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

LinkBack  Thread Tools  Display Modes 
September 26, 2011, 05:25 
How to choose or find Permeability for a Porous screen model using darcy's law....

#1 
Member
Arun raj.S
Join Date: Jul 2011
Posts: 53
Rep Power: 6 
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, 06:17 

#2 
Super Moderator
Glenn Horrocks
Join Date: Mar 2009
Location: Sydney, Australia
Posts: 11,050
Rep Power: 86 
What pressure loss does it give you when you run it?


September 26, 2011, 07:22 

#3 
Member
Arun raj.S
Join Date: Jul 2011
Posts: 53
Rep Power: 6 
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, 00:02 

#4 
Member
Arun raj.S
Join Date: Jul 2011
Posts: 53
Rep Power: 6 
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, 00:20 

#5 
Member
Arun raj.S
Join Date: Jul 2011
Posts: 53
Rep Power: 6 
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?


Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Modelling Combustion in Porous Zone  tanjinjack  FLUENT  1  August 31, 2015 04:56 
Errors running allwmake in OpenFOAM141dev with WM_COMPILE_OPTION%3ddebug  unoder  OpenFOAM Installation  11  January 30, 2008 21:30 
Permeability for porous zones  A8anato_psofimi  FLUENT  1  February 17, 2004 14:25 
Please help a newbie find the drag on a 3d model  David Amer  Main CFD Forum  1  March 6, 2002 04:33 
Darcy's Law flow into and out of a box  Paul Missel  Main CFD Forum  6  July 20, 2001 05:26 