hayat, khan, ayub some analytical solutions for second grade fluid flows for

8/10/2019 Hayat, Khan, Ayub Some Analytical Solutions for Second Grade Fluid Flows For

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Mathematical and Computer Modelling 43 (2006) 1629

www.elsevier.com/locate/mcm

Some analytical solutions for second grade fluid flows forcylindrical geometries

T. Hayat, M. Khan, M. Ayub

Department of Mathematics, Quaid-i-Azam University, 45320, Islamabad, Pakistan

Received 8 January 2003; received in revised form 11 April 2005; accepted 19 April 2005

Abstract

This paper deals with some unsteady flow problems of a second grade fluid. The flows are generated by the sudden application

of a constant pressure gradient or by the impulsive motion of a boundary. The velocities of the flows are described by the

partial differential equations. Exact analytic solutions of these differential equations are obtained. The well known solutions for a

NavierStokes fluid in the hydrodynamic case appear as the limiting cases of our solutions.

c 2005 Elsevier Ltd. All rights reserved.

Keywords:Second grade fluid; Exact solutions; Cylindrical coordinates; Transient flows; Hydrodynamic fluid

1. Introduction

The inadequacy of the classical NavierStokes theory for describing rheological complex fluids has led to the

development of several theories of non-Newtonian fluids. Rheological properties of materials are specified in general

by their so-called constitutive equations. The mechanical behavior of many real fluids, especially those of low

molecular weight, is well described by the NavierStokes theory. There are, however, many rheological complex

fluids such as polymer solutions, soaps, blood, paints, shampoo, ketchup, certain oils and greases that are not well

described by a Newtonian constitutive equation which do not show any relaxation and retardation phenomena.

Among the many models that have been used to describe the non-Newtonian behavior exhibited by these fluids,

the fluids of differential type [1]have received special attention. Here, we shall consider a model of second grade.

For a second grade fluid, the equations of motion are of a higher order than the NavierStokes equations and thus, in

general, one needs conditions in addition to the usual adherence boundary condition. For a detailed discussion of thisissue and for some interesting examples, we refer the reader to[24].

In recent years, interest in the unsteady flow has increased considerably. Rajagopal [5] has studied exact solutions

for a class of unsteady unidirectional flows of a second grade fluid under four different flow situations. Bandelli and

Rajagopal [6] examined a number of unidirectional transient flows of a second grade fluid in a domain with one

finite dimension. Bandelli [7] also studied the heat transfer analysis on some unsteady flows of a second grade fluid.

Corresponding author. Fax: +92 51 9219888.E-mail address:t [email protected] (T. Hayat).

0895-7177/$ - see front matter c 2005 Elsevier Ltd. All rights reserved.

doi:10.1016/j.mcm.2005.04.009
http://www.elsevier.com/locate/mcmhttp://www.elsevier.com/locate/mcm


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T. Hayat et al. / Mathematical and Computer Modelling 43 (2006) 1629 17

Bandelli et al. [8]also addressed the Rayleigh problem in a second grade fluid. Puri [9] has analyzed the impulsive

motion of a flat plate in a RivlinEricksen fluid. Unsteady flows of a second order fluid in a bounded region have

been discussed by Ting [10]. Hayat et al.[1115] discussed periodic and transient flows of a second grade fluid with

different geometrical configurations. Erdogan [16]addressed the problem of unsteady flow of a viscous fluid on an

oscillating plate. In cylindrical regions, the unsteady flows of a viscous fluid have been discussed by Batchelor [ 17],

Muller[18], Nanda [19] and Szmansky [20]. In[21], Erdogan discussed the viscous flows produced by the sudden

application of a constant pressure gradient or by the impulsive motion of a boundary. In another paper, Erdogan [ 22]

considered three types of unsteady flows, namely flow due to the impulsive motion of a flat plate, flow induced by a

constantly accelerating plate, and flow produced by a flat plate that applies a constant tangential stress to the fluid.

Moreover, Erdogan [23]examined five unsteady flows of a viscous fluid in a cylindrical regions.The present paper is concerned with some unsteady flows of a second grade fluid in cylindrical polar coordinates.

The extensive study of such flows is motivated by both their fundamental interest and their practical importance [6].

The arrangement of the paper is as follows. InSection 2,we determine the flow equations.Section 3contains the exact

solutions corresponding to the unsteady flow in a circular cylinder. Some solutions for starting flow in a circular pipe,

generalized starting flow in a circular pipe, and unsteady flow in a rotating cylinder are presented in Sections 46,

respectively. Finally, inSection 7,we give concluding remarks.

2. Constitutive equations

The constitutive equation for the fluids of second grade is in the following form [5,1114]:

T = pI + A1 + 1A2 + 2A21, (1)

in whichTis the Cauchy stress tensor,pIdenotes the indeterminate spherical stress,is the coefficient of viscosity,

1and 2are normal stress moduli, and A1and A2are the kinematic tensors defined through

A1 = (grad V)+ (grad V)T , (2)

A2 =dA1

dt+A1(grad V)+ (grad V)

T A1. (3)

In the above equations, V is the velocity, grad the gradient operator, and d/dtdenotes the material time derivative.Since the fluid is incompressible, it can undergo only isochoric motion and hence

div V = 0, (4)

and the equation of motion is

dV

dt= div T + B, (5)

where is the density of the fluid and B is the body force. In our analysis, we shall consider the model represented

by Eq.(1)as an exact model. For some comments regarding this problem, we refer the reader to [1]. If this model is

required to be compatible with thermodynamics, then the material moduli must meet the following restrictions [24]:

0, 1 0, 1 + 2 = 0. (6)

In the next sections, we discuss the four unsteady flows of a second grade fluid through two methods followed by

Erdogan [21,23]. The solutions for small and large times are obtained. The Laplace transform technique is particularly

well suited for small time solutions. However, it is not a trivial matter to invert the Laplace transform. Bandelli

et al. [8] have already shown that the Laplace transform treatment does not work for the Rayleigh problem (the

obtained solution does not satisfy the initial condition). They showed that this is due to an incompatibility between

the prescribed data. A comprehensive discussion on the issue has been given in great length by Bandelli [25].

3. Starting flow in a circular cylinder moving parallel to its length

Suppose that the second grade fluid is in a circular cylinder and is initially at rest, and the fluid starts suddenly due

to the motion of the cylinder parallel to its length. The axis of the cylinder is chosen as the z -axis. Using cylindrical


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18 T. Hayat et al. / Mathematical and Computer Modelling 43 (2006) 1629

polar coordinates, the governing partial differential equation is

w

t=

+

t

2w

r2 +

1

r

w

r

Nw, (7)

wherew (r, t)is the velocity along thez-axis, = /is the kinematic viscosity, = 1/is the material parameter,

andN= B20 /is the imposed magnetic field. The boundary and initial conditions are

w (a, t) = W fort >0,

w (r, 0) = 0 for 0 r


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Fig. 1. The variations ofV= w/ Wwith= r/a, M (=ma) = 0.5 and (=1/a2) = 0.005 for various values of= t/a2.

Fig. 2. The variations ofV= w/ W with= r/a, M= 0.5 and= 1 for various values of .

wheren are zeros of the above equation. The values of An can be obtained from the initial condition. Hence, the

velocity becomes

w

W=

I0(mr)

I0(ma ) 2

n=1

nJ0

nra

en t/a

2m2a2 + 2n

J1(n)

. (11)

The volume flux Qacross a plane normal to the flow is

Q = 2 a

0

wrdr.

Using Eq.(11)in above expression, we get

Q

2 (a/m) W=

I1(ma )

I0(ma ) 2ma

n=1

en t/a2

m2a2 + 2n. (12)

The values ofQfor various values of= t/a2 when (=1/a2) = 0.005 and M (=ma ) = 0.5 are

0.01 0.1 0.2 0.3 0.4 0.5

Q/2(a/m)W 0.09062 0.14796 0.19104 0.21395 0.22662 0.23367

The required time for Q to attain the asymptotic value is about = 0.5.


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The frictional force per unit area exerted by the fluid on the surface of the cylinder at r= ais

a =

w

r

r=a

+ 1

2w

rt

r=a

,

or

a

mW= I

1(ma )I0(ma )

+ 2ma

n=1

2nen t/a2

m2a2 + 2n[1 n] . (13)

The values of skin friction for various values of when = 0.005 and M= 0.5 are

0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6

a /mW 11.9423 4.1678 2.5652 1.4112 0.8846 0.5992 0.4409 0.3529

The required time to attain the asymptotic value of the skin friction is about = 0.6. It will be seen later that the

expressions given in Eqs.(11)(13)obtained for large values of time can also be used for small values of time.

Small time solutions

For small time, the Laplace transform method is used. If the Laplace transform ofw is w, then Eqs.(7) and (8)

take the following form:

w +1

rw q2w = 0, (14)

w (a, s) =W

s, (15)

where

q = N+ s

+ s1/2

and primes denote the differentiation with respect to r.The solution of Eq.(14)satisfying condition(15)is

w

W=

I0(qr)

s I0(qa). (16)

Laplace inverse of the above equation yields

w

W=

1

2 i

+ii

I0(qr) est

s I0(qa)ds. (17)

In Eq.(17),s = 0 is a simple pole. Therefore, the residue at s = 0 is

Res (0) =I0(mr)

I0(ma ).

The other singular points of Eq.(17)are the zeros of

I0(qa) = 0.

Settingq = i, we find that

J0(a) = 0. (18)

Ifn ,n = 1, 2, 3, . . . , are the zeros of Eq.(18),then

sn= N+ 2n

1+ 2n ,


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wheren = 1, 2, 3, . . . , are the poles. These are the simple poles and the residue at all these poles can be obtained

as

Res (sn) =

2

a

n( + sn) e

sn tJ0(nr)

sn

1+ 2n

J1(na)

.

Adding Res (0)and Res (sn

), a complete solution is obtained as

w

W=

I0(mr)

I0(ma )+

2

a

n=1

n( + sn ) esn t

sn

1+ 2n J0(nr)

J1(na). (19)

The volume flux Qacross a plane normal to the flow is given by

Q

2(a/m) W=

I1(ma )

I0(ma )+

2m

a

n=1

( + sn ) esn t

sn

1+ 2n . (20)

The values ofQfor various values of= t/a2 when = 0.005 and M= 0.5 are

0.01 0.1 0.2 0.3 0.4 0.5

Q/2(a/m)W 0.03044 0.14494 0.18951 0.21311 0.22615 0.23341

The required time for Q to attain the asymptotic value is about= 0.5. This table shows that the values of the flux

obtained for small values of time can be compared with those of large values of time.


a

mW=

I1(ma )

I0(ma )

2

ma

n=1

2n( + sn ) esn t

1+ 2n 1

sn+

1

. (21)

The values of skin friction for various values of when = 0.005 and M= 0.5 are

0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6

a /mW 15.3702 4.4467 2.6614 1.4472 0.9035 0.6096 0.4467 0.3561

The required time to attain the asymptotic value of the skin friction is about = 0.6. This table shows that the

values of the skin friction obtained for small values of time can be compared with those of large values of time.

4. Starting flow in a circular pipe

Suppose that the fluid is in a circular cylinder and is initially at rest, and the fluid starts suddenly due to a constant

pressure gradient. The governing equation and the boundary and initial conditions are

wt

= 1

dpdz+

+ t

2wr2

+ 1r

wr

Nw, (22)

w (a, t) = 0 for allt,

w (r, 0) = 0 for 0 r a. (23)

Employing the same procedure as inSection 3,the solutions are given by

w1/m2

dp/dz

= 1I0(mr)

I0(ma ) 2m2a2

n=1

J0

nra

en t/a

2

n

m2a2 + 2n

J1(n), (24)

Q

a21/m2

dp/dz

= 12

ma

I1(ma )

I0(ma )

4m2a2

n=1

en t/a2

2n

m2a2 + 2n (25)


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Fig. 3. The variations ofV= w/(1/m2) dp/dz with= r/a, M= 2 and = 0.05 for various values of.

and

a

(1/m) dp/dz =I1(ma )

I0(ma ) 2ma

n=1

en t/a2

m2a2 + 2n

[1 n] . (26)

The values ofQfor various values of when = 0.05 and M= 2 are

0.01 0.1 0.2 0.3 0.4 0.5 0.6

Q/ a2(1/m2) dp/dz 0.02328 0.16519 0.23911 0.27293 0.28857 0.29585 0.29924

The required time for Q to attain the asymptotic value is about = 0.6.The values of skin friction for various values of when = 0.05 and M= 2 are

0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6

a /(1/m) dp/dz 0.4188 0.4997 0.5671 0.6393 0.6711 0.6854 0.6920 0.6951

The required time to attain the asymptotic value of the skin friction is about = 0.6.


After taking the Laplace transform of Eqs.(22)and(23), we have

w +1

rw q2w =

1

s( + s)

dp

dz, (27)

w (a, s) = 0. (28)

For the solution of Eqs.(27)and(28),we employ the procedure ofSection 3.In order to avoid the details, the solutionsare of the following forms:

w1/m2

dp/dz

= 1I0(mr)

I0(ma )

2N

a

n=1

n( + sn) esn t

sn(N+ sn)

1+ 2n J0(nr)

J1(na), (29)

Q

a21/m2

dp/dz

= 1

2

ma

I1(ma )

I0(ma )

4N

a2

n=1

( + sn) esn t

sn(N+ sn)

1+ 2n , (30)

and

a

(1/m) dp/dz=

I1(ma )

I0(ma )

2N

ma

n=1

2n( + sn ) esn t

(N+ sn)

1+ 2n 1

sn+

1

. (31)


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Fig. 4. The variations ofV= w/(1/m2) dp/dzwith= r/a, M= 2 and= 0.5 for various values of .

The values ofQfor various values of when = 0.05 and M= 2 are

0.01 0.1 0.2 0.3 0.4 0.5 0.6

Q/ a2(1/m2) dp/dz 0.11983 0.21334 0.26148 0.28336 0.29344 0.29813 0.30031

The required time for Q to attain the asymptotic value is about= 0.6. This table shows that the values of the flux

obtained for small values of time can be compared with those of large values of time.

The values of skin friction for various values of when = 0.05 and M= 2 are

0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6

a /(1/m) dp/dz 0.6039 0.6318 0.6547 0.6787 0.6891 0.6938 0.6959 0.6969

The required time to attain the asymptotic value of the skin friction is about = 0.6. This table shows that thevalues of the flux obtained for small values of time can be compared with those of large values of time.

5. Generalized starting flow in a circular pipe

Suppose that the fluid is in a circular cylinder and is initially at rest. The fluid starts suddenly due to a constant

pressure gradient and the motion of the cylinder parallel to its length. The governing partial differential equation is

(22)and the boundary and initial conditions are

w (a, t) = W fort >0,

w (r, 0) = 0 for 0 r


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and

a

m= [W ]

I1(ma )

I0(ma )+

2

ma

n=1

en t/a2

m2a2 + 2n

Wn + m

2a2

[1 n] (35)

in which

=dp/dz

m2 .


The problem in the transformed s -plane becomes

w +1

rw q2w =

1

s( + s)

dp

dz, (36)

w (a, s) =W

s

. (37)

The solution of Eq.(36)satisfying the boundary condition(37)is of the following form:

w = WI0(qr)

s I0(qa)+

N

s(N+ s)

1

I0(qr)

I0(qa)

. (38)

Laplace inversion of the above equation yields

w = WI0(mr)

I0(ma )+

1

I0(mr)

I0(ma )

+2

a

n=1

n( + sn) esn t

sn1+ 2

n W

N

(N+ sn ) J0(nr)

J1(na). (39)

The volume flux Qacross a plane normal to the flow is

Q

2 a/m= W

I1(ma )

I0(ma )+

ma

2

1

2

ma

I1(ma )

I0(ma )

+2m

a

n=1

( + sn ) esn t

sn

1+ 2n W N

(N+ sn)

. (40)


a

m

= [W ] I1(ma )

I0(ma )

2

ma

n=1

2n( + sn) esn t

1+ 2n

W

N

(N+ sn )

1

sn+

1

. (41)

6. Starting flow in a rotating cylinder

Suppose that the fluid is in a circular cylinder and is initially at rest and the fluid sets in motion suddenly due to

rotation of the cylinder. The governing partial differential equation is

v

t

= + t

2v

r2+

1

r

v

r

v

r2 Nv, (42)


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wherev(r, t)is the rotating velocity. The boundary and initial conditions are

v(a, t) = a fort>0,

v(r, 0) = 0 for 0 r


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Fig. 5. The variations ofV= v/awith= r/a, M= 0.5 and = 0.005 for various values of.


a =

r r

vr

r=a

+ 1

r t

r

vr

r=a

,

or

a

ma=

I2(ma )

I1(ma )+

2

ma

n=1

2

n en t/a

2m2a2 +

2

n

1+ n . (47)The values of skin friction for various values of when = 0.005 and M= 2 are

0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6

a /ma 16.8647 0.9880 0.5999 0.4596 0.4377 0.4339 0.4333 0.4332

The required time to attain the asymptotic value of the skin friction is about = 0.5.


After taking the Laplace transform, the governing problem becomes

v +1

rv

q2 +

1

r2

v = 0, (48)

v(a, s) =a

s. (49)

The solution of Eqs.(48)and(49)can be written as

v

a=

I1(qr)

s I1(qa). (50)

Laplace inversion of Eq.(50)yields

v

a=

1

2 i

+ii

I1(qr) est

s I1(qa)ds. (51)

In Eq.(51),s = 0 is a simple pole. Therefore, the residue at s = 0 is

Res (0) =

I1(mr)

I1(ma ) .


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Fig. 6. The variations ofV= v/awith= r/a, M= 0.5 and= 0.5 for various values of .

The other singular points of Eq.(51)are the zeros of

I1(qa) = 0.

Settingq = i, we find that

J0

a= 0. (52)

If,n = 1, 2, 3, . . . , are the zeros of Eq.(52),then

sn =

N+

2

n

1+ 2

n

,

wheren = 1, 2, 3, . . . , are the poles. These are simple poles and the residue at all these poles can be obtained as

Res

sn=4

a

n + s

n esn tJ

1

nr

sn

1+

2

n

J0

n a J2

n a

.Adding Res (0)and Res

sn

, a complete solution is obtained as

v

a=

I1(mr)

I1(ma )

4

a

n=1

n

+ sn

esn tJ1

nr

sn

1+

2

n

J0

n a

J2

n a

. (53)The frictional force per unit area exerted by the fluid on the surface of the cylinder at r= ais

a

ma

=I2(ma )

I1(ma )

2

ma

n=1

2

n + sn e

sn t1+ 2n

1sn+

1

. (54)

The values of skin friction for various values of when = 0.005 and M= 2 are

0.01 0.05 0.1 0.2 0.3 0.4 0.5 0.6

a /ma 1.7911 0.7896 0.5564 0.4534 0.4367 0.4338 0.4333 0.43315

The required time to attain the asymptotic value of the skin friction is about = 0.5. This table shows that the

values of the flux obtained for small values of time can be compared with those of large values of time.

It is worthwhile to see how convergent the series solutions are in the present analysis. We use numerical

investigation of the rate of convergence of the series solutions presented. Let us first consider the series solutions

in Eqs.(11)and(19).It may be noted that both series solutions are the same in non-dimensional form. In the present


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analysis, we have taken 50 terms of the series in Section 3and found that the relative error compared to the value

obtained by taking five terms is about 0.01%. The value of the series is 0.15601 from the 5th term to the 50th term.

Similarly, it is noted that five terms of the series are sufficient for the problems discussed in Sections 4and 6.For

the second problem (inSection 4), the value of the series is 0.02335, which is valid from the 5th to the 50th terms,

whereas this value is 0.30704 for the problem considered inSection 6.Note that, in the numerical investigations of

the series ofSections 3and4,the values of = 0.005, M= 0.5,= 0.1 and= 0 have been taken into account,

while forSection 6these values are = 0.005, M= 0.5,= 0.1 and= 0.5.

7. Concluding remarks

In this paper, four types of unsteady magnetohydrodynamic flows of second grade fluids are presented, namely

unsteady flow in a circular cylinder moving parallel to its length, unsteady flow in a circular pipe, generalized

unsteady flow in a circular pipe, and unsteady flow in a rotating circular cylinder. The expressions for velocity profile,

volume flux, and frictional force in each case are constructed. These solutions have been established using the Laplace

transform method. It is found that the steady solution for a second grade fluid is identical to that of a NavierStokes

fluid. From the governing equations, we deduce that the effect of the viscoelastic parameter on the flow profile appears

in the unsteady state only. From the (Figs. 16), it can be seen that slight oscillations occur in the flow for very small

time. This indicates that, in the beginning, the flow is slightly unstable and, after some time, fluid viscosity takeover and stabilizes it. It is further observed that magnetic effects arise in both steady and unsteady solutions. When1 = 0 the solutions reduce to a magnetohydrodynamic NavierStokes fluid, and when 1 = 0 = N these reduce

to a classical hydrodynamic NavierStokes fluid. It is further interesting to note that, as the viscoelastic parameter

increases, the value of velocity (in each case) at the same distance r decreases. That is, increasing the viscoelastic

coefficient has the effect of increasing the boundary layer thickness. The results of Erdogan [23] for a NavierStokes

fluid can be recovered easily when N, 1 0.

Acknowledgements

The authors wish to express their gratitude to the anonymous reviewers for their valuable suggestions.

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hayat, khan, ayub some analytical solutions for second grade fluid flows for

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