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August 16, 2007, 05:58 
Hydraulic Diameter  Series of Holes

#1 
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Hello there,
I am simulating a case where I defined a series of holes which lie around a tube as one inlet area. Now I am wondering which hydraulic diameter I have to enter as a boundary condition. Do I have to enter the hydraulic diameter of one hole or the hydraulic diameter of all holes combined? Or maybe this is just the same, right? Hmm, maybe this is the same, but would be nice, if someone could verify for me. Greetigns, Jop. 

August 16, 2007, 08:01 
Re: Hydraulic Diameter  Series of Holes

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total inlet area / total wetted perimeter


August 16, 2007, 08:29 
Re: Hydraulic Diameter  Series of Holes

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Thx Joe. Another problem is, that I am simulating a rototional periodic system. So I have to use the total inlet area / total wetted perimeter which are actually in the computational area?


December 13, 2018, 04:56 

#4  
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Mustapha Mukhtar
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Quote:
I am simulating also a quencher with 192 hole (100 micrometer of diameter each), so I defined them all as one outlet (I won't do it individually, otherwise I will get old), so I foundthe same problem as you, how should I define the hydraulic diameter in the boundary conditions? and second quetion is about the mass fow rate. The mass flow rate in my case (whcih I got from experimental data) is not messured in the holes, but from a steam supply, and is set to 0.0219 Kg/s . the question is should I set this value of should I distribute it to the 192 hole? I hope it's clear what I just posted... 

December 13, 2018, 08:50 

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Lucky
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Quote:
You give the total mass flow rate of all 192 holes. But the hydraulic diameter should be the 100 micrometer. 

January 14, 2019, 07:54 

#6 
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Mustapha Mukhtar
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Thank you for your answer. Could you give me a reference for your reply?


January 14, 2019, 10:41 

#7  
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Lucky
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Quote:
No need a reference for this, it's how the settings work. Just read the Fluent manual. You are applying the total mass flow rate on the surface. In this case, your surface happens to have 192 holes but Fluent doesn't know that it is 192 holes, it's just a 1 surface because you defined it to be 1 surface. So therefore you apply the total mass flow rate of this outlet surface. The hydraulic diameter is the same whether you use 1 hole or 100 holes, it always comes out to 100 micrometer. If you don't get 100 micrometers, then you made a mistake somewhere punching buttons into your calculator. So why am I 100% sure you use 100 micrometer and not some bigger number? At boundaries you need to specify boundary conditions for the flow and the turbulent transport variables (k and epsilon, k & omega). For people that have no clue what they are doing, there are cute formulas that estimates the k and epsilon or k and omega from the turbulence intensity and turbulent length scale. For people that are too lazy or have no clue how to estimate a turbulent length scale, there is a hydraulic diameter option.It is important to point out that these formulas are based on centerline fully developed pipe flows and are incorrect for the vast majority of cases. But due to statistical dispersion, it's usually not that important as long as you are in the right ball park with your length scale. So if you use a number bigger than 100 microns, you will end up with properties for an eddy that is bigger than any of your holes. Btw, at outlets if there is no reversed flow, you can put in anything for the hydraulic diameter and it will not matter. The turbulence BC's are only needed at inlets (or outlets when there is reversed flow). 

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