2.2 Conservation Laws 守恒定律[cfd-2-2]

2.2.1 The Continuity Equation 连续方程[cfd-2-2-1]

如果把注意力限定在单相流体上，质量守恒定律表达的是这样一个事实：在这样的流体系统中，质量既不能被创造，也不能消失。连续方程中也没有扩散通量的贡献，因为对静止流体而言，质量的任何变化都意味着流体粒子的位移。en

If we restrict our attention to single-phase fluids, the law of mass conservation expresses the fact that mass cannot be created in such a fluid system, nor it can disappear. There is also no diffusive flux contribution to the continuity equation, since for a fluid at rest, any variation of mass would imply a displacement of the fluid particles.

为了导出连续方程，考虑如图2.1所示的固定在空间中的有限控制体模型。在控制面上的某一点处，流动速度为 $\vec{v}$ ，单位法向量为 $\vec{n}$ ， $$dS$$ 表示一个微元面积。此情形下的守恒量是密度 $\rho$ 。对有限体积 $\Omega$ 内部总质量的时间变化率，我们有en

In order to derive the continuity equation, consider the model of a finite control volume fixed in space, as sketched in Fig. 2.1. At a point on the control surface, the flow velocity is $\vec{v}$ , the unit normal vector is $\vec{n}$ , and $$dS$$ denotes an elemental surface area. The conserved quantity in this case is the density $\rho$ . For the time rate of change of the total mass inside the finite volume $\Omega$ we have

\frac{\partial}{\partial t}\int_{\Omega}\rho\,d\Omega. \tag{1}

流体通过某个固定在空间中的表面的质量流量，等于（密度）×（表面面积）×（垂直于表面的速度分量）的乘积。因此，对流通过每个面元 $$dS$$ 的贡献便为en

The mass flow of a fluid through some surface fixed in space equals to the product of (density) × (surface area) × (velocity component perpendicular to the surface). Therefore, the contribution from the convective flux across each surface element $$dS$$ becomes

\rho\left(\vec{v}\cdot\vec{n}\right)dS. \tag{2}

由于在对流中 $\vec{n}$ 总是指向控制体外，当乘积 $\left(\vec{v}\cdot\vec{n}\right)$ 为负时我们称之为流入（inflow），为正时则称为流出（outflow），此时质量离开控制体。en

Since by convection $\vec{n}$ always points out of the control volume, we speak of inflow if the product $\left(\vec{v}\cdot\vec{n}\right)$ is negative, and of outflow if it is positive and hence the mass leaves the control volume.

如上所述，此时不存在任何体积源或面源。于是，考虑到方程(2.1)的一般形式，我们可以写出en

As stated above, there are no volume or surface sources present. Thus, by taking into account the general formulation of Eq. (2.1), we can write

\frac{\partial}{\partial t}\int_{\Omega}\rho\,d\Omega + \oint_{\partial\Omega}\rho\left(\vec{v}\cdot\vec{n}\right)dS = 0. \tag{2.3}

这就是连续方程的积分形式——质量守恒定律。en

This represents the integral form of the continuity equation - the conservation law of mass.

2.2.2 The Momentum Equation 动量方程[cfd-2-2-2]

动量方程的推导可以从牛顿第二定律的一种特定形式入手，该定律指出，动量的变化是由作用在质量微元上的净力引起的。对控制体 $\Omega$ 中无限小部分（见图2.1）的动量，我们有en

We may start the derivation of the momentum equation by recalling the particular form of Newton's second law which states that the variation of momentum is caused by the net force acting on an mass element. For the momentum of an infinitesimally small portion of the control volume $\Omega$ (see Fig. 2.1) we have

\rho\vec{v}\,d\Omega. \tag{1}

控制体内动量随时间的变化等于en

The variation in time of momentum within the control volume equals

\frac{\partial}{\partial t}\int_{\Omega}\rho\vec{v}\,d\Omega. \tag{2}

因此，这里的守恒量是密度与速度的乘积，即en

Hence, the conserved quantity is here the product of the density and the velocity, i.e.,

\rho\vec{v} = \left[\,\rho u,\ \rho v,\ \rho w\,\right]^{T}. \tag{3}

描述动量越过控制体边界输运的对流通量张量，在笛卡尔坐标系中由以下三个分量组成en

The convective flux tensor, which describes the transfer of momentum across the boundary of the control volume, consists in the Cartesian coordinate system of the following three components

\begin{aligned} x\text{-component}&:\quad \rho u\,\vec{v}\\ y\text{-component}&:\quad \rho v\,\vec{v}\\ z\text{-component}&:\quad \rho w\,\vec{v}. \end{aligned} \tag{4}

对流通量张量对动量守恒的贡献则由下式给出en

The contribution of the convective flux tensor to the conservation of momentum is then given by

-\oint_{\partial\Omega}\rho\vec{v}\left(\vec{v}\cdot\vec{n}\right)dS. \tag{5}

扩散通量为零，因为对静止流体而言不可能存在动量的扩散。于是，剩下的问题是：流体微元受到哪些力的作用？我们可以辨识出作用在控制体上的两类力：

外部体积力或体力（body forces），直接作用在体积的质量上。例如重力、浮力、科里奥利力或离心力。在某些情形下，还可能存在电磁力。
表面力（surface forces），直接作用在控制体的表面上。它们仅来自两个来源：
- 由包围该体积的外部流体施加的压力分布；
- 由流体与体积表面之间的摩擦产生的切向应力和法向应力。

The diffusive flux is zero since there is no diffusion of momentum possible for a fluid at rest. Thus, the remaining question is now, what are the forces the fluid element is exposed to? We can identify two kinds of forces acting on the control volume:

External volume or body forces, which act directly on the mass of the volume. These are for example gravitational, buoyancy, Coriolis or centrifugal forces. In some cases, there can be electromagnetic forces present as well.
Surface forces, which act directly on the surface of the control volume. They result from only two sources:
- the pressure distribution, imposed by the outside fluid surrounding the volume,
- the shear and normal stresses, resulting from the friction between the fluid and the surface of the volume.

由此可以看出，单位体积上的体力（下面记作 $\rho\vec{f}_e$ ）对应于方程(2.2)中的体积源。因此，体力（外力）对动量守恒的贡献为en

From the above, we can see that the body force per unit volume, further denoted as $\rho\vec{f}_e$ , corresponds to the volume sources in Eq. (2.2). Thus, the contribution of the body (external) force to the momentum conservation is

\int_{\Omega}\rho\vec{f}_e\,d\Omega. \tag{6}

面源则由两部分组成——各向同性的压力分量和黏性应力（viscous stress）张量 $\overline{\overline{\tau}}$ ，即en

The surface sources consist then of two parts - of an isotropic pressure component and of a viscous stress tensor $\overline{\overline{\tau}}$ , i.e.,

\vec{Q}_S = -p\,\overline{\overline{I}} + \overline{\overline{\tau}} \tag{2.4}

其中 $\overline{\overline{I}}$ 为单位张量（关于张量可参见例如[2]）。面源对控制体的作用示于图2.2。在2.3节中，我们将更详细地阐述应力张量的形式，特别是说明法向应力和切向应力与流动速度之间的联系。en

with $\overline{\overline{I}}$ being the unit tensor (for tensors see, e.g., [2]). The effect of the surface sources on the control volume is sketched in Fig. 2.2. In Section 2.3, we shall elaborate the form of the stress tensor in more detail, and in particular show how the normal and the shear stresses are connected to the flow velocity.

图2.2:作用在控制体表面元上的表面力

图2.2:作用在控制体表面元上的表面力。图例： $\overline{\overline{\tau}}\cdot\vec{n}\,dS$ ——黏性应力； $p\vec{n}\,dS$ ——压力； $$dS$$ ——面元； $\Omega$ ——控制体。

于是，如果按照一般守恒定律（方程(2.2)）把上述所有贡献求和，我们最终得到表达式en

Hence, if we now sum up all the above contributions according to the general conservation law (Eq. (2.2)), we finally obtain the expression

\begin{aligned} \frac{\partial}{\partial t}\int_{\Omega}\rho\vec{v}\,d\Omega + \oint_{\partial\Omega}\rho\vec{v}\left(\vec{v}\cdot\vec{n}\right)dS\\ \quad = \int_{\Omega}\rho\vec{f}_e\,d\Omega - \oint_{\partial\Omega}p\,\vec{n}\,dS + \oint_{\partial\Omega}\left(\overline{\overline{\tau}}\cdot\vec{n}\right)dS \end{aligned} \tag{2.5}

即固定在空间中的任意控制体 $\Omega$ 内的动量守恒。en

for the momentum conservation inside an arbitrary control volume $\Omega$ which is fixed in space.

2.2.3 The Energy Equation 能量方程[cfd-2-2-3]

我们在推导能量方程时所依据的基本原理是热力学第一定律。把它应用于图2.1所示的控制体，可以表述为：体积内总能量随时间的任何变化，都是由作用在该体积上的力的做功速率以及流入该体积的净热流引起的。流体单位质量的总能量 $$E$$ ，由其单位质量内能 $$e$$ 加上单位质量动能 $|\vec{v}|^{2}/2$ 而得到。因此，总能量可以写为en

The underlying principle that we will apply in the derivation of the energy equation, is the first law of thermodynamics. Applied to the control volume displayed in Fig. 2.1, it states that any changes in time of the total energy inside the volume are caused by the rate of work of forces acting on the volume and by the net heat flux into it. The total energy per unit mass $$E$$ of a fluid is obtained by adding its internal energy per unit mass, $$e$$ , to its kinetic energy per unit mass $|\vec{v}|^{2}/2$ . Thus, we can write for the total energy

E = e + \frac{|\vec{v}|^{2}}{2} = e + \frac{u^{2}+v^{2}+w^{2}}{2}. \tag{2.6}

此情形下的守恒量是单位体积的总能量，即 $\rho E$ 。它在体积 $\Omega$ 内随时间的变化可表示为en

The conserved quantity is in this case the total energy per unit volume, i.e., $\rho E$ . Its variation in time within the volume $\Omega$ can be expressed as

\frac{\partial}{\partial t}\int_{\Omega}\rho E\,d\Omega. \tag{2}

按照推导一般守恒定律（方程(2.1)）时的讨论，我们可以直接写出对流通量的贡献为en

Following the discussion in course of the derivation of the general conservation law (Eq. (2.1)), we can readily specify the contribution of the convective flux as

-\oint_{\partial\Omega}\rho E\left(\vec{v}\cdot\vec{n}\right)dS. \tag{3}

与连续方程和动量方程不同，这里出现了扩散通量。如前所述，它与单位质量守恒量的梯度成正比（Fick定律）。由于扩散通量 $\vec{F}_D$ 是针对静止流体定义的，只有内能起作用，于是得到en

In contrast to the continuity and the momentum equation, there is now a diffusive flux. As we already stated, it is proportional to the gradient of the conserved quantity per unit mass (Fick's law). Since the diffusive flux $\vec{F}_D$ is defined for a fluid at rest, only the internal energy becomes effective and we obtain

\vec{F}_D = -\gamma\rho\kappa\,\nabla e. \tag{2.7}

上式中， $\gamma = c_p/c_v$ 为比热系数之比， $\kappa$ 为热扩散率系数（thermal diffusivity coefficient）。扩散通量代表进入控制体热流的一部分，即由分子热传导引起的热扩散——由温度梯度导致的传热。因此，方程(2.7)通常写成Fourier热传导定律的形式，即en

In the above, $\gamma = c_p/c_v$ is the ratio of specific heat coefficients, and $\kappa$ denotes the thermal diffusivity coefficient. The diffusion flux represents one part of the heat flux into the control volume, namely the diffusion of heat due to molecular thermal conduction - heat transfer due to temperature gradients. Therefore, Equation (2.7) is in general written in the form of Fourier's law of heat conduction, i.e.,

\vec{F}_D = -k\,\nabla T \tag{2.8}

其中 $$k$$ 为热导率系数（thermal conductivity coefficient）， $$T$$ 为绝对静温。en

with $$k$$ standing for the thermal conductivity coefficient and $$T$$ for the absolute static temperature.

进入有限控制体的净热流的另一部分，是由辐射的吸收或发射、或化学反应引起的体积加热。我们把热源——单位质量传热的时间变化率——记作 $\dot{q}_h$ 。它与为动量方程引入的体力 $\vec{f}_e$ 的做功速率一起，构成完整的体积源en

The other part of the net heat flux into the finite control volume consists of volumetric heating due to the absorption or emission of radiation, or due to chemical reactions. We will denote the heat sources - the time rate of heat transfer per unit mass - as $\dot{q}_h$ . Together with the rate of work done by the body forces $\vec{f}_e$ , which we have introduced for the momentum equation, it completes the volume sources

Q_V = \rho\,\vec{f}_e\cdot\vec{v} + \dot{q}_h. \tag{2.9}

能量守恒中尚待确定的最后一项贡献是面源 $$Q_S$$ 。它对应于压力以及切向和法向应力对流体微元做功的时间变化率（见图2.2），即en

The last contribution to the conservation of energy, which we have yet to determine, are the surface sources $$Q_S$$ . They correspond to the time rate of work done by the pressure as well as the shear and normal stresses on the fluid element (see Fig. 2.2), i.e.,

\vec{Q}_S = -p\,\vec{v} + \overline{\overline{\tau}}\cdot\vec{v}. \tag{2.10}

把上述所有贡献和各项整理起来，便得到能量守恒方程的表达式en

Sorting now all the above contributions and terms, we obtain for the energy conservation equation the expression

\begin{aligned} \frac{\partial}{\partial t}\int_{\Omega}\rho E\,d\Omega + \oint_{\partial\Omega}\rho E\left(\vec{v}\cdot\vec{n}\right)dS = \oint_{\partial\Omega}k\left(\nabla T\cdot\vec{n}\right)dS\\ + \int_{\Omega}\left(\rho\vec{f}_e\cdot\vec{v} + \dot{q}_h\right)d\Omega - \oint_{\partial\Omega}p\left(\vec{v}\cdot\vec{n}\right)dS + \oint_{\partial\Omega}\left(\overline{\overline{\tau}}\cdot\vec{v}\right)\cdot\vec{n}\,dS. \end{aligned} \tag{2.11}

能量方程(2.11)通常写成略有不同的形式。为此，我们将利用总焓、总能量与压力之间的如下一般关系en

The energy equation (2.11) is usually written in a slightly different form. For that purpose, we will utilise the following general relation between the total enthalpy, the total energy and the pressure

H = h + \frac{|\vec{v}|^{2}}{2} = E + \frac{p}{\rho}. \tag{2.12}

当我们在能量守恒定律(2.11)中把对流项（ $\rho E\vec{v}$ ）与压力项（ $p\vec{v}$ ）合并，并应用公式(2.12)时，最终可把能量方程写成en

When we now gather the convective ( $\rho E\vec{v}$ ) and the pressure term ( $p\vec{v}$ ) in the energy conservation law (2.11), and apply the formula (2.12), we can finally write the energy equation in the form

\begin{aligned} \frac{\partial}{\partial t}\int_{\Omega}\rho E\,d\Omega + \oint_{\partial\Omega}\rho H\left(\vec{v}\cdot\vec{n}\right)dS = \oint_{\partial\Omega}k\left(\nabla T\cdot\vec{n}\right)dS\\ + \int_{\Omega}\left(\rho\vec{f}_e\cdot\vec{v} + \dot{q}_h\right)d\Omega + \oint_{\partial\Omega}\left(\overline{\overline{\tau}}\cdot\vec{v}\right)\cdot\vec{n}\,dS. \end{aligned} \tag{2.13}

至此，我们已经导出了三个守恒定律的积分形式：质量守恒(2.3)、动量守恒(2.5)和能量守恒(2.13)。下一节中，我们将更详细地给出法向应力和切向应力的表述。en

Herewith, we have derived integral formulations of the three conservation laws: the conservation of mass (2.3), of momentum (2.5), and of energy (2.13). In the next section, we shall work out the formulation of the normal and the shear stresses in more detail.