Page 321 - Intro to Tensor Calculus

315

2
2
2
2
2
2
2
since uU r = vV r = wW r =0. Let V 2 = u + v + w and C = U + V + W and write equation
r r r r
(2.5.104) in the form
m 2
2
φ = C + V . (2.5.105)
r
2
Also note that
nm h i
2
2
nU r φ = U r (U r + u) + U r (V r + v) + U r (W r + w) 2
2
" #
nm U r C r 2
2
= + uU + vU r V r + wU r W r (2.5.106)
r
2 2
and that
nm h i
2
2
nV r φ = V r C + uV r U r + vV + wV r W r (2.5.107)
r
r
2
nm h i
2
nW r φ = W r C + uW r U r + vW r V r + wW r 2 (2.5.108)
r
2
are similar results.
We use ∂ (φ)= mV 1i together with the previous results substituted into the equation (2.5.95), and
∂V 1i
find that the Maxwell transport equation can be expressed in the form
!
D C 2 r V 2 ∂
2
ρ + = − ρ[uU + vU r V r + wU r W r ]
r
Dt 2 2 ∂x
∂
2
− ρ[uV r U r + vV + wV r W r ]
r
∂y
(2.5.109)
∂
2
− ρ[uW r U r + vW r V r + wW ]
r
∂z
! ! !
∂ U r C 2 r ∂ V r C r 2 ∂ W r C 2 r
~ ~
− ρ − ρ − ρ + nF · V.
∂x 2 ∂y 2 ∂z 2
Compare the equation (2.5.109) with the energy equation (2.5.48)
2
De D V
~
~ ~
ρ + ρ = ∇(σ · V ) −∇ · ~q + ρb · V (2.5.110)
Dt Dt 2
C 2
where the internal heat energy has been set equal to zero. Let e = r denote the internal energy due to
2
~
~
random motion of the gas particles, F = mb, and let
! ! !
∂ U r C 2 r ∂ V r C 2 r ∂ W r C r 2
q
∇· ~ = − ρ − ρ − ρ
∂x 2 ∂y 2 ∂z 2
(2.5.111)

∂ ∂T ∂ ∂T ∂ ∂T
= − k − k − k
∂x ∂x ∂y ∂y ∂z ∂z
2
mC
represent the heat conduction terms due to the transport of particle energy r by way of the random
2
particle motion. The remaining terms are related to the rate of change of work and surface stresses giving
∂ ∂
2
− ρ[uU + vU r V r + wU r W r ] = (uσ xx + vσ xy + wσ xz )
r
∂x ∂x
∂ ∂
2
− ρ[uV r U r + vV + wV r W r ] = (uσ yx + vσ yy + wσ yz ) (2.5.112)
r
∂y ∂y
∂ ∂
2
− ρ[uW r U r + vW r V r + wW ] = (uσ zx + vσ zy + wσ zz ) .
r
∂z ∂z