>[!info] Poiseuille Flow: flow in circular pipe, diameter $D$, circular cross sections normal to $z$ >- Pressure-driven flow: $\frac{dp}{dz}=\text{const.}$ >- Ignore gravity >- Fully developed flow in z-dir: $\frac{\partial u}{\partial z}=0$ >- Steady state: $\frac{\partial}{\partial t}(\dots)=0$ >- Axi-symmetric flow: $\frac{\partial}{\partial \theta}(\dots)=0$ >- Incompressible, Newtonian fluid See notebook pgs. 19 - 21 for full derivation. Condensed version below: **Conservation of Mass:** $ \cancel{ \frac{d}{dt}\int _{CV} }\rho dV+\int_{CS}\rho u_{rel}\cdot d\vec{A}=0 $ $ \int_{CS}\rho \vec{u}_{rel}\cdot d\vec{A}= -\rho\left[ u_{z} \frac{ \partial u_{z} }{ \partial z } \left( -\frac{dz}{2} \right) \right](2\pi rdr)+\rho\left[ u_{z}+\frac{ \partial u_{z} }{ \partial z } \left( \frac{dz}{2} \right) \right](2\pi r dr) $ $ -\rho(u_{r})(2\pi r dz)+\rho\left[ u_{r}+ \frac{ \partial u_{r} }{ \partial r } (dr) \right](2\pi(r+dr)dz) $ $ \dots~ r \cancel{ \frac{ \partial u }{ \partial z } } +u_{r}+r \frac{ \partial u_{r} }{ \partial r } =0~\dots~u_{r}=0 $ **LME (x-dir):** $ \cancel{ \frac{d}{dt} \int_{CV} u_{z}\rho dV }+\int_{CS}u_{z}(\rho u_{rel}\cdot dA)=\cancel{ F_{bz} }+F_{sz} $ $ \int_{CS}u_{z}(\rho u_{rel}\cdot dA)=u_{z}(-\dot{m})+u_{z}(\dot{m})=0 $ $ F_{sz}=0=\left[ p+ \frac{ \partial p }{ \partial z } \left( -\frac{dz}{2} \right) \right](2\pi rdr)-\left[ p+\frac{ \partial p }{ \partial z } \left( \frac{dz}{2} \right) \right](2\pi r dr) $ $ -\tau_{rz}(2\pi rd r)+\left[ \tau_{rz}+\frac{ \partial \tau_{rz} }{ \partial r } (dr) \right](2 \pi(r+dr)dz) $ $ \dots~u_{z}= \frac{1}{4\mu} \frac{dp}{dz} r^2+ \frac{c_{1}}{\mu}\ln(r)+c_{2}~\dots~ \boxed{ u_{z}=-\frac{1}{4\mu} \frac{dp}{dz}(R^2-r^2) } $