# Equations (not in formula sheet) - $\tau=\mu \frac{\partial u}{\partial y}$ where $y$ is perpendicular to the direction of flow - $dF=\tau dA$ - $\dot{m}_{in}=\dot{m}_{out}\to \rho_{1} Q_{in}=\rho_{2} Q_{out}$ where volumetric flow rate $Q=\int udA$ - If cross-sectional area is not normal to the velocity vector, $Q=\int_{A}u\cos\theta dA$ - $\bar{u}=\frac{Q}{A}$ - $Q=\frac{\dot{m}}{\rho}$ # Notes - It's often smart to find an equation for the final quantity and find the required values for that first. - If you use gauge pressure, atmospheric pressure is defined as zero - Be careful of what units you choose for $\rho$ in US/English - For $\int_{CS}\vec{v}_{x}(\rho \vec{V}_{xyz}\cdot d\vec{A}),$ $v_{x}\neq V_{xyz}$ > [!info] Application of the Energy Equation > The way we normally treat the energy equation is through simplifications. In this course, most of the time we will deal with steady flows that can be considered quasi 1-D. For **steady 1D flow**, the left-hand side is: > > $ > \oint \rho \left( u + \frac{p}{\rho} + \frac{V^2}{2} + gz \right)\,\vec{V}\cdot\vec{n}\,dA > $ > > This term can only be non-zero if $\vec{V}\cdot\vec{n} \neq 0$. Basically, the fluid has to be crossing that particular surface area. If we assume a quasi 1D flow where properties are considered constant within the entrance and exit areas of the flow, then: > > $ > \oint \rho \left( u + \frac{p}{\rho} + \frac{V^2}{2} + gz \right)\,\vec{V}\cdot\vec{n}\,dA > $ > > $ > = > \sum_{flow\ out} > \left( > u + \frac{p}{\rho} + \frac{V^2}{2} + gz > \right)\dot{m} > - > \sum_{flow\ in} > \left( > u + \frac{p}{\rho} + \frac{V^2}{2} + gz > \right)\dot{m} > $ ![[ME 308 Exam 1 Notes#^3433fe]] - In the full energy equation above, you need to include the $\rho u_{rel}\cdot dA$ term if the flow is non-uniform across the in/outlet area, for example if velocity varies with position. If flow is treated as 1D & uniform over each section, the integral simplifies to $\dot{m}\left( u+\frac{p}{\rho}+\frac{V^2}{2}+gz \right)$ - If there's no pump/turbine/shaft,... device doing work on the water between 2 points, $\dot{W}=0$ - If there's no pressure/temperature change, $\Delta u\approx 0$ - Can neglect pressure term sometimes in $F_{B}$ if there's no pressure change across the CV - If exposed to atmospheric pressure, use gauge pressure s.t there's no pressure change within the CV - If mass is accumulating within the CV then $\frac{d}{dt}\int_{CV}\rho dV\neq0$ - If fluid is leaving a moving CV, $\dot{m}=-\rho Av$ rather than $-\rho A(v-u);$ same applies to LME >[!info] Where to start on problems >1. **Start with COM** when mass flow/accumulation matters *(tank filling/draining, changing mass, multiple in/outlets, unknown $\dot{m}$)* >$ > \frac{d}{dt} \int_{CV} \rho dV+\int_{CS}\rho \vec{V}_{rel}\cdot d\vec{A}=0 >$ >2. **Start with LME** when problem asks for for force, acceleration, reaction force, or momentum change *(jet hitting vane/cart, nozzle force, bend force, hydraulic jump)* >$ > \sum \vec{F}=\text{ momentum accumulation }+ \text{ momentum flux} >$ >3. **Start with energy** when the unknown is pressure, velocity, elevation, pump/turbine work, or head *(siphon, nozzle velocity, pressure difference, height change, Bernoulli-type setup)* >$ > \frac{p}{\rho}+\frac{V^2}{2}+gz=\dots >$ - Force $F_{R}=p_{c}A$ acts through the center of pressure, $p_{c}$ is calculated using the centroid depth below free surface - Center of pressure $y'=y_{c}+\frac{I_{xc}}{Ay_{c}}$ where $y_{c}$ is the centroid depth below the free surface # Book Notes ## Engineering Fluid Mechanics ### [[Engineering Fluid Mechanics - Roberson, Crowe.pdf#page=192|The Control Volume Approach]] $B$ represents general extensive property and $\beta$ be the symbol for corresponding intensive property $ B=\int\beta dm=\int\beta \rho dV $ $ \text{Flow rate out - flow rate in }=u_{2}A_{2}-u_{1}A_{1}=\vec{u}_{2}\cdot \vec{A}_{2}+\vec{u}_{1}\vec{A}_{1}=\sum_{cs}\vec{u}\cdot \vec{A} $ $ \dot{m}=\sum_{cs}\rho \vec{u}\cdot A $ $ \dot{B}=\int_{cs}\beta \rho \vec{u}\cdot d\vec{A} $ $ \text{Well-defined, discrete in/outlets: } \frac{dB_{sys}}{dt}=\frac{d}{dt} \int_{CV}\beta \rho dV+\sum_{CS}\beta \rho \vec{u}\cdot A $ $ \text{When velocity is variable across a section, }\frac{dB_{sys}}{dt}=\frac{d}{dt} \int_{CV}\beta \rho dV+\int _{CS} \beta \rho \vec{u} \cdot d\vec{A} $ ### [[Engineering Fluid Mechanics - Roberson, Crowe.pdf#page=224|Momentum Equation]] $ \int_{CS}\vec{V}_{xyz}\rho \vec{V}_{xyz}\cdot d\vec{A}+\frac{d}{dt} \int_{CV}\vec{V}_{xyz}\rho dV=\vec{F}_{S}+\vec{F}_{B}-\int_{CV}\vec{a}_{rf}\rho dV$ First term (on the left) is the net flow of momentum out of the control volume. Therefore, if the flow through the control volume is uniform $\Delta MV=0$, the flow of momentum entering the CV is the same as the flow of momentum leaving the CV $\therefore$ this term will be zero If flow is nonuniform, change in the flow of momentum will exist between in/outflow sections, and magnitude of the term must be evaluated (common applications include flow in bends & thru nozzles) Control surface is chosen s.t the unknown(s) you're seeking in the solution are isolated by your choice There are two different velocity values used in the term on the left. The first is the velocity carried w/ momentum and the second is velocity thru the control surface, used for $\dot{m}$. To calculate $F_{x},$ you'd use the term $ \sum F_{x}=\frac{d}{dt} \int_{CS}V_{x}(\rho V\cdot dA)+\sum_{CS} v_{x} \rho V\cdot A $ **See Examples 6.2, 6.7** ### [[Engineering Fluid Mechanics - Roberson, Crowe.pdf#page=268|The Energy Equation]] From eqn. sheet we have the form of the energy equation $ \frac{d}{dt}\int_{CV} e\rho dV+\int_{CS}\left( h+\frac{1}{2}U^2+gz \right)(\rho u_{rel}\cdot dA)=\dot{Q}_{\text{into CV}}+\dot{W}_{\text{out of CV}} $ ^3433fe where $e=u+\frac{1}{2}U^2+gz,h=u+\frac{p}{\rho}$ ($h$ is specific enthalpy) Flow work is the work done by pressure forces as the system moves thru the space $ \Delta W_{f_{2}}=U_{2}p_{2}A_{2}\Delta t\to\dot{W}_{f_{2}}=U_{2}p_{2}A_{2} $ $ \dot{W}_{f}=\sum_{CS} \frac{p}{\rho}\rho \vec{U}\cdot \vec{A} $ Shaft work is defined as nay work other than flow work - usually the form of work done through a shaft that takes energy in/out of system Basic form of energy equation: $ \dot{Q}-\dot{W}_{s}-\sum_{CS} \frac{p}{\rho}\rho \vec{U}\cdot \vec{A}=\frac{d}{dt} \int_{CV}\left( \frac{U^2}{2} +gz+u\right)\rho dV+\sum_{CS}\left( \frac{U^2}{2}+gz+u \right)\rho \vec{U}\cdot \vec{A} $ More general form of energy equation: $ \dot{Q}-\dot{W}_{s}=\frac{d}{dt} \int_{CV} \left( \frac{U^2}{2}+gz+u \right)\rho dV+\int_{CS}\left( \frac{p}{\rho}+\frac{V^2}{2}+gz+u \right) $ See p. 278, ## [[Introduction to Fluid Mechanics - Fox, McDonald.pdf#page=125|Introduction to Fluid Mechanics - Momentum Eqn. for Inertial CV]] For inertial CV: $ \sum F_{x}=\frac{d}{dt} \int_{CS}V_{x}(\rho V\cdot dA)+\sum_{CS} v_{x} \rho V\cdot A $ Assuming steady state: $ \sum F_{x}=\sum_{CS}v_{x}\rho V\cdot A=\dot{m}(V_{2}'-V_{1}')=\rho AV'(V_{2}'-V_{1}') $ For non-inertial reference frames, the eqns. above will not work -> must derive momentum eqn. for linearly accelerating CV. Inertial reference frame is denoted by $XYZ$ $ \vec{P}_{sys}=\int_{M(\text{system})}\vec{u}dm=\int_{CV(\text{system})}\vec{u}\rho dV, \vec{F}= \frac{d\vec{P}}{dt}\Biggr)_{\text{system}} $ $ \vec{F}=\frac{d\vec{P}_{XYZ}}{dt}=\frac{d}{dt}\int_{M\text{(system)}}\vec{u}_{XYZ}dm=\int_{M\text{(system)}} \frac{d\vec{u}_{xyz}}{dt}dm $ $ \vec{u}_{XYZ}=\vec{u}_{xyz}+\vec{u}_{rf} $ $ \frac{d\vec{u}_{XYZ}}{dt}=\vec{a}_{XYZ}=\frac{d\vec{u}_{xyz}}{dt}+\frac{d\vec{u}_{rf}}{dt}=\vec{a}_{xyz}+\vec{a}_{rf} $ - $\vec{a}_{XYZ}$ is rectilinear acceleration relative to the inertial reference frame $XYZ$ - $\vec{a}_{xyz}$ is the rectilinear acceleration of the system relative to the noninitial reference frame $xyz$ - $\vec{a}_{rf}$ is the rectilinear acceleration of the noninitial reference frame $xyz$ (i.e. of the control volume relative to the inertial frame $XYZ$) $ \vec{F}=\int_{M\text{(system)}}\vec{a}_{rf}dm+\int_{M\text{(system)}} \frac{d\vec{u}_{xyz}}{dt}dm $ where the linear momentum of the system is given by $ \vec{P}_{xyz})_{\text{system}}=\int_{M\text{(system)}}\vec{u}_{XYZ}dm=\int_{V\text{(system)}}\vec{u}_{xyz}\rho dV $ $ \vec{F}-\int_{CV}\vec{a}_{rf}\rho dV=\frac{ \partial }{ \partial t } \int_{CV}\vec{u}_{xyz}\rho dV+\int_{CS}\vec{u}_{xyz}\rho \vec{u}_{xyz}\cdot d\vec{A} $ $ \boxed{ \vec{F}_{S}+\vec{F}_{B}-\int_{CV}\vec{a}_{rf}\rho dV=\frac{ \partial }{ \partial t } \int_{CV} \vec{u}_{xyz}\rho dV+\int_{CS}\vec{u}_{xyz}\rho \vec{u}_{xyz}\cdot d\vec{A} } $ **See Examples [[Introduction to Fluid Mechanics - Fox, McDonald.pdf#page=125&selection=187,0,192,1|4.10]], [[Introduction to Fluid Mechanics - Fox, McDonald.pdf#page=129&selection=313,0,313,4|4.11]]**