# 物理代写|流体力学代写Fluid Mechanics代考|CIVL3612

#### Doug I. Jones

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## 物理代写|流体力学代写Fluid Mechanics代考|Blasius Theorem

In this section, we utilize the complex analysis to provide the equation structure that is needed to derive the Kutta-Joukowsky lift equation from a potential theoretical point of view. As we saw in Chap. 5, any force exerted on any body of any shape that is subjected to a viscous or inviscid flow can be calculated using the integral balance of linear momentum. Thus, the following procedure is an alternative that can be applied only to potential flow. Equation (5.25) applied to a two-dimensional body results in:

$$\boldsymbol{R}=\int_{S_B} \boldsymbol{n} p d S+\int_{S_B} \boldsymbol{t} \tau d S$$
with $\boldsymbol{n}$ and $\boldsymbol{t}$ as the normal and tangential unit vectors shown in Fig. 6.16. For an an inviscid irrotational flow, Eq. (6.80) is reduced to:
$$\boldsymbol{R}=\int_{S_B} n p d S .$$
The integration has to be performed over the entire body surface. For a twodimensional body, the differential surface element is $d S=w d C$ with $w$ as the width of the body and $d C$ as a differential element of the body contour C, Fig. 6.16.
By moving along the contour, the normal unit vector changes the direction. However, it can be related to the velocity direction along the body contour. For the normal and tangential unit vectors, we may write:
$$\boldsymbol{n}=\cos \theta+i \sin \theta ; \boldsymbol{t}=\cos (\theta+\pi / 2)+i \sin (\theta+\pi / 2) .$$
Expanding $t$ in Eq. (6.82) and multiplying the results with $-i$, we find:
$$-i t=n .$$
Inserting Eq. (6.83) into Eq. (6.81), we get:
$$\boldsymbol{R}=-i \int_{S_B} t p d S .$$

## 物理代写|流体力学代写Fluid Mechanics代考|Kutta-Joukowski Theorem

With Eq. (6.89) we are now able to calculate the force acting on a cylinder of arbitrary contour using the Kutta-Joukowsky lift equation. We assume the cylinder is exposed to a flow with the velocity vector $V_{\infty}$ and the components $V_{\infty x}+i V_{\infty y}$ at infinity. We further assume that there are no singularities outside the body, although there will be inside in order to represent the body and to produce the lift. The velocity field can be represented by a Laurent series of the form \begin{aligned} \frac{d F}{d z}=& \bar{V}=u-i v=A_0+A_1\left(\frac{1}{z}\right)+A_2\left(\frac{1}{z^2}\right)+A_3\left(\frac{1}{z^3}\right) \ &+\ldots=\sum_{n=0}^{\infty} A_n\left(\frac{1}{z^n}\right) \end{aligned}
with $\bar{V}$ as the conjugate velocity vector. The integration of Eq. (6.90) yields the complex potential
$$F(z)=A_0 z+A_1 1 n z-\sum_{n=2}^{\infty} \frac{A_n}{n-1}\left(\frac{1}{z^{n-1}}\right)+\text { const. }$$
The boundary condition at infinity requires
$$\left.\frac{d F}{d z}\right|{\infty}=V{\infty x}-i V_{\infty y}$$
which determines the coefficient $A_0$
$$A_0=U_{\infty x}-i V_{\infty y}=\overline{\boldsymbol{V}}{\infty} .$$ To calculate the coefficient $A_1$ we integrate $(u-i v)$ around the contour of the body: $$\oint{(C)} \frac{d F}{d z} d z=\oint_{(C)}(u-i v) d z=\oint_{(C)}(u-i v)(d x+i d y) .$$
Performing the multiplication of the right-hand side integrand of Eq. (6.94) leads to:
$$\oint_{(C)} \frac{d F}{d z} d z=\oint_{(C)}(u d x+v d y)+i \oint_{(C)}(u d y-v d x) .$$

## 物理代写|流体力学代写流体力学代考|Blasius定理

$$\boldsymbol{R}=\int_{S_B} \boldsymbol{n} p d S+\int_{S_B} \boldsymbol{t} \tau d S$$
，其中$\boldsymbol{n}$和$\boldsymbol{t}$为图6.16所示的法向和切向单位向量。对于无粘无旋流，式(6.80)简化为:
$$\boldsymbol{R}=\int_{S_B} n p d S .$$

$$\boldsymbol{n}=\cos \theta+i \sin \theta ; \boldsymbol{t}=\cos (\theta+\pi / 2)+i \sin (\theta+\pi / 2) .$$

$$-i t=n .$$

$$\boldsymbol{R}=-i \int_{S_B} t p d S .$$

## 物理代写|流体力学代写流体力学代考|库塔-朱可夫斯基定理

$$F(z)=A_0 z+A_1 1 n z-\sum_{n=2}^{\infty} \frac{A_n}{n-1}\left(\frac{1}{z^{n-1}}\right)+\text { const. }$$

$$\left.\frac{d F}{d z}\right|{\infty}=V{\infty x}-i V_{\infty y}$$
，这决定了系数$A_0$
$$A_0=U_{\infty x}-i V_{\infty y}=\overline{\boldsymbol{V}}{\infty} .$$为了计算系数$A_1$，我们在物体的轮廓上对$(u-i v)$积分:$$\oint{(C)} \frac{d F}{d z} d z=\oint_{(C)}(u-i v) d z=\oint_{(C)}(u-i v)(d x+i d y) .$$

$$\oint_{(C)} \frac{d F}{d z} d z=\oint_{(C)}(u d x+v d y)+i \oint_{(C)}(u d y-v d x) .$$

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