QUESTION
What is the formula for estimating the rise in temperature within an enclosure, having no internal airflow and mounted outdoors?
- The enclosure is constructed of aluminum; having dimensions of 0.7mX0.5mX0.4m and a wall thickness of 0.005m
- The power dissipation inside the enclosure is 50w
- The ambient (outside) temperature is 40°C
I was hoping that that there was a straightforward first-order estimate equation that one might use to acquire a ROM estimate for the temperature rise in an enclosure (knowing the size of the enclosure, the material of the enclosure, that the fluid within the enclosure was air, the power dissipated in the enclosure and the ambient temperature outside the enclosure). At least to the point that I could begin to understand if it was 2 degrees or 20 degrees.
REPLIES
amorrison
This is like a single glazed house window where the glass is aluminum.
The aluminum will have an effective “R” value of zero -ie. no thermal resistance.
All the thermal resistance will come from the air films. There is one on the inside of the box probably about 0.68 in Imperial units (btu/hr/ft2/degF). The thermal resistance on the outside of the box will ~ the same unless there is a fan to provide air movement.
Unknowns
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There will be “natural convection” of air inside the box - but imo won’t change things much.
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The viscosity of the air inside the box will change with the heating - but imo this won’t affect much either -unless things are really hot.
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You are unlikely to get a film coefficient accuracy better than ±20% so consider this.
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There will be another film coefficient to consider - the one around your heat source in the box - but if you are only interested in the air temperature of the box provides no information of utility.
Good textbooks can provide a more “accurate” answer for these “heat source in an enclosure problems” but with the ±20 % in the “R” value - is it worth the effort for a standard analysis where nothing is “strange”.
Then its: Btu/hr = area x deltaT/(Total R)
SOURCE
https://www.eng-tips.com/viewthread.cfm?qid=290195
Above is a snippet.