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

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## 物理代写|流体力学代写Fluid Mechanics代考|Euler’s Equations

As we have already seen in Sect. 4.1.3, Euler’s equation emerges from the Navier-Stokes equation (4.8a, $4.8 \mathrm{~b})$ for $R e=\infty$. However Euler’s equation is also a special case of Cauchy’s equation (2.38) if we use the particular constitutive relation for inviscid fluids (3.9). Euler’s equation then reads
$$\varrho \frac{\mathrm{D} u_{i}}{\mathrm{D} t}=\varrho k_{i}+\frac{\partial}{\partial x_{j}}\left(-p \delta_{i j}\right)$$
or

$$\varrho \frac{\mathrm{D} u_{i}}{\mathrm{D} t}=\varrho k_{i}-\frac{\partial p}{\partial x_{i}}$$
and it holds without restriction for all inviscid flows. In symbolic notation we write
$$\varrho \frac{\mathrm{D} \vec{u}}{\mathrm{D} t}=\varrho \vec{k}-\nabla p$$
We derive Euler’s equations in natural coordinates from (4.40b) by inserting the acceleration in the form (1.24). Relative to the basis vectors $\vec{t}$ in the direction of the pathline, $\vec{n}{\sigma}$ in the principle normal direction and $\vec{b}{\sigma}$ in the binormal direction, the vectors $\nabla p$ and $\vec{k}$ are
$$\begin{gathered} \nabla p=\frac{\partial p}{\partial \sigma} \vec{t}+\frac{\partial p}{\partial n} \vec{n}{\sigma}+\frac{\partial p}{\partial b} \vec{b}{\sigma} \ \vec{k}=k_{\sigma} \vec{t}+k_{n} \vec{n}{\sigma}+k{b} \vec{b}{\sigma} \end{gathered}$$ and the component form of Euler’s equation in natural coordinates, with $u=|\vec{u}|$, becomes \begin{aligned} \frac{\partial u}{\partial t}+u \frac{\partial u}{\partial \sigma} &=k{\sigma}-\frac{1}{\varrho} \frac{\partial p}{\partial \sigma} \ \frac{u^{2}}{R} &=k_{n}-\frac{1}{\varrho} \frac{\partial p}{\partial n} \ 0 &=k_{b}-\frac{1}{\varrho} \frac{\partial p}{\partial b} \end{aligned}

## 物理代写|流体力学代写Fluid Mechanics代考|Bernoulli’s Equation

Under mildly restricting assumptions it is possible to find so-called first integrals of Euler’s equations, which then represent conservation laws. The most important first integral of Euler’s equations is Bernoulli’s equation. We assume that the mass body force has a potential $(\vec{k}=-\nabla \psi)$, i.e., $\psi=-g_{i} x_{i}$ for the gravitational force. We multiply Euler’s equation (4.40a) by $u_{i}$, thus forming the inner product with $\vec{u}$, and obtain the relation
$$u_{i} \frac{\partial u_{i}}{\partial t}+u_{i} u_{j} \frac{\partial u_{i}}{\partial x_{j}}=-\frac{1}{\varrho} u_{i} \frac{\partial p}{\partial x_{i}}-u_{i} \frac{\partial \psi}{\partial x_{i}}$$
After transforming the second term on the left-hand side and relabelling the dummy indices, this becomes
$$u_{j} \frac{\partial u_{j}}{\partial t}+u_{j} \frac{\partial}{\partial x_{j}}\left[\frac{u_{i} u_{i}}{2}\right]=-\frac{1}{\varrho} u_{j} \frac{\partial p}{\partial x_{j}}-u_{j} \frac{\partial \psi}{\partial x_{j}}$$
We could, in principle, integrate this equation along an arbitrary smooth curve, but we arrive at a particularly simple and important result if we integrate along a streamline. With $u=|\vec{u}|$, from the differential equation for the streamline (1.11), we have

$$u_{j}=u \mathrm{~d} x_{j} / \mathrm{d} s,$$
so that
$$u_{j} \frac{\partial}{\partial x_{j}}=u \frac{\mathrm{d} x_{j}}{\mathrm{~d} s} \frac{\partial}{\partial x_{j}}=u \frac{\mathrm{d}}{\mathrm{d} s}$$
holds, and because $u_{j} \partial u_{j} / \partial t=u \partial u / \partial t$ we can write for (4.53)
$$\frac{\partial u}{\partial t}+\frac{\mathrm{d}}{\mathrm{d} s}\left[\frac{u^{2}}{2}\right]=-\frac{1}{\varrho} \frac{\mathrm{d} p}{\mathrm{~d} s}-\frac{\mathrm{d} \psi}{\mathrm{d} s}$$
Integration along the arc length of the streamline leads us to Bernoulli’s equation in the form
$$\int \frac{\partial u}{\partial t} \mathrm{~d} s+\frac{u^{2}}{2}+\int \frac{\mathrm{d} p}{\varrho}+\psi=C$$
or integrating from the initial point $A$ to the final point $B$ we get the definite integral
$$\int_{A}^{B} \frac{\partial u}{\partial t} \mathrm{~d} s+\frac{1}{2} u_{B}^{2}+\int_{A}^{B} \frac{1}{\varrho} \frac{\mathrm{d} p}{\mathrm{~d} s} \mathrm{~d} s+\psi_{B}=\frac{1}{2} u_{A}^{2}+\psi_{A}$$

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

We shall now consider the circulation of a closed material line as it was introduced by (1.105)
$$\Gamma=\oint_{(C(t))} \vec{u} \cdot \mathrm{d} \vec{x} .$$
Its rate of change is calculated using (1.101) to give
$$\frac{\mathrm{D} \Gamma}{\mathrm{D} t}=\frac{\mathrm{D}}{\mathrm{D} t} \oint_{(C(t))} \vec{u} \cdot \mathrm{d} \vec{x}=\oint_{(C)} \frac{\mathrm{D} \vec{u}}{\mathrm{D} t} \cdot \mathrm{d} \vec{x}+\oint_{(C)} \vec{u} \cdot \mathrm{d} \vec{u} .$$
The last closed integral vanishes, since $\vec{u} \cdot \mathrm{d} \vec{u}=\mathrm{d}(\vec{u} \cdot \vec{u} / 2)$ is a total differential of a single valued function, and the starting point of integration coincides with the end point.

We now follow on with the discussion in connection with Eq. (1.102), and seek the conditions for the time derivative of the circulation to vanish. It has already been shown that in these circumstances the acceleration $\mathrm{D} \vec{i} / \mathrm{D} t$ must have a potential $I$, but this is not the central point of our current discussion.

Using Euler’s equation $(4.40 \mathrm{a}, 4.40 \mathrm{~b})$ we acquire the rate of change of the line integral over the velocity vector in the form
$$\frac{\mathrm{D} \Gamma}{\mathrm{D} t}=\oint_{(C)} \vec{k} \cdot \mathrm{d} \vec{x}-\oint_{(C)} \frac{\nabla p}{\varrho} \cdot \mathrm{d} \vec{x}$$
and conclude from this that $\mathrm{D} \Gamma / \mathrm{D} t$ vanishes if $\vec{k} \cdot \mathrm{d} \vec{x}$ and $(\nabla p / \varrho) \cdot \mathrm{d} \vec{x}$ can be written as total differentials. If the mass body force $\vec{k}$ has a potential the first closed integral is zero because
$$\vec{k} \cdot \mathrm{d} \vec{x}=-\nabla \dot{\psi} \cdot \mathrm{d} \vec{x}=-\mathrm{d} \psi .$$
In a homogeneous density field or in barotropic flow, because of
$$\frac{\nabla p}{\varrho} \cdot \mathrm{d} \vec{x}=\frac{\mathrm{d} p}{\varrho(p)}=\mathrm{d} P$$

the second integral also vanishes. The last three equations form the content of Thomson’s vortex theorem or Kelvin’s circulation theorem
$$\frac{\mathrm{D} \Gamma}{\mathrm{D} t}=0 .$$

## 物理代写|流体力学代写Fluid Mechanics代考|Euler’s Equations

ϱD在一世D吨=ϱķ一世+∂∂Xj(−pd一世j)

ϱD在一世D吨=ϱķ一世−∂p∂X一世

ϱD在→D吨=ϱķ→−∇p

∇p=∂p∂σ吨→+∂p∂nn→σ+∂p∂bb→σ ķ→=ķσ吨→+ķnn→σ+ķbb→σ和欧拉方程在自然坐标中的分量形式，有在=|在→|, 变成

∂在∂吨+在∂在∂σ=ķσ−1ϱ∂p∂σ 在2R=ķn−1ϱ∂p∂n 0=ķb−1ϱ∂p∂b

## 物理代写|流体力学代写Fluid Mechanics代考|Bernoulli’s Equation

∂在∂吨+dds[在22]=−1ϱdp ds−dψds

∫∂在∂吨 ds+在22+∫dpϱ+ψ=C

∫一个乙∂在∂吨 ds+12在乙2+∫一个乙1ϱdp ds ds+ψ乙=12在一个2+ψ一个

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

Γ=∮(C(吨))在→⋅dX→.

DΓD吨=DD吨∮(C(吨))在→⋅dX→=∮(C)D在→D吨⋅dX→+∮(C)在→⋅d在→.

DΓD吨=∮(C)ķ→⋅dX→−∮(C)∇pϱ⋅dX→

ķ→⋅dX→=−∇ψ˙⋅dX→=−dψ.

∇pϱ⋅dX→=dpϱ(p)=d磷

DΓD吨=0.

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