You could try putting the port on the bottom of the cabinet. Feet 1 inch (2.5 cm) high are enough to clear a 4 inch (10 cm) duct. A spot of number crunching:
The area of a duct is πr^2 (pi x radius squared). We want to find how far the opening needs to be from a wall to extend the duct. We know the circumference is 2πr so the area would be 2πrd, where d is the distance from the wall. Equating the two 2πrh = πr^2 and canceling the redundant πr terms yields h = r/2, so for a 4 inch duct the opening needs to be an inch from the wall/floor to lengthen the duct appreciably. Rounding over the duct on the bottom of the cabinet increases the circumference so the duct/floor interface would act as an extension of the flare.
The area of a duct is πr^2 (pi x radius squared). We want to find how far the opening needs to be from a wall to extend the duct. We know the circumference is 2πr so the area would be 2πrd, where d is the distance from the wall. Equating the two 2πrh = πr^2 and canceling the redundant πr terms yields h = r/2, so for a 4 inch duct the opening needs to be an inch from the wall/floor to lengthen the duct appreciably. Rounding over the duct on the bottom of the cabinet increases the circumference so the duct/floor interface would act as an extension of the flare.
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The second link to curved pipes wants a hard look: the dynamic viscosity of air is about 18 x 10^-6 kg-s/m^2 [1][2][3], not 1002 x 10^-6, so the Reynolds number calculation [4] is quite wrong. If we use the correct value and solve for fluid velocity at a Reynolds number of 2300 for laminar flow, the transition from laminar to turbulent flow happens at around 0.45 m/s. As a giggle test, that's about the originally posited 25 m/s multiplied by 18/1002, the ratio of correct to incorrect viscosity values.
In other words, unless you're listening to Ulm Stelmgest And Her Happy Dulcimer at very discreet levels, your ducts will be turbulent, no getting around it.
As for the rest, the derivations are... obscure. I'll look at them later, possibly considerably later given how wrong that one was. One thing I noticed in the third paper was a reference to a "Higher Mode Cut On Frequency", variously represented as K or Kc. It's supposedly 0.5, but where did this number come from, and how does it enter the room apart from being a fudge factor?
[1] https://theengineeringmindset.com/properties-of-air-at-atmospheric-pressure/
[2] https://resources.system-analysis.c...he-kinematic-viscosity-of-air-and-temperature
[3] https://www.engineeringtoolbox.com/air-absolute-kinematic-viscosity-d_601.html
[4] (fluid velocity x pipe inner diameter) / (kinematic viscosity) and kinematic viscosity = dynamic viscosity / density, from
https://www.pipeflow.com/pipe-pressure-drop-calculations/reynolds-numbers
https://en.wikipedia.org/wiki/Reynolds_number
In other words, unless you're listening to Ulm Stelmgest And Her Happy Dulcimer at very discreet levels, your ducts will be turbulent, no getting around it.
As for the rest, the derivations are... obscure. I'll look at them later, possibly considerably later given how wrong that one was. One thing I noticed in the third paper was a reference to a "Higher Mode Cut On Frequency", variously represented as K or Kc. It's supposedly 0.5, but where did this number come from, and how does it enter the room apart from being a fudge factor?
[1] https://theengineeringmindset.com/properties-of-air-at-atmospheric-pressure/
[2] https://resources.system-analysis.c...he-kinematic-viscosity-of-air-and-temperature
[3] https://www.engineeringtoolbox.com/air-absolute-kinematic-viscosity-d_601.html
[4] (fluid velocity x pipe inner diameter) / (kinematic viscosity) and kinematic viscosity = dynamic viscosity / density, from
https://www.pipeflow.com/pipe-pressure-drop-calculations/reynolds-numbers
https://en.wikipedia.org/wiki/Reynolds_number
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You could try putting the port on the bottom of the cabinet. Feet 1 inch (2.5 cm) high are enough to clear a 4 inch (10 cm) duct. A spot of number crunching:
The area of a duct is πr^2 (pi x radius squared). We want to find how far the opening needs to be from a wall to extend the duct. We know the circumference is 2πr so the area would be 2πrd, where d is the distance from the wall. Equating the two 2πrh = πr^2 and canceling the redundant πr terms yields h = r/2, so for a 4 inch duct the opening needs to be an inch from the wall/floor to lengthen the duct appreciably. Rounding over the duct on the bottom of the cabinet increases the circumference so the duct/floor interface would act as an extension of the flare.
One more thing: for flared ducts the calculation should use the radius at the outside edge of the flare, not the radius of the duct itself, for a smooth transition between duct and surface.
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