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粘性层流精确解

理想流体：忽略粘性力

为什么研究粘性流？

Coanda effect: 流体（水流或气流）偏离原本的直线路径，转而沿着一个曲面流动的现象。

粘性流的一般属性

有旋性

一般情况下，有粘必有旋（除了点涡的情况）。

耗散性

动量方程

\[ \frac{\partial u_i}{\partial t} + u_j \frac{\partial u_i}{\partial x_j} = - \frac{1}{\rho} \frac{\partial p}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_j^2} \]

动能为 \(E_k = \frac{1}{2} \rho u_i u_i\)，上式乘以 \(u_i\)

\[ u_i \frac{\partial u_i}{\partial t} + u_i u_j \frac{\partial u_i}{\partial x_j} = - \frac{1}{\rho} u_i \frac{\partial p}{\partial x_i} + \nu u_i \frac{\partial^2 u_i}{\partial x_j^2} \]
  • \(u_i \dfrac{\partial u_i}{\partial t} = \dfrac{\partial (\frac{1}{2} u_i^2)}{\partial t}\)
  • \(u_i u_j \dfrac{\partial u_i}{\partial x_j} = u_j \dfrac{\partial (\frac{1}{2} u_i^2)}{\partial x_j}\)
  • $u_i \dfrac{\partial^2 u_i}{\partial x_j^2}

涡旋的扩散性

NS-方程（不可压）

\[ \begin{aligned} \frac{\partial u_i}{\partial t} + u_k \frac{\partial u_i}{\partial x_k} &= - \frac{1}{\rho} \frac{\partial p}{\partial x_i} + \nu \frac{\partial^2 u_i}{\partial x_k^2} \\ \frac{\partial u_k}{\partial x_k} &= 0 \end{aligned} \]
  • 对流项（惯性项）是非线性的：\(u_k \dfrac{\partial u_i}{\partial x_k}\)
  • 理想不可压问题中，流动很多时候是无旋的，方程可以转化为线性的拉普拉斯方程！
  • 粘性流动中，流动一般是有旋的，不存在速度势

求解：

  • 精确解：在某些问题中，非线性的惯性项等于0或者形式非常简单，此时方程可以简化，得到简化条件下的精确解
  • 近似解
    • 小 \(\mathrm{Re}\)：粘性力主导，可以全部或部分忽略非线性项
    • 大 \(\mathrm{Re}\)：边界层问题（不能全部忽略粘性项！）

平行层流

在笛卡尔坐标系下，假设 \(v = w = 0\).

连续性方程

\[ \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0 \implies \frac{\partial u}{\partial x} = 0 \]

动量方程

\[ \left \{ \begin{aligned} \frac{\partial u}{\partial t} + u \cancel{\frac{\partial u}{\partial x}} + \cancel{v \frac{\partial u}{\partial y} + w \frac{\partial u}{\partial z}} &= - \frac{1}{\rho} \frac{\partial p}{\partial x} + \nu \left( \cancel{\frac{\partial^2 u}{\partial x^2}} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \right) \\ \cancel{\frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} + w \frac{\partial v}{\partial z}} &= - \frac{1}{\rho} \frac{\partial p}{\partial y} + \cancel{\nu \left( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} + \frac{\partial^2 v}{\partial z^2} \right)} \\ \cancel{\frac{\partial w}{\partial t} + u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y} + w \frac{\partial w}{\partial z}} &= - \frac{1}{\rho} \frac{\partial p}{\partial z} + \cancel{\nu \left( \frac{\partial^2 w}{\partial x^2} + \frac{\partial^2 w}{\partial y^2} + \frac{\partial^2 w}{\partial z^2} \right)} \end{aligned} \right. \]

得到

\[ \begin{equation} \tag{1} \label{eq:laminar-parallel} \frac{\partial u}{\partial t} = - \frac{1}{\rho} \frac{\mathrm{d} p}{\mathrm{d} x} + \nu \left( \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} \right) \end{equation} \]

考虑二维定常流动，\(\frac{\partial u}{\partial t} = 0\)，且 \(u = u(y)\)，则

\[ \frac{\mathrm{d}^2 u}{\mathrm{d} y^2} = \frac{1}{\mu} \frac{\mathrm{d} p}{\mathrm{d} x} = \text{const.} \]

解得

\[ \begin{equation} \label{eq:laminar-parallel-sol} \tag{2} u = \frac{U_0}{h} y - \frac{1}{2 \mu} \frac{\mathrm{d} p}{\mathrm{d} x} y (h - y) \end{equation} \]

经典库页特流（Plane Couette Flow）

两板间无压强梯度，上板以速度 \(U_0\) 平行运动，下板静止：

\[ \frac{\mathrm{d} p}{\mathrm{d} x} = 0 \implies u = \frac{U_0}{h} y \]

平面泊肃叶流动（Plane Poiseuille Flow）

\[ U_0 = 0 \implies u = - \frac{1}{2 \mu} \frac{\mathrm{d} p}{\mathrm{d} x} y (h - y) \]

Stratified Couette Flow

边界条件：

\[ \begin{aligned} \text{At walls: } & \left \{ \begin{aligned} u(0) &= 0 \\ u(h) &= U_0 \end{aligned} \right. \\ \text{At interface: } & \left \{ \begin{aligned} u(h_1^-) &= u(h_1^+) \\ \mu_1 \left. \frac{\mathrm{d} u}{\mathrm{d} y} \right|_{h_1^-} &= \mu_2 \left. \frac{\mathrm{d} u}{\mathrm{d} y} \right|_{h_1^+} \end{aligned} \right. \\ \end{aligned} \]

Square Duct Poiseuille Flow

边界条件：\(u = 0\) 在 \(y = \pm a\) 和 \(z = \pm a\) 处

假设 \(u = \frac{1}{2 \mu} \frac{\mathrm{d} p}{\mathrm{d} x} (y^2 - a^2) + f(y,z)\)，对 \(f(y,z)\) 使用分离变量法

Hagen-Poiseuille 流动（圆管哈根-泊肃叶流）

柱坐标系，假设流动轴对称

椭圆截面管中的 Poiseuille 流动

边界条件：\(u = 0\) 在 \(\dfrac{y^2}{a^2} + \dfrac{z^2}{b^2} = 1\) 处

假设

\[ u = C \left(\frac{y^2}{a^2} + \frac{z^2}{b^2} - 1\right)^n \]

代入 \eqref{eq:laminar-parallel} 得到 \(n = 1\)

解是否唯一？

环形 Couette 流动

\[ v_{\theta}(r) = \frac{1}{r_2^2 - r_1^2} \left[(\omega_2 r_2^2 - \omega_1 r_1^2) r + (\omega_1 - \omega_2) \frac{r_1^2 r_2^2}{r} \right] \]
  • \(\omega_2 = 0, \omega_1 \neq 0\)：
\[ v_{\theta}(r) = \omega_1 r_1^2 \frac{r_2/r - r/r_2}{r_2/r_1 - r_1/r_2} \]
  • 涡量：
\[ \omega_z = \frac{1}{r} \frac{\partial (r v_{\theta})}{\partial r} = \frac{2 (\omega_2 r_2^2 - \omega_1 r_1^2)}{r_2^2 - r_1^2} \]
  • 若无旋
    • 速度环量相等：\(\omega_1 r_1^2 = \omega_2 r_2^2 = K\)
    • 则 \(v_{\theta} = \dfrac{K}{r}\)
    • 类似于点涡！

平板突然起动（Stokes' First Problem）

一块无限大平板置于静止的粘性流体之中，在 \(t = 0\) 突然起动，以恒定速度 \(U\) 平行运动。

式 \eqref{eq:laminar-parallel} 中，压力梯度项为 0，忽略 \(z\) 方向变化，

\[ \frac{\partial u}{\partial t} = \nu \frac{\partial^2 u}{\partial y^2} \]

平板振荡流（Stokes' Second Problem）

\[ \begin{equation} u = U e^{-k y} \cos(\omega t - k y) \end{equation} \]

缝隙中的流动

小雷诺数近似解

当 \(\mathrm{Re}\) 很小时，惯性力与粘性力相比可忽略，动量方程简化为

\[ \begin{equation} \label{eq:stokes-flow} \tag{3} \nabla p = \mu \nabla^2 \mathbf{u} \end{equation} \]

NS 方程化为线性方程，结合连续性方程 \(\nabla \cdot \mathbf{u} = 0\)，可得到封闭解。

空中的云朵为什么不会掉下来？

液滴尺寸 \(\sim 10^{-5} \,\mathrm{m}\)

考虑球状小液滴，它受到重力、浮力和空气阻力。假设液滴受力平衡后，以恒定速度 \(U\) 下落，以液滴为参考系，相当于小球绕流问题。

小 \(\mathrm{Re}\) 绕流问题 Stokes 解

球坐标下的小球绕流

轴对称：

\[ \frac{\partial}{\partial \varphi} = 0, \quad u_{\varphi} = 0 \]

连续性方程：

\[ \frac{1}{r^2} \frac{\partial}{\partial r} (r^2 u_r) + \frac{1}{r \sin \theta} \frac{\partial}{\partial \theta} (u_{\theta} \sin \theta) = 0 \]

动量方程：

\[ \begin{aligned} \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial u_r}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial u_r}{\partial \theta} \right) - \frac{2 u_r}{r^2} - \frac{2}{r^2} \frac{\partial u_{\theta}}{\partial \theta} - \frac{2 u_{\theta} \cot \theta}{r^2} &= \frac{1}{\mu} \frac{\partial p}{\partial r} \\ \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \frac{\partial u_{\theta}}{\partial r} \right) + \frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta} \left( \sin \theta \frac{\partial u_{\theta}}{\partial \theta} \right) - \frac{u_{\theta}}{r^2 \sin^2 \theta} + \frac{2}{r^2} \frac{\partial u_r}{\partial \theta} - \frac{2 u_r \cot \theta}{r^2} &= \frac{1}{\mu r} \frac{\partial p}{\partial \theta} \end{aligned} \]

边界条件：

\[ \left \{ \begin{aligned} r = a: & \quad u_r = u_{\theta} = 0 \\ r \to \infty: & \quad u_r = U \cos \theta, \, u_{\theta} = - U \sin \theta \end{aligned} \right. \]

使用分离变量法。设

\[ u_r = f_1(r) \cos \theta, \quad u_{\theta} = - f_2(r) \sin \theta \]

解为

\[ \left \{ \begin{aligned} u_r &= U \left( 1 - \frac{3 a}{2 r} + \frac{a^3}{2 r^3} \right) \cos \theta \\ u_{\theta} &= - U \left( 1 - \frac{3 a}{4 r} - \frac{a^3}{4 r^3} \right) \sin \theta \\ p &= p_{\infty} - \frac{3 a \mu U}{2 r^2} \cos \theta \end{aligned} \right. \]

下面求小球受到的阻力。表面应力分量：

\[ \begin{aligned} (\sigma_{rr})_{r=a} &= \left(- p + 2 \mu \frac{\partial u_r}{\partial r} \right)_{r=a} = - p_{\infty} + \frac{3}{2} \mu U \cos \theta \\ (\sigma_{r \theta})_{r=a} &= \mu \left( \frac{1}{r} \frac{\partial u_r}{\partial \theta} + \frac{\partial u_{\theta}}{\partial r} - \frac{u_{\theta}}{r} \right)_{r=a} = - \frac{3}{2} \mu U \sin \theta \end{aligned} \]

投影到 \(z\) 轴方向，

\[ \mathrm{d} F_z = \left[ (\sigma_{rr})_{r=a} \cos \theta - (\sigma_{r \theta})_{r=a} \sin \theta \right] \mathrm{d} S \]

积分得阻力

\[ F = 6 \pi \mu a U \]

终极速度

\[ U = \frac{2}{9} \frac{\rho' - \rho}{\mu}g a^2 \]

回到云层的问题，假设空气静止，液滴以终极速度下落，液滴半径 \(a \sim 10^{-5} \,\mathrm{m}\)，代入上式计算得到 \(U \sim 10^{-2} \,\mathrm{m/s}\)

小 \(\mathrm{Re}\) 绕流问题 Oseen 解

问题：在远离圆球的地方，Stokes 解是否合理？

惯性力项和粘性力项的比值

\[\frac{|\vec{u} \cdot \nabla \vec{u}|}{|\nu \nabla^2 \vec{u}|} \propto \frac{u_r \frac{\partial u_r}{\partial r}}{\nu \frac{\partial^2 u_r}{\partial r^2}} \sim \frac{\frac{a U^2}{r^2}}{\nu \frac{U a}{r^3}} = \frac{U r}{\nu} = \mathrm{Re} \frac{r}{a}\]

Oseen 解的阻力和阻力系数

\[ F = 6 \pi \mu a U \left( 1 + \frac{3}{8} \mathrm{Re} \right), \quad C_D = \frac{24}{\mathrm{Re}} \left( 1 + \frac{3}{16} \mathrm{Re} \right) \]

小球在不同 \(\mathrm{Re}\) 下的受力情况

  • \(\mathrm{Re} \ll 1\)：蠕流
  • 驻涡
  • 卡门涡街

阻力危机：\(\mathrm{Re} > 10^5\)，阻力突然下降！

小球受力的一点延伸

  • 1851，Stokes 忽略流体惯性，得到了牛顿不可压流体在定常均匀流动中小球受到的阻力
  • 1888，Basset 考虑了非定常项 \(\partial \mathbf{u} / \partial t\)
  • Boussinesq (1903) 和 Oseen (1927)，求解含有非定常项的 N-S 方程，得到小球受力方程（BBO 方程）
  • 蔡树棠 (1957), Maxey & Riley (1983)，非均匀、非定常小雷诺数不可压牛顿流动中，小球的受力方程为 Tsai-Maxey-Riley 方程
\[ m_p \frac{\mathrm{d} \mathbf{V}}{\mathrm{d} t} = (m_p - m_f) \mathbf{g} + m_f \frac{\mathrm{D} \mathbf{u}}{\mathrm{D} t} - \frac{1}{2} m_f \left( \frac{\mathrm{d} \mathbf{V}}{\mathrm{d} t} - \frac{\mathrm{D} \mathbf{u}}{\mathrm{D} t} \right) - 6 \pi \mu a (\mathbf{V} - \mathbf{u}) - 6 a^2 \sqrt{\pi \rho_f \mu} \int_0^t \frac{1}{\sqrt{t - \tau}} \frac{\mathrm{d}}{\mathrm{d} \tau} (\mathbf{V} - \mathbf{u}) \, \mathrm{d} \tau \]

圆柱绕流

满足柱面无滑移条件的解

\[ u = \frac{A}{2 \mu} \left[\ln \frac{r}{a} + \frac{y^2}{2 r^2} \left( \]

倾斜平板缝隙流动

间隙很小：\(h(x) \ll l\)，故 \(w \ll u\)，近似平行流动

更严谨的做法：量级分析

\[\frac{\partial u}{\partial x} + \frac{\partial w}{\partial z} = 0\]

量级估计：\(\dfrac{U}{l} \sim \dfrac{W}{h(x)} \implies W \sim \dfrac{h(x)}{l} U \ll U\)

\[ \underset{\frac{u^2}{l}}{u \frac{\partial u}{\partial x}} + \underset{\frac{w u}{h}}{w \frac{\partial u}{\partial z}} = - \frac{1}{\rho} \frac{\mathrm{d} p}{\mathrm{d} x} + \nu \left(\underset{\frac{u}{l^2}}{\frac{\partial^2 u}{\partial x^2}} + \underset{\frac{u}{h^2}}{\frac{\partial^2 u}{\partial z^2}} \right) \]