Hagen-Poiseuille流れ

そのうち確認しよう

導出

条件は以下の通り

定常流れ

$ \frac{\partial \pmb{v}}{\partial t}=\pmb{0}

x軸方向に一様な流れ

$ \frac{\partial \pmb{v}}{\partial x}=\pmb{0}

x軸方向のみに流れが存在する

$ \pmb{v}=v_x\pmb{e}_x

幾何学的対称条件

$ \frac{\partial \pmb{v}}{\partial\theta}=\pmb{0}

外力なし

$ \pmb{f}=\pmb{0}

壁面の滑りなし条件

$ \left.\pmb{v}\right|_{r=a}=\pmb{0}

圧縮性Newton流れにおける運動方程式に、以上の条件を代入する

$ \underbrace{\cancel{\frac{\partial \pmb{v}}{\partial t}}}_{\frac{\partial \pmb{v}}{\partial t}=\pmb{0}}+(\pmb{v}\cdot\pmb{\nabla})\pmb{v}=\underbrace{\cancel{\frac{\pmb{f}}\rho}}_{\pmb{f}=\pmb{0}}-\frac1\rho\pmb{\nabla}P+\nu\pmb{\nabla}^2\pmb{v}+\left(\frac\lambda\rho+\nu\right)\pmb{\nabla}(\pmb{\nabla}\cdot\pmb{v})

$ \implies v_x\underbrace{\cancel{\frac{\partial v_x}{\partial x}}}_{\frac{\partial \pmb{v}}{\partial x}=\pmb{0}}\pmb{e}_x=-\frac1\rho\pmb{\nabla}P+\nu\pmb{\nabla}^2v_x\pmb{e}_x+\left(\frac\lambda\rho+\nu\right)\pmb{\nabla}\underbrace{\cancel{\frac{\partial v_x}{\partial x}}}_{\frac{\partial \pmb{v}}{\partial x}=\pmb{0}}

$ \implies\pmb{0}=-\frac1\rho\pmb{\nabla}P+\nu\pmb{\nabla}^2v_x\pmb{e}_x

$ = -\frac1\rho\pmb{\nabla}P+\nu\pmb{\nabla}\cdot\left(\frac{\partial v_x}{\partial r}\pmb{e}_r+\underbrace{\cancel{\frac1r\frac{\partial v_x}{\partial\theta}\pmb{e}_\theta}}_{\frac{\partial \pmb{v}}{\partial\theta}=\pmb{0}}\right)\pmb{e}_x

$ = -\frac1\rho\pmb{\nabla}P+\nu\pmb{\nabla}\cdot\left(\frac{\partial v_x}{\partial r}\pmb{e}_r\right)\pmb{e}_x

$ = -\frac1\rho\pmb{\nabla}P+\nu\left(\frac{\partial^2 v_x}{{\partial r}^2}+\frac1r\frac{\partial v_x}{\partial r}\right)\pmb{e}_x

ここの発散の計算は、以下の結果を用いた

$ \pmb{\nabla}\cdot(a_r\pmb{e}_r+a_\theta\pmb{e}_\theta)=\pmb{e}_r\cdot\frac{\partial(a_r\pmb{e}_r+a_\theta\pmb{e}_\theta)}{\partial r}+\pmb{e}_\theta\cdot\frac{\partial(a_r\pmb{e}_r+a_\theta\pmb{e}_\theta)}{r\partial\theta}

$ =\frac{\partial a_r}{\partial r}+\pmb{e}_r\cdot\frac{\partial(a_\theta\pmb{e}_\theta)}{\partial r}+\frac{a_r}{r}+\pmb{e}_\theta\cdot\frac{\partial(a_\theta\pmb{e}_\theta)}{r\partial\theta}

基底を偏微分した

$ =\frac{\partial a_r}{\partial r}+0+\frac{a_r}{r}+\frac1r\frac{\partial a_\theta}{\partial\theta}+0

$ =\frac{\partial a_r}{\partial r}+\frac{a_r}{r}+\frac1r\frac{\partial a_\theta}{\partial\theta}

$ \implies \pmb{\nabla}P=\mu\left(\frac{\partial^2 v_x}{{\partial r}^2}+\frac1r\frac{\partial v_x}{\partial r}\right)\pmb{e}_x

$ \iff \frac{\partial P}{\partial x}=\mu\left(\frac{\partial^2 v_x}{{\partial r}^2}+\frac1r\frac{\partial v_x}{\partial r}\right)\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0

$ \iff \frac{\partial P}{\partial x}=\frac{\mu}r\frac{\partial}{\partial r}\left(r\frac{\partial v_x}{\partial r}\right)\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0

$ \iff \frac1\mu\frac{\partial P}{\partial x}r=\frac{\partial}{\partial r}\left(r\frac{\partial v_x}{\partial r}\right)\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0

$ \iff \frac1{2\mu}\frac{\partial P}{\partial x}r^2+C=r\frac{\partial v_x}{\partial r}\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0\quad\text{.for}\exists C\in\R

$ \iff \frac1{2\mu}\frac{\partial P}{\partial x}r^2+C=r\frac{\partial v_x}{\partial r}\land0+C=0\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0\quad\text{.for}\exists C\in\R

$ \iff \frac1{2\mu}\frac{\partial P}{\partial x}r^2=r\frac{\partial v_x}{\partial r}\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0

$ \iff v_x=\frac1{4\mu}\frac{\partial P}{\partial x}r^2+C\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0\quad\text{.for}\exists C\in\R

$ \underline{\iff v_x=\frac1{4\mu}\frac{\partial P}{\partial x}(r^2-a^2)\land\frac{\partial P}{\partial r}=\frac{\partial P}{\partial\theta}=0\quad}_\blacksquare

$ \because\left.\pmb{v}\right|_{r=a}=\pmb{0}