Page 307 - Intro to Tensor Calculus

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 
ρV x
2
x
 ρV + p − τ xx 
 
E =  ρV x V y − τ xy  (2.5.57)
 
ρV x V z − τ xz
(e t + p)V x − V x τ xx − V y τ xy − V z τ xz + q x
 
ρV y
ρV x V y − τ xy
 
 2 
y 
F =  ρV + p − τ yy (2.5.58)
 
ρV y V z − τ yz
(e t + p)V y − V x τ yx − V y τ yy − V z τ yz + q y
 
ρV z
ρV x V z − τ xz
 
 
ρV y V z − τ yz (2.5.59)
G =  
 2 
ρV + p − τ zz
z
(e t + p)V z − V x τ zx − V y τ zy − V z τ zz + q z
where the shear stresses are τ ij = µ (V i,j + V j,i )+ δ ij λ V k,k for i, j, k =1, 2, 3.
∗
∗
Computational Coordinates
To transform the conservative system (2.5.55) from a physical (x, y, z) domain to a computational (ξ, η, ζ)
domain requires that a general change of variables take place. Consider the following general transformation
of the independent variables
ξ = ξ(x, y, z) η = η(x, y, z) ζ = ζ(x, y, z) (2.5.60)
with Jacobian different from zero. The chain rule for changing variables in equation (2.5.55) requires the
operators
∂() ∂() ∂() ∂()
= ξ x + η x + ζ x
∂x ∂ξ ∂η ∂ζ
∂() ∂() ∂() ∂()
= ξ y + η y + ζ y (2.5.61)
∂y ∂ξ ∂η ∂ζ
∂() ∂() ∂() ∂()
= ξ z + η z + ζ z
∂z ∂ξ ∂η ∂ζ
The partial derivatives in these equations occur in the differential expressions

dξ =ξ x dx + ξ y dy + ξ z dz      
dξ ξ x ξ y ξ z dx
dη =η x dx + η y dy + η z dz or  dη  =  η x η y η z   dy  (2.5.62)
dζ ζ x ζ y ζ z dz
dζ =ζ x dx + ζ y dy + ζ z dz

In a similar mannaer from the inverse transformation equations

x = x(ξ, η, ζ) y = y(ξ, η, ζ) z = z(ξ, η, ζ) (2.5.63)

we can write the differentials

dx =x ξ dξ + x η dη + x ζ dζ      
dx x ξ x η x ζ dξ
dy =y ξ dξ + y η dη + y ζ dζ or  dy  =  y ξ y η y ζ   dη  (2.5.64)
dz z ξ z η z ζ dζ
dz =z ξ dξ + z ζ dζ + z ζ dζ