296 and 0.4^2 = 0.16 m^2. (Key point: Opening areas)

4. The air flow Q comes to about:
   A. 0.0448 m^3/s
   B. 0.448 m^3/s
   C. 0.0100 m^3/s
   D. 0.1007 m^3/s
   Answer: A. A = 0.1007 m^2, so Q = 0.32 x 0.1007 x 1.389 = 0.0448 m^3/s. (Key point: Final answer)

## Close variants set in other exams

- [Air flow through openings numerical (0.3 m, 0.4 m)](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Aairflow-079a) — Building Technology, chapter 1, asked 2×

## Related in Building Technology

- [Building types, mechanical ventilation, air flow](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Aairflow-078c) — Building Technology, chapter 1

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- [Production room 27 × 45 m twin 36 W tubes](https://www.bctengineer.com/question.md?key=semester_6%2Findustrial_power_distribution_and_illumination%3Aproduction-room-27x45) — Industrial Power Distribution and Illumination, chapter 7, asked 4×