• 学問 -top-

A.2.1 エネルギー保存式での計算の途中式

計算１

$$\rho T \bm{\nabla} \cdot \bm{v} + \rho \bm{v} \cdot \bm{\nabla} T + T \bm{v} \cdot \bm{\nabla} \rho$$

$$= \rho T \bigg( \dfrac{\partial u}{\partial x} + \dfrac{\partial ... ... \dfrac{\partial \rho}{\partial y} + w \dfrac{\partial \rho}{\partial z} \bigg)$$

$$= \rho T \dfrac{\partial u}{\partial x} + \rho u \dfrac{\partial ... ...+ \rho w \dfrac{\partial T}{\partial z} + T w \dfrac{\partial \rho}{\partial z}$$

$$= \dfrac{\partial (\rho T u)}{\partial x} + \dfrac{\partial (\rho T v)}{\partial y} + \dfrac{\partial (\rho T w)}{\partial z}$$

$$= \left( \begin{array}{c} \dfrac{\partial }{\partial x} \vspace{.... ...rho T u \vspace{.5em} \\ \rho T v \vspace{.5em} \\ \rho T w \end{array} \right)$$

$$= \bm{\nabla} \cdot (\rho T \bm{v})$$ (A.8)

計算２

$$\rho \bm{v} \cdot \left( \begin{array}{c} \bm{v} \cdot \bm{\nabla} u \\ \bm{v} \cdot \bm{\nabla} v \\ \bm{v} \cdot \bm{\nabla} w \\ \end{array} \right)$$

$$= \rho \{ u ( \bm{v} \cdot \bm{\nabla}u) + v ( \bm{v} \cdot \bm{\nabla}v) + w ( \bm{v} \cdot \bm{\nabla}w) \}$$

$$= \rho \bigg\{ u \bigg( u \dfrac{\partial u}{\partial x} + v \dfrac... ...dfrac{\partial w}{\partial y} + w \dfrac{\partial w}{\partial z} \bigg) \bigg\}$$

$$= \frac{1} {2} \rho \bigg\{ 2 \bigg( u^2 \dfrac{\partial u}{\partia... ...rac{\partial v}{\partial z} + w^2 \dfrac{\partial w}{\partial z} \bigg) \bigg\}$$

$$= \frac{1} {2} \rho \bigg\{ u \bigg( 2 u \dfrac{\partial u}{\partia... ...rac{\partial v}{\partial z} + 2 w \dfrac{\partial w}{\partial z} \bigg) \bigg\}$$

$$= \frac{1} {2} \rho \bigg\{ u \bigg( \dfrac{\partial u^2 }{\partial... ...c{\partial v^2 }{\partial z} + \dfrac{\partial w^2 }{\partial z} \bigg) \bigg\}$$

$$= \frac{1}{2} \rho \left( \begin{array}{c} u \\ v \\ w \\ \end{arra... ...5 em} \\ \dfrac{\partial }{\partial z} (u^2 + v^2 + w^2) \\ \end{array} \right)$$

$$= \frac{1}{2} \rho \bm{v} \cdot \bm{\nabla} (\bm{v} \cdot \bm{v})$$ (A.9)

• 学問 -top-

この図を含む文章の著作権は著者にあり、クリエイティブ・コモンズ 表示 - 非営利 - 改変禁止 3.0 非移植 ライセンスの下に公開する。