### 物理代写|流体力学代写Fluid Mechanics代考|Fundamental Laws of Continuum

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## 物理代写|流体力学代写Fluid Mechanics代考|Conservation of Mass, Equation of Continuity

Conservation of mass has already been postulated in the last chapter, and now we will make use of our earlier results and employ (1.83) and (1.93) to change the conservation law (1.85) to the form
$$\frac{\mathrm{D}}{\mathrm{D} t} \iiint_{(V(t))} \varrho \mathrm{d} V=\iiint_{(V)}\left[\frac{\partial \varrho}{\partial t}+\frac{\partial}{\partial x_{i}}\left(\varrho u_{i}\right)\right] \mathrm{d} V=0$$
This equation holds for every volume that could be occupied by the fluid, that is, for arbitrary choice of the integration region $(V)$. We could therefore shrink the integration region to a point, and we conclude that the continuous integrand must itself vanish at every $\vec{x}$. Thus we are led to the local or differential form of the law of conservation of mass
$$\frac{\partial \varrho}{\partial t}+\frac{\partial}{\partial x_{i}}\left(\varrho u_{i}\right)=0$$
This is the continuity equation. If we use the material derivative (1.20) we obtain
$$\frac{\mathrm{D} \varrho}{\mathrm{D} t}+\varrho \frac{\partial u_{i}}{\partial x_{i}}=0$$
or written symbolically
$$\frac{\mathrm{D} \varrho}{\mathrm{D} t}+\varrho \nabla \cdot \vec{u}=0$$

## 物理代写|流体力学代写Fluid Mechanics代考|Balance of Momentum

As the first law (axiom) of classical mechanics, accepted to be true without proof but embracing our experience, we state the momentum balance: in an inertial frame the rate of change of the momentum of a body is balanced by the force applied on this body
$$\frac{\mathrm{D} \vec{P}}{\mathrm{D} t}=\vec{F}$$
What follows now only amounts to rearranging this axiom explicitly. The body is still a part of the fluid which always consists of the same material points. Analogous to ( $1.83)$, we calculate the momentum of the body as the integral over the region occupied by the body
$$\vec{P}=\iiint_{(V(t))} \varrho \vec{u} \mathrm{~d} V$$
The forces affecting the body basically fall into two classes, body forces, and surface or contact forces. Body forces are forces with a long range of influence which act on all the material particles in the body and which, as a rule, have their source in fields of force. The most important example we come across is the earth’s gravity field. The gravitational field strength $\vec{g}$ acts on every molecule in the fluid particle, and the sum of all the forces acting on the particle represents the actual gravitational force

$$\Delta \vec{F}=\vec{g} \sum_{i} m_{i}=\vec{g} \Delta m$$
The force of gravity is therefore proportional to the mass of the fluid particle. As before, in the framework of the continuum hypothesis, we consider the body force as a continuous function of mass or volume and call
$$\vec{k}=\lim {\Delta m \rightarrow 0} \frac{\Delta \vec{F}}{\Delta m}$$ the mass body force; in the special case of the earth’s gravitational field $\vec{k}=\vec{g}$, we call it the gravitational force. The volume body force is the force referred to the volume, thus $$\vec{f}=\lim {\Delta V \rightarrow 0} \frac{\Delta \vec{F}}{\Delta V}$$
(cf. Fig. 2.1), and in the special case of the gravitational force we get
$$\vec{f}=\lim _{\Delta V \rightarrow 0} \vec{g} \frac{\Delta m}{\Delta V}=\vec{g} \varrho .$$

## 物理代写|流体力学代写Fluid Mechanics代考|Balance of Angular Momentum

As the second general axiom of classical mechanics we shall discuss the angular momentum balance. This is independent of the balance of linear momentum. In an inertial frame, the rate of change of the angular momentum is equal to the moment of the external forces acting on the body
$$\frac{\mathrm{D}}{\mathrm{D} t}(\vec{L})=\vec{M} .$$
We calculate the angular momentum $\vec{L}$ as the integral over the region occupied by the fluid body
$$\vec{L}=\iiint_{(V(t))} \vec{x} \times(\varrho \vec{u}) \mathrm{d} V$$
The angular momentum in $(2.45)$ is taken about the origin such that the position vector is $\vec{x}$, and so we must use the same reference point to calculate the moment of the applied forces
$$\vec{M}=\iiint_{(V(t))} \vec{x} \times(\varrho \vec{k}) \mathrm{d} V+\iint_{(S(t))} \vec{x} \times \vec{t} \mathrm{~d} S$$

recalling, however, that the choice of reference point is up to us. Therefore the law of angular momentum takes the form
$$\frac{\mathrm{D}}{\mathrm{D} t} \iiint_{(V(t))} \vec{x} \times(\varrho \vec{u}) \mathrm{d} V=\iiint_{(V(t))} \vec{x} \times(\varrho \vec{k}) \mathrm{d} V+\iint_{(S)} \vec{x} \times \vec{t} \mathrm{~d} S$$
where, for the same reasons as before, we have replaced the time varying domain of integration on the right with a fixed domain. Now we wish to show that the differential form of the balance of angular momentum implies the symmetry of the stress tensor. We introduce the expression (2.29a, 2.29b) into the surface integral, which can then be written as a volume integral. In index notation this becomes
$$\iint_{(S)} \epsilon_{i j k} x_{j} \tau_{l k} n_{l} \mathrm{~d} S=\iiint_{(V)} \epsilon_{i j k} \frac{\partial}{\partial x_{l}}\left(x_{j} \tau_{l k}\right) \mathrm{d} V$$
and after applying (1.88) to the left-hand side of (2.47) we get first
$$\iiint_{(V)} \epsilon_{i j k}\left(\varrho \frac{\mathrm{D}}{\mathrm{D} t}\left(x_{j} u_{k}\right)-\frac{\partial}{\partial x_{l}}\left(x_{j} \tau_{l k}\right)-x_{j} \varrho k_{k}\right) \mathrm{d} V=0$$
and after differentiation and combining terms
$$\iiint_{(V)}\left[\epsilon_{i j k} x_{j}\left(\varrho \frac{\mathrm{D} u_{k}}{\mathrm{D} t}-\frac{\partial \tau_{l k}}{\partial x_{l}}-\varrho k_{k}\right)+\varrho \epsilon_{i j k} u_{j} u_{k}-\epsilon_{i j k} \tau_{j k}\right] \mathrm{d} V=0$$

## 物理代写|流体力学代写Fluid Mechanics代考|Conservation of Mass, Equation of Continuity

DD吨∭(在(吨))ϱd在=∭(在)[∂ϱ∂吨+∂∂X一世(ϱ在一世)]d在=0

∂ϱ∂吨+∂∂X一世(ϱ在一世)=0

DϱD吨+ϱ∂在一世∂X一世=0

DϱD吨+ϱ∇⋅在→=0

## 物理代写|流体力学代写Fluid Mechanics代考|Balance of Momentum

D磷→D吨=F→

ΔF→=G→∑一世米一世=G→Δ米

ķ→=林Δ米→0ΔF→Δ米质量体力；在地球引力场的特殊情况下ķ→=G→，我们称之为万有引力。体积体力是指体积的力，因此

F→=林Δ在→0ΔF→Δ在
（参见图 2.1），在引力的特殊情况下，我们得到

F→=林Δ在→0G→Δ米Δ在=G→ϱ.

## 物理代写|流体力学代写Fluid Mechanics代考|Balance of Angular Momentum

DD吨(大号→)=米→.

DD吨∭(在(吨))X→×(ϱ在→)d在=∭(在(吨))X→×(ϱķ→)d在+∬(小号)X→×吨→ d小号

∬(小号)ε一世jķXjτlķnl d小号=∭(在)ε一世jķ∂∂Xl(Xjτlķ)d在

∭(在)ε一世jķ(ϱDD吨(Xj在ķ)−∂∂Xl(Xjτlķ)−Xjϱķķ)d在=0

∭(在)[ε一世jķXj(ϱD在ķD吨−∂τlķ∂Xl−ϱķķ)+ϱε一世jķ在j在ķ−ε一世jķτjķ]d在=0

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