Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2010, Article ID 898720, 17 pages
doi:10.1155/2010/898720
Research Article
Unsteady Flow and Heat Transfer of a Dusty Fluid
through a Rectangular Channel
B. J. Gireesha, G. S. Roopa, and C. S. Bagewadi
Department of Studies and Research in Mathematics, Kuvempu University, Shankaraghatta, Shimoga,
Karnataka 577 451, India
Correspondence should be addressed to B. J. Gireesha, bjgireesu@rediffmail.com
Received 3 August 2010; Accepted 30 December 2010
Academic Editor: J. Jiang
Copyright q 2010 B. J. Gireesha et al. This is an open access article distributed under the Creative
Commons Attribution License, which permits unrestricted use, distribution, and reproduction in
any medium, provided the original work is properly cited.
The present discussion deals with the study of an unsteady flow and heat transfer of a dusty
fluid through a rectangular channel under the influence of pulsatile pressure gradient along with
the effect of a uniform magnetic field. The analytical solutions of the problem are obtained using
variable separable and Fourier transform techniques. The graphs are drawn for the velocity fields
of both fluid and dust phases under the effect of Reynolds number. Further, changes in the Nusselt
number are shown graphically, and, on the basis of these, the conclusions and discussions are
given.
1. Introduction
The concept of an unsteady flow and heat transfer of a dusty fluid has a wide range of
applications in refrigeration, air conditioning, space heating, power generation, chemical
processing, pumps, accelerators, nuclear reactors, filtration and geothermal systems, and so
forth. One common example of heat transfer is the radiator in a car, in which the hot radiator
fluid is cooled by the flow of air over the radiator surface. On this basis many mathematicians
were attracted by this field.
Saffman 1 has formulated the governing equations for the flow of dusty fluid and
has discussed the stability of the laminar flow of a dusty gas in which the dust particles are
uniformly distributed. Datta et al. 2 have obtained the solution of unsteady heat transfer to
pulsatile flow of a dusty viscous incompressible fluid in a channel. Heat transfer in unsteady
laminar flow through a channel was analyzed by Ariel 3. Ghosh et al. 4 have made the
solution for hall effects on MHD flow in a rotating system with heat transfer characteristics.
Ezzat et al. 5 analyzed a space approach to the hydromagnetic flow of a dusty fluid through
a porous medium.
2
Mathematical Problems in Engineering
Some researchers like Anjali Devi and Jothimani 6 have discussed the heat transfer
in unsteady MHD oscillatory flow. Further, Malashetty et al. 7 have investigated the
convective magnetohydrodynamic two phase flow and heat transfer of a fluid in an inclined
channel. Palani and Ganesan 8 have discussed the heat transfer effects on dusty gas flow
past a semi-infinite inclined plate. Attia 9 has investigated an unsteady MHD Couette flow
and heat transfer of dusty fluid with variable physical properties.
Unsteady hydromagnetic flow and heat transfer from a nonisothermal stretching
sheet immersed in a porous medium was discussed by Chamkha 10. Mishra et al. 11
have studied the two-dimensional transient conduction and radiation heat transfer with
temperature-dependent thermal conductivity. MHD flow and heat transfer of a dusty viscoelastic stratified fluid down an inclined channel in porous medium under variable viscosity
was analyzed by Chakraborty 12. Shawky 13 has investigated the solution for pulsatile
flow with heat transfer of dusty magnetohydrodynamic Ree-Eyring fluid through a channel.
Gireesha et al. 14 have obtained the analytical solutions for velocity fields using
variable separable method for an unsteady flow of dusty fluid through a rectangular channel
under the influence of pulsatile pressure gradients and in the absence of a magnetic field.
In continuation of this paper and with the help of the above cited papers we have studied
an unsteady flow and heat transport in a dusty fluid through a rectangular channel under
the influence of a pulsatile pressure gradient in the presence of uniform magnetic field
and viscous dissipation term. Further, heat transfer analysis and the effect of Reynolds
number, Prandtl number, and Nusselt number have been considered. This paper presents
three methods of solution, namely, perturbation technique, Fourier decomposition, and
finite Fourier transform, to obtain useful results on the problem. Finally, the graphical
representation of velocity fields of both fluid and dust phases and changes in the Nusselt
number are drawn for different values of Reynolds number and Prandtl number.
2. Equations of Motion
The governing equations of motion and energy for two phases are given by 1 the following.
For fluid phase,
−
∇ ·→
u 0,
−
−
σB02→
f →
1
∂→
u →
−
−
−
−
−
u ·∇ →
u − ∇p ν∇2→
u
v −→
u −
u,
∂t
ρ
τv
ρ
− →
∂E →
−
u · ∇E
Q →
v −−
u · F k∇ · ∇T Φf .
ρ
∂t
2.1
For dust phase,
−
∇ ·→
v 0,
−
−
1 →
∂→
v →
−
−
v
v ·∇ →
u −→
v ,
−
∂t
τv
∂Ep →
−
−Q − Φd .
N
v · ∇Ep
∂t
2.2
Mathematical Problems in Engineering
3
We have following nomenclature. E cp T , Ep cm Tp , Q Ncp Tp − T /τT is the thermal
−
−
interaction between fluid and dust particle phases, F N→
v −→
u/τv is the velocity interaction
force between the fluid and dust particle phase, τv m/6πaμ m/K is the velocity
relaxation time of the dust particles, τT mcp /4πak is the thermal relaxation time of the
dust particles, k∇. ∇T is the rate of heat added to the fluid by conduction in unit volume,
−
u, ρ, p, ν, T , cp , and
Φf and Φd are the viscous dissipation of fluid and dust particles. →
k are, respectively, the velocity vector, density, pressure, kinematic viscosity, temperature,
−
specific heat, and thermal conductivity of the fluid, →
v , N, Tp , cm , and m are, respectively,
the velocity vector, number density, temperature, specific heat, mass concentration of dust
particles, K 6πaμ is the Stoke’s resistance coefficient, and t is the time.
3. Formulation of the Problem
Consider an unsteady flow of an incompressible, viscous, electrically conducting fluid with
uniform distribution of dust particles through a rectangular channel. It is assumed that the
flow is due to the time-dependent pressure gradient and applied uniform magnetic field.
Both the fluid and the dust particle clouds are supposed to be static at the beginning. The
dust particles are assumed to be spherical in shape and uniform in size. The number density
of the dust particles is taken as a constant throughout the flow. The flow is taken along z-axis,
and it is as shown in Figure 1. For the above described flow the velocities of both fluid and
dust particles are given by
→
−
→
−
3.1
u u x, y, t k,
v v x, y, t k.
4. Solution of the Problem
The governing equations from 2.1 and 2.2 can be decomposed as follows.
For fluid phase,
σB02
f
u,
v − u −
τv
ρ
Ncp N
∂T
∂2 T ∂2 T
2
ρcp
Tp − T v − u k
∂t
τT
τv
∂x2 ∂y2
2 2 ∂u
4 ∂u
.
μ
3 ∂x
∂y
1 ∂p
∂u
−
ν
∂t
ρ ∂z
∂2 u ∂2 u
∂x2 ∂y2
4.1
For dust phase,
∂v
1
u − v,
∂t
τv
2 ∂Tp cp μp 4 ∂v 2
∂v
,
cm
T − Tp −
∂t
τT
N 3 ∂x
∂y
where u and v denote the velocity of the fluid and the dust phases, respectively.
4.2
4
Mathematical Problems in Engineering
y
x=d
x = −d
y=h
x
z
y=0
Figure 1: Schematic diagram of dusty fluid flow in a rectangular channel.
The boundary conditions of the given problem are taken as
u 0,
v 0,
T T0
at y 0,
u 0,
v 0,
T T1
at y h,
u 0,
v 0,
u 0,
v 0,
4.3
at x −d,
T T2
T T3
at x d.
Since we have assumed that the pulsatile pressure gradient has influence on the flow, we
have that
−
1 ∂p
A 1 eiωt ,
ρ ∂z
4.4
where is a small quantity and A and ω are constants.
To make the above system dimensionless, introduce the following nondimensional
variables:
u
ξ
Pr x
,
h
μcp
,
k
uω
,
A
η
v
y
,
h
Re vω
,
A
ψ
ωh2
,
ν
z
,
h
t tω,
p
Rep θ
p
,
Aρh
ωh2
,
νp
T − T0
,
T1 − T0
θp Ec Tp − T0
,
T1 − T0
ω2 c
4.5
A2
,
p T1 − T0 where h is the distance between plates, Ec the Eckert number, Pr the Prandtl number, Re and
Rep the Reynolds numbers of fluid and dust phases, x and y the space coordinates along and
perpendicular to the plates, θ and θp the dimensionless fluid and dust phase temperatures, μ
the viscosity of fluid, and νp and ρp the kinematic viscosity and density of the dust particles.
Mathematical Problems in Engineering
5
Using the above nondimensional variables in 4.1 and 4.2 and dropping the bars,
one can get that
∂p
1 ∂2 u ∂2 u
∂u
fαv − u − βu,
−
∂t
∂ψ Re ∂ξ2 ∂η2
∂θ
∂2 θ ∂2 θ
1
α1 β1 θp − θ αβ1 Ecv − u2
∂t
Re Pr ∂ξ2 ∂η2
2 Ec 4 ∂u 2
∂u
,
Re 3 ∂ξ
∂η
4.6
∂v
αu − v,
∂t
2 ∂θp
γEc 4 ∂v 2
∂v
,
γα1 θ − θp −
∂t
Rep β2 3 ∂ξ
∂η
where α 1/ωτv , β σB02 /ρω, α1 1/ωτT , β1 N/ρ, β2 N/ρp , and γ cp /cm .
The dimensionless boundary conditions are
u 0,
v 0,
θ 0 at η 0,
u 0,
v 0,
θ 1 at η 1,
u 0,
u 0,
v 0,
v 0,
θ Ta
at ξ −r,
θ Tb
at ξ r,
4.7
where Ta T2 − T0 /T1 − T0 , Tb T3 − T0 /T1 − T0 , and r d/h.
The nondimensional form of pressure gradient is given by
−
∂p 1 eit .
∂ψ
4.8
Now assume the solutions of the velocities and temperature of both fluid and dust phases as
u ξ, η, t u0 ξ, η u1 ξ, η eit ,
v ξ, η, t v0 ξ, η v1 ξ, η eit ,
θ ξ, η, t θ0 ξ, η θ1 ξ, η eit 2 θ2 ξ, η e2it ,
θp ξ, η, t θp0 ξ, η θp1 ξ, η eit 2 θp2 ξ, η e2it .
4.9
Substituting 4.8 and 4.9 in 4.6 and equating the coefficient of the similar powers
of on both sides, then we obtain the following set of equations.
6
Mathematical Problems in Engineering
Steady part coefficient of 0 :
1 ∂2 u0 ∂2 u0
fαv0 − u0 − βu0 −1,
Re ∂ξ2
∂η2
4.10
αu0 − v0 0,
1
Re Pr
4.11
∂2 θ0 ∂2 θ0
Ec 4 ∂u0 2
∂u0 2
2
α
0,
β
−
θ
Ecv
−
u
θ
αβ
1 1
p0
0
1
0
0
Re 3 ∂ξ
∂η
∂ξ2
∂η2
γEc 4 ∂v0 2
∂v0 2
γα1 θ0 − θp0 −
0.
Rep β2 3 ∂ξ
∂η
4.12
Unsteady part coefficient of :
1 ∂2 u1 ∂2 u1
fαv1 − u1 − β i u1 −1,
2
2
Re ∂ξ
∂η
4.13
αu1 − v1 − iv1 0,
∂2 θ1 ∂2 θ1
1
α1 β1 θp1 − θ1 2αβ1 Ecv0 − u0 v1 − u1 Re Pr ∂ξ2
∂η2
2Ec 4 ∂u0 ∂u1 ∂u0 ∂u1
− iθ1 0,
Re 3 ∂ξ ∂ξ
∂η ∂η
2γEc 4 ∂v0 ∂v1 ∂v0 ∂v1
− iθp1 0.
γα1 θ1 − θp1 −
Rep β2 3 ∂ξ ∂ξ
∂η ∂η
4.14
4.15
Unsteady part coefficient of 2 :
1
∂2 θ2 ∂2 θ2
α1 β1 θp2 − θ2 αβ1 Ecv1 − u1 2
2
2
Re Pr ∂ξ
∂η
Ec 4 ∂u1 2
∂u1 2
− 2iθ2 0,
Re 3 ∂ξ
∂η
γEc 4 ∂v1 2
∂v1 2
− 2iθp2 0.
γα1 θ2 − θp2 −
Rep β2 3 ∂ξ
∂η
4.16
4.17
Mathematical Problems in Engineering
7
Now, the corresponding dimensionless boundary conditions are as follows:
u0 0,
v0 0,
θ0 Ta
at ξ −r,
u0 0,
v0 0,
θ0 Tb
at ξ r,
u0 0,
v0 0,
θ0 0
at η 0,
u0 0,
v0 0,
θ0 1
at η 1;
u1 0,
v1 0,
θ1 0 at ξ −r,
u1 0,
v1 0,
θ1 0
at ξ r,
u1 0,
v1 0,
θ1 0 at η 0,
u1 0,
v1 0,
θ1 0 at η 1;
4.18
4.19
θ2 0 at ξ −r,
θ2 0
at ξ r,
θ2 0 at η 0,
4.20
θ2 0 at η 1.
By substituting 4.11 in 4.10, one can get
∂2 u0 ∂2 u0
− β Re u0 − Re.
∂ξ2
∂η2
4.21
To solve 4.21, we assume that the solution is in the form
u0 ξ, η X ξ, η Y ξ.
4.22
Substituting u0 ξ, η in 4.21, then we obtain that
∂2 X ∂2 Y ∂2 X
− β ReX Y − Re.
∂ξ2
∂η2
∂η2
4.23
∂2 Y
− β Re Y Re 0,
∂η2
4.24
∂2 X ∂2 X
− β Re X 0.
∂ξ2
∂η2
4.25
so that
8
Mathematical Problems in Engineering
The corresponding boundary conditions will become
u0 −r, η X −r, η Y −r 0,
u0 ξ, 0 Xξ, 0 Y ξ 0,
u0 r, η X r, η Y r 0,
u0 ξ, 1 Xξ, 0 Y ξ 0.
4.26
By solving 4.24 and using the method of separation of variables to 4.25, we obtain the
solution in the form
⎡
⎤
∞
β Re ξ
cosh
sinh A1 η − 1 − sinh A1 η
1⎢
nπ
⎥ 2
ξ
sin
u0 ξ, η ⎣1 −
⎦ β
β n1
r
sinh A1
cosh
β Re r
⎧
⎫
⎨ 1
⎬
β Re
nπ
n
×
−
,
−
−1
⎩ nπ r 2 A2 cosh
⎭
nπA21
β
Re
r
1
v0 ξ, η u0 ξ, η ,
4.27
%
where A1 β Re r 2 n2 π 2 /r 2 .
Here one can observe that the velocity of steady part of the fluid and the dust phases
is the same.
In a similar manner, by the method of separation of variables, the solution of 4.13,
and using the boundary conditions 4.19, one can obtain that
∞
sinh B η − 1 − sinh Bη
2 Re coshQξ
nπ
Re
ξ
2
sin
u1 ξ, η 2 1 −
coshQr
r
sinh B
Q
Q n1
1
nπ
Q2
×
− 2 2
−
−1n ,
nπ r B coshQr nπB2
α v1 u1 ,
αi
4.28
where Q2 Refαi/α i β i and B Q2 r 2 n2 π 2 /r 2 .
We define the finite Fourier sine transform of θ and θp as
&r
nπ
Fs θ ξ ,
θ ξ, η sin
r
−r
F s θp &r
nπ
ξ .
θp ξ, η sin
r
−r
4.29
Eliminating θp0 from 4.12, we get that
4 ∂u0 2
∂2 θ0 ∂2 θ0
∂u0 2
,
H1
3 ∂ξ
∂η
∂ξ2
∂η2
4 ∂v0 2
Ec
∂v0 2
θp0 ξ, η θ0 −
,
Rep α1 β2 3 ∂ξ
∂η
where H1 Ec Re Pr β1 /Rep β2 − 1/ Re.
4.30
4.31
Mathematical Problems in Engineering
9
Applying the finite fourier sine transform to 4.30 with respect to the variable ξ and
to boundary conditions, one obtains that
∞
d2 Fs θ0 n2 π 2
nπ 4H1 8π 2 β Re
n
− 2 Fs θ0 Tb − Ta −1 r
3
dη2
r
β2
n1
β
Re
r
n
×
2 4n2 π 2
β
Re
r
cosh
β Re r
n1
∞
2
sinh
⎞
⎛
4.32
β Re −1n ⎟
nπ
⎜ 1
−
×⎝
−
⎠
nπ r 2 A2 cosh
nπA21
β Re r
1
sinh A1 η − 1 − sinh A1 η
,
×
sinh A1
Fs θ0 ξ, 0 0,
Fs θ0 ξ, 1 0.
4.33
The temperature of fluid is obtained by solving 4.32 with the help of boundary conditions
4.33 as
⎡
∞
∞
32π 2 β ReH1 n2
⎢ r n
Fs θ0 ⎣
Tb − Ta −1 nπ
3β2
β Re r 2 4n2 π 2
n1
n1
⎞⎤
⎛
n
sinh
β Re r
β
Re
−1
nπ
r2
1
⎟⎥
⎜
−
× 2
⎝
−
⎠⎦
2
2
nπ
2
A1 r 2 − n2 π 2 cosh
nπA
β Re r
r A1 cosh
β Re r
1
nπ 1 − cosh nπ/rη
nπ
η sinh
η
× cosh
r
sinhnπ/r
r
∞
∞
32π 2 β ReH1 r n2
n
−
Tb − Ta −1 nπ
3β2
β Re r 2 4n2 π 2
n1
n1
⎞
⎛
n
β
Re
r
β Re −1 ⎟
nπ
r
⎜ 1
−
× 2
⎝
−
⎠
2
2
2
nπ r 2 A2 cosh
A1 r − n π cosh
nπA21
β Re r
β Re r
1
2
sinh
sinh A1 η − 1 − sinh A1 η
.
×
sinh A1
4.34
10
Mathematical Problems in Engineering
Now taking the inverse finite Fourier sine transform to 4.34, one can obtain that
⎧⎡
⎪ ∞
∞
2 ⎨⎢
32π 2 β Re H1
r n
θ0 ξ, η Tb − Ta −1 ⎣
r⎪
nπ
3β2
r1 ⎩ n1
sinh
β Re r
n2
r2
×
β Re r 2 4n2 π 2 A21 r 2 − n2 π 2 cosh β Re r
n1
∞
⎞⎤
n
β
Re
−1
nπ
1
⎟⎥
⎜
−
×⎝
−
⎠⎦
nπ r 2 A2 cosh β Re r
nπA21
1
⎛
1 − cosh nπ/rη
nπ
nπ
η sinh
η
× cosh
r
sinhnπ/r
r
∞
∞
32π 2 β Re H1 r n2
n
−
Tb − Ta −1 2
2
nπ
3β
β Re r 4n2 π 2
n1
n1
sinh
β
Re
r
2
sinh A1 η − 1 − sinh A1 η
r
× 2
sinh A1
A1 r 2 − n2 π 2 cosh
β Re r
4.35
⎞⎫
⎪ β Re −1 ⎟⎬
nπ
nπ
⎜ 1
−
ξ .
×⎝
−
⎠ sin
⎪
nπ r 2 A2 cosh β Re r
r
nπA21
⎭
1
⎛
n
The temperature of dust θp0 is obtained by substituting θ0 in 4.31.
Using 4.15 in 4.14 and boundary conditions 4.19, with the help of finite fourier
sine transform technique, one can get the solution for θ1 as
θ1 ξ, η
∞
∞
2 16 Re π 2 H3 1 n2
1
sinhQr
2
2
2
2
2
2
r
3β
Q n1 A1 − q1 Q r 4n π coshQr
r1
⎛
⎞
n
β Re −1 ⎟
nπ
⎜ 1
×⎝
−
−
⎠
nπ r 2 A2 cosh β Re r
nπA21
1
∞
β Re 1
n2
2
2
Q n1 B − q12 β Re r 2 4n2 π 2
nπ
Q2 −1n
1
−
−
×
nπ Q2 r 2 n2 π 2 coshQr
nπB2
1 − cos hq1
× cosh q1 η sinh q1 η
sinh q1
β Re r
cosh β Re r
sinh
Mathematical Problems in Engineering
∞
n2
1
16 Re π 2 H3 1 3β
Q n1 A21 − q12 Q2 r 2 4n2 π 2
⎞
⎛
n
β Re −1 ⎟ sinhQr
nπ
⎜ 1
−
×⎝
−
⎠
nπ r 2 A2 cosh β Re r
coshQr
nπA21
11
1
%
∞
β Re sinh A1 η−1 −sinh A1 η
1
n2
×
sinh A1
Q2 n1 B2 −q12 β Re r 2 4n2 π 2
nπ
Q2 −1n sinh β Re r
1
−
−
×
nπ Q2 r 2 n2 π 2 coshQr
nπB2
cosh β Re r
sinh B η − 1 − sinh Bη
nπ
ξ .
sin
×
sinh B
r
4.36
Using θ1 , we get the expression for θp1 as
θp1 ξ, η 2γEc
γα1
4 ∂v0 ∂v1 ∂v0 ∂v1
θ1 −
,
γα1 i
∂η ∂η
Rep β2 γα1 i 3 ∂ξ ∂ξ
4.37
where
.
q1 H2 α1 β1 i
i ,
γα1 i
α1 β1 γ
α
1
.
−
Rep β2 γα1 i α i Re
n2 π 2
,
r2
H3 2 Ec Re P r
H2 Re Pr
4.38
Similarly, the solutions of 4.16 and 4.17 using boundary conditions 4.20 are
obtained as
θ2 ξ, η ∞
2
r1
r
∞ ∞
1
8Re2 H5 4π 2 H6 n2
1
r−
3
2
2
2
2
2
Q n1
Q
3 n1 Q r 4n π B − q22
Q
1
nπ
Q2 −1n
1
sinhQr
−
−
×
coshQr nπ Q2 r 2 n2 π 2 coshQr
nπB2
sinh B η − 1 − sinh Bη
1 − cosh q2
sinh q2 η cosh q2 η
sinh q2
sinh B
nπ
ξ .
× sin
r
4.39
Mathematical Problems in Engineering
10
10
5
5
u0 (x, h)
u0 (x, h)
12
0
−5
0
−5
−10
−10
300
200
h
400
100
200
0 0
600
300
200
h
x
a
400
100
0 0
200
600
x
b
Figure 2: Variation of fluid velocity u0 ξ, η with ξ and η for Re 1 and Re 10.
From 4.17, one can get that
θp2 ξ, η γα1
γEc
4 ∂v1 2
∂v1 2
,
θ2 −
γα1 2i
∂η
Rep β2 γα1 2i 3 ∂ξ
4.40
where
.
q2 H5 −
H4 Ec Re P r αβ1
α i2
2iα1 β1
2i ,
γα1 2i
α 2
α1 β1 γ
1
.
H6 Ec Re Pr
−
Re
Rep β2 γα1 2i α i
n2 π 2
,
r2
H4 Re Pr
,
4.41
5. Results and Discussion
Figures 2, 3, 4, 5, 6, 7, and 8 represent the velocity and temperature fields, respectively, for
the fluid and dust particles, which are parabolic in nature. Here we can see that the path of
fluid particles is much steeper than that of dust particles. Further, one can see that if the dust
is very fine, that is, the mass of the dust particles is negligibly small, then the relaxation time
of dust particle decreases and ultimately as τv → 0 the velocities of fluid and dust particles
will be the same. Also we see that the fluid particles will reach the steady state earlier than
the dust particles. Further, one can observe the impressive effect of Reynolds number on the
velocity fields. It means that the Reynolds number is favorable to the velocity fields, that
is, the velocity profiles for both fluid and dust particles increases as the Reynolds number
increases.
The graphs are drawn for the following values: ω 0.5, N 0.4, β 2, σ 1, Ec 0.02,
Pr 0.72, T0 0.5, T1 1, T2 1.5, T3 2, τv τT 0.15, and γ 1.4.
Mathematical Problems in Engineering
13
2
5
1
v0 (x, h)
v0 (x, h)
×10 4
10
0
−5
0
−1
−10
−2
300
200
h
400
100
0 0
200
300
600
200
h
x
400
100
0 0
a
200
600
x
b
5
800
4
600
u1 (x, h)
u1 (x, h)
Figure 3: Variation of dust velocity v0 ξ, η with ξ and η for Re 1 and Re 10.
3
2
400
200
1
0
300
200
400
100
h
0 0
200
0
300
600
200
h
x
400
100
0 0
a
200
600
x
b
5
800
4
600
v1 (x, h)
v1 (x, h)
Figure 4: Variation of fluid velocity u1 ξ, η with ξ and η for Re 1 and Re 10.
3
2
400
200
1
0
300
200
h
400
100
0 0
a
200
x
600
0
300
200
h
400
100
0 0
200
x
b
Figure 5: Variation of dust velocity v1 ξ, η with ξ and η for Re 1 and Re 10.
600
14
Mathematical Problems in Engineering
6
4
Re = 1
5
Re = 1
Re = 3
4
Re = 3
3
3
2
Re = 5
Re = 5
Nu0
Nu0
5
1
−1
1
Re = 1
Re = 3
Re = 5
0
0
0.2
0.4
2
Re = 1
Re = 3
Re = 5
0
0.6
0.8
−1
1
0
0.2
0.4
0.6
ξ, η
0.8
1
ξ, η
Varies with ξ
Varies with η
Varies with ξ
Varies with η
a
b
Figure 6: Steady part of the Nusselt number Nu0 versus ξ and η for ξ r and η 0 and ξ −r and
η 1.
×10−3
×10−3
6
6
Pr = 3
4
Pr = 2
3
Pr = 3
2
Pr = 1
1
Pr = 2
0
−1
Re = 5
5
Amplitude of Nu1
Amplitude of Nu1
5
4
Re = 3
3
2
Re = 5
Re = 3
Re = 1
1
0
Pr = 1
0
0.2
0.4
0.6
0.8
1
−1
Re = 1
0
0.2
0.4
ξ, η
Varies with ξ
Varies with η
a
0.6
0.8
1
ξ, η
Varies with ξ
Varies with η
b
Figure 7: Unsteady part of Nusselt number Nu1 versus ξ and η for ξ r and η 0 and ξ −r and
η 1.
Now we discuss the heat transfer at the vertical walls, so we consider the Nusselt
number Nu of the fluid as
/
∂θ //
Nu − /
∂ξ at ξr or ξ−r
dθ1
dθ2
dθ0
5.1
eit
2 e2it
−
dξ
dξ
dξ at ξr or ξ−r
− Nu0 eit Nu1 2 e2it Nu2
.
at ξr or ξ−r
Mathematical Problems in Engineering
15
×10−4
Re = 1
Re = 1
Re = 3
Re = 3
Amplitude of Nu2
Amplitude of Nu2
×10−4
Re = 5
1
Re = 1
Re = 3
Re = 5
1
Re = 1
Re = 3
Re = 5
0
0
0.2
0.4
Re = 5
0.6
0.8
1
0
0
0.2
0.4
ξ, η
0.6
0.8
1
ξ, η
Varies with ξ
Varies with η
Varies with ξ
Varies with η
a
b
Figure 8: Unsteady part of Nusselt number Nu2 versus ξ and η for ξ r and η 0 and ξ −r and
η 1.
Next to discuss is the heat transfer at the horizontal walls, so we consider the Nusselt
number Nu of the fluid as
/
∂θ //
Nu − /
∂η at η0 or η1
dθ1
dθ2
dθ0
eit
2 e2it
−
dη
dη
dη
− Nu0 eit Nu1 2 e2it Nu2
5.2
at η0 or η1
at η0 or η1
,
where Nu0 , Nu1 , and Nu2 denote the Nusselt number for steady part, unsteady part for
coefficient of , and unsteady part for coefficient of 2 , respectively.
The graphs of steady part of Nusselt number Nu0 against ξ and η at η 0 and ξ r
or at η 1 and ξ −r has been drawn in Figure 6. It shows that for different values of
Prandtl number, Nusselt number increases with increase in ξ and η.
Figure 7 shows the unsteady part of amplitude of Nusselt number Nu1 against ξ and
η at η 0 and ξ r or at η 1 and ξ −r. It reveals that for different values of Prandtl
number, amplitude of Nusselt number increases with increase of ξ and η. The unsteady part
of the amplitude of Nusselt number Nu2 against ξ and η has been drawn in Figure 8. Here,
one can see that the amplitude of Nusselt number increases with increase of ξ and η for
different values of Prandtl number.
16
Mathematical Problems in Engineering
6. Conclusions
A detailed analytical study has been carried out for the unsteady flow and heat transfer of a
dusty fluid through a rectangular channel. Here, one can see that the flow of fluid particles is
parallel to that of dust. Further, one can see that the fluid particles will reach the steady state
earlier than the dust particles. From the graphs the impressive effect of Reynolds number
on the velocity fields of both fluid and dust phases is evident. It is clear that the effect of
Reynolds number on velocity fields is favorable, that is, the velocity profiles for both fluid
and dust particles increase as Reynolds number increases.
Further, one can observe the changes in the steady and unsteady parts of amplitude of
Nusselt number. It is clear that for different values of Prandtl number steady part of Nusselt
number Nu0 increases with increase of ξ and η. In the same manner unsteady parts of
amplitude of Nusselt number Nu1 and Nu2 increases with increase of ξ and η for different
values of Prandtl number.
Acknowledgments
The authors wish to express their thanks to DST Department of Science and Technology,
india for granting a Major Research Project no: SR/S4/MS: 470/07, 25-08-2008. They are
also grateful to the referees for useful comments.
References
1 P. G. Saffman, “On the stability of laminar flow of a dusty gas,” Journal of Fluid Mechanics, vol. 13, pp.
120–128, 1962.
2 N. Datta, D. C. Dalal, and S. K. Mishra, “Unsteady heat transfer to pulsatile flow of a dusty viscous
incompressible fluid in a channel,” International Journal of Heat and Mass Transfer, vol. 36, no. 7, pp.
1783–1788, 1993.
3 P. D. Ariel, “Heat transfer in unsteady laminar flow through a channel,” Applied Scientific Research,
vol. 47, no. 4, pp. 287–300, 1990.
4 S. K. Ghosh, O. Anwar Bég, and M. Narahari, “Hall effects on MHD flow in a rotating system with
heat transfer characteristics,” Meccanica, vol. 44, no. 6, pp. 741–765, 2009.
5 M. A. Ezzat, A. A. El-Bary, and M. M. Morsey, “Space approach to the hydro-magnetic flow of a
dusty fluid through a porous medium,” Computers & Mathematics with Applications, vol. 59, no. 8, pp.
2868–2879, 2010.
6 S. P. Anjali Devi and S. Jothimani, “Heat transfer in unsteady MHD oscillatory flow,” Czechoslovak
Journal of Physics, vol. 46, no. 9, pp. 825–838, 1996.
7 M. S. Malashetty, J. C. Umavathi, and J. Prathap Kumar, “Convective magnetohydrodynamic
two fluid flow and heat transfer in an inclined channel,” Heat and Mass Transfer/Waerme- und
Stoffuebertragung, vol. 37, no. 2-3, pp. 259–264, 2001.
8 G. Palani and P. Ganesan, “Heat transfer effects on dusty gas flow past a semi-infinite inclined plate,”
Forschung im Ingenieurwesen, vol. 71, no. 3-4, pp. 223–230, 2007.
9 H. A. Attia, “Unsteady MHD couette flow and heat transfer of dusty fluid with variable physical
properties,” Applied Mathematics and Computation, vol. 177, no. 1, pp. 308–318, 2006.
10 A. J. Chamkha, “Unsteady hydromagnetic flow and heat transfer from a non-isothermal stretching
sheet immersed in a porous medium,” International Communications in Heat and Mass Transfer, vol. 25,
no. 6, pp. 899–906, 1998.
11 S. C. Mishra, P. Talukdar, D. Trimis, and F. Durst, “Two-dimensional transient conduction and radiation heat transfer with temperature dependent thermal conductivity,” International Communications in
Heat and Mass Transfer, vol. 32, no. 3-4, pp. 305–314, 2005.
12 S. Chakraborty, “MHD flow and heat transfer of a dusty visco-elastic stratified fluid down an inclined
channel in porous medium under variable viscosity,” Theoretical and Applied Mechanics, vol. 26, pp. 1–
14, 2001.
Mathematical Problems in Engineering
17
13 H. M. Shawky, “Pulsatile flow with heat transfer of dusty magnetohydrodynamic Ree-Eyring fluid
through a channel,” Heat and Mass Transfer, vol. 45, no. 10, pp. 1261–1269, 2009.
14 B. J. Gireesha, C. S. Bagewadi, and B. C. Prasannakumar, “Pulsatile flow of an unsteady dusty fluid
through rectangular channel,” Communications in Nonlinear Science and Numerical Simulation, vol. 14,
no. 5, pp. 2103–2110, 2009.