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

A Novel Study for A Model of Diffusion and Reaction in Porous Catalysts

1. Filobello-Nino, H. Vazquez-Leal *, V.M. Jimenez-Fernandez, D. Pereyra-Diaz,

   C.E. Sampieri-Gonzalez. F.J. Gonzalez-Martinez, A.D. Contreras-Hernandez, O. Alvarez-Gasca,

   B.E. Palma-Grayeb, J. Matias-Perez, L. Cuellar-Hernandez, J.E. Pretelin-Canela,

   C. Hoyos-Reyes, L.J. Varela-Lara, J.L. Vazquez-Aguirre, L. Gil-Adalid.

   Facultad de Instrumentación Electrónica, Universidad Veracruzana, Circuito Gonzalo Aguirre Beltrán S/N, 9100,

   Xalapa, Veracruz, México.

   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 (LT-HPM). 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 [10E-18, 10E-6] and this requires only fourth order approximation of the proposed method.

   Keywords Homotopy perturbation metho; Laplace

   this work presents LT-HPM 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 semi-infinite 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)

   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 [2-5], tanh method [6], exp-function [7,

8], Adomians decomposition method [1,9-14], parameter or

p [0,1],

r (4)

expansion [15], homotopy perturbation method [16-42],

homotopy analysis method [43-45], 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 been

p [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 LT-HPM, with the purpose to get precise solutions for these equations. Nevertheless, LT-HPM 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

1. 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…

   1. LT-HPM METHOD

      (7)

   2. 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, LT-HPM 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 LT-HPM 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 :

1. )

   A

   , (35)

   s

   1. APPLICATION OF LT-HPM 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

      1. 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 .

      2. 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)

      3. 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 LT-HPM. 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

   2. DISCUSSION

      4.29941296548 1018

      respectively.

      and 5.79207356623 1012

      This article introduced LT-HPM 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 LT-HPM 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 LT-HPM 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, and

      50th 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 LT-HPM. In fact, Figure (1), Figure (2), and Figure (3) show the accuracy of LT-HPM 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 LT-HPM 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 LT-HPM 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 LT-HPM

      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 LT-HPM

      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 LT-HPM

      approximations (79) and (81).

   3. CONCLUSION

   From this study we conclude that LT-HPM 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 LT-HPM compared to other methods, like HAM and Adomian Decomposition Method. We emphasize that one advantage of LT-HPM 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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