# Heat Flux in thermal simulation

Dear friends,

I want to apply heat flux inthermal simulation using LBM…

I’ve used this formulae:

T(i,1)= (q"/k)2/3deltax+4/3T(i,2)-1/3T(i,3)

in which i is the directions and T(i,1) is the directional temperatue on wall, T(i,2)&T(i,3) are the directional temperatures on one node and two nodes above the that respectively. I distribute the heat flux along the directions based on their weights:

for example:

T(0,1)= (4/9q"/k)2/3deltax+4/3T(0,2)-1/3T(0,3)
T(1,1)= (1/9
q"/k)2/3deltax+4/3T(1,2)-1/3T(1,3)
T(2,1)= (1/9q"/k)2/3deltax+4/3T(2,2)-1/3T(2,3)
T(4,1)= (1/9
q"/k)2/3deltax+4/3T(4,2)-1/3T(4,3)
T(5,1)= (1/36q"/k)2/3deltax+4/3T(5,2)-1/3T(5,3)
T(7,1)= (1/36
q"/k)2/3deltax+4/3T(7,2)-1/3T(7,3)

However, when I want to calculate the local Nusselt number which calculates through the following formulae:

Nu=(T-Tref)/(q"*l/k)

the Nu number is obtained high number which does not corroborate the classical CFD…

is this algorithm correct!!!
I’d really appreciate if you can give me a good refrence or article pertaining to modeling heat flux using LBM…

Or
help me…

cheers

Dear nanolb,

Probably I’m not right, but when you want to to impose the Neumann condition on the wall, the following should be satisfied:
dT/dx=q’’, but if T(i,1) is the temperature on the wall then through the Taylor expansion the coefficients are different:
(T(i,1)-4/3T(i,2)+1/3T(i,3))/(1/3 deltax)=q’’ - that means you need to change all the coefficients with 2/3 to 1/3. Also, as far as the temperature is the summation of T_i - you need to interpolate the 3,6,8th populations.

Hopefully it will help,
Alex

Dear Alex.

SIMULATION OF NATURAL CONVECTION IN A SQUARE
CAVITY BY TAYLOR SERIES EXPANSION- AND LEAST
SQUARES-BASED LATTICE BOLTZMANN METHOD

C. SHU, Y. PENG and Y. T. CHEW International Journal of Modern Physics C, Vol. 13, No. 10 (2002) 1399{1414

Based on this article, the coefficient I’ve mentioned before is correct. However, rather than distributing the heat flux between the all directions(0,1,2,3,4,5,6,7,8) according to their weight, I’ve distributed this flux onlu on the direction which cause the heat flux: for example at bottom wall:

on the directions(2,5,6) Or on the directions (4,7,8) like this:

Do j=33,96
g( 0,j,1) = (4.0d0g( 0,j,2)-g( 0,j,3))/3.0d0
g( 1,j,1) = (4.0d0
g( 1,j,2)-g( 1,j,3))/3.0d0
g( 2,j,1) = (-8.0d0/6.0d0+4.0d0g( 2,j,2)-g( 2,j,3))/3.0d0
g( 3,j,1) = (4.0d0
g( 3,j,2)-g( 3,j,3))/3.0d0
g( 4,j,1) = (4.0d0g( 4,j,2)-g( 4,j,3))/3.0d0
g( 5,j,1) = (-2.0d0/6.0d0+4.0d0
g( 5,j,2)-g( 5,j,3))/3.0d0
g( 6,j,1) = (-2.0d0/6.0d0+4.0d0g( 6,j,2)-g( 6,j,3))/3.0d0
g( 7,j,1) = (4.0d0
g( 7,j,2)-g( 7,j,3))/3.0d0
g( 8,j,1) = (4.0d0*g( 8,j,2)-g( 8,j,3))/3.0d0

`````` End Do
``````

this change leads me to the fairly Nu number; however, I cannot get the good shape for streamlines. I am uncertain as to whether if I iterate more, it will be corrected. Or I should change the algorithm again…

I’d really appreciate if you share with your idea…

cheers

Dear nanolb,

I’ve never done it. But if I were you I would do the following:

1. The boundary condition should state that that dT/dx=q’’ on the wall. In terms of Taylor expansion if the T(i,1) is the temperature on the wall (in other words “1” is the superscript of the node located on the wall), one can obtain the following:
dT/dx_{wall}=(T(i,1)-4/3T(i,2)+1/3T(i,3))/(1/3 dx)=q’’ (Please check the Taylor expansion).

Therefore the boundary condition becomes for the temperature as follows:
T(i,1)= q’’ 1/3 dx+4/3 T(i,2) -1/3 T(i,3) (Note the coefficient is different)

The temperature is by definition is T(i,1)=g(0,i,1)+g(1,i,1)+g(2,i,1)+g(3,i,1)+g(4,i,1)+g(5,i,1)+g(6,i,1)+g(7,i,1)+g(8,i,1). The question is how to distribute the boundary condition criteria among the populations. My guess is according to the weights, i.e.

g(0,i,1)=4/9 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(1,i,1)=1/9 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(2,i,1)=1/9 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(3,i,1)=1/9 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(4,i,1)=1/9 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(5,i,1)=1/36 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(6,i,1)=1/36 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(7,i,1)=1/36 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))
g(8,i,1)=1/3 *(q’’ 1/3 dx +4/3 T(i,2)-1/3 T(i,3))

You can distribute the temperature among the populations coming inside the domain. Please ensure that macroscopic law is fulfilled T(i,1)= q’’ 1/3 dx+4/3 T(i,2) -1/3 T(i,3)

1. For the cavity we did a few simulations time ago and you need to take around 200000 steps for the system to be in steady-state condition.

Hopefully it will help,
Alex