The effects of shear stress on a pendular liquid ring

J.

-‘.

SWEENEY

ofPhysics.

Deportment

The

University,

L.ee&.LSZ

?JT

1

(U.K.)

and K. V. S. SASTRY

of

Deportment

(Received

Maten-&

December

Science

and

16, 1983;

Mineral

in revised

Engineering,

form

February

SUMMARY

Calculations

have

been

made

to determine

the shape and volume of the pendular bridge between a pair of contacting spheres shear

when stress.

simulate uacuum

the liquid

surface

The purpose

the situation or pressure

of

liquid equal

is subject

this study

to

is to

in a filter cake during deliquouring after air

breakthrough has taken place, and so determine theoretical limits to the saturation leuets which can be achieved. Such saturations are found to be strongly dependent on of particles (lower saturations being able with larger spheres), but weakly dent

OR the magnitude

the range

of

the shear

the size obtaindepenstress

in

considered_

The problem of the pendular liquid ring is of some practical interest, as it relates to physical situations in which liquids are held in beds of solid particles_ The basic problem in which two solid spheres of equal size entrap a liquid ring, with no external forces acting, has been analysed, both in the special case of contacting spheres [l, 21 and the more general case of non-contacting spheres [3]. A summary of research investigations in this field is available [4]_ These studies treat the problem by analysing the equilibrium of the pendular liquid.surface. In this paper, we present an extension of this work to the case in which shear stress acts upon the liquid surface; this is of relevance to filtration proQO32-591Qf85/$3.3Q

Uniuersity

if

California,

Berkeley.-

CA

94720

(U.S.A.)

9,1984)

cesses in which a gas is used to displace liquid from a cake of solid particles. The theoretical study presented here has been motivated by the-problem of coal dewatering using vacuum filtration_ In this process, water in a bed of particulate solids is displaced by air through a filter medium. Usually, a stage (known as air breakthrough) is reached at which both air and water are passing through the filter medium- Prior to air breakthrough, water is displaced by normal pressure, ti_ a manner akin to its being forced through a set of capillary tubes. After this stage, there are air pathways established through the wet filter cake, and it seems reasonable to assume tha5 a mechanism of water removal begins to operate which involves shear forces imposed on water droplets by flowing air. As more and more air paths are established, this mechanism will become increasingly important_ Calculations based on this mechanism should thus serve to set theoretical lower limits for water contents The above considerations lead us to set up a model of dewatering consisting of a pair of contacting spheres holding a pendular liquid ring, with a constant shear stress acting on the exposed liquid surface, such as to displace it in a direction parallel to the axis of symmetry of the sphere pair. This representi a water droplet in a filter cake susceptible to the influence of shear forces imposed on it by the surrounding airflow. The maximum volume of liquid as a proportion of sphere volume which is compatible in the model with a particular value of shear stress is related to the ultimate saturation in a filter cake. Both these quantities are dependent on sphere (particle) size_ 0 Elsevier Sequoia/Printed

in The

Netherlands

70 MATHEMATICAL ICAL SOLUTION

FORMULATION

AND

NUMER-

Theoretical considerations The imposition of a shear stress on the

liquid surface produces a qualitative change in the mathematical formulation of the problem, as compared with the problem with zero shear stress. In the latter case, the pressure difference P between the interior of the liquid ring and the exterior is spacewise constant, and can be assigned an infinite number of values, each corresponding to a physical solution having a particular pair of boundary conditions. (The boundary conditions are then the solid/liquid contact angles at each sphere.) For non-zero shear stress, P is determined once one boundary condition and the shear stress are specified, and varies with axial distance X. This relates to there being a pair of equilibrium equations to solve for the surface profile with non-zero shear, whereas one equation suffices for zero shear. For simplicity, the cases discussed here are restricted to solid/liquid contact angles of zero_ For a positive contact angle, it is feasible that the value of the contact angle at the ‘leading’ edge of the liquid surface would be reduced by the action of the shear stress, and it is not easily established what the resulting angle would be_ The restriction of the contact angles to zero enables us to discuss physically meaningful problems while avoiding this difficulty_ The maximum volume solutions discussed are obtained by increasing the angle + (see Fig_ 1) to as high a value as remains compatibIe with the generation of a solution which intersects the second sphere. This is equivalent to imposing a zero contact angle boundary condition at the second sphere also. The required pair of equations for this problem correspond to horizontal (axial) and hear

stress

vertical force balances, as they relate to both Figs. 1 and 2. The force balances are performed on a sectional element of the liquid bridge, with two plane boundaries perpendicular to the axis of the sphere pair and one plane boundary containing the axis. The horizontal force balance is Py’

+ 2r(y

-PlyI

+27

i

Quid s&ace l_ General

arrangement

of pendular

ring.

yi cos I9,)

y&=0

(I)

=1

and the vertical force balance

-22y(ysin8-y,~inf?,)-r(y~-y~~)=O (2)

Fig.

2. Section

of liquid

bridge.

The circular functions could be replaced by expressions in the derivative dy/dw; eqns. (1) and (2) are thus differential equations, and eqn. (1) is a straightforward generalization of that of Erle, Dyson and Morrow [S] for the zero shear case. Putting P = PI and 7 = 0 in eqns. (1) and (2) results in their ceasing to be independent equations, such that differentiating eqn. (1) with respect to y gives the same result as differentiating eqn.. (2) with respect to x. This is how the above system of equations reduces to the single equation of the zero shear stress problem _ Numerical solution The liquid bridge

Fig.

cos e -

is built up as a finite number of the sections illustrated in Fig. 2, with eqns. (1) and (2) being solved simulta- , neously for each element. The solution is

obtained for’successive increments H of tthe distance x_ ‘Contim& of the pressure differential P or Pi is enforced -at elements’ interfaces; hence, P, in each successive ele .ment is equated to P’of the previous element. The pressure-differential for the firstelement is assumed to be coustant throughout its thickness_ The solution is implemented by eliminating P algebraically between eqns. (1) and (Z), and solving the resulting equation numerically to give y and 8 at each step. The equation derived thus from eqns. (1) and (2) is

axial

7-y4+2”1sin0y3-

(

P,I+2yj.

n~erical Soiui;io’~ ok e-~n:, c6~. suggestsitselfi..uskg _a~ co~$&fitio& ‘&&fiod::._z’mc& a.:-~

&

2yy,

c~s 8, -P,~,~)T

=o

(3)

Here, the approximation made and I=

2P = Pi + P has been

JYdr 11

It is necessary to assume a particular shape of the surface profile, Le. a functional dependence of the radius y on the axial distance x, within each element. We have assumed that y is a quadratic function of x, and the integral I and the relationship between y and 0 are derived on this basis This results in I=@___

(

tae 6

+

be1 _ 3

2(Y -Y1) H

tme=

>

+HY,

_ttan

e

1

(4)

We also make the approximation =dx

-

/ cam e x1

=

w ln(cos

c0se

-112

1

=1

2Y-YI p-t~e, rr

+ (27+

e/cos 8 1) -c0se,

2

)1 This results in a polynomial of order ten. Newton’s method [5] was used with this polynomial, and the coefficients updated using the y-value obtained from the converged Newton’s method routine in the manner outlined above. The coefficients of the tenth order polynomial are given in the Appendix. The global error in the vertical force bal axe, obtained by comparing the two sides of an integrated version of eqn. (2), was used to check on the accuracy of the solution and determine suitable values of H%.The solution method breaks down at small angles 8 near to minima in y_ This problem is overcome by changing the sign of 8 when it has decreased to a prescribed finite value. This causes an increase in the global error, though it still stays within acceptable limits. It is clear that there must be some finite value of r below which this solution scheme will not operate, since the solution consists of only a single force balance equation when the shear stress is zero. The generation of invalid results due to this fact is avoided by ensuring that the shear-stress dependent term in eqns. (1) or (2) is large enough to be of significance relative to the other terms.

(5) Equation

(3) can be written

Ay4+Bsiney3+Cy*+Dy+E=0

RESULTSANDDISCUSSION (6)

This is quartic in form, but some of the coefficients A, B sin 8, C, D and E depend on Y_ The following method of approaching a

.:I‘

Newton’s, tomsolve it as :a ‘qua&with con:. Istant coefficients;- updat~e~t~~.coeffi~ients on the basis&f the,y.value thus obtained, and. proceeding with. the~it&tion_ However, convergence w,as not &hieved ~with-this method. A.convergent method wasobtained-by the same basic-procedure, but~usmg an eguatibn. derived from eqn. (6) by decreasing the nonlinearity of the coefficients at the expense of increasing the order of the polynomial_ This was done by eliminating the’ sin 0 term in eqn (6) using the relationship

Input data and numerical results Calculations have been made for sphere diameters in the range 50 - 400 mr~ and shear stress levels between O-1 and 1.0 kN/m2_

These values were chosen as being relevant to filtration processes (see next section)_ A typical liquid surface profile is shown in Fig. 3. The surface tension of the liquid is assumed in all cases to be that of water, i.e. 0.073 N/m. The maximum permissible volumes at each of the shear stress levels are plotted in dimensionless form V/r3 in Fig. 4. As expected, the dimensionless volumes increase with decreasing sphere size, reflecting the well-known increased difficulty in deliquouring as particles become finer. An unexpected feature of the results is that the volumes do not decrease with increasing shear stress, at least in the range for which results have been generated_ However, there is a decrease in maximum permitted volume with respect to zero shear stress, when an infinity of liquid bridges are permissible. As a logical consequence of this, there must be minima in the volume us. shear stress curves at a value of below the smallest value (0.1 kN/m’) for which calculations have been made. Unfortunately, it was not possible to explore these minima with the

01

.

.

.

123456789

.

.

.

.

.

l0

,

II

Fig. 3. Bridge profile (sphere diameter 50 pm, shear stress 0.5 kN/m’).

present numerical methods, as the solution method begins to break down at stress levels below 0.1 kN/m2 for the reason discussed in the section on Numerical solution. In view of the weak dependence of the maximum theoretical liquid volume apparent from Fig. 4, values at 0.1 kN/m’ can be taken to a good approximation as the smallest which can be achieved. Under this aesumption, we plot Fig. 5, the minimum pendular liquid content as a function of sphere size. Relationship with filtration variables While it is not possible to relate these theoretical results to saturations and pressure drops in real filter cakes in any rigorous way, some model estimates can be made. The first of these concerns the relationship between the shear stress on the pendular ring and the normal gas pressure difference across a filter cake, and the second between calculated liquid volumes and cake saturations. Pressure drop across cake It is possible to make some estimate of the pressure difference required to provide a given shear stress on the liquid surface by assuming the following conditions: air flow is cylindrically symmetric, and is confmed within a hollow cylinder of radius r (that of the spheres), with axis coincident with the axis of symmetry of the sphere pair. This gives a rough representation of the restriction of the air flow due to surrounding adjacent particles. Then a force balance relates the pressure drop AZ’, length and mean radius of the pendular ring, respectively I and y, and shear stress r by

a

-001.

+

-f

-‘cm2 0

a

I OO

I

5

S-bexslwOMNlm?

Fig. 4. Liquid volume as a function

of shear stress.

10

1 am

I

i

i/i

i

-.71

20

:

."

" . " • . . :".... .. :./.

. i"-/.: " :. ~i :i. ..--":.1.0~":~~ ...-i::.-..i..:-i.-i-!?~:.'. ?i.:!-~i..:"-.:..~:..-"~.::,.:::. • ......... ~

" .... -"-":~": :.. i..'. ~.. "~" " " " . " " " "" " " " " ' '

I£

...

1

i...:-.:..~...

".."::''"i~"-; ._.:".:i:./".'.::..:...."......!."":.. . . .:"""..: " '.

: ". '"-~.u. 0-02

3

2

i-. SO~":...../".::.,.....:./-.:.:.?-....:...~

.

.

..

4

Fig. 5. Liquid v o l u m e as a f u n c t i o n o f sphere size

~ p ( r 2 - - 9 2) 1

+ 2~r = 0

T o give a n i m p r e s s i o n o f t o t a l c a k e p r e s s u r e differences, some specific results are quoted. These are based on th e assumption that the p r e s s u r e g r a d i e n t is c o n s t a n t t h r o u g h o u t t h e cake. They are given in the Table in the form of pressure gradients.

t r a p w a t e r m o r e . e f _ f i c i e n t l y ; a n d t h a t air .breakthrough occurs through the cake in isolated pathways, so.that only a proportion o f t h e l i q u i d p r e s e n t is s u b j e c t e d t o t h e s h e a r m e c h ~ d i s c u s s e d h e r e . H o w e v e r , *.hese c o n s i d e r a t i o n s r e l a t e t o t h e c a k e . s tyr_.~cture, which may be umenable to mz_nipulation to reduce the saturation, while the figures given h e r e a r e o2 a m o r e a b s o l u t e n a t u r e a n d c a n s e r v e as t h e o r e t i c a l l o w e r l i m i t s .

TABLE Pressure gradients (MN/m 3) Sphere

diameter

Shear st-re_~ ( k N / m 2 )

(ram)

0.1

0.4 0.2 0.1 0.05

0.45 1_1 3.0 8.5

0.2 0.93 2.2 6.1 15.3

CONCLUSIONS 0.5

1.0

2.4 .5.8 15.1 39.1

5.0 11.7 30.6 92.0

Cake saturations These are estimated on the basis of a cubic array of equal spheres, with pendular rings between pairs of spheres with axes parallel to the air flow (envisaged to be parallel to one of the lattice directions).. Thesaturations are g i v e n as t h e r i g h t - h a n d v e r t i c a l a x e s o f F i g s . 4 a n d 5. The most striking feature of thesaturations is t h e i r s m a l l n e s s c o m p a r e d w i t h t h o s e ob~ served in practice. Tw0 principal explanations for this suggest themselves: that a filter cake is m u c h m o r e c o m p l e x t h a n a r e g u l a r a r r a y o f spheres, and may contain features that can

1 Calculatiorm have been made on the effect of shear stress on a pendular liquid drop, by solving numerically the equilibri, lm e q u a t i o n s o f t h e l i q u i d gUrface. T h e s e h a v e been related to the behaviour of liquid droplets in a filter cake undergoing vacuum or pressure filtration. 2 The. q u a l i t a t i v e e f f e c t o f p a r t i c l e size, in that smaller particles are associated with h i g h e r l e v e l s o f s a t u r a t i o n , is as e x p e c t e d f r o m practical observations. 3 Increasing the shear stress does .not decrease the maximl,m permissible liquid volume, but increases it slightly. : 4 The predicted levels of saturation, based on a cubic array of equal spheres, are generally lower than those obsel-eed in practice; r e a s o n s f o r t h i s a r e d i s c u s s e d u n d e r C a k e saturations.

.

74

ACKNOWLEDGEMENTS

-112

1 2(Y --Y1)

We wish to acknowledge the Electrical Power Research Institute, Palo Alto, CA (U.S.A.) for financial support during the course of this investigation.

Substitution

LIST OF SYMBOLS

5 a,y’=O i= cl

(A2)

_ttane

1

H

coefficients of tenth-order polynomial thickness of i;r?g element total length of ring capillary pressure mean capillary pressure total external pressure drop across ring axial distance distance of ring surface from axis of symmetry mean distance of ring surface from axis of symmetry surface tension angle of ring surface to axis of symmetry shear stress filling angle

of this into eqn. (Al)

yields (A3)

The coefficients

are defined

as follows:

a

2Ylm+tane,

z

a g=-4A

H 4(2AC

a,=A2(1+K)+

a 7=

-

B2)

H2

8AD ~ HZ

+

(4Bz -

8AC)(2y,/H

+ tan 0 1)

H

Subscript

pertaining tc vertical boundary of ring element at which solution has been found

1

SAD(2y

cl,=2AC(l+K)+

4(2AE

,/H + tan 6 1) H

+ C2)

Hz-KB2

REFERENCES R. A. Fisher, J. Agr. Sci.. I6 (1926) 492. J. C. Melrose and G. C. Wallick, J Phys. Chem.. 71 (1967) 3676. M. A. Erie. D. C. Dyson and N. R. Morrow,

AIChEJ,

a 5=2AD(1+K) __ 4(=

+ C2)(2y

17(1971)115.

P. Mehrotra and K. V. S_ Sast.ry. Powder Technol.. 25 (1980) 203.

V.

C. F. Gerald, dison--Wesley,

Applied Wesley,

+-

8CD w2

Analysis, Ad2nd edn., 1978.

Numerical

MA,

a4=(2AE

+ C2)(1+

8CD(2y

Je

K)

+ tan 0 1) +

4(2CE

+ D2)

H

H

APPENDIX

For completeness, eqn_ (6):

I/H + tan 0 1)

H

we begin by rewriting

Ay4+BsinBy3+Cy2+Dy+E=0 Using the assumption of quadratic dence of y on X, we have

(AlI depen-

a3 = 2CD(l+ 4(2CE

K) + D2)(2y,/H +-I

+ tan 0,)

-

~.

I-

SED H2

::

:

:

-

. :~:= ~ .... - ~: : 7:: :~=!= ~ ~ t : ~ : ~ : : ~ { :~:::7~=:-=---~: - ~=~77- -~ ~ ;:"/..~.~ - ~ : ! : ~ i ~ -

;!-~/~-=?:.~: ~ . 7 ~ ' ! ~ i ! ~ : ~ . ~ . ` . ~ - ~ ` ~ - ~ ! ~ / ~ , ~ 1 ~ L ~ : ~ z . ~ i - ~ : ~ : ~ . . ~ - ` . } ~ i

<,2<= (~.c+++ v=)(1 +~);} : ~: 7::~: ;~:;~ ;-:::!~:-::i:: ':~::/7-, Ma;>~2i~7;--~):

aI=2ED(I+K)--

i

: g

•

::::

; :1~=~:~:

:~~::+~?i~!Y-~-:-:"Tt.';++~+:~)7-:77;i:-:L:~7:~7..:i~:M:L; 7

-.: H

"

m:i~i;:;}::.

::-:-::i.:

:;::