Re: Re: heat diffusion calculation

From: QJos=E9_R=2E_Sabino=22?= (jrsabino_at_if.sc.usp.br)
Date: Mon Jun 18 2007 - 14:11:17 CDT

Dear Jonathan,

I just want to share my understanding of your question and try to give
an answer. An outer shell has its temperature kept constant, it is in a
thermal bath, I imagine the system is imersed in an heat repository,
when the system needs to withdraw or to release energy it is able to
give and receive without temperature variation of the bath. The inner
volume starts with a different temperature so you are going to measure
the heat diffusion to it. The temperature you measure in a simulation
(or in the physical system) is the time average of the system, it
converges to the outer shell temperature in a characteristic time. The
heating/cooling velocity should depend on the temperature gradient and
on the system heat conductivity.

Hope it helps,

Jose

Jonathan Lee wrote:

> Anybody? Thanks.
>
> Jonathan
>
>
> Jonathan Lee wrote:
>
>> Hello all,
>> I have some questions about the heat diffusion calculation in the
>> NAMD tutorial (page 53).
>>
>> 1) The shell is maintained at a temperature of 200, right? (As
>> opposed to just initialized to 200.)
>> 2) What is the temperature that is output (step 11, page 55)? Is
>> that the temperature of everything excluding the 200K shell?
>> 3) It seems to me that the calculation should take into account the
>> inner radius of the shell. If the radius is much larger, shouldn't
>> it take a longer time for the diffusion to occur?
>> 4) Can I do a similar calculation but with a rectangular prism domain
>> (i.e. fix the temperature at one end and calculate the heat diffusion
>> to the other end of the box)?
>>
>> Basically, my understanding is that the temperature should be
>> maintained in one region and measured in another region a finite
>> distance away. That distance (and the time of diffusion) should be
>> taken into consideration when finding the diffusivity. Am I
>> overlooking something? Thanks.
>>
>> Jonathan
>>
>>
>>
>>
>
>

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