Equilibrium droplet radius

The partial derivative in Equation 9.5 is taken at constant temperature, pressure, and number of moles per unit volume n, n, and n, of continuous phase, dispersed phase, and surfactant in the microemulsion. However, the number of droplets N may vary. This equation applies even when no excess phase is present and would be used to find the equilibrium droplet radius in that case. In Equation 9.6, the partial derivative is taken at constant N but with variable n. It is applicable only when excess disperse phase is present. Using both these equations, the above authors found that with a given amount of surfactant present, more droplets of smaller size are predicted than would be expected based on the pertinent natural radius. That is, solubilization with an excess of the dispersed phase is somewhat less than if droplets assumed their natural radius. In this case the energy required to bend the film beyond its natural radius is offset by the decrease in free energy associated with the increased entropy of the more numerous droplets. 

The minimization of the total free energy F -f F2) with the radius gives the equilibrium droplet radius / , which depends on the natural radius Ro, the interfacial elasticity k and the energy of inter-droplet interactions F2, as follows ... 

Figure 10.3 Equilibrium droplet radius vs. electrolyte concentration. The full line is for a = 1 pm, the dashed is for a = 0.5 pm and the dotted is for a = 2 pm. The remaining parameters used in the calculations are given in the text (Reproduced by permission of Academic Press from ref. 27)...

Water-to-droplet MT of 2-ferrocenyl-2-propanol (FeCp-PrOH) has been also studied in detail, as a t dependence of Q(t) (Figure 19). The Q(t) value almost saturates to 0.75 nC at t 30 s for the droplet (C0 = 0.047 M) with r = 4.3 fim. Q(t) at the distribution reequilibrium (Qeq), calculated on the basis of a droplet radius (Keq = 700) and the contact angle of an NB droplet on an Au electrode ( 120°), is 1.3 nC, so that the value is larger than the observed saturated Q(t) value in Figure 19. Liquid/liquid extraction experiments of FeCp-PrOH for a large number of NB droplets indicate that the time required for the distribution equilibrium is > 103 s. Clearly, water-to-... 

Wolf et al. calculated the equilibrium size of droplets formed in a phase-separated system. From a force balance, he derived an expression Indicating that the equilibrium droplet size r is a decreasing function of shear rate and that when r approaches the radius of gyration of the polymer molecules, redissolution will have occurred. Recently Kramer and Wolf have generalized the approach and formulated simple criteria for solution, respectively demixing (16). 

If we further raise the relative humidity after the phase change (see Fig. 37) the radius of the droplet increases and the solution becomes weaker and weaker. This means that at higher relative humidity a more dilute solution is in dynamic equilibrium with the vapour environment. It should be mentioned that the equilibrium radius is governed also by the curvature of the droplet. Since the relation between the curvature and droplet radius is given by the well-known Thomson equation, we may write (Dufour and Defay, 1963 E. Meszaros, 1969) ... 

The lower diagram of Fig.9 shows the size distribution of all droplets passing a certain cross section it was calculated from the results for droplet growth together with the corresponding nucleation rates. In the range where the two-phase equilibrium is nearly attained, (between x 3 mm and x 4 mm) the calculation yields a most probable droplet radius of about 40 nm and a maximum radius of about 70 nm. 

We now consider the structure of stationary, finite amplitude waves in one-dimensional steady flow. Far upstream of the wave the flow is assumed to be in thermodynamic and velocity equilibrium with a specified pressure, wetness fraction and droplet radius. Far downstream of the wave a new equilibrium condition is re-established. The far upstream... 

FIGURE 2 Equilibrium saturation ratio versus droplet radius for different masses of sodium chloride nuclei (solid lines). Asterisks show droplet radius where solution drop containing indicated mass of sodium chloride will continue to grow without further increase in the saturation ratio. Dashed curve is for a pure water droplet. RH, Relative humidity. [Adapted from Byers, H. R. (1965). Elements of Cloud Physics, courtesy of the University of Chicago Press and the author.]... 

FIGURE 27.7 Number of moles of ammonia vapor and liquid versus time predicted by AERCLOUD, for 85% liquid releases in dry air and in 99.99% humid air allowing for ammonia-water interactions. The initial droplet radius ranges from 1000 /xm to 100 /xm. The curves have been computed using three model options including and excluding droplet ventilation and in the homogeneous equilibrium limit. 

Let be the radius of the equilibrium droplet. According to the definition of the capillary pressure. 

Equation 5.205 allows determining the final droplet radius, r , at the end of the spreading process using the known dependence G(pjJ. To this end, it is necessary to substitute the expression for the equilibrium fraction of overturned molecules under droplet from relationship (5.198) into Equation 5.205, which yields... 

A homogeneous metastable phase is always stable with respect to the fonnation of infinitesimal droplets, provided the surface tension a is positive. Between this extreme and the other thennodynamic equilibrium state, which is inhomogeneous and consists of two coexisting phases, a critical size droplet state exists, which is in unstable equilibrium. In the classical theory, one makes the capillarity approxunation the critical droplet is assumed homogeneous up to the boundary separating it from the metastable background and is assumed to be the same as the new phase in the bulk. Then the work of fonnation W R) of such a droplet of arbitrary radius R is the sum of the... 

Let us first consider interfaces at equilibrium. Any stress (osmotic or shear stress) imposed to the emulsion increases the amount of interface, leading to a modification of the free energy. For a monodisperse collection of N droplets of radius a, the total interfacial area of the undeformed droplets is So = 47t No. If the emulsion is compressed up to each droplet is pressing against its neighbors through... 

Take as an example, a small dry particle of NaCl of a given mass (mu) that is introduced into air at a water vapor pressure corresponding to SA in Fig. 14.38a. Assuming that the RH is above the deliquescence point of NaCl, 75% at 25°C, the particle will take up water, dissolve, and form a stable droplet of radius rA. Similarly, if the air saturation ratio increases to Su, the particle will, under equilibrium conditions, take up water and grow to radius ru. 

Consider, however, a particle at point D in Fig. 14.38a. If it gains some water molecules and the radius starts to increase, the surrounding air will have a larger supersaturation than the required equilibrium value of S for this larger particle. As a result, water will condense out on the droplet, causing it to grow further. Particles that lie to the right of the peak are thus in an unstable equilibrium (e.g., see Reiss and Koper, 1995) and can therefore activate into cloud droplets from the condensation of water. 

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