# Notes # [[ME 308 HW 04.pdf#page=1|Problem 1]] # [[ME 308 HW 04.pdf#page=2|Problem 2]] $ Q=\int udA=u_{max}\int_{0}^.075 \left( 1-\left( \frac{r}{R} \right)^2 \right)(2\pi r)dr\implies \boxed{ u_{max}=11.31m/s } $ $ u_{in}=\frac{Q}{A}=5.66m/s $ $ \cancelto{ \text{stationary CV} }{ \frac{d}{dt} \int_{CV}\vec{u}_{XYZ}\rho dV } +{ \int_{CS}\vec{u}_{XYZ}(\rho \vec{u}_{rel}\cdot d\vec{A}) =F_{B}+F_{S} }-\cancelto{ \text{non accelerating CV} }{ \int_{CV}\vec{a}_{xyz/XYZ}\rho dV } $ $ \int u(\rho \vec{u}\cdot dA)=F_{S}=\int_{A_{in}}u\rho(\vec{u}\cdot dA)+\int_{A_{out}}u\rho(\vec{u}\cdot dA)=\rho\left( \int_{A_{in}} u^2dA +\int_{A_{out}}u^2dA \right) $ $ \text{in: }\vec{u}\cdot dA=-u_{in}A\qquad\text{out: }\vec{u}\cdot dA=u(r)dA $ $ F_{S}=-\rho u_{in}^2A+\rho \int_{0}^.075 u^2dA=-\rho u_{in}^2A+\rho \int_{0}^{.075}u_{max}^2\left[ 1-\left( \frac{r}{R} \right)^2 \right]^2(2\pi r)dr=160N $ $ p=\frac{F}{A}=9073Pa=\boxed{ 9.07kPa } $ # [[ME 308 HW 04.pdf#page=3|Problem 3]] $\rho=1.94slug/ft^3$ $ \frac{p_{1}}{\rho}+\frac{V_{1}^2}{2}=\cancel{ \frac{p_{2}}{\rho} }+\frac{V_{2}^2}{2}\to p_{1}=\frac{\rho}{2}(V_{2}^2-V_{1}^2) $ $ Q=A_{1}V_{1}=A_{2}V_{2}\to V_{2}=\frac{A_{1}V_{1}}{A_{2}}=45.71ft/s~~(?) $ $ p_{1}=\frac{\rho}{2}(V_{2}^2-V_{1}^2)=1639.39lbf/ft^2 $ $ \sum F_{x}=\dot{m}(V_{2}-V_{1})=pA-F\to F=p_{1}A_{1}-\dot{m}(V_{2}-V_{1}) $ $ \dot{m}=\rho AV=.846lb/s\text{ so }\boxed{ F=14lbf } $ # [[ME 308 HW 04.pdf#page=4|Problem 4]] # [[ME 308 HW 04.pdf#page=5|Problem 5]]