Port Size/Length with Multiple Ports
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 15th March 2020, 02:15 AM #1 AbilityBuild 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 PB16-Ultra. I am told the relationship between the various quantities that allow one to tune a port in a sub-woofer 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 I am told that the effective diameter of multiple ports is found by: Code: d=sqrt(d1^2+d2^2+...+dn^2) where d is the effective area di is the diameter of the i'th port In the SVS PB16-ULTRA (see a cutaway here and official webpage here ) to solve the above equation we need the following information: Port Diameter: The subwoofer has three 3.5 inch (.089 m) diameter ports Volume: The subwoofer has external dimensions 25" (H) x 21.7" (W) x 28.3" (D), lets say that 30% of that is the ports, walls, and driver, so that would leave 0.176 m^3 of air volume. Port Length: Eyeballing the cut-away from the video I would guess that the port is about 20% longer than the longest dimension of the enclosure. Lets call that 28.3" * 1.2 = 0.86 meters Solving for effective area we get: Code: d=sqrt(3 * .089 ^2) d=0.154 m Solving for the tuning frequency: 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 That seems a bit high for something that is trying to extend down to 14 hertz. Perhaps if we change the length by adding some port end corrections 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 17.7 still seems pretty high. I find that I have to tune down near 14Hz in my design that's quite a lot lower. 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 I most often see people suggest that v must be less than 0.1 and should ideally be closer to 0.05. Solving for the SVS subwoofer we need an estimate for W. Lets assume the sensitivity of the driver is 98 db per 1W @ 1 Meter, and its rated electrical power is 1500 RMS. 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 Solving the above Code: W = 1500 watts * 10^((98 – 112)/10) W = 59.7 watts Solving for velocity at the tuning frequency I would expect and the one I calculated above: 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 These are clearly way too high to be realistic. They would be chuffing constantly if the air speed was ~20% the speed of sound. What gives? Perhaps there is something about having three ports that is not captured by this equation? Last edited by AbilityBuild; 15th March 2020 at 02:32 AM. Reason: Fixed air speed equation
diyuser2010
diyAudio Member

Join Date: Jun 2018
Quote:
 Originally Posted by AbilityBuild 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 PB16-Ultra. I am told the relationship between the various quantities that allow one to tune a port in a sub-woofer 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 I am told that the effective diameter of multiple ports is found by: Code: d=sqrt(d1^2+d2^2+...+dn^2) where d is the effective area di is the diameter of the i'th port In the SVS PB16-ULTRA (see a cutaway here and official webpage here ) to solve the above equation we need the following information: Port Diameter: The subwoofer has three 3.5 inch (.089 m) diameter ports Volume: The subwoofer has external dimensions 25" (H) x 21.7" (W) x 28.3" (D), lets say that 30% of that is the ports, walls, and driver, so that would leave 0.176 m^3 of air volume. Port Length: Eyeballing the cut-away from the video I would guess that the port is about 20% longer than the longest dimension of the enclosure. Lets call that 28.3" * 1.2 = 0.86 meters Solving for effective area we get: Code: d=sqrt(3 * .089 ^2) d=0.154 m Solving for the tuning frequency: 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 That seems a bit high for something that is trying to extend down to 14 hertz. Perhaps if we change the length by adding some port end corrections 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 17.7 still seems pretty high. I find that I have to tune down near 14Hz in my design that's quite a lot lower. 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 I most often see people suggest that v must be less than 0.1 and should ideally be closer to 0.05. Solving for the SVS subwoofer we need an estimate for W. Lets assume the sensitivity of the driver is 98 db per 1W @ 1 Meter, and its rated electrical power is 1500 RMS. 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 Solving the above Code: W = 1500 watts * 10^((98 – 112)/10) W = 59.7 watts Solving for velocity at the tuning frequency I would expect and the one I calculated above: 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 These are clearly way too high to be realistic. They would be chuffing constantly if the air speed was ~20% the speed of sound. What gives? Perhaps there is something about having three ports that is not captured by this equation?

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.

David Morison
diyAudio Member

Join Date: Nov 2013
Location: Aberdeen, Scotland
Quote:
 Originally Posted by AbilityBuild In the SVS PB16-ULTRA (see a cutaway here and official webpage here ) to solve the above equation we need the following information: Port Diameter: The subwoofer has three 3.5 inch (.089 m) diameter ports Volume: The subwoofer has external dimensions 25" (H) x 21.7" (W) x 28.3" (D), lets say that 30% of that is the ports, walls, and driver, so that would leave 0.176 m^3 of air volume. Port Length: Eyeballing the cut-away from the video I would guess that the port is about 20% longer than the longest dimension of the enclosure. Lets call that 28.3" * 1.2 = 0.86 meters Solving for effective area we get: Code: d=sqrt(3 * .089 ^2) d=0.154 m
First, the area of a circle is Pi*r^2, not what is in your code.
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:
 I most often see people suggest that v must be less than 0.1 and should ideally be closer to 0.05. Solving for the SVS subwoofer we need an estimate for W. Lets assume the sensitivity of the driver is 98 db per 1W @ 1 Meter, and its rated electrical power is 1500 RMS.
Sorry, but there's no way the sensitivity of a driver designed to get down below 20Hz in boxes of that kind of size is anywhere near 98dB/W/m.
Take at least 10, maybe 15dB off that as a starting point.

HTH,
David.

AbilityBuild
diyAudio Member

Join Date: Mar 2020
@diyuser2010 Thank you for pointing out some real-world 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:
• Boundary effects of ports near walls/floors
• Acoustic Length of Tube does not equal measured length of tube depending on tube bends/gemoetry
• Use of polyfil may make the box appear larger, changing the effective internal area
• Sealing the box very well (As SVS must to get that nice piano black finish) changes its Q. (I need to read more about how this effects the tuning, as Q was not a variable in my simplified equations)

Quote:
 Originally Posted by diyuser2010 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.
Yeah, you are right. I am going to build an enclosure for experimenting with some of these questions, and not obsess too much on the math.

Quote:
 Originally Posted by diyuser2010 However, as you block a port, the effective enclosure volume goes up slightly (by the interior volume of the port up to the plug)
I never thought about why they do the port blocking thing in such clear terms. I have seen other articles give much more confusing and wrong explanations about the pros and cons. By blocking you can extend the low end frequency response, but sacrifice dynamics because chuffing will be much more noticeable.

@David Morison
Quote:
Quote:
 Originally Posted by AbilityBuild Solving for effective area we get: Code: d=sqrt(3 * .089 ^2) d=0.154
Originally Posted by David Morison
First, the area of a circle is Pi*r^2, not what is in your code.
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....
I was solving for the effective diameter in the equation you suggested may be wrong. I double checked the math and it is all right. If you take my effective diameter of 0.154 meters and solve for the area:
Code:
A = pi * (d/2)^2
A = 3.145159 * (0.154/2)^2
A = 0.0186 meters^2
You get the same result as with your approach.

Quote:
 Originally Posted by David Morison Sorry, but there's no way the sensitivity of a driver designed to get down below 20Hz in boxes of that kind of size is anywhere near 98dB/W/m. Take at least 10, maybe 15dB off that as a starting point.
Great point! While some 18" drivers used for this purpose have nominal sensitives of 98dB/W@1m, that is not important. I must look at the sensitivity at this extreme low end frequencies where the velocity of air is greatest. In my design I am starting with a driver with a nominal sensitivity of 89.7dB/W@1M. However at the peak air-velocity frequency (which happens to be 13.33 Hz in my design) its sensitivity is only 76.5 dB. That's a 13.2 dB drop. Lets rerun the SVS numbers assuming a 13.2 dB less efficient low end.

Code:
W = We * 10^((S – 112)/10) where
W = 1500 watts * 10^(((98-13.2) – 112)/10)
W = 2.858 watts
Solving for velocity at the two frequencies
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
Which falls under the guideline of 0.05 mach for chuffing free base. However if you block one port:
Code:
for f = 14 hertz
V = 0.00225 * sqrt(2.858 watts) / (14 hertz * (2/3) * (0.154 m/2)^2)
V = 0.068
Which is acceptable but more borderline, and if you block 2 ports (which SVS does not recommend):
Code:
for f = 14 hertz
V = 0.00225 * sqrt(2.858 watts) / (14 hertz * (2/3) * (0.154 m/2)^2)
V = 0.138
Which is beyond the limit, and shows why SVS tells people not to do this.

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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