- Email: [email protected]
Stokes paradox for power-law flow around a cylinder
Jo~rnai of ~o~-~e~~on~n
Fluid ~ec~~~c~,
50 ( 1993) 217-224
217
Elsevier Science Publishers B.V., Amsterdam
Stokes paradox for power-law flow around a cylinder R.I. Tanner Department of ~ec~~~ic~I Engineering,
(Received February
University of Sydney, NS. W. 2006 ~Au~~r~~ia~
15, 1993; in revised form June 8, 1993)
Abstract A simple analysis for power-law fluids shows that the Stokes paradox for creeping flow around a cylinder is removed for shear-thinning (n < 1) but not for shear-thickening (n > 1) fluids. An approximate drag value is found for n =C1 and is compared with computed results. Keywords: creeping flow; cylinder; power-law Aow; Stokes paradox
1. Introduction The creeping flow of an incompressible Newtonian fluid around a cylinder in an unbounded fluid has no solution [l] due to the form of the stream function at large distances from the cylinder. This is the Stokes paradox. Intuitively, one might expect that this behaviour would disappear for shear-thinning fluids of the power-law type; for more realistic models where the viscosity becomes constant (Newtonian) as the shear rate approaches zero one expects that the paradox will remain. Here we consider the flow of incompressible power-law fluids. Let the stream function tl/(~,8) be defined in a plane flow so that
(1) where U, v are the r and 13components of the velocity field in plane polar coordinates. Let the material behaviour be described by c = -pI +
2kj”-
‘d -
-PI+
z,
(2)
where c is the total stress, p is the pressure, I is the unit tensor and k is a consistency parameter. The non-zero components of the rate of defo~ation tensor d are 0377-0257/93/$06.00 0 1993 - Elsevier Science Publishers B.V. All rights reserved
218
R.I. Tanner / J. Non-Newtonian Fluid Mech. 50 (1993) 217-224
(3) (4)
and the shear rate f is defined as f2 = 2dodij = 4(d$ + d$).
(5)
From the equations of motion, in the absence of inertia and body forces [2], one can eliminate the pressure to find an equation for the extra stress components z,,, roe and zr6:
(6) Using (2) -(4) in (6) one finds an equation for tfi:
(7) which needs to be solved with zero velocity conditions on the cylinder r = 1 and an outer condition, as r + 03, that the velocity becomes a uniform stream of unit speed in the x direction so that $-+rsin6
as r+co
(8)
and u -+cos 8, v 3 -sin 8 as r + co. When n = 1 (Devonian case) (7) reduces to
(9) From the form (8) we set
(10)
$ =f(r) sin 0, where f satisfies j-iv
I
2:
3f” 2f’ -7+7-7=0.
3f
(11)
The general solution of (I 1) is then [l] f=Ar3+BrInr+Cr+f.
(12)
R.I. Tanner / J. Non-Ne~toni~~
Fluid Mech. 50 (1993) 217-224
219
The velocity component u derived from ( 12) behaves like Ar2 as Y--) 00, whereas we would like the velocity to approach a constant value. Following van Dyke [l], we set the coefficient of the most singular term to zero; that is, we set A = 0. The velocities at Y = 1 (on the cylinder) must be zero and hence for f’( 1) =f( 1) = 0, 14=-2C
rlnr-i+i
sine.
(13)
1
I
The constant C should be used to satisfy the condition $ -+ r sin 8 as r -+ co, but no choice of C can satisfy this condition, and no solution exists to the problem, which is the Stokes paradox. For large r we note that the rates of deformation auf& and &?u/& + l/r &./&3 - v/r) are, from (13) -do0 =;
=d,,=
-2c;se(l
_r-“>
and d _2Csin 0 rBY3 ’
(15)
so that the shear stress z,~ is negligible (for most 8 values) in comparison with r,, at large r. If we make this approximation in eqn. (6) and note that res = -r,,, we find from (6):
(16) The general solution of ( 16) vanishing as r + co is, for Newtonian power-law fluids,
and
(17) where g(0) is an even function of 19and G(r) vanishes as Y-+ co. Since the velocity u is connected by the constitutive equation (2) to z,, we also have 2ky” - ’ au/i% = z,, , and since we can compute, for large Y,i, z 2l&/&l, then
(18) Equation ( 18) can be integrated to find u. In the Newtonian case one finds ~=g~nr/2k+~(~)+
(19)
220
R.I. Tmmr / J. Non-Newtonian Fluid Me&. 50 (1993) 217-224
Now there is no source at the origin, and so c” UTd8 = 0 (for any r 5; 1). Applying this to (29) we see that g(S) must be of the form g(S) = -gfn - 3) and that G(r) must be zero. ff we take g/2k = - 2C cos 0, h(8) = C cos B and set G = 0 then one has a stream function of the form ( 13) given above for large Y.This shows that the Stokes paradox appears in this simplified analysis. For the power-law case, setting G(r) = 0 again, one has
If 12< 1 (shear-thinning) then u -+h(B) ( =cos 0) as r --) cc and there is no Stokes paradox; if n > 1 then the first term on the right-hand side of (20) dominates as I + 03 and the Stokes paradox continues to frustrate attempts to satisfy the far-field veto&y conditions, 2. Alternative approach One can go back to (5) and (7) and repeat the analysis by assuming a simple furm for 1(/(with a uniform speed of U at larger r and a cylinder of radius R) $ = U sin 8(Ar” + ’ + r),
(21)
where A is an amplitude factor and m is an unknown constant. Computing *+’ we find Y
F
= (UAm)“-
The trigonometric be expanded as
ly.fm-m- ‘“(4cos2fl + m2 sin2@)b- 11/2_
(22)
part of this expression varies from 4 to mn2in size and can
At least for n close to 1, we may ignore the term cos 28 (and higher terms) in (231, and approximate 9” - ’ as *Jz-1 Y
M or’“-
IXn-
0,
where K = (Udm)“- ‘[(4 + nt3/2’J(“-“‘2, a constant. forward to show that m must satisfy 0 = m(a + 1)[4 - m(rx+ l)],
(24)
It is then straight(25)
R.I. fanner / J. ~on”~e~~~~ia~ Fluid M&t.
50 (1993) 217-224
221
where
The roots of (25) are (i) m =0 (ii) a = - I or m = 1 --p6-‘,
(261
(iii) na = ((PI - 1) + [(n - 1)2 + 16~11’23/2~.
(27)
The root PN= 0 is the uniform stream and if n = 1 (Newtonian) (ii) also gives m = 0, which leads to the r In Y form for $. These roots otherwise conform to those found in eqn. (20); the roots (27) give the values which must be excluded (positive sign) and a rapidly dying function (negative sign). The results confirm the simple analysis above, but, due to the nonlinear viscosity function the sum of the individual solutions is not a solution. We may, however, assume an approximate form (which does not necessarily satisfy the partial di~erential equation (7)) rl, = Ur sin 8 1 + [ *,“‘,,
(;>“I
-,,“‘,,
(~)m,]_
(28)
which satisfies all the velocity boundary conditions. When m2-+ -GO we return to eqn. (21) with u(R) = 0; v(R) is not zero in this case. To compute an approximate value for the drag f) we need to evaluate, for very large r, D=r
s0
==KG. -P)
cos B - zre
sin 01 do.
(29)
To compute G, ( EZ,, -p) we can use the radial equilibrium equation for large Y, finding {noting ml B mz) 4c cos 8
f4m - 1)
0 (-- > r &ml-
z
where C = ~[U~,~2~R(~,
‘) m, 4
1 ,
(301
- ??22)]“[(4i- m~)/2]?-‘“*
Since m, = 1 -a-’ (311 and
222
RI.
Tanner / J. No~-Ne~~on~~
Thus we find an approximate
FoundMe&. 50 (1993) 217-224
value for the drag of 4nRC, or (33)
where ml = 1 --n-l and m2 is taken as some negative number. This analysis has assumed that the form (28) is useful; the approximate solution form (21) was assumed to begin with. While (28) satisfies the complete set of velocity boundary conditions, but not the differential equation, (21) satisfies the differential equations (approximately) but cannot satisfy all the inner velocity boundary conditions; if we let m2 -+- co in (28) we revert to (21). For very Iarge m2 the inner cylinder does not appear to satisfy the no-slip condition, hence the values of m2 from eqn. (27) were used. The resulting drag values are shown in Fig. 1 and Table 1, together with some computed values obtained using a boundary element method, There are difficulties with the computations at n M 1.0 (need for a very large field) and at small n (lack of convergence). The finest mesh used was as shown in Fig. 2. On the left- and right-hand boundaries the conditions B = i were imposed; the centreline and no-slip conditions were imposed on the bottom surface. On the top boundary (y = 8OR) zero x and y tractions were imposed. Examination of the velocity field shows that for y1= 0.9 it is not sticient to go out to 80R in the y direction; the full free-stream velocity has not yet
15.0,,,,,,,,,,,,,,,,"',,"","","',,"","',,"", L
12.5 -
10.0
Q
f F
7.5 -
3
5.0 -
2.5 -
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Fig. 1. Dimensionless cylinder drag as a function of the power-law index n. Dots are the boundary element solutions on a 16OR x 80R field; full line is the approximate theory (33).
223
R.I. Tanner / J. Non-Newtonian Fluid Mech. 50 (1993) 217-224
TABLE 1 Eigenvalues and drag values n
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
-ml
--m2
D* = D/kU”R’-”
0.111 0.250 0.429 0.667 1.OOo 1.500 2.333 4.000 9.000
2.164 2.365 2.614 2.937 3.372 4.000 5.OQO 6.899 12.262
1.761 4.223 7.052 9.802 11.916 12.672 11.364 7.851 3.306
(33)
Computed (b.e.m.) D*
3.552 4.876 6.696 9.094 12.154 15.481 -
Fig. 2. Mesh used in boundary element computations. Left and right boundaries force u, = 1, uv = 0; bottom boundary is a centreline (no shear, uY = 0) except for the no-slip condition on the sphere; the top surface is a no-traction surface.
been reached. On the other hand it seems likely field is sufficient since the boundary velocity error case. It is difficult to satisfy the double demands cylinder with a very large field here, and it would
that for p1= 0.4 the 80R is of order 10T3U in this of a fine mesh near the probably’be better to use
R.I. Tanner 1 J. Non-Newtonian Fluid Mech. 50 (1993) 217-224
224
a computational method based on mapping. Therefore, although the computed and analytical drag values are comparable, it is expected that the computed values are less accurate near y1= 1 than the analytical values. For low y1( ~0.5) the approximations made in the analysis become less true, and the computations are expected to be more reliable. For small ~1,as in the sphere case [3], the true drag and the approximate theory may be quite divergent. The assumption that ldrO) 4 Id,, 1 for Y--, co made in ( 16) appears to be justified from looking at the b.e.m. results, at least for n = 0.5. When a small amount of inertia is present, it is possible to conclude from dimensional analysis that the drag D (per unit length) on a cylinder of radius R in a stream of speed U is of the form, for y1< 1, D kR’ -“U”
= a + bRe’,
(34)
where a and b are constants and the modified Reynolds number Re’ is given by pU2-nR”/k. The results for the Newtonian case (n = 1) are more complex. Lamb’s result is [l] D =4rc[ln(4/Re) --y +;I-‘, VU where Re = pUR/q and y is Euler’s constant (0.5722). The results for n > 1 seem not to be known, and a correlation of the form (34) would not be expected in view of the continued paradox. In summary, although the above results may be of some theoretical and philosophical interest, it must be borne in mind that most real fluids behave like Newtonian fluids for very small stresses and hence the Lamb solution will generally be needed to resolve the paradox if the fluid is unbounded. Similar paradoxes also apply to the unbounded flow around a sphere when seeking secondary inertial corrections to drag [l] and one expects similar results to apply in this case also. When one deals with bounded fluid problems, then the arguments presented no longer apply, and solutions of plane creeping power-law flows exist for all n values. Acknowledgements
This study was supported by the Australian Research Council. I am grateful to Dr. Rong Zheng for making the boundary element computations. References 1 M. van Dyke, Perturbation Methods in Fluid Mechanics, Academic Press, New York, 1964. 2 R.I. Tanner, Engineering Rheology, Revised Edition, Oxford University Press, 1988. 3 G. Dazhi and R.I. Tanner, J. Non-Newtonian Fluid Mech., 17 (1985) l-12.
Fluid ~ec~~~c~,
50 ( 1993) 217-224
217
Elsevier Science Publishers B.V., Amsterdam
Stokes paradox for power-law flow around a cylinder R.I. Tanner Department of ~ec~~~ic~I Engineering,
(Received February
University of Sydney, NS. W. 2006 ~Au~~r~~ia~
15, 1993; in revised form June 8, 1993)
Abstract A simple analysis for power-law fluids shows that the Stokes paradox for creeping flow around a cylinder is removed for shear-thinning (n < 1) but not for shear-thickening (n > 1) fluids. An approximate drag value is found for n =C1 and is compared with computed results. Keywords: creeping flow; cylinder; power-law Aow; Stokes paradox
1. Introduction The creeping flow of an incompressible Newtonian fluid around a cylinder in an unbounded fluid has no solution [l] due to the form of the stream function at large distances from the cylinder. This is the Stokes paradox. Intuitively, one might expect that this behaviour would disappear for shear-thinning fluids of the power-law type; for more realistic models where the viscosity becomes constant (Newtonian) as the shear rate approaches zero one expects that the paradox will remain. Here we consider the flow of incompressible power-law fluids. Let the stream function tl/(~,8) be defined in a plane flow so that
(1) where U, v are the r and 13components of the velocity field in plane polar coordinates. Let the material behaviour be described by c = -pI +
2kj”-
‘d -
-PI+
z,
(2)
where c is the total stress, p is the pressure, I is the unit tensor and k is a consistency parameter. The non-zero components of the rate of defo~ation tensor d are 0377-0257/93/$06.00 0 1993 - Elsevier Science Publishers B.V. All rights reserved
218
R.I. Tanner / J. Non-Newtonian Fluid Mech. 50 (1993) 217-224
(3) (4)
and the shear rate f is defined as f2 = 2dodij = 4(d$ + d$).
(5)
From the equations of motion, in the absence of inertia and body forces [2], one can eliminate the pressure to find an equation for the extra stress components z,,, roe and zr6:
(6) Using (2) -(4) in (6) one finds an equation for tfi:
(7) which needs to be solved with zero velocity conditions on the cylinder r = 1 and an outer condition, as r + 03, that the velocity becomes a uniform stream of unit speed in the x direction so that $-+rsin6
as r+co
(8)
and u -+cos 8, v 3 -sin 8 as r + co. When n = 1 (Devonian case) (7) reduces to
(9) From the form (8) we set
(10)
$ =f(r) sin 0, where f satisfies j-iv
I
2:
3f” 2f’ -7+7-7=0.
3f
(11)
The general solution of (I 1) is then [l] f=Ar3+BrInr+Cr+f.
(12)
R.I. Tanner / J. Non-Ne~toni~~
Fluid Mech. 50 (1993) 217-224
219
The velocity component u derived from ( 12) behaves like Ar2 as Y--) 00, whereas we would like the velocity to approach a constant value. Following van Dyke [l], we set the coefficient of the most singular term to zero; that is, we set A = 0. The velocities at Y = 1 (on the cylinder) must be zero and hence for f’( 1) =f( 1) = 0, 14=-2C
rlnr-i+i
sine.
(13)
1
I
The constant C should be used to satisfy the condition $ -+ r sin 8 as r -+ co, but no choice of C can satisfy this condition, and no solution exists to the problem, which is the Stokes paradox. For large r we note that the rates of deformation auf& and &?u/& + l/r &./&3 - v/r) are, from (13) -do0 =;
=d,,=
-2c;se(l
_r-“>
and d _2Csin 0 rBY3 ’
(15)
so that the shear stress z,~ is negligible (for most 8 values) in comparison with r,, at large r. If we make this approximation in eqn. (6) and note that res = -r,,, we find from (6):
(16) The general solution of ( 16) vanishing as r + co is, for Newtonian power-law fluids,
and
(17) where g(0) is an even function of 19and G(r) vanishes as Y-+ co. Since the velocity u is connected by the constitutive equation (2) to z,, we also have 2ky” - ’ au/i% = z,, , and since we can compute, for large Y,i, z 2l&/&l, then
(18) Equation ( 18) can be integrated to find u. In the Newtonian case one finds ~=g~nr/2k+~(~)+
(19)
220
R.I. Tmmr / J. Non-Newtonian Fluid Me&. 50 (1993) 217-224
Now there is no source at the origin, and so c” UTd8 = 0 (for any r 5; 1). Applying this to (29) we see that g(S) must be of the form g(S) = -gfn - 3) and that G(r) must be zero. ff we take g/2k = - 2C cos 0, h(8) = C cos B and set G = 0 then one has a stream function of the form ( 13) given above for large Y.This shows that the Stokes paradox appears in this simplified analysis. For the power-law case, setting G(r) = 0 again, one has
If 12< 1 (shear-thinning) then u -+h(B) ( =cos 0) as r --) cc and there is no Stokes paradox; if n > 1 then the first term on the right-hand side of (20) dominates as I + 03 and the Stokes paradox continues to frustrate attempts to satisfy the far-field veto&y conditions, 2. Alternative approach One can go back to (5) and (7) and repeat the analysis by assuming a simple furm for 1(/(with a uniform speed of U at larger r and a cylinder of radius R) $ = U sin 8(Ar” + ’ + r),
(21)
where A is an amplitude factor and m is an unknown constant. Computing *+’ we find Y
F
= (UAm)“-
The trigonometric be expanded as
ly.fm-m- ‘“(4cos2fl + m2 sin2@)b- 11/2_
(22)
part of this expression varies from 4 to mn2in size and can
At least for n close to 1, we may ignore the term cos 28 (and higher terms) in (231, and approximate 9” - ’ as *Jz-1 Y
M or’“-
IXn-
0,
where K = (Udm)“- ‘[(4 + nt3/2’J(“-“‘2, a constant. forward to show that m must satisfy 0 = m(a + 1)[4 - m(rx+ l)],
(24)
It is then straight(25)
R.I. fanner / J. ~on”~e~~~~ia~ Fluid M&t.
50 (1993) 217-224
221
where
The roots of (25) are (i) m =0 (ii) a = - I or m = 1 --p6-‘,
(261
(iii) na = ((PI - 1) + [(n - 1)2 + 16~11’23/2~.
(27)
The root PN= 0 is the uniform stream and if n = 1 (Newtonian) (ii) also gives m = 0, which leads to the r In Y form for $. These roots otherwise conform to those found in eqn. (20); the roots (27) give the values which must be excluded (positive sign) and a rapidly dying function (negative sign). The results confirm the simple analysis above, but, due to the nonlinear viscosity function the sum of the individual solutions is not a solution. We may, however, assume an approximate form (which does not necessarily satisfy the partial di~erential equation (7)) rl, = Ur sin 8 1 + [ *,“‘,,
(;>“I
-,,“‘,,
(~)m,]_
(28)
which satisfies all the velocity boundary conditions. When m2-+ -GO we return to eqn. (21) with u(R) = 0; v(R) is not zero in this case. To compute an approximate value for the drag f) we need to evaluate, for very large r, D=r
s0
==KG. -P)
cos B - zre
sin 01 do.
(29)
To compute G, ( EZ,, -p) we can use the radial equilibrium equation for large Y, finding {noting ml B mz) 4c cos 8
f4m - 1)
0 (-- > r &ml-
z
where C = ~[U~,~2~R(~,
‘) m, 4
1 ,
(301
- ??22)]“[(4i- m~)/2]?-‘“*
Since m, = 1 -a-’ (311 and
222
RI.
Tanner / J. No~-Ne~~on~~
Thus we find an approximate
FoundMe&. 50 (1993) 217-224
value for the drag of 4nRC, or (33)
where ml = 1 --n-l and m2 is taken as some negative number. This analysis has assumed that the form (28) is useful; the approximate solution form (21) was assumed to begin with. While (28) satisfies the complete set of velocity boundary conditions, but not the differential equation, (21) satisfies the differential equations (approximately) but cannot satisfy all the inner velocity boundary conditions; if we let m2 -+- co in (28) we revert to (21). For very Iarge m2 the inner cylinder does not appear to satisfy the no-slip condition, hence the values of m2 from eqn. (27) were used. The resulting drag values are shown in Fig. 1 and Table 1, together with some computed values obtained using a boundary element method, There are difficulties with the computations at n M 1.0 (need for a very large field) and at small n (lack of convergence). The finest mesh used was as shown in Fig. 2. On the left- and right-hand boundaries the conditions B = i were imposed; the centreline and no-slip conditions were imposed on the bottom surface. On the top boundary (y = 8OR) zero x and y tractions were imposed. Examination of the velocity field shows that for y1= 0.9 it is not sticient to go out to 80R in the y direction; the full free-stream velocity has not yet
15.0,,,,,,,,,,,,,,,,"',,"","","',,"","',,"", L
12.5 -
10.0
Q
f F
7.5 -
3
5.0 -
2.5 -
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Fig. 1. Dimensionless cylinder drag as a function of the power-law index n. Dots are the boundary element solutions on a 16OR x 80R field; full line is the approximate theory (33).
223
R.I. Tanner / J. Non-Newtonian Fluid Mech. 50 (1993) 217-224
TABLE 1 Eigenvalues and drag values n
0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1
-ml
--m2
D* = D/kU”R’-”
0.111 0.250 0.429 0.667 1.OOo 1.500 2.333 4.000 9.000
2.164 2.365 2.614 2.937 3.372 4.000 5.OQO 6.899 12.262
1.761 4.223 7.052 9.802 11.916 12.672 11.364 7.851 3.306
(33)
Computed (b.e.m.) D*
3.552 4.876 6.696 9.094 12.154 15.481 -
Fig. 2. Mesh used in boundary element computations. Left and right boundaries force u, = 1, uv = 0; bottom boundary is a centreline (no shear, uY = 0) except for the no-slip condition on the sphere; the top surface is a no-traction surface.
been reached. On the other hand it seems likely field is sufficient since the boundary velocity error case. It is difficult to satisfy the double demands cylinder with a very large field here, and it would
that for p1= 0.4 the 80R is of order 10T3U in this of a fine mesh near the probably’be better to use
R.I. Tanner 1 J. Non-Newtonian Fluid Mech. 50 (1993) 217-224
224
a computational method based on mapping. Therefore, although the computed and analytical drag values are comparable, it is expected that the computed values are less accurate near y1= 1 than the analytical values. For low y1( ~0.5) the approximations made in the analysis become less true, and the computations are expected to be more reliable. For small ~1,as in the sphere case [3], the true drag and the approximate theory may be quite divergent. The assumption that ldrO) 4 Id,, 1 for Y--, co made in ( 16) appears to be justified from looking at the b.e.m. results, at least for n = 0.5. When a small amount of inertia is present, it is possible to conclude from dimensional analysis that the drag D (per unit length) on a cylinder of radius R in a stream of speed U is of the form, for y1< 1, D kR’ -“U”
= a + bRe’,
(34)
where a and b are constants and the modified Reynolds number Re’ is given by pU2-nR”/k. The results for the Newtonian case (n = 1) are more complex. Lamb’s result is [l] D =4rc[ln(4/Re) --y +;I-‘, VU where Re = pUR/q and y is Euler’s constant (0.5722). The results for n > 1 seem not to be known, and a correlation of the form (34) would not be expected in view of the continued paradox. In summary, although the above results may be of some theoretical and philosophical interest, it must be borne in mind that most real fluids behave like Newtonian fluids for very small stresses and hence the Lamb solution will generally be needed to resolve the paradox if the fluid is unbounded. Similar paradoxes also apply to the unbounded flow around a sphere when seeking secondary inertial corrections to drag [l] and one expects similar results to apply in this case also. When one deals with bounded fluid problems, then the arguments presented no longer apply, and solutions of plane creeping power-law flows exist for all n values. Acknowledgements
This study was supported by the Australian Research Council. I am grateful to Dr. Rong Zheng for making the boundary element computations. References 1 M. van Dyke, Perturbation Methods in Fluid Mechanics, Academic Press, New York, 1964. 2 R.I. Tanner, Engineering Rheology, Revised Edition, Oxford University Press, 1988. 3 G. Dazhi and R.I. Tanner, J. Non-Newtonian Fluid Mech., 17 (1985) l-12.