ABSTRACT : In the first part of the study, several mathematical methods for solving differentialequations systems describing a process which are most commonly used in engineering theory andpractice, were analyzed, for a characteristic example of a complex chemical reaction. The analysiswas performed from the aspect of required mathematical knowledge, the time necessary for solvingthe problem, the possibility of error, as well as accuracy the obtained results. The second part ofthe study gives a suggestion of a more efficient method for solving the observed problem, whichapplies the principle of reducing the system to one differential equation which is most convenientlyregarded as a nonhomogeneous first order equation with constant coefficients, dependent on time.The study shows that the suggested method is simpler then existing technique.Key words: chemical reactor, chemical kinetics, complex differential equations systems,nonhomogeneous equations with constant coefficients, reduction of differential equations,mathematical methods in chemistry
INTRODUCTION
In chemical engineering, many problems of both theoretical and practical nature, are solved by usingvarious mathematical methods. In this respect, although it is only auxiliary means, mathematical formalismbecomes a decisive factor in obtaining correct results that would be applied later on. Mathematical approach tothe same problem can be different, which can be seen in literature (Levenspiel, 1982; Gorbačev, 1994;Spiridonov, 1980; Coulson, 1984). Authors do not usually perform a more detailed analysis of the problem fromthe aspect of mathematics; instead, they mainly use several standard, common methods which are often verycomplicated and long. Most commonly, the accepted methods are not elaborated in detail; instead, final resultsare given directly after setting up the problem, which cannot always be justifiable. Mathematical formalismshould be simplified as much as possible. Primary concern would be to obtain as much information on theprocess observed as possible using the simplest possible mathematical technique. Mathematical procedureshould be represented fully, without ‘gaps’ which are often present in literature, mainly due to inappropriatemethods.
Defining of problemAs a characteristic example (Levenspiel, 1982; Coulson, 1984; Skala et al., 1979; Fogler, 2006), which
occurs in the field of designing chemical reactors, the successive first order reaction, where substance Agradually converts into an intermediary substance R which further converts into a substance S as shown below:→ →
The description of the kinetics of this complex reaction is achieved through relations of the substancesrate of reaction,= = − (1) == − (2)= = (3)
where: - rate of formation of A; - rate of formation intermediate of R; - rate of formation productof S; , - rate constants
Broader analysis of the information in literature which investigates and describes successive reactionskinetics, (Martin, 1986; Skala et al., 1979; Fogler, 2006; Steinfeld et al., 1999; Masel, 2002) reveals that themathematical part of the problem uses practically several standard methods, and they will be represented andanalyzed in detail in further sections of this study. After a critical review of the applied methods, in the secondpart of the paper, a suggestion of a more efficient method for solving the problem is made.3.1. Integral Approach (Coulson, 1984; Skala et al., 1979; Fogler, 2006)Integration of (1) will result in :∫ dt = −∫ (4)From it follows:c = c ∙ e (5)Substituting (5) into (2) the following equations are obtained:+ k c = k ∙ c e (6)c ′ + k c = k ∙ c e (7)Multiplication equation (7) on both sides by integration factor e∫ , result is:c ′ e∫ + k c e∫ = k c e ∙ e∫ (8)The first derivative of c can be represented as: c ′ =According to the rule for the derivative of the complex function product, equation (9) is obtained:(c e∫ )′ = c ′ e∫ + c ∙ k ∙ e∫ (9)Here the following was used:(e∫ )′ = (∫k dt)′ e∫ = k e∫ (10)Now, according to equation (8):(c e∫ )′ = k c ∙ e ∙ e∫ (11)The left side of equation (11) can be written as:∫(c e∫ )′ = ∫k c ∙ e ∙ e∫ dt (12)From it follows that:c e∫ = ∫k c ∙ e e∫ dt + C (13)Based on equation (13), we have:c = e ∫ ∙ (∫ k c e ∙ e∫ dt + C) (14)Relation (14) can be written as:c = e ∫ ∙ (k c ∫ e ∙ e dt + C) (15)c = e (k c ∙ ∫ e( ) dt + C) (16)Finally:c = e ∙ (k c ∙ e( ) + C) (17)Using the initial conditions for: t = 0, c = 0And according to equation (17) it follows that:e ∙ e + C = 0 (18)From this it follows that constant of integration is:C = − (19)By replacing (19) with (17) we will have:c = e ∙ ( ∙ e − ) (20)c = ∙ e − ∙ e (21)Or finally:c = c ∙ k ∙ ( + ) (22)Since the total amount of matter expressed in moles does not change, we will have:c = c + c + c (23)It follows that:c = c − c − c (24)Using the obtained values according to (5) and (22) for c ,we finally obtain:c = c ∙ (1 + e + e ) (25)In this way all the variables c , c , i c ,are determined depending of time, which was the aim of the process.
It is obvious that the integration factor method takes a long time and requires greater knowledge ofdifferential and integration calculations, especially of theorems which associate differential and integral.Furthermore, there is quite a large possibilty of error in the application of this method.3.2. Substitution Method (Coulson, 1984; Fogler, 2006)Let us introduce the substitution with two variables u and v:c = u ∙ v (26)Differentiation of relation (26) gives:c ′ = u′ ∙ v + u ∙ v′ u′ = v′ = (27)Now according to (6):u′v + uv′ + k ∙ uv = k c ∙ e (28)Equation (28) can be written as:v u′ + k u + uv′ = k c e (29)Equation (29) can be full filled if:u′ + k u = 0 (30)uv′ = k c ∙ e (31)Now relation (30), can be written as= −k u = −k dt (32)From this it follows that:ln u = −k t (33)i.e., finally:u = e (34)According to (31) we will have:e = k c ∙ e (35)Thus := k c ∙ e( ) (36)Integration of (36) gives:v = k c ∙ ∫ e( ) dt + C (37)Here we obtain another variable:v = e( ) + C (38)According to the shift (26), in respect to (34) and(38) we will have:c = e ∙ ( e( ) + C) (39)Using the initial values:0 = 1 ∙ ( ∙ 1 + C) C = − (40)Now, according to (39), we finally obtain:c = c ∙ k ∙ ( + ) (41)
The substitution (shift) method requires slightly lower mathematical knowledge than the previous one;however the time necessary for its completion is rather long due to the fact that two functions that are to bedetermined are involved in the procedure. The analysis of the two previously described methods commonlyencountered in books reaveals that reduction of the system of starting equations to one differential equation,which is regarded as one linear first order differential equation, equation (6), causes certain mathematicaldifficulties during the process of solving.3.3. Reduction to the second order equation, case I (Steinfeld et al., 1999; Masel, 2002)Differentiation of equation (3), gives the equation of order II:= k (42)By replacing (2) with (42), we will have:= k k c − k c (43)If we add member (k k c ) with different signs on both sides of equation (43), then we have:= k k c − k c + k k c − k k c (44)By grouping the members, (44) changes to:= k k (c + c ) − (k + k )k c (45)According to (23) and(3) we will have:c + c = c − c (46)
k c = (47)By replacing (46) and (47) with (45), we have:= k k (c − c ) − (k + k ) (48)Here:+ (k + k ) + k k c = k k c (49)Characteristic equation of the problem, according to (48), for the homogeneous part is:λ + (k + k )λ + k k = 0 (50)Quadratic equation solutions (50) are as follows: λ = −k ; λ = −kHomogeneous part of the equation (49) is:+ (k + k ) + k k c = 0 (51)Solution of equation (51), considering the roots of the characteristic equation is as follows :c = c ∙ e + c ∙ e (52)Particular solution of equation (49), is saught in the form of :c = A, where: = = 0 (53)By replacing (53) with (49), we have: 0 + (k + k ) ∙ 0 + k k A = k k cHere the unknown constant is:A = c c = c (54)Where as the general solution of the equation (49) is the sum of homogeneous and particular parts:c = c + c (55)By replacing (52) and (54) with (55) we have:c = c e + c e + c (56)According to the initial conditions, we have: t = 0; c = 0; c = 0 = 0Differentiation of equation (56) gives:= −k c ∙ e − k c ∙ e (57)By replacing, according to the initial conditions, equations (56) and (57) change to the system:c + c + c = 0 (58)−k c − k c = 0 (59)Solving the system (58) – (59) gives the following constants of integration:C = ∙ C = ∙ (60)By replacing constants (60) with (56) we finally obtain :c = ∙ e − ∙ e + c (61)c = c ∙ (1 + e + e ) (62)According to (46) it follows that:c = c − c − c (63)
The reduction to the second order equation that we showed here and which is commonly encounteredin books, takes relatively long time due to grouping of certain values of the system, in order to eliminate twoindependent variables. This grouping is not simple at times, therefore its application involves a certainpossibility of error.3.4. Reduction to the second order equation, case II (Martin, 1986; Steinfeld et al., 1999)Similarly to item 3.3., according to (2) we will have:= k c − k c (64)Differentiation of (64) gives the second order equation:= k − k (65)According to (5) we will have:c = c e = −c ∙ k e (66)Replacing (66) with (65) results in:+ k = −c ∙ k e (67)General solution (67) is the sum of homogeneous and particular parts:c = c + c (68)Characteristic equation for the homogeneous part, i.e. its solution will be as follows:λ + k λ = 0 λ = 0 λ = −k
Solution of homogeneous part will be:c = C + C e (69)Particular solution according to the right side (67), is sought in the form:c = B ∙ e (70)It follows that:= −k ∙ Be = k ∙ Be (71)Replacing (71) with (67)k ∙ Be − k k ∙ Be = −c ∙ k e (72)It follows that the unknown constant is:B = − ∙ (73)Particular solution , according to (70) is finally:c = − ∙ e (74)General solution is obtained by replacing (69) and (74) with (68)c = c + c e − ∙ e (75)Differentiation of (75) will give:= −c ∙ k e + ∙ e (76)Using the initial conditions, according to (75) and (76) we obtain:0 = c + c − ∙ (77)k c = −c k + ∙ (78)Here we have used, according to (64) the fact that:= k c − k c (79)For t = 0 c = 0 c = c= k c (80)Solving the system (77) – (78), for constants of integration, we obtain the following values:c = 0 c = ∙ (81)Now, replacing (81) with (75) we finally obtained:c = ∙ e − ∙ e (82)That is:c = c ∙ k ( + ) (83)
In comparison to the previous method, the reduction to the second order equation takes less time dueto simpler process of eliminating variables. Characteristic equation, as is the case in the previous method, isnot linear but quadratic, so two integration constants appear that have to be determined.
Apart from the two shown procedures (items 3.3. and 3.4.), it is also possible to reduce the firstequation of the system (1) to the second order equation (case III), by means of its differention:= −k (84)That is:+ k = 0 (85)
In order to solve the obtained equation, it is possible to apply the same procedure as with the twoprevious cases. Obviously, due to the possibility of direct solution of the equation (1), this does not simplify theprocedure. It should be mentioned at the end that it is also possible to solve the set up system of differentialequations by using the method of variation of constants (La Grange’s method). This method is rathercomplicated, long and requires a broader knowledge in the theory of differential calculation, i.e. of differentialequations, (Gorbačev, 1994; Spiridonov, 1980; Hardy, 1980). There is quite a large possibility of error in theapplication of this method.
The solution can also be achieved by using a completed formula derived in the theory of differentialequations (Banax, 1986; Ponomarev, 1983; Allendoerfer, 1983; Mitrinović, 1993). Using these formule inengineering theory and practice is not recommendable for a well-known reasons.3.5. Suggested method for solving the problemThe complex chemical reaction shown in item 2., will be expressed by introducing mathematical notations x, y,z:x y → z
These notations will give:x + y + z = c (86)Differentiation of (86) gives:+ + = 0 (87)
The rates of reactions, i.e. , the change of concentration with time, according to the new notations willbe as follows:= −k x (88)= k y (89)= k x − k y (90)
which represents a system of differential equations of the set up problem. Equation (90) is obtainedaccording to (87), (88) and (89):= − − = k x − k y (91)From (88) it follows that:dt = − (92)∫ dt = ∫ (93)Integration in the given limits from (93) will result in:t = − ln (94)Here we finally have:x = c e (95)
Variable x, according to all the methods is obtained in a simple way considering it can be reduced todifferential equation with the separation of variables.
Changing the solution (95) to (90) results in the equation that can be regarded as differentialnonhomogeneous first order equation with constant coefficients (Banax, 1986; Ponomarev, 1983; Allendoerfer,1983; Mitrinović, 1993).+ k y = c ∙ k e (96)Obviously, the reduction to the first order equation is rather simple here. Characteristic equation for thehomogeneous part ( on the left side) is linear and thus it can be solved easily:λ + k = 0 λ = − k (97)Solution λ is a real number, so the general solution of the homogeneous part is as follows:y = C e = C e (98)Solution of the particular part, considering the shape of the function of the right side of the equation (96), issaught in the form:y = C ∙ e (99)The shape (99) stands only if the exponent (−k ) is not the root of the characteristic equation of the left side(k ≠ k ), which is the commonest case in practice.Differentiation of exponent function (99) results in:= −C ∙ k e (100)Replacing (99) and (100) with (96), we obtain:−C ∙ k e + C ∙ k e = c ∙ k e (101)Thus it is simple to obtain the integration constant:C = ∙ (102)According to (99), particular solution will finally be:y = ∙ e (103)General solution of equation (96), is obtained by superponing the solution of the homogeneous and particularparts (98), (103):y = y + y (104)that is:y = C ∙ e + ∙ e (105)Using the initial conditions t = 0 y = 0,According to (105) we obtain:
C + ∙ = 0 C = − ∙ = ∙ (106)Now, according to (105) the final solution of the equation (96) is:y = ∙ e + ∙ e (107)That is:y = c ∙ k ( + ) (108)
The method of replacement can serve to prove that solution of equation (108), complies with equation(96).
Considering that according to (95) and (108) x and y are completely determined, from (86) it followsthat z is determined, too. In order to control the procedure, z will be determined in another way, by replacing(108) with (89):= c ∙ k k ( + ) (109)From this it follows that:dz = ∙ e dt + ∙ e dt (110)Integration of (110) will result in:z = ∙ ∫ e dt + ∙ ∫ e dt + C (111)That is:z = ∙ e + ∙ e + C (112)According to the initial conditions, from (112) it follows that: t = 0 z = 00= ∙ + ∙ + C C = c (113)By replacing the integration constant C in (112), we finally obtain:z = c ∙ (1 + e + e ) (114)The rates of reactions are determined by differentiation of (95), (108) and (114):= −k c ∙ e (115)= c ∙ ( e + e ) (116)= c ∙ (e − e ) (117)
Equations (86) and (87) were not used in this procedure, therefore they can be used in simple controlof the obtained results, by replacing (95), (108), (114), (115), (116), (117), in them.3.6. Analisys of reviewed mathematical methodsComparison of the existing procedures with the suggested one, according to the adopted criteria, is shown inTable 1. In this way it is possible to determine the most appropriate method for setting up a problem.
Table 1. Comparison of various mathematical methodsNumber Chapter Method Name Mathematical
knowledgeSolvingtime
Errorpossibility
Resultscontrol
1. 3.1
Usu
al (
stan
dard
)
Integration factor great quite long quite large partial2. 3.2 Substitution medium long medium partial3. 3.3 Reduction to order
II(case I)
considerable medium large partial
4. 3.4 Reduction to orderII(case II)
medium medium medium partial
5. 4. Suggested Nonhomogeneousof order I
elementary short minimal complete
CONCLUSION
Kinetics of complex chemical reactions is described, most commomly, by n differential equations andthe same number of variables. The obtained equations system is solved, most efficiently, by reducing it to onedifferential nonhomogeneous equation of the lowest possible order, preferably that of the first, dependent ontime. In that case a general solution can be obtained, by superponing the solution of the homogeneous andparticular part of the equation. In this respect, characteristic equation is also of the lowest order, and thenumber of integration constants is the lowest, too, which makes the process of solving the problem quicker and
more efficient. At the same time, using the initial conditions in this case is simplified. A problem should alwaysbe regarded purely mathematically, including the change of notations, because it minimizes the error possibilityand results are controlled most efficiently. As we have shown, the same differential equation can be observedin different ways. Both procedure and efficiency depend on that. Among many available procedures for solvinga problem one should always choose the most efficient one, i.e. which is the most appropriate according to theadopted criteria (Table 1). As it has been shown here, the most appropriate procedure need not be one of thetraditional ones that are used most commonly. This means that mathematical formalism can be representedcompletely, considering the fact that it will not take too much space.
The method we have suggested here, in comparison with those most commonly encountered, thusproves to be quicker, simpler with better layout and it does not require greater mathematical knowledge.Reduction to the equation of order I is relatively simple. At the same time, complete control of the obtainedresults, by means of independent equations, from the aspect of detecting a potential error is possible. Theprocess of reaching the solution is not complicated when compared to other methods, even in the case whenthe equation system cannot be reduced to the first order equation, but only to that of the second or higher.
As it has been shown here, the method was successfully applied in the field of chemical reactorsdesign in industry. Apart from that, the method can be applied in many areas of chemical technology, i.e.physical chemistry.
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