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Chapter%204%20Basic%20Eqns

Chapter%204%20Basic%20Eqns - IV Basic Equations of Fluid...

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IV. Basic Equations of Fluid Motion General comments: System vs control volume Laws of physics are often applied to points or systems of fixed mass . Newton's 2 nd law, F = ma , is applicable to a fixed collection of mass. System : a fixed collection of mass ; used in Lagrangian approach. Control Volume : a region in space where fluid flows through it; often used in Eulerian approach; fluid inside c.v. does not remain to be the same fluid. Control volume at time t. Control volume at t+ t System at t. System at time t+ t Control volume and system . We will use Control Volume : to derive basic laws for fluid moion. 4.1. Conservation of Mass v 0 n S u Fig. 4.1 Control volume v 0 for a portion of fluid 0
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Total mass of the fluid in v 0 is M = 0 dv v ρ ∫∫∫ Rate of increase of mass = 0 dv v ρ ∫∫∫ . Total mass of the fluid flowing out of v 0 through S 0 per unit time = 0 u nds S ρ ∫∫ Ò . No sources of mass, the conservation of mass requires 0 dv v ρ ∫∫∫ + 0 u nds S ρ ∫∫ Ò = 0 (4.1) Divergence theorem, Total mass of the fluid in v0 is 0 dv v ρ ∫∫∫ Rate of increase of mass = 0 dv v ρ ∫∫∫ Total mass of the fluid flowing out of v0 through S0 per unit time = 0 u nds S ρ ∫∫ Ò Divergence theorem 0 u nds S ρ ∫∫ Ò 0 ( ) u dv v ρ ∇⋅ ∫∫∫ (4.1) 0 [ ( )] u dv v t ρ ρ + ∇⋅ ∫∫∫ = 0 ( ) 0 t u ρ ρ + ∇⋅ = (4.2) Since ( ) u u u ρ ρ ρ ⋅∇ ∇⋅ = + ∇ ⋅ (4.2) 0 u u t ρ ρ ρ + ⋅∇ + ∇⋅ = Or 0 D Dt u ρ ρ + ∇⋅ = (4.3)
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If the flow is incompressible , ρ does not change along its pathline, i.e. 0 D Dt ρ = . (def. of incompressibility) Then (4.3) gives 0 u ∇⋅ = . (4.4) (result of incompressibility + continuity eqn.) 4.2 Momentum Equation Newton's second law: total rate of change of the momentum = the resultant force applied. Considering the fluid in the control volume v 0. f (body force) σ n σ n u C.V. u Fig.4.2 Forces acting on the fluid in a c.v. 0 u nds S ρ ∫∫ Ò 0 ( ) u dv v ρ ∇⋅ ∫∫∫ a. The total fluid momentum in v 0 is 0 udv v ρ ∫∫∫ . b. The total momentum flowing out of v 0 through S 0 per unit time is 0 ( ) u u n ds S ρ ∫∫ Ò Hence the net rate of change of the fluid momentum in v 0 is
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0 udv v ρ ∫∫∫ + 0 ( ) u u n ds S ρ ∫∫ Ò There are both body force and surface force on the fluid. c. The body force is 0 fdv v ρ ∫∫∫ in which f is the force per unit mass. d. The surfaces force is 0 n ds S σ ∫∫ Ò based on the discussion in § 2.2. Thus Newton's 2 nd law results in 0 udv v ρ ∫∫∫ + 0 ( ) u u n ds S ρ ∫∫ Ò = 0 fdv v ρ ∫∫∫ + 0 n ds S σ ∫∫ Ò (4.5) Divergence theorem 0 ( ) n d u u s S ρ ∫∫ Ò 0 ( ) u dv u v ρ ∇⋅ ∫∫∫ (you can verify it component by component for u ) 0 n ds S σ ∫∫ Ò 0 ( ) dv v σ ∇⋅ ∫∫∫ & then get rid of 0 v ∫∫∫ + ∇⋅ ( ρ u u ) = ρ f + ∇⋅ . (4.6) ρ u = momentum per unit mass dyadic product ρ u u = momentum flux tensor per unit volume.
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