Page 298 - Introduction to Computational Fluid Dynamics

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May 20, 2005
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APPENDIX A. DERIVATION OF TRANSPORT EQUATIONS
Thus, with respect to Figure A.1, we can write the contributions in the x 1 direction 13:6 277
as
∂(ρ m Vu 1 )
Mom ac = , (A.8)
∂t
Mom in = (ρ m x 2 x 3 u 1 )u 1 | x 1 + (ρ m x 3 x 1 u 2 )u 1 | x 2
, (A.9)
+ (ρ m x 1 x 2 u 3 )u 1 | x 3
Mom out = (ρ m x 2 x 3 u 1 )u 1 | x 1 + x 1 + (ρ m x 3 x 1 u 2 )u 1 | x 2 + x 2
,
+ (ρ m x 1 x 2 u 3 )u 1 | x 3 + x 3
F cv =−(σ 1 | x 1 − σ 1 | x 1 + x 1 ) x 2 x 3 + (τ 21 | x 2 + x 2 − τ 21 | x 2 ) x 3 x 1
) x 1 x 2 + ρ m B 1 V, (A.10)
+ (τ 31 | x 3 + x 3 − τ 31 | x 3
where B 1 is the body force per unit mass, the σs are tensile normal stresses, and τs
are shear stresses. Now, dividing by V and letting x 1 , x 2 , x 3 → 0, it can be
shown that

x 1 Direction Momentum Equation
∂(ρ m u 1 ) ∂(ρ m u 1 u 1 ) ∂(ρ m u 2 u 1 ) ∂(ρ m u 3 u 1 )
+ + +
∂t ∂x 1 ∂x 2 ∂x 3
∂(σ 1 ) ∂(τ 21 ) ∂(τ 31 )
= + + + ρ m B 1 . (A.11)
∂x 1 ∂x 2 ∂x 3
A similar exercise in the x 2 and x 3 directions will yield

x 2 Direction Momentum Equation
∂(ρ m u 2 ) ∂(ρ m u 1 u 2 ) ∂(ρ m u 2 u 2 ) ∂(ρ m u 3 u 2 )
+ + +
∂t ∂x 1 ∂x 2 ∂x 3
∂(τ 12 ) ∂(σ 2 ) ∂(τ 32 )
= + + + ρ m B 2 . (A.12)
∂x 1 ∂x 2 ∂x 3

x 3 Direction Momentum Equation
∂(ρ m u 3 ) ∂(ρ m u 1 u 3 ) ∂(ρ m u 2 u 3 ) ∂(ρ m u 3 u 3 )
+ + +
∂t ∂x 1 ∂x 2 ∂x 3
∂(τ 13 ) ∂(τ 23 ) ∂(σ 3 )
= + + + ρ m B 3 . (A.13)
∂x 1 ∂x 2 ∂x 3
A few comments on these equations are now in order:

1. By making use of Equation A.5, the left-hand sides of Equations A.11, A.12,
and A.13 can be replaced by ρ m D(u 1 )/Dt, ρ m D(u 2 )/Dt, and ρ m D(u 3 )/Dt,