Untitled - Aerobib - Universidad Politécnica de Madrid

3.4. ENERGY EQUATION 67
The energy equation is now obtained by expressing that the energy variation,
given by Eq. (3.21), must be equal to the work done by the exterior forces, given by
Eq. (3.23), plus the heat received, given by Eq. (3.24), thus obtaining
ρ D Dt (u + 1 2 v2 ) = ∇ · (¯v · τ e ) + ρ ¯F · ¯v − ∇ · ¯q. (3.25)
Expanding the first term of the right hand side of this expression, there results
∇ · (¯v · τ e ) = ¯v · (∇ · τ e ) + τ e : τ v , (3.26)
where τ v is the deformation velocity tensor, of components γ ij 10 defined by
γ ij = 1 2
( ∂vi
+ ∂v )
j
. (3.27)
∂x j ∂x i
In expression (3.26), the first term of the right hand side measures the work
done by the surface forces that act upon the unit volume when the surface moves with
no deformation with velocity ¯v. The second term measures the work done by the
surface forces in order to produce the deformation τ v .
Bringing forth the pressure and the viscous stresses in τ e by using expression
(3.17) for τ e , it can be verified that the deformation work (τ e : τ v ) consists of the
pressure work (−p∇ · ¯v) done during the expansion of the gas, and of the work
Φ = τ ev : τ v (3.28)
dissipated by the viscosity to produce the deformation τ v . Function Φ, which as it can
be easily proved is always positive, is the dissipation function of Lord Rayleigh. That
is
τ e : τ v = −p∇ · ¯v + Φ. (3.29)
In order to bring forth the variation of the internal energy u, the energy equation
(3.25) can be simplified by making use of the equation of motion (3.14). In fact, by
subtracting from Eq. (3.25) the scalar product of Eq. (3.14) by ¯v, taking into account
Eq. (3.26), there results
or else, making use of Eq. (3.29),
10 See chapter 2.
ρ Du
Dt
ρ Du
Dt = τ e : τ v − ∇ · ¯q, (3.30)
= −p∇ · ¯v + Φ − ∇ · ¯q. (3.31)