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ChapterII.PartIII - AE 3003 Chapter 2 Continued PDE form of...

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AE 3003 Chapter 2 - Continued... PDE form of Momentum Equations In the previous handout, we derived the integral form of the momentum equations. You used these in homework set I, and the infamous quiz I. In this section, we look at turning these integral expressions into PDEs. We start with the u- momentum equation in integral form: ( 29 ∫ ∫ ∫ ∫ ∫ ∫ ∫ = + + V S S dS n i p dS n V u dV t u 0 ρ ρ (1) Divergence theorem states that any surface integral can be converted into a volume integral using: ∫ ∫ ∫ ∫ ∫ = S V dV A dS n A (2) where A is any vector. Using this relationship to the surface integrals in the u-momentum equation, we get ( 29 ( 29 ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫ = = S V S V dV i p dS n i p dV V u dS n V u ρ ρ (3) The u-momentum equation thus becomes: ( 29 ( 29 ( 29 ∫ ∫ ∫ = + + V dV i p V u t u 0 ρ ρ (4) The only way this volume integral will be equal to zero for all arbitrary control volumes at all time levels is if the integrand is zero. We therefore get: ( 29 ( 29 ( 29 0 = + + i p V u t u ρ ρ (5)
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