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-   -   How to choose or find Permeability for a Porous screen model using darcy's law.... (http://www.cfd-online.com/Forums/cfx/92810-how-choose-find-permeability-porous-screen-model-using-darcys-law.html)

arunraj September 26, 2011 05:25

How to choose or find Permeability for a Porous screen model using darcy's law....
 
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 sub-domain in-order 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 m-2 ( I GUESS IT'S A WRONG VALUE)
RESISTANCE loss COEFFICIENT : 81.6 m-1

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*(1-porosity^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 m-1 = 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*(0.406^2-0.2^2))/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

-25166-25167=(2.6434*10^-4)/(K_Permeability)

k_permeability = - 5.25*10^-9 m-2

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....

ghorrocks September 26, 2011 06:17

What pressure loss does it give you when you run it?

arunraj September 26, 2011 07:22

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..

arunraj September 27, 2011 00:02

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 m-1 = 32.85/width of screen
width of screen=0.005 m
so loss coeff in m-1 becomes 32.85/0.005=6570.44

2. Loss coeff.=0.52*(1-porosity^2)
Porosity

32.85= 0.52*(1-porosity^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*(0.406^2-0.2^2)/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..

arunraj September 27, 2011 00:20

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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