# 物理代写|流体力学代写Fluid Mechanics代考|Averaging the Mechanical Energy Equation

## 物理代写|流体力学代写Fluid Mechanics代考|Averaging the Mechanical Energy Equation

The mechanical energy equation for a turbulent flow is obtained using Eq. (4.71), which includes the compressibility term $(\nabla \cdot V \not \equiv 0)$. Dividing the involved flow quantities into the mean and the fluctuating parts and applying the averaging procedure outlined in Sect. 9.2, results in a complex equation. To reduce the degree of complexity, we consider an incompressible flow with the mechanical energy equation given by Eq. (4.72) and also below:
$$\frac{D}{D t}\left(\frac{V^2}{2}\right)=\nabla \cdot\left(-\frac{p}{\rho} \mathbf{V}+2 \nu \mathbf{V} \cdot \mathbf{D}\right)-2 \nu \mathbf{D}: \mathbf{D}+\mathbf{V} \cdot \mathbf{g} .$$
Using the identity $\mathbf{V} \cdot \nabla\left(V^2\right)=\nabla \cdot\left(\mathbf{V} V^2\right)$ for an incompressible flow and its index notation $V_i \frac{\partial}{\partial x_i}\left(V_j V_j\right)=\frac{\partial}{\partial x_i}\left(V_i V_j V_j\right)$, we find the index notation of Eq. (9.61):
\begin{aligned} \frac{\partial}{\partial t}\left(\frac{V_j V_j}{2}\right)= & -\frac{\partial}{\partial x_i} V_i\left(\frac{p}{\rho}+\frac{V_j V_j}{2}\right)+v \frac{\partial}{\partial x_i} V_j\left(\frac{\partial V_i}{\partial x_j}+\frac{\partial V_j}{\partial x_i}\right) \ & -v\left(\frac{\partial V_i}{\partial x_j}+\frac{\partial V_j}{\partial x_i}\right) \frac{\partial V_j}{\partial x_i}+V_i g_i \end{aligned}
We introduce the following decompositions:
\begin{aligned} & \mathbf{V}=\overline{\mathbf{V}+\mathbf{V}^{\prime}} \text { with } \mathbf{V}{\mathbf{i}}=\overline{\mathbf{V}}{\mathbf{i}}+\mathbf{V}_{\mathbf{i}}^{\prime}, \ & V^2=V_i V_i=\bar{V}_i \bar{V}_i+2 \bar{V}_i V_i^{\prime}+V_i^{\prime} V_i^{\prime} \ & p=\bar{p}+p^{\prime} \end{aligned}
and substitute the quantities in Eq. (9.61) by Eq. (9.63) and average the results, we find:

## 物理代写|流体力学代写Fluid Mechanics代考|Averaging the Thermal Energy Equation

A thermal energy equation can be expressed in terms of specific internal energy $u$ or specific static enthalpy $h$. In both cases, the specific internal energy and specific static enthalpy can be expressed in terms of temperature $u=c_v T$ and $h=c_p T$. Both forms are fully equivalent and one can be converted into the other by $h=u+p v$, which is the defining equation for the specific static enthalpy. For averaging the thermal energy equation in terms of specific static enthalpy which we replace by the static temperature, we resort to Eq. (4.94)
$$c_p \frac{D T}{D t}=\frac{k}{\varrho} \nabla^2 T+\frac{1}{\varrho} \frac{D p}{D t}+\frac{1}{\rho} \mathbf{T}: \mathbf{D}$$
with the friction stress tensor $\mathbf{T}$ from Eq. (4.36):
$$\mathbf{T}=\lambda(\nabla \cdot \mathbf{V}) \mathbf{I}+2 \mu \mathbf{D}$$
Decomposing, in Eq. (9.65), the temperature and pressure $T$ and $p$ as well as the friction and deformation tensors $\mathbf{T}$ and $\mathbf{D}$ while neglecting the density fluctuation, we find:
\begin{aligned} c_p \frac{D\left(\bar{T}+T^{\prime}\right)}{D t}= & \frac{k}{\varrho} \nabla^2\left(\bar{T}+T^{\prime}\right)+\frac{1}{\varrho} \frac{D\left(p+p^{\prime}\right)}{D t} \ & +\frac{1}{\varrho}\left(\overline{\mathbf{T}}+\mathbf{T}^{\prime}\right):\left(\overline{\mathbf{D}}+\mathbf{D}^{\prime}\right) . \end{aligned}

# 流体力学代写

## 物理代写|流体力学代写Fluid Mechanics代考|Averaging the Mechanical Energy Equation

$$\frac{D}{D t}\left(\frac{V^2}{2}\right)=\nabla \cdot\left(-\frac{p}{\rho} \mathbf{V}+2 \nu \mathbf{V} \cdot \mathbf{D}\right)-2 \nu \mathbf{D}: \mathbf{D}+\mathbf{V} \cdot \mathbf{g} .$$

$$\frac{\partial}{\partial t}\left(\frac{V_j V_j}{2}\right)=-\frac{\partial}{\partial x_i} V_i\left(\frac{p}{\rho}+\frac{V_j V_j}{2}\right)+v \frac{\partial}{\partial x_i} V_j\left(\frac{\partial V_i}{\partial x_j}+\frac{\partial V_j}{\partial x_i}\right) \quad-v\left(\frac{\partial V_i}{\partial x_j}+\frac{\partial t}{\partial{ }^i}\right.$$

$\mathbf{V}=\overline{\mathbf{V}+\mathbf{V}^{\prime}}$ with $\mathbf{V i}=\overline{\mathbf{V}} \mathbf{i}+\mathbf{V}_{\mathbf{i}}^{\prime}, \quad V^2=V_i V_i=\bar{V}_i \bar{V}_i+2 \bar{V}_i V_i^{\prime}+V_i^{\prime} V_i^{\prime} p=\bar{p}+$

## 物理代写|流体力学代写Fluid Mechanics代考|Averaging the Thermal Energy Equation

$$c_p \frac{D T}{D t}=\frac{k}{\varrho} \nabla^2 T+\frac{1}{\varrho} \frac{D p}{D t}+\frac{1}{\rho} \mathbf{T}: \mathbf{D}$$

$$\mathbf{T}=\lambda(\nabla \cdot \mathbf{V}) \mathbf{I}+2 \mu \mathbf{D}$$

$$c_p \frac{D\left(\bar{T}+T^{\prime}\right)}{D t}=\frac{k}{\varrho} \nabla^2\left(\bar{T}+T^{\prime}\right)+\frac{1}{\varrho} \frac{D\left(p+p^{\prime}\right)}{D t} \quad+\frac{1}{\varrho}\left(\overline{\mathbf{T}}+\mathbf{T}^{\prime}\right):\left(\overline{\mathbf{D}}+\mathbf{D}^{\prime}\right)$$

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