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Subject:
Thermodynamics
Category: Science > Physics Asked by: sanjoser-ga List Price: $100.00 |
Posted:
05 Nov 2005 13:14 PST
Expires: 05 Dec 2005 13:14 PST Question ID: 589499 |
A thermodynamics question how can I determine if the Helmholtz Free Energy can depend on the gradient of temperature? I think the answer depends on violating the second law. |
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There is no answer at this time. |
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Subject:
Re: Thermodynamics
From: 88tuner-ga on 06 Nov 2005 18:07 PST |
http://arxiv.org/PS_cache/astro-ph/pdf/0008/0008399.pdf A method is presented for the calculation of the gradient of the free energy with respect to all the internal and external degrees of freedom of a periodic crystal. This gradient can be used in conjunction with a static-energy Hessian for efficient geometrical optimization of systems with large unit cells. The free energy is calculated using lattice statics and lattice dynamics in the quasiharmonic approximation, and its derivatives by means of first-order perturbation theory. In the present application of the method, particles are assumed to interact via arbitrary short-ranged spherically-symmetric pair potentials and long-ranged Coulomb forces, and polarizability effects are accounted for by use of the shell model. The method can be used directly as the basis for a computer program which makes efficient use of both storage and CPU time, especially for large unit cells. Detailed expressions for all the lattice sums are presented. The inclusion of anisotropic surface free energy and anisotropic linear interface kinetics in phase-field models is studied for the solidification of a pure material. The formulation is described for a two-dimensional system with a smooth crystal-melt interface and for a surface free energy that varies smoothly with orientation, in which case a quite general dependence of the surface free energy and kinetic coefficient on orientation can be treated; it is assumed that the anisotropy is mild enough that missing orientations do not occur. The method of matched asymptotic expansions is used to recover the appropriate anisotropic form of the Gibbs-Thomson equation in the sharp-interface limit in which the width of the diffuse interface is thin compared to its local radius of curvature. It is found that the surface free energy and the thickness of the diffuse interface have the same anisotropy, whereas the kinetic coefficient has an anisotropy characterized by the product of the interface thickness with the intrinsic mobility of the phase field. |
Subject:
Re: Thermodynamics
From: sanjoser-ga on 06 Nov 2005 19:23 PST |
thanks, but I'm asking a much simpler question. can I include the gradient of temperature in the Helmholtz Free Energy without violating the second law? I have a feeling this is a textbook type problem. It's not my area, and I don't know how to proceed. |
Subject:
Re: Thermodynamics
From: theeldest-ga on 09 Nov 2005 23:14 PST |
No. The Helmholtz Free Energy can not depend on the gradient of Temperature. (I think). http://tinyurl.com/byr6s The author discusses the Helmholtz F.E. then looks at the constraints place upon it by the Second Law of Thermodynamics, and comes to the conclusion that for a thermoelastic system, the Helmholtz is independant of the Temperature Gradient. (thermoelastic is a system where the only work exchanged with the surroundings is that of pressure forces -> http://tinyurl.com/d6v7g Now. I'm not completely sure if this is right, as it's kinda out of my league (not to mention the fact that I hate thermodynamics and Statistical Mechanics...) |
Subject:
Re: Thermodynamics
From: sanjoser-ga on 10 Nov 2005 07:28 PST |
to theeldest: ok, I looked at that textbook you suggsted http://tinyurl.com/byr6s and I think my answer is in there, so I'm willing to cough up the dough. I think you have to post it as the answer first. FYI, the question came up in a course I'm taking thru distance learning on continuum mechanics, specifically on thermoelasticity. It's a brand new area for me, and I also hate thermodynamics. thanks again sanjoser |
Subject:
Re: Thermodynamics
From: theeldest-ga on 12 Nov 2005 22:01 PST |
Oh. I'm not an Answerer. I was just browsing around here while I was looking for help on my Thermodynamics homework, and thought I could help out. (we had just gotten done with Helmholtz and a bunch of associated derivations). |
Subject:
Re: Thermodynamics
From: sanjoser-ga on 12 Nov 2005 23:55 PST |
to theeldest well then, how can I do something for you? I'm actually doing my continuum mechanics homework. I took your suggestion and worked out a pretty good answer to the problem I was doing. Is there something I can help you do? I know a fair bit of stuff. best sanjoser |
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