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2nd March 2007, 09:38 PM  #1 
diyAudio Member
Join Date: Nov 2004
Location: Suomi, Finland

Mach formula based on vent area?
I've been searching for a formula that would define the port mach according to port area. I seem to find only formulas that are based on port diameter (now these are ok only for calculating mach in round ports). Also, most of these formulas do not specify units they use so one basically has to guess, which variables are constants and which are result from unit conversion.
Is there an equation that uses standard SIunits and works also with "odd" shaped ports; in other words, bases the calculation on port area and not on it's diameter? If someone can provide a concerned equation I'd appreciate if he/she would also site the reference for it so I can study how the equation is derived. 
2nd March 2007, 10:26 PM  #2 
diyAudio Member
Join Date: Jun 2002
Location: USA, MN

Minimum Port Diameter
velocity in an odd shaped port is the same as a round one of the same area. You might run into issues if your port is 1cm high and 400cm wide.... extreme aspect ratios.
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3rd March 2007, 09:48 AM  #3  
diyAudio Member
Join Date: Nov 2004
Location: Suomi, Finland

Thanks. I did a search on this and saw the linked thread. I assume you are refering to
Sp_min=0.24*10^(SPL_max/20)/(Fb*up_max) in one of your posts. It indeed seems like what I need. However, I'm a bit sceptic with these "ruleofthumb" equations. From where is this equation derived from and what are the constants 0,24, 10 and 20? up_max is in m/s but sp_min in cm^2; what makes the unit convert from meter to centimeter? You see, I also want to know how this works. Quote:
mach = (13,7*sqrt(Wa))/(fb*r^2), where Wa is the acoustic power. Radius is given in inches (again a unit conversion). However, I find it hard to believe that all the initial flow calculations would be based on assumption that ports are round. The source where I found this equation failed to site a reference to it as well. None of the equations calculating mach in round port I have seen so far incorporate Pi although that is a variable needed in calculating radius from a circle area (or area based on radius). Now I assume this means that they either derived the equation differently or simplified the formula merging Pi in one of the constants? 

3rd March 2007, 10:54 AM  #4  
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

Hi,
Quote:
Quote:
The speed through the port is the displaced volume (per second) divided by the port area. What's so difficult with that? What is Sd? What is the maximum frequency that you can achieve Xmax? What is Xmax? What is the port area? Calculate the port air speed.
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regards Andrew T. 

3rd March 2007, 06:12 PM  #5  
diyAudio Member
Join Date: Nov 2004
Location: Suomi, Finland

Quote:


3rd March 2007, 06:30 PM  #6 
diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders

Hi,
I think it is that simple for average velocity. I believe the Mach ratio is based on average velocity, but if it's based on peak velocity then just multiply by sqrt2. It is only sinusoidal waveform where zero velocity occurs at the peaks of the movement and maximum velocity is at the midpoint of the movement. The mach ratio is just a recomendation to aim for or not to exceed. It matters little whether you are 10% or 7%, it's only the size of the hole (area of the port) that needs altering. What is much more important is setting the port tuning to best suit the speaker and the room. The variables affecting the tuning and it's frequency are much less tolerant of small errors and corrections.
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regards Andrew T. 
3rd March 2007, 08:11 PM  #7 
diyAudio Member
Join Date: Feb 2004
Location: Stockholm

So let's derive it...
The sound pressure from a point source is: p=Q*rho0*w/(4*pi*r) Q = volume flow, rho0=1.2kg/m3, r = distance. The volume flow is Q=v*S v= air velocity in the port, S= crosssectional area (independent of shape). Fiddling a bit with the equations results in v=p*4*pi*r/(S*rho0*w) So, for example, to acheive a sound pressure level of 100 dB (the sound pressure is p=2e5*10^(100/10)=2 Pa) at a distance of 1 m, when the port is 2x10 cm (=0.02*0.1=0.002 m2) at the f=30 Hz, the port velocity becomes v=2*pi*1/(0.002*1.2*2*pi*30)=13.9 m/s which corresponds to 13.9/345= 0.04 mach. 
3rd March 2007, 11:46 PM  #8 
diyAudio Member
Join Date: Nov 2004
Location: Suomi, Finland

Wow, thanks.

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