流体力学

Introduction

• how to describe fluid?

density $\rho$ , pressure $p$ , temperature $T$ , velosity $\mathit{u}$

• steady & unsteady

$$\frac{\partial }{\partial t}=0?$$

• 2D flow

two components (plane flow)

two coordinates (e.g. $\partial /\partial \phi =0$ )

streamline

the equation of streamline at a FIXED time:

$$\frac{\mathrm{d}x}{u_x}=\frac{\mathrm{d}y}{u_y}=\frac{\mathrm{d}z}{u_z}$$

e.g.

for a fluid rotate at $\mathbf{\Omega }=\mathrm{\Omega }\hat{z}$ , velocity $\mathit{u}=\mathbf{\Omega }\times \mathit{r}$

$$\mathit{u}=\left[\begin{array}{c}-y\mathrm{\Omega }\\ x\mathrm{\Omega }\\ 0\end{array}\right]$$

streamline equation:

$$\frac{\mathrm{d}x}{-y\mathrm{\Omega }}=\frac{\mathrm{d}y}{x\mathrm{\Omega }}\Rightarrow x^2+y^2=\mathrm{const}$$

material derivative and advection

given $\mathit{u}=(u_x,u_y,u_z)$ and function $f(x,y,z,t)$

we have

$$\begin{array}{rl}\frac{\mathrm{d}f}{\mathrm{d}t}& =\frac{\partial f}{\partial t}+\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t}+\frac{\partial f}{\partial z}\frac{\mathrm{d}z}{\mathrm{d}t}\\ & =\frac{\partial f}{\partial t}+u_x\frac{\partial f}{\partial x}+u_y\frac{\partial f}{\partial y}+u_z\frac{\partial f}{\partial z}\\ & =\frac{\partial f}{\partial t}+(\mathit{u}\cdot \mathrm{\nabla })f\end{array}$$

then we call

$$\underset{\text{material derivative}}{\underbrace{\frac{\mathrm{d}f}{\mathrm{d}t}}}=\underset{\text{local rate}}{\underbrace{\frac{\partial f}{\partial t}}}+\underset{\text{advection rate}}{\underbrace{(\mathit{u}\cdot \mathrm{\nabla })f}}$$

when $f$ is $\mathit{u}$

$$\underset{\text{Lagrangian description}}{\underbrace{\frac{\mathrm{d}\mathit{u}}{\mathrm{d}t}}}=\underset{\text{Eulerian description}}{\underbrace{\frac{\partial \mathit{u}}{\partial t}+(\mathit{u}\cdot \mathrm{\nabla })\mathit{u}}}$$

incompressible fluid

$$\iint \mathit{u}\cdot \mathit{n}\mathrm{d}S=0$$$$\mathrm{\nabla }\cdot \mathit{u}=0$$

mass conservation

control volume FIXED in space (Eulerian description)

$$\frac{\partial }{\partial t}\iiint \rho \mathrm{d}V=-\iint \rho \mathit{u}\cdot \mathit{n}\mathrm{d}S=-\iiint \mathrm{\nabla }\cdot (\rho \mathit{u})\mathrm{d}V$$$$\Rightarrow \frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho \mathit{u})=0$$$$\iff \frac{\partial \rho }{\partial t}+(\mathit{u}\cdot \mathrm{\nabla })\rho +\rho (\mathrm{\nabla }\cdot \mathit{u})=0$$$$\iff \frac{\mathrm{d}\rho }{\mathrm{d}t}+\rho (\mathrm{\nabla }\cdot \mathit{u})=0$$

• incompressible: $\mathrm{\nabla }\cdot \mathit{u}=0$, ${\displaystyle \frac{\mathrm{d}\rho }{\mathrm{d}t}}=0$
• compressible: ${\displaystyle \frac{\mathrm{d}}{\mathrm{d}t}}\ln \rho =-\mathrm{\nabla }\cdot \mathit{u}$

vorticity

• vorticity $\mathit{\omega }=\mathrm{\nabla }\times \mathit{u}$

e.g.

take a fluid rotate at $\mathbf{\Omega }=\mathrm{\Omega }\hat{z}$ again.

consider a circle ($\mathrm{\Gamma }$) in x-y plane with radius $R$ , so we can calculate the vorticity by:

$${\oint }_{\mathrm{\Gamma }}\mathit{u}\mathrm{d}\mathit{l}=\iint (\mathrm{\nabla }\times \mathit{u})\mathrm{d}S$$$$\Rightarrow \mathrm{\Omega }R\cdot 2\pi R=\omega \cdot \pi R^2$$$$\Rightarrow \omega =2\mathrm{\Omega }$$

thus we know vorticity is twice of local rotation rate.

potential flow

$$\mathit{\omega }=0\mathit{u}=\mathrm{\nabla }\mathrm{\Psi }$$

Inviscid (ideal) fluid

Euler equation

control volume FIXED in space, consider the momentum:

$$\begin{array}{rl}\frac{\partial }{\partial t}\iiint \rho u_i\mathrm{d}V& =-\iint \rho u_i(\mathit{u}\cdot \mathit{n})\mathrm{d}S-\iint p{n}_i\mathrm{d}S+\iiint \rho f_i\mathrm{d}V\\ & =-\iiint \mathrm{\nabla }(\rho u_i\mathit{u})\mathrm{d}V-\iiint \frac{\partial p}{\partial x_i}\mathrm{d}V+\iiint \rho f_i\mathrm{d}V\end{array}$$

(here $f_i$ means the volume force e.g. gravity or magnetic)

so we have,

$$\frac{\partial }{\partial t}(\rho u_i)+\mathrm{\nabla }(\rho u_i\mathit{u})=-\frac{\partial p}{\partial x_i}+\rho f_i$$

for incompressible fluid, we have ${\displaystyle \frac{\partial \rho }{\partial t}}=0$ and $\mathrm{\nabla }\cdot \mathit{u}=0$

$$\begin{array}{rl}\mathrm{LHS}& =\rho \frac{\partial u_i}{\partial t}+u_i\frac{\partial \rho }{\partial t}+\rho (\mathit{u}\cdot \mathrm{\nabla })u_i+u_i\mathrm{\nabla }\cdot (\rho \mathit{u})\\ & =\rho (\frac{\partial u_i}{\partial t}+(\mathit{u}\cdot \mathrm{\nabla })u_i)\end{array}$$

then we get the Euler equation (4 equivalent forms):

$$\frac{\partial u_i}{\partial t}+(\mathit{u}\cdot \mathrm{\nabla })u_i=-\frac{1}{\rho }\frac{\partial p}{\partial x_i}+f_i$$$$\frac{\mathrm{d}{u}_i}{\mathrm{d}t}=-\frac{1}{\rho }\frac{\partial p}{\partial x_i}+f_i$$$$\frac{\partial \mathit{u}}{\partial t}+(\mathit{u}\cdot \mathrm{\nabla })\mathit{u}=-\frac{1}{\rho }\mathrm{\nabla }p+\mathit{f}$$$$\frac{\mathrm{d}\mathit{u}}{\mathrm{d}t}=-\frac{1}{\rho }\mathrm{\nabla }p+\mathit{f}$$

Close the equation

now we have mass conservation equation and Euler equation:

$$\{\begin{array}{l}{\displaystyle \frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho \mathit{u})=0}\\ {\displaystyle \frac{\mathrm{d}\mathit{u}}{\mathrm{d}t}=-\frac{1}{\rho }\mathrm{\nabla }p+\mathit{f}}\end{array}$$

here we have 5 variables ( $\rho ,\mathit{u},p$ ), but only 4 equations.

• incompressible fluid:

  $\rho =\mathrm{const}$ closed.

• compressible fluid:

  need Equation of State (EOS)

  e.g. ideal gas $p=\rho RT$

  polytropic relation $p\propto {\rho }^{\gamma }$

Bernoulli theorem

for steady flow ( $\partial \mathit{u}/\partial t=0$ )

Euler equation reads,

$$\mathit{u}\cdot \mathrm{\nabla }\mathit{u}=-\frac{1}{\rho }\mathrm{\nabla }p+\mathit{g}$$

apply the relation

$$\mathit{u}\cdot \mathrm{\nabla }\mathit{u}=\mathrm{\nabla }\left(\frac{u^2}{2}\right)-\mathit{u}\times \mathit{\omega }$$

we have

$$-\mathrm{\nabla }(\frac{p}{\rho }+\frac{u^2}{2}+gz)+\mathit{u}\times \mathit{\omega }=0$$

let

$$H=\frac{u^2}{2}+\frac{p}{\rho }+gz$$

yields

$$-\mathrm{\nabla }H+\mathit{u}\times \mathit{\omega }=0$$

By performing $\mathit{u}\cdot$ on the above equation and applying $\mathit{u}\cdot (\mathit{u}\times \mathit{\omega })=0$ , we derive

$$\mathit{u}\cdot \mathrm{\nabla }H=0$$

this means $H={\displaystyle \frac{u^2}{2}}+{\displaystyle \frac{p}{\rho }}+gz=\mathrm{const}$ along a streamline

Navier-Stokes equation

$$\frac{\mathrm{d}\mathit{u}}{\mathrm{d}t}=-\frac{1}{\rho }\mathrm{\nabla }p+\frac{\mu }{\rho }{\mathrm{\nabla }}^2\mathit{u}+\frac{{\mu }^{\prime }+\mu /3}{\rho }\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathit{u})+\mathit{f}$$

$\mu$ : first viscosity; ${\mu }^{\prime }$ : second viscosity

$\mu /\rho \equiv \nu ;{\mu }^{\prime }/\rho \equiv {\nu }^{\prime }$

$\mu ,{\mu }^{\prime }$ : dynamic viscosity

$\nu ,{\nu }^{\prime }$ : kinematic viscosity

$[\nu ]=m^2{s}^{-1}$

$$\frac{\mathrm{d}\mathit{u}}{\mathrm{d}t}=-\frac{1}{\rho }\mathrm{\nabla }p+\nu {\mathrm{\nabla }}^2\mathit{u}+({\nu }^{\prime }+\frac{1}{3}\nu )\mathrm{\nabla }(\mathrm{\nabla }\cdot \mathit{u})+\mathit{f}$$

for incompressible fluid

$$\frac{\mathrm{d}\mathit{u}}{\mathrm{d}t}=-\frac{1}{\rho }\mathrm{\nabla }p+\nu {\mathrm{\nabla }}^2\mathit{u}+\mathit{f}$$

Stellar structure

for compressible fluid, assume spherical symmetric $\rho (r),p(r)$

hydrostatic equilibrium gives

$$-\mathrm{\nabla }p+\rho \mathit{g}=0$$$$\Rightarrow -\frac{\mathrm{d}p}{\mathrm{d}r}-\rho g=0$$

here

$$g=\frac{G}{r^2}{\int }_0^r\rho 4\pi r^{\prime 2}\mathrm{d}{r}^{\prime }$$

so we have

$$\frac{\mathrm{d}p}{\mathrm{d}r}+\rho \frac{G}{r^2}{\int }_0^{r}4\pi \rho r^{\prime 2}\mathrm{d}{r}^{\prime }=0$$$$\Rightarrow \frac{\mathrm{d}}{\mathrm{d}r}\left(\frac{\mathrm{d}p}{\mathrm{d}r}{r}^2\frac{1}{\rho }\right)+4\pi G\rho r^2=0$$$$\Rightarrow \frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(r^2\frac{1}{\rho }\frac{\mathrm{d}p}{\mathrm{d}r}\right)=-4\pi G\rho$$

Another approach

$$-\mathrm{\nabla }p+\rho \mathit{g}=0,\mathit{g}=-\mathrm{\nabla }\mathrm{\Phi }$$$$\mathrm{\nabla }\left(\frac{1}{\rho }\mathrm{\nabla }p\right)=\mathrm{\nabla }\cdot \mathit{g}=-{\mathrm{\nabla }}^2\mathrm{\Phi }$$

here

$$\iint \mathit{g}\cdot \mathrm{d}\mathit{S}=\iiint -{\mathrm{\nabla }}^2\mathrm{\Phi }\mathrm{d}V$$$$\Rightarrow \frac{G\rho V}{r^2}4\pi r^2={\mathrm{\nabla }}^2\mathrm{\Phi }V$$$$\Rightarrow {\mathrm{\nabla }}^2\mathrm{\Phi }=4\pi G\rho$$

polytropic gas

$$p=K{\rho }^{\gamma }=K{\rho }^{(1+1/n)}$$

here $n$ is polytropic index

• convection: atom H, $\gamma =5/3$, $n=1.5$
• radiation: photon, $\gamma =4/3$, $n=3$
• non-relativistic degenerate electron, $\gamma =5/3$, $n=1.5$
• relativistic degenerate electron, $\gamma =4/3$, $n=3$

$$\Rightarrow K(1+\frac{1}{n})\frac{1}{r^2}\frac{\mathrm{d}}{\mathrm{d}r}\left(r^2{\rho }^{-1+1/n}\frac{\mathrm{d}\rho }{\mathrm{d}r}\right)+4\pi G\rho =0$$

at $r=0$, $\rho ={\rho }_c$, $\mathrm{d}\rho /\mathrm{d}r=0$

at $r=R$, $\rho =0$ $\Rightarrow R$

let $a={\displaystyle \frac{4\pi G}{K(1+n)}}{\rho }_c^{1-1/n}$, $z=ar$, $u=(\rho /{\rho }_c{)}^{1/n}$, we have

$$\frac{1}{z^2}\frac{\mathrm{d}}{\mathrm{d}z}\left(z^2\frac{\mathrm{d}u}{\mathrm{d}z}\right)+u^n=0$$

at $z=0$, $u=1$, $u^{\prime }=0$

$u=0\Rightarrow z=aR\Rightarrow R$

the solve of Lane-Emden equation

• $n=0$: $u=1-{\displaystyle \frac{1}{6}}{z}^2$ ($z=\sqrt{6},u=0$)
• $n=1$: $u={\displaystyle \frac{\sin z}{z}}$ ($z=\pi ,u=0$)
• $n=5$: $u={\displaystyle \frac{1}{\sqrt{1+z^2/3}}}$ ($z\to \mathrm{\infty },u\to 0$)

for $n<5$, $R$ finite, $u\to 0$

for $n\ge 5$, $R\to \mathrm{\infty }$, $u\to 0$

estimation

$$\frac{\mathrm{d}p}{\mathrm{d}r}\simeq \rho g,\frac{p}{R}\simeq \rho \frac{GM}{R^2}\simeq \frac{G{M}^2}{R^5}$$$$\Rightarrow p\simeq \frac{G{M}^2}{R^4}$$

we know

$$p=K{\rho }^{\gamma }\simeq K{\left(\frac{M}{R^3}\right)}^{\gamma }$$

combine two relation yields,

$$\frac{G}{K}{M}^{2-\gamma }{R}^{3\gamma -4}\simeq 1$$

when $\gamma =5/3$: $M^{1/3}R\simeq \mathrm{const}$, $R\propto M^{-1/3}$

when $\gamma =4/3$: $M=\mathrm{const}$

Gas dynamics

sound wave (acoustic wave)

for air, we have equations:

$$\{\begin{array}{l}{\displaystyle \frac{\partial \rho }{\partial t}+\mathrm{\nabla }(\rho \mathit{u})=0}\\ {\displaystyle \frac{\partial \mathit{u}}{\partial t}+\mathit{u}\cdot \mathrm{\nabla }\mathit{u}=-\frac{1}{\rho }\mathrm{\nabla }p}\end{array}$$

now sound wave generate, we let $\rho ={\rho }_0+{\rho }_1,$ $p=p_0+p_1,$ $\mathit{u}=\mathbf{0}+{\mathit{u}}_1$

$$\{\begin{array}{l}{\displaystyle \frac{\partial {\rho }_1}{\partial t}+{\rho }_0\mathrm{\nabla }{\mathit{u}}_1=0}\\ {\displaystyle \frac{\partial {\mathit{u}}_1}{\partial t}=-\frac{1}{{\rho }_0}\mathrm{\nabla }{p}_1}\end{array}$$$$\Rightarrow \frac{{\partial }^2{\rho }_1}{\partial t^2}-{\mathrm{\nabla }}^2{p}_1=0$$

we know $p(\rho )$ , $p_1={\displaystyle \frac{\mathrm{d}p}{\mathrm{d}\rho }}{\rho }_1$, let ${\displaystyle \frac{\mathrm{d}p}{\mathrm{d}\rho }}=c_s^2$ yields,

$$\frac{{\partial }^2{\rho }_1}{\partial t^2}-c_s^2{\mathrm{\nabla }}^2{\rho }_1=0$$

assume that ${\rho }_1=\hat{\rho }{\mathrm{e}}^{i(\mathit{k}\cdot \mathit{x}-\omega t)},$ $p_1=\hat{p}{\mathrm{e}}^{i(\mathit{k}\cdot \mathit{x}-\omega t)},$ ${\mathit{u}}_1=\hat{\mathit{u}}{\mathrm{e}}^{i(\mathit{k}\cdot \mathit{x}-\omega t)}$

$$\Rightarrow -{\omega }^2+c_s^2{k}^2=0\Rightarrow \omega =\pm c_{s}k$$

so we have

• phase velosity ${\mathit{c}}_p={\displaystyle \frac{\omega }{k}}\hat{\mathit{k}}=\pm c_s\hat{\mathit{k}}$
• group velosity ${\mathit{c}}_g={\mathrm{\nabla }}_k\omega =\pm c_s\hat{\mathit{k}}$