written and published by Renzo Diomedi

the Fluid is incompressible, with variations of the density without appreciable effects and density approximately Constant.

Reminding that

Continuity + Momentum = +

Conservative continuity equation (scalar value) where u, v, w lie on x, y, z and u^x v^y w^z have cosine = 1.

Non-Conservative Momentum equation (not exactly measurable as a scalar field)

the non-conservative Momentum equation are however decomposable in 3 scalar equations laid along the directions of

Then considering

; = Density , = Viscosity , = standard gravity (acceleration)

where the independent variables are 4 :

The dependent variables are 10 , of which 3 velocity components + the pressure

and 6 dependent variables given by the Stress Tensor:

So, as seen above, to get the value of the unknowns, i.e. the 10 dependent variables, we have 4 scalar equations only.

expansed Continuity eq =

momentum eq is also

where the Stress Tensor

is applied to NS eq

The components of the

The Principal stresses and Shear stresses act in a 3d space. But if they act in a deformable and curvable sheet, they can act in a 2-dimensional space rather than 3d. The lattice deforms by the shift of its axes and coordinates

Let a lattice composed of equidistant points subjected to stresses and crossed by a flow to analyze. The Turbulence produces lattice distortions, but the points remain at a constant distance among them because only the axial coordinates of reference vary.

The Metric Tensor

Considering = metric abscissa and = metric ordinate

= distance of point from origin, whatever the inclination of the reference axes

where

(note that : , , ,

contravariant metric tensor: , covariant metric tensor: )

2-dimensional viscous stress tensor :

Let the Point lies in a Plane of a Volume used to calculate the turbulence. Then the Deformation of this plane determines the Turbulence (Reynolds number > 4000 )

calculating the variation of the angle between the axes of the plane which is 90 in the state of quietness, we can extract a value of the deformation and relative turbulence

We need to know the physical properties of the micro volume considered , eg the developed heat , so we use the co-variant coordinates only.

the angles of curvature in the volume defined by the points chosen, can be extracted using the Mohr formulas:

Sigma = Normal tension

Tau = Tangential tension

that derived is

that derived is

= angle beetween normal axis of the square and the oblique plane + the same angle in the opposite side

(note that in Mohr circle , the angle beetween x and y has 180 degrees instead of 90)

then

then applying this value of this double angle to and we get

then

and

where the Tension values we need, they are given by Stoke's relations

on 2-dimesional plane of the Volume.

So, first we must find the value of the tensions and then we apply these values on mohr cicle with metric axes

The half-values of the angles got by is the key to calculate the covariant coordinates

we can choose an appropriate series of planes inside the volume to calculate the turbulence in the different points inside the volume

hence we can calculate the amount of energy created over a certain period of time by 2-dimensional version of N-S equations:

Continuity:

Momentum :

Now, the independent variables are 3, the dependent variables are 6 and the output equations are 3. So the imbalance has been reduced from 6 to 3

The flow density is unsteady.

Reminding that

,

where V = average speed , D = system dimension , ,

Linear calculus of the Energy developed (output must be a scalar value) :

then assuming and using the lattice deformable by metric coordinates variation

TO BE CONTINUED........

TO REDEFINE...................................

We need to avoid non-linearity of NS. We need an unidimensional output. A lagrangian particle in its path creates a String. Nambu-Goto equation analyzes the behavior of the string and the energy produced by it, proportional to the minimum area of the worldsheet area. So we apply the N-G's Action :

where

the position of the lagrangian particle, in the A, is given by the

= String viscous Tension :

where , , then ,

END REDEFINE