I nternat. J. Math. & Math. Sci.
(1981) 181-200
Vol. 4 No.
181
FLOW BETWEEN COAXIAL ROTATING DISKS:
WITH AND WITHOUT EXTERNALLY APPLIED
MAGNETIC FIELD*
R.K. BHATNAGAR
Instituto de Matemtica
Estatstica e Cincia da Computao
Universidade Estadual de Campinas, 13. I00
Campinas
So
Paulo- Brasil
(Received May 28, 1979)
ABSTRACT. The problem of flow of a Rivlin-Ericksen type of viscoelastic fluid is discussed when such a fluid is confined between two infinite rotating coaxial disks. The
governing system of a pair of non-linear ordinary differential equation is solved
by
treating Reynolds number to small. The three cases discussed are: (I)
is
one
disks
held at rest while other rotates with a constant angular velocity, (ii) one disk
ro-
rates faster than the other but in the same sense and (iii) the disks rotate in
op-
posite senses and with different angular velocities. The radial, tranverse and
components of the velocity field are plotted for the above three cases for
axial
different
values of the Reynolds number. The results obtained for a viscoelastic fluid are corn-
pared with those for a Newtonian fluid. The velocity field for case (i) is also com-
puted when a magnetic field is applied in a direction perpendicular to the discs
and
the results are compared with the case when magnetic field is absent. Some interesting
features are observed for a viscoelastic fluid.
R.K. BHATNAGAR
182
KEY WORDS AND PHRASES.
perturbation s olon.
Viscoelastic fluid, rotng system, Newtonian
1980 MATHEMATICS SUBJECT CLASSIFICATION CODES.
i.
fluid motion,
76A10, 76A05.
INTRODUCTION.
In a recent investigation we have discussed the flow of a viscoelastic fluid
of Rivlln-Erlcksen type between a pair of infinite, coaxial rotating disks (see ref.
and all the other references therein).
The disks are taken to rotate with
different constant angular velocities, either, in the same senses or opposite
senses, or, one disk is held at rest and the other is taken to rotate with a constant
angular velocity.
The system of non-linear ordinary differential equations governing
the flow was obtained and solved
numerically under the appropriate boundary
conditions of the problem using flnite-dlfference technique and successive overrelaxation procedure.
The solutions were given for values of the Reynolds
number up to I000 and some interesting features for the flow of a viscoelastic
fluid were reported.
In the present investigation, therefore, it is our aim to discuss the problem of
flow of a Rivlin-Ericksen fluid between coaxial rotating disks but for small
of the Reynolds number. As the basic system of equations governing such
been already derived in
[I]
a
values
flow
has
we shall briefly formulate the problem in the next sec-
tion. In section 3, we obtain the pertubation solutions for small values
the
of
Reynolds number. In section 4, we treat the flow in the presence of an externally ap-
plied magnetic field and provide pertubation solutions for the case of one disk
rest. At the end in section 5, we give some interesting illustrative examples
radial, transverse and axial components
of the flow field.
at
for
of the velocity and discuss characteristics
FLOW BETWEEN COAXIAL ROTATING DISKS
183
-2. FORMULATION OF THE PROBLEM.
Let us consider the steady flow of a Rivlin
Ericksen type of non-Newtonian
fluid occupying the space between a pair of infinite parallel disks. In a cylindrical
polar co-ordinate system (r, 8, z), let the lower disk situated at
angular velocity
locity
,
S
while the other disk, situated at
d
z
z
0
have
have an angular
an
ve-
and S being constants. Let u,v and w represent the components of velocity
in the increasing directions of r,8
and z respectively.
We now write
I r fH’(q), v
u
So that the
Here
r G(q),
w
d H(r)
equation 6f continuity is identically satisfied for an axi-symmetric flow.
.
z/d
is a non-dimensional axial co-ordinate given by
differentiation with respect to
and a dash denotes
Substituting from (I) into the equations of motion (as was done in If]and elimi-nating the pressure from the equations in the radial and axial directions respectively, we obtain the following system of non-linear ordinary differential equations
the functions
G
and
G" + R(H’G
H:
H’’G’)
HG’) + R(H’G’’
+ KR(HG’’’
H’’G’)
(2)
0
iv
R(HH’’’ + 4GG’) + 0LR(12G’G’’ + H’H + 2H’’H’’’) +
+ I HHv)
0
+ 2KR(H’’H’’’ + 8G’G’’ + I H,
H
for.
iv
HiV
(3)
where
R
d2qp
1
is the Reynolds number and e--
@3
pd
2
K
@2
pd
2
are the non-dimensional parameters characterising the viscoelasticity of the fluid.
The equations (I) and (3) have to be solved under the following non-dimensional-
R.K. BHATNAGAR
184
ized boundary conditions:
G(0)
S
G(t)
,
H(0)
0
H()
0,
H’(0)
(4)
0.
0, the above equations (2) and (3) reduce to the well-known equations
0, K
When a
H’I)
0
to the corresponding problem for a classical viscous fluid.
R.
3. SOLUTION FOR SMALL VALUES OF REYNOLDS NUMBER
We now develop regular pertubation solutions for the functions G and H in the
following manner:
G(n)
H(n)
R2G2(n) + R3G3(n) + ...,
-H0(n) + RH l(n) + R2H2(n) + R3H3(n) +
G0(n)
+
RGI(n)
+
order to obtain the various order solutions for the functions G and H
we
substitute their expressions in the equations (2) and (3) and the boundary conditions
(4).
(i) ZERO-ORDER SOLUTION
The equations for
G
o
and H
have the form
0
G ’’
(6)
0
O
and
H
to be solved under the
C0(0)
S,
G0()
iv
o
(7)
0
boundary conditions
, H0(0)
It can be easily seen that
H0()
0
(0)
H0’(t)
0
(8)
FLOW BETWEEN COAXIAL ROTATING DISKS
G0(n)
185
S + (- S) U
0()
H
(I0)
0
(ii) FIRST-ORDER SOLUTION
Making use of (9) and (I0), the equations for
G
I
and
H
I
take the form
and
Hiiv
S)[S +(I- S)].
4(1
The conditions satisfied by
G
I(0)
G
I(I)
G
and
I
0, H I(0)
H
112)
H
are
I
I(1)
H (0)
I
H ’()
I
(13)
0
Solving (II) and (12) and making use of (13), we get,
G(n)
(4)
0
and
H
I(N)
(l-S) (2+3S)N
+ I s(-s)
2
n4 + 3 (-s) 2
I (l-S) (3+7S) 3
5
(5)
It may be noted that the zero and first order solutions do not involve the viscoelastic parameters u and
K.
(iii) SECOND-ORDER SOLUTION
It can be easily seen that the equations determining the functions
G
2
and
have respectively the forms
G
and
z’’
(HIG -H{G0)
+ (u + K)H
I
G
O
(16)
H
2
186
R.K. BHATNAGAR
(17)
to be solved under the boundary conditions
G
2(0)
G
2(I)
H 2(I)
H 2(0)
0
H
2(0)
H 2(I)
(18)
0
Thus we get,
l-S
G2()
(27 S
6300
I-S
12S
-i--(I-4S
I
4-
S(l-S)
-
+ ( + K)
+
and
2
n
2
2) N4
I
6
3
(2 + sS)D
8)’- S(I-S)
90
+ 16 S
315
+
I-S
(l-S)
3
2
s)2
-------(2 + 3s)n
[(0
I S(I-S) 2
17S3)n 5
(3 + 4S
3--
n
7
(l-s)
30
2
(3 + 7S)
3
n4 + 3 (l-S) 3 n5
H
2(n)
(19)
(2o)
0
(iv) THIRD-ORDER SOLUTION
Substituting from (5) into the equations (2) and (3), it may be easily verified
that
G
and
3
G ’’
3
H
iv
3
H
3
satisfy the following equations respectively:
(21)
0
(HIH{’’
+
< 12GoG 2’’
+
2K(8G G2’’
+
+ I
+
4GoG
+
4G G 2)
iv +
2G’’G 2 + HIH I
8G’’G2
2H
+ H ,, H ,,, + 1
I I
H ’’’)
I
iv
HiH I
+
I H
I
HI
0
(22)
FLOW BETWEEN COAXIAL ROTATING DISKS
GO,
Using the expressions for
G
H
and
2
187
in the above equations and
I
them under the prescribed boundary conditions on
G3,
H
and
3
I-13
we firid
solving
that,
third-order solution is given by,
G
3(r])
0
H
3(r)
X
+ X
+ X
where
X.
i
1
Xo
0
(23)
r]
4
r
8
to
s(-s) (461
=-[ 2’2’680"0
I011 S + 2466 S
2361 S
+
X1
I
31’50 (e
--2S(I-S)26800
+ 3117
283500
2
1
4725
2
+ X
6
+ X
I0
i
9
922
s
9
+ X
2
]
r
II
(24)
are given by
s 2)
1219
+ 913
K(I-S)
283500
r]
5
+ X
D4 + X3 D 5
9
7
+ X n8 + X
6
7
]3
I
+
-s
-26’80bb (332
o(I-S)
S3)]
2
(474
567000
2
(552
1059 S
1743 S
1713 S
2)
3
+ K) (3e + 4K)(l-S) (16 + 19S)
(856- 1712S- 2504 S
$2
K(I-S)
X
2)
2
0 D
+ 1781 S
3) (I-S)2
567000
(7 + 4636 S + 5257 S
I (c, + K)(3e +
+
I- S
2268000" (579-
1277 S
(839 + 8072 S + 8189 S
2
S(I- S)(8- 16S-
2)
(25)
2)
+
27S2)
4K)(I-S)3(2+3S)
31 (e
2)
+ K)(3o + 4K)(I-S)
4,
(26)
2
+ (z+K) (l-S) (2+3S)(3+7S)
900
(27)
the
R.K. BHATNAGAR
188
I
X3
+ 117 S
X
(I-S)2(8-16S
"500
2)
+ K(27+86S+ 87 S
+ 57S
2)
+ K(4 + 20S +
I
X5
(lS,)
2
2-
-0+4S’-
315
2)
36S2)]
(l-S) (27 + 19S +
’3"7800b
+
2)
-
i
4500
+
161S2
X6--
2
1
+ 17S + 20S
151200 (I-S)(2
X7
453600
2
(l-S) (3 + 4S
i
II
X8
226800
X9
226800
S(l-S)
I
(I-S)
107S
2)
2
[(27 + 106 S
4K_) (l-S) 3 (3+7S)
(’I-S’2
+
( + K)(3 + 4K)S(I-S)
3)
3
(29)
3)
36
(l-S)
3
S(I-$)
S(l-S) 3_K
630
(30)
(31)
4
22680
(7 + 4K)
3
(32)
(33)
4
(34)
Thus the components of the velocity field
third power of the Reynolds number
(28)
(4 + 29S
2700
2
-(c + K)(3or + 4K)(l-S)
99S
2)
/
150
+ 393S
67S2) 2KS2
(l-S)
(_+..K).(3
(l-S)(2 + 3S)(3 + 4S + 13S
54000
4
27S
u,v
and
w
have been computed
to
the
R.
4. EFFECTS OF TRANSVERSE MAGNETIC FIELD ON THE FLOW FIELD :CASE OF ONE DISK HELD
AT
REST.
We now consider the problem of flow of a viscoelastic fluid when a constant magnetic field
H
0
is applied in a direction perpendicular to the discs. Making use
of
the steady state Maxwell’s equations and adding the contributions due to Lorentz
force in the equations of motion, it can be verified that the equations (2) and
(3)
189
FLOW BETWEEN COAXIAL ROTATING DISKS
for the functions G and H are modified to have the following forms respectively:
G’’ + R(H’G
H’’G’)
HG’) + aR(H’G’’
+ KR(HG’’’
M2G
H’’G’)
(35)
0
and
H
iv
iv
R(HH’’’ + 4GG’) + eR(12G’G’’ + H’H + 2H’’H’’’)
H’’H’’’
+ 2KR(8G’G’’’ +
M
where the Hartmann number
In the above expression for
+ I
H,Hiv
+
M2H ,,
I HHv)
0
(36)
is defined as
M,
0
o the
represents the magnetic permeability and
conductivity.
The boundary conditions satisfied by
G() and
H() in equations (35) and (36)
are the same as in (4).
We shall now obtain solutions for the case of one disk at rest, namely, the case
with
S
R
0. Once again treating Reynolds number
to be samall, we expand the func-
tions G(N) and H() in a series in R as was done in (5). The zeroth, first
and
the
second order solutions are then given as follows:
(i) ZERO-ORDER SOLUTION
When
M # 0
the equations for
M2G0
G ’’
O
iv
H0
G
O
and
H
0
are
0
(37)
M H’’= 0
(38)
Solving (37) and (38) and using the boundary conditions (8), it can be easily seen
that
G0(B)
I
S’inh M
Sinh (M)
(39)
190
R.K. BHATNAGAR
and
H
0()
(4O)
0
(ii) FIRST-ORDER SOLUTION
The functions G
I()
M2GI
GI’’
H
M2H
iv
I
satisfy the following equations:
Hi()
and
(41)
0
4(3C + 4K)
4GoG
(42)
GoG0
Solving them under the boundary conditions (13), we get
G ()
0
I
H I()
C +
DN
+
a
A
a0
O
M
2
B
Cosh(M) +
M
M
i
3
0
6M s inh
a0
00
a
C
0
0
M
2
2M
sinh(M) +
au
[2(1
(Cosh M
sinh(2M)
2
M (3c + 4K)]
[i
Cosh M) sinh(2M) + 2(M Cosh M
(i
+ 2(sinh(M)
B
-
A
where
(43)
(45)
sinh M)
M) Cosh(2M)]
I)
Cosh 2M)(Cosh M
I) sinh 2M
2 Cosh(2M)(Sinh M
(44)
(46)
sinh M(2M
2(M Cosh M
sinh 2M)]
(47)
Sinh M)
M)]
(48)
a
0
D --r--M [- sinh(M)sinh(2M) + 2(Cosh M-17(I + Cosh 2M)]
0
(49)
FLOW BETWEEN COAXIAL ROTATING DISKS
191
and
b
(5O)
M sinh(M) + 2(1"- Cosh M)
0
(iii) SECOND-ORDER SOLUTION
G2()
The equation satisfied by
is
(51)
Solving (51) under the appropriate boundary conditions, we have
PI { Cosh(M)-l} P2
G2()
+
where
P
P
P4(2CN
D 2) sinh(MN)
+
PI’ P2’ P3’ P4
D
4M
2
M K)
(3
P5
<52)
sinh(3MN)
are given by
P5
and
(5)
-
Cosh M
2
Cosh M)
(1
sinh M
+
P3 Csh(M)
(1 + M-<)
M3sinh (M)
2
sinh(M) +
#
a0sinh(3M) "el
2
16Msinh (M)
3a0.
:4
P3
I
sinh(M)
P4
I
4Sinh(M’) (I
sinh (M)
M
2
M K)
2C+D
4 sinh M
MDc
2(3K
+ 2()
2
{i + M (2( + K)}-
M2K)
4
(3
M2K)
a
(55)
(56)
and
a
0
16M sinh (M)
{I
M
2(3K
+
2a)}
(57)
192
R.K. BHATNAGAR
As for the case with M
0, here also it can be verified that
M ()
2
(58)
0
This completes the solution for the case of one disk at rest in the presence of tranverse magnetic field. The solutions for the case
S # 0
will be presented separately.
5. DISCUSSION OF THE RESULTS
ret or
(i)-oer dlk hd
Throughou
be
rog ft
R
our discussion we have taken the values of the Reynolds number
to
R --0.2, 0.4, 0.6 and 0.8. For the sake of comparison, results are depicted for a
(s
classical viscous fluid
0.I
and
K
0) and a typical viscoelastic fluid for which
0, K
0.05.
for S
H when the Hartmann number
function
H
M
K
0,
0
0, comparison is also made between the curves
S
and 5.0. For the case
0
e
for a classical viscous fluid
H
Fig. i depicts the curves of
of
than the upper dik:
and 1.0. It is noted that when
0
representing the axial velocity is always positive whether
S
0,
M--0
the
or 1.0.
The effect of the tranverse magnetic field on the flow is to decrease the axial velo-
city for each value of the Reynolds number from R
are nearly parabolic in character. When
body notation and as S
S
0.2
to
R
I, the whole system undergoes
increases further, the function
and
0.8
a
S
5.0 and R=0.2,
in Fig. i.
In Fig. 2, we hsve drawn the axial velocity profils for a viscoelastic
0.I, K
0.05
rigid
changes its sign. Ths may
M
be observed from the curves depicting axial velocity profiles for
0.4, 0.6
profiles
0.8. The
for
S
0
and
5.0. The values of
The .}Eacter of the profiles is similar
R
are same as in
to those for a classical Newtonian
fluid
Fig. I.
fluid.
It is interesting to note that in the presence of viscoelasticity in the fluid,
the
effect of tranverse magnetic field is to decrease the axial velocity much more than
for a Newtonian liquid. This may be easily oSserved from the curves of H for
M
1.0
FLOW BETWEEN COAXIAL ROTATING DISKS
193
0, K
and comparing them with the corresponding curves in Fig. 1 for
0. Here also
5.0, i.e. when the lower disk rotates five times faster the upper disk,
for S
the
axial velocity becomes completely negative.
Fig. 3 represents the curves for the function H’ for S
0.I, K--0.05.
0 and
-H’. For
It may be recalled that the radial velocity is actually designated by
value of the Reynolds number
R
from
R
0.8, H’ is positive in
0.2 to
each
the
lower
half region whereas it is negative in the upper half region between the two disks. The
increase in
R
gives rise to increase in
true in the upper half
H’
region. Once again, it is observed that the presence of
verse magnetic field decreases
H’
in the lower half region and
R
for each
is
in the lower half region. The reverse
tran-
the
that
reverse holds for the upper half region.
H’
In Fig. 4. we have drawn the profiles of the function
0.I, K
for
S
5.0. Comparing these curves with the corresponding curves for
and S
we note that the flow field is reversed and now
H’
-0.05
0 in Fig.4,
is negative in the lower half region
whereas it is positive in the upper, half region between the two disks.
Fig, 5 represents the curves of the tranverse velocity function G for th case of
lower disk held at rest i.e. S
Choosing the value of the Reynolds number
number
M
0.I, K
0, the fluid parameters being
-0.05.
0.2, comparison is made for Hartmanr
R
0 and I. In the absence of magnetic field, G varies linearly from
at the lower disk to G
G
0
1 at the upper disk. The tranverse velocity is increased when
magnetic field is applied as it is clear from the curve of G for M
is contrary in Fig. 6, which shows curves of
G
for
R
0.4 and
1.0. The situation
-0.05,
0.I, K
the lower disk rotating five times faster than the upper disk, G now decreases linear-
ly between the two disks from G
5 at the lower disk to G
(ii) two disks rotating in opmosite decto and with
We now discuss some results for a viscoelastic fluid
1 at the upper disk.
different angula
i, K
veloci-
-0.05 when
the
194
R.K. BHATNAGAR
two disks rotate in opposite directions and with different angular velocities i.e.
S is non-zero negative.
Fig. 7 depicts the curves of the axial velocity function H for R
0.6, 0.8 and different negative values of S.
When S
-0.5, H is completely posi-
However, when S
tive and increases in the entire region with increase in R.
i.e.
0.2, 0.4,
-i.0,
when the two disks rotate with same angular velocity but in opposite senses,
H is positive but only in the upper half region.
takes negative values.
smallness.
In the lower half region, it
They are not shown in the figure because of their extreme
For S =-0.5, the region of positive H completely disappears and the
In
axial velocity becomes negative in the entire region between the two disks.
this case, therefore H decreases with increase in R.
In Fig. 8, we have drawn the curves of H’ for S
H’ takes both positive and negative values.
the upper half region (n
0.57).
The point where
H’
The increase in R increases
region and reverse is true for the upper region.
so when S
-0.5 and -i.0.
When S
-0.5,
vanishes lies in
H’
in the lower
The situation no longer remains
The flow region is now divided into three parts with the appear-
-i.0.
ance of a central core in which
H’
is positive for all values of R considered.
In
the other two regions, one near the lower disk and the other near the upper disk,
H’
is negative.
Thus for this case, we have two points where
H’ vanishes;
one lies
in the lower half plane whereas the other lies in the upper half plane.
From Fig. 9 showing the curves of H’ for S
core has disappeared.
in the upper region.
-0.5, we observe that the central
Now H’ is negative in the lower region whereas it is positive
The point where
H’
vanishes lies in the lower half plane.
The increase in R causes increase in H’ in the upper half region while the reverse
is true for the lower half plane between the two disks.
Fig. I0 represents the graph for the function G depicting the tranverse velocity for R
0.6, S
linearly from-5.0 at
for S
0 as S
-5.0 and
0.i, K
0 to 1.0 at
-0.05.
1.0.
It is observed that G varies
This behaviour is similar to those
5.0 in the absence of magnetic field.
The numerical computations were carried out on the University of Campinas
PDP i0 computer.
195
FLOW BETWEEN COAXIAL ROTATING DISKS
S-O, =0
-0.19
-0.15
-0.05
-0.10
--H
0.0015
0
I.1 Cur,e, ofllfor==O,K--O;$--O, II-0.2, 0.4, 0.6 and 0.8 for M 0 and
for M
1.0.
K-O
0.0045
0.0050
H-’-
I.O
S-O
S-5.0
K--O.05
<..R-0.4
-018
-0.15
-0.05
-OlO
H FI. 2
0
Curel of II fr o, 0.1, K
O, and $.0, II 0.2, 0.4 0.8 and 0.8
and--- for i 1.0.
00015
-O.OS; $
fr M 0
0.0050
H
0.0045
0.0060
196
R.K. BHATNAGAR
0.1, K- -0 05
S O,
\\
-0.05
-0010
-0.005
H’
0
0.005
F. 3 Cu of H’ for
0.1, K
0.05; S O, R 0.2, 0.4, 0.6 and 0.8
for M 0 and ---for M 1.0.
0.010
0015
H’
1.0
S-- 5.0
x 0.1, K
0.05
R-02
-0.60 -0.48 -036 -0.24 -0.12
H’
0
0]2 0.24
O.l, K -0.05; S
FIG. 4 Curves of H’ for cx
S.O, R 0.2, 0.4, 0.6 and 0.8.
0.:56
H’
Q48
Q60
FLOW BETWEEN COAXIAL ROTATING DISKS
J
s.o
a- 0.1, K- -0.05
FIG. S Curves of G for a 0.1, K
for M 0 and ---for M |.0.
x,
1.0
1.0
0.5
0
1.0
/////’
-0.05; S
0 and R
0.2.
S-50
-0.1, K --0.05
2.0
.3.0
4.0
5.0
198
R.K. BHATNAGAR
S ---Q5
S---1.0
S --5,0
/
R-0.8
\
"\
-0.20
-QI5
-0 I0
H
-0.05
-0.015 -0.012 -0.009-0.006-0.00:5
H’
FIlL 8 :ll,es of
0.001
0
rlQ. 7 Cures of II for
varbu$ vale$ of R,
fer $
for $ -5.0.
0
H’ for
0.1, K -0.05 for
-0.5,--- $ -1.0 and
0.00 2
0005
H’-"
0.005 0006 0009 0.012 0.015
oI
0.05 for varkm$ valuo$ of R,
-0.5 and
for S -1.0.
0.1_, K
for $
H
FLOW BETWEEN COAXIAL ROTATING DISKS
|99
1.0
S’-5.0
0.1, K---0.05
01"
-0.56
-0.40
H"
-0.08 0
-0.24
0.24
0.08
FIG. 9 Curves of H’ for (x 0.1, K
and S -5.0 for various values of R.
-0.05
0.40
H’
1.0
S
oi--
.0
-4.0
-5.0
0.1, K -005
30
FIG. 10 Curve of G for (:x
-1.0
-2.0
0.1, K
-0.05, S
-5.0 and R
0.0
0.6.
0.56
200
R.K. BHATNAGAR
*The paper was presented at the IUTAM Symposium on Non-Newtonian fluid Mechanics
held at Louvain-la-Neuve, Belgium from Aug. 28 to Sept. i, 1978.
REFERENCES
[i]
Bhatnagar, R.K. and J.V. Zago. Numerical Investigations of Flow of a Viscoelastic fluid between Rotating Coaxial Disks, Rheologica Acta 17, (1978),
557-567.