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Average absorption coefficient of a hall

Short answer

Using Sabine’s formula T = 0.161V/(āS), the average absorption coefficient is ā = 0.161V/(TS) = (0.161×5500)/(2.3×750) ≈ 0.51.

A reverberation time of 2.3 sec is observed in a hall of volume 5500 m³. The sound absorbing surface of the hall has an area of 750 m². Calculate the average absorption coefficient.

Given: reverberation time T = 2.3 s, volume V = 5500 m³, sound-absorbing surface area S = 750 m².

Formula (Sabine):

T = 0.161 V / (ā S)     →     ā = 0.161 V / (T S)

(V in m³, S in m², T in s; 0.161 s/m is 4 ln(10⁶)/c with c = 343 m/s.)

Steps

  1. Numerator: 0.161 × 5500 = 885.5 s·m².
  2. Denominator: T × S = 2.3 × 750 = 1725 s·m².
  3. ā = 885.5 / 1725 = 0.513.

(Equivalently, the total absorption is A = ā S = 0.161 V/T = 385 m² of open window, i.e. 385 sabins.)

Answer: the average absorption coefficient is ā ≈ 0.51 (dimensionless; about 51% of the incident sound is absorbed).

✓ Answer checked by Claude Opus 5.5 (high effort)

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