1
$\begingroup$

I don't see anything in Wikipedia about it, but (from this question) when I try to reproduce the equations in this 1933 paper, I get values 106 away from the values in the paper. When I reproduce this equation, for instance:

The r.m.s. pressure P in the interval between f1 and f2, is ... For example, if f1=1000 and f2=6000 c.p.s., P=5×10<sup>−5</sup> bars

I get 5×10−11 bar

Likewise, it says:

At 1000 c.p.s the average M.A.F observed is 1.9×10<sup>−16</sup> watts per cm², corresponding to a pressure 71 db below 1 bar.

Sound intensity and pressure are related by acoustic impedance of air Z0 = 400 N·s/m3.

So

$\endgroup$
  • 2
    $\begingroup$ I wonder how long it will take for "bars" as a unit of mobile phone signal strength to outweigh the unit of pressure in common usage? $\endgroup$ – Jonathan R Swift Mar 8 '18 at 9:07
  • 2
    $\begingroup$ Or how many in the street... when going drinking :) $\endgroup$ – Solar Mike Mar 8 '18 at 10:42
  • 1
    $\begingroup$ At STP 1 bar is 101325 pa if I remember correctly. $\endgroup$ – Solar Mike Mar 8 '18 at 10:42
  • 2
    $\begingroup$ interestingly the "standard pressure" in STP actually varies by context, see en.wikipedia.org/wiki/…, it is 101325 or 100000 Pa depending on who you ask. In any case a bar is always 10^5 pa $\endgroup$ – agentp Mar 8 '18 at 22:22
  • 1
    $\begingroup$ I think 1 bar = 100,000 Pa and 1 atm = 1.01325 bar = 101325 Pa $\endgroup$ – Ohio ChemE Mar 9 '18 at 1:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.