The model is the same as Exercise 8: A heat source is placed on a substrate, and there is a forced air flow for cooling in parallel to the substrate. The heat radiation is analyzed under the transient condition.
The heat source is temperaturedependent.
The heat transfer coefficient is acquired manually.
To acquire it automatically, see “Ex.1 of Simple FluidThermal Analysis”.
The temperature distribution and the heat flux vectors are solved.
Unless specified in the list below, the default conditions will be applied.
Item 
Settings 
Analysis Space 
3D 
Model unit 
mm 
Item 
Settings 
Solvers 
Thermal Analysis [Watt] 
Analysis Type 
Transient analysis 
Options 
N/A 
The transient analysis is set up in transient tab as follows. The total number of steps is 20. The time step is 30 second.
Therefore, the temperature distributions for 600 seconds are solved.
Tab 
Setting Item 
Settings 

Transient analysis 
Table 


Initial Temperature 
25[deg] 
The same as Exercise 7. The material properties and the boundary conditions are the same as well.
The substrate (VOL1) and the heat source (VOL2) are created as solid body box, and the heat source is defined in the body attribute of VOL2.
The heat transfer coefficients for the top and bottom faces of the substrate and the top face of the heat source are calculated based on the simplified equation.
Body Number/Type 
Body Attribute Name 
Material Name 
0/Solid 
VOL1 
006_Glass_epoxy * 
1/Solid 
VOL2 
001_Alumina * 
* Available from the Material DB
The heat source of VOL2 is set up as follows.
Body Attribute Name 
Tab 
Settings 
VOL2 
Heat Source 
Heat density Temperature Dependency: Yes 
Heat source is specified by heat density in the analyses where temperaturedependent heating materials are involved.
On Exercise 8, we set 1[w] for the heat source of VOL2. As the volume of VOL2 is 800*10^9[m^3], the heat density is equal to
1.25 x 10^6 [W/m^3].
This is the value for 25[deg], so
P(25) = 1.25 x 10^6 [W/m^3]
in the Arrhenius equation which is given by the following:
P(T) = 1.25 x 10^6 * exp(0.15/(k*(T+273))) / exp (0.15/(k*(25+273)))
where 0.15[eV] is the activation energy and k is the Boltzmann constant.
Body Attribute Name 
Item 
Settings 
VOL2 
Nonlinearity Table 
Select “Smooth interpolation” 
Temperature 
Heat Source 
Temperature 
Heat Source 
Temperature 
Heat Source 
25 
1.25 
155 
7.365383041 
605 
59.18105993 
35 
1.51094343 
205 
11.26925562 
655 
65.85048973 
45 
1.804711948 
255 
15.90794507 
705 
72.47585198 
55 
2.132370783 
305 
21.15582761 
755 
79.02738627 
65 
2.494768691 
355 
26.88623825 
805 
85.48214754 
75 
2.892545312 
405 
32.98193112 
855 
91.82276684 
85 
3.326141169 
455 
39.339762 
905 
98.03640656 
95 
3.795809582 
505 
45.87196227 
955 
104.1138907 
105 
4.301629871 
555 
52.5055812 
1005 
110.0489868 
Heat density’s temperature plot is shown below.
The heat transfer coefficients for the forced convection are calculated as follows. The equation is given in Exercise 8: Heat Radiation by Forced Convection (Transient Analysis).
For the details, please refer to the Heat Transfer Coefficient for Forced Convection
To acquire it automatically, see “Ex.1 of Simple FluidThermal Analysis”.
Boundary Condition Name/Topology 
Tab 
Boundary Condition Type 
Settings 
BC1/Face 
Thermal 
Heat Transfer/Ambient Radiation 
Heat transfer coefficient: 17.26[W/m2/deg] Room Temperature : 25[deg] 
BC2/Face 
Thermal 
Heat Transfer/Ambient Radiation 
Heat transfer coefficient: 27.3[W/m2/deg] Room Temperature : 25[deg] 
The temperature distributions for Exercise 8 (right figures) and this exercise (left figures) are shown below for the elapsed time of 60, 300 and 600 seconds.
The unit of the color scale is [deg].
At Minimum/Maximum Value on the Contour tab of [Graphics Setup], deselect “Automatic” and set 25 => 150.
There is almost no difference in 60 seconds.
However, it becomes obvious that the temperature increases much higher than Exercise 8 in 300 and 450 seconds.
The temperature vs. time is plotted below.