CFD Online Logo CFD Online URL
www.cfd-online.com
[Sponsors]
Home > Forums > Software User Forums > OpenFOAM

buoyantBoussinesqSimpleFoam

Register Blogs Community New Posts Updated Threads Search

Reply
 
LinkBack Thread Tools Search this Thread Display Modes
Old   December 2, 2011, 09:41
Default buoyantBoussinesqSimpleFoam
  #1
Member
 
Join Date: Jun 2011
Posts: 42
Rep Power: 14
mikeP is on a distinguished road
In the tutorials of buoyantBoussinesqSimpleFoam, Laminar Prandtl number is defined as 0.9 and turbulent Prandtl number is defined as 0.7.

Why is the difference? Shouldn't laminar Prandtl number for air be something around 0.7?

In Fluent, default settings for air are:
Cp: 1006.43
mu: 1.79e-5
k = 0.0242

which give Pr = 0.744
mikeP is offline   Reply With Quote

Old   December 2, 2011, 10:35
Default
  #2
Senior Member
 
Bernhard
Join Date: Sep 2009
Location: Delft
Posts: 790
Rep Power: 21
Bernhard is on a distinguished road
Does is really matter for a tutorial case?
Bernhard is offline   Reply With Quote

Old   December 8, 2011, 06:23
Default
  #3
Member
 
Join Date: Jun 2011
Posts: 42
Rep Power: 14
mikeP is on a distinguished road
well not for the tutorial case, no.

I modified buoyantBoussinesqSimpleFOAM and added a heat source term into the equation.

my TEqn:

fvScalarMatrix TEqn
(
fvm::div(phi, T)
- fvm::Sp(fvc::div(phi), T)
- fvm::laplacian(kappaEff, T)
==
heatSource // heatSource is Q/(Cp*rho) - in units [K/s] where Q is [W/m3]
);

now I use a heatSource file in my 0/ directory and using setFields modify the internal values where I want my heat sources. However, the value of my heatSource input is Q/(Cp*rho). The value of heat capacity is defined in Prandtl number. That is why I asked.

Right now I assume rho = 1.225 and Cp = 1006.43 but my temperature results are not always the same as I get in Fluent.

If I define a boussinesq force - they are quite different
If I turn the boussinesq force off, then temperature results are similar
why would that be the case?
mikeP is offline   Reply With Quote

Old   December 9, 2011, 11:56
Default
  #4
Member
 
Join Date: Jun 2011
Posts: 42
Rep Power: 14
mikeP is on a distinguished road
The thing is I create a simple room with a size x,y,z= 1x1x1m and put a heatSource with a value of 1 at the cells where x=[0.2,0.4], y=[0.4,0.6], z=[0.4,0.6].

There is a flow from left to right with a velocity of (0.1,0,0) at the inlet.

so, the volumetric flow rate = 0.1 m3/s

The temperature increase at the outlet is less than I expect.

Using Q=m*c*deltaT, I expect a temperature increase of 0.08 K but I get 0.037 K (those are outlet average T - inlet average T)

There must be something wrong with my heatSource input, I guess?

Here is the domain and heatSource input: http://i.imgur.com/qDrBd.png

Last edited by mikeP; December 9, 2011 at 12:00. Reason: I have uploaded the image showing the heat source
mikeP is offline   Reply With Quote

Old   July 27, 2013, 09:50
Default Incident Solar Radiation
  #5
New Member
 
Alex Lee
Join Date: Sep 2012
Posts: 15
Rep Power: 13
Alex Lee is on a distinguished road
Hi, I am working on a similar problem and would like to seek your help.

We are able to map the solar incident radiation (G, w/m^2) on the floor and wall of a air tight room. We are looking for way to use these radiative flux as BCs to drive the flow field inside the room.

We are wondering the use of bouyantBoussinesaRadiationFoam allow the use of G as wall heat flux?

Thanks for your help.

Alex
Alex Lee is offline   Reply With Quote

Reply


Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are On
Refbacks are On



All times are GMT -4. The time now is 19:09.