# 物理代写|空气动力学代写Aerodynamics代考|ME439

## 物理代写|空气动力学代写Aerodynamics代考|Incompressible Flow Navier-Stokes Equations

In a wide region of aerodynamical applications low subsonic speeds are encountered. Since the free stream Mach number for these types of are very low, the flow is assumed incompressible. The continuity equation for the incompressible flow becomes
$$\vec{\nabla} \cdot \vec{V}=0$$
Equation $2.65$ implies that the flow is divergenless which in turn simplifies the constitutive relations, Eq. $2.51 \mathrm{a}$, b. In addition, because of low speeds the temperature changes in the flow field will also be low which makes the viscosity remain constant. Since the viscosity is constant, the momentum equation is simplified also to take the following form
$$\rho \frac{D \vec{V}}{D t}=-\vec{\nabla} p+\mu \nabla^2 \vec{V}$$

In case of turbulent flows, we use the effective viscosity: $\mu_e=\mu+\mu_T$ in Eq. $2.66$ which undergoes an averaging process after Reynolds decomposition which makes the final form of the equations to be called ‘Reynolds Averaged Navier-Stokes Equations’.

Another convenient form of incompressible Navier-Stokes equations is written in terms of a new variable called vorticity. The vorticity vector is derived from the velocity vector as
$$\vec{\omega}=\vec{\nabla} x \vec{V}$$
The vorticity transport equation obtained from two dimensional version of Eq. $2.66$ reads as
$$\frac{\partial \omega}{\partial t}+(\vec{V} \cdot \vec{\nabla}) \omega=\nabla^2 \omega$$
Here, $\omega$ as the third component of the vorticity appears as a scalar quantity in Eq. 2.68, which does not have any pressure term involved. The integral form of Eqs. $2.65$ and $2.67$ reads as, (Wu and Gulcat 1981),
$$\vec{V}(\vec{r}, t)=-\frac{1}{2 \pi} \int_R \frac{\vec{\omega}_o x\left(\vec{r}_o-\vec{r}\right)}{\left|\vec{r}_o-\vec{r}\right|^2} d R_0+\frac{1}{2 \pi} \int_B \frac{\left(\vec{V}_0 \cdot \vec{n}_0\right)\left(\vec{r}_o-\vec{r}\right)-\left(\vec{V}_0 x \vec{n}_0\right) x\left(\vec{r}_o-\vec{r}\right)}{\left|\vec{r}_o-\vec{r}\right|^2} d B_0$$

## 物理代写|空气动力学代写Aerodynamics代考|Aerodynamic Forces and Moments

The aim in performing the real gas flow analysis over bodies is to determine the aerodynamic forces, moments and the heat loads acting. For this purpose the computed pressure and stress fields are integrated over whole surface of the body. The surface stresses are ohtained from the velocity gradients calculated at the surface. Let us now write down the $x, y$ and $z$ components of the infinitesimal surface force $\mathrm{dF}$ acting on the infinitesimal area $\mathrm{dA}$ of the surface

Here, $n_x, n_y$ and $n_z$ are the direction cosines of the vector normal to the infinitesimal surface $\mathrm{dA}$ Let us now express the area $\mathrm{dA}$ in curvilinear coordinates. We can express the integral relations which give the total force components in xyz in terms of the differential area given in curvilinear coordinates $\xi \eta$ as shown in Fig. 2.8.

As seen in Fig. $2.8$ the differential area dA can be computed in terms of the product of two infinitesimal vectors given as the changes of the position vector $\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}$ in directions of $\xi$ and $\eta$ coordinates as $\mathrm{dA}=|(\mathrm{d} \mathbf{r} / \mathrm{d} \xi) \mathrm{x}(\mathrm{dr} / \mathrm{d} \eta)| \mathrm{d} \xi$ di . The vector product of these two vectors also give the direction of the unit normal $\mathbf{n}$ of dA. In explicit form we find
\begin{aligned} d A &=\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \ x_{\xi} & y_{\xi} & z_{\xi} \ x_\eta & y_\eta & z_\eta \end{array} \mid\right| \xi d \eta \ &=\sqrt{\left(y_{\xi} z_\eta-z_{\xi} y_\eta\right)^2+\left(x_{\xi} z_\eta-z_{\xi} x_\eta\right)^2+\left(y_{\xi} y_\eta-x_{\xi} y_\eta\right)^2 d \xi d \eta} \end{aligned}
Here, the term under the square root is named reduced Jacobian I. The unit normal vector in open form becomes
$$\vec{n}=\left[\left(y_{\xi} z_{\eta \eta}-z_{\xi} y_\eta\right) \vec{i}-\left(x_{\xi} z_{\eta \eta}-z_{\xi} x_\eta\right) \vec{j}+\left(y_{\xi} y_{\eta \eta}-x_{\xi} y_\eta\right) \vec{k}\right]$$

## 物理代写|空气动力学代写空气动力学代考|不可压缩流Navier-Stokes方程

$$\vec{\nabla} \cdot \vec{V}=0$$

$$\rho \frac{D \vec{V}}{D t}=-\vec{\nabla} p+\mu \nabla^2 \vec{V}$$

$$\vec{\omega}=\vec{\nabla} x \vec{V}$$

$$\frac{\partial \omega}{\partial t}+(\vec{V} \cdot \vec{\nabla}) \omega=\nabla^2 \omega$$

$$\vec{V}(\vec{r}, t)=-\frac{1}{2 \pi} \int_R \frac{\vec{\omega}_o x\left(\vec{r}_o-\vec{r}\right)}{\left|\vec{r}_o-\vec{r}\right|^2} d R_0+\frac{1}{2 \pi} \int_B \frac{\left(\vec{V}_0 \cdot \vec{n}_0\right)\left(\vec{r}_o-\vec{r}\right)-\left(\vec{V}_0 x \vec{n}_0\right) x\left(\vec{r}_o-\vec{r}\right)}{\left|\vec{r}_o-\vec{r}\right|^2} d B_0$$

## 物理代写|空气动力学代写空气动力学代考|空气动力和力矩

\begin{aligned} d A &=\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \ x_{\xi} & y_{\xi} & z_{\xi} \ x_\eta & y_\eta & z_\eta \end{array} \mid\right| \xi d \eta \ &=\sqrt{\left(y_{\xi} z_\eta-z_{\xi} y_\eta\right)^2+\left(x_{\xi} z_\eta-z_{\xi} x_\eta\right)^2+\left(y_{\xi} y_\eta-x_{\xi} y_\eta\right)^2 d \xi d \eta} \end{aligned}

$$\vec{n}=\left[\left(y_{\xi} z_{\eta \eta}-z_{\xi} y_\eta\right) \vec{i}-\left(x_{\xi} z_{\eta \eta}-z_{\xi} x_\eta\right) \vec{j}+\left(y_{\xi} y_{\eta \eta}-x_{\xi} y_\eta\right) \vec{k}\right]$$

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