A hall has a volume of 1,25,000 m3. It has reverberation time of 1.3 seconds. What is the average absorbing power of the surface, if the total absorbing surface area is 20,000 m2.
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Answered by
5
Answer:
Reverberation time=0.161
a×c
v
=
2×10
4
×0.2
0.161×10
5
=4.025seconds
Answered by
1
Given:
Volume of hall (V) = 1,25,000 m³
Reverberation time (T) = 1.3 seconds
Total absorbing surface area (∑s) = 20,000 m²
To Find:
The average absorbing power of the surface
Formula: Reverberation Time
T = 0.167v/∑as
Total absorption coefficient ∑αs =0.167v/T
∑αs =0.167×1,25,000 /1.3
∑αs = 16057.6 OWU
Average power of the surface α = 0.167/ ∑ST
= 16.57.6/20000
= 0.802
Answer = 0.802 is the average absorbing power.
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