Geoscience Reference
In-Depth Information
The mean shear can be rewritten
∂U
i
∂x
j
+
∂U
i
∂x
j
−
∂U
i
∂x
j
=
1
2
∂U
j
∂x
i
1
2
∂U
j
∂x
i
+
=
S
ij
+
R
ij
,
(5.44)
the sumof mean strain-rate andmean rotation-rate tensors. Since
u
i
u
j
is symmetric,
its contraction with the antisymmetric tensor
R
ij
vanishes
(Problem 1.15)
and the
shear-production term becomes
u
i
u
j
S
ij
+
R
ij
=
u
i
u
j
∂U
i
∂x
j
=
u
i
u
j
S
ij
,
(5.45)
the contraction of kinematic stress and strain-rate tensors.
The viscous dissipation term can be interpreted similarly. By definition it is the
average rate, per unit mass, of working against viscous stresses. Thus, it is the mean
value of the contraction of the instantaneous viscous stress tensor
σ
ij
(
Chapter 1
)
and the instantaneous strain-rate tensor
s
ij
, divided by density. Using the notation
∂u
i
/∂x
j
≡
u
i,j
,thisis
2
u
i,j
+
u
j,i
u
i,j
+
u
j,i
=
ν
u
i,j
u
i,j
+
u
i,j
u
j,i
σ
ij
s
ij
ρ
ν
≡
=
ν
u
i,j
u
i,j
+
(u
i
u
j
)
,jj
=
ν u
i,j
u
i,j
,
(5.46)
the term involving the second derivative of
u
i
u
j
being negligible by our scaling
guidelines
(Problem 5.13)
.
In a steady flow the right side of the TKE
equation (5.42)
sums to zero. The
mean advection, turbulent transport, and pressure transport terms are divergences;
if we integrate
(5.42)
over the entire flow volume, on the bounding surface of which
the velocity vanishes, these divergence terms integrate to zero. That means these
divergence terms only move TKE around in space. Then we have, using
Eq. (5.45)
,
∂
∂t
u
i
u
i
2
dV
=
0
=−
u
i
u
j
S
ij
dV
−
dV.
(5.47)
volume
volume
volume
The first term on the right side of
(5.47)
represents the volume-integrated mean rate
of production of TKE through the interaction of the Reynolds stress and the mean
strain rate. The second term is the mean rate at which the volume-integrated TKE
is dissipated - irreversibly converted into internal energy through viscous forces.
Since
is positive definite, we can write
Eq. (5.47)
as
−
u
i
u
j
S
ij
dV
=
dV >
0
.
(5.48)
volume
volume
