Need Round Port Displacement Assistance

[Slammed 5,000+ posts]

Port area for a circular port = (1/2 the outside diameter X 2) X 3.14 X length of

the port divided by 1728?

example:

port tube width = 2.5"

port tube length = 8.5"

(outside diameter of the 2.5" tube is roughly 3")

1.5 x 2 x 3.14 x 8.5 / 1728 = 0.046

does it make sense?

(I **** at math lol)

how many cu. ft. does the port displace?

thanks //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

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[blackbonnie 5,000+ posts]

(outside diameter/2)^2 * 3.14 *(length of port inside said enclosure)= in^3

assuming it doesnt have flanged ends, but don’t worry about the internal flanges of an aero as they are minuscule in the whole scheme of things

answer = .0347 ft^3

hope that helps

 

[Slammed 5,000+ posts]

thanks man //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

 

[blackbonnie 5,000+ posts]

yes sir, help one help all