Somewhat Complicated Thermal Problem

[unix101os]

Problem Description:
I have a solar panel of some surface area, material, and thickness mounted to an enclosure. The panel is isolated from the enclosure at some distance with a multitude of materials (air, insulation, plastic, metal) between the back surface of the panel and interior volume (which is air tight). Given the temperature of the backside of the panel and that the materials behind the panel will be shaded (ie only receive IR light radiating from the back of the panel) how can I calculate the temperature of the volume inside the enclosure?

 

[Babadag]

s=source r=receiver T[oK=273.4+oC]
Transferred by radiation heat will be:
qrad=ks(Ts/100)^4-kr*(Tr/100)^4)
k=4.96 kcal/m^2*h*(100/oK)^4 for black material.
For aluminum surface -for instance- could be 0.26.
Heat transfer could be done by convection also.
qconv=a*A*qa where:
a=Nu*lk/D A=panel surface area[m^2] Nu=Nussfeld factor
lk=air thermal conductivity[at 40oC =0.027 W/m/oC
D=panel height]m]
Nu[Nussfeld factor] depends on two other factors :Grashof and Prandtl[0.74 for air].
qa=qA-(qo+Dq/2) where:
qA=panel surface temperature qo=ambient air temperature
Dq=air temperature rise
The air will receive at first the convection heat. If you don't know which part of this will be transferred to the enclosure take it all.
If the panel is insulated we have a temperature drop through the insulated material by conduction.
q=l/s*A*ql
s=insulation thickness [m] l=material thermal conductivity ql=dT/dx
This calculation is iterative [you have to recalculate the heat transfer many time].
If the enclosure does not contain any heat sources a convection heat transfer may be enough for calculation of cooling .

 

[unix101os]

View attachment 211552
s=source r=receiver T[oK=273.4+oC]
Transferred by radiation heat will be:
qrad=ks(Ts/100)^4-kr*(Tr/100)^4)
k=4.96 kcal/m^2*h*(100/oK)^4 for black material.
For aluminum surface -for instance- could be 0.26.
Heat transfer could be done by convection also.
qconv=a*A*qa where:
a=Nu*lk/D A=panel surface area[m^2] Nu=Nussfeld factor
lk=air thermal conductivity[at 40oC =0.027 W/m/oC
D=panel height]m]
Nu[Nussfeld factor] depends on two other factors :Grashof and Prandtl[0.74 for air].
qa=qA-(qo+Dq/2) where:
qA=panel surface temperature qo=ambient air temperature
Dq=air temperature rise
The air will receive at first the convection heat. If you don't know which part of this will be transferred to the enclosure take it all.
If the panel is insulated we have a temperature drop through the insulated material by conduction.
q=l/s*A*ql
s=insulation thickness [m] l=material thermal conductivity ql=dT/dx
This calculation is iterative [you have to recalculate the heat transfer many time].
If the enclosure does not contain any heat sources a convection heat transfer may be enough for calculation of cooling .

Thank you so much this is very helpful!

 

[unix101os]

some questions though,

What is meant my air temp rise (Dq)?
Would I add qrad and qconv to get total heat transfer through air?
Do I use qrad+qconv to solve for Thot?

 

[Babadag]

I am sorry. Since I did not succeed to write the Greek letter q and theta was the same.
Actually, the text has to be as in image. The radiation and convection transferred
heat will reach together the enclosure. However, the air between will get only the convection.

 

[Nidum]

Problem Description:
I have a solar panel of some surface area, material, and thickness mounted to an enclosure. The panel is isolated from the enclosure at some distance with a multitude of materials (air, insulation, plastic, metal) between the back surface of the panel and interior volume (which is air tight). Given the temperature of the backside of the panel and that the materials behind the panel will be shaded (ie only receive IR light radiating from the back of the panel) how can I calculate the temperature of the volume inside the enclosure?

Realistically there is no way to obtain a definite answer to this problem by calculation . As in many similar problems it will be easier to just build the assembly and see how hot it gets . If it gets too hot then modify the assembly .

 

[Babadag]

I agree with you, Nidum. However, the temperature rise calculation this is the way for electrical machine design. Finally the rated temperature has to be stated by measurement, of course. Some articles could be more effective. See-for instance:
http://www.engineersedge.com/heat_transfer/isothermal_vertical_plate_natural_convection_13964.htm
https://www.electronics-cooling.com...on-heat-transfer-coefficient-on-a-flat-plate/