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15th March 2020, 02:15 AM  #1 
diyAudio Member
Join Date: Mar 2020

Port Size/Length with Multiple Ports
I am trying to design my first loudspeaker project inspired by SVS subwoofers, but I find their port configuration does does not match the theory I have been told. What am I missing? If I try to follow WinISD blindly it is going to make my ports multiple times longer than what I see in a real life SVS. Now just to keep things simple, lets not talk about my design, but just try to apply the theory to the SVS PB16Ultra.
I am told the relationship between the various quantities that allow one to tune a port in a subwoofer enclosure are described by the Helmholtz equation: Code:
f=v/(2*pi)*sqrt(A/(VL)) where v is the speed of sound 340 meters/second pi is the constant 3.14159 A is the effective cross sectional area of the port V is the volume of the enclosure L is the length of the ports Code:
d=sqrt(d1^2+d2^2+...+dn^2) where d is the effective area di is the diameter of the i'th port
Solving for effective area we get: Code:
d=sqrt(3 * .089 ^2) d=0.154 m Code:
f=v/(2*pi)*sqrt(A/(VL)) f = 340 m/s /(2*pi) * sqrt(3.14159*(0.154 m /2)^2/(0.176 m^3 * 0.86 m)) f = 18.98 hertz Code:
L' = L + N * K * d/2 where L' is the effective port length L is the physical port length N is the number of flared ports, in this case 2 K is the coefficient of a flared port, in this case assume 0.85 d is the effective diameter L' = 0.86m + 2 * 0.85 * 0.154m /2 L' = 0.99 m f= v/(2*pi)*sqrt(A/(VL')) f = 340 m/s /(2*pi) * sqrt(3.14159*(0.154 m /2)^2/(0.176 m^3 * 0.99 m)) f = 17.7 hertz Lets also consider port velocity (equation is from here, constant differ to convert to metric units): Code:
V = 0.00225 * sqrt(W) / (f * (d/2)^2) where v = port air velocity in mach (e.g. % of speed of sound) W = acoustic power in watts F = frequency d = effective diameter in meters Code:
W = We * 10^((S – 112)/10) where W is the acoustic power in watts We is the electric power in watts S is the sensitivity in db per 1W @ 1 Meter Code:
W = 1500 watts * 10^((98 – 112)/10) W = 59.7 watts Code:
V = 0.00225 * sqrt(W) / (f * (d/2)^2) for f = 17.2 hertz V = 0.00225 * sqrt(59.7 watts) / (17.2 hertz * (0.154 m/2)^2) V = 0.17 for f = 14 hertz V = 0.00225 * sqrt(59.7 watts) / (14 hertz * (0.154 m/2)^2) V = 0.21 Last edited by AbilityBuild; 15th March 2020 at 02:32 AM. Reason: Fixed air speed equation 
15th March 2020, 04:03 AM  #2  
diyAudio Member
Join Date: Jun 2018

Quote:
Guidelines and formulas that were created by pioneers in the field were relevant to the projects and experiments at the time. There are cases where simulation and real world do not correlate directly. When the port outlet is near a boundary inside the box ( or outside at the floor too ) it acts as if it were physically longer than it measures. Having a turn in the port will also make the port act as if it were physically longer. Approximations are used in the HVAC and plumbing world, each turn and fitting adds an equivalent pipe length, and the best case is a large, sweeping bend for good fluid dynamics behavior. Here is one source: Friction Loss In PVC Pipe Fittings Building a subwoofer and measuring the impedance will nail down your tuning with little ambiguity. One can also use single frequency test tones with a generator and a sound card to measure for excursion minima. Having multiple ports is not as efficient as a single round port, but can give you some flexibility in tuning by blocking ports. However, as you block a port, the effective enclosure volume goes up slightly ( by the interior volume of the port up to the plug ) and velocity through the remaining port or ports increases. 

15th March 2020, 02:19 PM  #3  
diyAudio Member
Join Date: Nov 2013
Location: Aberdeen, Scotland

Quote:
Second, you need to multiply by 3 for the multiple ports after the squaring otherwise you skew the result completely. So, the effective area for those 3 ports is 0.0186m^2. That's nearly an order of magnitude smaller than your calculation.... Quote:
Take at least 10, maybe 15dB off that as a starting point. HTH, David. 

15th March 2020, 04:54 PM  #4  
diyAudio Member
Join Date: Mar 2020

@diyuser2010 Thank you for pointing out some realworld effects I should be aware of. In addition to the ones you notes I thought of a few more, here is my working list so far:
Quote:
Quote:
@David Morison Quote:
Code:
A = pi * (d/2)^2 A = 3.145159 * (0.154/2)^2 A = 0.0186 meters^2 Quote:
Code:
W = We * 10^((S – 112)/10) where W = 1500 watts * 10^(((9813.2) – 112)/10) W = 2.858 watts Code:
V = 0.00225 * sqrt(W) / (f * (d/2)^2) for f = 17.2 hertz V = 0.00225 * sqrt(2.858 watts) / (17.2 hertz * (0.154 m/2)^2) V = 0.037 for f = 14 hertz V = 0.00225 * sqrt(2.858 watts) / (14 hertz * (0.154 m/2)^2) V = 0.045 Code:
for f = 14 hertz V = 0.00225 * sqrt(2.858 watts) / (14 hertz * (2/3) * (0.154 m/2)^2) V = 0.068 Code:
for f = 14 hertz V = 0.00225 * sqrt(2.858 watts) / (14 hertz * (2/3) * (0.154 m/2)^2) V = 0.138 Thanks everyone! I think the SVS design now matches my theoretical understanding. The difference between the two shows how hard the SVS team worked to get all these smaller real world effects working in their favor. 

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