# chap05 - Finite Control Volume Analysis Application of...

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Finite Control Volume Analysis CVEN 311 Application of Reynolds Transport Theorem

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Moving from a System to a Finite Control Volume Mass Linear Momentum Moment of Momentum Energy Putting it all together!
Conservation of Mass B B = Total amount of ____ in the system = Total amount of ____ in the system b b = ____ per unit mass = __ = ____ per unit mass = __ ˆ sys cv cs DM dV dA Dt t ρ = + ⋅ ∫ ∫ V n ˆ cs cv dA dV t ⋅ =- ∫ ∫ V n mass 1 mass But DM sys / Dt = 0! cv equation mass leaving - mass entering = - rate of increase of mass in cv ˆ sys cv cs DB bdV b dA Dt t = + ⋅ ∫ ∫ V n Continuity Equation

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Conservation of Mass 1 2 1 1 1 2 2 2 ˆ ˆ 0 cs cs dA dA ρ ⋅ + ⋅ = ∫ ∫ V n V n 1 1 2 2 V V 1 A A 1 If mass in cv is constant Unit vector is ______ to surface and pointed ____ of cv ˆ cs cv dA dV t ⋅ =- ∫ ∫ V n ˆ n normal out ˆ cs dA ρ⋅ = V n m & ˆ cs dA V A = V n VA = ˆ n We assumed uniform ___ on the control surface is the spatially averaged velocity normal to the cs V [M/T]
Continuity Equation for Constant Density and Uniform Velocity 1 2 1 1 2 2 0 V A V A ρ - + = 1 2 1 2 V A V A Q = = 1 2 1 1 1 2 2 2 ˆ ˆ 0 cs cs dA dA ⋅ + ⋅ = ∫ ∫ V n V n Density is constant across cs Density is the same at cs 1 and cs 2 [L 3 /T] Simple version of the continuity equation for conditions of constant density. It is understood that the velocities are either ________ or _______ ________. 1 1 2 2 V A V A Q = = uniform spatially averaged

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Example: Conservation of Mass? The flow out of a reservoir is 2 L/s. The flow out of a reservoir is 2 L/s. The reservoir surface is 5 m x 5 m. The reservoir surface is 5 m x 5 m. How fast is the reservoir surface How fast is the reservoir surface dropping? dropping? res A Q dt dh - = h h dt dh A Q res out - = Example Example Constant density ˆ cs cv dA dV t ρ ⋅ =- ∫ ∫ V n ˆ cs V dA t V n out in dV Q Q dt - =- Velocity of the reservoir surface
Linear Momentum Equation m = B V m m = V b R.T.T. momentum momentum/unit mass Steady state ˆ sys cv cs DB bdV b dA Dt t ρ = + ⋅ ∫ ∫ V n ˆ cv cs Dm dV dA Dt t = + ⋅ ∫ ∫ V V V V n ˆ cs Dm dA Dt = ⋅ V V V n 0 F

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Linear Momentum Equation ( 29 ( 29 1 1 1 1 2 2 2 2 Dm V A V A Dt ρ =- + V V V ( 29 ( 29 1 1 1 1 1 1 V V M Q A V - = - = ( 29 ( 29 2 2 2 2 2 2 V V M Q A V = = Assumptions Assumptions Vectors!!! Uniform density Uniform velocity V A Steady ˆ cs Dm dA Dt = ⋅ V V V n 1 2 1 1 1 1 2 2 2 2 ˆ ˆ cs cs Dm dA dA Dt = ⋅ + ⋅ ∫ ∫ V V V n V V n V fluid velocity relative to cv
( 29 1 2 D m Dt = = + V F M M Steady Control Volume Form of Newton’s Second Law What are the forces acting on the fluid in the control volume? 2 1 M M F + = 1 2 wall wall p p p τ = + + + + F F F F F W Gravity Shear forces at the walls Pressure forces at the walls Pressure forces on the ends Why no shear on control surfaces? _______________________________ No velocity gradient normal to surface

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Resultant Force on the Solid Surfaces The shear forces on the walls and the pressure forces on the walls are generally the unknowns
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## This note was uploaded on 10/16/2010 for the course L.C.SMITH MAE643 taught by Professor Xx during the Spring '10 term at Syracuse.

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chap05 - Finite Control Volume Analysis Application of...

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