Question

A hall of dimension 25×18×12m³ has an average absorption coefficient 0.2. find the reverberation time. If a curtain cloth of area 150m² is suspended at the  centre of Hall with coefficient of absorption 0.75, What will be the reverberation
time?

Answer 1

Given Data  :- volume of hall = 25 ×18 ×12 = 5400

α = 0.75  α_av = 0.2  S_curtain = 100 m²

Formula :-  `T_1= 0.161 × V/(α_(av)  × S`)`

`T_2 = 0.161 ×  V/(α_(av) × S + α_("curtain ") × S`

Calculate :- S = 2[(20 ×15)+(15 ×10)+(10 ×20)] = 1300m² 
V = 20×15×10 = 3000 m² 

`T_1 = 0.161 ×3000/(0.1×1300) = 3.7 sec`.

∴ Absorption takes place by both the surfaces of the curtain
S’ = 2×100 m² = 200 m²

`T_2 = 0.161 × 3000/ ((0.1 ×1300)+(0.66 ×200))= 1.84 sec`

Answer :- Change in reverberation time = 1.85 sec.