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Mathematically, it obviously is, but I would like to have the physical justification also, and I’m stuck on one certain step.

The solution to the CSTR problem begins like this:

$$f’(t) = \frac{qw – qf(t)}{V}$$

where $f(t)$ is the concentration at time $t$, $q$ is the flow rate, $w$ is the concentration of the inlet flow, and $V$ is the volume.

Applying this to the situation of a cooling body, it is easy to identify $qw$ as the amount of “coolness” arriving, but I am unable to justify the amount of “hotness leaving” as $qf(t)$, because the “mixing” present in the CSTR problem seems absent in the case of a cooling body. It seems like the cooling body is losing heat much like an onion might be peeled. So, a picture or analogy that would justify the $qf(t)$ term would be appreciated.

Background: CSTR stands for “Continuous Stirred Tank Reactor”, which is an abbreviation for “Continuous Flow Stirred Tank Reactor”. Finding an expression for the concentration in the tank at time $t$ is a textbook example in Differential Equations.

Regards,

Mike Jones

22.May.2011 (Beijing time)

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up vote 3 down vote accepted

Physically there is no such thing as "coolness", but there is heat. Heat leaves the cooling body at a rate proportional to the temperature of the body (that corresponds to the term $-q f(t)/V$) and enters the body at a rate proportional to the temperature of the environment (that corresponds to the term $qw/V$).

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Totally cool! (pun intented) I love MSE! –  Mike Jones May 22 '11 at 20:56
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