Mathematical Analysis of the Free-Falling Velocity

We start by considering the free-falling velocity of a single particle of diameter DWhen the particle reaches the free falling-velocity, tfso, the gravitational force and the drag force are in equilibrium, after which the falling velocity is constant.

The basic problem in determining the falling velocity lies in the evaluation of the drag force that a particle experiences as a result of motion relative to the surrounding gas. The gas flow and resulting drag force, Fd, have been characterized most thoroughly for the motion of a smooth nonrotating sphere moving at constant velocity through an otherwise undisturbed and un-

Bounded gas. The result obtained under these highly idealized conditions is normally represented by a standard drag coefficient curve, which expresses the relationship between the coefficient of drag, Cd, defined by

Q — —7T~~^—— -, (14.19)

7T_dЈ wl0

4 ‘Pc ‘ 2 and the particle Reynolds number, defined by

R (14.20)

“ V ‘ ‘

Where V is the kinematic viscosity of the gas. With the dimensionless parame­ters Cd and Rerf the problem is to find the correlation

Cd = f(Ked), (14.21)

Which then, together with the force equilibrium condition that the drag force Fd and the gravitational force minus hydrostatic force are equal, i. e.,

Id*

Fd =

(Ps~Pg)S, (14.22)

Gives us the free-falling velocity W&0. Here G is the acceleration due to gravity, 9.82 m/s2. ‘

Despite the long history of determinations of the standard drag curve, Eq. (14.21), values for the drag coefficient under idealized conditions are still a matter of dispute. A comprehensive review has been represented by R. Clift and W. H. Gauvin.6

Here we consider the problem from another point of view, namely how a set of particles of different sizes fall in a gas. In other words, we wish to find a simple way to calculate the free-falling velocity of a group of particles when we know the distribution of the size of the particles. But before we can do this, some basic widely-used formulas for calculating the free-falling velocity of a separate single particle are introduced.

The drag coefficient based on the theoretical analysis of Stokes is

Cd = F(Red) = (14.23)

Substituting this into Eq. (14.19) and then combining this with the definition of Re, Eq. (14.20), and the force equilibrium condition, Eq. (14.22), we ob­tain the following equation for the free-falling velocity:

= (14-24) If the gas is air at atmospheric pressure (p = 1.023 bar) and at tempera­ture 20 °C, then PG = 1.20 kg/m3 and V = 15 x 10-6 m2/s. In the case that Ps = 1500 kg/m3 we get from Eq. (14.24) the values shown in Table 14.1. The greater the Reynolds number becomes, the more inaccurate Eq.

(14.24) will be. For Rej < 5 x 10 2 the drag coefficient may be calculated with sufficient accuracy from Stokes’s law (Eq. (14.23)) and in these cases Eq.

(14.24) is adequate.

TABLE 14.1 The Free-Falling Velocity of a Spherical Particle in Air (20 °C, 1.023 bar) based on Stokes’s Theory D„ jim WSo*m S’1 Red Cd B 1.6 x 10"3 6.5 X НИ 3.69 X К. И 10 4.5 x K)"3 3.0 x IQ-5 8.0 x lip 15 1.0 x 10— 1.0 x 1,0— 2.4 x 10 ’

The density of the solid particle is 1500 kg/m

Beard and Pruppacher have examined the free-falling velocity for small water droplets in saturated air.7 These results could be correlated as follows

24Re^1 + 2.4Re^0’05, 0.001 < Red < 2

24Re;/ +2.76Re/20, 2 < Rerf < 21 (14.25}

24Re;/ + 2.45 Re/368, 2 1 < Ked < 200

Combining this with Eqs. (14.19) and (14.22), the corresponding free-falling velocity WSq can be determined.

The old empirical equation of Schiller and Nauman,8

Cd = 24 Re,/ + 3.60 Re/313, (14.26)

Has been observed to fit the recent results, too, over a wide range of Reynolds number (approximately 1 Ј Re^ < 1000).

Clift and Gauvin9 modified Schiller and Nauman’s Eq. (14.26) to repre­sent the drag force throughout the transitional and turbulent regimes:

Cd = 24 Rej1 + 3.60 Rea-°313 +———————— ГТГ‘ (14.27)

(1 + 4.25 x 10 Rej )

The region 1000 sRej< 200 000, sometimes called Newton’s regime, is characterized by relatively constant drag coefficient, indicating that there are no major changes occurring in the flow pattern. The most significant change is the transition from laminar to turbulent flow in the detached boundary layer. From Eq. (14.27) w7e get

Cd = 0.47 for Rerf= 1000

And

Cd = 0.49 for Red = 200 000.

Substituting

Cd = const. = 0.5 (14.28)

Into Eq. (14.19) and then using Eq. (14.22), we get the weIl-knowrn Newton’s formula for the free-falling velocity,

WSo

S 1/2

La the pneumatic conveying process the flow around the particle is not uni­form, the particle is not in steady-state motion, and the flow contains turbulence which is not merely generated by the particles. Thus the use of Eqs. (14.23)- (14.29) is of course rather restricted. Despite these limitations we will now esti­mate the free-falling velocity of a set of different-sized particles based on the as­sumption that we know the free-falling velocity of each single particle.

Particles of different sizes fall at different velocities. When a set of differ­ent-sized particles falls in a group, the particles collide with each other and the faster ones tend to accelerate the slower ones. In all collisions the linear mo­mentum is conserved, so that if all particles collide with each other sufficiently many times, the set of particles will achieve one mean free-falling velocity. Thus the mean free-falling velocity of the set of particles can be defined by

W

So

I MIu;soi (14.30)

Where M,-is the mass of a particle whose diameter is Dsi and whose free-falling velocity is correspondingly wSO|. Since

M=’y ^ nij (14.31}

I = 1

Is the total mass of falling particles, we see from Eq. (14.30) that Ntwso is equal to the total linear momentum of the particle set. The identification index / is allowed to take any large values without any limitations, i. e., the number of different particles can be infinire.

In order to use Eq. (14.30) we need to know the particle size distribution. In many cases it has been observed that the size distribution obeys normal probability distribution, or at least can be well approximated by it. In fact, the number of particles DN whose logarithm of diameter

X = In Ds (14.32)

Lies in the interval X… x + dx is

= T=^e ^ 2tr2 ^ dx, (14.33)

N(f) V ‘

Where Fx is the mean diameter of the particles (in logarithmic scale, Eq. (14.32)), A is the mean variance, and N(t) is the total number of the particles.

It will be useful to give a geometrical illustration of the concepts /jl and A Included in Eq. (14.33). First, we will construct a special coordinate system.

The coordinates, X- and y-axes, are provided with nonuniform scales, de­noted by U and V, as follows:

U(x) = E (14.34)

1 —s’

Is(y) =—== e2 ds. (14.35)

Note that the U scale is precisely the same as the D, scale (Eq. (14.32)).

We will now show that the distribution Eq. (14.33) will be a straight line in the coordinate system X-y, whose axes are provided with the nonuniform scales of Eqs. (14.34)-(14.35). To show this we first write Eq. (14,33) in the integrated form

(14.36)

Where N(x) describes the number of those particles whose diameters are in the

Region

— co < In Ds —- X, i. e., 0 < ds< ex. Changing the integration variable T to

— CO

(14.37)

We get from Eq. (14.36)

(14.38)

Comparing this with Eq. (14.35), we see that

(14.39)

Where the value of Y is a linear function of X,

(14.40)

These results are illustrated in Fig. 14.4. The value of Ft gives a mean diameter

(14.41)

Which divides the set of particles into two groups. Half of the particles are smaller than or equal to Dm and the other half are larger than Dm. The variance A gives the slope of the line in Fig. 14.4, i. e.,

(14.42)

From Fig. 14.4 we may, as an example, read that 40% of the particles (measured in numbers) are smaller than 2.2 |xm.

Before we go back to Eq. (14.30), we shall evaluate the mass distribution function for the particles whose size distribution is of the form (14.33), i. e., normal probability size distribution.

The mass of the particles DN is

3′ Чу)

1.0

0.5

U(x) = Ds

0.0

-0.5

FIGURE 14.4 Log-normal probability size distribution.

Which can, on the basis of Eq. (14.33), be written in the form

(14.44)

Dm = ^ps^W — e dx. b ‘ J2tt a

Taking into account that Ds = ex, Eq. (14.44) can be written as

Dx.

(14.45)

We define the mass distribution function as follows:

(14.46)

Z»Ј3 = — i — fx Dm. m(t) m(t) j_ m

The function M(x) gives the mass of all those particles whose diameter Ds satis­fies the condition Ds Ј Ex. The total mass of the particles is denoted by M{t).

Substituting Eq. (14.45) into Eq. (14.46) the function M(x) can be eval­uated. Note that the only x-dependent part of the integrand is

2 it

All the other terms being constant. Since

We get

M(t) — ps^]V(J) E

And hence

2′ Ds

114.47)

Mix) — 1

M{t) JTv

Using the same reasoning as with the particle number distribution above, we observe that if the X- and y-axes are provided with the nonlinear scales, U and

V, defined by Eqs. (14.34) and (14.35), the mass distribution M(x)/m(t) can be

Described by a straight line

Y = ~(x — (/jl + 3a2)). (14.48)

The only difference compared to the number distribution Eq. (14.40) is that the medium point is transferred to the right by the amount 3<r2. The slope of the line is the same. The result is illustrated in Fig. 14.5.

Now we go back to Eq. (14.30). The free-falling velocity (wso) of a single particle, with diameter Ds can be estimated as

Wso = k d", (14.49)

Where, applying Stokes’s law (Eq. (14.24)) for small particles,

N = 2

(14.50)

V(y)

1.0

0.5

Ulx) = D.

-0,5

-1.0

FIGURE 14.5 Log — normal probability size distributions, illustrating geometrical transposition Between number and mass curves.

And

* = 2^-if. (14.51)

P g 1» V

Correspondingly, applying Newton’s formula (Eq. (14.29)) for larger particles,

* = i (14.52)

And

8 P. s — Pa ‘ 3’ Pc

1/2

. (14.53)

All other cases are between the extreme limits of Stokes’s and Newton’s formulas. So we may say, that modeling the free-falling velocity of any single particle by the formula (14.49), the exponent N varies in the region

0. 5 sns2. In the following we shall assume that K and N are fixed, which means that we consider a certain size-class of particles.

The linear momentum DL of particles whose logarithm of diameter, X = In Ds, lies in the interval X. . . x + dx is

DL = dm k d", (14.54)

WherE dm is given by Eq. (14.45). Since Dns = EXn, we get on the basis of Eq.

(14.45) ‘

■n — N(t)k -7^2 (3 +K)cOf (fi2-(M+(3 +„},/)*)

DL = Ps f E 2- C 2- Dx.

(14.55)

Equation (14.55) contains same kind of algebraic manipulation with the term Exn as Eq. (14.45).

The linear momentum distribution function is

Ґ? g = y^, (14.56)

LJ .„dL

Where L(x) gives the linear momentum of all the particles whose diameter Ds is Ds The total linear momentum of all particles is denoted by L(t).

Substituting Eq. (14.55) into (14.56) the function L(x) can be evaluated.

Since

1 -^2 (x-(V-+0+n)tr))2

1 E 2a dx = 1 ,

We get from Eq. (14.56)

L(x) _ 1 F ~(*-(M + (3 + h))o-2) -is2

E

Ds. (14.57)

L(t) JTrr .

Here we have changed the integration variable X to s, which is defined by

The result, Eq. (14.57), can be interpreted in a similar way to Eq. (14.47). If x — and y-axes are provided with nonlinear scales U and V, defined by Eqs. (14.34) and (14.35), the linear momentum distribution L(x) / Lit) can be de­scribed by a straight line

Y — ~(x — {> + (3 + N)(f’)) (14,59)

And then

7^7) = "(y)- (14.60)

On the other hand the total mass of the particles is

-zpi’X-O-‘ + 3o-‘)’)

And correspondingly the total linear momentum is Ut) =

M(t) = dm = ps^N(t)e ~‘T (14.61)

—^r(n ‘ — ((t + (3 + )

DL = kps%N(t)e 2<r . (14.62

S6

Replacing the infinite sums in Eq. (14.30) by the integrals in Eqs. (14.61) and (14.62), we get for the mean free-falling velocity of the set of particles

N [1 + —.

= = ke[ " (14.63)

So M(t)

The corresponding diameter of a single particle with free-falling velocity ma,0 is, according to Eq. (14.49),

Wso = kdsm. (14.64)

Combining Eqs. (14.63) and (14.64), we obtain

. ^ + N 1 FS + — T-a

Dsm = E . (14,65)

This gives us a mean particle diameter, that has the same free-falling velocity as the set of particles of different sizes. Taking a logarithm of Eq. (14.65), we get

Xm = ix + 3a2 + , (14.66)

Where

Xm lntism. (14.6/)

Comparing this result with Eq. (14.48) and Fig. 14.5, we see that the

Mean size of the particle in the sense of linear momentum, and therefore also

In the sense of free-falling velocity, is always greater than the mean mass size

FIGURE 14.6 Log-normal probability size distributions, illustrating geometrical transposition between number, mass, and linear momentum curves and the mean size particle dsm, which can be used in estimating the free-falling velocity of the particle group.

X = In Ds

Particle. The difference between these two values is (w/2)cr. The result is il­lustrated in Fig. 14.6.

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