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APPENDIX A. DERIVATION OF TRANSPORT EQUATIONS
Thus, with reference to Figure A.1, we have May 20, 2005 13:6
∂(ρ m V )
˙
M ac = , (A.1)
∂t
˙
M in = ρ m x 2 x 3 u 1 | x 1 + ρ m x 3 x 1 u 2 | x 2
, (A.2)
+ ρ m x 1 x 2 u 3 | x 3
˙
M out = ρ m x 2 x 3 u 1 | x 1 + x 1 + ρ m x 3 x 1 u 2 | x 2 + x 2
. (A.3)
+ ρ m x 1 x 2 u 3 | x 3 + x 3
Dividing each term by V (constant) = x 1 x 2 x 3 ,wehave
∂ρ m (ρ m u 1 | x 1 − ρ m u 1 | x 1 + x 1 )
=
∂t x 1
) )
(ρ m u 2 | x 2 − ρ m u 2 | x 2 + x 2 (ρ m u 3 | x 3 − ρ m u 3 | x 3 + x 3
+ + . (A.4)
x 2 x 3
Now, letting x 1 , x 2 , x 3 → 0, this equation can be written as
∂ρ m ∂(ρ m u 1 ) ∂(ρ m u 2 ) ∂(ρ m u 3 )
+ + + = 0. (A.5)
∂t ∂x 1 ∂x 2 ∂x 3
Equation A.5 represents the mass conservation law in conservative differential
form. When the spatial derivatives are expanded, the equation can be written in the
following nonconservative form:
∂ρ m ∂ρ m ∂ρ m ∂ρ m ∂u 1 ∂u 2 ∂u 3
+ u 1 + u 2 + u 3 =−ρ m + + , (A.6)
∂t ∂x 1 ∂x 2 ∂x 3 ∂x 1 ∂x 2 ∂x 3
or
D ρ m
=−ρ m · V. (A.7)
Dt
For a single-component fluid, the mixture density ρ m may be replaced by ρ.
A.3 Momentum Equations
Newton’s second law of motion states that for a given direction
Rate of accumulation of momentum (Mom ac )
= Rate of momentum in (Mom in )
− Rate of momentum out (Mom out )
+ Sum of forces acting on the CV (F cv ).