Up to this point we have considered spherical particles of a known diameter Dp and density pp. Atmospheric particles are sometimes nonspherical and we seldom have information about their density. Also a number of techniques used for atomospheric aerosol size measurement actually measure the particle's terminal velocity or its electrical mobility. In these cases we need to define an equivalent diameter for the nonspherical particles or even for the spherical particles of unknown density or charge. These equivalent diameters are defined as the diameter of a sphere, which, for a given instrument, would yield the same size measurement as the particle under consideration. A series of diameters have been defined and are used for such particles.
The volume equivalent diameter Dve is the diameter of a sphere having the same volume as the given nonspherical particle. If the volume Vp of the nonspherical particle is know then:
TT r
For a spherical particle the volume equivalent diameter is equal to its physical diameter,
To account for the shape effects during the flow of nonspherical particles, Fuchs (1964) defined the shape factor % as the ratio of the actual drag force on the particle FD to the drag force Fp on a sphere with diameter equal to the volume equivalent diameter of the particle:
The dynamic shape factor is almost always greater than 1.0 for irregular particles and flows at small Reynolds numbers and is equal to 1.0 for spheres. For a nonspherical particle of a given shape % is not a constant but changes with pressure, particle size, and as a result of particle orientation in electric or aerodynamic flow fields.
The dynamic shape factor for flow in the continuum regime is equal to 1.08 for a cube, 1.12 for a 2-sphere cluster, 1.15 for a compact 3-sphere cluster, and 1.17 for a
550 nm | ||
491 nm |
i |
vlaC! |
PSL | ||
|- |
■4 500 nm |
FIGURE 9.15 A micrograph of a single NaCl particle with electrical mobility equivalent diameter of 550 nm. The dry NaCl is aimosl cubic with rounded edges. Also shown is a polystyrene latex (PSL) particle with diameter of 491 nm (Zclcnyuk ct al. 2006).
FIGURE 9.15 A micrograph of a single NaCl particle with electrical mobility equivalent diameter of 550 nm. The dry NaCl is aimosl cubic with rounded edges. Also shown is a polystyrene latex (PSL) particle with diameter of 491 nm (Zclcnyuk ct al. 2006).
compact 4-sphere cluster (Hinds 1999). These values are averaged over all orientations of the particle, which is the usual situation for atmospheric aerosol flows (Re < 0.1) because of the Brownian motion of the particles. Liquid and organic atmospheric particles are spherical for all practical purposes. Dry NaCl crystals have cubic shape (Figure 9.15), while dry (NHj^SC^ is approximately but not exactly spherical (Figure 9.16). Dynamic shape factors ranging from 1.03 to 1.07 have been measured in the laboratory (Zelenyuk et al. 2005) with the higher values observed for larger particles with diameters of ~500nm.
NonsphericaS particles are subjected to a iarger drag force compared to their volume equivalent spheres because % > 1 and therefore settle more slowly. The terminal settling velocity of a nonspherical particle is then [following the same approach as in the derivation of (9.42)1
Volume Equivalent Diameter An approximately cubic NaCl particle with density 2.2gem-3 has a terminal settling velocity of 1 mm s-1 in air at ambient conditions. Calculate its volume equivalent diameter and its physical size using the continuum regime shape factor.
The terminal settling velocity is given by (9.104), so after some rearrangement; we obtain
and substituting v, — 10-3ms-1, ^ = 1.8 x 10~5 kgm~! s~\ % = 1.08 for a cube, pp = 2200 kg nT3 for NaCl we find that
This equation needs to be solved numerically using (9.34) for calculation of the slip correction, Cr(Dve), to obtain Dve = 0.328 ^im. One can also estimate the value iteratively assuming as a first guess that Cc = 1 and then Dve = 0.402 |im. This suggests that for this particle Dve » 2X and the exponential term in (9.34) will be much smaller than the 1.257 term. With this simplification we are left with the following quadratic equation for Dvc
with Dvl, = 0.329 Jim as the positive solution. Our simplification of the slip correction expression resulted in an error of only 1 nm.
To calculate the physical size L of the particle, we only need to equate the volume of the cube to the volume of the sphere:
1} = (it/6)E?„ and L = 0.264 pm This calculation requires knowledge about the shape of the particle and its density.
The diameter of a sphere having the same terminal settling velocity and density as the particle is defined as its Stokes diameter DS1. For irregularly shaped particles DSt is the diameter of a sphere that would have the same terminal velocity. The Stokes diameter for Re <0.1 can then be calculated using (9.42) as
where v, is the terminal velocity of the particle, p is viscosity of air, and pp is the density of the particle that needs to be known. Evaluation of (9.106) because of the dependence of Cc on Ds, requires in general the solution of a nonlinear algebraic equation with one unknown. For spherical particles by its definition Dst = Dp and the Stokes diameter is equal to the physical diameter.
Stokes Diameter What is she relationship connecting the volume equivalent diameter and the Stokes diameter of a nonspherical particle with dynamic shape factor X for Re <0.1?
Calculate the Stokes diameter of the NaCl panicle of the previous example. The two approaches (dynamic shape factor combined with the volume equivalent diameter and the Stokes diameter) are different ways to describe the drag force and terminal settling velocity of a nonspherical particle. The terminal velocity of a nonspherical particle with a volume equivalent diameter Dve is given by (9.104),
By definition this settling velocity can be also written as a function of its Stokes diameter as
1 D2SlPpgCc{DSt)
Combining these two and simplifying wc find that
The Stokes diameter of the NaCl particle can be calculated from (9.106):
and solving numerically Dsi = 0.313 ¡.tm. This value corresponds to the volume equivalent diameter that we would calculate based on the terminal velocity of the panicle if we did not know that the particle was nonspherical.
For most atmospheric aerosol measurements the density of the particles is not known and the Stokes diameter cannot be calculated from the measured particle terminal velocity. This makes the use of the Stokes diameter more difficult and requires the introduction of other equivalent diameters that do not require knowledge of the particle density.
The diameter of a unit density sphere, pJJ = 1 gem-3, having the same terminal velocity as the particle is defined as its classical aerodynamic diameter, Dca. The classical aerodynamic diameter is then given by
Dividing (9.i08) by (9.106), one can then find the relationship between the classical aerodynamic and the Stokes diameter as
For spherical particles we replace DS[ with Dp in this equation to find that
For a spherical particle of nonunit density the classical aerodynamic diameter is different from its physical diameter and it depends on its density. Aerosol instruments like the cascade impact or and aerodynamic particle sizer measure the classical aerodynamic diameter of atmospheric particles, which is in general different from the physical diameter of the particles even if they are spherical.
Aerodynamic Diameter Calculate the aerodynamic diameter of spherical particles of diameters equal to 0.01, 0.1, and 1 pm. Assume that their density is 1.5 gem-3, which is a typical average density for multicomponent atmospheric particles. Using (9.110), we obtain
for Dp = 0.01 ¡im, Cc = 22.2, and D2caCc{Dcu) = 3.33 x 10"3 pm2 with solution Dcg = 0.015pm. Repeating the same calculation we find that for Dp = 0.1 |im, Ce = 2.85, and D2cuCc(Dca) = 0.043 pm2, resulting in Dca = 0.135 pm, Finally, for Dp = I pm we find that Dia — 1.242 pm. For typical atmospheric particles the aerodynamic and physical diameters are quite different (more than 20% in this case) with the discrepancy increasing for smaller particles.
Vacuum Aerodynamic Diameter Calculate the ratio of the aerodynamic to the physical diameter of a spherical particle of density pp in the continuum and the free molecular regimes.
Using equation (9.110) yields
For conditions in the continuum regime the slip correction factor is practically unity and
If the particle is in the free molecular regime, then 2a 3> Dp and the second term dominates the RHS of (9.34), which simplifies to
Combining this and simplifying, we find that in the free molecular regime:
The aerodynamic diameter of a spherical panicle with diameter Dp and nonunit density will therefore depend on the mean free path of the air molecules around it and thus also on pressure [see (9.6)]. For low pressures resulting in high Knudsen numbers, the particle will be in the free molecular regime and the aerodynamic diameter will be proportional to the density of the particle and given by (9,113). This diameter is often called the vacuum aerodynamic diameter or the free molecular regime aerodynamic diameter of the particle. A number of aerosol instruments that operate at low pressures such as the aerosol mass spectrometer measure the vacuum aerodynamic diameter of particles. In the other extreme, for high pressures resulting in low Knudsen numbers the particle will be in the continuum regime and its aerodynamic diameter will be proportional to the square root of its density (9.112). This is known as the continuum regime aerodynamic diameter.
The aerodynamic diameter of the particle changes smoothly from its vacuum to its continuum value as the Knudsen number decreases.
Electrical mobility analyzers, like the differential mobility analyzer, classify particles according to their electrical mobility Be given by (9.50). The electrical mobility equivalent diameter De„, is defined as the diameter of a particle of unit density having the same electrical mobility as the given particle. Particles with the same Dem have the same migration velocity in an electric field. Particles with equal Stokes diameters that carry the same electrical charge will have the same electrical mobility.
For spherical particles assuming that the particle and its mobility equivalent sphere have the same charge then Dem ~ Dp = Dve. For nonspherical particles one can show that
Instruments such as the differential mobility analyzer (DMA) (Liu et al. 1979) size particles according to their electrical mobility equivalent diameter.
9.1A (a) Knowing a particle's density pp and its settling velocity vt, show how to determine its diameter. Consider both the non-Stokes and the Stokes law regions, (b) Determine the size of water droplet that has v, = 1 cms"' al T = 20°C, 1 atm.
9.2a (a) A unit density sphere of diameter 100 pm moves through air with a velocity of 25 cms"1. Compute the drag force offered by the air. (b) A unit density sphere of diameter 1 pm moves through air with a velocity of 25 cm s~'. Compute the drag force offered by the air.
9.3a Calculate the terminal settling velocities of silica particles (p = 2.65 gem 3) of 0.05 pm, 0.1 pm, and 0.5 pm, and 1.0 pm diameters.
9.4a Calculate the terminal settling velocities of 0.001, 0.1, 1, 10, and 100 pm diameter water droplets in air at a pressure of 0.1 atm.
9.5a Develop a table of terminal settling velocities of water drops in still air at T = 20°C, 1 atm. Consider drop diameters ranging from 1.0 to 1000 pm. (Note that for drop diameters exceeding about 1 mm the drops can no longer be considered spherical as they fall. In this case one must resort to empirical correlations. We do not consider that complication here.)
9.6A What is the stop distance of a spherical particle of 1 pm diameter and density 1.5 g cm 3 moving in still air at 298 K with a velocity of lms 1 ?
9.7B A 0.2-pm-diameter particle of density 1 g cm 3 is being carried by an airstream at 1 atm and 298 K in the y direction with a velocity of 100 cm s_1. The particle enters a charging device and aquires a charge of two electrons (the charge of a single electron is 1.6 x 10~19 C) and moves into an electric field of constant potential gradient Ex = 1000 V cm 1 perpendicular to the direction of flow.
a. Determine the characteristic relaxation time of the particle.
b. Determine the particle trajectory assuming that it starts at the origin at time zero.
c. Repeat the calculation for a 50-nm-diameter particle.
9.8C At i = 0 a uniform concentration Nq of monodisperse particles exists between two horizontal plates separated by a distance h. Assuming that both plates are perfect absorbers of particles and the particles settle with a settling velocity v„ determine the number concentration of particles as a function of time and position. The Brownian diffusivity of the particles is D.
9.9B The dynamic shape factor of a chain that consists of 4 spheres is 1.32. The diameter of each sphere is 0.1 pm. Calculate the terminal settling velocity of the particle in air at 298 K and 1 atm. What is the error if the shape factor is neglected? Assume the density of the spheres is 2 g cm 3
9.10B Derive (9.110) using the appropriate force balance for the motion of a charged particle in an electric field.
9.11B Spherical particles with different diameters can have the same electrical mobility if they have a different number of elementary charges. Calculate the diameters of particles that have an electrical mobility equal to that of a singly charged particle with Dp = 100 nm assuming that they have 2, 3, or 4 charges. Assume T= 298 K and 1 atm.
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