[L78toLT1]

Originally Posted by Norm Peterson

I think your 3.5" wide rectangle would only be 0.82" high; a 3" wide rectangle would be 3 x 1.32.

You should probably be thinking a bit deeper than simply outside cross-sectional area (you did consider that inside areas are not based directly on the OD measurement?). But even running with outside 2.75" OD and 8.64" circumference dimensions, a 3 x 1.32 rectangle provides (3 + 3 + 1.32 + 1.32) = 3.96 in^2, or about 85% of the 2.16 x 2.16 square.

I don't blame you for not wading through the round to flat oval math, but I think that the inside area reductions don't happen as rapidly (I think the area loss going from a 2.75" circle down to a 2.2.5" x 3.75" flat oval runs a lot closer to 5%). Feel free to check this.


Beyond that, flow restriction is also a function of how long the restriction is, relative to the unrestricted length. This is at least part of the reason you can get away with dimpling header primary tubes for clearance without losing power until you've really choked the inside area down.

Norm

Thanks, Norm for correcting my math! And yes, I understand that flow restriction is a function of several variables, including the geometry of the transition, constricted length, and undoubtedly several others that I don’t recall as it’s been forty years since I was in Je Han’s Fluids class. A few of the preceding posts, if I’m interpreting them correctly, indicate that some folks are under the impression that crimping, squashing, or otherwise deforming a tube has no effect on the X-section. My post was a poorly constructed attempt to show that deforming a tube does affect the area, and the more you crush it the worse it gets.