ah, something to work with...
So I multiply .0625 X 25" and I get 1.5625 square inches cross sectional flow area. Times 4=6.25
divided by the wetted perimeter of 50.125" and I get a hydraulic diameter of .0312
How do I use that figure relative to to my problem? (Layman)
sush54
Thanks quark;
I think I wasn't exactly clear in describing the situation. The cross sectional configuration is constant throughout. 1/16" X 25" We have a very tight size restriction in one plane so we have to build our own flat "pipe" which will probably wind up having that...
Gentlemen:
I'm more than a little out of my field here so please bear with me. We have a proprietary job that requires some unusal fixes.
By way of example, let's assume we have a 5' high vertical pipe with inner dimensions of 1-1/4" X 1-1/4" (1.5625" square inch cross...
