# Notes - Shear stress $\tau=\mu \frac{\partial u}{\partial y}$ - $\dot{m}_{in}=\dot{m}_{out}\to \rho_{1} Q_{in}=\rho_{2} Q_{out}$ where volumetric flow rate $Q=vA$ # [[ME 308 HW 03.pdf#page=1|Problem 1]] $ \tau(r)=\mu \frac{du}{dr}=\mu\left( \frac{\beta}{4\mu}(-2r) \right)=-\frac{\beta r}{2} $ $ \tau\left( \frac{D}{2} \right)=\boxed{ -\frac{\beta D}{4} } $ $ \tau\left( \frac{D}{4} \right)=-\frac{\beta D}{8} $ $ F_{D}=|\tau|A=\frac{\beta D}{4}(\pi dL)=\boxed{ \frac{\beta DL^2}{4} } $ # [[ME 308 HW 03.pdf#page=2|Problem 2]] $ \text{At terminal velocity, }\sum F=0\therefore F_{V}=W_{P}+W_{m} $ $ \rho =\frac{m}{V}\to m_{p}=\rho V\implies F_{V}=g(\rho V+m) $ $ F_{V}=\tau A=\mu \frac{du}{dx}A $ $ g(\rho V+m)=\mu \frac{V}{x}(\pi dL)\to V=\frac{gx(\rho V+m)}{\mu \pi dL}=\boxed{ 5.55m/s } \text{ where }x=\frac{d_{o}-d_{i}}{2} $ # [[ME 308 HW 03.pdf#page=3|Problem 3]] $ 0=\cancelto{ \text{SF} }{ \frac{d}{dt}\int_{CV}\rho dV }+\int_{CS}\rho(\vec{v}\cdot d\vec{A})=\int_{in}\rho(\vec{v}\cdot d\vec{A})+\int_{out}\rho(\vec{v}\cdot d\vec{A}) \to-\dot{m}_{in}+\dot{m}_{out}=0$ $ \cancel{ \rho } Q_{out}=\cancel{ \rho } Q_{in}=3m^3/s $ $ Q_{out}=\int \vec{v}\cdot d\vec{A}=vA\cos 45 $ $ v=\frac{Q_{in}}{A\cos45}=\frac{Q_{in}}{\pi(r_{o}^2-r_{i}^2)\cos 45}=\boxed{ 1.78 m/s } $ # [[ME 308 HW 03.pdf#page=4|Problem 4]] $ \int_{ab}\vec{V}\cdot d\vec{A}+\int_{ad}\vec{V}\cdot d\vec{A}+\int_{cd}\vec{V}\cdot d\vec{A}+Q_{bc}=0 $ $ u_{ab}\delta w-V_{ad}Lw-w\int_{0}^{\delta}u_{cd}(y)\,dy=Q_{bc} $ $ Q_{bc}=1.425E-3 m^3/s $ $ \dot{m}_{bc}=\rho Q_{bc}=\boxed{ 1.425 kg/s } $