# 物理代写|热力学代写Thermodynamics代考|ENME485 What if Joule Heating is Negligible?

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## 物理代写|热力学代写Thermodynamics代考|What if Joule Heating is Negligible?

The Joule power density $T \sigma=\frac{\left|\mathbf{j}{e l}\right|^2}{\sigma{\Omega}}$ becomes negligible either if $\mathbf{j}{e l} \rightarrow 0$ or if $\sigma{\Omega} \rightarrow$ $\infty$, i.e. whenever Ohm’s law $\mathbf{E}+\mathbf{v} \wedge \mathbf{B}=\frac{\mathbf{j}{e l}}{\sigma{\Omega}}$ reduces to
$$\mathbf{E}+\mathbf{v} \wedge \mathbf{B}=0$$
i.e. the electric field vanishes in the local frame of reference moving at speed $\mathbf{v}$. We refer to this relationship as ‘Ohm’s law for negligible Joule heating’ below. Let us take the curl of both sides of this relationship. Maxwell’s equation $\nabla \wedge \mathbf{E}=-\frac{\partial \mathbf{B}}{\partial t}$ leads, therefore, to
$$\frac{\partial \mathbf{B}}{\partial t}-\nabla \wedge(\mathbf{v} \wedge \mathbf{B})=0$$
which is to be solved together with the usual condition $\nabla \cdot \mathbf{B}=0$. The latter two relationships are identical to the relationships $\frac{\partial(\nabla \wedge \mathbf{v})}{\partial t}-\nabla \wedge[\mathbf{v} \wedge(\nabla \wedge \mathbf{v})]=0$ and $\nabla \cdot(\nabla \wedge \mathbf{v})=0$ satisfied by the vorticity $\nabla \wedge \mathbf{v}$ in an inviscid, unmagnetized fluid. ${ }^{18}$ It is well known that these equations for vorticity lead to well-known Kelvin’s theorem of circuitation, i.e. the circuitation $\int_\gamma \mathbf{v} \cdot d \mathbf{l}$ of $\mathbf{v}$ along any closed line $\gamma$ in the fluid is constant. If $\gamma$ encloses a closed surface $\Lambda$, then Stokes’ theorem ensures the circuitation to be equal to the flux of vorticity across $\Lambda: \int_\gamma \mathbf{v} \cdot d \mathbf{l}=\oint_{\Lambda}(\nabla \wedge \mathbf{v}) \cdot d \mathbf{a} .^{19}$ The analogy displayed above between the equations for vorticity and the equations for the magnetic field imply that the magnetic flux across a closed surface in the fluid is constant: $\oint_{\Lambda} \mathbf{B} \cdot d \mathbf{a}=$ const.

Moreover, if the fluid is incompressible $(\nabla \cdot \mathbf{v}=0)$ then $\frac{\partial \mathbf{B}}{\partial t}-\nabla \wedge(\mathbf{v} \wedge \mathbf{B})=0$ gives $^{20}$ :
$$\frac{d \mathbf{B}}{d t}=(\mathbf{B} \cdot \nabla) \mathbf{v}$$
We are going to proof that a solution of this equation satisfies the so-called ‘frozenfield-lines’ condition ( $\alpha_B$ suitably defined scalar quantity with $\nabla \alpha_B=0$ ):
$$\mathbf{B}=\alpha_B \cdot \mathbf{v}$$

## 物理代写|热力学代写Thermodynamics代考|Viscosity: Korteweg-Helmholtz’ Principle

The entropy production density due to viscosity is $(i, k=1,2,3)$
$$\sigma=\frac{\sigma_{i k}^{\prime}}{T} \frac{\partial v_i}{\partial x_k}$$
(we refer to $[6,16,25,26])$. Following [16] , here we assume that both $\eta$ and $\zeta$ do not depend on the stresses applied to the fluid. This is the case of ‘Newtonian’ fluids. ${ }^{22}$ It can be shown that $\sigma_{i k}^{\prime}=\eta\left(\frac{\partial v_i}{\partial x_k}+\frac{\partial v_k}{\partial x_i}-\frac{2}{3} \delta_{i k} \nabla \cdot \mathbf{v}\right)+\zeta \delta_{i k} \nabla \cdot \mathbf{v}$ and that the viscous heating power density is [26]: $T \sigma=\frac{\eta}{2}\left(\frac{\partial v_i}{\partial x_k}+\frac{\partial v_k}{\partial x_i}-\frac{2}{3} \delta_{i k} \nabla \cdot \mathbf{v}\right)^2+\zeta(\nabla \cdot \mathbf{v})^2=$ $\left(\frac{4}{3} \eta+\zeta\right)(\nabla \cdot \mathbf{v})^2+\eta|\nabla \wedge \mathbf{v}|^2+2 \nabla \cdot[(\mathbf{v} \cdot \nabla) \mathbf{v}-\mathbf{v}(\nabla \cdot \mathbf{v})]$. The coefficients $\eta$ and $\zeta$ are usually referred to as ‘dynamic viscosity’ and ‘bulk viscosity’, respectively in the literature.

Since viscous heating is an irreversible process, $\sigma>0$, hence both $\eta>0$ and $\zeta>$ 0 . The viscous term in the equation of motion is $\frac{\partial \sigma_{i k}^{\prime}}{\partial x_k}=\Delta v_i+\left(\zeta+\frac{\eta}{3}\right) \frac{\partial}{\partial x_i}(\nabla \cdot \mathbf{v})+$ $O(\nabla \eta, \nabla \zeta)$. If $\nabla \eta=0$ and $\nabla \zeta=0$ then Navier-Stokes’ equation with no electromagnetic fields reads:
$$\rho \frac{\partial \mathbf{v}}{\partial t}+\rho(\mathbf{v} \cdot \nabla) \mathbf{v}=-\nabla p+\eta \Delta \mathbf{v}+\left(\zeta+\frac{\eta}{3}\right) \nabla(\nabla \cdot \mathbf{v})$$
If the fluid is incompressible $(\nabla \cdot \mathbf{v}=0)$ then Navier-Stokes’ equation reduces to:
$$\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \nabla) \mathbf{v}=-\frac{\nabla p}{\rho}+\nu \Delta \mathbf{v}$$
where the quantity $\nu \equiv \eta$ is called ‘kinematic viscosity’.

## 物理代写|热力学代写THERMODYNAMICS代考|WHAT IF JOULE HEATING IS NEGLIGIBLE?

$$\mathbf{E}+\mathbf{v} \wedge \mathbf{B}=0$$

$$\frac{\partial \mathbf{B}}{\partial t}-\nabla \wedge(\mathbf{v} \wedge \mathbf{B})=0$$

$$\frac{d \mathbf{B}}{d t}=(\mathbf{B} \cdot \nabla) \mathbf{v}$$

\mathbf{B}=\alpha_B \cdot \mathbf{v}
$$## 物理代写|热力学代写THERMODYNAMICS代考|VISCOSITY: KORTEWEG-HELMHOLTZ’ PRINCIPLE 由粘度引起的熵产生密度为 (i, k=1,2,3)$$
\sigma=\frac{\sigma_{i k}^{\prime}}{T} \frac{\partial v_i}{\partial x_k}
$$wereferto \[6,16,25,26]. Following [16], hereweassumethatboth 和and \backslash zeta and thattheviscousheatingpowerdensityis [26] :T \sigma =\langle fracfleta}{2}}left |wedge \backslash mathbf \left.{v}\right|^{\wedge} 2+2 \backslash draw \backslash cdot$$
(\mathbf{v} \cdot \nabla) \mathbf{v}-\mathbf{v}(\nabla \cdot \mathbf{v})
$$.Thecoefficients 和 and \backslash zetas 在文献中通常分别称为“动态粘度”和“体积粘度”。 由于粘性加热是一个不可逆的过程， \sigma>0, 因此两者 \eta>0 和 \zeta>0. 运动方程中的粘性项是 \frac{\partial \sigma_{1 k}^{\prime}}{\partial x_k}=\Delta v_i+\left(\zeta+\frac{\eta}{3}\right) \frac{\partial}{\partial x_i}(\nabla \cdot \mathbf{v})+O(\nabla \eta, \nabla \zeta). 如果 \nabla \eta=0 和 \nabla \zeta=0 那么没有电磁场的 Navier-Stokes 方程为:$$
\rho \frac{\partial \mathbf{v}}{\partial t}+\rho(\mathbf{v} \cdot \nabla) \mathbf{v}=-\nabla p+\eta \Delta \mathbf{v}+\left(\zeta+\frac{\eta}{3}\right) \nabla(\nabla \cdot \mathbf{v})
$$如果流体不可压缩 (\nabla \cdot \mathbf{v}=0) 然后 Navier-Stokes 方程简化为:$$
\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v} \cdot \nabla) \mathbf{v}=-\frac{\nabla p}{\rho}+\nu \Delta \mathbf{v}


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