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Collision, diameter radius

Force fields for biological macromolecules fall into two classes united-atom, and all-atom force fields. All-atom force fields, as the name suggests, represent all atoms in a protein explicitly. United-atom force fields, on the other hand, treat only heavy (non-hydrogen) atoms and polar hydrogen atoms explicitly, while nonpolar hydrogen atoms are not represented explicitly, but rather represented as part of the carbon atom to which they are bonded (which will have an enlarged van der Waals radius (Lennard-Jones collision diameter)). [Pg.26]

Figure 1. The central ion K of radius b and the collision diameter a which represents the distance of closest approach between the centers of two ions. Figure 1. The central ion K of radius b and the collision diameter a which represents the distance of closest approach between the centers of two ions.
Considering the collisions between the part ides AB in configuration space, we must replace the actual collision diameter d by an effective collision diameter d = 2d, where d is the radius of a particle AB corresponding to the collision complex in the physical space. Each particle AB with an effective mass... [Pg.244]

W is the adsorption Vm is the liquid molar volume pf is the fluid density in occupied pores pni is the density of the multilayered adsorbate in pores is the radius of pores occupied at the pressure p Oss is the collision diameter of the surface atoms... [Pg.930]

If one sphere is much larger than the other one (f 2 R ), it is found again in the expression for a sphere near a plane surface. Equation 6.20 if both spheres have the same radius, half that value is obtained Fn(d) = nRJJi2 (d). For two spheres in contact (d = d, the collision diameter), the interaction energy may be equated to -2y, where y is the surface energy (Chapter 2),... [Pg.167]

The Kihara potential function [12] is used as described in McKoy and Sinanoglu [13]. The Kihara potential parameters, a (the radius of the spherical molecular core), a (the collision diameter), and e (the characteristic energy) are taken from Tohidi-Kalorazi [14], The fugacity of water in the empty hydrate lattice, // in Equation , can be calculated by ... [Pg.370]

Here the symbols have the same meaning as before, except that o, the collision diameter, is equal to b plus the mean radius of the ions in the solution, p is the ratio b/a, and a (p) is a function, tabulated by Kirkwood, equal to 1.21 when p = 0.6, and to 1.96 when p = 0.9. The first term in parentheses in Equation (8) reflects a solvent effect ( salting in ), due to electrostatic forces the second term is a salting out term which increases essentially in proportion to the volume of the molecule. [Pg.416]

In Equation 11.43 and Equation 11.44, the term f is the average pore radius in the dried layer (m). In Equation 11.45 and Equation 11.46, the terms o-win and (Tinw are the average collision diameters, and fiwin and fiinw are the collision integrals. [Pg.296]

In solutions, the distances between the centers of ions and of the nearest atoms of the surrounding solvent molecules can also be measured by x-ray and neutron diffraction, but with a somewhat larger uncertainty, 2pm. In aqueous solutions, if the water molecule is assigned a constant radius r = 138 pm (one half of the experimental collision diameter), then the distances t((I -0 )/pm = 138-fr,/pm have been established by Marcus within the experimental uncertainty, with the same ionic radii as in the crystals [45, 46], These radii, as selected in Ref 6 and annotated there, are listed in Table 2.8. The distances between the centers of ions in solutions in solvents other than water and of the nearest atoms of the solvents have also been determined in some cases reported by Ohtaki and Radnai [50] and confirm the portability of the Tj values among solvents, provided the mean solvent coordination number is near that in water. [Pg.30]

Suppose we consider a sample of a pure gas. (We will consider gas mixtures later, briefly.) How often does any one gas particle collide with other gas particles, and how far does the particle travel between collisions We can answer these questions by considering the hypothetical situation of one gas particle moving while all other particles are stationary. As the moving particle P travels through space, it will collide with any gas particle whose center gets within 2r (twice the radius) of the center of particle P. This is illustrated two-dimensionally in Figure 19.7. In three dimensions, the path of particle P sweeps out a cylinder of space, and any other particle whose center is in that space will collide with particle P. The radius of that cylinder, which is equal to twice the radius (2r) or the diameter (d) of the particle, is called the collision diameter of the particle. In three dimensions, the cross section of this cylinder is a circle whose area is this area is called the collision cross section of the gas particle. [Pg.681]

The molecular simulation method employed in this paper is the Grand Canonical Monte Carlo (GCMC) simulation. The parameters associated with the MC simulation used in this paper are (i) the linear dimension of the simulation box in the X- and y-diiections is at least 10 times the collision diameter, (ii) the cut-off radius is taken to be half of the box length, (iii) the number of cycles for equilibration and statistical collection step is 50,000 and (iv) in each cycle, there are N displacement moves and N rotations (in the case of 5-Site model) where N... [Pg.156]

As the moving particle travels along, it sweeps out a cylindrical volume as shown in Figure 9.19. The radius of this collision cylinder is equal to twice the radius of the molecules and is equal to d. We call d the collision diameter. The cross-sectional area of the collision cylinder is called the collision cross section... [Pg.426]

We now consider the rate of collisions of unlike molecules. The radius of the collision cylinder (the collision diameter) for collisions between molecules of substance 1 and substance 2 is denoted by d 2 and is equal to the sum of the radii of the molecules, or half the sum of their diameters ... [Pg.430]

Solution For the Stokes-Einstein equation, the chief problem is to estimate the radius of the oxygen molecule. If we assume that this is half the collision diameter in the gas, then from Table 5.1-2,... [Pg.132]

H. Termolecular Collisions Collision Complexes. Having defined a collision radius o-a for the hard sphere molecule M of core diameter o-r, it is possible to define a collision complex M M as a pair of molecules whose centers are a distance o- apart, where o-a [Pg.156]

In this section we will consider the collision frequency of the particles in a gas. We will start by considering a single spherical gas particle with diameter d (in meters) that is moving with velocity wavg. As this particle moves through the gas in a straight line, it will collide with another particle only if the other particle has its center in a cylinder with radius d, as shown in Fig. 5.21. [Pg.168]

The properties of gas flow in porous media depend on the ratio of the number of molecule-molecule collisions to that of the molecule-wall collisions. The Knudsen number Kn is a characteristic parameter defining different regions of this ratio. Its value is defined by Kn = Xl dp with X being the average free path length of the gas molecules and dp the characteristic pore diameter (sometimes the hydraulic pore radius is taken). [Pg.337]

The difficulties of a theoretical explanation of these facts arise not only from our ignorance of the forces which act between particles at distances approaching values smaller than the electronic radius, but from certain other circumstances which cannot be understood without explanations to be given in later chapters. One point is evident at once. Since the diameter of the nucleus has been shown by collision experiments to be of the same order of magnitude as the diameter of the electron, it follows that the obviously suggested and long-accepted idea that nuclei are built up of protons and electrons alone. [Pg.52]

Collision efficiency was calculated by the method proposed for the first time by Dukhin Derjaguin (1958). To calculate the integral in Eq. (10.25) it is necessary to know the distribution of the radial velocity of particles whose centre are located at a distance equal to their radius from the bubble surface. The latter is presented as superposition of the rate of particle sedimentation on a bubble surface and radial components of liquid velocity calculated for the position of particle centres. Such an approximation is possibly true for moderate Reynolds numbers until the boundary hydrodynamic layer arises. At a particle size commensurable with the hydrodynamic layer thickness, the differential of the radial liquid velocity at a distance equal to the particle diameter is a double liquid velocity which corresponds to the position of the particle centre. Such a situation radically differs from the situation at Reynolds numbers of the order of unity and less when the velocity in the hydrodynamic field of a bubble varies at a distance of the order ab ap. At a distance of the order of the particle diameter it varies by less than about 10%. Just for such conditions the identification of particle velocity and liquid local velocity was proposed and seems to be sufficiently exact. In situations of commensurability of the size of particle and hydrodynamic boundary layer thickness at strongly retarded surface such identification leads to an error and nothing is known about its magnitude. [Pg.407]

The collision frequency t must be obtained from some model for the liquid. If an effective hard sphere radius for the molecules can be chosen, the collision rate can then be obtained from Enskog theory or from molecular dynamics simulations of the hard sphere fluid. An alternative that has been popular is to use a cell model for the liquid." In its simplest form a molecule moves in a cell created by fixed neighbors located on a lattice. For a cubic lattice, the distance between neighbors is where p is the hquid-state number density. The central molecule then moves a distance 2p / —2a between collisions, if its effective diameter is a. The time between collision is then... [Pg.490]


See other pages where Collision, diameter radius is mentioned: [Pg.413]    [Pg.41]    [Pg.413]    [Pg.523]    [Pg.477]    [Pg.46]    [Pg.141]    [Pg.672]    [Pg.248]    [Pg.277]    [Pg.264]    [Pg.442]    [Pg.26]    [Pg.56]    [Pg.388]    [Pg.142]    [Pg.209]    [Pg.133]    [Pg.258]    [Pg.275]    [Pg.116]    [Pg.8]    [Pg.296]    [Pg.95]    [Pg.500]    [Pg.258]    [Pg.369]    [Pg.408]    [Pg.398]   
See also in sourсe #XX -- [ Pg.156 ]

See also in sourсe #XX -- [ Pg.156 ]




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Collision radius

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