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# Open Physics

### formerly Central European Journal of Physics

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Volume 16, Issue 1

# Unsteady flow of fractional Oldroyd-B fluids through rotating annulus

/ Muhammad Nawaz Naeem
/ Maria Javaid
/ Rabia Safdar
Published Online: 2018-04-18 | DOI: https://doi.org/10.1515/phys-2018-0028

## Abstract

In this paper exact solutions corresponding to the rotational flow of a fractional Oldroyd-B fluid, in an annulus, are determined by applying integral transforms. The fluid starts moving after t = 0+ when pipes start rotating about their axis. The final solutions are presented in the form of usual Bessel and hypergeometric functions, true for initial and boundary conditions. The limiting cases for the solutions for ordinary Oldroyd-B, fractional Maxwell and Maxwell and Newtonian fluids are obtained. Moreover, the solution is obtained for the fluid when one pipe is rotating and the other one is at rest. At the end of this paper some characteristics of fluid motion, the effect of the physical parameters on the flow and a correlation between different fluid models are discussed. Finally, graphical representations confirm the above affirmation.

PACS: 47.10.A-; 47.27.nf

## 1 Introduction

Humans benefit from fluid flow but also need to understand the negative effects in nature, e.g. hurricanes and cyclones. The topographic conditions and the sufficient supply of rain flow is important in the cultivation of crops. Cultivation in deserts can also be achieved with artificial irrigation. Artificial breathing machines, artificial hearts and dialysis systems are designed using fluid dynamics. Hence fluids are vital.

The stress and rate of strain of a fluid is related: fluid is Newtonian if this relationship is linear, if not, the fluid is non-Newtonian. The non-Newtonian fluids, like polymers, plasma, paste and pulps are very important in industry. Therefore, the discussion of non-Newtonian fluids in fluid mechanics is important. All existing fluids cannot be described by the Navier-stocks equations. By categorizing different fluids, we have different mathematical models to represent fluids. Oldroyd-B fluid is one of these rate-type models.

The non-Newtonian fluids possessing shear dependent viscosity are known as viscoelastic fluids. The first rate type fluid model is a Maxwell model that is still in used. Fetecau et. al [1] derived the results of the velocity field and shear stress of an Oldroyd-B fluid on a plate, accelerating constantly. Hayat et. al [2] determined the solutions for MHD flow over an infinite oscillatory plate about an axis that is normal to the oscillatory plate. Some authors determined solutions for a viscoelastic fluid model by reproducing a kernel method (RKM) [3,4,5,6]. The unsteady incompressible Oldroyd-B fluid is analyzed by Jamil et. al [7] and special cases are also obtained for Maxwell, second grade and Newtonian fluids. Similarly, Burdujan [8] used this idea on Taylor-Couette flow and calculated the velocity field and shear stress of a rotating fluid between annuls of double cylinders. Numerical solutions of fractional fluids are also under attention [9, 10].

A lot of interest is given to a rotating circular domain corresponding to helical flow for second grade, Maxwell fluids and Oldroyd-B [11,12,13,14]. The flow of a polymer in a circular cylinder by considering pulsatile APG was analyzed by Barnes et. al [15, 16]. Davies et. al [17] and Phan-Thien [18] presented the same problem for a White-Metzner fluid. To solve the viscoelastic motion of different fluid models, one needs to learn about fractional calculus [19, 20]. Fractional calculus is very useful in fields of mathematics and physics [21,22,23]. Furthermore, the interested reader can consult the references [24,25,26,27,28,29,30].

In this proposal, we consider the fractional Oldroyd-B fluid model and study the flow due to the circulation of two pipes around its axial. Integral transforms are used to obtain the general solutions and the final result is in the form of hypergeometric functions. There are some limiting cases by applying the limits on physical parameters, i.e ς →1, λr →0 and λ → 0, solutions for the ordinary Oldroyd-B, fractional and ordinary Maxwell and Newtonian fluids are discussed respectively.

## 2 Development of the flow

Consider a fractional Oldroyd-B fluid (OBFFD) between two coaxial pipes. The radii of the pipes are R1 and R2(>R1), after time t = 0+, both pipes and the fluid between them start rotating along their axis. For the considered problem, the z-axis is along the axis of the pipes, the r-axis is perpendicular to the axis of the pipe and θ is along the tangent to the boundary. Let vr, vθ and vz be the velocity components in the direction of the r, θ and z axes respectively.

The continuity and linear momentum equations for incompressible flow is

$divV=0,$(1)

$dVdt=1ρdivT+b,$(2)

where V, t, ρ, b, d/dt is velocity, time, density, body force and the material time derivative respectively.

For an incompressible Oldroyd-B fluid [36]

$T=−pI+S,S+λ(S˙−LS−SLT)=μ[A+λr(A˙−LA−ALT)],$(3)

where p is the hydrostatic pressure, I denotes the identity tensor, S is the extra-stress tensor, λ and λr are relaxation and retardation times, superposed dot indicates the material time derivative, L represents the velocity gradient, A = L + LT is the first Rivlin Ericsen tensor, μ is the dynamic viscosity of the fluid and the superscript T denotes the transpose operation. For the problem in consideration, the velocity field and the extra stress tensor has the following form [31]

$V=V(r,t)=ω(r,t)eθ,S=S(r,t).$(4)

For such flows, (1) is true.

The governing equations for a Oldroyd-B fluid, corresponding to the above defined motions are [31]

$1+λ∂∂tτ(r,t)=μ(1+λr∂∂t)∂∂r−1rω(r,t),$(5)

$1+λ∂∂t∂w(r,t)∂t=ν(1+λr∂∂t)(∂2∂r2+1r∂∂r−1r2)ω(r,t),$(6)

where ν is the kinematic viscosity and τ(r, t) is the non zero shear stress.

The governing equations for a fractional Oldroyd-B fluid is obtained by using a fractional differential operator $\begin{array}{}{D}_{t}^{\varsigma },\end{array}$ defined by [32, 33]

$Dtςf(t)=1Γ(1−ς)ddt∫0tf(τ)(t−τ)ςdτ;0≤ς<1,$(7)

in governing equations (5) and (6) we have

$1+λςDςτ(r,t)=μ(1+λrςDς)∂∂r−1rω(r,t),$(8)

$1+λςDς∂w(r,t)∂t=ν(1+λrςDς)(∂2∂r2+1r∂∂r−1r2)ω(r,t).$(9)

## 3 Initial and boundary conditions

We need three conditions to solve a fractional partial differential equation (9). The relevant initial and boundary conditions are

$ω(r,t)|t=0=0,∂ω(r,t)∂t|t=0=0,τ(r,t)|t=0=0,r∈[R1,R2],$(10)

$ω(R1,t)=R1℧1tκ,ω(R2,t)=R2℧2tκfort>0,$(11)

where 1, 2 and κ are real numbers.

## 3.1 Analytic solutions for the velocity field

Using the Laplace transform Eqs. (9) and (11) implies that

$q+λςqς+1ω¯(r,q)=ν1+λrςqς(∂2∂r2+1r∂∂r−1r)ω¯(r,q);r∈(R1,R2),$(12)

$ω¯(R1,q)=R1℧1qκ+1,ω¯(R2,q)=R2℧2qκ+1,$(13)

where the Laplace transform of the function ω(r, t) is ω(r, q). The finite Hankel transform [34] of the function ω(r, q), can be expressed as

$ω¯H(rn,q)=∫R1R2rω¯(r,q)A(r,rn)dr,n=1,2,3,...$(14)

where

$A(r,rn)=J1(rrn)Y1(R2rn)−J1(R2rn)Y1(rrn),$(15)

rn are the positive roots of the transcendental equation A(R1, r) = 0, and Jp(·) is Bessel functions of the first kind and Yp(·) is Bessel functions of second kind of order p.

Taking the Hankel transform of the differential equation (12) and keeping in mind the boundary condition (13) and following identities

$ddrA(r,rn)=rn[J0(rrn)Y1(R2rn)−J1(R2rn)Y0(rrn)]−1rA(r,rn),$(16)

$J0(z)Y1(z)−J1(z)Y0(z)=−2πz,$(17)

we get

$ω¯H(rn,q)=2ν(1+λrςqς)[R2℧2J1(R1rn)−R1℧1J1(R2rn)]πJ1(R1rn)1qκ+1(q+νrn2+λςqς+1+νλrςrn2qς).$(18)

Writing the above equations in a suitable form

$ω¯H(rn,q)=2[R2℧2J1(R1rn)−R1℧1J1(R2rn)]πrn2J1(R1rn)1qκ+1−1+λςqςqκ(q+νrn2+λςqς+1+νλrςrn2qς),$(19)

The inverse Hankel transform of ωH(rn, q) [34]

$ω¯(r,q)=π22∑n=1∞rn2J12(R1rn)A(r,rn)J12(R1rn)−J12(R2rn)ω¯H(rn,q).$(20)

Using the inverse Hankel transform on Eq. (19), and the identity

$R2(r2−R12)(R22−R12)r=π∑n=1∞J12(R1rn)A(r,rn)J12(R1rn)−J12(R2rn),$

we obtain

$ω¯(r,q)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)1qκ+1−π∑n=1∞J1(R1rn)A(r,rn)J12(R1rn)−J12(R2rn)1+λςqςqκ×℧2R2J1(R1rn)−℧1R1J1(R2rn)(q+νrn2+λςqς+1+νλrςrn2qς).$(21)

Using the identity

$1q+νrn2+λςqς+1+νλrςrn2qς=1λς∑k=0∞∑m=0kk!m!(k−m)!×−νrn2λςkλrςmqςm−k−1qς+1λςk+1,$(22)

Eq. (21) can be further simplified to give

$ω¯(r,q)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)1qκ+1−πλς×∑n=1∞J1(R1rn)A(r,rn)℧2R2J1(R1rn)−℧1R1J1(R2rn)J12(R1rn)−J12(R2rn)×∑k=0∞∑m=0kk!m!(k−m)!−νrn2λςkλrςm×qςm−k−κ−1+λqς+ςm−k−κ−1qς+1λςk+1.$(23)

Now using inverse Laplace transform on Eq. (23) and taking into account (A1) [35] from the Appendix, we find that

$ω(r,t)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)tκ−πλς∑n=1∞J1(R1rn)A(r,rn)℧2R2J1(R1rn)−℧1R1J1(R2rn)J12(R1rn)−J12(R2rn)∑k=0∞∑m=0kk!m!(k−m)!−νrn2λςkλrςm×[Gς,ςm−k−κ−1,k+1−λ−1,t+λGς,ς+ςm−k−κ−1,k+1−λ−1,t].$(24)

## 3.2 Calculation of the shear stress

Using the Laplace transform, Eq. (5) implies that

$τ¯(r,q)=μ1+λrςqς1+λςqς∂∂r−1rω¯(r,q).$(25)

Using Eqs. (21) in the above equation we get

$τ¯(r,q)=μ1+λrςqς1+λςqς2R12R22(R22−R12)r2℧2−℧1qκ+1+μπλς∑n=1∞2rA(r,rn)−rnA∗(r,rn)J1(R1rn)J12(R1rn)−J12(R2rn)×℧2R2J1(R1rn)−℧1R1J1(R2rn)∑k=0∞∑m=0kk!m!(k−m)!×−νrn2λςkλrςm(qςm−k−κ−1+λrςqς+ςm−k−κ−1)qς+1λςk+1],$(26)

where

$A∗(r,rn)=J1(rrn)Y0(R2rn)−J0(R2rn)Y1(rrn).$(27)

Now, using the inverse Laplace transform on Eq. (26), finally the shear stress

$τ(r,t)=2μR12R22(℧2−℧1)λς(R22−R12)r2[Rς,−κ−1−λ−ς,t+λrςRς,ς−κ−1−λ−ς,t]+μπλς×∑n=1∞2rA(r,rn)−rnA∗(r,rn)J1(R1rn)J12(R1rn)−J12(R2rn)(℧2R2J1(R1rn)−℧1R1J1(R2rn))∑k=0∞∑m=0kk!m!(k−m)!−νrn2λςk×λrςm[Gς,ςm−k−κ−1,k+1−λ−ς,t+λrςGς,ς+ςm−k−κ−1,k+1−λ−ς,t].$(28)

## 4 Limiting cases

1. Letting ς → 1 into Eqs. ((24)) and ((28)) the velocity field

$ω(r,t)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)tκ−πλ×∑n=1∞J12(R1rn)A(r,rn)℧2R2J1(R1rn)−℧1R1J1(R2rn)J12(R1rn)−J12(R2rn)×∑k=0∞∑m=0kk!m!(k−m)!−νrn2λk×λrm[G1,m−k−κ−1,k+1−λ−1,t+λG1,1+m−k−κ−1,k+1−λ−1,t],$(29)

and the shear stress

$τ(r,t)=2μR12R22(℧2−℧1)λ(R22−R12)r2[R1,−κ−1−λ−1,t+λr×R1,−κ−λ−1,t]+μπλ×∑n=1∞2rA(r,rn)−rnA∗(r,rn)J1(R1rn)J12(R1rn)−J12(R2rn)×℧2R2J1(R1rn)−℧1R1J1(R2rn)∑k=0∞∑m=0kk!m!(k−m)!×−νrn2λkλrm[G1,m−k−κ−1,k+1−λ−1,t+λrG1,1+m−k−κ−1,k+1−λ−1,t],$(30)

for the ordinary Oldroyd-B fluid.

2. Making λr →0 in Eqs. ((24)) and ((28)), the velocity field

$ω(r,t)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)tκ−πλς∑n=1∞J1(R1rn)A(r,rn)J12(R1rn)−J12(R2rn){℧2R2J1(R1rn)−℧1R1J1(R2rn)}∑k=0∞−νrn2λςk×{Gς,−k−κ−1,k+1−λ−ς,t+λGς,ς−k−κ−1,k+1−λ−ς,t},$(31)

and the shear stress

$τ(r,t)=2μR12R22(℧2−℧1)λr2(R22−R12)Rς,−2−λ−1,t+πμλς∑n=1∞J1(R1rn)2rA(r,rn)−rnA∗(r,rn)J12(R1rn)−J12(R2rn)××℧2R2J1(R1rn)−℧1R1J1(R2rn)×∑k=0∞−νrn2λςkGς,−k−2,k+1−λ−ς,t,$(32)

corresponding to a fractional Maxwell fluid.

3. Making ς → 1 in Eqs. ((31)) and ((32)), the velocity field is

$ω(r,t)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)tκ−πλς∑n=1∞J1(R1rn)A(r,rn)J12(R1rn)−J12(R2rn){℧2R2J1(R1rn)−℧1R1J1(R2rn)}∑k=0∞−νrn2λk×{G1,−k−κ−1,k+1−λ−1,t+λG1,1−k−κ−1,k+1−λ−1,t},$(33)

and the shear stress is

$τ(r,t)=2μR12R22(℧2−℧1)λr2(R22−R12)Rς,−2−λ−1,t+πμλς∑n=1∞J1(R1rn)2rA(r,rn)−rnA∗(r,rn)J12(R1rn)−J12(R2rn)××℧2R2J1(R1rn)−℧1R1J1(R2rn)×∑k=0∞−νrn2λkG1,−k−2,k+1−λ−1,t,$(34)

for the ordinary Maxwell fluid.

Using Appendix A2, the expressions (33) and (34) in the simplified form for κ = 1, in terms of expositional functions, are

$ωM(r,t)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)t−πν∑n=1∞J1(R1rn)A(r,rn)rn2[J12(R1rn)−J12(R2rn)]{℧2R2J1(R1rn)−℧1R1J1(R2rn)}1−λq1n2eq2nt−q2n2eq1ntq2n−q1n,$(35)

and

$τM(r,t)=2μR12R22(℧2−℧1)r2(R22−R12)t−λ1−e−t/λ+πρ∑n=1∞J1(R1rn)2rA(r,rn)−rnA∗(r,rn)rn2[J12(R1rn)−J12(R2rn)]×℧2R2J1(R1rn)−℧1R1J1(R2rn)×1+q1neq2nt−q2neq1ntq2n−q1n.$(36)

4. By now letting λ→ 0 in Eqs. (35) and (36) or ς → 1 and λ→ 0 in Eqs. (31) and (32) and using $\begin{array}{}\underset{\lambda \to 0}{lim}\frac{1}{{\lambda }^{k}}{G}_{1,\phantom{\rule{thinmathspace}{0ex}}b,\phantom{\rule{thinmathspace}{0ex}}k}\left(\frac{-1}{\lambda },t\right)=\frac{{t}^{-b-1}}{\mathit{\Gamma }\left(-b\right)};\end{array}$ b < 0, the velocity field is

$ωN(r,t)=℧1R12(R22−r2)+℧2R22(r2−R12)r(R22−R12)t−πν∑n=1∞J1(R1rn)A(r,rn)rn2[J12(R1rn)−J12(R2rn)]{℧2R2J1(R1rn)−℧1R1J1(R2rn)}1−e−νrn2t,$(37)

and the shear stress is

$τN(r,t)=2μR12R22(℧2−℧1)r2(R22−R12)t+πρ∑n=1∞J1(R1rn)2rA(r,rn)−rnA∗(r,rn)rn2[J12(R1rn)−J12(R2rn)]××℧2R2J1(R1rn)−℧1R1J1(R2rn)1−e−νrn2t,$(38)

corresponding to a Newtonian fluid.

## 5 Flow through a circular pipe

The finite Hankel transform of the function f(r, t) is

$fH(rn′)=∫0R2rω(r)J1(rrn′)dr,$(39)

and the inverse Hankel transform of fH $\begin{array}{}\left({r}_{n}^{\mathrm{\prime }}\right)\end{array}$ is

$ω(r)=2R22∑n=1∞J1(rrn′)J22(R2rn′)fH(rn′).$(40)

where J1(R2r) = 0 as rn are the positive roots of the equation.

By making J1(R2rn) = 0, and using the identities J0(z) + J2(z) = $\begin{array}{}\frac{2}{z}{J}_{1}\end{array}$ (z) in Eq. (15) we get

$A(r,rn)=2J1(rrn)πR2rnJ2(R2rn),$(41)

Writing the Integral

$∫R1R2rf(r,t)A(r,rn)dr=∫0R2rf(r,t)A(r,rn)dr−∫0R1rf(r,t)A(r,rn)dr.$(42)

Introducing Eq. (41) and taking R1 = 0, the above integral takes the form

$∫R1R2rf(r,t)A(r,rn)dr=2πR2rnJ2(R2rn)∫0R2rf(r,t)J1(rrn)dr,$(43)

as a result from (24) and (28) we recovered the expansions for the velocity field

$ω(r,t)=℧2rt−2℧2λ∑n=1∞J1(rrn)rnJ2(R2rn)∑k=0∞∑m=0kk!m!(k−m)!×−νrn2λkλrm[Gς,ςm−k−2,k+1−λ−1,t+λGς,ς+ςm−k−2,k+1−λ−1,t],$(44)

and

$τ(r,t)=2μπ℧2λ∑n=1∞J2(rrn)J2(R2rn)∑k=0∞∑m=0kk!m!(k−m)!×−νrn2λkλrm[Gς,ςm−k−2,k+1−λ−1,t+λrGς,ς+ςm−k−2,k+1−λ−1,t],$(45)

through a cylinder already obtained by Kamran et al. [31] in Equation (21) and (27) for p = 1.

## 6 Conclusions and results

In this article, exact solutions for a fractional Oldroyd-B fluid between two rotating pipes are determined. Fluid motion is produced due to rotation of the pipes around their axis with time dependent angular velocities. The solutions determined by the use of integral transforms and presented in terms of Bessel functions and hypergeometric functions which are free of integrals. The final result satisfies the initial and boundary conditions. In the limiting cases, the corresponding results for Oldroyd-B fluid, Maxwell and Newtonian fluid, are obtained from general results. Moreover, the solution for the fluid through a circular pipe is obtained as a special case.

The velocity profiles and shear stress are illustrated in Figures 27 for different values of parameters. Fig. 2 for the effect of t, Fig. 3 for the effect of ν, Fig. 4 for the effect of λ, Fig. 5 for the effect of λr and Fig. 6 shows the effect of the fractional parameter ς on the fluid motion. Clearly, the velocity as well as shear stress are increasing as a function of t and ν, The effect is qualitatively the same for fractional parameter ς and λ, more exactly the velocity ω(r, t) is a decreasing function with regards to ς.

Figure 1

Problem Geometry.

Figure 2

Profiles of the velocity field ω(r, t) given by Eqs. (24) and (28) for R1 = 1, R2 = 7.4, ℧1 = 1, ℧2 = −1, λ = 4, λr = 1, ν = 0.002, ς = 0.1 and different values of time

Figure 3

Profiles of the velocity field ω(r, t) given by Eqs. (24) and (28) for t = 7, R1 = 1, R2 = 7.4, ℧1 = 1,℧2 = −1, λ = 20, λr = 5, ς = 0.9 and different values of ν

Figure 4

Profiles of the velocity field ω(r, t) given by Eqs. (24) and (28) for t = 7, R1 = 1, R2 = 7.4,℧1 = 1, ℧2 = −1, ν = 0.2,λr = 1, ς = 0.5 and different values of λ

Figure 5

Profiles of the velocity field ω(r, t) given by Eqs. (24) and (28) for t = 8, R1 = 1, R2 = 7.4, ℧1 = 1, ℧2 = −1,ν = 0.2,λ = 10,ς = 0.5 and different values of λr

Figure 6

Profiles of the velocity field ω(r, t) given by Eqs. (24) and (28) for t = 20, R1 = 1, R2 = 7.4,℧1 = 1, ℧2 = −1, ν = 0.001, λ = 4,λr = 1 and different values of ς

Finally, Fig. 7 is the comparison of ω(r, t) corresponding to the motion of the Newtonian, fractional and ordinary Maxwell, and fractional and ordinary Oldroyd-B for the same values of time and for the same material and fractional parameters. From this comparison, the Oldroyd-B fluid is the slowest and the Newtonian fluid is the fastest on the whole flow domain. SI units are used for all figures, and the roots $\begin{array}{}{r}_{n}=\frac{n\pi }{\left({R}_{2}-{R}_{1}\right)}.\end{array}$.

Figure 7

Profiles of the velocity field ω(r, t) corresponding to the Newtonian, Maxwell, Fractional Maxwell, Oldroyd-B and fractional Oldroyd-B fluids, for t = 8, R1 = 1, R2 = 7.4, ℧1 = 1, ℧2 = −1,ν = 0.001,λ = 3.6,λr = 3 and ς = 0.8

## Appendix

$L−1{qb(qa−d)c}=Ga,b,c(d,t)=∑j=0∞djΓ(c+j)Γ(c)Γ(j+1)×t(c+j)a−b−1Γ[(c+j)a−b],Re(ac−b)>0,|dqa|<1,$(A1)

$∑k=0∞−νrn2λkG1,−k−1,k+1−λ−1,t=eq2nt−eq1ntq2n−q1n,$(A2)

$∑k=0∞−νrn2λkG1,−k−2,k+1−λ−1,t×=λνrn21+q1neq2nt−q2neq1ntq2n−q1n,$(A3)

$R1,−2−λ−1,t=λt−λ21−e−t/λ;q1n,q2n=−1±1−4νλrn22λ.$(A4)

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## About the article

Accepted: 2018-01-15

Published Online: 2018-04-18

Funding: This research is supported by the Government College University, Faisalabad, Pakistan and the Higher Education Commission Pakistan.

Competing interestCompeting interests: The authors declare that they have no competing interests.

Citation Information: Open Physics, Volume 16, Issue 1, Pages 193–200, ISSN (Online) 2391-5471,

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