Posted on Categories:Thermodynamics, 热力学, 物理代写

# 物理代写|热力学代写Thermodynamics代考|PHY360 Entropy Production Due to Diffusion

avatest™

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 物理代写|热力学代写Thermodynamics代考|Entropy Production Due to Diffusion

We describe the diffusion ${ }^{49}$ from region 1 to region 2 located at different lengths in a region of cross section $\Sigma$ as a ‘reaction’ $1 \rightarrow 2$ where one particle disappears from region 1 (with length $l$ and volume $l \cdot \Sigma$ ) and appears in region 2-see Fig. 4.8.
Our formalism gives then $-v_1=\nu_2=1,-d n_1=d n_2=d \xi^{\prime}, T_1=T_2=T$ and $A=-v_1 \mu_1-v_2 \mu_2=\mu_1-\mu_2$. If diffusion is one-dimensional and occurs only in the $x$ direction perpendicular to the cross section with $j_{p x} x$-th component of the particle flow $\mathbf{j}p$, then volume integration on region 1 of the the balance of mass leads to: $\frac{d n_1}{d t} \cdot l \cdot \Sigma+j{p x} \cdot \Sigma=0$, hence $\frac{d n_1}{d t}=-\frac{j_{p x}}{l}$. It follows that $T \sigma=\Upsilon^{\prime} A=\frac{d \xi^{\prime}}{d t} A=-\frac{d n_1}{d t} A=\frac{j_{p x} A}{l}=-\frac{j_{p x}\left(\mu_2-\mu_1\right)}{l}$. Straightforward generalization to three dimensions for arbitarily small $l$ leads to $\sigma=-\mathbf{j}p \cdot \nabla\left(\frac{\mu}{T}\right)$. Moreover, if many species are present and the particle flow of the $k$-th species is $\mathbf{j}{p k} \equiv \frac{\mathbf{j}k}{m_k}$, further generalization gives: $$\sigma=-\mathbf{j}{p k} \cdot \nabla\left(\frac{\mu_k}{T}\right)=-\mathbf{j}_k \cdot \nabla\left(\frac{\mu_k^0}{T}\right)$$

What about LNET? The quantity $\nabla\left(\frac{\mu_k}{T}\right)$ is a gradient, and the divergence of $\mathbf{j}_{p k}$ is proportional to a time derivative. Moreover, Fick’s law ${ }^{50}$ is a linear relationship between these quantities. Finally, $\tau$ is related to the diffusion time-scale. We conclude that LNET holds. It is therefore far from surprising that Prigogine has proposed a MinEP for this problem [19]-see Sect. 4.3.10.

## 物理代写|热力学代写Thermodynamics代考|Fick’s Law

The thermodynamic fluxes and forces in the diffusion-related entropy production density $\sigma=-\mathbf{j}{p k} \cdot \nabla\left(\frac{\mu_k}{T}\right)$ are $\mathbf{j}{p k}$ and $-\nabla\left(\frac{\mu_k}{T}\right)$ respectively. Accordingly, if $\nabla T=$ 0 then the linear phenomenological laws ${ }^{52}$ are of the type $\mathbf{j}{p k}=-\frac{L{p k}}{T} \nabla \mu_k$. In a linear treatment of a system with $N$ species it is usually assumed that:
$$\mathbf{j}{p k}=-D{i k} \nabla n_k \quad ; \quad i, k=1 \ldots N$$

(‘Fick’s law’). If $N=1$ then $\nabla \mu=\frac{\partial \mu}{\partial n} \nabla n$ and Fick’s law implies $\mathbf{j}p=-D \nabla n$, hence $L{11}=\frac{T D}{\frac{\partial \mu}{\partial n}}$. If $N=2$ then it is customary to deal with the case $\nabla T=0$, $\nabla p=0$ for simplicity. As usual in such cases, we start with the condition of extremum of Gibbs’ free energy $G ; d G=0$ implies $\mu_k d n_k=0$. But $G=\mu_k n_k$, then $\mu_k d n_k=0$ implies $n_k d \mu_k=0$. Moreover, LTE at all times implies $d a=$ $\frac{d a}{d t} \cdot d t=\left(\frac{\partial a}{\partial t}+\mathbf{v} \cdot \nabla a\right) \cdot d t=\left(\frac{\partial a}{\partial t}+\frac{d \mathbf{r}}{d t} \cdot \nabla a\right) \cdot d t$ for the generic quantity $a$. In steady state $\frac{\partial}{\partial t}=0$, then $d a=d \mathbf{r} \cdot \nabla a$. Let $a=\mu_k$. Accordingly, $n_k d \mu_k=0$ implies $\left(n_k \nabla \mu_k\right) \cdot d \mathbf{r}=0$ for arbitrary $d \mathbf{r}$, hence:
$$0=n_k \nabla \mu_k=n_1 \nabla \mu_1+n_2 \nabla \mu_2$$
Correspondingly:
$$\sigma=-\mathbf{j}{p k} \cdot \nabla\left(\frac{\mu_k}{T}\right)=-\frac{\mathbf{j}{p k}}{T} \cdot \nabla \mu_k=-\frac{\mathbf{j}{p 1}}{T} \cdot \nabla \mu_1-\frac{\mathbf{j}{p 2}}{T} \cdot \nabla \mu_2=-\frac{1}{T}\left(\mathbf{j}{p 1}-\frac{n_1}{n_2} \mathbf{j}{p 2}\right) \cdot \nabla \mu_1$$
Further simplification follows from the fact that $d\left(\frac{1}{\rho}\right)=v_k d c_k$ for $\nabla T=0$ and $\nabla p=0$, as the definition of the mass flow $\mathbf{j}k$ for the $k$-th species implies $\rho \frac{d}{d t}\left(\frac{1}{\rho}\right)=$ $v_k \rho \frac{d c_k}{d t}=v_k \nabla \cdot \mathbf{j}_k$. Usually, diffusion leaves $\rho$ unaffected, as it is a slow, incompressible process. Then $v_k \nabla \cdot \mathbf{j}_k=0$. Moreover, $v_k \nabla \cdot \mathbf{j}_k=\nabla \cdot\left(v_k \mathbf{j}_k\right)-\mathbf{j}_k \cdot \nabla v_k=\nabla$. $\left(v_k \mathbf{j}_k\right)-\mathbf{j}_k \cdot \nabla\left(\frac{\partial \mu_k^0}{\partial p}\right)_T$ and $\mathbf{j}_k \cdot \nabla\left(\frac{\partial \mu_k^0}{\partial p}\right)_T=\mathbf{j}{p k} \cdot \nabla\left(\frac{\partial \mu_k}{\partial p}\right)T=\left[\frac{\partial\left(\mathbf{j}{p k} \cdot \nabla \mu_k\right)}{\partial p}\right]T \propto$ $\left[\frac{\partial\left(\mathbf{j}{p k} \cdot \nabla \mu_k\right)}{\partial p}\right]T=O\left(n_k \nabla \mu_k\right)=0$ because of $n_k \nabla \mu_k=0$. Consequently, incompressibility implies $\nabla \cdot\left(v_k \mathbf{j}_k\right)=0$. Together with the definitions of $\mathbf{j}{p k} \equiv \frac{\mathbf{j}k}{m_k}$ and of $v_k^{\prime} \equiv m_k v_k$, this gives ${ }^{53} \mathbf{j}{p 1} v_1^{\prime}+\mathbf{j}{p 2} v_2^{\prime}=0$. Then $$\sigma=-\frac{1}{T}\left(1+\frac{n_1 v_1^{\prime}}{n_2 v_2^{\prime}}\right) \mathbf{j}{p 1} \cdot \nabla \mu_1=-\frac{1}{T}\left(1+\frac{n_1 v_1^{\prime}}{n_2 v_2^{\prime}}\right)\left(\frac{\partial \mu_1}{\partial n_1}\right){p, T} \mathbf{j}{p 1} \cdot \nabla n_1$$

## 物理代写|热力学代写Thermodynamics代考|Entropy Production Due to Diffusion

$\frac{d n_1}{d t} \cdot l \cdot \Sigma+j p x \cdot \Sigma=0$, 因此 $\frac{d n_1}{d t}=-\frac{j_{\mu x}}{l}$. 它邅循 $T \sigma=\Upsilon^{\prime} A=\frac{d \xi^{\prime}}{d t} A=-\frac{d n_1}{d t} A=\frac{j_{\mu x} A}{l}=-\frac{j_{p x}\left(\mu_2-\mu_1\right)}{l}$. 任意小到三 个维度的简单概括 $l$ 导致 $\sigma=-\mathbf{j} p \cdot \nabla\left(\frac{\mu}{T}\right)$. 此外，如果存在许多物种并且粒子流 $k$-th 种是 $\mathbf{j} p k \equiv \frac{\mathrm{j} k}{m_k}$ ，进一步概括给出:
$$\sigma=-\mathbf{j} p k \cdot \nabla\left(\frac{\mu_k}{T}\right)=-\mathbf{j}k \cdot \nabla\left(\frac{\mu_k^0}{T}\right)$$ LNET呢? 数量 $\nabla\left(\frac{\mu_k}{T}\right)$ 是一个梯度，而散度 $\mathbf{j}{p k}$ 与时间导数成正比。此外，菲克定律 ${ }^{50}$ 是这些量之间的线性关系。最后， $\tau$ 与扩散时 间尺度有关。我们得出结论，LNET 成立。因此，Prigogine 为这个问题提出了一个MinEP 就不足为奇了 [19] – 见 Sect. 4.3.10。

## 物理代写|热力学代写Thermodynamics代考|Fick’s Law

$\frac{d a}{d t} \cdot d t=\left(\frac{\partial a}{\partial t}+\mathbf{v} \cdot \nabla a\right) \cdot d t=\left(\frac{\partial a}{\partial t}+\frac{d \mathbf{r}}{d t} \cdot \nabla a\right) \cdot d t$ 对于通用数量 $a$. 处于稳定状态 $\frac{\partial}{\partial t}=0$ ，然后 $d a=d \mathbf{r} \cdot \nabla a$. 让 $a=\mu_k$. 因此， $n_k d \mu_k=0$ 暗示 $\left(n_k \nabla \mu_k\right) \cdot d \mathbf{r}=0$ 对于任意 $d \mathbf{r}$ ， 因此:
$$0=n_k \nabla \mu_k=n_1 \nabla \mu_1+n_2 \nabla \mu_2$$

$$\sigma=-\mathbf{j} p k \cdot \nabla\left(\frac{\mu_k}{T}\right)=-\frac{\mathbf{j} p k}{T} \cdot \nabla \mu_k=-\frac{\mathbf{j} p 1}{T} \cdot \nabla \mu_1-\frac{\mathbf{j} p 2}{T} \cdot \nabla \mu_2=-\frac{1}{T}\left(\mathbf{j} p 1-\frac{n_1}{n_2} \mathbf{j} p 2\right) \cdot \nabla \mu_1$$

$v_k \nabla \cdot \mathbf{j}_k=\nabla \cdot\left(v_k \mathbf{j}_k\right)-\mathbf{j}_k \cdot \nabla v_k=\nabla \cdot\left(v_k \mathbf{j}_k\right)-\mathbf{j} k \cdot \nabla\left(\frac{\partial \mu_k}{\partial p}\right)_T$ 和
$\mathbf{j}_k \cdot \nabla\left(\frac{\partial \mu_k^0}{\partial p}\right)_T=\mathbf{j} p k \cdot \nabla\left(\frac{\partial \mu_k}{\partial p}\right) T=\left[\frac{\partial\left(\mathbf{j} p k \cdot \nabla \mu_k\right)}{\partial p}\right] T \propto\left[\frac{\partial\left(\mathbf{j} p k \cdot \nabla \mu_k\right)}{\partial p}\right] T=O\left(n_k \nabla \mu_k\right)=0$ 因为 $n_k \nabla \mu_k=0$. 因此，不可 压㿟性意味着 $\nabla \cdot\left(v_k \mathbf{j}_k\right)=0$. 连同定义 $\mathbf{j} p k \equiv \frac{\mathrm{j} k}{m_k}$ 和 $v_k^{\prime} \equiv m_k v_k$ ，这给出 ${ }^{53} \mathbf{j} p 1 v_1^{\prime}+\mathbf{j} p 2 v_2^{\prime}=0$. 然后
$$\sigma=-\frac{1}{T}\left(1+\frac{n_1 v_1^{\prime}}{n_2 v_2^{\prime}}\right) \mathbf{j} p 1 \cdot \nabla \mu_1=-\frac{1}{T}\left(1+\frac{n_1 v_1^{\prime}}{n_2 v_2^{\prime}}\right)\left(\frac{\partial \mu_1}{\partial n_1}\right) p, T \mathbf{j} p 1 \cdot \nabla n_1$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。