A NEW PRESSURE LOSS EQUATION A Theoretical Approach for Calculating the Pressure Drop in Pneumatic Transport

We consider the pipe inclined upward at an angle 8 from the horizontal as shown in Fig. 14.15.

The mixture element shown in Fig. 14.15 contains the flowing gas and solid particles. The partial densities of these two elements are PR and PS, re­spectively. The void fraction is 4> and this can be interpreted as the partial cross-sectional area for gas flow (see Eq. (14.13)). This means that if the pres­sure of the gas is P, then the pressure force per unit area of the total mixture affecting the flow of gas is <j>p and the pressure force affecting the flow of sol­ids is (1 — (f>)p.

A NEW PRESSURE LOSS EQUATION A Theoretical Approach for Calculating the Pressure Drop in Pneumatic Transport

Psjj = ~ 4>)p)~psgsn8-fsw + fd, (14.96)

Where Fsw contains the interactive force due to the different velocities of parti­cles and the friction force caused by the walls. The force Fd is the drag force caused by the surrounding gas flow. The drag force Fi{ is an interactive force between gas and solids, this means that the opposite of the force, -fj, is the force that affects the gas flow.

The momentum balance equation for the gas flow in the direction of x — axis is

Pg^fj. = -^(4>P)-P*SsinS………….. F%v/-fd, (14.97)

Where F is the friction force caused by the walls and Fti is the same drag force as in Eq. (14.96).

The solid particles vibrating up and down along the y-axis, perpendicular to the x-axis, change the internal velocity profile of the gas, so that the friction force is not the same as in an empty tube. We may divide the friction force F Into two parts:

X(-Pv —

F —— ——————————————— 1 ’ — -+- F (14 9 8 ^

/ gw 0 2 ‘mf > l-r.-M) L

Where the first part is the normal friction due to the walls, assuming the ab­sence of the particles. The force Fnt arises from the fact that neither the veloc­ity nor the pressure distribution is uniform along the y-axis and this creates a complicated internal fluid flow pattern, which also implies an additional fric­tion force in the direction of the ac-axis.

Due to the nonuniform velocity and pressure distribution along the y-axis, the particles remain separate and floating in the gas stream. In a vertical trans­portation the force Fint is obviously zero, because then the particles do not tend to fall and gather on the bottom of the tube. The force Fm cannot be included in the drag force Fd, because the drag force Fd pushes the particles forward in the di­rection of the x-axis, whereas Fim does not affect the particles but the gas itself.

The way in which the force Fmt is modeled clearly determines the type of the pneumatic flow; this has been discussed earlier in Section 14.2.2, where we considered the classification of different types of flow. In the following we will give a detailed description for the force Fim in a way that suits a par­ticular type of flow. This approach will be adequate for so-called dilute — phase flow or, more generally speaking, for homogeneous flow where the particles move separately.

We wish to solve the following problem: how to model the force Fim so that it implies the floating effect which prevents the particles from falling down to the bottom of the tube? The question itself already suggests the an­swer. The idea is based on the so-called virtual power method. An excellent re­view of this topic has been represented by G. A. Maugin.1

The power per unit volume (W/m3) needed to keep the particles floating in the direction of the y-axis is

On the other hand this power is, on the basis of the ideal of the virtual power method, the same as the force Fiat multiplied by the velocity of the gas //, i. e.,

P"’ = fimv. (14.100)

Combining Eqs. (14.99) and (14.100), we get a constitutive equation for the

Force /inr:

It)

/int = Psg^WS. (14.101)

Equation (14.101) was first derived by Weber,3 although it is formally repre­sented in a different way.

Summing Eqs. (14.96) and (14.97), we get

Dv dc _ dp Pg j. . . ,

Pg Ps ~dx T) "2" ^ *" """" si n 5 — Psg ■ sin 3 — /jnt — /sw. lI 4.102)

For the force Fmt we have a model, Eq. (14,101), but we also need a formula

For the force Fsw. The following model is based on our own idea and is not

Presented elsewhere. In Section 14.3.2 we will briefly discuss another way of modeling the force Fsw, but without any detailed mathematical analysis.

The friction force Fsw Can formally be expressed with the aid of the friction coefficient

Fsw = P-fPsgcos8. (14.103)

The friction coefficient ^ is a complicated function of the flow conditions (i. e., of the velocities V and C) and of the angle 8. In a horizontal flow (<5 = 0) we get from substituting Eq. (14.86) into Eq. (14.103)

V — c(S = 0)

Ws

2

Fs w(S = 0) =

Psg, (14.104)

A NEW PRESSURE LOSS EQUATION A Theoretical Approach for Calculating the Pressure Drop in Pneumatic Transport

Where we have replaced Wso by Ws, i. e., the falling velocity of the mixture of solid particles rather than the Wso falling velocity of a single particle.

F = j V^з 2

WS J

подпись: f = j v^з 2
ws j
Next we make a very straightforward approximation. Equation (14.104) gives the force Fsv/ For the angle 5 = 0. To get the force at other an­gles, all we do is replace c(S = 0) by the corresponding velocity c = C(8), i. e., we write

Psg• (14.105)

Substituting the specific constitutive Eqs. (14.101) and (14.105) into the gen­eral force balance Eq. (14.102) we get

Dv dc dp ^GPg 2 ■ s

PsTt + p*Tt = -Tx’ D2v — PSgsmS

Next we develop further the left-hand side of Eq. (14.106). The total deriva ­tives, also called material derivatives, are

 

Dv __ Dv. ,.dv Dt dt + Vdx

Dc_.bc. dc Dt Dt dx’

 

(14.107)

(14.108)

 

The derivation of Eqs. (14.107) and (14.108) can be found in any textbook of continuum mechanics, e. g., Mayer.2

In a stationary flow, the partial derivatives with respect to time vanish, i. e., V = v(x) and C = C(x), and

 

Dv = Vdv.

Dt dx

Dc = rdc dt dx’

 

(14.109;

(14.110)

 

Dv , Dx

Dc ^dx

Л Q pQ

+ V2 + p^gsinS

+ JX-

(

WS 2 о V

— cos-8 + —

C c

W c

■Wg8

On the other hand, in the steady state the mass balance for the gas in a tube with a constant cross-sectional area is simply

 

Constant = Rh„

 

(14,111/

 

PgI

 

And similarly the mass balance for the material flow is (see Eq. (14.17))

Psc — constant = TxpgV = ixritg. (14.112)

 

Substituting Eqs. (14.109)-(14.112) into Eq. (14.106) and using Eq. (14.1′ we obtain

 

A NEW PRESSURE LOSS EQUATION A Theoretical Approach for Calculating the Pressure Drop in Pneumatic Transport

(14.113)

 

With the aid of this equation the pressure loss in a pneumatic conveying sys­tem can be calculated. All we need to know are the velocity difference between gas and solid particles,

V — c = F{pci S) 5 (14.114)

And the falling velocity of solid particles in the vertical tube,

Ws^giPc)- (14.115)

Both of these functions F and G are specific for the solid material in question

And to some extent also for the diameter of the tube and the mixture ratio /a.

The great advantage of Eq. (14.113) is that no material or particle friction fac­tors are needed.

Based on the models and discussions presented in Sections 14.2.3-14.2.5 we may now write Eqs. (14.114) and(14.115) in more concrete forms. Since P. s>:>PG) it Follows from Eqs. (14.29), (14.63), and (14.71) that

1

 

Jpc

 

M’s = g(PC) = ™iA L&A = »SA (14.117)

A/ Pc Af r

Hence, using this approximation the only experimental data required is a fail­ing velocity WsA with a known gas density PCA.

On the other hand, the velocity difference in the vertical pipe is the same as the falling velocity (see Eq. (14.84)), and hence

V-cv = f pG,8=^=(v-cv)AjЈЈA, (14.118)

Where Cv is the velocity of solid particles in the vertical pipe.

Correspondingly for the horizontal pipe we may write

V-ch=f(pc,8=0) = (v-ch)A (14.119)

V Pc

Where Ch is the velocity of solid particles in the horizontal pipe. Note that V in Eq. (14.118) and in Eq. (14.119) is the velocity of gas in the vertical and in the horizontal pipe line, respectively.

Finally, the angle dependence can be estimated by Eq. (14.83), i. e.,

V-c = F{pc,8) = /^SA[(t’-Cf,)ACos8 + (i’-c,,)^sin5]I/- , (14.120)

j Pc

Which is, of course, as its derivation revealed, a very rough estimation and should only be used if no experimental data on the angle dependence are available.

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