How Thick Should My Thermal Barrier Be to Protect Against Heat?

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SUMMARY

Tina seeks to determine the appropriate thickness of a thermal barrier to maintain a component's temperature below 50°C when exposed to a heat source at 150°C. The material identified has a thermal conductivity of 0.03 W/mK. The discussion highlights the importance of knowing the heat transfer rate (Q) and the thermal properties of the material on the other side of the barrier to accurately calculate the necessary thickness using the equation Q = kA (ΔT/d).

PREREQUISITES
  • Understanding of thermal conductivity and its units (W/mK)
  • Familiarity with the heat transfer equation Q = kA (ΔT/d)
  • Knowledge of thermal conduction principles
  • Basic concepts of heat sources and ambient temperature effects
NEXT STEPS
  • Calculate the heat transfer rate (Q) for different heat sources
  • Research thermal barrier materials and their thermal conductivities
  • Explore methods for estimating the effective thermal resistance of layered materials
  • Learn about thermal insulation design principles and applications
USEFUL FOR

Engineers, materials scientists, and anyone involved in thermal management and insulation design will benefit from this discussion.

tina-duncan
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Hello friends,

I am having a hard time understanding thermal conduction and was hoping someone could shed some light for me.

I am looking to protect a component from heat by placing a thermal barrier between it and the heat source, the heat applied to the surface of this barrier would be 150°C. I have identified a material with a thermal conductivity of 0.03W/mK, however I am unsure what thickness I would require to prevent any heat of greater than 50°C reaching my component surface.

I have been trying to use the equation Q=kA deltaT/d, however I feel this does not suit what I am trying to accomplish due to not knowing Q?

So my question is, is there another way I should be approaching this problem? Is there a way of calculating Q and then rearranging the above equation to solve for thickness?

Any help would be greatly appreciated.

Thanks,
Tina
 
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I believe you need to know Q and the properties of whatever is on the other side of the object to be protected. For example..

Heat source (150C) -> Insulation -> object -> something -> Ambient Air (20C?)

If the "something" is a good thermal conductor then the temperature of the object will be nearer "ambient" than the heat source temperature.

What is the heat source? If it has a known power you might be able to estimate what percentage goes in the critical direction.
 

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