How Do You Calculate Reverberation Time and Sound Levels in an Auditorium?

[roldy]

Homework Statement

The surfaces of an auditorium are 200ft x 50ft x 30ft and have an average absorptivity a=0.29.

(a) What is the reverberation time?
(b) What must be the power output of a source if it is to produce a steady-state sound pressure level of 65 dB reference 20 [tex]\mu[/tex]Pa?
(c)What average absorptivity would be required if a speaker having an acoustic output of 100 [tex]\mu[/tex]W is to produce a steady-state level of 65dB reference 20[tex]\mu[/tex]Pa?
(d) Calculate the resulting reverberation time, and comment on its influence on the intelligibility of the speaker

Homework Equations

Reverb time=0.191V/(Sa)
S=surface area
a is the average absorptivity coefficient
V is the volume of the room
A is the absorptivity

[tex]W=10^{\frac{SWL}{10}}*W_o[/tex]
W is the power output (Watts)
SWL is the sound power level
[tex]W_o=10^{-12}[/tex]

[tex]SIL=10^{-12}*10^{\frac{SPL}{10}}[/tex]
SIL=sound intensity level
SPL=sound pressure level

[tex]SWL=SIL-10log\left(\frac{4}{A}\right)[/tex]

a=A/S

[tex]W=10^{\frac{SWL}{10}}*W_o[/tex]

The Attempt at a Solution

(a)S=2(50*30)+2(200*30)+2(200*50)=35000

Reverb time=[tex]\frac{0.101*200*50*30}{35000*.29}=4.759[/tex]sec


(b)First I found A
A=a*S=0.29*35000=1015

Then I found SIL from the SPL given

SIL=[tex]10^{-12}*10^{\frac{65}{10}}=3.16*10^{-6}W/m^2[/tex]

Then I found the SWL.

[tex]SWL=SIL-10log\left(\frac{4}{A}\right)=3.16*10^{-6}-10log\left(\frac{4}{1015}\right)=24 W/m^2[/tex]

Then I calculated the power out put from [tex]W=10^{\frac{SWL}{10}}*W_o[/tex]

W=[tex]2.51*10^{-10}[/tex] Watts

Did I do this right? This seems like a really small number for this situation

(c) I just solved [tex]SWL=SIL-10log\left(\frac{4}{A}\right)[/tex] in terms of A

[tex]A=\frac{4}{10^{\frac{SIL-SWL}{10}}}=4*10^9 m^2[/tex]

So then the average absorptivity is [tex]a=\frac{4*10^9}{35000}=1.14*10^6[/tex]

Now this doesn't seem right if I think about it logically. If my power is really small (part b), why would I need a tremendous amount of absorptivity? I can solve part d easily but if I solved either part b or c then d will be wrong. Can anyone see if I made a mistake if any?

 

[jbsteele]

(d)Reverb time=\frac{0.101*200*50*30}{35000*1.14*10^6}=4.2secThis reverberation time is on the higher end and will significantly reduce the intelligibility of the speaker.