Abstract
Using the Ostrogradsky-Gauss theorem to construct the laws of conservation and replacement of the integral over the surface by the integral over the volume, we neglect the integral term outside, i.e. neglect the circulation on the sides of the elementary volume (in the two-dimensional case, this is clearly visible). Circulation means the presence of rotation, which in turn means the presence of a moment of force (angular momentum). As a result, we have a symmetric stress tensor, a symmetric velocity tensor, etc. Static pressure, as follows from kinetic theory, there is a zero-order quantity; the terms associated with dissipative effects are first-order quantities. It does not follow from the Boltzmann equation and from the phenomenological theory that the pressure in the Euler equation is equal to one third of the sum of the pressures on the corresponding coordinate axes. The inaccuracy of determining the velocities in the stress tensor in the stress tensor does not strongly affect the results at low speeds. All these issues are discussed in the work. As example in this paper suggests task of flowing liquid at little distance of two parallel plates.
Highlights
The aim of the work is to study the consequences of using the Ostrogradsky-Gauss theorem in continuum mechanics in deriving conservation laws and numerically solving received equations
Differential laws are obtained in two ways: using the finite volume method for an elementary volume and using the Ostrogradsky Gauss theorem by replacing the surface integral to the volume integral, that is, taking the integral by parts with further use of the theorems on the conditions Integral turning in zero
The whole theory is based on the action of the law of conservation of moments [1], [2] partially the moment is used when considering stresses in beams
Summary
The aim of the work is to study the consequences of using the Ostrogradsky-Gauss theorem in continuum mechanics in deriving conservation laws and numerically solving received equations. Hydrostatic pressure is a zero-order value, but the theory remains the same when determining different pressures at each surface, i.e. ppxx, ppyy, ppzz The use of one pressure is possible under equilibrium conditions (Pascal's law), but for nonequilibrium conditions the fact is not obvious. This is highlighted in the textbook [1]. For Euler's equations, this means that Pascal's law in the non-equilibrium case does not work and it is necessary to consider separately not pressure pp, but ppxx, ppyy, ppzz. The work discusses the listed issues and, if possible, gives answers to some of them
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