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 Authors : U. FilobelloNino , H. VazquezLeal , V. M. JimenezFernandez , D. PereyraDiaz, C. E. SampieriGonzalez. F. J. GonzalezMartinez, A. D. ContrerasHernandez, O. AlvarezGasca, B. E. PalmaGrayeb, J. MatiasPerez, L. CuellarHernandez, J. E. PretelinCan
 Paper ID : IJERTV7IS110100
 Volume & Issue : Volume 07, Issue 11 (November – 2018)
 Published (First Online): 05012019
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
A Novel Study for A Model of Diffusion and Reaction in Porous Catalysts
A Novel Study for A Model of Diffusion and Reaction in Porous Catalysts

FilobelloNino, H. VazquezLeal *, V.M. JimenezFernandez, D. PereyraDiaz,
C.E. SampieriGonzalez. F.J. GonzalezMartinez, A.D. ContrerasHernandez, O. AlvarezGasca,
B.E. PalmaGrayeb, J. MatiasPerez, L. CuellarHernandez, J.E. PretelinCanela,
C. HoyosReyes, L.J. VarelaLara, J.L. VazquezAguirre, L. GilAdalid.
Facultad de InstrumentaciÃ³n ElectrÃ³nica, Universidad Veracruzana, Circuito Gonzalo Aguirre BeltrÃ¡n S/N, 9100,
Xalapa, Veracruz, MÃ©xico.
Abstract This work studies the nonlinear equation that models the diffusion and reaction in porous catalysts, which is of great importance in chemical engineering. The proposal is to obtain an approximate analytical expression that adequately describes the phenomenon considered. In order to find such approximation, we propose to use Laplace transform homotopy perturbation method (LTHPM). It is observed in this analysis that the proposed solution is compact and easy to evaluate and involves polynomial functions of only five terms, which is ideal for practical applications. We find that the square residual error (S.R.E) of our solutions is in the range [10E18, 10E6] and this requires only fourth order approximation of the proposed method.
Keywords Homotopy perturbation metho; Laplace
this work presents LTHPM method, in order to find approximate solutions for the second order nonlinear ordinary differential equation, that models a relevant problem in chemical engineering with, mixed boundary conditions [41,42].
Problems with boundary conditions on infinite intervals belong to problems on semiinfinite domains [29,40]; nevertheless, in this article we present a different approach to solve them.
II. HPM METHOD
With the purpose to understand how HPM works, we will present the following nonlinear problem [17,18]
transform; nonlinear differential equations; porous catalysts
diffusion and reaction.
A(u) f (r) 0,
which obeys the boundary conditions
r , (1)

INTRODUCTION

B(u, u / n) 0,
r .
(2)
A relevant problem is the prediction of diffusion and
Where A symbolizes a differential operator, B is a
reaction rates in porous catalysts, in the general case which the
reaction rate depends nonlinearly on concentration [1].
boundary operator,
f (r) is a known function and is the
Problems like the one mentioned, give rise to the search of solutions to nonlinear differential equations but unfortunately, solving this kind of equations is a difficult task. As a matter of fact, most of the times, it can only be get an approximate solution to such problems. With the end to approach various types of nonlinear problems, have been proposed several methods as an alternative to classical methods, such as
domain boundary for . Besides A can be divided into two
operators L and N , where L is linear and N nonlinear; in such a way that (1) can be expressed as follows
L(u) N (u) f (r) 0 . (3)
In general terms, a homotopy is constructed in accordance with [17,18]
H (U, p) (1 p)[L(U ) L(u0 )] p[L(U ) N(U ) f (r)] 0,
variational approaches [25], tanh method [6], expfunction [7,
8], Adomians decomposition method [1,914], parameter or
p [0,1],
r (4)
expansion [15], homotopy perturbation method [1642],
homotopy analysis method [4345], and perturbation method
0 0
H (U , p) L(U ) L(u ) p[L(u ) N (U ) f (r)] 0,
[46,47] among others. Also, some exact solutions have beenp [0,1],
r ,
(5)
reported in [48].
where p is known as homotopy parameter, with values
Laplace Transform (LT) has been relevant in mathematics, both for its theoretical and practical interest, in particular
within of interval 0 to 1, while
u0 approximates the solution
because LT let to solve many problems in science and engineering, in a simpler way in comparison with other techniques [41,42,49]. The use of LT for nonlinear ordinary
of (3) regarding the boundary conditions of the problem.
Considering that solution for (4) or (5) is expressed in terms of p as
differential equations has focused on approximate solutions;
2
U v0 v1 p v2 p
…
(6)
reference [38] combined Homotopy Perturbation Method
Then, after substituting (6) into (5) and equating terms
(HPM) and LT and denominated this method as LTHPM, with the purpose to get precise solutions for these equations. Nevertheless, LTHPM has been employed above all to solve problems with initial conditions [29,38,39]. For the above,
with identical powers of functions0 ,1,2 ,…
p, there can be evaluated the
Finally, considering the limit value p 1, a solution for

is obtained as follows
solution is obtained in accordance with the following limit value u limU 0 1 2 …
p1
U v0 v1 v2 v3…

LTHPM METHOD
(7)

PROBLEM FORMULATION
In accordance with [1,44], let us consider a coupled diffusion and reaction porous catalyst pellets. We will study
Next, it is shown how to employ the proposed method to
calculate approximate solutions of Differential Equations such as (3) [29,38,39,41,42].
With this end, LTHPM uses the same steps of basic HPM until (5); next it is applied LT on both sides of (5), to get
the case where the reaction rate depends nonlinearly on concentration; in such a way that it is possible to envisage the system as a solid material with pores through which the reactants and products diffuse. For the sake of simplicity, the system is conceived as simple diffusion by employing an
effective constant diffusion coefficient D .
L(U ) L(u0 ) p[L(u0 ) N (U ) f (r)] 0.
(8)
e
The mass balance on a volume of the aforementioned
After employing the differential property of LT, it is
obtained [49]
medium is mathematically expressed by [1,44].
Y
sn U sn1U (0) sn2U (0) … U (n1) (0)
(9)
e
D 2Y r(Y ) , (14)
t
L(u0 ) pL(u0 ) p N (U ) f (r),
or
where, Y is the chemical reactant concentration, rate of reaction per unit volume, and t is the time.
r(Y )
the
1 sn1U (0) sn2U (0) .. U (n1) (0)
(10)
Next [1,44,50], we will study the steady one dimensional
n
(U )
s L(u0 ) pL(u0 ) p N (U ) f (r)
case, Y 0 , in such a way that (14) is expressed as:
t
After employing inverse Laplace transform to previous
equation, we get
D
d 2Y
e 2
r(Y ) 0 , (15)
1 sn1U (0) sn2U (0) .. U (n1) (0)
(11) dX
U 1
where X is the diffusion distance.
n
s
L(u0 ) pL(u0 ) pN (U ) f (r)
We will assume that the system is limited by plane
If the solutions of (3) can be written as
boundaries at
X 0 and X L, so that the side
X 0 , is
impermeable (vanishing mass flux) and that X L is held a
n
U
n0
p vn ,
(12)
constant concentration
Y ys
, therefore
then after substitting (12) into (11), yields in
1 sn1U (0) sn2U (0) .. U (n1) (0)
(13)
dY 0,
dX X 0
Y (L) ys ,
(16)
pn
sn
1
Next, we consider the case where the reaction rate per
.
n
unit volume r,
is given as a power law function of the
n0
1
L(u ) pL(u ) p N ( pn ) f (r)
s
n 0 0
n
n=0
concentration [1,44,50].
The comparison of coefficients with the same power of p results in
r kY n , (17)
in this equation n is called the reaction order and the constant
0 : 1 1 n1
(0)
n2
(0) ..
(n1) (0))
( ),
k is a function of temperature (the admitted range of the
sn
p 0
s U

s U

U


L u0
reaction order is n 1).
p1 : 1 1 N( ) Lu f (r),
Finally, for a suitable solution, we express (15)(17), in
1 n 0 0
s
terms of the following dimensionless variables.
p2 :
1 1 N ( , ),
x X ,
y(x) Y ( X ) ,
(18)
2 n
s
0 1
L ys
p3 :
1 1 N ( , , ),
so that we get the following differential equation
sn
3
0 1 2
y m2 yn 0, (19) with boundary conditions
j : 1 1
( , , ,…, ),
y(0) 0,
y(1) 1.
(20)
sn
p j
N 0 1 2
j
where prime denotes from here on, differentiation respect to x , and the Thiele modulus m is defined by
2 n1
1/ 2 .
m kL ys De
(21)
(14)
Considering the following values for the initial
Next, it is possible to express the boundary value problem (19)(20) in terms of the initial value problem [45].
approximation:
U (0) u0 0 ,U(0) 1,..,U
n1
(0) n1;
the exact
y m2 yn
0,
y(0) A,
y(0) 0,
(22)
liable to the additional condition y(1) 1,
besides A denotes
the concentration of the reactant on the boundary at
x 0 and
it is an unknown parameter to be determined as a part of the
pn
1 A
1
y p
y m2
p p2
..2
solution of the problem.
n 0
n
0 0 0 1 2
s s2
(34)
As a matter of fact, the problem as it is expressed in (22) will result particularly useful for application of LTHPM algorithm. We will see that, this method is able to obtain very
accurate and handy approximations. Also it calculates A ,
After equating terms with the same powers of p , it is obtained
1
from the condition
y(1, A) 1.
0
p0 :

)
A
, (35)
s

APPLICATION OF LTHPM TO OBTAIN HANDY
1 2 0
p1 : ( x) m2 1
1 2
, (36)
SOLUTIONS FOR THE NONLINEAR CHEMICAL
s
By convenience of this study, we will consider
2 2
0 1 , (37)
EQUATION UNDER STUDY
p2 :
( x) m2 1
1 2
equivalently the system (22) , expressed in the most complete
s
form
p3 : ( x) m2 1 1 2 2
, (38)
y m2 yn 0, (n 1),
(23)
3 2
s
1 0 2
Next, we will analyze the following representative case
p : ( x) m
2 2 0 3 21 2 , (39)
studies.
y(0) A,
y(0) 0,
y(1) 1.
4
4 2 1
1
s

Case studies 1 and 2
0
The solutions for
1
2
(x), (x),
(x),.. are
Reaction order n 2 ,
Thiele modulus m 0.3 and m 0.6 .
After identifying
L( y) y(x) , (24)
0
p0 : (x) A , (40)
p1 : (x) m A x (41)
2 2 2
1
,
2
m A
4 3
2 4
N ( y) m2 y2 , (25) we propose the following homotopy equation
2
p : (x)
x
12
m6 A4
, (42)
2 2 3 6
(1 p)( y y0) p y m y 0 , (26)
or
2 2
3
p : (x)
4
x
72
m8 A5
, (43)
y y0 p y0 m y
Applying LT to (27) we obtain
.
(27)
4
p : (x)
8
x
504
, (44)
( y )
0 0
y p y m2 y2
.
(28)
By substituting solutions (40)(44) into (32) results in a
Next, in accordance with [49], we rewrite (28) as follows
fourth order approximation
s2Y (s) sy(0) y(0)
y0 p
0
y m2 y2
, (29)
y(x)
m2 A2 x2
A
m4 A3
x4
m6 A4
x6
m8 A5
x8 .
where Y (s) ( y(x)) .
2 12 72 504
(45)
Taking into account that y(0) 0 , it is possible to rewrite
(29) as
Next, we will consider separately the cases of Thiele modulus m 0.3 and m 0.6 .
s2Y (s) sA
y0 p
0
y m2 y2
, (30)
With the purpose to find A , we require that (45) satisfies y(1, A) 1, for m 0.3 , so that we obtain
where A y(0) .
After solving for Y (s) and applying 1 we get
y(x) 1 A 1 y p y m2 y2 . (31)
A 0.958090536681.
In the same way for m 0.6 , we get
A 0.859724737059.
(46)
(47)
s s2 0 0
In according with the propose method, we will assume that the solution for (23) is expressed as
Substituting (46) into (45), we obtain
y(x) 0.958090536681 0.0413071864415×2
y(x) pn
, (32)
0.000593640366398x
4 0.00000853141825856×6
(48)
n
n 0
1.05092628403 10
7 x8 .
Next, we will choose
0 (x) A , (33)
On the other hand, by substituting (47) into (45), we get
y(x) 0.859724737059 0.133042792232x
as the first approximation for the solution of (23) that
2
0.00686281077415×4 0.000354007691298×6
(49)
fulfills the condition
y(0) 0.
8
Substituting (32) and (33) into (31), we obtain
0.0000156522429935x .

Case studies 3 and 4
Next we will study the cases Reaction order n 1 ,
Thiele modulus m 0.3 and m 0.6 .
0
p0 : (x) A , (65)
2 2
1
p1 : (x) m x , (66)
2 A
4
In this case we identify from (23)
L( y) y(x) , (50)
N ( y) m2 y1 , (51)
m
p2 : (x)
2 24 A3
6 6
x4 , (67)
Next, we propose
m x
p3 : (x) , (68)
0
(1 p)( y y ) p y m2 y1 0 , (52)
3 144 A5
8 8
or
y y p y m2 y1 .
(53)
4
p4 :
25m x
(x) , (69)
8064 A7
0 0
Applying L.T. to (53)
( y ) y p y m2 y1 .
0 0
Using the differential property of LT
and so on.
By substituting (65)(69) into (57) we obtain a handy eight order approximation
0 0
s2Y (s) sy(0) y(0) y p y m2 y1 , (54)
y(x)
m2 x2 m4
A
m6
x4
x6
25m8
x8 . (70
p>3 5 7
where Y (s) ( y(x)) .
From condition y(0) 0 , (54) is simplified as
2 A 24 A
144A
8064A
)
0 0
s2Y (s) sA y p y m2 y1 , (55) with A y(0).
We will consider separately the cases of Thiele modulus
m 0.3 and m 0.6 .
To calculate the value of A , we solve the algebraic
After solving for Y (s) and applying 1 we get
equation y(1) 1
obtain
from (70), and use
m 0.3
so that we
y( x) 1 A 1 y p y m2 y 1 . (56)
A 0.953172744786
(71)
s s2 0 0
Next, we will assume that
In the same way for m 0.6 , we get
A 0.779378540012.
Substituting (71) into (70), we obtain
(72)
n
y(x) pn
n0
, (57)
y(x) 0.953172744786 0.0472107498314×2
4 6
(73)
and
0.000389725596578x
0.00000643438374400x
0
(x) A.
(58)
2.03404017857 10
7 x8 .
After substituting (58), (57) into (56)
After substituting (72) into (70), we obtain
y(x) 0.779378540012 0.230953241281×2
n0
n
pn
0.0114063955916×4 0.00112668572799×6
0.0000520714285714×8.
(74)
1 2
A 1 y p y m

p
2

p
..1
2
0 0 0 1 2
s s
(59)


Case studies 5 and 6
Reaction order n 1/ 2 ,
Equating terms with the same powers of p terms we get
Thiele modulus m 0.3 and m 0.6 .
0
p0 :
( x) 1 A, (60)
s
In this case, from (23)
L( y) y(x) . (75)
N ( y) m2 y1/ 2 , (76)
p1 : ( x) m2 1 1 1 , (61)
1 2 0
( x) m2 1 1 , (62)
s
Since a similar procedure is followed to the previous cases, we only present the relevant results
2
p2 :
1
2 2
s A
In this case we obtain the following handy eight order approximation.
m2 Ax2 m4 m6 m8
2 y( x) A
x4 x6
x8 .
p3 : (x) m2 1 1 1
2 , (63)
2 48
1440 A
11520 A
3
s2
A3
A2
(77)
3
Case m 0.3
p4 :
(x) m2 1 1 3
2
1 2 1
(64)
A 0.955836657231.
(78)
4
s2
A2
A3 A4
By substituting (78) into (77) results in
The solution for equations (60)(64) yields in
y(x) 0.955836657231 0.0439951046242×2
respectively of 7.75287785167 1016
and
0.00016875×4 5.17813292968107 x6
5.95845791947 109 x8 .
(79)
2.37672485412 1010 , which confirms the high accuracy of LTHPM. Figure 2, compare numerical solution of (23)
Case m 0.6 .
for n 1 , m 0.3 and n 1 , m 0.6
with (73) and (74)
A 0.833045373527.
After substituting (80) into (77), we get
(80)
for the same values. We note that the figures appear indeed
overlapping, while the S.R.E of (73) and (74) are of
10
y(x) 0.833045373527 0.164288374824×2
6.10465343999 10
, and 0.00000335024984825.
Finally, the cases
n 0.5 ,
m 0.3
and n 0.5 ,
m 0.6 ,
0.0027×4 0.0000354985555505×6
0.00000175020478636×8 .
(81)
are shown in Figure 3, and the corresponding S.R.E for approximations (79) and (81) are scarcely of


DISCUSSION
4.29941296548 1018
respectively.
and 5.79207356623 1012
This article introduced LTHPM in order to get handy
accurate approximate solutions for the problem with mixed boundary conditions that describes the problem of the diffusion and reaction in porous catalysts. Such as it is explained in [1], the understanding of this process it turns out relevant for the chemical engineer due to its applications in the design and operation of catalytic reactors. It should be mentioned that this problem has been successfully attacked for several authors. Thus, [1] found an approximate solution for (23), by using the Adomian decomposition method for some values of n order and Thiele modulus m . Although Adomian is a powerful tool, the process of obtaining its polynomial solutions are not straightforward for practical
It is clear from the above discussion that LTHPM describes a highly accurate way for solving, the nonlinear problem (23). On the other hand, is worth to note that our proposed solutions (48), (49), (73), (74), (79) and (81) are short and simple polynomial functions, ideal for practical applications. In all the cases considered we keep the order of approximation as four. If more accuracy for solutions is even required, one can go on with higher orders in a straightforward fashion, following LTHPM algorithm.
s
Our results y(x) , indicate the concentration under steady conditions. From definition of Thiele modulus m (21), we identify the quantity 1LKyn 1 as a characteristic property
applications. On the other hand [44] showed the application
of homotopy analysis method HAM, to get the approximate
for reaction, and
e
D L as a characteristic property for
solution of the nonlinear model (23), for the cases of
n 0.5, 2, and 4, for several values of m . Although in general HAM is very accurate, its expressions use to be long and cumbersome, as to be used in practical applications thus,
[44] proposed approximations of 6th, 15th, 20th, 35th, and50th order. What is more, [50] went beyond, this article showed that, this model is exactly solvable in terms of Gauss hypergeometric function. The main advantage of this paper is that the hypergeometric function is well known, although its study it is not elementary. [45] employed HAM to investigate in all detail the case n 1. The solutions obtained are very long, and correspond to 30 th and 50 th order approximations of HAM, even though this paper obtained multiple solutions for this case. Unlike of the above studies, the goal of this article is to show the manner of getting handy approximate solutions for nonlinear problems like (23), through the use of LTHPM. In fact, Figure (1), Figure (2), and Figure (3) show the accuracy of LTHPM for the problem under study.
In more precise terms, Figure 1 compares the numerical
diffusion [1,44]. Therefore, the bigger the value of m , the
bigger in proportion the diffusion respect to reaction phenomena and viceversa. From values of S.R.E one deduce that LTHPM is more accurate for the cases, where the diffusion is relatively less important than reaction for a given value of n , that is for small values of m . Figure 1, Figure 2, and Figure 3, explain the above, noting that the curves with larger value of m present a major curvature than those, with smaller values, and for the same reason, they are more complicated to model. The simplicity and accuracy of the proposed method indicates that, unlike other methods, our solutions keep the nature of the studied phenomena and on the other hand, the reliability of the obtained results for the initially unknown values of the concentration of the reactant A .
Finally, future investigations of LTHPM should follow the aim of [45], where the authors obtained multiple solutions of (23), for the case n= – 1.
solution of (23) for cases study:
n 2 ,
m 0.3 and
n 2 ,
m 0.6
and approximations (48) and (49). Although the
mentioned figure, shows the high accuracy of the proposed solutions, it was verified by evaluating the square residual
b
error (S.R.E) of (48) and (49) given by
R2 u(t) dt ,
a
where a and b are two values depending on the given problem, the residual is defined by
R u(t ) L u(t ) N u(t ) f (t ) , and
u(t)
is an
approximate solution to (3) [16]. The resulting values were
Figure 1 Comparison between numerical solution of (23) for cases study: n 2 , m 0.3 and n 2 , m 0.6 and LTHPM
approximations (48) and (49).
Figure 2 Comparison between numerical solution of (23) for cases study: n 1 , m 0.3 and n 1 , m 0.6 and LTHPM
approximations (73) and (74).
Figure 3 Comparison between numerical solution of (23) for cases study: n 0.5 , m 0.3 and n 0.5 , m 0.6 and LTHPM
approximations (79) and (81).

CONCLUSION
From this study we conclude that LTHPM is a useful tool to get accurate and handy solutions for the considered
problem, which describes the phenomenon of the diffusion and reaction in porous catalyst. One of the highlights of this work lies in the practical and precise solutions obtained by LTHPM compared to other methods, like HAM and Adomian Decomposition Method. We emphasize that one advantage of LTHPM is that it does not require to solve several recurrence differential equations like other perturbative methods. Finally it is clear that the proposed methodology can be applied equally to other nonlinear problems, especially to heat diffusion problems.
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