A note on sinusoidal motion of a viscoelastic non-Newtonian fluid

Arch Appl Mech (2012) 82:659–667DOI 10.1007/s00419-011-0581-5

ORIGINAL

A. Mahmood · N. A. Khan · I. Siddique · S. Nazir

A note on sinusoidal motion of a viscoelasticnon-Newtonian fluid

Received: 27 May 2011 / Accepted: 11 August 2011 / Published online: 6 September 2011© Springer-Verlag 2011

Abstract In this note, the exact solutions of velocity field and associated shear stress corresponding to theflow of second-grade fluid in a cylindrical pipe, subject to a sinusoidal shear stress, are determined by means ofLaplace and finite Hankel transform. These solutions are written as sum of steady-state and transient solutions,and they satisfy governing equations and all imposed initial and boundary conditions. The correspondingsolutions for the Newtonian fluid, performing the same motion, can be obtained from our general solutions. Atthe end of this note, the effects of different parameters are presented and discussed by showing flow profilesgraphically.

Keywords Second-grade fluid · Shear stress · Steady-state and transient solutions

1 Introduction

The motion of a fluid in a rotating or translating in a cylinder is of interest to both theoretical and practicaldomains. It is of very important significance to study the mechanism of viscoelastic fluids flow in many indus-try fields, such as oil exploitation, chemical and food industry, and bio-engineering. For Newtonian fluids,the transient velocity distribution for the flow within a circular cylinder may be found in [1]. The first exactsolutions for flows of non-Newtonian fluids in such a domain seem to be those of Ting [2] correspondingto second-grade fluids and Srivastava [3] for Maxwell fluids. Later, Waters and King [4] studied the start-upPoiseuille flow of an Oldroyd-B fluid in a straight circular tube and its decay from the steady-state conditionwhen the pressure gradient is removed. During recent years, quite many papers of this type have been pub-lished. The most general solutions corresponding to the helical flow of a second-grade fluid seem to be those ofFetecau and Fetecau [5], in which the cylinder is rotating around its axis and sliding along the same axis withtime-dependent velocities. Others interesting solutions for different flows of the same fluids have been alsoobtained by Hayat et al. [6]. Exact solutions for the helical flows of Oldroyd-B fluids in cylindrical domainshave been obtained by Wood [7] and Fetecau et al. [8]. In the meantime, a lot of papers regarding such motionshave been published. The interested readers can see for instance the papers [9–13] and their related references.

However, it is important to point out that all above-mentioned works dealt with problems in which thevelocity is given on the boundary. To the best of our knowledge, the first exact solutions for flows into cylindri-cal domains when the shear stress is given on the boundary are those obtained by Bandelli and Rajagopal [14]for second-grade fluids. Similar solutions, corresponding to a time-dependent shear stress on the boundary,have been recently obtained in [15–19].

A. Mahmood (B) · I. Siddique · S. NazirDepartment of Mathematics, COMSATS Institute of Information Technology, Lahore 54000, PakistanE-mail: [email protected]

N. A. KhanDepartment of Mathematics, University of Karachi, Karachi 75270, Pakistan

660 A. Mahmood et al.

In this note, exact solutions corresponding to the flow of an second-grade fluid in an infinite circular cylinderare studied. In order to produce the flow of fluid, the boundary of cylinder is subject to an axial sinusoidal shearstress. Exact solutions of such kind of flows, in addition to serving as approximations to some specific initialvalue problems, also serve to a very important purpose, namely they can be used as tests to verify numericalschemes that are developed to study complex unsteady flow problems. Just as in the case of Newtonian fluids,it is necessary to develop a large class of exact and approximate solutions for fluids such as the second-gradefluid. The solutions that have been obtained here are written as sum of steady-state and transient solutions andtend to the similar solutions of Newtonian fluid by considering appropriate limiting case. Finally, the influ-ence of physical parameters on the velocity profile and the shear stress is shown and discussed by graphicalillustrations.

2 Governing equations

The flow of an incompressible viscoelastic fluid is governed by a set of conservation and constitutive equations.The mass and momentum equations can be written as follows:

∇.v = 0, (1)

ρ

[∂v∂t

+ (v · ∇)v]

= ∇ · T + ρb, (2)

where ρ is the fluid density, v the velocity vector, b is the body force per unit mass, and T the Cauchy stresstensor.

The constitutive equation for an incompressible second-grade fluid is given by

T = −pI + μA1 + α1A2 + α2A21, (3)

where −p is the pressure, I is the unit tensor, μ is the coefficient of viscosity, α1 and α2 are the normal stressmoduli, and A1 and A2 are the kinematic tensors defined through

A1 = ∇v + (∇v)T , A2 = ∂A1

∂t+ (v · ∇)A1 + A1(∇v) + (∇v)T A1, (4)

where v is the velocity vector. If this model is required to be compatible with thermodynamics, then the materialmoduli must meet the following restrictions

μ ≥ 0, α1 ≥ 0 and α1 + α2 = 0. (5)

The sign of the material moduli α1 and α2 has been the subject of much controversy. A comprehensive discus-sion on the restrictions given in (4), as well as a critical review on the fluids of differential type, can be foundin the extensive work of Dunn and Rajagopal [20].

In this work, We will consider the flow whose velocity vector v has the following form

v = v(r, t) = v(r, t)ez, (6)

where ez is the unit vector in the z-direction of cylindrical coordinate system (r, θ, z). For such flows, theconstraint of incompressibility (1) is automatically satisfied.

Substituting Eq. (6) in (3), we find following meaningful equation:

τ(r, t) =(

μ + α1∂

∂t

)∂v(r, t)

∂r, (7)

where τ(r, t) = Srz(r, t) is the shear stress that is different from zero. In the absence of body forces andpressure gradient in the axial direction, the balance of the linear momentum (2) leads to the relevant equation

ρ∂v(r, t)

∂t=

(∂

∂r+ 1

r

)τ(r, t). (8)

Eliminating τ(r, t) between Eqs. (7) and (8), we get the governing differential equation of our problem, asfollows:

∂v(r, t)

∂t=

(ν + α

∂

∂t

)(∂2

∂r2 + 1

r

∂

∂r

)v(r, t). (9)

where α = α1/ρ and ν = μ/ρ are the kinematic viscosity of the fluid.

A note on sinusoidal motion of a viscoelastic non-Newtonian fluid 661

3 Flow induced by cylinder with prescribed shear on its boundary

Suppose that an incompressible second-grade fluid is situated in an infinite circular cylinders of radius R. Attime t = 0, the fluid is at rest. At time t = 0+, the boundary of cylinder applies a sinusoidal shear stress onfluid along its axis (r = 0). Owing to the shear, the fluid is gradually moved, its velocity is of the form (6),and governing equation is (9). The appropriate initial and boundary conditions are

v(r, 0) = 0; r ∈ (0, R), (10)

τ(R, t) =(

μ + α1∂

∂t

)∂v(R, t)

∂r= A sin(ωt); t > 0, (11)

3.1 Calculation of the velocity field

Applying the Laplace transform to Eq. (9) and having the initial and boundary conditions (10) and (11) inmind, we find that

q v(r, q) = (αq + ν)

(∂2

∂r2 + 1

r

∂

∂r

)v(r, q), (12)

where the image function v(r, q) = L{v(r, t)} has to satisfy the condition

∂v(R, q)

∂r= Aω

(α1q + μ)(q2 + ω2). (13)

In the following, we denote by

vH (rn, q) =R∫

0

rv(r, q)J0(rrn)dr, (14)

the Hankel transform of v(r, q), where J0(·) is the Bessel function of first kind of order zero and rn; n =1, 2, 3, . . . are the positive roots of the transcendental equations J1(Rrn) = 0.

Multiplying now both sides of Eq. (12) by r J0(rrn), integrating with respect to r from 0 to R and takinginto account the condition (13) and the equality

R∫0

r[∂2v(r, q)

∂r2 + 1

r

∂v(r, q)

∂r

]J0(rrn)dr = R J0(Rrn)

∂v(R, q)

∂r− r2

n vH (rn, q) , (15)

we find that

vH (rn, q) = ARωJ0(Rrn)

ρ

1[(1 + αr2

n

)q + νr2

n

](q2 + ω2)

. (16)

Now, for a more suitable presentation of the final results, we rewrite Eq. (16) in the following equivalent form

vH (rn, q) = v1H (rn, q) + v2H (rn, q), (17)

where

v1H (rn, q) = ARωJ0(Rrn)

r2n

1

(α1q + μ)(q2 + ω2), (18)

and

v2H (rn, q) = − ARωJ0(Rrn)

r2n

q[(1 + αr2

n

)q + νr2

n

](α1q + μ)(q2 + ω2)

. (19)


Using the formula

R∫0

r3 J0(rrn)dr = 2R2

r2n

J0(Rrn) , (20)

we find that inverse Hankel transform of v1H (rn, q) is

v1(r, q) = Aωr2

2R

1

(α1q + μ)(q2 + ω2), (21)

while the inverse Hankel transforms of v2H (rn, q) is given by

v2(r, q) = 2

R2

∞∑n=1

J0(rrn)

J 20 (Rrn)

v2H (rn, q). (22)

From Eqs. (17), (21), and (22), we find that v(r, q) has the following form

v(r, q) = Aωr2

2R

1

(α1q + μ)(q2 + ω2)

−2Aω

R

∞∑n=1

J0(rrn)

r2n J0(Rrn)

q[(1 + αr2

n

)q + νr2

n

](α1q + μ)(q2 + ω2)

. (23)

Applying inverse Laplace transform to Eq. (23), we find for v(r, t) the following expression

v(r, t) = Ar2

2R(μ2 + α2

1ω2)

{ωα1 exp

(−μt

α1

)+ μ sin(ωt) − ωα1 cos(ωt)

}

+2A

R

∞∑n=1

J0(rrn)

r2n J0(Rrn)

{ω

[ω2α1

(1 + αr2

n

) − μνr2n

]cos(ωt) − ω2μ

(1 + 2αr2

n

)sin(ωt)(

μ2 + α21ω2

) [ω2

(1 + αr2

n

)2 + ν2r4n

]

+ ωr2n

(1 + αr2

n

)ρ

[ω2

(1 + αr2

n

)2 + ν2r4n

] exp

( −νr2n t

1 + αr2n

)− ωα1

μ2 + α21ω2

exp

(−μt

α1

)}. (24)

v(r, t) = vs(r, t) + vt (r, t), (25)

where

vs(r, t) = 2A

ρR

∞∑n=1

J0(rrn)[ω2

(1 + αr2

n

)2 + ν2r4n

]J0(Rrn)

{νr2

n sin(ωt) − ω(1 + αr2

n

)cos(ωt)

}, (26)

is the steady-state part of the velocity and

vt (r, t) = 2Aω

ρR

∞∑n=1

(1 + αr2

n

)J0(rrn)[

ω2(1 + αr2

n

)2 + ν2r4n

]J0(Rrn)

exp

( −νr2n t

1 + αr2n

), (27)

is the transient part of the velocity field.


3.2 Calculation of the shear stress

Applying the Laplace transform to Eq. (7), we find that

τ(r, q) = (α1q + μ)∂v(r, q)

∂r. (28)

Now, differentiating Eq. (23) with respect to r , we get

∂v(r, q)

∂r= Aωr

R

1

(α1q + μ)(q2 + ω2)

+2Aω

R

∞∑n=1

J1(rrn)

rn J0(Rrn)

q[(1 + αr2

n

)q + νr2

n

](α1q + μ)(q2 + ω2)

. (29)

Introducing (27) into (24) and then applying the inverse Laplace transform, we find that the shear stressτ(r, t) has the following form

τ(r, t) = τs(r, t) + τt (r, t), (30)

where

τs(r, t) = Ar

Rsin(ωt) + 2A

R

∞∑n=1

J1(rrn)

rn

[ω2

(1 + αr2

n

)2 + ν2r4n

]J0(Rrn)

×{ωνr2

n cos(ωt) + ω2 (1 + αr2

n

)sin(ωt)

}, (31)

and

τt (r, t) = −2ων A

R

∞∑n=1

rn J1(rrn)[ω2

(1 + αr2

n

)2 + ν2r4n

]J0(Rrn)

exp

( −νr2n t

1 + αr2n

). (32)

are, respectively, the steady-state and transient parts of the shear stress.

4 Concluding remarks and numerical results

The non-Newtonian fluids are increasingly being considered more important and appropriate in technologicalapplications than the Newtonian fluids. Strictly speaking, the linear relationship between stress and the rate ofstrain does not exist for a lot of real fluids, such as blood, oils, paints, and polymeric solutions. In general, theanalysis of the behavior of the fluid motion of the non-Newtonian fluids tends to be much more complicatedand subtle in comparison with that of the Newtonian fluids. There has been a fairly large number of flows ofNewtonian fluids for which a closed form analytical solution is possible. However, for non-Newtonian fluids,such exact solutions are rare.

The purpose of this paper was to establish exact solutions for the velocity field and shear stress correspond-ing to the flow of a second-grade fluid in an infinite circular cylinder, by using Laplace and Hankel transforms.The obtained solutions have been written as sum of steady-state and transient solutions. The motion of fluidwas due to the application of shear stress on the fluid by the boundary of cylinder. The solutions that have beenobtained satisfy the governing equation and all imposed initial and boundary conditions and for α → 0 reduceto the similar solutions for Newtonian fluid. Finally, we would like to examine the dependence of differentmaterial parameters on the time required to obtain steady-state regime of the oscillations.

In Figs. 1, 2, 3, 4, and 5, the behavior of transient part of the velocity with respect to different parametershas been shown. Following are the major concluding remarks from the graphical illustrations:

• Figure 1 shows that required time to get the steady-state increases if α increases.• Close view of Fig. 2 reveals that required time to get the steady-state decreases if ρ increases.


Fig. 1 Transient velocity profiles for different values of α. A = 1, ρ = 2, R = 1, ω = 2, ν = 0.1, r = 0.5

Fig. 2 Transient velocity profiles for different values of ρ. A = 1, α = 1, R = 1, ω = 2, ν = 0.1, r = 0.5


Fig. 3 Transient velocity profiles for different values of ω. A = 1, α = 1, R = 1, ρ = 2, ν = 0.1, r = 0.5

Fig. 4 Transient velocity profiles for different values of ν. A = 1, α = 1, R = 1, ρ = 2, ω = 2, r = 0.5


Fig. 5 Transient velocity profiles for different values of A. α = 1, R = 1, ρ = 2, ω = 2, ν = 0.1, r = 0.5

• Figure 3 shows required time to get the steady-state decreases if angular frequency ω increases.• Figure 4 shows that time required to get steady-state decreases if kinematic viscosity ν increases.• Figure 5 shows required time to get the steady-state increases if amplitude of shear stress A increases.

References

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A note on sinusoidal motion of a viscoelastic non-Newtonian fluid

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