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Posted
Hello,


i have the following problem.


Option 1 : i can generate 7 kW inside 1 m x 0.5 m x 1 m cubical. that cubical placed in 1.5 m x 2 m x 2 m compartment.


Option 2 : i can generate 7 kW inside 0.5 m x 0.5 m x 0.5 m cubical inside the same compartment


can i expect same temperature at the walls of the compartment in both options?


Posted

If this is a homework problem you should start by copying it out properly as written.

 

That means accurately and completely and including any diagrams provided.

 

I cannot believe even the most disreputable educational establishment would have written such a question.

 

1m x 0.5m x 1m is not cubical.

 

There is nowhere near sufficient information to even begin to guess at the problem.

Posted

 

 

I believe there is.

 

There must be more implied in the question to make (allow) us to assume Gauss's Law applies to the heat flux from the emitter through to the outside of the container.

Posted

 

Hello,
i have the following problem.
Option 1 : i can generate 7 kW inside 1 m x 0.5 m x 1 m cubical. that cubical placed in 1.5 m x 2 m x 2 m compartment.
Option 2 : i can generate 7 kW inside 0.5 m x 0.5 m x 0.5 m cubical inside the same compartment
can i expect same temperature at the walls of the compartment in both options?

 

Eventually you can expect so.

Posted

It has long been known by electronics engineers that audio amplifiers and other heat producing circuitry that have undersized heatsinks and situated within enclosed cases fail prematurely through overheat stress.

 

We do not know the mechanism(s) by which the heat is transferred from the source to the outer walls or the available rate of removal of heat through those outer walls..

Posted

 

We do not know the mechanism(s) by which the heat is transferred from the source to the outer walls or the available rate of removal of heat through those outer walls..

 

 

But you can assume the outer wall removal is identical in either case. Is there a way of telling the configuration of the 7 kW source once you've reached steady-state, only looking at the outer walls?

Posted

Steady state was not mentioned in the OP

 

Yes the outer walls are the same but why should the heat reaching them be identical?

 

Therefore thermal runaway is not precluded

Posted

Steady state was not mentioned in the OP

 

Yes the outer walls are the same but why should the heat reaching them be identical?

 

Therefore thermal runaway is not precluded

The heat reaching them should be identical because energy is conserved. True thermal runaway requires perfect insulation.

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