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Reverberation time of small hall

Short answer

Total absorption ΣaS = 0.04×150 + 0.04×90 + 0.08×95 = 17.2 m². Average absorption coefficient ā = 17.2/335 ≈ 0.051. Reverberation time T = 0.161×900/17.2 ≈ 8.4 s.

The volume of a small hall in BICC is 900m³. The wall area of the hall is 150 m², ceiling area 90 m² and floor area is 95 m². The sound absorption coefficient for wall is 0.04, for ceiling is 0.04 and for the floor is 0.08. Calculate the average sound absorption coefficient and reverberation time.

Use the average absorption coefficient ā = ΣaS / ΣS and Sabine's formula T = 0.161 V / ΣaS.

Given: V = 900 m³; walls: S₁ = 150 m², a₁ = 0.04; ceiling: S₂ = 90 m², a₂ = 0.04; floor: S₃ = 95 m², a₃ = 0.08.

  1. Total absorption:
ΣaS = 0.04×150 + 0.04×90 + 0.08×95
    = 6.0 + 3.6 + 7.6
    = 17.2 m² (open-window units)
  1. Total surface area ΣS = 150 + 90 + 95 = 335 m²
  2. Average absorption coefficient: ā = 17.2 / 335 = 0.0513
  3. Reverberation time:
T = 0.161 × 900 / 17.2
  = 144.9 / 17.2
  = 8.42 s

Answer: ā ≈ 0.051; T ≈ 8.4 s (very long, so the hall needs more absorbing material).

✓ Answer checked by Claude Opus 5.5 (high effort)

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