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How to choose or find Permeability for a Porous screen model using darcy's law.... 

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September 26, 2011, 06:25 
How to choose or find Permeability for a Porous screen model using darcy's law....

#1 
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Arun raj.S
Join Date: Jul 2011
Posts: 191
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Hi all,
I m trying to simulate Inlet flow distortion on axial compressor. So for I have created the screens which u can see in the attachments(samples not original). I have created 1* 90 degree circumferential screens. This screen i have created it as a subdomain inorder to place in front of rotor. The boundary conditions to the inlet of the screen is mass flow rate= 2.7 kg/s. And the rotor exit boundary condition I have given as Pressure = 101667. The design pressure rise of the rotor is 1000 pa. The design mass flowa rate is 3.5 kg/s. Hub dia = 0.2 m , shroud dia = 0.406 m. width of the screen is 0.005 m. TURULENCE MODEL FOR ROTOR AND SCREEN:SST In Sub domain: Basic Settings: step1 : First I have given the location step 2: I have set the coordinate frame Sources: step1: tick mark the sources step2: LOSS model: Directional loss step 3: for Loss option I have specified Steam wise Direction Options : Cartesian components x component: 1 y component: 1.05 z component:1.05 Stream wise loss options: Permeability and Resistance loss co eff. Permeability : 5.25*10^9 m2 ( I GUESS IT'S A WRONG VALUE) RESISTANCE loss COEFFICIENT : 81.6 m1 Transverse loss: Stream wise coefficient Multiplier : 70 My questions are: 1. I haven't tick mark the General Momentum Source option Under this option I haven't specify any values Is that ok for this screen model? 2. For calculating Resistance Loss coefficient and Permeability , I have followed the approach step 1: Loss coeff.= 0.52*(1porosity^2) porosity ((( empirical relation got from journals on screens))) Porosity i have chosen as 0.7 then when i calculated loss coefficient, it comes as 0.408 Since the unit of the loss coefficient for loss model tab in ansys cfx is m^1 For entering the value of loss coefficient in that option , I have divided it with width of the screen which is 0.005 m so finally loss coeff in m1 = 81.6 m^1 step 2: calculation of del_p Here rho has been assumed as 1.225 normal velocity for my case is calculated by m_dot= rho*a*U_normal here the area i have taken is annulus area u_normal=2.7/(1.225*pi*(0.406^20.2^2))/4 U_normal= 22.44 m/s For calculating del_p I have used an loss coeff = del_p/(0.5*rho*U_normal ^2) So del_p I got as del_p = (0.408*0.5*1.225*22.44^2) del_p = 125.83 pa stpe 3: calculation of k_permeability then using Darcy's law i have calculated dp/dx = ((mu*u_normal)/K_permeability)+(K_loss coeff.* rho*u*u)/2) LHS =125.83/0.005 RHS= ((1.178*10^5*22.44)/k_permeabiltiy)+(81.6*1.225*22.44^2)/2) when i calculated from this formula 2516625167=(2.6434*10^4)/(K_Permeability) k_permeability =  5.25*10^9 m2 I think this value is wrong .. I don't know where I am going wrong ..Plz help me... All ur suggestions i much needed here ... Hi Glenn Horrocks, I think you can help me well in this issue..PLEASE HELP ME IN SORTING OUT I think there will be a shortcut for giving these values in cfx. But i Don wat to do... Also Please explain where i m going wrong.... 

September 26, 2011, 07:17 

#2 
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Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
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What pressure loss does it give you when you run it?


September 26, 2011, 08:22 

#3 
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Arun raj.S
Join Date: Jul 2011
Posts: 191
Rep Power: 14 
Hi Glenn,
Tomo only i have to start my run..I want to know the procedure to find the permeability and loss coefficient for screen model.. If i find FROM THIS LAW then i m getting permeability as a negative value..It can't be a negative value rite.. 

September 27, 2011, 01:02 

#4 
Senior Member
Arun raj.S
Join Date: Jul 2011
Posts: 191
Rep Power: 14 
Hi Glenn Horrocks,
When I consult with one of my staff who s a experimental fluid mechanics specialist. He told permeability coefficient and porosity are specifying the same things only.Is that rite?..Now i change my calculated like dis.. Also I have to produce del_p=10132.5 pa.so my calculation becomes like dis 1. loss coeff = del_p/(0.5*rho*U_normal ^2) Loss coeff.= 10132.5/(0.5*1.225*22.44^2) =32.85 Loss coeff in m1 = 32.85/width of screen width of screen=0.005 m so loss coeff in m1 becomes 32.85/0.005=6570.44 2. Loss coeff.=0.52*(1porosity^2) Porosity 32.85= 0.52*(1porosity^2) Porosity (from a journal named Upstream Influence of a Porous Screen on the Flow Field of a Free Jet) by solving this eqn this i got porosity as 0.000079143 3. From this i have calculated peeability as permeabiltiy coeff= area which the flow is allowed in the porous medium total cross section area Total cross sectional area= pi/4*(0.406^20.2^2)/4 = 0.02455368 m^2 here one of the four is due to 90 degree extension of the screen so area which the flow is allowed in porous medium which is permeability in m^2 = 0.000079143*0.02455368 = 0.000001943251956 Now this concept is correct rite? Also why the porosity is very less ..Bcos of the pressure loss is very high only rite? .. if i am doing any mistake plz tel me.. 

September 27, 2011, 01:20 

#5 
Senior Member
Arun raj.S
Join Date: Jul 2011
Posts: 191
Rep Power: 14 
Also why the unit of the permeability and loss coeff.. Is in m^2 and m^1...it shoud be some other thing rite..how can coeff be in m^1 or m^2...then wat they have mentioned is wrong in the software itself?


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