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in an au аж ах axi a UIT ().

in an au аж ах axi a UIT ().

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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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.

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: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.
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
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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