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Unread 10-15-2002, 11:44 AM   #6
utabintarbo
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Join Date: Jul 2002
Location: Sterling Hts., MI
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Quote:
Originally posted by LiquidRulez
can you tell me how you deterrmined that for future reference??
thanks again
area of channel = x * .125 (1/8")
area of inlet = pi * r^2 = 3.1416 * (.1875^2)

It's all algebra from here:

x = (pi * (.1875^2))/(2 * .125) = .11045/.25 = .4418

Hope that helps

Bob
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