Pressure of a Drop

Calculate the vapor pressure of a liquid drop as a function of its radius. Eind the increase in vapor pressure for water at 100°C for drop radii of 100 / m, 1 and 0.01 pim. Take the surface tension to be 60 mN/m and the density to be 994 kg/m for water at this temperature. 

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In this chapter we get to know the second essential equation of surface science — the Kelvin5 equation. Like the Young-Laplace equation it is based on thermodynamic principles and does not refer to a special material or special conditions. The subject of the Kelvin equation is the vapor pressure of a liquid. Tables of vapor pressures for various liquids and different temperatures can be found in common textbooks or handbooks of physical chemistry. These vapor pressures are reported for vapors which are in thermodynamic equilibrium with liquids having planar surfaces. When the liquid surface is curved, the vapor pressure changes. The vapor pressure of a drop is higher than that of a flat, planar surface. In a bubble the vapor pressure is reduced. The Kelvin equation tells us how the vapor pressure depends on the curvature of the liquid. 

The Young-Laplace equation (3.4/3.5) shows that, pA>pB, the pressure inside a bubble or drop exceeds that outside. For a sphere, Ap=pA - Pb = 2y/R, so that Ap varies with the radius, R. Thus the vapour pressure of a drop should be higher, the smaller the drop. This is shown by a related equation, the Kelvin equation [13,26], which is described here. 

Now, conversely, if we consider a spherical liquid drop in air, having a radius of r, the vapor pressure of a drop, Pcv > P that is Pv is higher than that of the same liquid with a flat surface, Pv (the superscript c indicates a curved surface). If d mol of liquid evaporates from the drop and condenses onto the bulk flat liquid under isothermal and reversible conditions, the free-energy change of this process can be written by differentiating Equation (155) as... 

Since for a drop C is positive, the vapour pressure of a drop is greater than that of the bulk liquid conversely for liquid con-... 

This is Kelvin s equation for the vapour pressure of a drop of radius R the second (approximate) form of whidi holds only if NjkT v Piy. It is inapplicable to drops containing only a few molecides since they have no uniform interior for whidi the calculation of p is meaningful. iW a drop of water of radius 1 mm, ln(p /p ) is lOA and for a drop of radios 1 pm, it is 10. Hence in a mist the very small drops evaporate and the larger drops grow. All are unstable with respect to the pool of liquid that ultimately forms. 

If Ap is larger than the Laplace pressure of a drop (eqn. (2.1) pt = 1-yl near the eddy, the drop would be broken up. Break-up would be most effective if rf = Z. Putting eqn. (2.15) equal to eqn. (2.1) and putting x = dm , there results the following expression for the largest droplets that are not broken up in the turbulent field... 

Some pre-requisites are that (i) the membrane is hydrophilic for a hydrophobic disperse phase and vice versa, since otherwise the drops cannot be detached (ii) the pores must be sufficiently far apart from each other to prevent the drops that come out touching each other and coalescing and (iii) the pressure over the membrane should be sufficient to achieve drop formation. The latter pressure should be at least of the order of the Laplace pressure of a drop of diameter equal to the pore diameter. For pores of 0.4 pm and y = 5 mNm , this would amount to 50 kPa, but larger pressures are needed in practice, for instance 3 X 10 Pa, also to obtain a significant flow rate of the disperse phase through the membrane. 

The vapor pressure of a drop is higher than that of a liquid with a planar surface. One consequence is that an aerosol of drops (fog) should be unstable. To see this, let us assume that we have a box filled with many drops in a gaseous environment. Some drops are larger than others. The small drops have a higher vapor pressure than the... 

When fluid flow in the reservoir is considered, it is necessary to estimate the viscosity of the fluid, since viscosity represents an internal resistance force to flow given a pressure drop across the fluid. Unlike liquids, when the temperature and pressure of a gas is increased the viscosity increases as the molecules move closer together and collide more frequently. 

Equip a 1 litre bolt-head flask with dropi)ing fuuncl and a double surface reflux condenser to the top of the latter attach a device (e.g.. Fig. II, 8, 1. c) for the absorption of the hydrogen bromide evolved. Place 100 g. (108 ml.) of dry iso-valeric acid (Section 111,80) and 12 g. of pmified red phosphorus (Section 11,50,5) in the flask. Add 255 g. (82 ml.) of dry bromine (Section 11,49,5) slowly through the dropping funnel at such a rate that little or no bromine is lost with the hydrogen bromide evolved the addition occupies 2-3 hours. Warm the reaction mixture on a water bath until the evolution of hydrogen bromide is complete and the colour of the bromine has disappeared. Pour off the liquid reaction product into a Claisen flask and distil mider the reduced pressure of a water pump. Collect the a-bromo-wo-valeryl bromide at 117-122°/25-30 mm. The yield is 150 g. 

Partial Pressure Pinch An example of the hmitations of the partial pressure pinch is the dehumidification of air by membrane. While O9 is the fast gas in air separation, in this apphcation H9O is faster still. Special dehydration membranes exhibit a = 20,000. As gas passes down the membrane, the pai-dal pressure of H9O drops rapidly in the feed. Since the H9O in the permeate is diluted only by the O9 and N9 permeating simultaneously, p oo rises rapidly in the permeate. Soon there is no driving force. The commercial solution is to take some of the diy air product and introduce it into the permeate side as a countercurrent sweep gas, to dilute the permeate and lower the H9O partial pressure. It is in effect the introduction of a leak into the membrane, but it is a controlled leak and it is introduced at the optimum position. 

The hydrostatic testing is the first test used on the lube system. The system is tested while assembled or partially assembled, based on the particular system. A test pressure of VA times the maximum allowable working pressure, a minimum of 20 psi for the oil side, is used for the test. For the oil-wetted parts, the test fluid should be light oil, which is normally the recommended lubricant for the compressor train. The test period is the length of time needed to inspect for leaks, or a minimum ol thirty minutes. Acceptance is based on the lack of leaks as visually observed or the lack of a drop in the test pressure. 

Lapple s method is useful when the upstream pressure of a header is known and the downstream pressure has to be calculated. However, it is often required to develop the pressure profile of the flare headers as a function of the distance from the stack. For this reason, it is more convenient to calculate the pressure drop backward, starting from the flare stack exit where the pressure is atmospheric. Figure 20 provides another plot which enables the pressure loss calculation when the downstream pressure is known. 

When the temperature drops from 20°C to 10°C, the pressure of a cylinder of compressed N2 drops by 3.4%. The same temperature change decreases the pressure of a propane (C3H8) cylinder by 42%. Explain the difference in behavior. 

Overall, the partial pressure of N204 drops, and the forward reaction slows down. Conversely, the partial pressure of N02 increases, so the rate of the reverse reaction increases. Soon these rates become equal. A dynamic equilibrium has been established. 

It is an immediate consequence of the result of the preceding paragraph that the vapour-pressure of a liquid in the form of drops... 

Notice that, for every 1000 mmHg drop in the pressure of A(g), there will be a... 

Here we are talking about evaporation under thermodynamic equilibrium. We can also have evaporation under nonequilibrium conditions. For example, if the pressure of a liquid is suddenly dropped below its saturation pressure, flash evaporation will occur. The resulting vapor will be at the boiling point or saturation temperature corresponding to the new pressure, but the bulk of the original liquid will remain (out of equilibrium) at the former higher temperature. Eventually, all of the liquid will become vapor at the lower pressure. The distinction between flash evaporation and equilibrium evaporation is illustrated in Figure 6.6 for water. 

The triple point of a substance is reached when the vapor pressure of the solid phase is equal to that of the liquid phase. If both solid and liquid are subjected to external pressure (which may be caused by capillary forces), their curves of vapor pressure versus temperature lie above those for uncompressed phases and intersect at a temperature different from the triple point. The melting point Tm observed at atmospheric pressure, as a rule, is very near to the triple point. Thus the freezing temperature Tmr of a drop of radius r should be different from Tm. 

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