[vibhs]

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"?

[eng_s_sadeghi]

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

[vibhs]

Originally Posted by eng_s_sadeghi

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".

[eng_s_sadeghi]

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

[vibhs]

Originally Posted by eng_s_sadeghi

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

[eng_s_sadeghi]

Is it an axisymmetic 2D problem?
If yes, It's correct.

[vibhs]

Originally Posted by eng_s_sadeghi

Is it an axisymmetic 2D problem?
If yes, It's correct.

How come?...Could you please explain?