Up to this point, we have considered the drag force on a particle moving at a steady velocity Mqo through a quiescent fluid. Recall that this case is equivalent to the flow of a fluid at velocity past the stationary particle. The motion of the particle, however, arises in the first place because of the action of some external force on the particle such as gravity. The drag force arises as soon as there is a difference between the velocity of the particle and that of the fluid. The basis of the description of the behavior of a particle in a fluid is an equation of motion. To derive the equation of motion for a particle of mass mp, let us begin with a force balance on the particle, which we write in vector form as i where v is the velocity of the particle and F, is the ;th force acting on the particle.

For a particle falling in a fluid there are two forces acting on it, the gravitational force mpg and the drag force Fdrag. Therefore, for Re < 0.1, the equation of motion becomes mP^=mP g + (9-36)

where the second term of (9.36) is the corrected Stokes drag force on a particle moving with velocity v in a fluid having velocity u. Equation (9.36) implicitly assumes that even though the particle motion is unsteady, this acceleration is slow enough that Stokes' law applies at any instant. This equation can be rewritten as dx

dt where is the characteristic relaxation time of the particle.

Let us consider the case of a particle in a quiescent fluid (u = 0) starting with zero velocity and let us take the z axis as positive downward. Then the equation of motion becomes d i)

T-ir = Tg-vz vz(0) = 0 (9.39) dt and its solution is vz(t) = xg[l - exp(—i/x)] (9.40)

For t» x, the particle attains a characterstic velocity, called its terminal settling velocity v, = xg or

3n\iDp

Dp, |im |
T, S |

0.05 |
4 x 10'8 |

0.1 |
9.2 x 10'8 |

0.5 |
1 x 10~6 |

1.0 |
3.6 x 10"6 |

5.0 |
7.9 x lO"5 |

10.0 |
3.14 x lO"4 |

50.0 |
7.7 x lO"3 |

For a spherical particle of density pp in a fluid of density p, mp — (n/6)Dip(pp — p), where the factor (pp — p) is needed to account for both gravity and buoyancy. However, since generally pp P,mp = (n/6)Dppp and (9.41) can be rewritten in the more convenient form:

The timescale x indicates the time required by the particle to reach this terminal settling velocity and is given in Table 9.4. The relaxation time t also describes the time required by a particle entering a fluid stream, to approach the velocity of the stream. Thus the characteristic time of most particles of interest to achieve steady motion in air is extremely short.

Settling velocities of unit density spheres in air at 1 atm and 298 K as computed from (9.42) are given in Figure 9.6. Submicrometer particles settle extremely slowly, only a few

FIGURE 9.6 Settling velocity of particles in air at 298 K as a function of their diameter.

FIGURE 9.6 Settling velocity of particles in air at 298 K as a function of their diameter.

centimeters per hour. Particles larger than 10pm settle with speeds exceeding 10mh~! and therefore are expected to have short atmospheric lifetimes.

Our analysis so far is applicable to Re < 0.1 or particles smaller than about 20 pm (Table 9.2). For larger particles, one needs to use the drag coefficient as an empirical means of representing the drag force for higher Reynolds numbers. The equation along the direction of motion of the particle in scalar form, assuming no gas velocity, is then

At steady-state = v,, the particle reaches its terminal velocity given by

However, as CD is a function of Re and therefore v„ we have only an implicit expression for v, in (9.44). One needs then to solve (9.44) numerically with CD calculated by (9.32) or one can use the following technique (Flagan and Seinfeld 1988), If we form the product

and substitute into this the v, given by (9.44), we obtain

CpRe2 can be calculated from (9.32) and one can prepare the plot of Q>Re2 versus Re shown in Figure 9.7, The terminal velocity can now be calculated as follows. First, we calculate G'nRe2 using (9.46). Then using Figure 9.7, we calculate Re. Then uRe

PDp and there is no need to solve the system of nonlinear algebraic equations.

Settling Velocity Calculate the settling velocity of a 200-pm-diameter droplet with density pp = 1 gem-3. What would be the value if one uses Stokes' law?

For a drop with Dp = 200 pm using (9.34), Cc = 1 and therefore from (9.46) Ca Re2 = 385. Using Figure 9.7, we find that the corresponding Reynolds number is roughly 10. Now the terminal velocity can be calculated from the definition of Re and it is approximately 75 cm s--*

Using Stokes' law given by (9,42) we calculate v, = 120cm s-1. Stokes' law overestimates the settling speed of such a droplet by 60%.

Reynolds Number, Re FIGURE 9.7 QjRc2 as a function of Re for a sphere.

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