# On Thermodynamics Problems in the Single-Phase-Lagging Heat Conduction Model

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## Abstract

**:**

## 1. Introduction

_{V}is the specific heat. Equation (3) is a hyperbolic heat conduction equation, which predicts wave-like transport in heat conduction processes, called thermal wave. In thermodynamics, researchers have discussed the violation of the second law of thermodynamics by the CV model [9,10,11,12,13,14]. It is found that the CV model could lead to a negative entropy production rate in the framework of classical irreversible thermodynamics. This problem can be overcome by extended irreversible thermodynamics [15], where the definitions of thermodynamics could be extended and appropriate constitutive assumptions are proposed. There are many non-Fourier heat conduction models which have close connection with the CV one, and the single-phase-lagging (SPL) model [16] is a very classical example of them:

## 2. Entropy Production Rate

_{eq}(u) is the entropy in equilibrium. Then the entropy production rate $\dot{S}$ is calculated as:

## 3. Thermal Relaxation Time

_{S}> 0, and when τ < τ

_{S}, q(x,y,z,t)q(x,y,z,t+τ) ≥ 0. Therefore, the entropy production rate will be positive or zero as long as the relaxation time is small enough, which means that reducing the thermal relaxation time (larger than zero) could be another way for avoiding the negative entropy production rate. As an example, we consider again the heat conduction problem in Section 2, where the initial condition is taken as ${T\mid}_{t=0}={T}_{0}\left(2+\mathrm{sin}\frac{\pi x}{l}\right),$the boundary conditions are taken ${T\mid}_{x=0,l}={T}_{0}$ and the physical properties satisfy $\frac{\lambda \tau \pi}{{c}_{V}{l}^{2}}=\frac{1}{2}$. If we reduce the thermal relaxation time τ to τ′, where ${\tau}^{\prime}=\frac{\tau}{\pi \sqrt{e}}$, and keep the other physical properties, the initial and boundary conditions unchanged, the classical solution for this new heat conduction problem is:

_{S}is determined by the initial and boundary conditions, which can be expressed as τ

_{S}= τ

_{S}(T|

_{Γ},T|

_{t}

_{=0}). Then if the initial or boundary conditions are influenced by the relaxation time (such as T|

_{Γ}= T|

_{Γ}(t,τ), T|

_{t}

_{=0}= T|

_{t}

_{=0}(x,y,z,τ)), τ

_{S}will also be influenced by the relaxation time, which means τ

_{S}= τ

_{S}(τ). Therefore, as a function of τ, it is not necessary that τ

_{S}(τ) satisfies τ < τ

_{S}(τ). Next, we will provide a simple example for this problem, where T|

_{Γ}= T|

_{Γ}(t,τ). Consider a one-dimensional problem in 0 ≤ x ≤ l, where the initial condition is taken as T|

_{t}

_{=0}= 2T

_{0}, and the boundary conditions are taken as ${T\mid}_{x=0}={T}_{0}\left(2+\mathrm{sin}\frac{3\pi t}{2\tau}\right)$ and ${T\mid}_{t=l}={T}_{0}\left(2+\mathrm{sin}\frac{3\pi t}{2\tau}{e}^{-l\sqrt{\frac{3\pi \rho {c}_{V}}{2\tau \lambda}}}\right)$. The classical solution for this problem is:

_{Γ}= T|

_{Γ}(t,τ), which will also change as the relaxation time changing. The thermal relaxation time is usually discussed in the CV model, and the Taylor series approximation expressed by Equation (6) seems that the SPL model has the same relaxation time as the CV model. However, the physical meanings of the thermal relaxation time in the two models are quite different. For the SPL model, the heat flux only depends on the previous instantaneous temperature gradient τ, but for the CV model, from Equation (2), it is not difficult to find that the temperature gradient at a certain moment must influence the heat flux at the same moment. Therefore, although the Taylor series approximation expressed by Equation (6) is often established between the two models, their thermal relaxation times have different physical meanings. For the SPL model, the thermal relaxation means that the speed of the heat flux response to temperature gradient is not infinite. For the CV model, whose heat conduction equation is a hyperbolic equation and predicts wave-like transport, the thermal relaxation means that the speed of temperature perturbation itself is not infinite.

## 4. Spontaneous Equilibrium

_{Γ}and the initial conditions T|

_{t}

_{=0}satisfy T|

_{Γ}= T|

_{t}

_{=0}= C

_{0}, where C

_{0}is a constant, the whole temperature T(x,y,z,t) must satisfy T(x,y,z,t) = C

_{0}. For the CV model, the heat conduction equation contains a second-order time derivative term. Therefore, we need two initial conditions T|

_{t}

_{=0}and ${\frac{\partial T}{\partial t}|}_{t=0}$ to determine the solution for the CV model. Then the equilibrious boundary and initial conditions for the CV model are T|

_{Γ}= T|

_{t}

_{=0}= C

_{0}and ${\frac{\partial T}{\partial t}|}_{t=0}$. If these conditions are satisfied, the whole temperature field must be C

_{0}. From the view point of physics, it is not difficult to understand these conclusions. Whether for diffusive transport or for wave transport, if the system has been in equilibrium, equilibrium will not be broken without perturbation. However, for the single-phase-lagging model, the conclusion is different. Even if the boundary and initial conditions are all in equilibrium, equilibrium could be broken in some special circumstances. Next, we will provide a simple example. Consider a one-dimensional problem where the physical properties satisfy $\frac{2\tau {n}^{2}\pi}{\rho {c}_{V}{l}^{2}}=1$ (n is an integer). The boundary conditions are taken T|

_{x}

_{=0,l}= C

_{0}, and the initial condition is taken T|

_{t}

_{=0}= C

_{0}. Then consider a solution:

_{1}is an arbitrary constant. This solution satisfies Equation (5), boundary conditions T|

_{Γ}= T|

_{t}

_{=0,l}= C

_{0}and initial condition T|

_{t}

_{=0}= C

_{0}. Obviously, ${C}_{1}\mathrm{sin}\frac{n\pi x}{l}\mathrm{sin}\frac{\pi t}{2\tau}\left({C}_{1}\ne 0\right)$ breaks equilibrium, and this means that for the SPL model, equilibrium could be broken spontaneously. In this special circumstance, there is no temperature difference at the initial time, but the SPL model could cause a temperature difference. Therefore, for the SPL model, although we can avoid the negative entropy production rate by extending the definition of entropy and proposing appropriate constitutive assumptions, the second law of thermodynamics could also be violated by breaking equilibrium spontaneously. From this view point, for the SPL model, modifying the entropy production rate is not enough to avoid the violation of the second law of thermodynamics. In addition, solution expressed by Equation (22) also shows that the solution for the SPL model may not be unique. For an arbitrary heat conduction problem, where the boundary conditions are taken T|

_{x}

_{=0,l}= f

_{Γ}and the initial conditions are taken as T|

_{t}

_{=0}= f

_{0}, if this problem has a solution ${T}_{\mathsf{\Lambda}}\left(x,t\right),$${{T}^{\prime}}_{\mathsf{\Lambda}}\left(x,t\right)={T}_{\mathsf{\Lambda}}\left(x,t\right)+{C}_{1}\mathrm{sin}\frac{n\pi x}{l}\mathrm{sin}\frac{\pi t}{2\tau}$ will also satisfy Equation (5) and all conditions. Therefore, ${{T}^{\prime}}_{\mathsf{\Lambda}}\left(x,t\right)$ is also a solution for this problem, which means that for the SPL model, one determined heat conduction problem could have infinitely many solutions. It is necessary to emphasize that these above non-physical phenomena only occur with special physical properties $\frac{2\tau {n}^{2}\pi}{\rho {c}_{V}{l}^{2}}=1$.

_{2}when t < 0. Let ϕ = T − C

_{2}and from Equation (5), ϕ satisfies:

_{Γ}= T|

_{Γ}− C

_{2}and ϕ|

_{t=0}= T|

_{t}

_{=0}− C

_{2}. From the second assumption, we can obtain that ϕ is also finite, and the Laplace integral $F={{\displaystyle \int}}_{0}^{+\infty}\phi {e}^{-pt}dt$ is convergent. Therefore, the Laplace transform $F={{\displaystyle \int}}_{0}^{+\infty}\phi {e}^{-pt}dt$ exists. Substituting it into Equation (23), and according to the Time-Shift Theorem, we can obtain:

_{Γ}= T|

_{t}

_{=0}= C

_{0}, it is not difficult to find that T

_{II}(x,t) = C

_{0}is a solution for Equation (5) with these conditions, and because of the uniqueness, there is no other solution besides C

_{0}. Then, we conclude that the two assumptions can avoid breaking equilibrium spontaneously.

## 5. Mathematical Energy Integral

_{Γ}= C

_{0}, the definition of energy integral is:

_{0}. The changing rate of the energy integral can be written as:

_{t=0}= C

_{0}, we can obtain E

_{F}(0) = 0, and from $\frac{d{E}_{F}\left(t\right)}{dt}\le 0$, E

_{F}(t) ≤ E

_{F}(0) = 0. However, from Equation (25), obviously E

_{F}(t) ≥ 0, and then we have 0 ≤ E

_{F}(t) ≤ 0. Therefore, we can obtain E

_{F}(t) = 0 and T(x,y,z,t) = C

_{0}. This shows that Fourier heat conduction will not break equilibrium when the boundary and initial conditions are all in equilibrium. For the CV model, the definition of energy integral is:

_{t=0}= C

_{0}, we can obtain ∇T|

_{t=0}= 0 and then we have E

_{CV}(0) = 0. Equation (28) shows that $\frac{d{E}_{CV}\left(t\right)}{dt}\le 0$, and therefore, E

_{CV}(t) ≤ E

_{CV}(0) = 0. From Equation (27), obviously E

_{CV}(t) ≥ 0, and then we have 0 ≤ E

_{CV}(t) ≤ 0. Therefore, we can obtain E

_{CV}(t) = 0, which shows that ∇T(x,y,z,t) = 0 and $\frac{\partial T\left(x,y,z,t\right)}{\partial t}=0$. Because $\frac{\partial T\left(x,y,z,t\right)}{\partial t}=0$ and ∇T(x,y,z,t) = 0, we can finally obtain T(x,y,z,t) = C

_{0}. Thus, it is shown that the CV model will not break equilibrium when the boundary and initial conditions are all in equilibrium. For the CV model, the energy integral Equation (27) doesn’t show the vibration amplitude from the equilibrium temperature but the size of the first-order differentials. In Equation (26), we find that the decay rate of E

_{F}(t) is determined by (∇T)

^{2}, which means the attenuation will not stop unless ∇T = 0. Then, we have q = 0 and T(x,y,z,t) = C

_{0}. Therefore, for Fourier’s law, the attenuation of the heat conduction process will not stop until the equilibrium is achieved, when E

_{F}(t) = 0. As comparison, in Equation (28), the decay rate of E

_{CV}(t) is determined by ${\left(\frac{\partial T}{\partial t}\right)}^{2}$, which means the attenuation will not stop unless $\frac{\partial T}{\partial t}=0$. Different from Fourier’s law, for the CV model, when the attenuation stops, it is possible that ∇T ≠ 0, q ≠ 0, T(x,y,z,t) ≠ C

_{0}and $\frac{\partial q}{\partial t}\ne 0$. As an example, for T(x,y,z,t) ≠ C

_{0}, ∇T ≠ 0 and q ≠ 0, T(x,t) = kx + b and q(x,t) = −λkx could satisfy the CV model and heat conduction equation Equation (3), where k and b are constants. As an example for $\frac{\partial q}{\partial t}\ne 0$, T(x,t) = C

_{0}and $q\left(\mathrm{x},\mathrm{t}\right)={q}_{0}{e}^{-\frac{t}{\tau}}$ could satisfy the CV model and heat conduction equation Equation (3), where q

_{0}is constant. Therefore, for the CV model, the attenuation of the heat conduction process stops as long as the temperature achieves stability, and it is possible that ∇T ≠ 0, q ≠ 0, T(x,y,z,t) ≠ C

_{0}and $\frac{\partial q}{\partial t}\ne 0$. In summary, from the analysis of mathematical energy integral Fourier’s law means that heat conduction must tend to equilibrium, and the CV model only requires that heat conduction tends to steady temperature.

## 6. Conclusions

- In the framework of classical irreversible thermodynamics, the SPL model could cause a negative entropy production rate problem. There are two perspectives for the SPL model to avoid the negative entropy production rate. One is extending the definition of entropy in classical irreversible thermodynamics, which is based on extended irreversible thermodynamics. The other is reducing the thermal relaxation time, which is based on the continuity of heat flux in time.
- It is shown that modifying the entropy production rate to a positive or zero value is not enough to avoid the violation of the second law of thermodynamics for the SPL model, because the SPL model could cause spontaneous equilibrium breaking under some special circumstances. What’s more, the SPL model could also lead to infinitely many solutions for determined heat conduction problems, which is also non-physical. To avoid these problems, two assumptions are proposed from the view point of engineering and physics. One assumption is that heat conduction process begins at t = 0, and the system is in thermal equilibrium when t < 0. The other assumption is that the whole temperature field is finite.
- It is proved that Fourier’s law and the CV model cannot break equilibrium spontaneously by analyzing the mathematical energy integral. The energy integral of Fourier’s law shows the deviation from equilibrium and the vibration amplitude from the equilibrium temperature, and the energy integral of the CV model shows the size of the first-order differentials. For Fourier’s law, the attenuation of the heat conduction process will not stop until the equilibrium is achieved, but for the CV model, the attenuation of the heat conduction process stops as long as the temperature achieves stability.

## Acknowledgments

## Author Contributions

## Conflicts of Interest

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**MDPI and ACS Style**

Li, S.-N.; Cao, B.-Y.
On Thermodynamics Problems in the Single-Phase-Lagging Heat Conduction Model. *Entropy* **2016**, *18*, 391.
https://doi.org/10.3390/e18110391

**AMA Style**

Li S-N, Cao B-Y.
On Thermodynamics Problems in the Single-Phase-Lagging Heat Conduction Model. *Entropy*. 2016; 18(11):391.
https://doi.org/10.3390/e18110391

**Chicago/Turabian Style**

Li, Shu-Nan, and Bing-Yang Cao.
2016. "On Thermodynamics Problems in the Single-Phase-Lagging Heat Conduction Model" *Entropy* 18, no. 11: 391.
https://doi.org/10.3390/e18110391