 Open Access
 Total Downloads : 384
 Authors : Dr. P. Venkat Raman, Mr. Sreepada Sathyendar
 Paper ID : IJERTV1IS10589
 Volume & Issue : Volume 01, Issue 10 (December 2012)
 Published (First Online): 28122012
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Dissipation Of Energy In Viscous Liquid Through Porous Region
Dr. P. Venkat Raman 
Mr. Sreepada Sathyendar 
Professor & Director of MCA, Alluri Institute of Management Sciences, Warangal 506001 (A.P), India 
Assistant Professor of Mathematics, Department of Mathematics, Vaagdevi College of Engineering, Bollikunta, Warangal 506005 (A.P), India 
Abstract
In the paper, the flow of a viscous liquid is considered through a cylinder containing porous region. The mechanical energy dissipated in the fluid is calculated. The boundary of the tube performs harmonic oscillations. The effect of the permeability coefficient on the flow and the dissipation of energy are examined and discussed completely.
Key words and phrases: Newtonian Fluid, dissipation of energy, Porous medium, Permeability.
developed and the internal stress generates in the fluid due to its viscous nature and produces distortions in the velocity field. In the case of highly porous medium such as fiberglass, papus of dandelion the flow occurs even in the absence of the pressure gradient.
Modifications of the classical Darcys law were considered by the Beverse and Joseph [1], saffman
[9] and other. A generalized Darcys law proposed by Brinkman [2] in given by
O p
K
2

Introduction
In the paper, the study of flow through porous medium has many interesting applications in the diverse fields of science, engineering and technology. The particular applications which are wellknown include the percolation of water through soil, extraction and filtration of oils from wells, the
Where and K are coefficient of viscosity of the fluid and permeability of the porous medium respectively.
The generalized equation of momentum for the flow through the porous medium is
drainage of water, irrigation and sanitary engineering
and also in the interdisciplinary fields such as
t
. p 2
K
medical and biomedical engineering etc. The lung
alveolar is an example that finds application in the animal body. The classical Darcys law musakat (3) states that the pressure gradient pushes the fluid against the body forces exerted by the medium which can be expressed as.
The classical Darcys law helps in studying flows through porous medium. In the case of highly porous medium such as papus of dandelion etc., The Darcys law fails to explain the flow near the surface in the absence of pressure gradient. The nonDarcian
K
p
(with usual notion)
approach is employed to study the problem of flow through highly porous medium by several
investigators. Narsimha charyulu and pattabhi Rama Charyulu [4, 5] Narsimha Charyulu [6] and singh [7]
The classical Darcys law gives good results in the situations when the flow is unidirectional or at low speed. In general, the specific discharge in the medium need not be always low. As the specific discharge increases, the convective forces get
etc, studied the flow employing Brinkman law [2] for the flow through highly porous medium.
The problem of flow of the Newtonian fluid in the presence of transverse magnetic field, find
application in nuclear engineering and other fields.
u(a,t) U cos pt ,
t 0
(2.4)
The rotatory flow of the fluid has special applications in various engineering field such as mechanical, petroleum and chemical in addition to the geophysical fluid dynamics, which helps in explaining the phenomena like oceanic circulation [8]. Several investigations are made in the study of flow of viscous fluid in the presence of transverse magnetic field under the assumption that the induced magnetic effect is negligible on the flow of the fluid
e.g. see Greenspan [10] and Herbut [10] etc. But some investigations are made by considering the effect of the induced magnetic field on the flow; e.g. see somdalgekar [12] and pop [13] etc.
The flow through porous medium in the presence of transverse magnetic field is studied in the past by several investigators. The nonDarcian flow in the presence of transverse magnetic field is investigated by Nassimha charyulu [14]. Venkat Raman and
where, U is the amplitude of the velocity fluctuation and p the frequency of the motion of the tube.
Flow model
raxis
0 2a 0 xaxis
Figure 1. Flow of Newtonian fluid through porous region
By applying Laplace transform on equations (2.2), (2.3) and (2.4), we get the transformed equations as
Narsimha Charyulu [15, 16, 17, and 18] have studied the flow employing Brinkmans law [2] for the flow through a rotating porous duct and highly porous medium.
and
d 2u dr 2

1 du r dr
2
u 0
(2.5)
In the paper, we considered the flow of a viscous
liquid through a cylinder containing porous region.
u (a, s)
Us
s 2 p 2
(2.6)
The mechanical energy dissipated in the fluid is
calculated. The boundary of the tube performs
where,
2
s 1 .
harmonic oscillations. The effect of the permeability k
coefficient on the flow and the dissipation of energy are examined and discussed completely.
The solution of equation (2.5) satisfying the condition at the origin and the boundary condition (2.6) is


Formulation and solution of the
u (r, s) Us
I0 (r)
(2.7)
0
problem
s2 p2 I
(a)
The inverse Laplace transform of u(r, s) , gives
Consider the flow of an incompressible, viscous liquid through porous region contained by an infinite circular tube of radius a. Let (r, , x) be the co
u(r,t)
i
U
s
i
2i s2 p
I0 (r)
0
2 I (a)
exp(st)ds
(2.8)
ordinate system such that the x coordinate is along the axis of the tube. Let the velocity of the fluid is
where, is a constant such that all poles lie to its left.
It can easily be verified that the integrand is a single valued function of s. the inversion may be performed
given by V (u,0,0)
continuity
which satisfies the equation of
by summing the residues at the simple poles s = Â± ip, and s + ( /k) = – 2 where Â± (n = 1, 2, 3 ) are
n n
V 0 . (2.1)
The physical quantities are independent of x and also independent of because of symmetry of the flow.
the roots of J0(n a) = 0. It can be shown that there are no branch points. Evaluating the residues and using Cauchys integral theorem gives,
The NavierStokes equation for the flow problem
U I [
(l)r]
U I [
(l)r]
will be
u(r,t) 0 exp(ipt ) 0
exp(ipt )
u 2u
t r2
1 u
r r k u
(2.2)
2 I0[
(l)a]
2 I0[
(l)a]
2 2U
3 J ( r) 2
where, the kinetic viscosity, t is the time and k is

n
0 n exp(n t)
(2.9)
a 4 p2 J ( a)
the permeability of the medium.
n 1 n 1 n
The boundary conditions are given by
where,
l i p 1 . (2.10)
u(a,t) 0 ,
t 0
p/>
k
(2.3)

Dissipation of Energy
The time rate of energy dissipated per unit
The energy dissipation per unit length of the tube at the end of the mth cycle is obtained by integrating the equation (2.13) over m cycles. The resulting expression is
2U 2
r 6
4r 2 m
length along the axis of the tube in viscous flow is
obtained from
E n exp n 1
p (r 4 2 )2
dE a u 2
(2.11)
n
dt 2 r r dr
. 2 2
0
U m i
(i )I0[ (i )]I1[ (i )] (i )I0[ (i )]I1[ (i )]
where, denotes the coefficient viscosity and the negative sign is inserted because the integral on the right represents a loss of energy. The velocity gradient obtained from the equation (2.9) as
p
I0[ (i )]I0[ (i )]
(2.15)
u U
r 2
2 2U
(l) I1[
I0[
n
(l)r] exp(ipt) U
(l)a] 2
4 J ( r)
(l) I1[
I0[
(l)r] exp(ipt) (l)a]
2


SPECIAL CASES

Case (i). For the highly porous medium (i.e. k is very large)
n 1 n exp( t) . (2.12)
a 4 p2 J ( a)
The energy dissipation is given by
n1 n 1 n
Using the equations (2.11), (2.12) and carrying out the integrations gives,
U 2m 2 8
r
E n
1 2
192
2 4
(3.1)
p n1 (r4 2 )2 81

dE U 2 (i ) exp(i2 pt)
0
n 32
4096

dt
8 I 2[ (i )]


Case (ii). For the flow through clear
I 2 (i ) 2 I0[ (i )]I1[ (i )] I 2[ (i )]
medium (i.e. k)
0 (i )
1
The dissipation energy is given by
8
I 2[ (i )]
U 2 (i ) exp(i2 pt)
2 4
2
0
E n 192
2
(3.2)
U 2m 2
r8
I [ (i )]I [ (i )]
p (r4 p2 a4 )2
1 p a
81 p2a4
p4a8
0
I 2 (i ) 2 0 1
I 2[ (i )]
n1 n 2
32 2
4096 4
1
(i )
r 2 pr 2t U 2 i

Case (iii). When the permeability of the
2U 2 n exp n
n
n1 (r 2 2 )2
(i )I0[ (i )]I1[
(i )]
4 I0[ (i )]I0[ (i )]
(i )I0[ (i )]I1[ (i )]
medium is very small (i.e. k0)
The energy dissipation is given by
2U 2 (i ) exp(ipt ) pr 2t
exp n
I0 [
(i )]
n1
E
i i
(i )J1 (rn )I2 [
(i )] rn I1[
4
rn
2
4
(rn

i )(rn
(i )]J 2 (rn )
2 )
1
3
64
1 8(
i i )
(3.3)
2U 2 (i ) exp(ipt ) pr 2t
1 1 8( i i ) 1
exp n
64
I0 [ (i )]
n1
r 4 (i )J (r )I [
(i )] r I [ (i )]J (r )

n 1 n 2
n 1 2 n
(2.13)
(r 2 i )(r 4 2 )
where,
n n
rn n a and
p 1 a2 . (2.14)
k
5
4
3
2
1
0
0 200 400 600 800 1000 1200
1
2
E
Figure.1 Variation of E for different values of . (rn=1)



Results and Conclusion
In the present problem, an attempt is made to estimate the mechanical energy dissipated in the fluid through a circular tube whose boundary performs harmonic motion. Expression for the energy is obtained when the tube is filled with highly porous medium. The energy dissipated per unit length of the tube is obtained in terms of which involves the permeability coefficient. The graph depicting the variation of E for different values of the permeability coefficient is drawn in fig.1.

References

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