Page 304 - Intro to Tensor Calculus

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we cannot interchange the order of differentiation and integration in this term. Here we must use the result
that Z Z
D ∂e t
~
e t dτ = + ∇· (e t V ) dτ.
Dt ∂t
V V
To prove this result we consider a more general problem. Let A denote the amount of some quantity per
unit mass. The quantity A can be a scalar, vector or tensor. The total amount of this quantity inside the
R
control volume is A = %A dτ and therefore the rate of change of this quantity is
V
Z Z Z
∂A ∂(%A) D
~
= dτ = %A dτ − %AV · ˆndS,
∂t ∂t Dt
V V S
which represents the rate of change of material within the control volume plus the influx into the control
volume. The minus sign is because ˆn is always a unit outward normal. By converting the surface integral to
a volume integral, by the Gauss divergence theorem, and rearranging terms we find that
Z Z
D ∂(%A)
~
%A dτ = + ∇· (%AV ) dτ.
Dt ∂t
V V
ij
i
In equation (2.5.46) we neglect all isolated external forces and substitute F i = σ n j , F i = b where
(s) (b)
σ ij = −pδ ij + τ ij . We then replace all surface integrals by volume integrals and find that the conservation of
energy can be represented in the form
∂e t ∂Q
~
~
~ ~
+ ∇· (e t V )= ∇(σ · V ) −∇· ~q + %b · V + (2.5.47)
∂t ∂t
2
2
2
e
where e t = %e + %(v + v + v )/2 is the total energy and σ = P 3 i=1 P 3 j=1 σ ij ˆ i ˆ j is the second order stress
e
3
2
1
tensor. Here
3 3 3
X X X
~
~
~
~
σ · V = −pV + τ 1j v j ˆ 1 + τ 2j v j ˆ 2 + τ 3j v j ˆ 3 = −pV + τ · V
e
e
e
j=1 j=1 j=1
∗
and τ ij = µ (v i,j + v j,i )+ λ δ ij v k,k is the viscous stress tensor. Using the identities
∗
2
D(e t /%) ∂e t D(e t /%) De D(V /2)
~
% = + ∇· (e t V ) and % = % + %
Dt ∂t Dt Dt Dt
~
together with the momentum equation (2.5.25) dotted with V as
~
DV
~
~
~
~ ~
% · V = %b · V −∇p · V +(∇·τ ) · V
Dt
the energy equation (2.5.47) can then be represented in the form
De ∂Q
~
q
% + p(∇· V )= −∇ · ~ + +Φ (2.5.48)
Dt ∂t
where Φ is the dissipation function and can be represented
~
~
Φ=(τ ij v i ) ,j − v i τ ij,j = ∇· (τ · V ) − (∇·τ ) · V.
∗
∗
As an exercise it can be shown that the dissipation function can also be represented as Φ = 2µ D ij D ij +λ Θ 2
where Θ is the dilatation. The heat flow vector is determined from the Fourier law of heat conduction in