I have a box (as shown below) with 3 measurement devices in it. They are all the same size and stacked on each other. The power consumption of the devices is 250W, 50W, and 50W. They all have internal fans that blow the hot air out to the left side. Since the box is sealed ( Only a small hole for cable inlets) the hot air collects and it gets very warm inside.

How can I calculate how warm it will get inside the box?

The box has a total volume of 239 L and is full-body aluminum with 1.5mm thickness. The devices are placed right in the middle of the box.

thanks :)

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  • $\begingroup$ sounds like you are describing a convection oven $\endgroup$
    – jsotola
    Commented Feb 17, 2023 at 8:34
  • $\begingroup$ balance the energy lost by the box (conduction, convection and radiation) with the energy input. $\endgroup$
    – Solar Mike
    Commented Feb 17, 2023 at 10:27

1 Answer 1


You can basically treat the box as a whole as a body with an internal source equal to the sum of the power of the electronis (350 W), and then calculate what would the equilibrium be based on the convection radiation (and conductive if it rests on another surface) losses from the box.

I.e. for each of the faces calculate:


$$P_{conv} = h \cdot A \cdot (T_{wall}-T_\infty)$$


  • $P_{conv}$: The power lost due to convection
  • $h$ : the convection coefficient
  • $A$ : is the surface area of the face of the box
  • $T_{wall}$: the wall surface (this is the value you are after)
  • $T_\infty$: is the ambient temperature


The radiative losses are:

$$P_{rad} = \epsilon A \cdot \sigma \cdot T_{wall}^4$$


  • $P_{rad}$ : the ratiative losses
  • $\epsilon$: the emissivity of aluminium
  • $\sigma$: Stefan-Boltzmann constant
  • $A$ and $T_{wall}$ are the same as above


(follows fouriers law), this should only be applied if a face is touching on another surface (ground).

$$P_{cond} = U \cdot A\cdot (T_{wall}- T_\infty)$$


  • U : Coefficinet of Heat transfer (units W/(m^2 K))
  • $A, T_{wall}, T_\infty$: the same as above.

total losses

The total losses should be equal to the thermal power within. Therefore

$$P_{conv} + P_{cond} + P_{rad} = 350$$

You can solve the above equation (numerically usually), and obtain the $T_{wall}$ which can be thought of as the temperature within the box (it's slightly higher actually because of the conductivity of aluminium).


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