Port length - multiple ports

[mashaffer]

I looked in the help files for both of the box design programs that I have and neither one could clarify something for me. When using multiple ports to reduce mach number the specified length more than doubles when the number of ports is doubled. What is not specified is whether the length listed is the length of each port or the sum of all port lengths. So which is it? And if it is the sum does the total length need to be divided evenly between the ports?

mike

[Nanook]

dt=sqrt(d1^2+d2^2+...+dn^2).. that's the effective diameter .
calculate the effective diameter
then calculate the total area and put her into Helmholtz equation.area

[Svante]

The simplified version is this: The (individual) port length is proportional to the sum of all cross-sectional areas.

In real life there is a bit of a difference due to the co-oscillating air outside and inside the tube (which makes the tube appear a bit longer). That makes an exact calculation of the port length a bit more difficult.

But roughly: add a second port - double their lengths.

[Nanook]

as an example , use 2 ports of 2" diameter..

dt=sqrt(2^2 + 2^2)
=sqrt(8)=2*(sqrt(2))=2*1.414=2.828

--->r=1.414

Area,S, is pi * r^2=2*pi=6.2832
so the area doubles, therefore the length needs to be expanded.

The Helmholtz equation is as follows

f=(c/2*pi)*sqrt(S/VL)

so if f is constant, V is constant (the enclosure), c is a constant as is 2*pi

f is proportional to sqrt(S/L), but 2 ports of 2" diameter does not = the same as 1 port of 4", it =2.824 . (which is 2*sqrt(2)), so the length must be multiplied by sqrt(2) or 1.414

now take it to 3" port.. L would have to be multiplied by sqrt(3) and so on...

[mashaffer]

OK. I think I have it. For a design with 1 4" port of 11.53" length the Mach # came out to .17. To reduce this I add one more 4" port. This gives a port length of 25.98 for each of the two ports and reduces Mach to .09. Does that sound about right.

mike

[Svante]

Originally posted by Nanook

The Helmholtz equation is as follows

f=(c/2*pi)*sqrt(S/VL)

...isn't the answer right there? Double S (by using two ports), then L also has do double if everything else is to be constant, right?

There is no need to involve equivalent diameters, unless the software requires it (if it does, the effective diameter is multiplied by sqrt(2) as you say).

[Nanook]

in this case yes, using 2 ports requires a doubling of the port length.


but what if you want a single 5" or 6" (unusual but not out of the question). Or a rectangular port? The answer provided is the most generalized one, else if others read into the explaination, they may not see the important relation and could make assumptions.

As the most generalized solution, it can be used in any case, not just simply putting in a second port, perhaps maybe 5 ports or some other geometry. Why not an oval?

It's like algerbra or calculus. Solve the equation in most general terms first, then find a specific solution based on provided information, boundary conditions, etc.

[Svante]

Originally posted by Nanook
in this case yes, using 2 ports requires a doubling of the port length.


but what if you want a single 5" or 6" (unusual but not out of the question). Or a rectangular port? The answer provided is the most generalized one, else if others read into the explaination, they may not see the important relation and could make assumptions.

As the most generalized solution, it can be used in any case, not just simply putting in a second port, perhaps maybe 5 ports or some other geometry. Why not an oval?

It's like algerbra or calculus. Solve the equation in most general terms first, then find a specific solution based on provided information, boundary conditions, etc.

Hmm, I don't see the problem... S/L is to be kept constant. Calculate the total new area, regardless of shape (almost), and change L by the same factor. I don't see how yuor formula could be used for rectangular or oval ports either.

[Nanook]

it is the Helmholtz equation.

I agree for the same tuning S/L is to be kept constant. How long should a 4" X 5" port be(rectangular)? Or a 7" diameter port. you get the idea, I hope. Understanding the whole equation and understanding where the tuning comes from is important as a whole. What if you change the Volume of the enclosure..?


Limiting the possibilites to only doubling the number of ports to attain the correct tuning and mach number ..is , well ...limiting.

In the example I am using ..a 4" X 5" rectangular port, the length needs to be increased by a factor of 1.592.

[Svante]

Originally posted by Nanook
In the example I am using ..a 4" X 5" rectangular port, the length needs to be increased by a factor of 1.592.

Hmm, S1/S2=4*5/(2*2*pi)=1.592, yes!