Civil Engineering Reference
In-Depth Information
and an antisymmetrical part
(
)
2
A
A ij
= ∂
U i
x j
− ∂
U j
x i
The terms on the left-hand side of equation [2.29] can
easily be decomposed into
a ij
U k
a ij
DS ij
Dt
+
x k +
a ik a kj
=
+
A ik A kj
+
S ik S kj
+
t
[2.34]
DA ij
Dt
+
A ik S kj +
S ik A kj
The first term in square brackets on the right of the above
equality is symmetrical, and the second term below it is
antisymmetrical. However, the Hessian
x j is
symmetrical. By identifying the symmetrical and
antisymmetrical parts of the transport equation [2.30] after
decomposition, we successively come to
2 P
−∂
x i
2 S ij
DS ij
Dt
2 P
1
ρ
[2.35]
+
A ik A kj
+
S ik S kj
=−
x j + ν
x i
x k
x k
and
2 A ij
DA ij
Dt
[2.36]
+
A ik S kj
+
S ik A kj
= ν
x k
x k
We can easily show that the antisymmetrical tensor A ij is
linked to the vorticity component
k by
Ω
= Ω
+ ω
k
k
1
2 ε ijk Ω k
[2.37]
A ij =−
The antisymmetrical tensor
ε ijk is defined such that:
-
0 if two of the indices are arbitrarily identical
ε
=
ijk
(
0 );
ε 112
= ε 122
= ε 313
=
Search WWH ::




Custom Search