A paradox about energy balance and steady state

  • #1
fluidistic
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I met a paradox I am unable to resolve.
Here it goes: a special material is placed in thermal contact with 2 reservoirs kept at different temperature. The boundaries of the material that aren't in contact with the reservoirs are thermally insulated. The material being special is able to generate internal currents, a bit like Eddy currents for conductors when placed in a time varying magnetic field. These internal currents generate a Joule heat worth ##\rho \vec J^2## (always positive, i.e. heating the material) in all points in the material where ##\vec J## the current density does not vanish. There is an additional heat source that is more complicated than Joule heat, locally it can be either a cooling or a heating. Note that the currents are internal to the material, they do not have any normal component on the boundaries.
The paradox is that the steady state condition leads to a non steady state condition.
We know for sure that ##\vec J \neq \vec 0## because ##\nabla \times \vec J \propto \Delta T>0##. The material acts therefore a bit like a heat engine, it seems to exploit a temperature difference and do work.... on itself rather than on a load.
The steady-state heat equation has the form ##\nabla \cdot (\kappa \nabla T) + q_\text{Joule} + q_\text{special}=0##, it is derived from non equilibrium thermodynamics and conservation of energy, ##\nabla \cdot \vec J_U=0## where ##\vec J_U=J_Q + V\vec J## is the flux of internal energy, ##\vec J_Q=-\kappa \nabla T +ST\vec J ## is the heat flux, ##S## being a tensor. So far so good, the heat equation tells us that the temperature distribution adapts itself such that the thermal gradient creates a flux that evacuates the generated heats.
However, here comes the problem.
From the steady-state condition ##\nabla \cdot \vec J_U=0##, if we integrate ##\vec J_U## on the whole volume and use Gauss law, we find that the input energy is made of ##Q_\text{in}=\int_{\Gamma_{hot}} -\kappa \nabla T \cdot d\vec A_\text{hot}## only, and the output energy is ##Q_\text{out}=\int_{\Gamma_{cold}} -\kappa \nabla T \cdot d\vec A_\text{cold}##. We get that ##Q_\text{in} - Q_\text{out}=0## must hold for the steady-state to hold.

If we integrate the heat equation on the volume of the material and use Gauss law for the conduction term, we find that ##Q_\text{in} - Q_\text{out}=Q_\text{Joule}+Q_\text{special}## where ##Q_\text{Joule}## and ##Q_\text{special}## are the volume integrals of their respective local formulations written above. At first glance it looks like the special heat evolved has to cancel out exactly Joule heat for the steady state condition to hold. However this is impossible to hold, because Joule heat is proportional to ##\Delta T ^2## whereas the special heat is proportional to ##T \Delta T##. If, by luck, the base temperature of the material has been adjusted so that the 2 heats exactly cancel out globally one another, then we could change the base temperature and the condition cannot hold. This means the steady state condition cannot be fulfilled. But we started assuming a steady state condition... I do not understand where I go wrong.

Note that to keep things simple, no physical property depend on temperature.
 
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  • #2
Do there even exist real materials which could have your special property or is this just theoretical? I suggest simulation with COMSOL or some such program.
 
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  • #3
bob012345 said:
Do there even exist real materials which could have your special property or is this just theoretical? I suggest simulation with COMSOL or some such program.
Yes they do exist, I already have a finite element code that computes those quantities but I need to make.sure the results make sense.

I believe my error is in computing the surface integral of (ST+V)J dA, where J is parallel to dA. I thought this integral vanishes because J is parallel to dA on all boundaries, but S is a 2x2 matrix, or a tensor, and I suspect this can make the surface integral not vanish. T and V are scalar fields. The VJ part should yield Joule heat whereas the STJ part should yield Qspecial. Can someone tell me whether the surface integral vanishes or not?
 
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