J. Comp. & Math. Sci. Vol.2 (2), 352-357 (2011)
MHD Flow of a Dusty Fluid MADAN LAL Department of Applied Maths, M.J.P. Rohilkhand University, Bareilly 243 006. India ABSTRACT In this paper, the unsteady flow of a viscous conducting dusty fluid is considered. Making use of Laplace and its inverse transformation, the expressions for velocities of fluid and dust particles are obtained in presence of constant and harmonically oscillating pressure gradient under the action of a uniform magnetic field. Keywords : MHD flow Incompressible fluid, Pressure gradient, Laplace transformation.
1. INTRODUCTION Due to importance of a viscous flows in petroleum industry, in the purification of crude oil, in physiological flows and in other technological fields, various studies have appeared in literature. The dispersion and fall out of pollutants in air or in water has necessitated the study of the flow of dusty fluids. Shaffman8 assumed that the dust particles were of uniform shape and size as well as uniformly distributed then he has formulated the basic equations for the flow of dusty fluid. Since then, many researchers have discussed the problems of dusty fluid. Michael and Miller5, Norey and Michael6, G.C. Chadda1, Krishna3, Purkait7, Kaur4 and Saxena and Sharma9 have studied various problems under different initial and boundary conditions. Recently, Ghosh and Sanyal2 solved the problem of unsteady motion of a dusty incompressible viscous
fluid through a long uniform rectangular tube. In the present chapter, we consider the unsteady motion of a dusty viscous incompressible conducting fluid through a long uniform tube of square cross section under arbitrary pressure gradient in presence of a uniform applied magnetic field perpendicular to the direction of flow. Some particular cases of pressure gradient are also considered. At last, the variations in velocity for constant and harmonically oscillating pressure gradient are shown graphically. 2. FORMULATION AND SOLUTION OF THE PROBLEM If q & V are the velocities of fluid and dust particles respectively, then the equations of motion of a dusty viscous incompressible fluid are
Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
353
Madan Lal, J. Comp. & Math. Sci. Vol.2 (2), 352-357 (2011)
∂q 1 KN J×B (V − q) + + ( q. ∇)q = − ∇p + ν∇ 2 q + ∂t ρ ρ ρ
(2.1)
∂V K + ( V. ∇) V = (q − V) ∂t m
(2.2)
div q = 0
(2.3)
∂N + div( NV) = 0 ∂t
(2.4) (2.5)
is the current density, m is the mass of the dust particle, B is the magnetic field, N is the number density field and v is the kinematic coefficient of viscosity. Let us consider an unsteady flow of an incompressible dusty viscous conducting fluid through a long uniform square tube which is along x-axis whose cross-section is given by the lines y = 0, z = 0, and y = a, z = a (2.6) q = (qx, 0, 0), V = (Vx, 0, 0)
For convenience, let us introduce the following non-dimensional quantities x' =
x a
U=
,
q xa ν
y' =
,
y a
,
u=
Z' =
Vx a ν
z a
,
,
(2.9) (2.10)
where
λ1 =
Where J = σ (E + q x B)
Let
∂u = ( U − u) ∂t ∂P ∂P = =0 ∂y ∂z
λ2
λ2 =
KNa 2 νρ
(2.11)
νm Ka 2
(2.12)
The boundary conditions in the nondimensional forms are U = 0, u=0 at y = 0 (2.13) and U = 0, u=0 at z = 0 The initial condition of the problem is U = 0, u=0 at t < 0 (2.14) ∂ P
Equations (2.10) represents that ∂ x is a function of t only hence we consider
−
∂P = φ( t ) dx
(2.15)
pa 2 ρν
In view of equation (2.15), equation (2.8) takes the form
νt a2 (2.7)
∂U ∂2 U ∂2 U = φ( t ) + 2 + 2 + λ1 ( u − U) − M 2 U ∂t ∂y ∂z (2.16) Eliminating u between (2.9) and (2.16), we have
p' =
t=
Then the equations (2.1) and (2.2) become
∂U ∂P ∂ 2 U ∂2 U =− + + + λ 1 ( u − U) − M 2 U ∂t ∂x ∂y 2 ∂z 2 (2.8)
∂2 U 1 ∂U ∂ 1 + (λ 1 + + M2 ) − + ∂t 2 λ2 ∂t ∂t λ 2
FG H
2
2
IJ FG ∂ U + ∂ U IJ K H ∂y ∂z K
Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
2
2
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Madan Lal, J. Comp. & Math. Sci. Vol.2 (2), 352-357 (2011)
+M2U −
FG ∂ + 1 IJ φ( t ) = 0 H ∂t λ K
+ M 2 D r ,s − S r ,s
2
2
(2.17) Consider the solution of above equation satisfying the boundary condition (2.13) as below ∞
where
S r ,s =
∞
U ( y , z, t ) = ∑
∑
r =0
s= 0
(2.18)
and
=
(r + ½) π
=
(s + ½)π
(2.19)
Dr,s(t) = 0
for t < 0
(2.20)
FG H
∂D r , s ∂ 1 − + ∂t ∂t λ 2
IJ K
FG H
IJ R K
r ,s
R r ,s = l2r + ls2
,
p 2 D r ,s + p ( λ 1 +
1 1 + M 2 ) D r ,s + ( p + ) R r ,s D r ,s + M 2 D r ,s λ2 λ2
−Sr ,s ( p + 1 / λ 2 ) φ ( p) = 0
(2.22)
Where p is the parameter of Lapace transformation, on simplifying (2.22)
putting the value of U from equation (2.18) in equation (2.17) and then multiplying by cos lr y cos ls z on both sides and integrating w.r.t. y and z from y = 0 to 1 and z = 0 to 1, we get ∂ 2 D r ,s 1 + λ1 + + M2 ∂t 2 λ2
16 l r ls
1 λ2
FG H
IJ φ(p) K F I F 1I 1 + pG λ + H λ + M JK + R GH p + λ JK + M S r ,s p +
D r ,s ( p ) = p2
2
1
b
S r ,s (λ 3 − λ 4
zg LMMNFGH t
0
2
(2.23)
D r ,s
1 1 − λ 4 e − λ 4λ − − λ3 λ2 λ2
IJ K
2
r ,s
2
Taking Inverse Laplace transform, equation becomes D r ,s ( t ) =
(2.21)
Taking Laplace transform on both sides of equation (2.21)
D r ,s ( t ) cos lr y cos l s z
where
lr ls
FG ∂ + 1 IJ φ( t ) = 0 H ∂t λ K
FG H
IJ K
e λ 3λ
OP φ ( t − λ ) dλ PQ
above
(2.24)
Where λ3 and λ4 are the roots of the equation
FG H
x 2 − λ1 +
1 + M 2 + R r ,s λ2
IJ x + 1 R K λ
r ,s
+ M2 = 0
2
(2.25)
Using equation (2.24) into equation (2.18), we have ∞
∞
U( y, z, t ) = ∑
∑
r =0
s= 0
LMF 1 − λ I e − F 1 − λ I e bλ − λ g z MNGH λ JK GH λ JK S r ,s
3
t
4
0
4
2
λ 4λ
3
2
λ 3λ
OP PQ
φ( t − λ)dλ cos l r y cos ls z Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
(2.26)
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Madan Lal, J. Comp. & Math. Sci. Vol.2 (2), 352-357 (2011)
The velocity of the dust particle is obtained by putting above value of U(y,z,t) in equation (2.16) ∞
U(y, z, t) = ∑ r =0
LMF 1 − λ I e − F 1 − λ I e F I 1 ∂ 1 1 + − R ∑ (λ − λ ) GH λ ∂t λ JK z MGH λ JK GH λ JK N ∞
Sr ,s
t
r ,s
s=0
3
4
1
4
0
1
λ 4λ
3
2
λ 3λ
2
φ( t − λ)dλ cos l r y cos ls z
OP PQ
(2.27)
Remark 1 : If the pressure gradient is constant : In this case, let us consider φ(t) = C H(t) where C is constant and H(t) is the Heaviside unit step function. Using above retation in equation (2.26) and (2.27), we have ∞
∞
U(y, z, t) = C∑
∑
r =0
s=0
LMR 1 F 1 − λ Ic1− e h − 1 F 1 − λ Ic1− e G JK λ Hλ MNSTλ GH λ JK
Sr ,s (λ3 − λ4 )
λ4 t
4
4
3
2
3
2
cos l r y cos ls z ∞
∞
and u( y, z, t ) = C∑
∑
r =0
s= 0
−
RS 1 F1− 1 R Tλ GH λ 3
S r ,s (λ 3 − λ 4 )
IJc1− e h + 1 e λ K λ 3t
r ,s
1
1
λ 3t
λ3t
hUVOPP WQ (2.28)
LMR 1 F1 − R I c1 − e h + 1 e Ub1 / λ VW λ MNST λ GH λ JK λ4t
r ,s
4
λ4t
1
2
UVF 1 − λ I OP cosl y cosl z WGH λ JK PQ 3
− λ4
g
1
r
s
2
(2.29)
If the fluid is dust free (i.e.) λ1 = 0 Then equation (2.28) gives ∞
∞
U( y , z, t ) = C ∑
∑
r =0
s= 0
S r ,s ' ' e λ 3 t − e λ 4 t cos l r y cos l s z ' (λ − λ 4 ) ' 3
e
j
where λ and λ are the roots of equation ' 3
' 4
x2 – (Rr,s + M2) x + M2 = 0 Remark 2 : If the pressure gradient is harmonically oscillating : Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
(2.30)
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Madan Lal, J. Comp. & Math. Sci. Vol.2 (2), 352-357 (2011)
In this case, we consider φ(t) = C0 cos ωt where C0 is constant then the equation (2.26) and (2.27) become ∞
LM FG 1 − λ IJ cλ S MM H λ K (λ − λ ) MMN
∞
U( y , z , t ) = C 0 ∑
∑
r =0
2
r ,s
s= 0
3
3
3
3
4
cos ωt + ω sin ωt − λ 4 e λ 4 t
2
λ3t
2 3
hOP cos l y cos l z PQ r
s
(2.31)
LM FG 1 − λ IJ S MM H λ K R|SFG λ − λ R + ω IJ and u( y, z, t ) = C ∑ ∑ λ K ( λ − λ ) ω + λ |H T λ MMN F ω λ ω IJ sin ωt − FG λ − λ R − λ IJ e U|V +G ω − R − λ K λ K H λ H λ |W FG 1 − λ IJ H λ K RSFG λ − λ R + ω IJ cos ωt + FG ω − ωR − λ ω IJ sin ωt − (ω + λ TH λ λ K H λ λ K 0
∞
∞
r =0
s= 0
3
4
4
1
3
2 3
3
3
1
r ,s
cos ωt
1
λ4t
1
2
r ,s
3
r ,s
1
F λR −G (λ − H λ
2 4
1
3
r ,s
1
r ,s
1
3
2 2
2
4
4
2 4
4
4
r ,s
4
2 2
r ,s
h
ω 2 + λ 24
4
F 1 I cλ cosωt + ω sin ωt − λ e −G − λ J ω +λ Hλ K 2
4
λ 23 − λ1
1
IJ e K
λ3 t
1
1
U|OP cos l y cos l z V|P WQ r
s
(2.32)
It the fluid is dust free (i.e.) λ1 = 0, then equation (2.31) reduces to ∞
∞
U( y, z, t ) = ∑
∑
r =0
s= 0
e
LM MN
2
2
S r ,s λ 3' cos ωt + λ 3' ω sin ωt − λ 3' e λ 3 t ) 2 (λ 3 − λ 4 ) (ω 2 + λ 3' )
2
2
λ'4 cosωt + λ'4ω sin ωt − λ'4 eλ4t 2
(ω2 + λ'4
'
'
jOP cosl y cosl z PQ r
s
Journal of Computer and Mathematical Sciences Vol. 2, Issue 2, 30 April, 2011 Pages (170-398)
(2.33)
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Madan Lal, J. Comp. & Math. Sci. Vol.2 (2), 352-357 (2011)
Fig. 1
Fig. 2
3. CONCLUSION The velocity distributions in both the cases (when the fluid is dusty and non nondusty) are given in equations (2.28), (2.30) and (2.31), (2.33). It we assume λ1 = 1, λ2 = 1.8, ω = .5 and z = .5, then from the following figures it can be observed that the dust particles in the fluid decreases its velocity in both the cases and also the velocity increases with increasing t and is maximum at the centre of the tube. REFERENCES 1. Chadda, G.C. Ph.D. thesis, University, Agra, , 75 (1977).
Agra
2. Ghosh, S. and Sanyal, D.C. Jour. Indian Acad. Math., 16, 181 (1994). (1994) 3. Krishna, B. Ph.D. thesis, Agra University, Agra, 122 (1982). (1982) 4. Kaur, J. Ph.D. thesis, Agra University, Agra, 111 (1985). 5. Michael, D.H. and Miller, D.A. Mathematica, 13, 97 (1966). 6. Michael, D.H. and Norey, ey, P.W. Jour. Mech. Appl. Math. 21, 375 (1968). (1968) 7. Purkait, P.K. Acta Ciencia Indica, Indica 3(3), 264 (1977). 8. Shaffman, P.G. Jour. Fluid Mech. Mech 13, 120 (1962). 9. Saxena, S. and Sharma, G.C. Ind. Jour. Pure and Appl. Math., ., 18(12), 18(12) 1131 (1987).
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