物理代写|流体力学代写Fluid Mechanics代考|AMME2261

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物理代写|流体力学代写Fluid Mechanics代考|Sucking problem

In the sucking problem, steam and liquid water are contained in a 1D tube with a heated wall (see Figure $2.15$ ). The liquid is superheated at $T_{\max }$ and the wall temperature is fixed at $T_{\text {wall }}=T_{\text {sat }}$, so that the steam is at the saturation temperature (see Figure 2.16). Therefore, there is no temperature gradient in the vapor. Both fields are at the rest at $\mathrm{t}=0 \mathrm{~s}$. For $\mathrm{t}>0 \mathrm{~s}$, the liquid begins to boil at the interface, which induces a displacement of the steam/water interface.

The theory of the sucking problem is given in Welch and Wilson (2000). First, let us make a transformation of the spatial coordinates, such as the steam/water is located at $\xi=0$ :
$$\xi=x-\int_{0}^{t} v_{S}\left(t^{\prime}\right) d t^{\prime}$$
where $v_{S}$ is the interface velocity. With this new coordinate, the temperature profile is given by:
$$T=T_{s a t}+B \sqrt{\frac{\pi}{2}} \phi^{\prime}(0) \exp \left(x_{0}^{2}\right) \operatorname{erf}\left(x_{0}, x\right)$$
where $B=\frac{\chi_{l} \rho_{l}}{C \rho_{v}}, C=\frac{\lambda_{l}}{L \rho_{v}}$, and $\chi_{l}$ are the thermal diffusivity of the liquid phase, $\mathrm{L}$ is the latent heat between steam and liquid water and $x=\frac{\eta+\phi^{\prime}(0)}{\sqrt{2}}$, $x_{0}=\frac{\phi^{\prime}(0)}{\sqrt{2}}, \quad \eta=\frac{\xi}{\sqrt{2 \chi_{l}}}$, and $\operatorname{erf}\left(x_{0}, x\right)=\frac{2}{\sqrt{\pi}} \int_{x_{0}}^{x} \exp \left(-t^{2}\right) d t$ are the error function. The evolution of the interface position has the following expression:
$$X(t)=\frac{\phi^{\prime}(0) \rho_{l}}{\rho_{v}} \sqrt{2 \chi_{l} t}$$
$\phi^{\prime}(0)$ can be obtained by solving the following equation:
$$\phi^{\prime}(0) \exp \left(\frac{\phi^{\prime}(0)^{2}}{2}\right)=\sqrt{\frac{2}{\pi}} \frac{T_{\max }-T_{s a t}}{B}$$
The study is based on Sato’s and Welch’s publications (Welch and Wilson 2000; Sato and Niceno 2013). The physical properties of steam and water are updated at each time step using the standard set of thermodynamic Equations Of State based on CATHARE functions (Emonot et al. 2011). The pressure of the system is equal to $1.013 \cdot 10^{5} \mathrm{~Pa}$ (the atmospheric pressure), and the liquid temperature in the bulk is $T_{\max }=378.15 \mathrm{~K}$.

物理代写|流体力学代写Fluid Mechanics代考|Stefan problem

The Stefan problem is very close to the sucking problem. Steam and liquid water are contained in a 1D tube with a heated wall under atmospheric pressure. However, in this case, steam is superheated, and the liquid water temperature is equal to the saturation temperature $T_{s a t}=373.15 \mathrm{~K}$. The wall temperature keeps a constant value during the simulation: $T_{\text {wall }}=398.15 \mathrm{~K}$. As a result, there is a temperature gradient in the vapor phase (see Figure 2.20).

The evolution of the interface position is given by the equation (Welch and Wilson 2000 ):
$$X(t)=2 \beta \sqrt{\chi_{v} t}$$
The temperature profile is equal to:
$$T(x, t)=T_{\text {wall }}+\frac{T_{\text {sat }}-T_{\text {wall }}}{\operatorname{erf}(\beta)} \operatorname{erf}\left(\frac{x}{2 \sqrt{\chi_{v}}}\right)$$
where $\beta$ is the solution of the equation:
$$\beta \exp \left(\beta^{2}\right) \operatorname{erf}(\beta)=\frac{C_{m v}\left(T_{\text {wall }}-T_{s a t}\right)}{\sqrt{\pi} L}$$
with $C_{p v}$ as the specific heat capacity of vapor.
The physical properties of steam and water are updated at each time step using the standard set of thermodynamic Equations Of State based on CATHARE functions (e.g. Emonot et al. 2011). The pressure of the system is equal to $1.013 \cdot 10^{5} \mathrm{~Pa}$. The same grids have been used with the same boundary conditions. The only difference is the wall temperature, which is equal to $398.15 \mathrm{~K}$, that is, $T_{\text {sat }}+25 \mathrm{~K}$, in the Stefan problem, instead of $T_{\text {sat }}$ in the sucking problem. Three different mesh refinements have been used with cell sizes equal to: $110^{-4} \mathrm{~m}, 510^{-5} \mathrm{~m}$ and $2.510^{-5} \mathrm{~m}$. The time steps are, respectively, equal to $1.610^{-3} \mathrm{~s}, 410^{-4} \mathrm{~s}$ and $1.10^{-4} \mathrm{~s}$.

物理代写|流体力学代写Fluid Mechanics代考|Sucking problem

Welch and Wilson (2000) 给出了吸哂问题的理论。首先，让我们对空间坐标进行变换，例 如承汽 $/$ 水位于 $\xi=0:$
$$\xi=x-\int_{0}^{t} v_{S}\left(t^{\prime}\right) d t^{\prime}$$

$$T=T_{\text {sat }}+B \sqrt{\frac{\pi}{2}} \phi^{\prime}(0) \exp \left(x_{0}^{2}\right) \operatorname{erf}\left(x_{0}, x\right)$$

$$X(t)=\frac{\phi^{\prime}(0) \rho_{l}}{\rho_{v}} \sqrt{2 \chi_{l} t}$$
$\phi^{\prime}(0)$ 可以通过求解以下方程得到:
$$\phi^{\prime}(0) \exp \left(\frac{\phi^{\prime}(0)^{2}}{2}\right)=\sqrt{\frac{2}{\pi}} \frac{T_{\max }-T_{\text {sat }}}{B}$$

(Emonot 等人，2011 年)。系统压力等于 $1.013 \cdot 10^{5} \mathrm{~Pa}$ (大气压)，本体中的液体温 度为 $T_{\max }=378.15 \mathrm{~K}$.

物理代写|流体力学代写Fluid Mechanics代考|Stefan problem

$\mathrm{~ S t e f a n ~ 问 题 非 常 接 近 于 吸 㕬 问 题 。 菡 汽 和 液 态 水 包 含 在 大 气 压 下 带 有 加 热 垚}$ 是，在这种情况下，㦯汽是过热的，液态水的温度等于饱和温度 $T_{s a t}=373.15 \mathrm{~K}$. 壁温在 模拟过程中保持恒定值: $T_{\text {wall }}=398.15 \mathrm{~K}$. 因此，气相中存在温度梯度（见图 2.20）。

$$X(t)=2 \beta \sqrt{\chi_{v} t}$$

$$T(x, t)=T_{\text {wall }}+\frac{T_{\text {sat }}-T_{\text {wall }}}{\operatorname{erf}(\beta)} \operatorname{erf}\left(\frac{x}{2 \sqrt{\chi_{v}}}\right)$$

$$\beta \exp \left(\beta^{2}\right) \operatorname{erf}(\beta)=\frac{C_{m v}\left(T_{\text {wall }}-T_{\text {sat }}\right)}{\sqrt{\pi} L}$$

$110^{-4} \mathrm{~m}, 510^{-5} \mathrm{~m}$ 和 $2.510^{-5} \mathrm{~m}$. 时间步长分别等于 $1.610^{-3} \mathrm{~s}, 410^{-4} \mathrm{~s}$ 和 $1.10^{-4} \mathrm{~s}$.

有限元方法代写

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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。

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