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An unsteady flow due to eccentrically rotating porous disk and a fluid at infinity
hr. 3. Engng Sci. Vol. 25, No. 11/12, pp. 1419-1425,1987
0020-7225/87 $3.00+ 0.00 Copyright 0 1987Pergamon Journals Ltd
Printed in Great Britain. All rights reserved
AN UNSTEADY ROTATING
FLOW DUE TO ECCENTRICALLY POROUS DISK AND A FLUID AT INFINITY
S. R. KASIVISWANATHAN Department
of Applied Mathematics,
and A. RAMACHANDRA
RAO
Indian Institute of Science, Bangalore 560012, India
Abstract-An exact solution of the unsteady Navier-Stokes equations is obtained for the flow due to non-coaxial rotations of a porous disk, executing non-torsional oscillations in its own plane, and a fluid at infinity. It is shown that the infinite number of solutions existing for a flow confined between two disks reduce to a single unique solution in the case of a single disk. The adjustment of the unsteady flow near the rotating disk to the flow at infinity rotating about a different axis is explained.
1. INTRODUCTION
The possibility of an exact solution of the Navier-Stokes equations for the flow due to non-coaxial rotations of a disk and a fluid at infinity has been implied by Berker [I]. Exact solutions for the flow due to a single disk in a variety of situations have been obtained by number of workers. Thornley [2] has studied the flow due to non-torsional oscillations of a single disk in semi-infinite expanse of fluid in a rotating frame of reference. The MHD effect on the Ekman layer over an infinite horizontal plate at rest relative to an electrically conducting liquid which is rotating with uniform angular velocity about a vertical axis has been studied by Gupta [3]. The flow due to rotations of a porous disk and a fluid at infinity rotating about a different axis has been investigated by Erdogan [4]. In this paper, an exact solution for the unsteady flow due to an eccentrically rotating porous disk oscillating in its own plane and a fluid at infinity is presented. It is shown that the infinite number of solutions existing for the flow in the geometry of two parallel disks given by Berker [5] reduce to a single unique solution for the case of a single disk. How the unsteady flow near the disk adjusts to the steady flow at infinity which rotates about a different axis is shown in the graphs. The effect of suction or injection on the flow field is also discussed. 2.
FORMULATION
AND
SOLUTION
Consider the flow due to an oscillating porous disk lying in the Oxy plane rotating about Oz axis perpendicular to the disk with uniform angular velocity 52 in Cartesin coordinate system. The fluid at z = m rotates about an axis parallel to Oz passing through the point (xl, yl). For this motion, the velocity field has the form, Rajagopal
[617
c4 = -Q[Y
-&,
01,
v = Q[x -f(z,
t)l,
w = -w,,
where W, is the suction velocity and it satisfies the constraint Substituting (1) in the Navier-Stokes equations, we obtain
0)
of incompressibility.
(3) and
1420
and A. R. RAO
S. R. KASI~ISWANA~AN
From (41, it is clear that p is independent of z. Eliminating p from (2) and (3) by differentiating with respect to z and combining them, we get
dZW d2W v*+w O-iiT-G-
id dW
o
(5)
dz-- ’
az”
where W = f + ig. Each point of the disk z = 0 is subjected to elliptic harmonic oscillations and the boundary condition corresponding to it is given by W = aein’ + bemint at r =: 0,
(6)
to the where a and b are complex constants. The boundary condition co~espon~ng fluid at infinity rotating about an axis passing through (x,, yi) parallel to z-axis is given bY
W=xl+iyl
at
z=m.
(7)
Further, we assume that the solutions are bounded at infinity. The bounda~ conditions suggest, for an os~iliatory flow with non-vanishing mean, a solution of the form W = F,(z) + al;,(z)e’“” -i- b_F2(z)emint,
rl
>o.
(8)
By substituting (8) in (S), we get three third order ordinary differential equations for 4, Fi and Fz and the solutions of these equations satisfying the corresponding boundary conditions derived from (6) and (7), for n
iyl)(l
- e-‘V)
+ ae-“Yein”
$ be-“8e-i”*,
(9)
where FYljZ
=
yj
=
r;=(-) 0
cj((yi
CXj=fiSj+
ifij),
+
{(s; + 1y + s;> 1’2,
pi = {(s;
yj
+ 1)‘”
_ $}1’2
I
i =
O*I, 2
l/2
$2 2Y
z,
s-2
Kl s"=2(iav)'"
!!.J2 = (y)l’zz,
[I = (y)1’2z,
Sl =so
f
IR
( 1 Q
+a
and
sz
s2=so
l/2
( 1 a_n
(10)
*
It is interesting to note that there are three linearly independent solutions for each of the differential equations but only two of them are bounded at inanity. Thus, the two boundary conditions given in (6) and (7) are enough to determine the solution completely. The first term in (9) corresponds to steady solution while the second and third terms correspond to unsteady solution due to non-torsional o~iilations of the disk. For an impermeable disk the suction parameter sg is equal to zero. Equating the real and imaginary parts of (9) by taking a = al + iuz, b = bl + ib2, we get f = e-“15~{alCOS(/~~~~ - nt) + a2 sin&c1 +
- nt)}
e-cr2s2{blCOS(~~~,+ nt) + bz sin(/32[2 + nt))
+x1(1 - e-luoCO cos &JO) - y,e-*050 sin PO&),
(111
g = e-I”i51(a2 cos(&Sl - at) - al W&C1 - nt)> f e--(u2c2{b2 cos(/3& + nt) - bl sir@& + nt)} + yl(l - e-nOi;ocos &JO) + xle-“ofo sin &&.
(12)
1421
Eccentrically rotating porous disk
The solution for the case when n > Q is given by f = e-n151{ul cos(/3,5, - nt) + a2 sin&f1
- nt)}
+ e--(Y3t3{b1 cos(/&& - nt) - b2 sin&&
- nt)}
+ x,(1 - eeaoCocos /3&J - y1e-no5‘osin PO&, g = e-&
(13)
(112cos(P11;1- nt) - 4 sin(PIC1- nt)>
+ e-+{b,
cos(~& - nt) + bl sin&&
+ y,(l - e-“OCO cos p&J
- nt)}
+ x1e-(yo50sin PO&,
(14)
where (y3
=
fis, + {(St:+ 1)“2 + s;}1’2,
p3 = {(s: + 1)“2 - $}l’2,
53 = (Y)
sg=so
(
n--Q
When the natural frequency of rotation is equal to the forced frequency resonates and in this case the solution is given by f = e-ru151{ulcos(plCl - nt) + u2 sin&f1
l/2
Q
lIzz,
>
*
(15)
n, the system
- nt)}
+ e-2v’050{b1 cos rtt + b2 sin nt} +x1(1 - emnoCo cos Bofo) - yle -ao5osin Pofo,
(16)
g = e-ff151{a2 cos(B& - nt) - al sin(P& + e - 2v’oCo{b2 cos nt - bl sin nt}
- nt)>
+ y,{l - eelrota cos polo} + xleCnosOsin Boco.
(17)
When suction W. = 0, the second terms on the right hand sides of eqns (16) and (17) do not satisfy the boundary condition at z = 03, Thornley [2], and an initial value formulation is required to obtain a solution. In the absence of suction and when x1 = y, = 0, our results are similar to those given by Thornley [2]. The results presented in (11) and (12) reduce to those given by Erdogan [4] in the absence of oscillations, x1 = 0 and y, = 1. 3.
DISCUSSION
OF
THE
RESULTS
Introducing a disk at z = d which rotates about an axis parallel to z-axis and passing through (x1, yl), with the same angular velocity, the boundary condition at z = CCis replaced by W =x1 + iyl at 2 = d. (18) We focus our attention on the steady solution F. in (8) for the discussions, as the essential nature of the solutions Fl and F2 for unsteady case in (8) are similar to Fo. The solution F. satisfying (5), (6) and (18) is given by F,=
x1 + e-wd
ih (eem@- 1) + A[ eyz-_i - 1
(1 - eemor) + em* - 11,
(19)
where m& =
5‘0(&
+
$01,
a;)
=
-es,
+
yo,
(20)
and A is an arbitrary constant which remains undetermined under the stated boundary conditions giving a possibility of infinite number of solutions. Similarly the solutions for Fl and F2 also contain one arbitrary constant each undetermined. Thus there exist three sets of infinite number of solutions for this problem in general (see Berker [5] and Ramachandra Rao and Kasiviswanathan [7]). If one prescribes the pressure gradient in Navier-Stokes equations, the unknown constants will be determined and one will get an unique solution. For example in the absence of pressure gradient the solution for F,
1422
S. R. KASIVISWANATHAN
and A. R. RAO
reduces to sinh nlz
F,(z) = (x1 + iyl)eCwo(L-d)‘2Y~,
(21)
1
where 11i= - Wo/2Y + m(J,
(22)
and this corresponds to the steady solution presented by Rajagopal [6]. In the absence of suction W, = 0, (21) reduces to the symmetric solution for the problem given by Berker [5]. For a single disk problem, by taking the limit d+ = in (19) and imposing the condition that the solution remains bounded at infinity, we get 4 = (xi + iyi)(l - e-moz), which coincides In order to expressions for the second and
(23)
with the solution given in (9). get the nature of the velocity distribution near the disk z = 0, the f and g given in (11) and (12) are expanded in powers of z. Neglecting higher order terms in z, they are given by f - {&(I - %C,) + +
{b20
+
Xl%fO
-
a2C2) -
b,(l -
a2521
-
-
a2(1
~lC~)
+
azP,f1+
b2P25‘2bOS
+
-
~IPI~;I
nt
hP2C2)sin
nt
YAL
(24)
g - (a~(1 - mis;J + &(I - w&) - aiPifr + {ai(I - alC1) -
bi(l - (~~52)+ &Cl
- MLJcos - W&]sin
nr nr
+ Y,W& + xi/M-0.
(25)
The velocity vector near the disk is inclined at an angle tan-‘(-f/g) to the disk. For a special case in which a, = u2 = b, = b2 = y, = 1, x1 = 0, so = 0, nt = rr/2, eqns (24) and (25) simplify to f -2(1;1 - L) - 50, (26) g--Cot (27) and the velocity vector is inclined at an angle tan-‘(1 - 2N), where N = (1 + n/Q)“’ (1 - n/&)“2, from the disk. This angle depends on the value of n/Q and when n + 0, the velocity vector is inclined at an angle of 45” to the disk, Erdogan [4]. The thickness of the boundary layer is of order (v/IS2 - n])“’ and the penetration depth of the boundary layer decreases with increase in the suction parameter so. This is in keeping with the fact that suction causes reduction in boundary layer thickness.
Fig. 1. Variations off and g for different values of s0 at nt = 0 and n/Q = 0.5.
1423
Eccentrically rotating porous disk
0
1.0
Fig. 2. Variations off and g for different values of s0 at nt = 1112and n/Q = 0.5.
The nature of the flow near the disk is seen from the variations of f and g as observed from the Figs l-5 for the case in which a, = a2 = br = b2 = y, = 1 and x1 = 0. The variations of f and g for different values of suction parameter for nt = 0 and n/Q = 0.5 are shown in Fig. 1, whereas for nt = ~rd/2and nlQ = 0.5 or 2.0 in Figs 2 and 3. Each point of the disk is subjected to an oscillatory motion together with the axis about which it is rotating. Whereas the fluid at infinity is rotating about an axis passing through the fixed point (0,l). Figures l-3 clearly indicate how the unsteady flow near the disk is adjusting to the steady flow at infinity. We observe that the boundary layer thickness decreases with the increase in either suction parameter so or the parameter
Fig. 3. Variations off and g for different values of s0 at nt = n/2 and n/S2 = 2.0.
1424
S. R. KASIVISWANATHAN
and A. R. RAO
Fig. 4. Variations off and g for different values of se at nt = n/2 and n/S2 = 0.5.
Fig. 5. Variations off and g for different values of se at nt = n/2 and n/Q = 2.0.
n/8. Figures 4 and 5 depict the variations of f and g for different values of blowing parameter (sO< 0) at nt = 36/2 for n/Q = 0.5 and 2.0 respectively. It is seen that blowing causes thickening of the boundary layer and this boundary layer thickness decreases with increase in the rotation. Further, we observe that the distance from the disk in which the unsteady rotating flow adjusts with steady rotating flow about a different axis at infinity is smaller for the case when suction is present compared to that of injection. Acknowledgement-The
authors are thankful to the referee for his helpful comments and suggestions.
REFERENCES R. BERKBR, Integrations des du mouvement dun fluide visqueux incompressible. Handbuch der Physik, Vol. VIII/2, p. 87. Springer-Verlag, (1963). C. THORNLEY, On Stokes and Rayleigh layers in a rotating system. Quart. /. Mech. Appl. Math. 21, 451-461
(1968).
A. S. GUITA, Magnetohydrodynamic Ekman Layer. Acta Mech. 13, 155-160 (1972). M. E. ERDOGAN, Flow due to eccentrically rotating a porous disk and a fluid at infinity. ASME /. Appl. Mech. 43, 203-204 (1976). R. BERKBR, An exact solution of the Navier-Stokes equation-The vortex with curvilinear axis. In?. J. Engng Sci. 20,217-230 (1982).
Eccentrically rotating porous disk
1425
[6] K. R. RAJAGOPAL, A class of exact solutions to the Navier-Stokes equations. Int. J. Engng Sci. 22, 451-458 (1984). [7] A. RAMACHANDRA RAO and S. R. KASIVISWANATHAN, On exact solutions of unsteady Navier-Stokes equations-The vortex with instantaneous curvilinear axis. Int. J. Engng Sci. 25, 337-349 (1987). (Revised version received 24 March 1987)
0020-7225/87 $3.00+ 0.00 Copyright 0 1987Pergamon Journals Ltd
Printed in Great Britain. All rights reserved
AN UNSTEADY ROTATING
FLOW DUE TO ECCENTRICALLY POROUS DISK AND A FLUID AT INFINITY
S. R. KASIVISWANATHAN Department
of Applied Mathematics,
and A. RAMACHANDRA
RAO
Indian Institute of Science, Bangalore 560012, India
Abstract-An exact solution of the unsteady Navier-Stokes equations is obtained for the flow due to non-coaxial rotations of a porous disk, executing non-torsional oscillations in its own plane, and a fluid at infinity. It is shown that the infinite number of solutions existing for a flow confined between two disks reduce to a single unique solution in the case of a single disk. The adjustment of the unsteady flow near the rotating disk to the flow at infinity rotating about a different axis is explained.
1. INTRODUCTION
The possibility of an exact solution of the Navier-Stokes equations for the flow due to non-coaxial rotations of a disk and a fluid at infinity has been implied by Berker [I]. Exact solutions for the flow due to a single disk in a variety of situations have been obtained by number of workers. Thornley [2] has studied the flow due to non-torsional oscillations of a single disk in semi-infinite expanse of fluid in a rotating frame of reference. The MHD effect on the Ekman layer over an infinite horizontal plate at rest relative to an electrically conducting liquid which is rotating with uniform angular velocity about a vertical axis has been studied by Gupta [3]. The flow due to rotations of a porous disk and a fluid at infinity rotating about a different axis has been investigated by Erdogan [4]. In this paper, an exact solution for the unsteady flow due to an eccentrically rotating porous disk oscillating in its own plane and a fluid at infinity is presented. It is shown that the infinite number of solutions existing for the flow in the geometry of two parallel disks given by Berker [5] reduce to a single unique solution for the case of a single disk. How the unsteady flow near the disk adjusts to the steady flow at infinity which rotates about a different axis is shown in the graphs. The effect of suction or injection on the flow field is also discussed. 2.
FORMULATION
AND
SOLUTION
Consider the flow due to an oscillating porous disk lying in the Oxy plane rotating about Oz axis perpendicular to the disk with uniform angular velocity 52 in Cartesin coordinate system. The fluid at z = m rotates about an axis parallel to Oz passing through the point (xl, yl). For this motion, the velocity field has the form, Rajagopal
[617
c4 = -Q[Y
-&,
01,
v = Q[x -f(z,
t)l,
w = -w,,
where W, is the suction velocity and it satisfies the constraint Substituting (1) in the Navier-Stokes equations, we obtain
0)
of incompressibility.
(3) and
1420
and A. R. RAO
S. R. KASI~ISWANA~AN
From (41, it is clear that p is independent of z. Eliminating p from (2) and (3) by differentiating with respect to z and combining them, we get
dZW d2W v*+w O-iiT-G-
id dW
o
(5)
dz-- ’
az”
where W = f + ig. Each point of the disk z = 0 is subjected to elliptic harmonic oscillations and the boundary condition corresponding to it is given by W = aein’ + bemint at r =: 0,
(6)
to the where a and b are complex constants. The boundary condition co~espon~ng fluid at infinity rotating about an axis passing through (x,, yi) parallel to z-axis is given bY
W=xl+iyl
at
z=m.
(7)
Further, we assume that the solutions are bounded at infinity. The bounda~ conditions suggest, for an os~iliatory flow with non-vanishing mean, a solution of the form W = F,(z) + al;,(z)e’“” -i- b_F2(z)emint,
rl
>o.
(8)
By substituting (8) in (S), we get three third order ordinary differential equations for 4, Fi and Fz and the solutions of these equations satisfying the corresponding boundary conditions derived from (6) and (7), for n
iyl)(l
- e-‘V)
+ ae-“Yein”
$ be-“8e-i”*,
(9)
where FYljZ
=
yj
=
r;=(-) 0
cj((yi
CXj=fiSj+
ifij),
+
{(s; + 1y + s;> 1’2,
pi = {(s;
yj
+ 1)‘”
_ $}1’2
I
i =
O*I, 2
l/2
$2 2Y
z,
s-2
Kl s"=2(iav)'"
!!.J2 = (y)l’zz,
[I = (y)1’2z,
Sl =so
f
IR
( 1 Q
+a
and
sz
s2=so
l/2
( 1 a_n
(10)
*
It is interesting to note that there are three linearly independent solutions for each of the differential equations but only two of them are bounded at inanity. Thus, the two boundary conditions given in (6) and (7) are enough to determine the solution completely. The first term in (9) corresponds to steady solution while the second and third terms correspond to unsteady solution due to non-torsional o~iilations of the disk. For an impermeable disk the suction parameter sg is equal to zero. Equating the real and imaginary parts of (9) by taking a = al + iuz, b = bl + ib2, we get f = e-“15~{alCOS(/~~~~ - nt) + a2 sin&c1 +
- nt)}
e-cr2s2{blCOS(~~~,+ nt) + bz sin(/32[2 + nt))
+x1(1 - e-luoCO cos &JO) - y,e-*050 sin PO&),
(111
g = e-I”i51(a2 cos(&Sl - at) - al W&C1 - nt)> f e--(u2c2{b2 cos(/3& + nt) - bl sir@& + nt)} + yl(l - e-nOi;ocos &JO) + xle-“ofo sin &&.
(12)
1421
Eccentrically rotating porous disk
The solution for the case when n > Q is given by f = e-n151{ul cos(/3,5, - nt) + a2 sin&f1
- nt)}
+ e--(Y3t3{b1 cos(/&& - nt) - b2 sin&&
- nt)}
+ x,(1 - eeaoCocos /3&J - y1e-no5‘osin PO&, g = e-&
(13)
(112cos(P11;1- nt) - 4 sin(PIC1- nt)>
+ e-+{b,
cos(~& - nt) + bl sin&&
+ y,(l - e-“OCO cos p&J
- nt)}
+ x1e-(yo50sin PO&,
(14)
where (y3
=
fis, + {(St:+ 1)“2 + s;}1’2,
p3 = {(s: + 1)“2 - $}l’2,
53 = (Y)
sg=so
(
n--Q
When the natural frequency of rotation is equal to the forced frequency resonates and in this case the solution is given by f = e-ru151{ulcos(plCl - nt) + u2 sin&f1
l/2
Q
lIzz,
>
*
(15)
n, the system
- nt)}
+ e-2v’050{b1 cos rtt + b2 sin nt} +x1(1 - emnoCo cos Bofo) - yle -ao5osin Pofo,
(16)
g = e-ff151{a2 cos(B& - nt) - al sin(P& + e - 2v’oCo{b2 cos nt - bl sin nt}
- nt)>
+ y,{l - eelrota cos polo} + xleCnosOsin Boco.
(17)
When suction W. = 0, the second terms on the right hand sides of eqns (16) and (17) do not satisfy the boundary condition at z = 03, Thornley [2], and an initial value formulation is required to obtain a solution. In the absence of suction and when x1 = y, = 0, our results are similar to those given by Thornley [2]. The results presented in (11) and (12) reduce to those given by Erdogan [4] in the absence of oscillations, x1 = 0 and y, = 1. 3.
DISCUSSION
OF
THE
RESULTS
Introducing a disk at z = d which rotates about an axis parallel to z-axis and passing through (x1, yl), with the same angular velocity, the boundary condition at z = CCis replaced by W =x1 + iyl at 2 = d. (18) We focus our attention on the steady solution F. in (8) for the discussions, as the essential nature of the solutions Fl and F2 for unsteady case in (8) are similar to Fo. The solution F. satisfying (5), (6) and (18) is given by F,=
x1 + e-wd
ih (eem@- 1) + A[ eyz-_i - 1
(1 - eemor) + em* - 11,
(19)
where m& =
5‘0(&
+
$01,
a;)
=
-es,
+
yo,
(20)
and A is an arbitrary constant which remains undetermined under the stated boundary conditions giving a possibility of infinite number of solutions. Similarly the solutions for Fl and F2 also contain one arbitrary constant each undetermined. Thus there exist three sets of infinite number of solutions for this problem in general (see Berker [5] and Ramachandra Rao and Kasiviswanathan [7]). If one prescribes the pressure gradient in Navier-Stokes equations, the unknown constants will be determined and one will get an unique solution. For example in the absence of pressure gradient the solution for F,
1422
S. R. KASIVISWANATHAN
and A. R. RAO
reduces to sinh nlz
F,(z) = (x1 + iyl)eCwo(L-d)‘2Y~,
(21)
1
where 11i= - Wo/2Y + m(J,
(22)
and this corresponds to the steady solution presented by Rajagopal [6]. In the absence of suction W, = 0, (21) reduces to the symmetric solution for the problem given by Berker [5]. For a single disk problem, by taking the limit d+ = in (19) and imposing the condition that the solution remains bounded at infinity, we get 4 = (xi + iyi)(l - e-moz), which coincides In order to expressions for the second and
(23)
with the solution given in (9). get the nature of the velocity distribution near the disk z = 0, the f and g given in (11) and (12) are expanded in powers of z. Neglecting higher order terms in z, they are given by f - {&(I - %C,) + +
{b20
+
Xl%fO
-
a2C2) -
b,(l -
a2521
-
-
a2(1
~lC~)
+
azP,f1+
b2P25‘2bOS
+
-
~IPI~;I
nt
hP2C2)sin
nt
YAL
(24)
g - (a~(1 - mis;J + &(I - w&) - aiPifr + {ai(I - alC1) -
bi(l - (~~52)+ &Cl
- MLJcos - W&]sin
nr nr
+ Y,W& + xi/M-0.
(25)
The velocity vector near the disk is inclined at an angle tan-‘(-f/g) to the disk. For a special case in which a, = u2 = b, = b2 = y, = 1, x1 = 0, so = 0, nt = rr/2, eqns (24) and (25) simplify to f -2(1;1 - L) - 50, (26) g--Cot (27) and the velocity vector is inclined at an angle tan-‘(1 - 2N), where N = (1 + n/Q)“’ (1 - n/&)“2, from the disk. This angle depends on the value of n/Q and when n + 0, the velocity vector is inclined at an angle of 45” to the disk, Erdogan [4]. The thickness of the boundary layer is of order (v/IS2 - n])“’ and the penetration depth of the boundary layer decreases with increase in the suction parameter so. This is in keeping with the fact that suction causes reduction in boundary layer thickness.
Fig. 1. Variations off and g for different values of s0 at nt = 0 and n/Q = 0.5.
1423
Eccentrically rotating porous disk
0
1.0
Fig. 2. Variations off and g for different values of s0 at nt = 1112and n/Q = 0.5.
The nature of the flow near the disk is seen from the variations of f and g as observed from the Figs l-5 for the case in which a, = a2 = br = b2 = y, = 1 and x1 = 0. The variations of f and g for different values of suction parameter for nt = 0 and n/Q = 0.5 are shown in Fig. 1, whereas for nt = ~rd/2and nlQ = 0.5 or 2.0 in Figs 2 and 3. Each point of the disk is subjected to an oscillatory motion together with the axis about which it is rotating. Whereas the fluid at infinity is rotating about an axis passing through the fixed point (0,l). Figures l-3 clearly indicate how the unsteady flow near the disk is adjusting to the steady flow at infinity. We observe that the boundary layer thickness decreases with the increase in either suction parameter so or the parameter
Fig. 3. Variations off and g for different values of s0 at nt = n/2 and n/S2 = 2.0.
1424
S. R. KASIVISWANATHAN
and A. R. RAO
Fig. 4. Variations off and g for different values of se at nt = n/2 and n/S2 = 0.5.
Fig. 5. Variations off and g for different values of se at nt = n/2 and n/Q = 2.0.
n/8. Figures 4 and 5 depict the variations of f and g for different values of blowing parameter (sO< 0) at nt = 36/2 for n/Q = 0.5 and 2.0 respectively. It is seen that blowing causes thickening of the boundary layer and this boundary layer thickness decreases with increase in the rotation. Further, we observe that the distance from the disk in which the unsteady rotating flow adjusts with steady rotating flow about a different axis at infinity is smaller for the case when suction is present compared to that of injection. Acknowledgement-The
authors are thankful to the referee for his helpful comments and suggestions.
REFERENCES R. BERKBR, Integrations des du mouvement dun fluide visqueux incompressible. Handbuch der Physik, Vol. VIII/2, p. 87. Springer-Verlag, (1963). C. THORNLEY, On Stokes and Rayleigh layers in a rotating system. Quart. /. Mech. Appl. Math. 21, 451-461
(1968).
A. S. GUITA, Magnetohydrodynamic Ekman Layer. Acta Mech. 13, 155-160 (1972). M. E. ERDOGAN, Flow due to eccentrically rotating a porous disk and a fluid at infinity. ASME /. Appl. Mech. 43, 203-204 (1976). R. BERKBR, An exact solution of the Navier-Stokes equation-The vortex with curvilinear axis. In?. J. Engng Sci. 20,217-230 (1982).
Eccentrically rotating porous disk
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[6] K. R. RAJAGOPAL, A class of exact solutions to the Navier-Stokes equations. Int. J. Engng Sci. 22, 451-458 (1984). [7] A. RAMACHANDRA RAO and S. R. KASIVISWANATHAN, On exact solutions of unsteady Navier-Stokes equations-The vortex with instantaneous curvilinear axis. Int. J. Engng Sci. 25, 337-349 (1987). (Revised version received 24 March 1987)