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物理代写|热力学代写Thermodynamics代考|MECH3720 Joule-Thomson Throttled Expansion

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物理代写|热力学代写Thermodynamics代考|Joule-Thomson Throttled Expansion

Let us discuss in more detail the result of fluid mechanics $\Gamma=\left[\frac{d\left(n_2\right)}{d t}\right]{\partial T_2=0}^{-1}$ $\left[\frac{d\left(E_2\right)}{d t}\right]{\partial T_2=0}=C_p T$ quoted in Sect. 4.1.12 [2]. In a perfect gas, $C_p T$ is just equal to the molar enthalpy $m_m h$, where $m_m$ is the molar mass and $h \equiv u+\frac{p}{\rho}$ is the enthalpy per unit mass, $u, \rho=m_m n_m$ being the internal energy per unit mass, the mass density and the molar density respectively. We write the mass balance (Sect.4.2.1):
$$\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{v})=0$$
and the energy balance (Sect.4.2.8, no electromagnetic field, no gravity, no heat flux):
$$\frac{\partial}{\partial t}\left(\rho \frac{|\mathbf{v}|^2}{2}+\rho u\right)+\nabla \cdot\left[\rho \mathbf{v}\left(\frac{|\mathbf{v}|^2}{2}+h\right)-\mathbf{v} \cdot \boldsymbol{\sigma}^{\prime}\right]=0$$
where the viscous stress tensor $\boldsymbol{\sigma}^{\prime}$ is linear in the components of the velocity $\mathbf{v}$.
If $|\mathbf{v}|$ is so small $^{30}$ that terms $\propto O\left(|\mathbf{v}|^2\right)$ are negligible, then the energy balance reduces to $\frac{\partial}{\partial t}(\rho u)+\nabla \cdot(\rho \mathbf{v} h)=0$ and this result, together with the balance of mass,

$$\frac{\partial}{\partial t}(\rho u)+\rho \mathbf{v} \cdot \nabla h-h \frac{\partial \rho}{\partial t}=0$$
Moreover, multiplication of $\Gamma=C_p T=h m_m$ by $\frac{\partial n_m}{\partial t}$ gives $\Gamma \frac{\partial n_m}{\partial t}=h \frac{\partial \rho}{\partial t}$ since $m_m$ is obviously constant. For the same reason we may also write $\rho \mathbf{v} \cdot \nabla h=n_m \mathbf{v}$. $\nabla \Gamma$. The energy balance reduces further to:
$$\frac{\partial}{\partial t}(\rho u)=\Gamma \frac{\partial n_m}{\partial t}-n_m \mathbf{v} \cdot \nabla \Gamma$$
Now, we rewrite $\Gamma$ as $\Gamma=\left[\frac{d\left(n_m\right)}{d t}\right]{T_2=\text { const. }}^{-1}\left[\frac{d(\rho u)}{d t}\right]{T_2=\text { const. }}$, which simplifies further to ${ }^{32}: \Gamma=\left[\frac{\partial\left(n_m\right)}{\partial t}\right]{T_2=\text { const. }}^{-1}\left[\frac{\partial(\rho u)}{\partial t}\right]{T_2=\text { const. }}$ as $\mathbf{v}$ is small. Furthermore, since $T_2$ is constant (a difference in $T$ is artificially maintained by the external world) then we may drop the subscript, and write: $\Gamma=\left(\frac{\partial n_m}{\partial t}\right)^{-1} \frac{\partial(\rho u)}{\partial t}$. Thus, the energy balance takes the simple form $n_m \mathbf{v} \cdot \nabla \Gamma=0$. Integration of both sides of this equation on a volume enclosing the hole connecting regions 1 and 2 leads to ${ }^{33}$ (this volume is displayed with dotted boundaries in Fig. 4.7):
$$\oint d \mathbf{a} \cdot \mathbf{v} n_m \Gamma=0$$
i.e. the net enthalpy flux vanishes.
Accordingly, total enthalpy is conserved across the hole. This is the well- known Joule-Thomson throttled expansion. Implicitly, however, we have extended our result to region 1 as well when integrating: we neglected both $|\mathbf{v}|^2$ and $\nabla T$.

物理代写|热力学代写Thermodynamics代考|Fluids with Electromagnetic Fields

Let us include electromagnetic fields [13]. Here we assume that there is no net electric charge $^{34}$ and refer to the electric field, the electric current density, the velocity of light in vacuum, the vacuum magnetic permittivity and the vacum dielectric constant as to

$\mathbf{E}, \mathbf{j}{e l}, c=3 \cdot 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}, \mu_0=4 \cdot \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{A}^{-1} \cdot \mathrm{m}$ and $\varepsilon_0=\left(\mu_0 c^2\right)^{-1}$ respectively. Lorenz force density $\mathbf{j}{e l} \wedge \mathbf{B}$ adds to the R.H.S. of Navier-Stokes’ equation. Then, we replace ${ }^{35}$ :
$$\mathbf{v} \cdot \rho \frac{\partial \mathbf{v}}{\partial t} \rightarrow \mathbf{v} \cdot \rho \frac{\partial \mathbf{v}}{\partial t}+\mathbf{v} \cdot \mathbf{j}{e l} \wedge \mathbf{B} \quad ; \quad-\mathbf{j}{e l} \wedge \mathbf{B} \rightarrow-\mathbf{j}{e l} \cdot(\mathbf{E}+\mathbf{v} \wedge \mathbf{B})+\mathbf{j}{e l} \cdot \mathbf{E}$$
where 2 of the 4 Maxwell’s equations of electromagnetism, namely $\nabla \wedge \mathbf{E}+$ $\frac{\partial \mathbf{B}}{\partial t}=0$ and $\nabla \wedge \mathbf{B}-\frac{1}{c^2} \frac{\partial \mathbf{E}}{\partial t}-\mu_0 \mathbf{j}{e l}=0$, lead to ${ }^{36} \mathbf{j}{e l} \cdot \mathbf{E}=-\frac{1}{\mu_0} \nabla \cdot(\mathbf{E} \wedge \mathbf{B})-$ $\frac{\partial}{\partial t}\left(\frac{1}{2 \mu_0}|\mathbf{B}|^2+\frac{1}{2} \varepsilon_0|\mathbf{E}|^2\right)$. Together with these substitutions, our discussion leads to the conclusion that for viscous fluids with electromagnetic fields the mass balance, the momentum balance and the approximation of LTE at all times give the following balances of energy and entropy:
$$\begin{gathered} \frac{\partial}{\partial t}\left(\rho \frac{|\mathbf{v}|^2}{2}+\rho u+\frac{1}{2 \mu_0}|\mathbf{B}|^2+\frac{1}{2} \varepsilon_0|\mathbf{E}|^2\right)+\nabla \cdot\left[\rho \mathbf{v}\left(\frac{|\mathbf{v}|^2}{2}+h\right)-\mathbf{v} \cdot \boldsymbol{\sigma}^{\prime}+\frac{\mathbf{E} \wedge \mathbf{B}}{\mu_0}\right]=0 \ \rho \frac{d s}{d t}=\frac{\sigma_{i k}^{\prime}}{T} \frac{\partial v_i}{\partial x_k}+\frac{\mathbf{j}{e l} \cdot(\mathbf{E}+\mathbf{v} \wedge \mathbf{B})}{T} \end{gathered}$$ Joule heating is irreversible for $\frac{\mathbf{j}{e l} \cdot(\mathbf{E}+\mathbf{v} \wedge \mathbf{B})}{T}>0$, with $\mathbf{j}_{e l} \cdot(\mathbf{E}+\mathbf{v} \wedge \mathbf{B})$ Joule power density. Both the balance of energy and the balance of entropy are affected.

物理代写|热力学代写Thermodynamics代考|Joule-Thomson Throttled Expansion

$$\frac{\partial \rho}{\partial t}+\nabla \cdot(\rho \mathbf{v})=0$$

$$\frac{\partial}{\partial t}\left(\rho \frac{|\mathbf{v}|^2}{2}+\rho u\right)+\nabla \cdot\left[\rho \mathbf{v}\left(\frac{|\mathbf{v}|^2}{2}+h\right)-\mathbf{v} \cdot \boldsymbol{\sigma}^{\prime}\right]=0$$

$$\frac{\partial}{\partial t}(\rho u)+\rho \mathbf{v} \cdot \nabla h-h \frac{\partial \rho}{\partial t}=0$$

$$\frac{\partial}{\partial t}(\rho u)=\Gamma \frac{\partial n_m}{\partial t}-n_m \mathbf{v} \cdot \nabla \Gamma$$

${ }^{32}: \Gamma=\left[\frac{\partial\left(n_m\right)}{\partial t}\right] T_2=$ const. $^{-1}\left[\frac{\partial(\rho u)}{\partial t}\right] T_2=$ const. 作为 $\mathbf{v}$ 是小。此外，由于 $T_2$ 是但定的（在 $T$ 由外部世界人为维

$$\oint d \mathbf{a} \cdot \mathbf{v} n_m \Gamma=0$$

物理代写|热力学代写Thermodynamics代考|Fluids with Electromagnetic Fields

$\mathbf{E}, \mathbf{j} e l, c=3 \cdot 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}, \mu_0=4 \cdot \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{A}^{-1} \cdot \mathrm{m}$ 和 $\varepsilon_0=\left(\mu_0 c^2\right)^{-1}$ 分别。洛伦兹力密度j $\mathbf{e l} \wedge \mathbf{B}$ 添加到 Navier $-$ Stokes 方程的 RHS。然后，我们蕏换 ${ }^{35}$ :
$$\mathbf{v} \cdot \rho \frac{\partial \mathbf{v}}{\partial t} \rightarrow \mathbf{v} \cdot \rho \frac{\partial \mathbf{v}}{\partial t}+\mathbf{v} \cdot \mathbf{j} e l \wedge \mathbf{B} \quad ; \quad-\mathbf{j} e l \wedge \mathbf{B} \rightarrow-\mathbf{j} e l \cdot(\mathbf{E}+\mathbf{v} \wedge \mathbf{B})+\mathbf{j} e l \cdot \mathbf{E}$$

${ }^{36} \mathbf{j} e l \cdot \mathbf{E}=-\frac{1}{\mu_0} \nabla \cdot(\mathbf{E} \wedge \mathbf{B})-\frac{\partial}{\partial t}\left(\frac{1}{2 \mu_0}|\mathbf{B}|^2+\frac{1}{2} \varepsilon_0|\mathbf{E}|^2\right)$. 连同这政替换，我们的讨论得出的结论是，对于具有电磁场的 粘性㧧体，质量平衡、动量平衡和 LTE 的近似值始絡给出以下能量和嫡的平衡:
$$\frac{\partial}{\partial t}\left(\rho \frac{|\mathbf{v}|^2}{2}+\rho u+\frac{1}{2 \mu_0}|\mathbf{B}|^2+\frac{1}{2} \varepsilon_0|\mathbf{E}|^2\right)+\nabla \cdot\left[\rho \mathbf{v}\left(\frac{|\mathbf{v}|^2}{2}+h\right)-\mathbf{v} \cdot \boldsymbol{\sigma}^{\prime}+\frac{\mathbf{E} \wedge \mathbf{B}}{\mu_0}\right]=0 \rho \frac{d s}{d t}=\frac{\sigma_{i k}^{\prime}}{T} \frac{\partial v_i}{\partial x_k}+\frac{\mathbf{j} e l \cdot(\mathbf{E}+\mathbf{v} \wedge \mathbf{B})}{T}$$

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