Scholarly article on topic 'Analytic solutions of Oldroyd-B fluid with fractional derivatives in a circular duct that applies a constant couple'

Analytic solutions of Oldroyd-B fluid with fractional derivatives in a circular duct that applies a constant couple Academic research paper on "Mathematics"

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Abstract of research paper on Mathematics, author of scientific article — M.B. Riaz, M.A. Imran, K. Shabbir

Abstract The aim of this article was to analyze the rotational flow of an Oldroyd-B fluid with fractional derivatives, induced by an infinite circular cylinder that applies a constant couple to the fluid. Such kind of problem in the settings of fractional derivatives has not been found in the literature. The solutions are based on an important remark regarding the governing equation for the non-trivial shear stress. The solutions that have been obtained satisfy all imposed initial and boundary conditions and can easily be reduced to the similar solutions corresponding to ordinary Oldroyd-B, fractional/ordinary Maxwell, fractional/ordinary second-grade, and Newtonian fluids performing the same motion. The obtained results are expressed in terms of Newtonian and non-Newtonian contributions. Finally, the influence of fractional parameters on the velocity, shear stress and a comparison between generalized and ordinary fluids is graphically underlined.

Academic research paper on topic "Analytic solutions of Oldroyd-B fluid with fractional derivatives in a circular duct that applies a constant couple"

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ORIGINAL ARTICLE

Analytic solutions of Oldroyd-B fluid with fractional derivatives in a circular duct that applies a constant couple

M.B. Riaz a, M.A. Imran a *, K. Shabbir b

a Department of Mathematics, University of Management and Technology, Lahore, Pakistan b Department of Mathematics, GC University Lahore, Pakistan

Received 31 May 2016; revised 26 July 2016; accepted 28 July 2016

KEYWORDS

Oldyrod-B fluid; Fractional calculus; Velocity field; Shear stress; Circular duct; Analytic solutions

Abstract The aim of this article was to analyze the rotational flow of an Oldroyd-B fluid with fractional derivatives, induced by an infinite circular cylinder that applies a constant couple to the fluid. Such kind of problem in the settings of fractional derivatives has not been found in the literature. The solutions are based on an important remark regarding the governing equation for the nontrivial shear stress. The solutions that have been obtained satisfy all imposed initial and boundary conditions and can easily be reduced to the similar solutions corresponding to ordinary Oldroyd-B, fractional/ordinary Maxwell, fractional/ordinary second-grade, and Newtonian fluids performing the same motion. The obtained results are expressed in terms of Newtonian and non-Newtonian contributions. Finally, the influence of fractional parameters on the velocity, shear stress and a comparison between generalized and ordinary fluids is graphically underlined.

© 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

1. Introduction

Non-Newtonian fluids play a paramount role in a large number of industries and the branches of knowledge concerned with the applied sciences. Among the numerous models that have been used to portray the behavior of non-Newtonian fluids, the Oldroyd-B model seems to be agreeable to examine and more prominently experiment. This model can describe

* Corresponding author.

E-mail addresses: imranasjad11@yahoo.com, Imran.asjad@umt.edu. pk (M.A. Imran).

Peer review under responsibility of Faculty of Engineering, Alexandria University.

stress-relaxation, creep and usual stress differences is the most successful model for relating the response of many dilute polymeric liquids. It contains, as special cases the Maxwell and viscous fluid models. The motion of a fluid in the neighborhood of a moving body is of great interest for industry. The flow between cylinders or through a rotating cylinder has applications in the food industry and being one of the most important and interesting problems of motion near rotating bodies. The velocity distribution for different motions of Newtonian fluids through a circular cylinder is given in [1]. Accurate solutions regarding motions of non-Newtonian fluids in cylindrical domains appear to be those of Ting [2], Srivastava [3] and Waters and King [4] for second grade, Maxwell and Oldroyd-B fluids respectively. Through the time numerous

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1110-0168 © 2016 Faculty of Engineering, Alexandria University. Production and hosting by Elsevier B.V. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/).

papers concerning such motions of non-Newtonian fluids have been published. Among them, we here recall only a small number of those regarding Oldroyd-B fluids [5-19].

Though, all above-mentioned papers incorporate motion problems in which the velocity is given on the boundary. Although in some realistic problems what is specified is the shear stress [20,21], more precisely the force with which the cylinder is moved. To restate, in Newtonian mechanics force is the reason and kinematics is the effect (see Rajagopal [23] for a comprehensive discussion on the same). Therefore, the boundary condition on stresses is significant, and Renardy [22] illustrated how well-posed boundary value problems can be formulated in this way. In the last time numerous solutions for such motions of rate type fluids have been recognized (see [24-29] and the references therein), but all these problems respond to differential expressions of the stress on the boundary. This is due to their governing equations that, unlike those corresponding to Newtonian and second grade fluids, contain differential expressions acting on the non-trivial shear stresses.

With these motivations, the aim of this article was to analyze the rotational flow of an Oldroyd-B fluid with fractional derivatives induced by an infinite circular cylinder that applies a constant couple to the fluid. These solutions are obtained by means of integral transforms. The obtained solutions satisfy all imposed initial and boundary conditions and expressed in terms of Newtonian and non-Newtonian contributions that can be reduced to the similar solutions corresponding to ordinary Oldroyd-B [30], fractional Maxwell, ordinary Maxwell, fractional/ordinary second-grade and Newtonian fluids.

2. Mathematical formulation of the problem

The Cauchy-stress tensor T corresponding to an incompressible Oldroyd-B fluid is related to the fluid motion by the relations

T --pI + S, S + k(S - LS - SLT = i[A + kr(A - LA - ALr)],

where —pI denotes the indeterminate spherical stress due the constraint of incompressibility, S is the extra-stress tensor, L is the velocity gradient, A = L + LT is the first Rivlin-Ericksen tensor, i is the dynamic viscosity, k and kr are the relaxation and retardation times respectively. The superscript T indicates the transpose operation and the superposed dot denotes the material time derivative. The model characterized by the constitutive Eq. (1) contains as special cases the upper convected Maxwell model for kr — 0 and the Newtonian fluid model for k — kr — 0. In some special cases like that to be here considered, the governing equations for Oldroyd-B fluids resemble to those for second-grade fluids.

Let us assume that an incompressible Oldroyd-B fluid is at rest in an infinite circular cylinder of radius R. At time t — 0+, the cylinder begins to turn about its axis due to constant torque per unit length 2pRf. Owing to the shear, the fluid is gradually moved and we are looking for a velocity field of the form

V = x(r, 6)ee,

where eh is the unit vector in the h-direction of the system of Cylindrical coordinates.

For such a flow, the constraint of incompressibility is identically satisfied. We also assume that the extra-stress tensor S, as well as the velocity V is a function of time and radius only. If the fluid is at rest up to the moment t — 0, then

V(r, 0) = 0, S(r, 0)=0,

and the constitutive Eq. (1) implies Srr — Srz — Shz — Szz — 0 and the meaningful partial differential equation

d\ id 1 + k dt) S(r, t)=H 1 + kr di

dr -- 'x(r't),

where s(r, t) —Srh(r, t) is the non-trivial stress. Neglecting body forces, the balance of linear momentum leads to the relevant equation due to rotation symmetry

dx(r, t) ' dt

2 I s(r, t),

where q is the constant density of the fluid.

Usually in the literature, the governing equation for the velocity is obtained by eliminating s(r, t) between Eqs. (4) and (5). Since our interest here is to solve a motion problem with shear stress on the boundary, we follow the pattern and eliminate rn(r, t) in order to get the governing equation

I) ^ ^ " '

b d d2 1 d 4

dt fe +1 dr- ^s(r,t),

for the shear stress s(r, t). Here v — q is the kinematic viscosity of the fluid.

Its corresponding fractional model is

(1 + k-D;) zm - m(i + w^£ +1 ! - i),(r.

' D?(t)- cshj /0 (t - s)m-;-xfm) (s)ds, m - 1 < ; < m, m 2 N

.D7f(t)-f, m - 1,2,...

is the Caputo fractional derivative operator [31-33]. The appropriate initial and boundary conditions are:

ds(r, t)

s(r, 0)--

s(R, t)-fH(t),

r 2 [0,R],

t > 0.

(9) (10)

3. Solution of the problem

Applying the Laplace transform to Eq. (7) and keeping in mind the initial conditions Eq. (9), we have

sir q)=mi!±M + 11 _ l)s(r q) K ; q) q + kaq«+i lyÔr2 + r dr r2 'S(r' q);

and the equation Eq. (10) becomes

S(R; q) =~ ; q

where s(R, q) and s(r, q) are the Laplace transform of the function s(R, t) and s(r, t) respectively. Multiplying Eq. (11) by rJ2(rrn) where rn are the positive roots of J2(Rr) — 0, integrating the result from 0 to R and use the identities

■ CO ÖJ2[rr,

fR t, 1 @ 4\_. \j

l rj2{rr-\dr2+r - V2)s(rq)dr

— -RrnJ2(Rrn)S(R, q) - r2nl(rn, q)

{ r v(1+ kßqß)

L rj2(rr2)s(r'q)dr — q + kV+1

fR w jtf 1 @ 4\_. \ j

X i + 7 dT - 72) T(r' q)dr-

Taking Hankel transform [34, Section 14, Eq. (59)], and using Eq. (12)

sn(rn, q) —-

vfRrn (1 + kßqß) J2(Rrn) q[q + k'q'+1 + v(1 + kßqß)rl] '

In order to inverting Eq. (14), it can be written in suitable form

sH(r2, q)

q q + vrl

- k? -qT-2 +

+kß-q2?-1

(-1)pp! fvr^ p

U и (p k

'-ß+ßj-p

q+vr„ q+vr„

(q'+k-a)

«\ p+i

2 (q'+k-a)

a\ P+1

Applying the inverse Laplace and the Hankel transforms and use the identity [34, the entry 1 of Table X],

—-2re

j2(rr2)

=1 rJi (Rrn) '

Lengthy but straightforward computations show that

s(r, 0=7»- f E^J^^t

v ' ' T>2 R / /

R 2—1 rnJ1 (Rrn)

2fv-1 J2(rrn) , 2 л

- 2R § r2j2RR222) k?G1?-11(-vr»;t)

J2(rrn) + R ¿j r2J1(Rr2]

+k?G1,2?-1,1(-vr2, s - s)]

i [G1,?-1,1(—vr2 ' s - s) ■Jo

VV (-1)pp! fvr2„\pk?j(r

p—o j—o k?(G--?+j?-p'p+1

+avr2„GxJß-ß-1p+1[- ка,s

The fractional model of Eq. (5), t).

By introducing Eq. (16) into Eq. (17), we obtain

PD>(r; t)—{ + 2 r

dJiirrn)

p t K ' ' R2 R*-(rnJx (Rrn)

2 f.,1 BJ2(rrn)

- 2R E ^ kßÖ1,ß-1,1(-Vr2, 0

n/;, 1 BJ2(rrn) r t

+RUR ^ r

R^ nJ1(Rfn)Jo

+kßG1,2ß-M(-VR2, S - S)]

[G1,ß-1,1 ( — vrl' s - s)

^^ (-1)Pp! fvr2AP.j(r ( 1 \

P—o j—o (P^lk^J kß ^ra,a-ß+jß-p,p+^- ka, SJ

+avr„rajß-ß-1p+1 [ - k?; S

2rfH(t) 44f ^ J2(rrn) e-vr2t R2 rRn=\ RnJ1(Rrn)e

4fv-1 J2(rRn) ,?n , 2 A

- RR!! RnJ1"(RRn)kß 'ß-1'1 ( n;t)

4ft-1 J2(rrn) it , , 2

+4RV-Jul [r1'?-1'1(-Vrn's-S)

+"?r1,2?-1,1(-vrl s-s)]

^ (-1)PP! fvR2APkßjfr ( 1

P—О j—О (p - ka J ^ra'a-ß+jß-p.+^ ka 'S)

+avrnrajß-ß-1p+1\ - k?'s

Datoj(r, t) —

4rfH(t) 2f 1 J1 (rr,

'n) _vr2 t

pR2 pR^ rnJ1(Rrn)

2fv^„ J1(rrn) / 2

pRn—i J1(RRn) ^^n't)

2fv^ J1 (rRn

pRn—(nJ1(Rrn, Jo +k"r1'2?-1,1 (-vr2' S( s)]

(-1)pp! (vji\p

j—O (p - j)(A k')

+avrn¡rttjß—ß—lp+A- jt ' s

/ [r1'?-1'1(-vr2's - s)

r'''-ß+jß-pp+1\ - kt 's

Applying Laplace transform to Eq. (19) and use the initial condition Eq. (3)1, we get

4rfH(t) 1 2f 1 Jl(rrn) q-a

x(r; q)= ,n2 — r

pR2 q'+1 pR RnJ1(Rrn ) q + _2f11 J1(rrn) .ß qß-'-1 pRn—1Г" J1 (Rrn) Rq + vr2

.R'l J1 (RRn) + pR^ rnJ1(Rrni

qß—a—1 q2ß—a—1

q + kßq

q + vrn q + vrn_

VV (—1)pp! (vRhpkj

p—О j—o (p - jm\k') R

q'.-ß+ßj-p qßj-p-1

—-— + avr2---

\p+1 n

(q' + k—

' (q' + k—

Applying inverse Laplace transform to Eq. (20), we have

, ï 4rf t

x(r,t) = —-

2f ^ J1(rrn) G , 9 A

-sl^Tn— G1._«. i(_vr» 't)

pR2 r(a + 1) pR^ J (Rrn) 2fm

JA rrn

prr s rn:J1(RRn) kbG1'b_a_11(_mr»'t)

/ [G1'b_a_1'1 ( mrn' s _ s)

J1 (rrn

pR~1r" ^(R^ Jo

+kfG1;2b_a_1'1(_Vrn; s _ s)] ,2\P

-a-1 1

EEf^kf^Ga

p=o j=0

(p _ j ka

+avr„Ga,/f_f_1'P+1 ( _ ka ' s

4. Limiting cases

4.1. Case a = f ! 1, solutions corresponding to ordinary Oldroyd-B fluid

By letting a = f ! 1 into Eqs. (16) and (21), we obtain

S OB (r ; t) =

r2fH(t) 2f1 J2

' n) -vrit

R n=1 rnJl (Rrn)

J2 rrn

2("n) -, n , 2 a

Rn-1 Jx (Rrn)

J2 (rrn)

+ R Un

i [G1'01(_Vr2n; s _ s) ■Jo

R J1(Rrni Jo +k?G1'1'1(_v^; s _ s)]

ËÊt% C4) p'-:(g,-,.J _ 1. s

lp=o j=o

(P _T)!A k .

+ J Vr2nG1j;P+^_ 1 ' s

s (rA r2fH(t) 2f1 J2(rrn) e_v21 SGM(r't)=-W- _ RE JR-) e n

J2 (rrn )

R n=i nJ1(Rrn)

x [' [G1 '_1'1(_vrn ' s _ s)] ■Jo

ë(_1)P( #) (G1 ' _p _ k"'s))

Wgm (r ' t) =

_p=o 4rf ta

PR r(a + 1)

^ J1 (rr„)

pR~=1 rnJ1(Rrn) / ^ J1 (rr«)

ds' (24)

G1 ' _a ' 1 (_vt2 't)_--

r(a + 1)

x i [G1 '_a_1'1(_vr;; 's _ s)] ■Jo

ds. (25)

pR ^=1 " J1 (Rr»

1 fvr2\ P / 1 \

E(_1)P y)?) (Ga'a_P'P+^ _ j"' sJ

a,p->\(GOB)

□□□ OB

h—► t = 2 1

Figure 1 Comparison of shear stress s(r , t) versus r with kr — 0.15 , k — 0.85 , t — 2 sec of Eq. (22) and [30, Eq. (15)].

4rft 2f 1 J1 (rrn) 2 a

Xob(t )=PR _ PRh JR-) G1'_1'1(_Vr2 't)

J1(rrn) . 2 a

_ fH "«jr«) krG1'_1'1 (_vr-'t)

f V- J1(rrn) + „rIJ2

pR ~1n J 1(Rr,

+KG1'o ' 1 (—vr;;' s _ s ) ]

(_1 )pp!

P=o j=o (P -j) !j! V k / / 1 \\/

i [G1 '_1'1(_vrn 's _s )

EE(^) g^_pP+1\ _ t 's

+ 7 vt«g1 jp+1[ _ p s

As a check of our results we plot the graphs for Eqs. (22) and (23) to justify [30, Eqs. (15) and (17)] (see Figs. 1 and 2).

4.2. Case kb ! 0, b ! 0, solutions corresponding to generalized Maxwell fluid

By now letting kb ! 0 , b ! 0 into Eqs. (16) and (21), we have

Figure 2 Comparison of velocity x(r t versus r with kr — 0.15 k — 0.85 , t — 2 sec of Eq. (22) and [30, Eq. (17)].

4.3. Case kr ! 0 or a ! 1, solutions corresponding ordinary Maxwell fluid

We can find shear stress and velocity profile of Maxwell fluid by using kr ! 0 into Eqs. (22) and (23), or by letting a ! 1 into Eqs. (24) and (25), we have

r2fH(t) 2f1 J

R 2-^г

n) „-vrii

2fm1 J2 (rrn

R^ rnJi(Rrn ) R 2=л nJi(Rrn

X i [Gi,0 ,i(-vr2 , s - s)] Jo

X (-i) 1i)'(«• 4 - i, s))

ds, (26)

4rft 2f ^ Ji(rrn

. . 4rjl 21 s—^

XM(r,0 =-R - -rE

:Gi, -i,i(-mrn ;t) -t

-R2 -R^ rnJi(Rrn)

x i' [Gi,-i,i(-mrn , s - s)] 0

2fv-1 Ji (rrn

+ -REr>

pr7=i Ji(Rr^ Jo

,2\ P ,

Ê (-i К - i. s))

ds. (27)

4.4. Case ka ! 0 ' a ! 0, solutions corresponding generalized second grade fluid

Putting limit ka ! o into Eqs. (16) and (21), we have shear stress and velocity profile of generalized second grade fluid

sgsg (r,t) r

r2fH(t) 2f1 J (rr,

n ) „-mr't

R2 R^i rnJi(Rrn)

2fm^ J2 (rrn) , 2 Л

R У nJi(Rrni

.f1 Ji(rrn ) + R U2

R ^ n Ji(Rrn) Jo X [Gi,b-i ,i(-mr2, s - s) + kfGi, 2b-i,i (-vr2n, s - s) ]

E(-iПm#b)P(Gipb(-mrn ,s)) ,p=o

ds, (28)

XGSGy , t ) =-^2 -

2f ^ Ji(rrn ) G , о

" Gi, -ai (-mrn 't)

-R2 Г(а + i) pRj={rnJi(Rrn)

2fm^.. Ji(rr»j_kbG

pRnri' nJi(Rrn 2fm1 Ji (rrn]

+ -REr

i, b-i,i(-mrn ,t)

-R^i nJi(Rrn)

/ [Gi,b-1 ,i (-mrn, s - s ) +kbGi,2b-i, Jo

i(-mr2 , s - s)

E(-i )P (m^kb) P(Gi,pAp+i(-vrn,s)) Pro

ds. (29)

Furthermore, when b ! 1 into Eqs. (28) and (29), we have shear stress and velocity fields for ordinary second grade fluid

sSG (r, t) =

r2fH(t) 2f^ J2(

'n) „-mrit

R2 Rj=i rnJi(Rrn ) J2(rrn) , ^ , ..2 л , f

-krGiAi(-vr2 , t) +-R-Er

J2 (rrn)

J i (Rrn )r ^'R^rf nJi(Rrn )

f' [Gi, o , i (-mr2 , s - s) + kbGiii (-mr2 , s - s ) ]

E(-i)P№k,)P(Gipp+i(-*i,s)) ds,

. . 4rf ta Xsg (r,t ) -

-R2 Г(а + i)

2f ^ Ji (rrn ) G , 9 Л 2fv^ Ji(rrn ) . , 2 л , f1 Ji(rrn )

pRj^i nJi(Rrn )

/ [Gi,o ,i(-mrn, s - s) + krGiii(-vrn,s - s ) ]

E(-i)P(mr2Jr)P(GiP,P+i(-mrn ,s)) ds. Pro

4.5. Case ka ! 0 and kf ! 0 solutions corresponding to Newtonian fluid

Finally by making ka ! o and kf ! o into Eqs. (16) and (21)

r2fH( t) , 2f1 J2 (rrn)

sN(r ,t) + IE

R2 R^ri rnJi(Rrn)

n ) e-mr^t

6 100 -

Figure 3 Shear stress s(r , t) versus r with kr — 0.15 , k — 0.85 , b — 0.8 , t — 2 sec and different values of a.

4rft , 2f 1 JiK

t)=—2+-y;

pR2 iR^ r2 J1(Rrn)

[1 - e-^] ,

we recovered the solutions which are identical to those obtained in [30, Eqs. (22) and (23)].

5. Conclusion

This study highlights the key features of the rotational flow of an Oldroyd-B fluid with fractional derivatives induced by an infinite circular cylinder that applies a constant couple to the fluid. Such kind of problem has not been solved in the literature. The present solutions are based on a simple but important remark regarding the governing equation for the

Figure 4 Shear stress s(r, t) versus r with kr = 0.15, k = 0.85, b = 0.8, t = 4 sec and different values of a.

[ 1 1 *** jS=0.3

- /?=0.6

- y y.-y

S ■' y

y ■ y /■■'s

- t = 2

- i i i

Figure 6 Shear stress s(r, t) versus r with kr = 0.15, k = 0.85, a = 0.2, t = 2 sec and different values of p.

,5=0.3 /?= 0.6 /.■y y./ ■ 'j*

0=0.8 / v>

- yy'x /--'y -

- y yy -

____ i i i i = 4

Figure 7 Shear stress s(r, t) versus r with kr = 0.15, k = 0.85, a = 0.2, t = 4 sec and different values of p.

i = 0.3

&&0 = 0.8 ^ >

i 1 1 t= 6

Figure 5 Shear stress s(r, t) versus r with kr = 0.15, Figure 8 Shear stress s(r, t) versus r with kr = 0.15, k = 0.85, b = 0.8, t = 6 sec and different values of a. k = 0.85, a = 0.2, t = 6 sec and different values of b.

non-trivial shear stress. The solutions that have been obtained satisfy all imposed initial and boundary conditions and can easily be reduced to the similar solutions corresponding to ordinary Oldroyd-B, fractional/ordinary Maxwell, fractional/ ordinary second-grade, and Newtonian fluids performing the similar motion. The obtained results are in the form of Newtonian and non-Newtonian parts. To study the influence of fractional parameters on the velocity and shear stress is plotted in the Figs. 3-15. From Figs. 3-5 it is observed that the s(r, t) increases with the increase in value of a and same behavior is observed for p cleared from Figs. 6-8. Figs. 10-15 show that the w(r, t) increases with the increase in the values of a and p. Finally, a comparison for time derivative of integer order versus fractional order is shown graphically for GOB, OB,

Figure 9 Comparison of shear stress among different fluids s(r, t) versus r with kr = 0.15, k = 0.85 and at t = 2 sec.

Figure 10 Profiles of velocity m(r, t) versus r with kr = 0.15, k = 0.85, p = 0.8, t = 2 sec and different values of a.

Figure 12 Profiles of velocity m(r, t) versus r with kr = 0.15, k = 0.85, p = 0.8, t = 6 sec and different values of a.

-0.015

ß=0.3

/?=0.6

\ \ a \\ / e*€>o ß= 0.8

" \i / {'\j V i i f = 2 i i

Figure 11 Profiles of velocity m(r, t) versus r with kr = 0.15, k = 0.85, p = 0.8, t = 4 sec and different values of a.

Figure 13 Profiles of velocity m(r, t) versus r with kr = 0.15, k = 0.85, a = 0.2, t = 2 sec and different values of p.

Figure 14 Profiles of velocity x(r, t) versus r with kr = 0.15, k = 0.85, a = 0.2, t = 4 sec and different values of p.

/3 = 0.3

«i /2=0.6

I a vi 9eo ß= 0.8

t\ v\ -■ v-X ^ f=6

Figure 15 Profiles of velocity x(r, t) versus r with kr = 0.15, k = 0.85, a = 0.2, t = 6 sec and different values of p.

GMF, MF, GSG, SG and N in Fig. 9. It is found that ordinary Maxwell fluid is swiftest and ordinary second grade fluid is slowest.

Acknowledgement

The authors highly appreciate the reviewers for their profound comments and feedback which enhance the quality of the paper. We are thankful to the Department of Mathematics, University of Management and Technology Lahore, Pakistan, for supporting and facilitating this research work.

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