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Fluid mechanics: Control volume analysis January 26, 2011

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Conservation of mass
\dot m = \int_{CS} \rho \left(\mathbf{V}\cdot\mathbf{n} \right) dA

For steady flow:
\int_{CS} \rho \left(\mathbf{V}\cdot\mathbf{n} \right) dA = 0

For incompressible flow:
\int_{CS} \left(    \mathbf{V}\cdot\mathbf{n} \right) dA = 0

Steady flow through a tube:
\rho_1A_1V_1 = \rho_2A_2V_2

and for incompressible steady flow:
A_1V_1 = A_2V_2

Conservation of momentum
\sum F = \frac{d}{dt} \left(\int_{CV} \mathbf{V} \mathbf{\rho}\right) + \int_{CS} \mathbf{V}\mathbf{\rho} \left(\mathbf{V}\cdot\mathbf{n} \right)dA

Conservation of energy
\frac{d\mathbf{Q}}{dt} + \frac{d\mathbf{W}}{dt} = \frac{d}{dt} \left(\int_{CV} e\mathbf{\rho} \right) + \int_{CS} e\mathbf{\rho} \left( \mathbf{V}\cdot\mathbf{n}  \right)dA
where ”e” is the energy per unit mass.

Conservation equations of Mass, Momentum and Energy
Equation of Continuity
\frac{\partial\rho}{\partial t} + \nabla\cdot \left(\rho \mathbf{V} \right) = 0

For incompressible fluids, the equation of continuity reduces to:
\rho \left(\nabla \cdot \mathbf{V} \right) = 0

Euler’s Equation
It applies conservation of momentum to inviscid, incompressible flow.
\rho \mathbf{g} - \nabla p = \rho\frac{d\mathbf{V}}{dt}

Stokes’ Equation
It applies conservation of momentum in creeping flow limit (low Reynold’s Number)
\nabla p = \mu \nabla^2 \mathbf{V}

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