FDM-1:量纲与无量纲

1. 两组方程的对比

第一组（无量纲形式）：

• x-动量方程： \(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = -\frac{\partial p}{\partial x} + \frac{1}{Re}\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right)\)
• y-动量方程： \(\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = -\frac{\partial p}{\partial y} + \frac{1}{Re}\left(\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}\right)\)

第二组（有量纲形式）：

• x-动量方程： \(\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = -\frac{1}{\rho}\frac{\partial p}{\partial x} + \nu \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right)\)
• y-动量方程： \(\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = -\frac{1}{\rho}\frac{\partial p}{\partial y} + \nu \left( \frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2} \right)\)

下面来一步步推导如何从有量纲的纳维-斯托克斯方程（第二组）得到无量纲的形式（第一组）。

为了区分，我们用带下标 d 的变量表示有量纲的物理量，不带下标的表示无量纲的量。

1. 定义特征尺度和无量纲变量

我们需要选择一些特征尺度来对变量进行无量纲化：

• 特征长度：\(L\) (例如，方腔的宽度或高度)
• 特征速度：\(U\) (例如，顶盖的驱动速度)
• 特征时间：\(T = L/U\)
• 特征压力：\(P_0 = \rho U^2\) (这是一个常见的选择，基于动压)

现在定义无量纲变量：

• 无量纲坐标： \(x = \frac{x_d}{L}\), \(y = \frac{y_d}{L}\)
• 无量纲速度： \(u = \frac{u_d}{U}\), \(v = \frac{v_d}{U}\)
• 无量纲时间： \(t = \frac{t_d}{T} = \frac{t_d U}{L}\)
• 无量纲压力： \(p = \frac{p_d}{P_0} = \frac{p_d}{\rho U^2}\)

2. 转换导数

我们需要用链式法则来转换偏导数：

• \[\frac{\partial}{\partial x_d} = \frac{\partial}{\partial x} \frac{\partial x}{\partial x_d} = \frac{\partial}{\partial x} \frac{1}{L}\]
• \[\frac{\partial}{\partial y_d} = \frac{\partial}{\partial y} \frac{\partial y}{\partial y_d} = \frac{\partial}{\partial y} \frac{1}{L}\]
• \[\frac{\partial}{\partial t_d} = \frac{\partial}{\partial t} \frac{\partial t}{\partial t_d} = \frac{\partial}{\partial t} \frac{U}{L}\]

对于二阶导数：

• \[\frac{\partial^2}{\partial x_d^2} = \frac{\partial}{\partial x_d} \left( \frac{1}{L} \frac{\partial}{\partial x} \right) = \frac{1}{L^2} \frac{\partial^2}{\partial x^2}\]
• \[\frac{\partial^2}{\partial y_d^2} = \frac{1}{L^2} \frac{\partial^2}{\partial y^2}\]

3. 代入有量纲的x-动量方程

我们从有量纲的x-动量方程开始：

\[\frac{\partial u_d}{\partial t_d} + u_d\frac{\partial u_d}{\partial x_d} + v_d\frac{\partial u_d}{\partial y_d} = -\frac{1}{\rho}\frac{\partial p_d}{\partial x_d} + \nu \left( \frac{\partial^2 u_d}{\partial x_d^2} + \frac{\partial^2 u_d}{\partial y_d^2} \right)\]

现在，我们将每个有量纲变量替换为其无量纲形式：
\(u_d = Uu\),
\(v_d = Uv\),
\(p_d = \rho U^2 p\),
\(x_d = Lx\),
\(y_d = Ly\),
\(t_d = (L/U)t\).

逐项替换：

• 第一项（时间导数项）： \(\frac{\partial u_d}{\partial t_d} = \frac{\partial (Uu)}{\partial t_d} = U \frac{\partial u}{\partial t_d} = U \left( \frac{U}{L} \frac{\partial u}{\partial t} \right) = \frac{U^2}{L} \frac{\partial u}{\partial t}\)

• 第二项（对流项1）： \(u_d\frac{\partial u_d}{\partial x_d} = (Uu) \frac{\partial (Uu)}{\partial x_d} = (Uu) U \frac{\partial u}{\partial x_d} = (Uu) U \left( \frac{1}{L} \frac{\partial u}{\partial x} \right) = \frac{U^2}{L} u \frac{\partial u}{\partial x}\)

• 第三项（对流项2）： \(v_d\frac{\partial u_d}{\partial y_d} = (Uv) \frac{\partial (Uu)}{\partial y_d} = (Uv) U \frac{\partial u}{\partial y_d} = (Uv) U \left( \frac{1}{L} \frac{\partial u}{\partial y} \right) = \frac{U^2}{L} v \frac{\partial u}{\partial y}\)

• 第四项（压力梯度项）： \(-\frac{1}{\rho}\frac{\partial p_d}{\partial x_d} = -\frac{1}{\rho}\frac{\partial (\rho U^2 p)}{\partial x_d} = -\frac{1}{\rho} (\rho U^2) \frac{\partial p}{\partial x_d} = -U^2 \left( \frac{1}{L} \frac{\partial p}{\partial x} \right) = -\frac{U^2}{L} \frac{\partial p}{\partial x}\)

• 第五项（扩散项1）： \(\nu \frac{\partial^2 u_d}{\partial x_d^2} = \nu \frac{\partial^2 (Uu)}{\partial x_d^2} = \nu U \frac{\partial^2 u}{\partial x_d^2} = \nu U \left( \frac{1}{L^2} \frac{\partial^2 u}{\partial x^2} \right) = \frac{\nu U}{L^2} \frac{\partial^2 u}{\partial x^2}\)

• 第六项（扩散项2）： \(\nu \frac{\partial^2 u_d}{\partial y_d^2} = \nu \frac{\partial^2 (Uu)}{\partial y_d^2} = \nu U \frac{\partial^2 u}{\partial y_d^2} = \nu U \left( \frac{1}{L^2} \frac{\partial^2 u}{\partial y^2} \right) = \frac{\nu U}{L^2} \frac{\partial^2 u}{\partial y^2}\)

4. 组合并简化

将所有替换后的项代回方程：

\[\frac{U^2}{L} \frac{\partial u}{\partial t} + \frac{U^2}{L} u \frac{\partial u}{\partial x} + \frac{U^2}{L} v \frac{\partial u}{\partial y} = -\frac{U^2}{L} \frac{\partial p}{\partial x} + \frac{\nu U}{L^2} \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right)\]

现在，我们将整个方程除以 \(\frac{U^2}{L}\)：

\[\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = -\frac{\partial p}{\partial x} + \frac{\nu U/L^2}{U^2/L} \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right)\]

简化扩散项的系数：

\[\frac{\nu U/L^2}{U^2/L} = \frac{\nu U L}{L^2 U^2} = \frac{\nu}{UL}\]

我们知道雷诺数 \(Re = \frac{UL}{\nu}\)。因此，\(\frac{\nu}{UL} = \frac{1}{Re}\)。

所以，方程变为：

\[\frac{\partial u}{\partial t} + u\frac{\partial u}{\partial x} + v\frac{\partial u}{\partial y} = -\frac{\partial p}{\partial x} + \frac{1}{Re}\left(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}\right)\]

这正是第一组方程中的x-动量方程。

5. y-动量方程

对于y-动量方程，推导过程完全类似，只需将 \(u_d\) 替换为 \(v_d\) (以及无量纲的 \(u\) 替换为 \(v\)) 即可得到：

\[\frac{\partial v}{\partial t} + u\frac{\partial v}{\partial x} + v\frac{\partial v}{\partial y} = -\frac{\partial p}{\partial y} + \frac{1}{Re}\left(\frac{\partial^2 v}{\partial x^2} + \frac{\partial^2 v}{\partial y^2}\right)\]

6. 连续性方程

有量纲的连续性方程：

\[\frac{\partial u_d}{\partial x_d} + \frac{\partial v_d}{\partial y_d} = 0\]

代入无量纲变量：

\[\frac{\partial (Uu)}{\partial (Lx)} + \frac{\partial (Uv)}{\partial (Ly)} = 0\] \[\frac{U}{L}\frac{\partial u}{\partial x} + \frac{U}{L}\frac{\partial v}{\partial y} = 0\]

两边同除以 \(\frac{U}{L}\) (假设 \(U \neq 0, L \neq 0\))：

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

连续性方程在无量纲化后形式保持不变。

总结 通过选择合适的特征尺度对有量纲的纳维-斯托克斯方程进行无量纲化，我们可以得到以雷诺数为唯一参数的无量纲形式。这个过程涉及到变量替换和导数的链式法则转换。这种无量纲形式在理论分析和数值模拟中非常有用，因为它减少了参数的数量，并使得不同尺度下的流动问题可以进行比较。