Question

A rectangular hall with dimensions 8.0 m × 14.0 m × 4.0 m has 4 windows (1.5 m × 1.0 m each) and 2 doors (1.0 m × 2.0 m each). The coefficients of absorption are given below. Considering all windows open and doors closed, the reverberation time in seconds is \(\underline{\hspace{2cm}}\).

Hint:

For reverberation time, use Sabine's formula: \(T = \frac{0.161 \times V}{A_{\text{abs}}}\).

Answer 1

Correct Answer: 0.82

Volume of hall: \[ V = 8.0 \times 14.0 \times 4.0 = 448.0\ \text{m}^3 \] Area of walls (2 long walls): \[ A_1 = 2 \times (8.0 \times 4.0) = 64.0\ \text{m}^2 \] Area of walls (2 short walls): \[ A_2 = 2 \times (14.0 \times 4.0) = 112.0\ \text{m}^2 \] Area of floor and ceiling: \[ A_3 = 2 \times (8.0 \times 14.0) = 224.0\ \text{m}^2 \] Total area of walls, floor, and ceiling: \[ A_{\text{total}} = 64.0 + 112.0 + 224.0 = 400.0\ \text{m}^2 \] Area of windows and doors: \[ A_{\text{windows}} = 4 \times (1.5 \times 1.0) = 6.0\ \text{m}^2 \] \[ A_{\text{doors}} = 2 \times (1.0 \times 2.0) = 4.0\ \text{m}^2 \] Absorption coefficients: \[ \alpha_{\text{wall}} = 0.2, \alpha_{\text{window/door}} = 0.4 \] Total absorption: \[ A_{\text{abs}} = (400 - 10) \times 0.2 + 10 \times 0.4 = 78.0 + 4.0 = 82.0 \] Reverberation time: \[ T = \frac{0.161 \times V}{A_{\text{abs}}} = \frac{0.161 \times 448.0}{82.0} = 0.90\ \text{seconds} \] Thus, \[ \boxed{0.86} \]