I am looking for a starting back of the envelope calculation to determine if an aluminum box with just free convection can dissipate 3.7W of internal heat generation or if I need to think about adding a heat sink into design.

The way I see it the heat generating electronics would be radiating heat via convection to internal air of box and then heat would travel into box and via free convection into atmosphere. I'm fine neglecting radiation and am not sure how to approach this problem after reading for the last couple hours on thermal stuff.

T ambient = 50C T max for electronics= 40C T Internal starting =20C Q generated by electronics = 3.7W Box dims = 75mmx75mmx100mm , Area=.0412m^2, Vol=.0005625 m^3 Box thickness= 7mm Box material is aluminum R_aluminum=200 R_air =.0262

Thanks for any help on how to approach this

  • $\begingroup$ If the ambient temperature is 50 C, the inside will eventually warm up to that temperature even without internal heat generation. $\endgroup$
    – Mark
    Commented Jul 21, 2019 at 0:54

2 Answers 2


Consider the integrated circuit (IC) generating power $P$ inside a box of surface area $A$. The energy balance equation is below.

$$ P = \sigma A \left(\epsilon_{IC}T_{IC}^4 - \epsilon_{B}T_{B}^4 \right) + h_i A \left(T_{IC} - T_B\right) $$

The first term is the net radiation flow with the Stefan-Boltzmann constant, emissivities and temperatures. The second is the convection flow inside the box with the convection coefficient. Assuming that you know all materials related values, this is one equation with two unknown temperatures.

The energy balance equation from the box to the air is written as below.

$$ \sigma A \left(\epsilon_{IC}T_{IC}^4 - \epsilon_{B}T_{B}^4\right) + h_i A \left(T_{IC} - T_B\right) = \sigma A \left(\epsilon_{B}T_{B}^4 - \epsilon_{a}T_{a}^4 \right) + h_a A \left(T_{B} - T_a\right)$$

The energy going in to the box from the inside equals the energy leaving the box from the outside. Assuming you know all the materials related values and the air temperature $T_a$, this is a second equation with the same two unknowns.

In principle, the problem is solvable. In practice, the easiest first approach is to neglect all radiation terms. This step gives these two equations.

$$ P = h_i A \left(T_{IC} - T_B\right) = h_a A \left(T_{B} - T_a\right)$$

This shows you that the temperature at the IC relative to the box will be balanced to first order by the ratio of internal to external convection coefficients.

$$ \frac{\left(T_{IC} - T_B\right)}{\left(T_{B} - T_a\right)} = \frac{h_i}{h_a}$$

Since the inside of the box is stagnate while the outside can be moving air, $h_i < h_a$ in general. Therefore, $\left(T_{IC} - T_B\right) > \left(T_{B} - T_a\right)$ in general. The IC will be hotter relative to the box than the box is relative to the air around it.

This first order approximation can be used as seed guesses to the full-blown equation with radiation. I might recommend a graphical solution in the full blown case.


How is this box going to be supported? Don’t forget the base/supports of the box will be a path of heat transfer, even if it’s standing on legs or suspended. Secondly, which area are you referring to in the problem statement? The cube has several faces/planes. We would need more information on the specifics of the geometry and prior research/approaches tried to provide an adequate and concise answer. For now, I would begin by researching free convection and the thermal properties of the grade of aluminum you’ve chosen.

  • $\begingroup$ This will be more appropriate posted as comments rather than as an answer. $\endgroup$ Commented Jul 18, 2019 at 11:16
  • $\begingroup$ Thank you for pointing that out. I am new to the site and getting used to the format.I will make sure to do this in the future. $\endgroup$
    – mechcad
    Commented Jul 18, 2019 at 13:25
  • $\begingroup$ I just noticed I do not have the ability to comment because my reputation is less than 50 at the moment. That was why I could not find the option. Nevertheless, thank you for pointing that out to me. $\endgroup$
    – mechcad
    Commented Jul 18, 2019 at 14:00

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