By Flügge S. (ed.)

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**Example text**

167) dT . 168) T where q stands for the heat input per unit mass of gas. According to Joule’s experiments, the internal energy of a perfect gas is contained entirely within the molecule itself, as energy of translation and rotation, and the intermolecular forces are not important. Therefore, for a perfect gas, we have ∂e ∂v = 0. 169) T Employing Eq. 169), directly from Eq. 168), we define the specific heat at constant volume Cv as Cv = ∂q ∂T = v ∂e ∂T . 170) v Hence, integrating Eq. 170), assuming that the internal energy vanishes at zero temperature, which is in accordance with the third law of thermodynamics, we derive e = Cv T .

In numerical approaches for FSI problems, if we are to follow the moving boundaries, in order to avoid excessive mesh distortions we often need to arbitrarily select proper mesh motions using the so-called arbitrary Lagrangian–Eulerian (ALE) kinematic description for continuum media, in which the mesh position for the description of the state variables is neither fixed in space nor attached with material particles. Mass point and rigid body A point mass is a single mass point whose spatial dimension is negligibly small in comparison with the distances involved in the problem under consideration.

Aspects of Mathematics and Mechanics 46 Using Eq. 165), from Eq. 172), we can express the specific heat at constant pressure Cp as follows: Cp = ∂q ∂T = p d(pv) + Cv = R + Cv . 175) where so , To , and po represent the initial entropy per unit mass, temperature, and pressure. 176) we have Cv = R/(γ − 1). Hence, we can easily derive from Eqs. 177) where C is a constant. According to the basic principle in classical kinetic theory, the internal energy of each degree of freedom of the individual molecules of a gas can be expressed as KT /2, where K represents the Boltzmann constant.