in an au аж ах axi a UIT ().
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
Related questions
Question
How do you get from equation 3.1.1 to 3.1.5? I understand that yoy mutiply both sides by Ui, but I'm confused on the math that is done to bring Ui into the partial derivative. Please show all intermediate steps.

Transcribed Image Text:n. The use of
much
small scal
and length
thought, in
length scale
strain-rate fluctuations play a role.
3.1
Kinetic energy of the mean flow
le found in Section 2.1 that the equations of motion for steady mean flow
located on
parable to
a qualita-
in an incompressible fluid are
(3.1.1)
au, a
!xe
ax ax
omentum
(3.1.2)
3D0.
The stress tensor T is
(3.1.3)
Tj = -Pôy + 2uSij -p ujuj.

Transcribed Image Text:The dynamics of turbulence
61
The mean rate of strain S, is defined by
1 ( au, au
(3.1.4)
Since the mean momentum u, of the turbulent velocity fluctuations is zero
we cannot discuss the effects of the mean flow on the turbulence very well in
terms of mean momentum. We shall study the equations for the kinetic
energy of the mean flow and of the turbulence instead. The equation govern-
ing the dynamics of the mean-flow energy U,U;, is obtained by multiplying
(3.1.1) by Uj. It is useful to split the stress term in the resulting equation into
two components. The energy equation becomes
'ne
(TU;) -T
axi
pUj
UU)=
(3.1.5)
%3D
Because T is a symmetric tensor, the product Tj aU;/ax, is equal to the
product of T and the symmetric part S of the deformation rate aU,/axi
(3.1.5) thus becomes
GU;U) =
pUi axi
(TUi) - TSi.
%3D
(3.1.6)
The first term on the right-hand side of (3.1.6) represents transport of
mean-flow energy by the stress T. Thie te
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