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Old   August 21, 2012, 00:34
Default hydraulic diameter
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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"?
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Old   August 21, 2012, 03:35
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Hydraulic diameter = 4A/P
A= pi*r^2
P= 2*pi*r
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Old   August 21, 2012, 05:36
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Quote:
Originally Posted by eng_s_sadeghi View Post
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".
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Old   August 21, 2012, 05:40
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In this case, width of the channel is your hydraulic diameter.
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Old   August 21, 2012, 05:46
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Quote:
Originally Posted by eng_s_sadeghi View Post
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......I am confused
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Old   August 21, 2012, 05:54
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Is it an axisymmetic 2D problem?
If yes, It's correct.
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Old   August 21, 2012, 06:31
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Quote:
Originally Posted by eng_s_sadeghi View Post
Is it an axisymmetic 2D problem?
If yes, It's correct.
How come?...Could you please explain?
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