[akiaqi]

hi, I am trying to simulate a case where an XPS slab is receiving solar radiation and losing heat to the external environment (25oC) by radiation and convection and to the room through conduction and convection (25oC). I thought it would be pretty simple but turned out something I need help. The temperature is crazy high (249oC). Both the radiation flux and the convection flux is about 100 W, which is a lot less than my calculation based on the temperature computed.
The solar radiation is set as an internal source of the boundary cells at the edge of the wall. From the result file, the 200 W solar radiation was correctly executed. The boundary source was set as instructed by the Immersol instruction:
PATCH(IMSW001,WWALL,1,1,1,1,1,1,1,1)
COVAL(IMSW001,T3,GRND4,GRND5)
COVAL(IMSW001,TEM1,wallco,GRND5)

The default values and coefficients were then modified
SPEDAT(SET,EXTRN_TEMP,OF IMSW001,R,25) -----modify the external temperature to 25 oC
SPEDAT(SET,EXTRN_CO1 ,OF IMSW001,R,4.536E-08) ---modify A=Sigma*E*Area
SPEDAT(SET,EXTRN_CO2 ,OF IMSW001,R,5) ---modify B=Area*h
SPEDAT(SET,EXTRN_PWR ,OF IMSW001,R,1) ----modify D=1
I was expecting the source would follow Q(W) = 5.67*10-8*ε*Area*(Te^4 – Tw^4) + Area*h*(Te - Tw)
But the output source for T3 is 98 W and for the TEM1 is 98W as well. The wall temperature at the boundary reaches 250 oC. With this value and the above formula, I calculated the radiation would be 760 W and the convection would be 1125 W. Why is that? I think I must have understand the formula and the patch-coval wrong. But t I could not find any other documents on how they are to be used.
BTW, the XPS is an 0.5*0.5*0.5m3 slab, but gridded as a 1-D problem. So the area above is 0.5*0.5m2.

Any help or hint is very much appreciated.
Thanks
Aki

[danbrown]

For the calculation of the drag coefficient, the reference area will be the projected frontal area of the vehicle.

[phoenics_cfd]

Originally Posted by akiaqi

hi, I am trying to simulate a case where an XPS slab is receiving solar radiation and losing heat to the external environment (25oC) by radiation and convection and to the room through conduction and convection (25oC). I thought it would be pretty simple but turned out something I need help. The temperature is crazy high (249oC). Both the radiation flux and the convection flux is about 100 W, which is a lot less than my calculation based on the temperature computed.
The solar radiation is set as an internal source of the boundary cells at the edge of the wall. From the result file, the 200 W solar radiation was correctly executed. The boundary source was set as instructed by the Immersol instruction:
PATCH(IMSW001,WWALL,1,1,1,1,1,1,1,1)
COVAL(IMSW001,T3,GRND4,GRND5)
COVAL(IMSW001,TEM1,wallco,GRND5)

The default values and coefficients were then modified
SPEDAT(SET,EXTRN_TEMP,OF IMSW001,R,25) -----modify the external temperature to 25 oC
SPEDAT(SET,EXTRN_CO1 ,OF IMSW001,R,4.536E-08) ---modify A=Sigma*E*Area
SPEDAT(SET,EXTRN_CO2 ,OF IMSW001,R,5) ---modify B=Area*h
SPEDAT(SET,EXTRN_PWR ,OF IMSW001,R,1) ----modify D=1
I was expecting the source would follow Q(W) = 5.67*10-8*ε*Area*(Te^4 – Tw^4) + Area*h*(Te - Tw)
But the output source for T3 is 98 W and for the TEM1 is 98W as well. The wall temperature at the boundary reaches 250 oC. With this value and the above formula, I calculated the radiation would be 760 W and the convection would be 1125 W. Why is that? I think I must have understand the formula and the patch-coval wrong. But t I could not find any other documents on how they are to be used.
BTW, the XPS is an 0.5*0.5*0.5m3 slab, but gridded as a 1-D problem. So the area above is 0.5*0.5m2.

Any help or hint is very much appreciated.
Thanks
Aki

Hi, Please send your Q1 input file to support@cham.co.uk, and we will take a look, and then we'll get back to you.