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Hall effects on unsteady duct flow of a non-Newtonian fluid in a porous medium
Applied Mathematics and Computation 157 (2004) 103–114 www.elsevier.com/locate/amc
Hall effects on unsteady duct flow of a non-Newtonian fluid in a porous medium T. Hayat
a,*
, R. Naz a, S. Asghar
b
a b
Department of Mathematics, Quaid-i-Azam University, Islamabad 45320, Pakistan Department of Mathematical Sciences, Comsats Institute of Information Technology, Abbottabad, Pakistan
Abstract The unsteady flow of an incompressible Oldroyd-B fluid in a circular duct embedded in a porous medium is studied. The fluid is conducting in the presence of a transverse magnetic field and the influence of Hall currents is investigated. The motion in the duct is generated by a given but arbitrary time dependent inlet volume flow rate. Laplace transform technique is used to determine the exact solutions for the four cases: (a) constantly accelerated flow, (b) sudden started flow, (c) the flow rate has a trapezoidal variation with time and (d) oscillatory flow. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Exact solutions; Porous medium; Hall current; Oldroyd-B fluid; Initial value problems
1. Introduction Porous materials are important in many scientific and engineering applications such as catalysis, hydrology tissue engineering, membrane separation process, wetting and drying processes, powder technology, contaminant transport in soil and aquifers, and enhanced oil recovery. The prediction of the effective or macroscopic transport properties in these porous materials is a longstanding problem of great fundamental and practical significance [1]. The
*
Corresponding author. E-mail address: [email protected] (T. Hayat).
0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.08.069
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theoretical prediction of macroscopic transport properties of porous media is made difficult by its complex dependence on composition and structure of pore space. In an ionized gas where the density is low and/or the magnetic field is very strong, the conductivity normal to the magnetic field is reduced due to the free spiraling of electrons and ions about the magnetic lines of force before suffering collisions; also a current is induced in a direction normal to both the electric and the magnetic fields. This phenomenon, well known in the literature, is called the Hall effect. The study of magnetohydrodynamic flows with Hall currents has important engineering applications in problems of magnetohydrodynamic generators and of Hall accelerators as well as in flight magnetohydrodynamics. The flow of fluid in a circular duct/pipe is of considerable interest in the technical field due to its frequent occurrence in industrial and technological applications. Many papers in this field have been published by Szymanski [2], Uchida [3], Weinbaum and Parker [4], Lefebvre and White [5], Das and Arakeri [6], and Chen et al. [7]. In all these investigations, the effects of Hall currents in a porous medium are not considered. In this work, we investigate the effect of Hall current on unsteady flow of an Oldroyd-B fluid in a circular duct. The magnetohydrodynamic fluid fills the porous space. The layout of the paper is as follows. In Section 2, we give the problem formulation. Section 3 contains the exact solution corresponding to the general case. Some special solutions for flows are presented in Section 3.1. Finally, in Section 4 the concluding remarks are given. 2. Mathematical formulation We consider a cylindrical coordinate system ðr; /; zÞ in which r is the radial direction, / is the circumferential direction and z-axis is taken as the center line direction of the circular duct. We consider an incompressible, unidirectional and electrically conducting fluid filling the porous space. The magnetic field B0 is applied in the z-direction. All the physical properties of the fluid are assumed to be constant. The fluid is taken as an Oldroyd-B fluid. The governing equations are given by [8,9]. Continuity equation: div V ¼ 0:
ð1Þ
Momentum equation: q
dV l/ ¼ div T þ ðJ BÞ V: dt k
ð2Þ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
Generalized OhmÕs law:
J ¼ rðE þ V BÞ
105
r ðJ B $pe Þ: ene
ð3Þ
oB ; ot
ð4Þ
MaxwellÕs equations: div B ¼ 0;
curl B ¼ e l j;
curl E ¼
V ¼ uðr; tÞk;
ð5Þ
s ¼ srz ðr; tÞk:
ð6Þ
In above equations V is the velocity vector, u and k are the velocity and unit vectors in the z-direction respectively, srz is the shear stress acting on the rplane toward the z-direction, B is the magnetic induction vector, E is the electric field vector, J is the current density vector, pe is the electron pressure, q is the fluid density, rðr ¼ e2 ne te =me Þ is the electrical conductivity, te is the electron collision time, e is the electron charge, ne is the electron number density, me is the mass of the electron, d=dt is the material derivative, l is the dynamic viscosity, kð> 0Þ and / ð0 < / < 1Þ, are respectively the (constant) permeability and porosity of the porous space [10] and e l is the magnetic permeability. The Cauchy stress tensor T for an Oldroyd-B fluid is given by [11] T ¼ pI þ s; Ds D ¼l 1þh sþk A1 ; Dt Dt
ð7Þ ð8Þ
in which p ¼ pðr; /; zÞ is the static fluid pressure, k and h are the relaxation and retardation times respectively and k P h P 0, I is the identity tensor, D=Dt is upper convected time derivative and the first Rivlin–Ericksen tensor A1 is given by A1 ¼ grad V þ ðgrad VÞT :
ð9Þ
Making use of Eq. (5), Eq. (1) is identically satisfied and following [12] we have from Eqs. (2)–(9) as follows: op 1 o rB20 l/ ou ðrsrz Þ ð10Þ þ u¼q ; þ e oz r or k ot 1 im op op ¼ ¼ 0; or o/
ð11Þ
where szz þ k
oszz ou 2srz or ot
¼ 2lh
ou or
2 ;
ð12Þ
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T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
srz þ k
osrz ou o2 u ; ¼ l þ lh or orot ot
ð13Þ
e ¼ xe te is the Hall parameter. Eq. (11) shows that pressure gradient is not and m a function of r and / and from Eqs. (10) and (13) we obtain o ou 1 o op o 1 o ou ¼ 1þk þm 1þh r 1þk ot ot q ot oz ot r or or o N 1þk u; ð14Þ ot where 1 N¼ q
rB20 l/ ; þ e k 1 im
l m¼ : q
The boundary and initial conditions for the problem are uðR; tÞ ¼ 0; ou ð0; tÞ ¼ 0; or Z R 2pruðr; tÞ dr ¼ up ðtÞpR2 ¼ QðtÞ;
ð15Þ
ð16Þ ð17Þ ð18Þ
o
where R is the duct radius, upðtÞ and QðtÞ are the known average inlet velocity and inlet flow rate respectively. 3. Exact solution The problem given by Eqs. (14) and (16)–(18) can be solved by the use of the Laplace transform method. If the Laplace transform of u is u then, Eq. (14) and conditions (16)–(18) take the following forms: d2 uðr; sÞ 1 duðr; sÞ 1 þ ks dpðz; sÞ m2 uðr; sÞ ¼ ; þ dr2 r dr lð1 þ hsÞ dz uðR; sÞ ¼ 0; Z
duð0; sÞ ¼ 0; dr
ð19Þ ð20Þ
R
2pruðr; sÞ dr ¼ up ðsÞpR2 ;
ð21Þ
0
in which sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 þ ksÞðs þ N Þ ; m¼ mð1 þ hsÞ and up ðsÞ is the Laplace transform of up ðtÞ.
ð22Þ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
107
The general solution to Eq. (19) is uðr; sÞ ¼ AI0 ðmrÞ þ BK0 ðmrÞ þ Cp ;
ð23Þ
where I0 and K0 are modified Bessel functions of order zero but first and second kind respectively, A and B are arbitrary constants and Cp is assumed particular solution. The solution (23) after using conditions (20) and (21) becomes uðr; sÞ ¼ up ðsÞGðr; sÞ:
ð24Þ
In above expression Gðr; sÞ ¼
I0 ðmRÞ I0 ðmrÞ : 2 I1 ðmRÞ I0 ðmRÞ mR
Laplace inversion of Eq. (24) gives Z cþi1 1 up ðsÞGðr; sÞest ds: uðr; tÞ ¼ 2pi ci1
ð25Þ
ð26Þ
The pressure gradient distribution can be obtained by taking inverse Laplace transform of the following expression dpðz; sÞ qðS þ N Þup ðsÞI0 ðmRÞ ; ¼ 2 dz I0 ðmRÞ mR I1 ðmRÞ
ð27Þ
where I1 is the modified Bessel function of the first kind of order zero. We will discuss now some interesting examples. 3.1. Constant acceleration case For this we take Up up ðtÞ ¼ ap t ¼ t; t0
ð28Þ
in which ap is the constant acceleration, Up is the velocity after acceleration and t0 is the time period of oscillation. Taking Laplace transform of Eq. (28) and then using in Eq. (26) we obtain Z cþi1 Up ½I0ðmRÞ I0 ðmrÞ 1 est dt: ð29Þ uðr; tÞ ¼ 2 2pi ci1 t0 s2 I0 ðmRÞ mR I1 ðmRÞ In order to evaluate the integration in above equation we need to find the residue at s ¼ 0 and is given by
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T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
Resðs ¼ 0Þ 2 ¼
r 2
3
2t 1 7 R Up 6 6 7 6 7: r 4 r 2 5 t 0 4 R2 1 1 1 þ fN ðt þ k hÞ þ 1g 1 þ 4 R 3 R 2m ð30Þ
The other singular points are the roots of the transcendental equation RI0 ðmRÞ
2 I1 ðmRÞ ¼ 0: m
ð31Þ
Using m ¼ ia in above equation we get aRJ0 ðaRÞ 2J1 ðaRÞ ¼ 0: If an ðn ¼ 1; 2; 3; . . . ; 1Þ are zeros of above equation then qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðkN þ 1 þ mha2n Þ þ ðkN þ 1 þ mha2n Þ 4kðN þ ma2n Þ ; s1n ¼ 2k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðkN þ 1 þ mha2n Þ ðkN þ 1 þ mha2n Þ 4kðN þ ma2n Þ ; s2n ¼ 2k
ð32Þ
ð33Þ
ð34Þ
are simple poles and the residue at these poles are given by Resðs1n Þ ¼
2man ð1 þ hs1n ÞUp es1n t ½J0 ðan RÞ J0 ðan rÞ ; t0 Rs21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ J1 ðan RÞ
ð35Þ
Resðs2n Þ ¼
2man ð1 þ hs2n ÞUp es2n t ½J0 ðan RÞ J0 ðan rÞ : t0 Rs22n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ J1 ðan RÞ
ð36Þ
From Eqs. (29), (30), (35) and (36) we have 3 r 2 2t 1 7 uðr; tÞ 1 6 R 7 ¼ 6 r 4 1 r 2 5 1 Up t0 4 R2 þ þ fN ðt þ k hÞ þ 1g 1 1 4 R 3 R 2m 2 3 2 sin t e ð1 þ hs1n Þ 1 6 2 7 2m X 6 s1n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7 2 5 Rt0 n¼1 4 es2n t ð1 þ hs2n Þ þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ 2
an ½J0 ðan RÞ J0 ðan rÞ : J1 ðan RÞ
ð37Þ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
109
In Eq. (37), the first term on right hand side is the steady state velocity, the second term, the transient response of the flow due to an abrupt change either in the boundary conditions or body forces or pressure gradient. The expression for the pressure gradient is dp qUp 4 8m ¼ ðNt þ 1Þ þ 2 ðt þ h kÞ þ dz 3 R t0 2 3 2 ðs1nþN Þð1 þ hs1n Þ es1n t 1 7 2 2 2lUp X J0 ðan RÞ 6 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 an 6 7: ð38Þ 2 5 Rt0 n¼1 J0 ðan RÞ 4 ðs2n þ N Þð1 þ hs2n Þ es2n t þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ 3.2. Impulsively started flow The expressions for velocity and pressure gradient of an impulsively started flow up ¼ 0; for t 6 0; ¼ Up ; for t > 0;
ð39Þ
are given by r 2 NR2 1 r 4 1 r 2 uðr; tÞ 1 1 ¼ 2 1 þ þ Up R R 3 R 2m 4 2 3 2 s1n t e ð1 þ hs1n Þ 1 6 7 2 X 2m 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7 2 5 R n¼1 4 es2n t ð1 þ hs2n Þ þ 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ an ½J0 ðan RÞ J0 ðan rÞ ; ð40Þ J1 ðan RÞ dpðz; tÞ 8lUp NR2 ¼ þ 1 þ dz R2 6m 2
3 2 es1n t ðs1n þ N Þð1 þ hs1n Þ 1 7 2 2lUp X an J0 ðan RÞ 6 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7: 2 5 R n¼1 J1 ðan RÞ 4 es2n t ðs2n þ N Þð1 þ hs2n Þ þ 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ ð41Þ
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3.3. Trapezoidal piston motion In this section we obtain the solution for a piston velocity which vary with time as Up t; for 0 6 t 6 t0 ; t0 ¼ Up ; for t0 6 t 6 t1 ; t2 t ; for t1 6 t 6 t2 ; ¼ Up t2 t1
up ðtÞ ¼
¼ 0;
ð42Þ
for t2 6 t 6 1:
Making use of Heaviside unit step function the piston motion becomes of the following form up ðtÞ ¼
Up Up tH ðtÞ tH ðt t0 Þ þ Up H ðt t0 Þ Up H ðt t1 Þ t0 t0 t2 t t2 t þ Up H ðt t1 Þ Up H ðt t2 Þ: t2 t1 t2 t1
ð43Þ
The solutions for different time periods of velocity and pressure gradient are as follows.
3.3.1. During constant acceleration piston motion (0 6 t 6 t0 ) up ðtÞ ¼
Up tH ðtÞ; t0
ð44Þ
and the results for velocity and pressure gradient are 2
3 r 2 R2 þ fN ðt þ k hÞ þ 1g 7 2t 1 uðr; tÞ 1 6 R 2m 6 7 ¼ 6 7
r 2 4 5 Up t0 4 1 r 1 1 1 þ 4 R 3 R 2 3 es1n t ð1 þ hs1n Þ2 1 6 2 7 2 2m X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 an ½J0 ðan RÞ J0 ðan rÞ þ ; 6 7 5 Rt0 n¼1 4 J1 ðan RÞ es2n t ð1 þ hs2n Þ2 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
ð45Þ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
111
dpðz; tÞ qUp 4 8m 2lUp ¼ ð1 þ NtÞ þ 2 ðt þ h kÞ þ dz 3 R t0 t0 R 2 3 2 s1n t ðs1n þ N Þð1 þ hs1n Þ e 1 7 2 2 X an J0 ðan RÞ 6 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 6 7: 2 5 J1 ðan RÞ 4 ðs2n þ N Þð1 þ hs2n Þ es2n t n¼1 þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ
ð46Þ
3.3.2. During the constant velocity period (t0 6 t 6 t1 ) We have up ðtÞ ¼
Up Up tH ðtÞ tH ðt t0 Þ þ Up H ðt t0 Þ; t0 t0
ð47Þ
r 2 NR2 1 r 4 1 r 2 uðr; tÞ 1 1 þ þ ¼ 2 1 Up R R 3 R 2m 4 2 3 ð1 þ hs1n Þ2 fes1n t es1n ðtt0 Þ g 1 6 7 2 2 2m X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 an ½J0 ðan RÞ J0 ðan rÞ ; þ 6 7 2 s t s ðtt Þ t0R n¼1 4 J1 ðan RÞ ð1 þ hs2n Þ fe 2n e 2n 0 g 5 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
ð48Þ 1 dpðz; tÞ 8lUp NR2 2lUp X an J0 ðan RÞ ¼ þ 1 þ dz R2 6m Rt0 n¼1 J1 ðan RÞ 2 3 ð1 þ hs1n Þ2 ðs1n þ N Þðes1n t es1n ðtt0 Þ Þ 6 7 6 s21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 7 6 7: 2 4 ð1 þ hs2n Þ ðs2n þ N Þðes2n t es2n ðtt0 Þ Þ 5 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
ð49Þ
3.3.3. For constant declaration period (t1 6 t 6 t2 ) We obtain up ðtÞ ¼
Up Up tH ðtÞ tH ðt t0 Þ þ Up H ðt t0 Þ Up H ðt t1 Þ t0 t0 t2 t þ Up H ðt t1 Þ; t2 t1
ð50Þ
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uðr;tÞ ¼ Up
r 2 NR2 1 r 4 t2 t 2 1 1 2m 4 t2 t1 R R
2 1 r 2 R 1 f1 þ N ðk hÞg 2mðt2 t1 Þ 3 R r 4 r 2 1 1 1 þ 1 4 R 3 R 2 s1n t 3 ð1 þ hs1n Þ2 e es1n ðtt0 Þ es1n ðtt1 Þ 7 1 6 t0 t2 t1 7 6 s21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 2m X 7 6 þ s2n t 7 2 s ðtt Þ s ðtt Þ R n¼1 6 2n 0 2n 1 5 4 ð1 þ hs2n Þ e e e þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ t0 t2 t1
an ½J0 ðan RÞ J0 ðan rÞ ; J1 ðan RÞ
ð51Þ
dpðz; tÞ 8lUp ðt t2 Þ Up q 4 8m 3ðt t2 Þ ¼ þ ðh kÞ þ þ Nqþ dz R2 ðt2 t1 Þ ðt2 t1 Þ 3 R2 4ðt2 t1 Þ 2 s1n t 3 2 ðs1n þ N Þð1 þ hs1n Þ e es1n ðtt0 Þ es1n ðtt1 Þ 7 1 6 t0 t2 t1 7 6 s21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 2Up l X 7 6 7 6 2 R n¼1 4 ðs2n þ N Þð1 þ hs2n Þ es2n t es2n ðtt0 Þ es2n ðtt1 Þ 5 þ 2 t0 t2 t1 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
an J0 ðan RÞ : J1 ðan RÞ
ð52Þ
3.3.4. After the piston has stopped ðt2 6 t 6 1Þ up ðtÞ is same as described in Eq. (43) and we have the following expressions for velocity and pressure gradient 2
s1n t 3 ð1 þ hs1n Þ2 e es1n ðtt0 Þ es1n ðtt1 Þ es1n ðtt2 Þ 1 6 2 7 2 t0 t2 t1 uðr; tÞ 2m X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 ¼ 6 7 2 s t s ðtt Þ s ðtt Þ s ðtt Þ Up R n¼1 4 ð1 þ hs2n Þ e 2n e 2n 0 e 2n 1 e 2n 2 5 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ t0 t2 t1 an ½J0 ðan RÞ J0 ðan rÞ ð53Þ ; J1 ðan RÞ 2
s1n t 3 ðs1n þ N Þð1 þ hs1n Þ2 e es1n ðtt0 Þ es1n ðtt1 Þ es1n ðtt2 Þ 1 6 2 7 2 t0 t2 t1 dpðz;tÞ 2lUp X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 ¼ 6 s2n t 7 2 s2n ðtt2Þ s2n ðtt0 Þ s2n ðtt1 Þ 5 dz R n¼1 4 ðs2n þ N Þð1 þ hs2n Þ e e e e þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ t0 t2 t1 an J0 ðan RÞ : ð54Þ J1 ðan RÞ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
113
3.4. Oscillatory flow For this case we define up ðtÞ ¼ 0;
for t 6 0;
¼ Up sin wt;
ð55Þ
for t > 0:
The results for velocity and pressure gradient are given by uðr; tÞ i iwt ¼ ½e Gðr; iwÞ eiwt Gðr;iwÞ Up 2
2
3 es1n t ð1 þ hs1n Þ2 1 7 2 2 2mw X an ½J0 ðan RÞ J1 ðan rÞ 6 6 ðs1n þ w2 Þ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7; 2 4 5 R n¼1 J1 ðan RÞ es2n t ð1 þ hs2n Þ þ 2 2 2 ðs2n þ w Þ½khs2n þ 2ks2n þ 1 þ N ðk hÞ
ð56Þ dpðz; tÞ iUp q ¼ ½ðN þ iwÞeiwt CðiwÞ ðN iwÞeiwt CðiwÞ dz 2 2 3 2 es1n t ðs1n þ N Þð1 þ hs1n Þ 1 7 2 2 2lUp w X an J0 ðan RÞ 6 6 ðs1n þ w2 Þ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 6 7; 5 R n¼1 J1 ðan RÞ 4 es2n t ðs2n þ N Þð1 þ hs2n Þ2 þ 2 2 2 ðs2n þ w Þ½khs2n þ 2ks2n þ 1 þ N ðk hÞ
ð57Þ where CðsÞ ¼
I0 ðmRÞ 2 I1 ðmRÞ I0 ðmRÞ mR
ð58Þ
and Gðr; sÞ is given by Eq. (25).
4. Concluding remarks We have examined the governing equation for an unsteady, incompressible, electrically conducting, Oldroyd-B fluid in circular duct in a porous medium. Analytical results up to oðN Þ are presented to illustrate the details of the flow and Hall and porosity effects and their depedence on the material parameters. We observe that in steady state the results are identical to that of Newtonian fluid. In the special case, when one of the two times of retardation or relaxation tends to zero, our solutions reduce to those corresponding to a Maxwell fluid or a second grade one. If both times k and h ! 0 the solution corresponding to a Newtonian fluid are obtained. The steady state flows are also obtained as a
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limiting case for t ! 1. The results in absence of Hall current and porosity are obtained by taking N ¼ 0.
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Hall effects on unsteady duct flow of a non-Newtonian fluid in a porous medium T. Hayat
a,*
, R. Naz a, S. Asghar
b
a b
Department of Mathematics, Quaid-i-Azam University, Islamabad 45320, Pakistan Department of Mathematical Sciences, Comsats Institute of Information Technology, Abbottabad, Pakistan
Abstract The unsteady flow of an incompressible Oldroyd-B fluid in a circular duct embedded in a porous medium is studied. The fluid is conducting in the presence of a transverse magnetic field and the influence of Hall currents is investigated. The motion in the duct is generated by a given but arbitrary time dependent inlet volume flow rate. Laplace transform technique is used to determine the exact solutions for the four cases: (a) constantly accelerated flow, (b) sudden started flow, (c) the flow rate has a trapezoidal variation with time and (d) oscillatory flow. Ó 2003 Elsevier Inc. All rights reserved. Keywords: Exact solutions; Porous medium; Hall current; Oldroyd-B fluid; Initial value problems
1. Introduction Porous materials are important in many scientific and engineering applications such as catalysis, hydrology tissue engineering, membrane separation process, wetting and drying processes, powder technology, contaminant transport in soil and aquifers, and enhanced oil recovery. The prediction of the effective or macroscopic transport properties in these porous materials is a longstanding problem of great fundamental and practical significance [1]. The
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Corresponding author. E-mail address: [email protected] (T. Hayat).
0096-3003/$ - see front matter Ó 2003 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2003.08.069
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theoretical prediction of macroscopic transport properties of porous media is made difficult by its complex dependence on composition and structure of pore space. In an ionized gas where the density is low and/or the magnetic field is very strong, the conductivity normal to the magnetic field is reduced due to the free spiraling of electrons and ions about the magnetic lines of force before suffering collisions; also a current is induced in a direction normal to both the electric and the magnetic fields. This phenomenon, well known in the literature, is called the Hall effect. The study of magnetohydrodynamic flows with Hall currents has important engineering applications in problems of magnetohydrodynamic generators and of Hall accelerators as well as in flight magnetohydrodynamics. The flow of fluid in a circular duct/pipe is of considerable interest in the technical field due to its frequent occurrence in industrial and technological applications. Many papers in this field have been published by Szymanski [2], Uchida [3], Weinbaum and Parker [4], Lefebvre and White [5], Das and Arakeri [6], and Chen et al. [7]. In all these investigations, the effects of Hall currents in a porous medium are not considered. In this work, we investigate the effect of Hall current on unsteady flow of an Oldroyd-B fluid in a circular duct. The magnetohydrodynamic fluid fills the porous space. The layout of the paper is as follows. In Section 2, we give the problem formulation. Section 3 contains the exact solution corresponding to the general case. Some special solutions for flows are presented in Section 3.1. Finally, in Section 4 the concluding remarks are given. 2. Mathematical formulation We consider a cylindrical coordinate system ðr; /; zÞ in which r is the radial direction, / is the circumferential direction and z-axis is taken as the center line direction of the circular duct. We consider an incompressible, unidirectional and electrically conducting fluid filling the porous space. The magnetic field B0 is applied in the z-direction. All the physical properties of the fluid are assumed to be constant. The fluid is taken as an Oldroyd-B fluid. The governing equations are given by [8,9]. Continuity equation: div V ¼ 0:
ð1Þ
Momentum equation: q
dV l/ ¼ div T þ ðJ BÞ V: dt k
ð2Þ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
Generalized OhmÕs law:
J ¼ rðE þ V BÞ
105
r ðJ B $pe Þ: ene
ð3Þ
oB ; ot
ð4Þ
MaxwellÕs equations: div B ¼ 0;
curl B ¼ e l j;
curl E ¼
V ¼ uðr; tÞk;
ð5Þ
s ¼ srz ðr; tÞk:
ð6Þ
In above equations V is the velocity vector, u and k are the velocity and unit vectors in the z-direction respectively, srz is the shear stress acting on the rplane toward the z-direction, B is the magnetic induction vector, E is the electric field vector, J is the current density vector, pe is the electron pressure, q is the fluid density, rðr ¼ e2 ne te =me Þ is the electrical conductivity, te is the electron collision time, e is the electron charge, ne is the electron number density, me is the mass of the electron, d=dt is the material derivative, l is the dynamic viscosity, kð> 0Þ and / ð0 < / < 1Þ, are respectively the (constant) permeability and porosity of the porous space [10] and e l is the magnetic permeability. The Cauchy stress tensor T for an Oldroyd-B fluid is given by [11] T ¼ pI þ s; Ds D ¼l 1þh sþk A1 ; Dt Dt
ð7Þ ð8Þ
in which p ¼ pðr; /; zÞ is the static fluid pressure, k and h are the relaxation and retardation times respectively and k P h P 0, I is the identity tensor, D=Dt is upper convected time derivative and the first Rivlin–Ericksen tensor A1 is given by A1 ¼ grad V þ ðgrad VÞT :
ð9Þ
Making use of Eq. (5), Eq. (1) is identically satisfied and following [12] we have from Eqs. (2)–(9) as follows: op 1 o rB20 l/ ou ðrsrz Þ ð10Þ þ u¼q ; þ e oz r or k ot 1 im op op ¼ ¼ 0; or o/
ð11Þ
where szz þ k
oszz ou 2srz or ot
¼ 2lh
ou or
2 ;
ð12Þ
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srz þ k
osrz ou o2 u ; ¼ l þ lh or orot ot
ð13Þ
e ¼ xe te is the Hall parameter. Eq. (11) shows that pressure gradient is not and m a function of r and / and from Eqs. (10) and (13) we obtain o ou 1 o op o 1 o ou ¼ 1þk þm 1þh r 1þk ot ot q ot oz ot r or or o N 1þk u; ð14Þ ot where 1 N¼ q
rB20 l/ ; þ e k 1 im
l m¼ : q
The boundary and initial conditions for the problem are uðR; tÞ ¼ 0; ou ð0; tÞ ¼ 0; or Z R 2pruðr; tÞ dr ¼ up ðtÞpR2 ¼ QðtÞ;
ð15Þ
ð16Þ ð17Þ ð18Þ
o
where R is the duct radius, upðtÞ and QðtÞ are the known average inlet velocity and inlet flow rate respectively. 3. Exact solution The problem given by Eqs. (14) and (16)–(18) can be solved by the use of the Laplace transform method. If the Laplace transform of u is u then, Eq. (14) and conditions (16)–(18) take the following forms: d2 uðr; sÞ 1 duðr; sÞ 1 þ ks dpðz; sÞ m2 uðr; sÞ ¼ ; þ dr2 r dr lð1 þ hsÞ dz uðR; sÞ ¼ 0; Z
duð0; sÞ ¼ 0; dr
ð19Þ ð20Þ
R
2pruðr; sÞ dr ¼ up ðsÞpR2 ;
ð21Þ
0
in which sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 þ ksÞðs þ N Þ ; m¼ mð1 þ hsÞ and up ðsÞ is the Laplace transform of up ðtÞ.
ð22Þ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
107
The general solution to Eq. (19) is uðr; sÞ ¼ AI0 ðmrÞ þ BK0 ðmrÞ þ Cp ;
ð23Þ
where I0 and K0 are modified Bessel functions of order zero but first and second kind respectively, A and B are arbitrary constants and Cp is assumed particular solution. The solution (23) after using conditions (20) and (21) becomes uðr; sÞ ¼ up ðsÞGðr; sÞ:
ð24Þ
In above expression Gðr; sÞ ¼
I0 ðmRÞ I0 ðmrÞ : 2 I1 ðmRÞ I0 ðmRÞ mR
Laplace inversion of Eq. (24) gives Z cþi1 1 up ðsÞGðr; sÞest ds: uðr; tÞ ¼ 2pi ci1
ð25Þ
ð26Þ
The pressure gradient distribution can be obtained by taking inverse Laplace transform of the following expression dpðz; sÞ qðS þ N Þup ðsÞI0 ðmRÞ ; ¼ 2 dz I0 ðmRÞ mR I1 ðmRÞ
ð27Þ
where I1 is the modified Bessel function of the first kind of order zero. We will discuss now some interesting examples. 3.1. Constant acceleration case For this we take Up up ðtÞ ¼ ap t ¼ t; t0
ð28Þ
in which ap is the constant acceleration, Up is the velocity after acceleration and t0 is the time period of oscillation. Taking Laplace transform of Eq. (28) and then using in Eq. (26) we obtain Z cþi1 Up ½I0ðmRÞ I0 ðmrÞ 1 est dt: ð29Þ uðr; tÞ ¼ 2 2pi ci1 t0 s2 I0 ðmRÞ mR I1 ðmRÞ In order to evaluate the integration in above equation we need to find the residue at s ¼ 0 and is given by
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Resðs ¼ 0Þ 2 ¼
r 2
3
2t 1 7 R Up 6 6 7 6 7: r 4 r 2 5 t 0 4 R2 1 1 1 þ fN ðt þ k hÞ þ 1g 1 þ 4 R 3 R 2m ð30Þ
The other singular points are the roots of the transcendental equation RI0 ðmRÞ
2 I1 ðmRÞ ¼ 0: m
ð31Þ
Using m ¼ ia in above equation we get aRJ0 ðaRÞ 2J1 ðaRÞ ¼ 0: If an ðn ¼ 1; 2; 3; . . . ; 1Þ are zeros of above equation then qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðkN þ 1 þ mha2n Þ þ ðkN þ 1 þ mha2n Þ 4kðN þ ma2n Þ ; s1n ¼ 2k qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ðkN þ 1 þ mha2n Þ ðkN þ 1 þ mha2n Þ 4kðN þ ma2n Þ ; s2n ¼ 2k
ð32Þ
ð33Þ
ð34Þ
are simple poles and the residue at these poles are given by Resðs1n Þ ¼
2man ð1 þ hs1n ÞUp es1n t ½J0 ðan RÞ J0 ðan rÞ ; t0 Rs21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ J1 ðan RÞ
ð35Þ
Resðs2n Þ ¼
2man ð1 þ hs2n ÞUp es2n t ½J0 ðan RÞ J0 ðan rÞ : t0 Rs22n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ J1 ðan RÞ
ð36Þ
From Eqs. (29), (30), (35) and (36) we have 3 r 2 2t 1 7 uðr; tÞ 1 6 R 7 ¼ 6 r 4 1 r 2 5 1 Up t0 4 R2 þ þ fN ðt þ k hÞ þ 1g 1 1 4 R 3 R 2m 2 3 2 sin t e ð1 þ hs1n Þ 1 6 2 7 2m X 6 s1n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7 2 5 Rt0 n¼1 4 es2n t ð1 þ hs2n Þ þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ 2
an ½J0 ðan RÞ J0 ðan rÞ : J1 ðan RÞ
ð37Þ
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109
In Eq. (37), the first term on right hand side is the steady state velocity, the second term, the transient response of the flow due to an abrupt change either in the boundary conditions or body forces or pressure gradient. The expression for the pressure gradient is dp qUp 4 8m ¼ ðNt þ 1Þ þ 2 ðt þ h kÞ þ dz 3 R t0 2 3 2 ðs1nþN Þð1 þ hs1n Þ es1n t 1 7 2 2 2lUp X J0 ðan RÞ 6 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 an 6 7: ð38Þ 2 5 Rt0 n¼1 J0 ðan RÞ 4 ðs2n þ N Þð1 þ hs2n Þ es2n t þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ 3.2. Impulsively started flow The expressions for velocity and pressure gradient of an impulsively started flow up ¼ 0; for t 6 0; ¼ Up ; for t > 0;
ð39Þ
are given by r 2 NR2 1 r 4 1 r 2 uðr; tÞ 1 1 ¼ 2 1 þ þ Up R R 3 R 2m 4 2 3 2 s1n t e ð1 þ hs1n Þ 1 6 7 2 X 2m 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7 2 5 R n¼1 4 es2n t ð1 þ hs2n Þ þ 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ an ½J0 ðan RÞ J0 ðan rÞ ; ð40Þ J1 ðan RÞ dpðz; tÞ 8lUp NR2 ¼ þ 1 þ dz R2 6m 2
3 2 es1n t ðs1n þ N Þð1 þ hs1n Þ 1 7 2 2lUp X an J0 ðan RÞ 6 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7: 2 5 R n¼1 J1 ðan RÞ 4 es2n t ðs2n þ N Þð1 þ hs2n Þ þ 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ ð41Þ
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3.3. Trapezoidal piston motion In this section we obtain the solution for a piston velocity which vary with time as Up t; for 0 6 t 6 t0 ; t0 ¼ Up ; for t0 6 t 6 t1 ; t2 t ; for t1 6 t 6 t2 ; ¼ Up t2 t1
up ðtÞ ¼
¼ 0;
ð42Þ
for t2 6 t 6 1:
Making use of Heaviside unit step function the piston motion becomes of the following form up ðtÞ ¼
Up Up tH ðtÞ tH ðt t0 Þ þ Up H ðt t0 Þ Up H ðt t1 Þ t0 t0 t2 t t2 t þ Up H ðt t1 Þ Up H ðt t2 Þ: t2 t1 t2 t1
ð43Þ
The solutions for different time periods of velocity and pressure gradient are as follows.
3.3.1. During constant acceleration piston motion (0 6 t 6 t0 ) up ðtÞ ¼
Up tH ðtÞ; t0
ð44Þ
and the results for velocity and pressure gradient are 2
3 r 2 R2 þ fN ðt þ k hÞ þ 1g 7 2t 1 uðr; tÞ 1 6 R 2m 6 7 ¼ 6 7
r 2 4 5 Up t0 4 1 r 1 1 1 þ 4 R 3 R 2 3 es1n t ð1 þ hs1n Þ2 1 6 2 7 2 2m X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 an ½J0 ðan RÞ J0 ðan rÞ þ ; 6 7 5 Rt0 n¼1 4 J1 ðan RÞ es2n t ð1 þ hs2n Þ2 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
ð45Þ
T. Hayat et al. / Appl. Math. Comput. 157 (2004) 103–114
111
dpðz; tÞ qUp 4 8m 2lUp ¼ ð1 þ NtÞ þ 2 ðt þ h kÞ þ dz 3 R t0 t0 R 2 3 2 s1n t ðs1n þ N Þð1 þ hs1n Þ e 1 7 2 2 X an J0 ðan RÞ 6 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 6 7: 2 5 J1 ðan RÞ 4 ðs2n þ N Þð1 þ hs2n Þ es2n t n¼1 þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ
ð46Þ
3.3.2. During the constant velocity period (t0 6 t 6 t1 ) We have up ðtÞ ¼
Up Up tH ðtÞ tH ðt t0 Þ þ Up H ðt t0 Þ; t0 t0
ð47Þ
r 2 NR2 1 r 4 1 r 2 uðr; tÞ 1 1 þ þ ¼ 2 1 Up R R 3 R 2m 4 2 3 ð1 þ hs1n Þ2 fes1n t es1n ðtt0 Þ g 1 6 7 2 2 2m X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 an ½J0 ðan RÞ J0 ðan rÞ ; þ 6 7 2 s t s ðtt Þ t0R n¼1 4 J1 ðan RÞ ð1 þ hs2n Þ fe 2n e 2n 0 g 5 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
ð48Þ 1 dpðz; tÞ 8lUp NR2 2lUp X an J0 ðan RÞ ¼ þ 1 þ dz R2 6m Rt0 n¼1 J1 ðan RÞ 2 3 ð1 þ hs1n Þ2 ðs1n þ N Þðes1n t es1n ðtt0 Þ Þ 6 7 6 s21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 7 6 7: 2 4 ð1 þ hs2n Þ ðs2n þ N Þðes2n t es2n ðtt0 Þ Þ 5 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
ð49Þ
3.3.3. For constant declaration period (t1 6 t 6 t2 ) We obtain up ðtÞ ¼
Up Up tH ðtÞ tH ðt t0 Þ þ Up H ðt t0 Þ Up H ðt t1 Þ t0 t0 t2 t þ Up H ðt t1 Þ; t2 t1
ð50Þ
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uðr;tÞ ¼ Up
r 2 NR2 1 r 4 t2 t 2 1 1 2m 4 t2 t1 R R
2 1 r 2 R 1 f1 þ N ðk hÞg 2mðt2 t1 Þ 3 R r 4 r 2 1 1 1 þ 1 4 R 3 R 2 s1n t 3 ð1 þ hs1n Þ2 e es1n ðtt0 Þ es1n ðtt1 Þ 7 1 6 t0 t2 t1 7 6 s21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 2m X 7 6 þ s2n t 7 2 s ðtt Þ s ðtt Þ R n¼1 6 2n 0 2n 1 5 4 ð1 þ hs2n Þ e e e þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ t0 t2 t1
an ½J0 ðan RÞ J0 ðan rÞ ; J1 ðan RÞ
ð51Þ
dpðz; tÞ 8lUp ðt t2 Þ Up q 4 8m 3ðt t2 Þ ¼ þ ðh kÞ þ þ Nqþ dz R2 ðt2 t1 Þ ðt2 t1 Þ 3 R2 4ðt2 t1 Þ 2 s1n t 3 2 ðs1n þ N Þð1 þ hs1n Þ e es1n ðtt0 Þ es1n ðtt1 Þ 7 1 6 t0 t2 t1 7 6 s21n ½khs21n þ 2ks1n þ 1 þ N ðk hÞ 2Up l X 7 6 7 6 2 R n¼1 4 ðs2n þ N Þð1 þ hs2n Þ es2n t es2n ðtt0 Þ es2n ðtt1 Þ 5 þ 2 t0 t2 t1 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ
an J0 ðan RÞ : J1 ðan RÞ
ð52Þ
3.3.4. After the piston has stopped ðt2 6 t 6 1Þ up ðtÞ is same as described in Eq. (43) and we have the following expressions for velocity and pressure gradient 2
s1n t 3 ð1 þ hs1n Þ2 e es1n ðtt0 Þ es1n ðtt1 Þ es1n ðtt2 Þ 1 6 2 7 2 t0 t2 t1 uðr; tÞ 2m X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 ¼ 6 7 2 s t s ðtt Þ s ðtt Þ s ðtt Þ Up R n¼1 4 ð1 þ hs2n Þ e 2n e 2n 0 e 2n 1 e 2n 2 5 þ 2 s2n ½khs22n þ 2ks2n þ 1 þ N ðk hÞ t0 t2 t1 an ½J0 ðan RÞ J0 ðan rÞ ð53Þ ; J1 ðan RÞ 2
s1n t 3 ðs1n þ N Þð1 þ hs1n Þ2 e es1n ðtt0 Þ es1n ðtt1 Þ es1n ðtt2 Þ 1 6 2 7 2 t0 t2 t1 dpðz;tÞ 2lUp X 6 s1n ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 ¼ 6 s2n t 7 2 s2n ðtt2Þ s2n ðtt0 Þ s2n ðtt1 Þ 5 dz R n¼1 4 ðs2n þ N Þð1 þ hs2n Þ e e e e þ 2 2 s2n ½khs2n þ 2ks2n þ 1 þ N ðk hÞ t0 t2 t1 an J0 ðan RÞ : ð54Þ J1 ðan RÞ
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113
3.4. Oscillatory flow For this case we define up ðtÞ ¼ 0;
for t 6 0;
¼ Up sin wt;
ð55Þ
for t > 0:
The results for velocity and pressure gradient are given by uðr; tÞ i iwt ¼ ½e Gðr; iwÞ eiwt Gðr;iwÞ Up 2
2
3 es1n t ð1 þ hs1n Þ2 1 7 2 2 2mw X an ½J0 ðan RÞ J1 ðan rÞ 6 6 ðs1n þ w2 Þ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 þ 6 7; 2 4 5 R n¼1 J1 ðan RÞ es2n t ð1 þ hs2n Þ þ 2 2 2 ðs2n þ w Þ½khs2n þ 2ks2n þ 1 þ N ðk hÞ
ð56Þ dpðz; tÞ iUp q ¼ ½ðN þ iwÞeiwt CðiwÞ ðN iwÞeiwt CðiwÞ dz 2 2 3 2 es1n t ðs1n þ N Þð1 þ hs1n Þ 1 7 2 2 2lUp w X an J0 ðan RÞ 6 6 ðs1n þ w2 Þ½khs1n þ 2ks1n þ 1 þ N ðk hÞ 7 6 7; 5 R n¼1 J1 ðan RÞ 4 es2n t ðs2n þ N Þð1 þ hs2n Þ2 þ 2 2 2 ðs2n þ w Þ½khs2n þ 2ks2n þ 1 þ N ðk hÞ
ð57Þ where CðsÞ ¼
I0 ðmRÞ 2 I1 ðmRÞ I0 ðmRÞ mR
ð58Þ
and Gðr; sÞ is given by Eq. (25).
4. Concluding remarks We have examined the governing equation for an unsteady, incompressible, electrically conducting, Oldroyd-B fluid in circular duct in a porous medium. Analytical results up to oðN Þ are presented to illustrate the details of the flow and Hall and porosity effects and their depedence on the material parameters. We observe that in steady state the results are identical to that of Newtonian fluid. In the special case, when one of the two times of retardation or relaxation tends to zero, our solutions reduce to those corresponding to a Maxwell fluid or a second grade one. If both times k and h ! 0 the solution corresponding to a Newtonian fluid are obtained. The steady state flows are also obtained as a
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limiting case for t ! 1. The results in absence of Hall current and porosity are obtained by taking N ¼ 0.
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