inkliing Posted December 23, 2015 Share Posted December 23, 2015 For Fourier's law [latex]\vec{J}=-k\nabla{T},[/latex] where [latex]\vec{J}=[/latex] conductive heat flux in [latex]\frac{W}{m^2}[/latex] k=thermal conductivity in W/m-K T=temperature, [latex]\frac{1}{k}\vec{J}[/latex] certainly appears to be a conservative field with temperature as potential and for any path from position a to position b, [latex]\int\limits_{path}\vec{J}\cdot d\vec{r}=-k(T_b-T_a)[/latex], i.e, conductive heat flows from an equipotential surface at [latex]T_a[/latex] to an equipotential surface at [lower] [latex]T_b[/latex] This is analogous to: [latex]\vec{F}_{grav.}=-\nabla{\mbox{[potential energy]}}[/latex] where [latex]\int\limits_{path}\vec{F}_{grav}\cdot d\vec{r}=-(P.E._b-P.E._a)[/latex] and mass flows (falls) from a to b. Note: The path from position a to position b is a path in three dimensional space with coords x,y,z, not a path in a thermodyn. phase space, such as p-V diagram, in which heat is always path dependent. But this path-independent heat seems to only be useful when heat is considered to be an idealized conserved quantity, i.e., 1. heat is a perfect fluid - no meaningful internal or thermal energies, no chemical interaction, nuclear interations, partical collisions, particle velocity distributions, etc. 2. heat is conserved - no unwanted heat losses. Which is a clearly limited model. On the other hand, isn't it true that, whenever we do any relatively-straightforward engineering analysis involving conductive heat flow, such as the very-common one dimensional application of Fourier's law, that we are treating heat as a ideal conserved quantity? But on the other hand, heat transfer requires [latex]\Delta T\Rightarrow[/latex] heat transfer never occurs in thermodyn. equilibrium [latex]\Rightarrow[/latex] heat transfer is always irreversible [latex]\Rightarrow[/latex] heat transfer always involves unwanted heat losses [latex]\Rightarrow[/latex] heat transfer can never involve a perfectly idealized conserved quantity. So I think the conclusion is: We do pretend that heat is an idealized conserved quantity when we do relatively-straightforward conductive heat transport analysis, even though the 2nd law [latex]\Rightarrow[/latex] there'r always unwanted heat losses. Does this seem reasonable, or am I missing something important? Link to comment Share on other sites More sharing options...
studiot Posted December 23, 2015 Share Posted December 23, 2015 (edited) I am not sure what point your musing is trying to make, but how does about latent heat and phase changes fit in? Do you not supply exactly the same heat to melt something as you get back when it refreezes? Fourier's Law is about energy (heat) transfer. The First and Second Laws are about energy interconversion. Edited December 23, 2015 by studiot Link to comment Share on other sites More sharing options...
swansont Posted December 24, 2015 Share Posted December 24, 2015 where would the heat go, where it would not be a conserved quantity? Link to comment Share on other sites More sharing options...
inkliing Posted December 24, 2015 Author Share Posted December 24, 2015 More than one person has objected to "unwanted heat loss", which is apparently the wrong thing to say in this context. I'm referring to the idea that heat transfer through a finite temperature difference can contribute to the irreversibility of processes (Moran, John (2008). "Fundamentals of Engineering Thermodynamics", p. 220. John Wiley & Sons). I should have been more clear in the OP. I'm talking about a real process (not idealized or quasistatic, etc.) of heat conduction across some finite temperature difference (not infinitesimal), and therefore not reversible. What I mean by unwanted heat losses is heat loss that causes an irreversible increases in entropy, either in the system or the surroundings, and that these heat losses that cause irreversible entropy increases imply that heat, even if it's being treated as some sort of idealized fluid, is not a conserved quantity. A long-needed review of thermodynamics started me thinking about this. Of the many ways that work and heat are different, one that struck me was that work in a conservative field could be path independent, whereas it seems that heat is never path independent in a real process. What seemed to me to be the only process which might come close to path independent heat was Fourier's conduction law: a vector quantity = -gradient (a scalar), which is similar in form to the definition of a conservative field: field vector=-gradient(potential). So I wondered if the vector quantity in Fourier's law, conductive heat flux, can be considered to be a conservative field with potential= the scalar quantity in Fourier's law =temperature. It seems to me that all heat transport texts, when they discuss conductive heat transport, treat heat in this fashion - as an idealized, conserved fluid that flows from higher temperature surfaces to lower temp. surfaces. But if real heat transport across a finite temp. diff. is irrev. then heat must be lost to irrev. increases in entropy (syst. or surr.), implying heat is never a conserved quantity, even when we try to pretend it's an idealized fluid. I realize that there are irrev. processes, which therefore cause irrev. increases in entropy, but involve no heat in/out of the system- e.g., free expansion vs a vacuum. But free expansion isn't a heat transfer process (conduction, convection, radiation). It seems to me that any real heat transfer process across a finite temp. diff. must be irrev., and therefore incur irrev. entropy losses (in the syst. or surr.), and that these irrev. entropy differences must be cause by heat in or out, which I was calling unwanted heat losses. I'm not stating these ideas as fact. In fact, I do not completely understand them, and I'm trying to understand them. That is why I posted. Link to comment Share on other sites More sharing options...
swansont Posted December 25, 2015 Share Posted December 25, 2015 Seems to me that unwanted heat loss, then, is for a real system, and the equations are for an ideal system. If the losses are small, then you can use the equations. If not, you need to make additions to the model. Link to comment Share on other sites More sharing options...
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