93
Journal of Non-Newtonian Fluid Mechanics, 15 (1984) 93-108 Elsevier Science Publishers B.V., Amsterdam - Printed in The Netherlands
FREE SURFACE ON A SIMPLE FLUID BETWEEN ROTATING ECCENTRIC CYLINDERS. PART I: ANALYTICAL SOLUTION
AYDENIZ Department (U.S.A.)
SIGINER * of kngineering
(Received October 19,1982;
Mechanics,
The University
oj Alabama,
University,
AL 35486
in revised form August 24,1983)
The steady motion of a simple fluid between vertical cylinders which rotate about non-concentric axes is solved by means of domain perturbations. The theory is developed as a perturbation of the rest state in powers of the angular frequency Q of the inner cylinder, and the solution is carried out to 0 (a’). The stress is expanded in a series of Rivlin-Ericksen tensors. At the second order only one material parameter, the climbing constant, enters the analysis. A procedure is developed for predicting the shape of the free surface on the fluid. Secondary motions generated by the eccentricity are shown to appear at the second order.
1. Introduction
The deformation of the free surface on a viscoelastic liquid which is subjected to a shearing motion can be used to generate information about the constitutive equation for the liquid. In a recent paper, Beavers, Yoo and Joseph [l] examined the potential for using a concentric cylinder geometry for such measurements. They found in their experiments that the shape of the free surface was extremely sensitive to smaIl eccentricities in the axes of the cylinders. These observations demonstrated that an analysis and systematic experimental investigation to document the influence of eccentricity on the free surface shape would be of value in determining the usefulness of the concentric cylinder configuration for rheological free surface measurements. This paper presents an analysis for the motion of a viscoelastic fluid * Now at the Mechanical Engineering Department,
Auburn University,
AL 36849, U.S.A.
94 between vertical eccentric cylinders undergoing differential rotation, and a companion paper [2] reports experimental observations. The region between two vertical eccentric cylinders is filled with a viscoelastic liquid. The top of the region is open to the atmosphere. When the liquid is sheared from the boundary, normal stresses are induced. These stresses are along and perpendicular to cylinder generators, on planes which are perpendicular to the planes of shear. The free surface on the liquid is deformed in the direction of the cylinder generators by the normal stresses. The idea of using the free surface on a simple fluid in motion to make rheological measurements with the aim of determining the material constants was first investigated by Wineman and Pipkin [3] and Tanner [4]. Later Joseph and Fosdick [5] observed that the free surface on a simple fluid in the Weissenberg phenomena reflects the effect of the internal pressures and thus can be used to yield viscometric functions. They solved the free surface problem between concentric cylinders rotating at different speeds using domain perturbation theory. Joseph and Fosdick set their treatment within the framework of Noll’s theory of simple fluids so that a fluid of grade N could be considered. They developed the analysis up to order four for narrow gaps. The flow field between concentric rotating cylinders with a large annular gap was solved by Yoo [6,7]. It was observed that, up to order four, the deformation of the free surface is completely governed by the rheological constants entering the viscometric functions. Thus the free surface in a non-viscometric flow is determined, at least up to order four, by the viscometric functions alone. Beavers, Yoo and Joseph [l] proceeded to determine numerical values for combinations of the constitutive constants entering viscometric functions. A potential difficulty in using a concentric cylinder geometry to measure normal stress effects is that any eccentricity in the axes of the cylinders will induce circumferential pressure gradients independent of the normal stress effects. These additional gradients are mainly first order effects and may be described as “lubrication effects”. These effects can completely dominate the dynamics of the flow system for fluids which exhibit only small normal stresses. In this paper the analytical problem is solved using the ideas developed by Coleman and Noll[8] and the theory of domain perturbations. It is assumed that the stress response functional of the simple fluid is Frechet differentiable at the zero history in an appropriate function space of fading memories. The stress is expanded in a perturbation series in terms of the angular velocity St. At first order, we obtain a plane flow with an asymmetrical free surface. At second order the problem is reduced to three sub-problems: (i) a plane Newtonian flow which collapses, as the eccentricity goes to zero, onto
95
Fig. 1. The free surface between eccentric rotating cylinders.
the axially symmetrical solution for the case of two concentric cylinders; (ii) a top-driven, three dimensional, Newtonian flow which tends to zero as the eccentricity tends to zero and which generates secondary flows due to eccentricity; and (iii) a non-Newtonian pressure gradient which is balanced by a gradient with zero flow. As the eccentricity gets larger, the flow dynamics start being dominated by the first order terms and the top driven Newtonian flow at second order. The plane Newtonian flow at second order is solved analytically, and the three dimensional top driven flow is treated by a finite difference method. 2. Statement of the problem We consider the problem of an incompressible simple fluid which initially occupies the space VO between two fixed eccentric cylinders. The fluid is exposed to the atmospheric pressure pa. The inner and outer cylinders are rotated about their respective axes with angular velocities fJ and AQ respec-
Fig. 2. The bipolar coordinate system.
96 tively. The final shape of the free surface is determined by a complex balance of forces, Fig. 1. We use a bipolar coordinate system such that the material surfaces become coordinate surfaces, Fig. 2. The following are valid, with the choice of 4, < 0, i = 1,2: Bsinht B sin q x, =z, x, = cash 5 - cos 9 ’ cash 5 - cos 17’
x1 = -
R,=
-
B ->O,R,=
-- B
sinh 5r
sinh tZ
> 0, R,
R,.
’
The flow takes place in Va %=
{(~,77,z)Is1~~~52,0~0<28,
-~
Q)},
and satisfies the following mathematical problem DU/Dt=
-v++
+(& 17, z) =A&
v-S,
in Vn,
v-U=0
11, z) + Pgz,
s=
;
si,
i=l u(&,
17, z)
=
e,AQR,,
U*n=S,[=S,,=O
u(&,
v,
z)
at z = J&77;
W= U~e,,Szb,Szll+O
asz+
=
e,,QR,,
Q),
-cc.
(2.1) (2.2) (2.3a) (2.3b) (24
The expression (2.2) for the extra-stress comes from the retardation expansion of Coleman and No11 [8,9]. The first three terms of the expansion are
S, = PA,,
S,
= alA
+ azAf,
S,=PlA3+~,(A,Al+AlA,)+~,
TJ = @(t,
rl; 02)+ X,, + P, - +(6, v, z),
kinematic
(2.5)
which expresses the balance of the normal component of the jump in stress across the free surface and the surface tension T. J is the mean curvature of
97
the asymmetric free surface. A lengthy computation
yields
cash t - cos q % = WE, + %P,, +
+
(h,EY -
B
B2
(cash 5 - cos q)2 I
cash t - cos .i~ h,,h,,S,z
B S,, = h,&,
+ h,,S,,, + ‘Osh ‘;
(2.6)
- h,ESrz 9 B2
‘OS ’
(cash [ - cos q)2 - cash 5 - cos n B s
_ s
nnJ=
S&
_
h,,h,,S,z
- (h,,j2
1 Szq
(2.7)
+ h,$L
cash <-- cos n
zz
B
(h,&Z
(2.8)
+ h,&)
-${B[(~,~)'+(~,~)~]+(cosh~-cos9)[h,a(..~,.+ (cash 6 - cos TI)~
+h,ll(n.s),l+ _
B
[(h,,)2b,,),
+(h.02(“,n)?l
(cash 5 - cos TJ)~
P-9)
B n=
-
hhE-cow)
A =
B (cosht-cos#
A
((cash 5 - cos n,‘[ hf6 + A:,] + Bz)1’2.
The second order, ordinary, inhomogeneous, non-linear differential equation (2.5) describing the shape of the free surface has to be solved subject to contact angle or contact line boundary conditions: h&q)
= 0,
h(5i, n>=O
i= 1,2.
(2.10)
It has been shown that for Newtonian fluids the solution is regular, even at the comers, if we assume contact angle type boundary conditions [lo]. Consequently we feel justified in discarding contact line type boundary conditions in this work. Finally, the plane on which z = 0 is fixed by the conservation of mass condition B2 2?r tf2h(t, rl; Q) /J (cash 5 - cos n)2 0
dtdq=O.
(2.11)
3. Perturbation equations 3.1 Dimensional form of the equations We apply the theory of domain perturbations and construct the solution as Taylor series [SJl],
where
(-)‘“‘(I, 9,4 =
f&q_,,
are partial derivatives evaluated in the rest state VO o
‘tro={5,rl,zl&~5QEz,
-cc
The series expansion when used together with (2.1, 2.2, 2.3a,b, 2.4) gives at first order: +(‘>
U(‘)(&,
v . u(l)
= v . SW,
U(‘)(L,
rl, z) = e,%,
(3.la)
in *tr,,
= 0
7, z) = e,&,
(3.lb)
The conditions on the flat top of VO are identically satisfied wq,o
= S,‘:)l,=,
= S;;)l,_o
= 0,
together with the asymptotic conditions w(t),
s,<,l), sg’
+ 0
aszd
-cc.
The problem is two dimensional and we can introduce a stream function ~(1) (6, q) =
cash E - cos q B
( #yi)ec
- JI$)e,)
=
V$(')A e,.
Further manipulation gives AN 0) = 0,
(3.2a) (3.2b)
+/!;‘>(b,d = -
cosh
#(‘)( tl, 7) = constant,
tacos 9 =
-J-M,
1c/“)(lZ, 7) = 0.
(3.2~) (3.2d)
99 This is a very difficult problem to solve as it stands. But it can be brought to a tractable form by the transformation A2+“> = A2( Fx’l)),
A2 = ( F2A,)‘,
F=B/(cosh[-cosn),
AP=&+a,,.
(3.3)
At second order &)u(‘>
. vu(‘)
= -
vp
+
v
.
s(2),
v * uC2)= 0
in q,
(3.4a)
UC2)(61, 11, z) = UC2)(&, 9, z) = 0,
(3.4b)
(U.n)(2)Iz_0
(3-k)
(U.e,)“),
= S~&o SJ:),
= s:$0,
S!,z)+O
Z+
(3.4d)
-00.
We note that S(l) has-to be linear in U(‘) and does not depend on U(‘+m), M 2 1. Then SC> = Sj2> + $1) = PA,(‘) + alAp) + a,( A,(1))2.
(3.5)
We can rewrite the first of (3.4a) as 2pU(1> . vu(l)
= - Vc#b’) + 2( a, + a2)
+pv
V -(A,(1))2 + ? v(tr( A,(1))2)
-A,(‘).
(3.6)
Using the stretching tensor in bipolar coordinates, we compute
A$+
[QegeDe(+c(eg8e,+e,8et)+de,8e,],
a = BF-‘U$) c = BF-l(
- Y(l) sin n,
d = BFelV,$)
(3.7) -
U(l) sinh 5,
V,, + U,s)(l) + Utl) sin n + Y(l) sinh t,
u(l) = (u, Y)“’ ) (3.8)
where F is defined in (3.3) and reduce (3.6) to /LAU(~)=~~(F-~[(~e,,2+(~~~‘~]} _ (30~ + B2
2a2)
v(4a2
+ c’) - 2pA~%~(‘)
+ v+(~).
(3 -9)
The boundary conditions (3.4~) on the top of V0 are evaluated using (2.6), (2.7) and the recurrence relation for Rivlin-Ericksen kinematic tensors.
100 3.2 Non-dimensional
representation
of the equations
We introduce dimensionless coordinates xi = X/R,, sinh 6 cash 5 - cos n ’
x1=-b
b
x2 =
i = 1, 2, 3,
Silllj
(3.10)
cash 5 - cos 7 ’ b
I=--
sinh [2 ’
A dimensionless first order stream function, velocity, potential and height correction coefficient are defined by \k(‘) = #“>/R;,
@)
q,Bw = qp>/p,
= @/R,,
H(‘) = pgh(l)/p,
(3.12)
and dimensionless second order variables are defined by @W
p
u(2)p
= -
PR;
(p(2) = -
’
4w = (u,
PR;
u,
’
wy,
i = 1, 2. (3.13)
We use superposition and express the 2nd order problem as the sum of three separate physicafly meaningful problems. The dimensionless perturbation equations up to and including 2nd order are: At first order, a plane problem which contribution goes to zero as the eccentricity goes to zero, (A; - 2a,, + 2& + 1)x(l) = 0,
(fx"'),clc=f, = -XRf,, x(l)([r,
q) = const.,
in V0
(3.14a)
(fx"'),&=<,= -5,
xY52,17)
(3.14b) (3.14c)
= 0,
where A,, x(l) and f are defined in (3.3) and (3.11) respectively. At second order, (i) A plane Newtonian
which collapses onto the corresponding solution at 2nd order for concentric cylinder geometry as the eccentricity goes to zero,
-2A’P(‘)v\k(‘)
problem,
+ v@(~), i=l,
@(12)(Si, 11,z) = 0
2,
V .4j2)=0in
VO,
(3.15a) (3.15b)
W1(2)lr_o= ul’,iqz=o = u1,2,)1,,0= 0,
(3.15c)
W$2)
(3.15d)
, u,(y,
u$‘J
+
0
z+
-00.
101 (ii) A top driven Newtonian problem, which contribution goes to zero as the eccentricity goes to zero and which models secondary flows generated by the eccentricity, A.4$2) = ~a,(~), 4$2’(5i,
v -4Y$‘) = 0 in VO,
17, z, = O
w(2qr_o= --2 pR;
i=l,
(P21
2,
(3.16b)
I$(‘> . @(‘),
v
P
(3.16a)
’
V,
=fl(e& + evaq),
(3.16~)
(3.16d)
r,$) +
x [
dl)f sinh 4 1 + H,$u(‘)
Wz”), U$?, v$;) + 0
z+
+ H,$$(‘)
, 1
--co.
(3.16e) (3.16h)
(iii) A plane non-Newtonian problem, which collapses onto the corresponding solution at 2nd order for concentric cylinder geometry as the eccentricity goes to zero, A@12)=
-(3cr1+2a2)~(4a2+c2)+~Qj2), pRZ,B’
v.@$‘)=Oin%, (3.17a)
*s2’(5i9 w3
(91
z=lJ
11, ‘> = O =
u3(2)1 ,z 2 =(J =
wi2), U3(,2!,u3(,2!+ 0
(3.17b)
i= 1,2, u~“!~,=,
z+
=
(3.17c)
0 ,
-co.
(3.17d)
4. Solution for the flow field 4. I First order problem To solve problem (3_14a,b,c) we look for Fourier solutions periodic in r), XW,
s) = c CA,(O@,(s), m n
O”(V) = O,(Tj + 2a).
102 We find *Yt,
(4.1)
7) =
A,(6) = (hl+ hot) cash E +(X, + X&) sinh 5,
(4.2a)
AI(E) =X1, cash 2E + X,, sinh 2E + X,& + X4r,
(4.2b)
(4.2~) @I(V) = cos r). The boundary conditions (3.14b,c) together with the uniqueness of cP(‘) determine the eight constants in (4.2a,b), [12]. The pressure distribution at 1st order can be obtained from (3.la) and (4.1) O(l) = 2b-l[ Qr(6) sin 7 + Q,( 6) sin 2q] + const., Q,( 6) = (A,
(4.3a)
- 2x1,) sinh t - 2A2, msh 5,
Q,( t> = A,, sinh
21+ A21 cash
26,
(4.3b)
The constant in (4.3a) is determined in section 5. 4.2 Second order plane Newtonian problem The problem (3.15a,b,c,d) is solved by introducing a dimensionless stream function qC2), AA\kc2)e, = 2 curl( A\k(l)v*(‘)).
(4.4) By a change of variables \k c2) = fxc2), we obtain the following boundary value problem (A;
- 2i3,, + 2C$, + 1)x t2) = 2f2( \k,(;)a, - 9$%,)Aq(‘),
.X,(2’(5i9V)+X~~~(li,
q)=O
(4.5)
i=l,2.
(4.6)
Upon substituting (4.1) into (4.9, we get (A; - 23,, + 28,, + l)xc2) = if20,(E)
sin rnt,
m= 1,...,4,
(4.7)
al(E)=A,(Q1,~si~5-Q2)-~~,~~Ql,~cosh~-~Q2.~) -Al(Q2
a2(t)=A,,(+Ql+Q2,,sinhE)+Aa~(iQl,t-
Q2,tcoshO
+ +A,( Q, cash 6 + Q,,, sinh t) - 3Al,t(Q1,t
a3( t) = A,Q, + iQ2,,A,,, + tA,,,(iQ,,, a&)
Qz,ECOS~O~
cosh.$-fQ2,EsinhE)+fAl.~(~Ql.E-
= ~Q~,cAI,,.
cash 5 - Q2,t)T
+ A,( Q2 cash 6 + iQ2,~ si*
- Q2,i cash 0,
t)
103
Qr and Q, are given by (4.3b). We now seek Fourier solutions of the form n = l,...,co.
x’2)(& 7) = L”(I) sin n12
(4.8)
Substitution and orthogonality of trigonometric basis together with residue calculus give the following set of ordinary differential equations (4.9a) (49b)
3 coth ((coth 5 - n) + k2 + n2 - l] sinh k( -k(3cotht-2n)cothk[}a&),
n >, 2,
(4.9c)
to be solved subject to L,(5i) =
Ln,t(5i)
=
(4.10)
i=l,2.
O
The problem (4.9a,b,c, 4.10) is solved by variation of parameters. The solution is long and tedious and the interested reader is referred to [12]. Expressions similar to (4.1) and (4.5) were formerly obtained by Kamal [13], and Ballal and Rivlin [14] in the related but completely different problem of the flow of a Newtonian liquid between infinitely long eccentric cylinders without a free surface. Ballal and Bivlin’s work is very extensive and corrects some mistakes in the analysis given by Kamal. The pressure distribution may be obtained from (3.15a) $‘)
= @r + 2 M,(t)
cos kq + constant,
(4.11a)
k=O @‘T =f2
[( q$‘)’
+ (q,$))‘]
-
2( +(‘))‘A+,$).
(4.11b)
The expressions for Mk’s may be found elsewhere [12] and the constant in (4.11a) is determined in section 5. 4.3 Second order top-driven Newtonian Problem The Newtonian, top driven, three dimensional problem (3.16a,b,c,d,e,h) gives rise to secondary motions driven by the eccentricity. The problem is solved using a three dimensional lattice of finite differences with unequal spacings. The field equations can be rewritten as, A4g2) = v@,<‘),
A@J2) = 0.
(4.12)
104 In d&retiring the problem (4.12), second order truncation error is used throughout. With the’ introduction of a relaxation factor a, every discretized equation corresponding to (4.12) may be expressed as
[1+Q$]a&
= Za,,T,,
y
f (’ - a) up7
(4.13)
+ t
where P = (i, j, k), T = ( ui2), uj’), wj2), c#B$‘))and (nb) denote respectively a node point, one of the flow variables and a neighbor point. TF is the previously calculated value of T at P = (i, j, k). When a > 1 we have over-relaxation whereas a < 1 induces under-relaxation [15]. A line by line iteration or traverse method is used. The convergence of the method is quite fast. Upon noting that the boundary conditions (3.16c,d) and (3.16e) are respectively even and odd in 17 and that the field equations (3.16a) or their equivalent (4.12) are satisfied by flow variables even in 1 except for the 9 component of the velocity which is odd in 77, the domain of the finite difference solution is taken as ~={E,77,z152~5~5~,0~17~~,Z~~zo}, where 2 is determined through exploratory computations such that the pressure field at z = 0 is not affected further by a change in the depth of field. The algorithm has several interesting features, chief among them the convergence properties and the treatment of the corner singularity. Details of the analysis will be published elsewhere [16]. 4.4 Plane non-Newtonian problem The non-linearity appears at 2nd order. The solution of the field equations (3.17a) subject to (3.17b,c,d) is straightforward. q)(2)
4q2)=o,
=
(3%
+
pR;B2
3
2a2) @a2
+
(4.14)
c2h
where a and b are defined in (3.8). 5. Solution of the free surface problem We introduce dimensionless variables q&l) @W = -, &‘>
T
O=_ p&
’
P
pgh(‘) =__,q)w=P
442) PR;
Hc2> ’
= -ht2)g
R;
’
and following [5,11] we obtain from (2.5), (2.8), (2.9) and (2.10) If-u
2AP- l] H (9 = -@CO,
where
f
H,(i’( 5j, 11) = 0
i,j=1,2
and AP are defined in (3.11) and (3.3) respectively.
7
(5.1)
105 5. I Newtonian contribution to the free surface deformation The elliptic equation (5.1) satisfies a variational principle. As a consequence we apply Kantorovich’s method. At first order the integral to be minimized is
The form of the driving term O(l) given by (4.3a) suggests HP) = @‘> (5) sin 7l-F H$“)( 5) sin 2q + Cl. The constant Cl is determined from (2.11) 2n %‘)( lJ0 Ez
6, q; 0) f2 d5 dq = 0,
c,=O.
It follows from the substitution of (5.3) and (4.3a,b) in (5.2) d’H,(‘) -+k2 dt2
;&[sinh(k-l)c+2cosh(k-l)t]
HLfi(Si)= -fQ,,,(si)
-l}@=O,
(5.4)
k,i=l,2.
Each of the equations (5.4) is solved by a combination of Runge-Kutta and shooting methods. At second order there are Newtonian and non-Newtonian contributions to the height correction coefficients. The Newtonian contribution is twofold, Hi2) from the Newtonian plane flow (3.15) and Hi2) from the top-driven Newtonian flow (3.16). Although the Newtonian plane flow problem can be solved analytically, the expression (4.11a,b) for the pressure field is complicated. In the top-driven Newtonian problem the pressure field is solved simultaneously with the velocity field by finite differences and the values of the pressure at z = 0 are then known at the grid points. A finite difference method with a 2nd order truncation error is then used to solve (5.1) and evaluate Hi’) and Hi2). 5.2 Non-Newtonian
contribution to the free surface deformation
The non-Newtonian contribution H 12) to the height rise coefficient at second order is the solution to an equation similar to (5.1) with @i2) given by (4.14). We note that 0i2) is even in q and assume Hi2> = Hj’oz>+ Hi;> cos q + Hj’22)cos 217 + C,.
(5.5)
106 The variational principle for elliptic equations gives the following integrals to be minimized
k=O,
- 2H,<,2)ipj2) cos kq)) d5‘ dq;
1,2;
(5 -6)
which upon defining a new dimensionless height rise coefficient and a new constant H,“jpR; HZ2) = (3a, + 2a2) ’
yield three corresponding
ordinary differential equations
d2H;,j2) + g -H;o(% cash t e sinh3< dt2
+&c-h5
1 (5.7a)
’ e sinh3t ’
1
H;,‘2’=ahl,C(ho-ho,(()+~~~:1
-_331(t)+1
sinh3t ’ (5 .w
d’H;,(‘) dt2 (5.7c) i=
H;,(i’( Et) = 0, 331(1)
&&)=
1, 2; k = 0, 1,2,
= 27r e’(etsinh[-cosh2[), s.31
2 sk3f
CdC(3eSE- 5)-2
cosht].
The constant C,* may be computed using (2.11). We get aAPH;(z) +f2ipd2)]
d[ dq = 0.
Substituting @j2) from (4.14) q
= - (R; x (sinh
1)
{2~~(E,-Ei)+(X220+h240+4h21h31)
2& - sinh t2) + 2(h,A,
+ 2( qr + &)(sinh
44, -
Shh
+ 2X,,h,,)(cosh 41,)
+
4X,,h,,(cosh
2& - cash 28,) 4& - cash 46,)) -
107 of Equations (5.7a,b,c) can now be solved using a combination set of systems of Runge-Kutta and shooting methods by reducing them to a first order ordinary differential equations [12]. 6. The rise in height at first and second orders The shape of the free surface up to and including second order is given by
h&q;
n>= Qth(‘)(&q; 0) +gh’2’(E, q;0) +0(Q3), =
Q
+
CL@‘) ,g(h;
o)+$y H,“)(t,
(3.Ip+Rfa2) H$')( 6, q; 0)
+ 0(
71; 0) + @‘)(S,
$2’).
q; 0) +
(6-l)
As E + 0, (H(l), Hi”)) + 0. The pressure field Q(l), antisymmetric with respect to q, goes to zero together with the corresponding velocity field. At second order, the pressure field @I’) corresponding to the plane Newtonian flow collapses onto the pressure field of the axisymmetric flow corresponding to the concentric cylinder geometry [5,12]. We checked these assertions numerically for several geometries. In the same way the pressure field of the top-driven Newtonian flow goes to zero and the pressure field of the non-Newtonian flow collapses onto the axially-symmetric case as the eccentricity goes to zero. The height rise coefficients follow the same pattern. At second order the analysis involves only one viscoelastic parameter, the climbing constant. Its value can be determined from rod-climbing experiments [17]. Acknowledgement
This work was supported in part by the U.S. Army Research Office. The author would like to thank Dr. D.D. Joseph for suggesting the problem treated in this paper. References G.S. Beavers, J.Y. Yoo and D.D. Joseph, The free surface on a liquid between cylinders rotating at different speeds, Part III. Rheol. Acta, 19 (1980) 19-31. A. Siginer and G.S. Beavers, Free surface on a simple fluid between rotating eccentric cylinders, Part II: Experiments. J. Non-Newt. FIuid Mech., 15 (1984) 109-126. A.S. Wineman and A.C. Pipkin, Slow viscoelastic flow in tilted troughs. Acta Mech., 2 (1966) 104-115.
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