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 vibhs August 21, 2012 00:34

hydraulic diameter

With reference to Fluent tutorial no. 1: Fluid flow and heat transfer in mixing elbow, could someone plz expalin me how the hydraulic diameter of an elbow in 2-D is 32", which has its width to be 16"?

Hydraulic diameter = 4A/P
A= pi*r^2
P= 2*pi*r

 vibhs August 21, 2012 05:36

Quote:
 Originally Posted by eng_s_sadeghi (Post 377917) Hydraulic diameter = 4A/P A= pi*r^2 P= 2*pi*r
Its a 2-d problem. i dont have any "r" value. I just have one dimension of the elbow i.e. 16".

In this case, width of the channel is your hydraulic diameter.

 vibhs August 21, 2012 05:46

Quote:
 Originally Posted by eng_s_sadeghi (Post 377946) In this case, width of the channel is your hydraulic diameter.
hydaulic diameter taken is 32". You can cross check in the first tutorial of the fluent...:confused:...I am confused