Kinetic Modeling

Authored by: Mustafa Özilgen

Handbook of Food Process Modeling and Statistical Quality Control

Print publication date:  March  2011
Online publication date:  March  2011

Print ISBN: 9781439814864
eBook ISBN: 9781439877678
Adobe ISBN:

10.1201/b10833-4

 

Abstract

Chemical reactions and microbial growth or product formation are among the most common causes of food spoilage. Typical examples to food deterioration reactions may include lipid oxidation, thermal, or photocatalytic degradation processes. Toxins, enzymes, or chemicals of microbial origin are the microbial products causing food deterioration. Microbial growth and product formation are achieved via metabolic activity involving sets of chemical reactions occurring mostly in the microbial cell. Some form of metabolic activity also continues in the plant tissues after harvesting or inside the animal cells after slaughtering. Slowing down the deteriorative chemical reactions or microbial, postmortem, or postharvest metabolic activity is indeed the goal of most food processing and preservation methods. The initial objective of the experimental kinetics studies is the development of a mathematical model to describe the reaction rate as a function of the experimental variables.

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Kinetic Modeling

3.1  Kinetics and Food Processing

Chemical reactions and microbial growth or product formation are among the most common causes of food spoilage. Typical examples to food deterioration reactions may include lipid oxidation, thermal, or photocatalytic degradation processes. Toxins, enzymes, or chemicals of microbial origin are the microbial products causing food deterioration. Microbial growth and product formation are achieved via metabolic activity involving sets of chemical reactions occurring mostly in the microbial cell. Some form of metabolic activity also continues in the plant tissues after harvesting or inside the animal cells after slaughtering. Slowing down the deteriorative chemical reactions or microbial, postmortem, or postharvest metabolic activity is indeed the goal of most food processing and preservation methods. The initial objective of the experimental kinetics studies is the development of a mathematical model to describe the reaction rate as a function of the experimental variables.

A chemical reaction may actually involve many elementary steps as explained in the following example:

A 2 2 A * elementary reaction 1 A * + B 2 A B + B * elementary reaction 2 A * + B * 2 A B elementary reaction 3 A 2 + B 2 2 A B overall reaction

In this example, the sum of the elementary reactions gives the overall reaction, A* and B* are the intermediates, which have a very short life and convert into other components immediately after being formed. The intermediates usually may not be detected; therefore, their concentrations may not be monitored with the conventional direct measurement techniques. Their presence may be understood after getting them reacted with the externally added chemicals. The overall reaction may also be called the observed reaction.

The order of an elementary reaction is equal to the number of molecules entering the reaction. The forward reaction A 2 → 2A * in the elementary step 1 is a first order with respect to A 2, because only one molecule is converted into 2A*, but the reverse reaction 2A *A 2 is a second order with respect to A*, since two molecules of A* are involved. The overall order of an elementary step is the sum of the orders with respect to each species. With the same reasoning we can easily find out that the overall orders of both the forward and reverse reactions of the elementary step 2 and that of step 3 are 2.

We have both forward and reverse reactions associated with the elementary steps 1 and 2, which are called reversible reactions. There is only a forward reaction associated with the elementary step 3, and the reaction may not be reversed under the pertinent experimental conditions. The elementary step 3 is called an irreversible reaction.

A chemical reaction may be prevented via eliminating one of the reactants. Dipping the potatoes in water, or waxing the apples after cutting into half may reduce the availability of the oxygen for the color change reactions. Fermenting eggs eliminate glucose, thus preventing it from entering into deteriorative reactions during the dried egg production (Hill and Sebring 1990). Eliminating intermediates from the stepwise propagating reactions may slow down spoilage. Antioxidants react with free radicals and slow down the lipid oxidation reactions.

Increasing product concentration may establish equilibrium between the reactants and products of a reaction, thus slowing down the metabolic activity. Carbon dioxide in a controlled atmosphere gas mixture may slow down the carbon dioxide producing reactions of the Krebs cycle and consequently the whole metabolic activity.

A chemical reaction may be slowed down via eliminating the enzyme catalyzing the reaction. In thermal processing, heat is applied to convert the enzymes into inactive proteins (Chapter 4). Off-odors in unblanched or under blanched frozen vegetables may be caused by enzymatic oxidation of lipids. Lipoxigenase is the catalyst involved in these oxidation reactions. It is inactivated upon blanching and development of off-odors during the subsequent storage is prevented (Fennema, Powrie, and Marth 1973). Ionizing radiation radiolizes water and may cause the formation of reactive intermediates including excited water (H2O)*, free radicals (OH∙ and H∙), ionized water molecules (H3O)+ , and a hydrated electron ( e a q )

, which may consequently react with food components including the proteins. Ionizing radiation may cause cross-linking or scission of the proteins (Karel 1975). When the enzymes undergo such changes, they lose their activity and may not be able to catalyze the deteriorative reactions.

Changing the pH of the food with an acid addition or fermentation may denature the enzymes. Drying the foods may eliminate water needed for the enzyme activity. Some chemical reactions occur with the effect of light. Opaque packaging prevents light reaching the light sensitive foods or beverages.

Refrigeration is among the most common food preservation methods and slows down spoilage via reducing the rate constants of the deterioration reactions.

Food deterioration usually occurs via a large number of chemical reactions. The slowest reaction involving into a process is referred to as the rate determining step and its rate may be equal to the spoilage rate. Preventing a single reaction in a set of deterioration reactions may create a new rate determining steps and slowing down the process and increasing the shelf life of food. Since the chemical reactions are among the major causes of food spoilage, their kinetics may deserve special attention.

3.2  Rate Expression

Consumption rate RA of the species A with reaction A k 1 2 B

may be expressed as RA = –dcA /dt = k 1 cA . The minus sign implies that A is a reactant, therefore its concentration cA is decreasing with time. The parameter k 1 is called the reaction rate constant. The term –k 1 cA implies that the higher the concentration of reactant A, the higher the rate. The reaction rate also increases with k 1. If this reaction should describe loss of a nutrient in storage, the preservation methods would try to minimize the value of k 1 to slow down the nutrient loss. The production rate of B is RB = dcB /dt = –2dcA /dt = 2k 1 cA implying that two molecules of B are produced for each molecule of A consumed.

With the reaction 2 A + B k 2 k 1 C

the consumption rates of A and B and the production rate of C are d c A / d t = 2 d c B / d t = 2 d c C / d t = k 1 C A 2 C B k 2 C C . There are actually two reactions represented here. The forward reaction implies that two molecules of A and one molecule of B react together with the rate constant k 1 to produce one molecule of C. The reverse reaction implies that one molecule of C disintegrates with the reaction rate constant k 2 to produce two molecules of A and one molecule of B. When the rate of the forward reaction, k 1 C A 2 C B , is larger than that of the reverse reaction, k 2 cC , components A and B are depleted and C is produced with the net rate predicted by the rate expression. When the rate of the forward reaction becomes the same as that of the reverse reaction, the production rate of each component becomes the same as its depletion rate, and their concentrations do not change with time. This is called a chemical equilibrium. Since two molecules of A are involved into a reaction, the power of cA is 2 on the right-hand side of the rate expression. When a reaction represents the actual mechanism at the molecular level, the power of the concentration term on the right-hand side of the rate expression is the number of the molecules involved into the reaction. It is also possible that a chemical conversion may involve many reactions and the net change may be expressed with an apparent expression summing up the involving steps. In the rate expression of such reactions, the powers of the species concentration may not be equal to the net number of the molecules involving into the apparent expression. The forward rate expression is second order in A and first order in B. The order of the overall forward rate expression is three (i.e., sum of the orders with respect to each components). There is only one mole of chemical species involved into the reverse reaction and it is a first order.

Deterioration of the foods may be simulated in analogy with first- or zero-order irreversible monatomic reactions (Labuza 1980):

3.1 d c A d t = k c A ,

or

3.2 d c A d t = - k .

This does not mean that the reactions taking place are very simple reactions, rather it shows that complex systems can be simulated with simple apparent mathematical models. Other simple chemical reactions that may involve into food processing and preservation with their differential and integrated rate expressions are given in Table 3.1.

Units of the rate constant are determined by the reaction. With any elementary reaction, units of k are expressed in (concentration)1–n /time, where n = overall reaction rate order. With the zero-order reaction (i.e., Equation 3.2), k has the units of concentration/time, with a first-order reaction (i.e., Equation 3.1), k has the units of 1/time.

The rate expressions are arranged such that the rate constants are evaluated from the slopes of the differential or integrated rate expressions. The differential methods are based on the rate expressions evaluated via differentiation of the experimentally determined concentration versus time data. The integral methods (Examples 3.1 through 3.3) are based on an integration of the reaction rate expression. The numerical differentiation techniques are usually unstable, therefore, the integral methods are usually preferred over the differential methods.

Table 3.1   Elementary Reactions and their Rate Expressions

 

Example 3.1  Ascorbic Acid Loss in Packaged and Nonpackaged Broccoli

Reduced ascorbic acid retention in packaged and nonpackaged broccoli were measured by Barth et al. (1993) as

t (h)

Packaged cA (mg/g)

Nonpackaged cA (mg/g)

0

5.6

5.6

24

5.38

4.82

48

4.76

3.86

72

4.59

3.75

96

4.42

3.25

a. Find out if the data may be represented by Equation 3.1 or Equation 3.2.

Solution: Integrating Equation 3.1 gives: ln(cA ) = ln(cA0 ), where c A0 is the initial concentration of the reduced ascorbic acid in broccoli. Substituting the data into the integrated equation gives: ln(cA ) = 1.69-5.54‗10-3 t with r = –0.98 and se = 0.037 with nonpack-aged broccoli and ln(cA ) = 1.72-2.58×10-3 t with r = –0.97 and se = 0.006 with packaged broccoli.

Integrating Equation 3.2 gives: cA = c A0kt. Substituting the data into the integrated equation gives: CA = 5.41 – 0.024t with r = –0.97 and se = 0.28 with nonpackaged broccoli and cA = 5.58 – 0.013t with r = –0.97 and se = 0.011 with packaged broccoli. Since the correlation coefficient and the standard error of both models are very good, we may conclude that both models may represent the data very well. It should be noted that these models are simple empirical equations and do not describe the actual mechanism of the reaction.

MATLAB® code E.3.1 plots the variation of ascorbic acid concentration with time. Generally, it is not expected to have two models to simulate the same model equally well especially when we have large changes during the experiments. The small range of the reduced ascorbic acid loss may contribute to our observation.

 

Figure E.3.1.1   Comparison of Equation 3.1 with the experimental data. (Adapted from Barth, M. M., Kerbel, E. L., Perry, A. K., and Schmidt, S. J., Journal of Food Science, 58, 140–43, 1993.)

Figure E.3.1.2   Comparison of Equation 3.2 with the experimental data. (Adapted from Barth, M. M., Kerbel, E. L., Perry, A. K., and Schmidt, S. J., Journal of Food Science, 58, 140–43, 1993.)

b. Use Equations 3.1 and 3.2 to find out what percentage of the reduced ascorbic acid will remain after 100 hours of storage of the packaged broccoli.

Solution: Equations 3.1 and 3.2 are empirical equations and expected to simulate the data within the range of the experiments only. Although 100 hours of storage time is not in this range, it is close enough not to expect a drastic change in the trend with packaged broccoli. An integrated form of Equation 3.1 is ln(CA ) = ln(CA0 )-kt, after substituting c A0 = 5.6 mg/g and k = 2.58 10–3 h–1 and t = 100 hours we obtain cA = 4.32 mg/g. An integrated form of Equation 2.2 is cA = c A0kt, after substituting c A0 = 5.6 mg/g and k = 0.013 mg/g h and t = 100 hours we obtain cA = 4.30 mg/g.

 

Example 3.2  Simultaneous Nutrients and Toxin Degradation During Thermal Processing

The half-life of a bacterial toxin and a nutrient are 3 and 180 minutes, respectively, in a food at 121°C. Degradation processes may be described by Equation 3.1. Four log cycles of reduction is required in a toxin for safe food production. How much of the nutrient survives the heat treatment?

Solution: We have two reactions occurring in the same medium. Although these reactions do not interfere with each other, they inevitably take the same time. The half-life (t 1/2) of the chemical species is the time required to lose half of the initial value. An integrated form of Equation 2.1 is ln(cA /c A0) = -kt after substituting cA /c A0 = 0.5 and t 1/2 = 3 minutes we obtain k = 0.231 min–1 for the toxin. Similarly with cA /c A0 = 0.5, and t 1/2 = 180 we find k = 0.004 min–1 for the nutrient. After substituting k = 0.231 min–1 and cA /c A0 = 10–4 in the integrated model we will get t = 40 minutes = time required to reduce the toxin content by four log cycles. After substituting k = 0.004 min–1 and t = 40 minutes in the same integrated model, we will calculate cA /c A0 = 0.85, implying that 85% of the nutrient will survive the heat treatment.

 

Example 3.3  Shelf Life Calculation Based on Nutrient Loss

A micronutrient A undergoes a reaction A + B k 1 C

during storage of a food. The initial concentration of A is 2 g/kg and that of B is 75 g/kg. When the micronutrient concentration falls 75% of its initial level the food becomes inedible. Calculate shelf life of the food with k 1 = 0.0001 week–1 (g/kg)–1.

Solution: The rate expression is dcA /dt = –kcAcB . Variation in B is negligible even after all of A is consumed with the reaction, since cB is much larger than cA . We may consider k = kcB = constant, and the rate expression becomes –dcA /dt = –k cA . This is called a pseudo first-order rate expression. An integrated pseudo first-order rate expression is ln(cA /c A0)=-k* t. After substituting k = 0.0075 week–1 and cA /c A0 = 0.75, we may calculate the shelf life of the food t = 38 weeks.

 

Example 3.4  Kinetics of Nutrient Loss With Sequential Chemical Reactions

Nutrient A undergoes a degradation reaction ABC. The rate expressions for this reaction are

E.3.4.1 d C A d t = - k 1 c A + k 2 c B ,
E.3.4.2 d C B d t = - k 1 c A ( k 2 + k 3 ) c B ,
E.3.4.3 d C c d t = k 3 c B ,

where the rate constants are k 1 = 0.6 weeks–1, k 2 = 0.2 weeks–1, and k 3 = 0.1 weeks–1. Initial substrate concentrations were c A0 = 35 g/L, c B0 = c C0 = 0 g/L. Solve these differential equations simultaneously and plot variations of cA , cB , and cC with time. When cC = 20 g/L, the food is considered inedible. Determine the shelf life of the food from the plot.

Solution: We may rearrange the equations and solve as described in Table 2.6:

( D + k 1 ) c A - k 2 c B = 0 - k 1 c A + ( D + k 2 + k 3 ) c B = 0 , - k 3 c B D c c = 0

where D = d/dt,

Δ = | D + k 1 - k 2 0 - k 1 D + k 2 + k 3 0 0 - k 3 D | = ( D + k 1 ) | D + k 2 + k 3 0 - k 3 D | - ( - k 1 ) | - k 2 0 - k 3 D | + ( 0 ) | - k 2 0 D + k 2 + k 3 0 | = ( D + k 1 ) [ ( D + k 2 + k 3 ) ( D ) - ( - k 3 ) ( 0 ) ] + ( k 1 ) [ ( - k 2 ) ( D ) - ( - k 3 ) ( 0 ) ] = D 3 + ( k 1 + k 2 + k 3 ) D 2 + k 1 k 2 D ,
D c A = | 0 - k 2 0 0 D + k 2 + k 3 0 0 - k 3 D | = 0 , D c B = | D + k 1 0 0 k 1 0 0 0 0 D | = 0 , D c C = | D + k 1 k 2 0 k 1 D + k 2 + k 3 0 0 k 3 0 | = 0 ,
  • cA = DcA
E.3.4.4 d 3 c A d t 3 + ( k 1 + k 2 + k 3 ) d 2 c A d t 2 + k 1 k 3 d c A d t = 0 ,
  • cB = DcB
E.3.4.5 d 3 c B d t 3 + ( k 1 + k 2 + k 3 ) d 2 c B d t 2 + k 1 k 3 d c B d t = 0 ,

and

  • cC = DcC
E.3.4.6 d 3 c c d t 3 + ( k 1 + k 2 + k 3 ) d 2 c c d t 2 + k 1 k 3 d c c d t = 0.

Assume a solution y = Ce λ t, therefore, dy/dt = λe λ t, d 2 y/dt 2 = λ3 d λ t, and d 3 y/dt 3 = λ3 e λ t substitutes all in the differential equation and obtains the characteristic equation:

E.3.4.7 λ 3 + ( k 1 + k 2 + k 3 ) λ 2 + k 1 k 3 λ = 0.

Solutions of the characteristic equation (after substituting values of the rate constants) λ1 = 0, λ2 = –0.072, and λ3 = –0.827, therefore

E.3.4.8 c A = K 1 + K 2 e - 0 0 72 t + K 3 e - 0 8 27 t ,
E.3.4.9 c B = K 4 + K 5 e - 0 0 72 t + K 6 e - 0 8 27 t ,
E.3.4.10 c c = K 7 + K 8 e - 0 0 72 t + K 9 e - 0 8 27 t .

We have three equations with nine unknown constants. We may determine the constants if we can reduce their number to three. After substituting the solutions for cA and cB in

d c A d t = - k 1 c A + k 2 c B ,

we will get

  • -0.072K 2 e -0.072t -0.827k 3 e -0.827t =-0.6(K 1+K 2 e -0.0727t +K 3 e -0.827t )
  • +0.2(K 4+K 5 e -0.072t +K 6 e -0.827t ).

The same exponential terms must have the same coefficients on both sides of the equation, therefore

Term

Equal Coefficients

Relation Between Constants

e 0

0 = 0.6 K 1 + 0.2 K 4

K 4 = –0.3 K 1

e –0.072t

–0.072 K 2 = –0.6 K 2 + 0.2 K 5

K 5 = 2.64 K 2

e –0.827t

–0.827 K 3 = –0.06 K 3 + 0.2 K 6

K 6 = –1.135 K 3

After substituting the solutions for cB and cC in dcc /dt = k 3 cB , we will get

  • -0.072K 8 e -0.072t - 0.827K 9 e-0.827t = 0.1(-0.3K 1+2.64K 2 e -0.072t -1.135K 3 e -0.827t ).

The same exponential terms must have the same coefficients on both sides of the equation, therefore

Term

Equal Coefficients

Relation Between Constants

e 0

0 = –0.03 K 1

K 1 = 0

e –0.072t

–0.072 K 8 = 0.264 K 2

K 8 = –3.67 K 2

e –0.827t

–0.827 K 9 = –0.114 K 3

K 9 = 0.137 K 3

After substituting the constants we will have

E.3.4.11 c A = K 2 e - 0 0 72 t + K 3 e - 0 8 27 t ,
E.3.4.12 c B = - 3 . 6 7 K 2 e - 0 0 72 t + 0 . 1 3 7 K 3 e - 0 8 27 t ,
E.3.4.13 c c = K 7 + 5 . 2 5 K 2 e - 0 0 72 t + 0 . 1 9 K 3 e - 0 8 27 t .

Equations E.3.4.11 through E.3.4.13 have three unknown constants (K 2, K 3, K 7); we may solve these constants by using the initial conditions:

  • 35=K 2+K 3,
  • 0=2.64K 2-1.135K 3,
  • 0=K 7-3.667K 2+0.137K 3.

Therefore, K 2 = 10.5, K 3 = 24.48, K 7 = 35.23, and

E.3.4.14 c A = 1 0 . 5 e - 0 0 72 t - 2 4 . 4 7 e - 0 8 2 7 t ,
E.3.4.15 c B = 27.7 2 e - 0 0 72 t - 27.7 8 e - 0 8 27 t ,
E.3.4.16 c C = 35.2 3 - 38.5 e - 0 0 72 t - 3 . 3 5 e - 0. 8 27 t .

An incomparably easy solution may also be obtained with MATLAB® code E.3.4. Variations of cA , cB , and cC during the storage period are shown in Figure E.3.4 It may also be seen from the figure that the shelf life of the food (cC = 20 g/L) is about 13 weeks.

 

3.3  Why Do Chemicals React?

In nature all the systems try to lower their energy and increase their disorder. The same rule is also valid for chemicals. A chemical reaction occurs spontaneously if the total Gibbs free energy of formation of the products is smaller than that of the reactants. When molecules collide, bonds of the molecules are broken first and new bonds are established to produce an activated complex. The lifetime of the activated complex is very short, it rearranges its molecular structure very rapidly to form the products. The rate of a chemical reaction may be computed by assuming either collision or dissociation of the activated complex as the rate-limiting steps.

When we consider the reaction (A and B are ideal gases):

  • A + B → products,

Figure E.3.4   Variation of cA , cB , and cC in storage.

with the rate

3.3a R = k c A c B .

The number of collisions (zAB ) of A with B in unit volume in unit time is

3.3b z A B = ( σ A + σ B 2 ) 2 N A v 2 1 0 6 8 π κ T ( 1 M A + 1 M B ) ,

where σ = diameter of a molecule; MA , MB = molecular weights, κ = Boltzmann constant. Only the collisions involving with more than a minimum activation energy Ea may lead to a reaction. The fraction of all bimolecular collisions involving more energy than Ea is exp(–Ea /RT); therefore, the rate of the reaction and the rate constant are

3.3c R = z A B exp ( E a R T ) c A c B ,
3.3d k = ( σ A + σ B 2 ) 2 N A v 2 1 0 6 8 π κ T ( 1 M A + 1 M B ) exp ( - E a R T ) .

When the dissociation of the activated complex is the rate-limiting step, we need to compute the activated complex AB* concentration from the chemical reaction:

3.4a A + B k 2 k 1 A B * .

There is an equilibrium between the reactants and the activated complex at all times:

3.4b K e q = k 1 k 2 = c A B * c A c B .

The activated complex undergoes decomposition as

3.4c A B * k 3 products.

The rate constant of decomposition is the same for all reactions:

3.4d k 3 = κ T h ,

where h = Planck constant.

The product formation rate and the rate constant are

3.4e R = k 3 c A B * = κ T h K e q c A c B ,
3.4f k = κ T h K e q .

Variation of the Gibbs free energy along the reaction path is described in Figure 3.1. The higher the Gibbs free energy of the activated complex, the smaller is the fraction of the molecules that can gain sufficient energy to exceed the energy barrier. A catalyst makes it possible to form an activated complex with a lower activation energy barrier, therefore, a higher fraction of the molecules may pass through it and the reaction rate increases (Figure 3.2).

Schematic description of the Gibbs free energy levels at different reaction coordinates. Δ

Figure 3.1   Schematic description of the Gibbs free energy levels at different reaction coordinates. ΔG activation is the activation energy barrier.

Figure 3.2   A different activated complex with smaller activation free energy forms when a catalyst is used. More molecules will be eligible to pass through the activation energy barrier when the ΔG activation is lower.

Collision theory and transition state theory permit computation of the rates of the reactions involving idealistic cases, which are beyond the scope of this book, but they form the basis for the analogy models, which will be discussed extensively under numerous titles.

3.4  Temperature Effects on Reaction Rates

Temperature effects on the rate constants may be described with the Arrhenius expression:

3.5 k = k 0 exp { - E a R T } ,

where k = rate constant, k 0 = preexponential constant, Ea = activation energy, R = gas constant, and T = absolute temperature.

 

Example 3.5  Vitamin Loss in a Snack Food

Loss of a vitamin in a snack food agrees with Equation 3.1. Estimate the time required to lose 15% of the initial vitamin content at 22°C if half-lives of the vitamin at different storage temperatures were

T(°C)

10

15

20

25

t 1/2 (days)

2900

1600

925

530

Solution: Half-life (t 1/2) is the time required to lose half of the initial vitamin content. An integrated form of Equation 3.1 is ln(CA /C A0)=-kt after substituting cA /c A0 = 0.5 and t = t 1/2 we obtain

T(°C)

10

15

20

25

K(day–1)

2.4 10–4

4.3 10–4

7.5 10–4

1.3 10–3

Equation 3.5 may be rewritten as ln(k) = ln (k 0) - E/RT, then data are

1/T(K–1)

3.53 10–3

3.47 10–3

3.41 10–3

3.36 10–3

ln k

–8.33

–7.75

–7.20

–6.65

After plotting ln k versus 1/T (Figure E.3.5) we may obtain the best line ln(k) = 26.23-9.8103/T (r = –1.0) with the intercept ln(k 0) = 26.23 and the slope Ea /R = 9.8 103 K –1. After substituting these parameters and T = 295 K in Equation 3.5, we obtain = 9.2 10–4 day–1 at 22°C. We may substitute the calculated value of k and cA /c A0 = 0.85 in ln(CA /C A0) = -kt to estimate the time required to lose 15% of the initial vitamin content at 22°C at 177 days. The details of the computations are available in MATLAB® code E.3.5.

Comparison of the best fitting line with the data as produced by the

Figure E.3.5   Comparison of the best fitting line with the data as produced by the MATLAB code E.3.5.

 

 

Example 3.6  Total Amounts of Nutrient Loss After Sequences of a Canning Process

Initial nutrient content of a fresh vegetable is 5 g/kg. The degradation rate of the nutrient and the temperature effects on the rate constant may be described with Equations 3.1 and 3.5, respectively, with frequency factor k 0 = 0.2 min–1 and activation energy Ea = 5030 J/mole. The following operations occur during processing at the given average temperatures: (i) blanching 5 minutes at 100°C, (ii) canning 15 minutes at 60°C, (iii) thermal processing 20 minutes at 121°C. What will be the remaining concentration of the nutrient at the end of processing?

Solution: Amounts of the nutrient surviving may be calculated from cA = c A0 exp{–kt}, where k = k 0 exp{–Ea /RT} and R = 8.314 J/mole K. After substituting the numbers,

  • Blanching T = 373 K, k = 0.040 min–1, c A0 = 5 g/kg, cA = 4.09 g/kg.
  • Canning T = 333 K, k = 0.033 min–1, c A0 = 4.09 g/kg, cA = 2.60 g/kg.
  • Thermal processing T = 394 K, k = 0.043 min–1, c A0 = 2.60 g/kg, cA = 1.10 g/kg = remaining concentration of the nutrient at the end of processing.

MATLAB® code E.3.6 carries out the computations.

3.5  Precision of Reaction Rate Constant and Activation Energy Determinations

A general nth order rate expression (n ≠ 1) is

3.6a d c A d t = k c A n ,

 

after rearrangement and integration the rate constant will be

3.6b k = c A 1 n - 1 - c A 2 n - 1 ( n - 1 ) ( t 2 - t 1 ) c A 2 n - 1 c A 1 n - 1 ,

where subscripts 1 and 2 denote the beginning and the end of an interval, respectively. In the completely general case of dependent variable y = f (x 1, x 2,...,xn ) the relative error in y due to the relative errors of x 1, x 2,...,xn is given by

3.6c ( Δ y y ) = Σ i = 1 n ( f x i ) 2 ( Δ x i x i ) 2 .

If we can assume that errors in c A1, c A2, t 1, and t 2 are independent we may calculate the relative error in k after using Equation 3.6c as (Hill and Grieger-Block 1980)

3.6d ( Δ k k ) 2 = ( Δ t 1 t 2 - t 1 ) 2 + ( Δ t 2 t 2 - t 1 ) 2 + ( ( n - 1 ) c A 2 n - 1 c A 1 n - 1 - c A 2 n - 1 ) 2 ( Δ c A 1 c A 1 ) 2 + ( ( n - 1 ) c A 1 n - 1 c A 1 n - 1 - c A 2 n - 1 ) 2 ( Δ c A 2 c A 2 ) 2 .

In the case of n = 1, the corresponding equation is:

3.6e ( Δ k k ) 2 = ( Δ t 1 t 2 - t 1 ) 2 + ( Δ t 2 t 2 - t 1 ) 2 + ( 1 ln ( c A 1 / c A 2 ) ) 2 ( Δ c A 1 c A 1 ) 2 + ( 1 ln ( c A 1 / c A 2 ) ) 2 ( Δ c A 2 c A 2 ) 2 .

Activation energy Ea may be calculated from the Arrhenius expression 3.5 as

3.6f E a = R T 1 T 2 T 2 - T 1 ln ( k 2 / k 1 ) ,

where subscripts 1 and 2 denote the beginning and the end of an interval, if the errors in each of the quantities k 1, k 2, T 1, and T 2 are random, the relative error in the Arrhenius activation energy is given after using Equation 3.6c as

3.6g ( Δ E E a ) 2 = ( T 2 T 2 - T 1 ) 2 ( Δ T 1 T 1 ) 2 + ( T 2 T 2 - T 1 ) 2 ( Δ T 2 T 2 ) 2 + { 1 ln ( k 1 / k 2 ) } 2 { ( Δ k 1 k 1 ) 2 + ( Δ k 2 k 2 ) 2 }

Equation 3.6g shows that the relative error in the activation energy is strongly dependent on the size of the temperature interval chosen and the error involved in the temperature measurements.

 

Example 3.7  Precision of the Rate Constant

Determine the uncertainty involved in the determination of the reaction rate constant and the activation energy (of the Arrhenius equation) of the following reaction:

E.3.7 d c A d t = - k c A 2 .

Data were recorded at t 1 = 0 and at t 2 = 100 minutes, uncertainty in each time measurement was Δt 1 = 1 s and Δt 2 = 1 s. At time t 2 = 100 minutes it was found that c A1 = 0.5c A0 and relative uncertainty (Δc/c) in each concentration measurement was 1%. It was also observed that the reaction rate constant increases by 5% when temperature increases from 40 to 60°C. MATLAB® code E.3.7 uses Equations 3.6d and 3.6g to compute the uncertainty in the reaction rate constant and the activation energy determinations.

 

Therefore the values of the reaction rate constant and the activation energy are k 1 = 0.60 ± 0.03 s–1; k 2 = 0.63 ± 0.03 s–1; Ea = 889120 ± 80 J/mol K.

3.6  Enzyme-Catalyzed Reaction Kinetics

Enzymes are the natural protein catalysts of the cellular reactions. They are usually very specific and catalyze only one reaction involving only one substrate. They lose their activity if the natural folding pattern of the protein changes. A single enzyme catalyzed one substrate reaction may be expressed as

3.7a S E P ,

where the reaction rate is

3.7b υ = - d c s d t = d c P d t .

The terms –dcS /dt is the substrate consumption rate and dcp /dt is the product formation rate. The rates of the enzyme catalyzed reactions were referred to as velocity in the pioneering biology literature, therefore they are conventionally denoted with the letter v. The mechanism for a single enzyme catalyzed one substrate reaction was first suggested by Michaelis and Menten (1913) as

3.7c S + E k 2 k 1 S E ,
3.7d S E k 3 P + E .

The first elementary reaction of this mechanism is considered as an equilibrium step with the dissociation constant

3.7e K M = c E c S c E S ,

where KM = Michaelis constant, cE = enzyme concentration, cS = substrate concentration, and cES = concentration of the ES complex. The total enzyme concentration was initially c E0, after making the ES complex, the concentration of the free enzyme cE may be calculated as

3.7f c E = c E 0 - c E S .

The second reaction is slow, therefore its rate is the same as the rate of the overall apparent reaction. The slowest reaction in such a mechanism is called the rate determining step. The rate of the second elementary reaction is

3.7g υ = d c P d t k 3 c E S .

It is not usually possible to measure cES , therefore we may use Equations 3.7a through f to rearrange Equation 3.7g as

3.8 υ = υ max C S K M + c s .

This is called the Michaelis–Menten equation, where

3.9 υ max = k 3 c E 0 .

It was later claimed by Briggs and Haldane (1925) that 3.7[c] may not be an equilibrium step, and the material balances for the substrate and the intermediary complex ES were expressed as

3.10a υ = - d c s d t = k 1 c S c E - k 2 c E S ,

Table 3.2   Linear Arrangements of the Michaelis–Menten Equation

Lineweaver–Burk arrangement (Lineweaver and Burk, 1934)

1 υ = 1 υ max + K M υ max 1 c S

Eadie–Hofstee arrangement (Eadie, 1942; Hofstee, 1959)

υ = υ max K M υ c S

Hanes–Woolf arrangement (Hanes, 1932; Haldane and Stern, 1932)

c s υ = K M υ max + 1 υ max c s

3.10b d c E S d t = k 1 c S c E - k 2 c E S - k 3 c E S .

The complex does not accumulate, i.e.,

3.10c d c E S d t = 0.

After using Equations 3.10a through c, Equation 3.8 is obtained with the Michaelis constant

3.10d K M = k 2 + k 3 k 1 .

Enzymes belonging to the classification of hydrolases are the typical examples to the single enzyme catalyzed one substrate reactions.

The Michaelis–Menten equation has a variable apparent order. When KM cS the apparent rate is v = kappcS (first order in cs ), where kapp = v max/KM ; when KM cS the apparent rate is v = v max (zero order in c s). The Michaelis–Menten equation is generally arranged in three different linear forms to evaluate the apparent constants vmax and KM from the slopes and the intercepts of the plots of the experimental data (Table 3.2). A number of advantages and disadvantages are associated with each type of plot. Even spacing of the data points along the line and best fit of the data points to a straight line are among the factors to be considered while making such a decision. More workers use a Lineweaver–Burk method than the other two combined.

 

Example 3.8  Kinetics of Linolenic Acid Peroxidation by Sunflower Lipoxygenase

Strong lipoxygenase activity is observed during the first days of sunflower seed germination, which may cause lipid peroxidation under unfavorable storage conditions. Linolenic acid is a substrate for sunflower lipoxygenase. The following data were evaluated from a publication by Leoni, Iori, and Palmeri (1985).

c (mM)

0.0025

0.0033

0.0052

0.0080

0.012

0.05

0.015

0.25

υ (U/mg protein)

21

27

33

43

50

60

53

54

If the apparent reaction agrees with the Michaelis–Menten scheme, determine the constants of the rate expression.

  • Double reciprocal (Lineweaver–Burk) plot. Equation 3.8 and the data may be rearranged as
    1 V = 1 V m a × + K M V max 1 C .

    1/c(mM)-1 400 303 192 125 83 20 6.6 4
    1/υ (U/mg protein)-1 0.0476 0.0370 0.0303 0.0232 0.0200 0.0167 0.0189 0.0185
    The best fitting line to the data is
    1 V = 0 . 0 1 6 1 + 7 . 3 5 × 1 0 - 5 1 c ( r = 0.98 ) .
    The intercept is 1/v max, therefore v max = 62 U/mg protein and the slope is KM /v max, therefore KM = 5 × 10–6 M. The Lineweaver–Burk plot is shown in Figure E.3.8.1.
  • Eadie–Hofstee plot. Equation 3.8 and the data may be rearranged as
    V = V max - K M V c .

    υ/c (U/mg protein mM) 8400 8181.8 6346.1 5375 4166.7 1200 3533.3 216
    υ (U/mg protein) 21 27 33 43 50 60 53 54
The Lineweaver–Burk plot to determine kinetic constants

Figure E.3.8.1   The Lineweaver–Burk plot to determine kinetic constants KM and υ max.

The best fitting line to the data is

V = 63 0.0044 V C ( r = 0.94 ) .

The intercept is v max, therefore v max = 63 U/mg protein and the slope is KM , therefore KM = 4.4 × 10–6 M. The Eadie–Hofstee plot (Eadie, 1942; Hofstee, 1959) is shown in Figure E.3.8.2. Details of the computations are given in MATLAB® code E.3.8.

The Eadie–Hofstee plot to determine kinetic constants

Figure E.3.8.2   The Eadie–Hofstee plot to determine kinetic constants KM and υ max.

 

Enzymatic reactions involved in food processing and preservation may also involve two substrates:

3.11 A + B E P + Q .

Three different mechanisms are suggested for these reactions (Whitaker 1994). The ordered and the random mechanisms suggest that the products may be released only after both of the substrates bound to the enzyme. In the ordered mechanism, it is always the same substrate bound to the enzyme first, and the same product is released first. In the random mechanism there is no priority in binding of the substrates or removal of the products. The ordered and random mechanisms result in the same rate expression:

3.12 υ = υ max C A C B ( c A + K A ) ( c B + K B ) .

In a ping pong mechanism the first product is released after binding the first substrate; then the second substrate is bound and subsequently the second product is formed, leading the rate expression:

3.13 υ = υ max C A C B c A c B + K A c B + K B c A .

The KAKB term is missing in the denominator of Equation 3.13 since there is no ternary complex in the mechanism. It should be noticed that Equations 3.12 and 3.13 are valid when cA and cB are maintained at constant levels and there is no product accumulation in the reaction medium. Kinetic constants of Equations 3.12 and 3.13 may be evaluated by performing two sets of experiments. In the first set of experiments where cA is constant and cB is variable, the first set of apparent kinetic constants may be obtained treating the data as explained for the single substrate reactions. The remaining kinetic constants may be obtained after making experiments with variable cA and constant cB (Whitaker 1994).

In aqueous solutions H+ and OH ions interact with the enzyme as

3.14a E + H + K E H E H + ,
3.14b E + O H K E O H E O H + .

When we combine the temperature and the pH effects we end up with the following equation:

3.15 c a c t i v e = A c t 0 ( 1 + c H + K E H + K W K E O H 1 c H + ) .

Where Act0 is the maximum attainable activity of the enzyme. MATLAB® code 3.1 simulates the pH effects on hazelnut lipase.

 

Figure 3.3   Effect of the pH variations on the hazelnut lipase. The first iso-enzyme has the maximum activity at about pH = 4.75, the second iso-enzyme has the maximum activity at about pH = 7.75. (Experimental data and constants of the model adapted from Seyhan, F., Tijskens, L. M. M., and Evranuz, O., Journal of Food Engineering, 52, 387–95, 2002.)

Any substance that reduces the rate of an enzyme catalyzed reaction is called an inhibitor. A competitive inhibitor competes with the substrate for the active site of the enzyme, whereas a noncompetitive inhibitor does not.

Under competitive inhibition the Michaelis–Menten mechanism prevails

3.16a 3.16a

where

3.16b K S = C E C S C E S ,

and

3.16c K i = C E C I C E I .

The product formation rate is

3.16d υ = d c p d t = k p c E S .

Equation 3.16b may be rearranged as

3.16e υ = k p c E 0 c E + c E S + c E I c E S ,

where c E0 = c E + cES + cEI . After substituting υ max = kpc E0 and using Equations 3.16b and 3.16c to eliminate c ES and cEI , Equation 3.16e becomes

3.16f υ = υ max C S c s + K s ( 1 + c I K i ) .

After comparing Equation 3.16f with Equation 3.8 we see that the competitive inhibitor increased the Michaelis constant KM by a factor of (1 + cI /Ki ). Equation 3.16f implies that v decreases with cI and increases with Ki . When cs Ks (1 + cI /Ki ), an inhibitor may not have a considerable effect on the reaction rate. After substituting KI = Ks (1 + cI /Ki ), Equation 3.16f may be rearranged as

3.17 1 υ = 1 υ max + K I υ max 1 C S .

The type of an inhibition where υ max remains approximately the same with all inhibitor concentrations is referred to as the noncompetitive. A noncompetitive inhibitor binds to a site different than that of the substrate. If the Michaelis–Menten mechanism prevails, after including the inhibitor we will have

3.18a 3.18a

where the dissociation constants are

3.18b K S = c E c S c E S = c E I c S c E S I ,
3.18c K i = c E c I c E I = c E S c I c E S I ,

The product formation rate is

3.18d υ = d c p d t = k p c E S .

Equation 3.18d may be rearranged as

3.18e υ = k p c E 0 c E + c E S + c E I + c E S I c E S ,

where cE0 = cE + cES + cEI + cESI . After substituting v max = k p c E0 and using Equations 3.18b and 3.18c to eliminate c ES and cEI , Equation 3.18e becomes

3.18f υ = υ a p p c s c s + K s ,

where υapp = υ max/(1 + cI /Ki ). After comparing Equation 3.18f with Equation 3.8 we see that the noncompetitive inhibitor decreased the apparent maximum rate υ max by a factor of (1 + cI /Ki ). Equation 3.18f implies that v decreases with cI . Increasing cS does not affect υapp . Equation 3.18f may be rearranged as

3.19 1 υ = 1 υ a p p + K s υ a p p 1 c S .

Parameters υapp and KS may be evaluated from intercept and slope.

MATLAB® code 3.2 describes determination of the type of inhibition and numerical values of parameters υ max and KI during inhibition of mushroom tyrosinase by o-toluic acid.

 

Parameters v max and KI may be evaluated from the intercept and slope as explained with a typical example in Figure 3.4a. Increasing the concentration of the inhibitor results in a family of lines with different slopes and intercepts. They also intercept each other when 1/c ≅ –0.75. Since both KI and v max varies with an o-toluic acid concentration (Figure 3.4b) inhibition is referred to as mixed type.

(a) Double reciprocal plot to evaluate kinetic constants under noncompetitive inhibition of mushroom tyrosinase by o-toulic acid. (b) Variations of the model parameters

Figure 3.4   (a) Double reciprocal plot to evaluate kinetic constants under noncompetitive inhibition of mushroom tyrosinase by o-toulic acid. (b) Variations of the model parameters v max and KM with inhibitor (o-toluic acid) concentration. (Adapted from Huang, X.-H., Chen, Q.-X., Wang, Q., Song, K.-K., Wang, J., Sha, L., and Guan, X., Food Chemistry, 94, 1–6, 2006.)

The maximum attainable rate υ max in Equation 3.8 was defined when the initial enzyme activity c E0 was constant as

3.9 υ max = k 3 c E 0 .

Temperature effects on the rate constant k 3 may be simulated with the Arrhenius expression:

3.5 k 3 = k 0 exp { - E a R T } .

Denaturation of the enzyme may be expressed with a first-order rate expression:

3.20 d c E d t = - k d c E .

At a constant temperature T when the enzyme undergoes denaturation, after integrating Equation 3.20 and combining with Equations 3.5 and 3.9 we may express the temperature dependence of the enzyme activity as

3.21 υ = c S c E 0 k 0 e - k d e - E a / R T t K M + c s .
 

Example 3.9  Kinetic Compensation Relations for Pectinesterase Inactivation During Pasteurization of Orange Juice

Enzyme inactivation during thermal processing of the foods may be described in analogy with unimolecular, irreversible, first-order chemical reaction (Ulgen and Özilgen 1991):

active enzyme E inactive enzyme E i
3.20 d c E d t = - k d c E .

After integrating Equation 3.20 we will have

E.3.9.1 1 n c E = 1 n c E 0 - k d t .

When we plot ln(cE ) versus time during inactivation at a constant temperature, the slope of the line gives kd as exemplified in Figure E.3.9.1. MATLAB® code E.3.9 evaluates the inactivation model constants from the data obtained at 60 and 70°C.

Kinetic constants k 0 and Ea of the Arrhenius expression are not usually independent of each other in a family of related systems where parameters k 0 and Ea change due to slight variations in the experimental conditions (like pH, sugar concentration, etc.). The variation in Ea may be compensated by the changes in k 0 with the relation (Figure E.3.9.2):

E.3.9.2 l n ( k 0 ) = α E a + β ,

where α and β are constants. MATLAB® code E.3.9.b evaluates the compensation relation with the given lnk 0 and Ea data.

Figure E.3.9.1   Typical plots for pectinesterase inactivation (observed as a decrease in enzyme activity) during pasteurization of orange juice. (From Ulgen, N., and Özilgen, M. Journal of the Science of Food and Agriculture, 57, 93–100, 1991.)

 

Figure E.3.9.2   Kinetic compensation relation for pectinesterase inactivation during pasteurization of orange juice. Equation of the line: ln(k 0)=0.3373E a -3.24. (From Ulgen, N., and Özilgen, M. Journal of the Science of Food and Agriculture, 57, 93–100, 1991.)

 

 

Example 3.10  Kinetics of Inactivation of the Peroxidase Iso Enzymes During Blanching of Potato Tuber

Peroxidase is usually present in the fruits and vegetables as a combination of various iso enzymes with different heat stabilities. During blanching of a spherical potato tuber the controlling equation of the temperature profile is (Example 2.9)

E.3.10.1 T - T 1 T 0 - T 1 = R r ( 2 π ) Σ n = 0 { ( - 1 ) n + 1 n e - ( π n ) 2 sin ( π n r R ) } .

Inactivation kinetics of the enzyme may be described with separate first-order reactions for heat stable and heat labile fractions (Sarikaya and Özilgen 1991):

E.3.10.2 d c E 1 d t = - k 1 c E 1 ,

and

E.3.10.3 d c E 2 d t = - k 2 c E 2 .

Total enzyme activity is

E.3.10.4 c E = c E 1 + c E 2 .

Temperature effects on the inactivation rate constants k 1 and k 2 were described with the Arrhenius expression:

E.3.10.5 k 1 = k 1 0 exp { - E a 1 R g T } ,

and

E.3.10.6 k 2 = k 20 exp { - E a 2 R g T } .

MATLAB® code E.3.10 plots the model to describe the variation of temperature and enzyme activity as a function of time.

 

Figure E.3.10.1   Temperature profiles at r = 0.6 cm distance from the center during thermal processing of whole potatoes. (Constants of the model adapted from Sarikaya, A., and Özilgen, M., Lebensmittel-Wissenschaft und Technologie, 24, 159–63, 1991.)

Figure E.3.10.2   Enzyme activity profiles at r = 0.6 cm distance from the center during thermal processing of whole potatoes. (Constants of the model adapted from Sarikaya, A., and Özilgen, M., Lebensmittel-Wissenschaft und Technologie, 24, 159–63, 1991.)

Enzyme activity may reappear some time after thermal processing, if heat treatment is not sufficient. The total Gibbs free energy of an enzyme suspension is (Shulz and Schirmer 1979)

3.22 Δ G t o t a 1 = Δ H c h a i n - T Δ S c h a i n + Δ G s o l v e n t ,

where ΔG total is the total Gibbs free energy of an enzyme plus the solvent, ΔHchain is the binding enthalpy of the chain in a vacuum provided mostly by hydrogen bonding and van der Waals interactions in the chain, ΔS chain is the chain entropy, T is the absolute temperature, and ΔGsolvent is the Gibbs free energy of solvent. The chain plus a solvent system attains the minimum ΔG total (and maximum ΔS chain) corresponding to the active folding pattern of the enzyme in its native environment (i.e., the vegetable or animal tissue). The secondary, tertiary, and the quaternary structure of the proteins are dictated by their primary structure. Destroying the protein structure at any level eliminates the enzyme activity, but the enzyme may fold back to the original structure and regain its activity due to the thermodynamic reasons, if the primary structure should not be destroyed. The primary structure of the proteins or their postproduction modifications, including the –S–S- bonds, are required to be destroyed irreversibly to prevent regaining of the enzyme activity.

In biological reactors or sensors, enzymes are frequently immobilized by using a carrier. Immobilization may be achieved via entrapment in the network or binding on the surface of a carrier. Either pure or crude enzyme preparations or the whole cell may be immobilized. When the enzymes are immobilized into the spherical particles Equation 2.14 (the equation of continuity in a spherical coordinate system) may be simplified under steady state conditions as

3.23a D e 1 r 2 d d r ( r 2 d c d r ) = υ a p p c c + K M ,

where c = c(r) = substrate concentration in the particle and De = effective diffusivity of the substrate within the particle, υapp is the maximum attainable apparent rate of the immobilized enzyme, and KM is the Michaelis constant of the free enzyme. The boundary conditions are

3.23b c = c s at r = R ,
3.23c d c d r = 0 at r = 0 ,

where c s is the substrate concentration on the surface of the particle. When cKM , Equation 3.23a becomes

3.24a 1 r 2 d d r ( r 2 d c d r ) = υ a p p c D e K M .

The solution to Equation 3.24a is (Bird, Stewart and Lightfoot 2007)

3.24b c ( r ) c b = r * sinh [ ϕ r * ] sinh [ ϕ ] ,

where r * = r/R and parameter ϕ = R υ max / D e K M

. Parameter ϕ is called the Thiele modulus.

The reaction rate on the surface of the particle is

3.24c r s u r f a c e A s u r f a c e = 4 π R 2 N s = - 4 π R 2 D e [ d c d r ] r = R .

Where r surface is the reaction rate per unit area of the surface and A surface is the surface area of the particle. We will obtain the following equation after substituting Equation 3.24b in Equation 3.24c (Bird, Stewart and Lightfoot 2007):

3.24d r s u r f a c e A s u r f a c e = 4 π R 2 D e c s [ 1 - ϕ coth ( ϕ ) ] .

If the enzymes would not have been immobilized, the total reaction rate would be

3.24e r v o l u m e V p a r t i c l e = 4 3 π R 3 [ υ m c s K M ] .

The effectiveness factor η is defined as (Bird, Stewart and Lightfoot 2007; Wang et al. 1979)

3.24f η = r s u r f a c e A p a r t i c le r v o 1 u m e V p a r t i c le = 3 ϕ [ 1 tanh ( ϕ ) - 1 ϕ ] .

Variation of the effectiveness factor with Thiele modulus is computed with MATLAB® code 3.3 and depicted in Figure 3.5.

The apparent value of Michaelis constant KM and υ max may change upon immobilization. The support may reduce the activity of an immobilized enzyme toward large molecules via steric hindrance of the active site. An immobilized enzyme is also in a different

 

Theoretical relationship between the Thiele modulus and the effectiveness factor.

Figure 3.5   Theoretical relationship between the Thiele modulus and the effectiveness factor.

medium than its native environment. When the support has a net charge, the pH of the microenvironment of the enzyme may be different than the pH of the bulk medium due to the electrostatic interactions (Goldstein, Levin, and Katchalski 1964):

3.25 Δ p H = p H i - p H b = 0 . 4 3 z F ψ R T ,

where pH i = pH of the microenvironment of the enzyme, pH b = pH of the bulk medium, z = net charge on the diffusing substrate, F = Faraday constant, y = electrostatic potential, R = gas constant. The intrinsic activity of the enzyme is altered by the local changes in pH and ionic constituents. Further alterations in the apparent kinetics are due to the repulsion or attraction of substrates or inhibitors. A similar expression may also be suggested for the variation of Km upon immobilization (Wang et al. 1979):

3.26 Δ p K M = p K M i - p K M b = log [ K M b K M i ] = 0 . 4 3 z F ψ R T ,

where K M b

= Michaelis constant in the bulk medium and K M i = Michaelis constant in the microenvironment.

When the enzymes are bound and evenly distributed on the surface of a nonporous support material, substrate diffuses through a thin liquid film surrounding the support to reach the active sites of the enzymes. Under steady state conditions the reaction rate equals the mass transfer rate:

3.27 k ( c A b - c A s ) = υ a p p c A s c A s + K a p p ,

where k = mass transfer coefficient, cAb = substrate concentration in the bulk liquid, cAs = substrate concentration on the surface, vapp and Kapp = apparent kinetic constants. When the system is strongly mass transfer limited cAs ≅ 0, since the reaction is rapid compared to mass transfer, and the system behaves as pseudo first order:

3.28 υ = k c A b ( when Da 1 ) ,

where a Damköhler number is defined as Da = υapp /kcAb = maximum attainable reaction rate/maximum attainable mass transfer rate. When the system is reaction limited the reaction rate is often expressed as

3.29 υ = υ a p p c A s c A s + K a p p ( when Da 1 ) .

Under these circumstances the apparent constants may be obtained from the double- reciprocal plot.

Variation of the sensory properties of the foods may be studied in analogy with chemical kinetics. There are numerous examples to this approach in the literature.

 

Example 3.11  Kinetics of Browning of Milk

The difference of color DC between samples and the raw milk reference were determined with the following equation:

E.3.11.1 Δ C = Δ a 2 + Δ b 2 + Δ L 2 ,

where a, b, and L are the hunter color scale parameters. Kinetics of browning was expressed in analogy with zero-order kinetics:

E.3.11.2 d ( Δ C ) d t = k .

After integration we will have

E.3.11.3 Δ C = Δ C 0 + k t .

Where ΔC 0 is the intercept with t = 0 axis. Comparison of the integrated equation with the experimental data is shown in Figure E.3.11

Further analysis of the data indicated that

E.3.11.4 Δ C 0 = - 1 3 . 8 9 0 9 + 0 . 0 3 8 5 7 8 5 T ,

where T is the heating temperature in K. The rate constant was expressed with the Arrhenius equation:

E.3.11.5 k = 1.2 × 1 0 1 3 exp { - 1 0 1 . 8 8 . 3 1 4 × 1 0 - 3 T }

Details of computations are depicted in MATLAB® code E.3.11.

Variation of Δ

Figure E.3.11   Variation of ΔC with time at different temperatures. (From Pagliarini, E., Vernille, M., and Peri, C., Journal of Food Science, 55, 1766–67, 1990.)

 

 

Example 3.12  Kinetics of the Color Change and Polyphenol Oxidase Inactivation During Blanching of the Sultana Grapes

Sultana grapes were blanched in water then dried in a tray dryer. The effect of the blanching time on Hunterlab L values of raisins was expressed as an analogy with zero-order rate expression

E.3.12.1 d L d t = k L ,

where kL is the apparent rate constant. After integration

E.3.12.2 L = L 0 + k L t ,

where L 0 is the initial Hunterlab L value. The inactivation of polyphenol oxidase (PPO) was modeled with zero-order rate expression (Aguilera, Oppermann, and Sanchez 1987):

E.3.12.3 d ( P P O ) d t = - k P P O ,

where k PPO is the apparent rate constant. After integration

E.3.12.4 PPO = P P O 0 - k P P O t

where PPO0 is the initial polyphenol oxidase activity. The percentage of inactivation of the PPO may be expressed as (Aguilera, Oppermann, and Sanchez 1987)

E.3.12.5 P P O 0 - P P O P P O 0 × 1 0 0 = k P P O 1 0 0 P P O 0 t .

Equations E.3.12.2 and E.3.12.5 were compared with the experimental data in Figures E.3.12.1, and E.3.12.2, respectively. The apparent reaction rate constants were also found to agree with the Arrhenius expression (Figure E.3.12.3). Details of the computations are shown in MATLAB® code E.3.12.

 

Example 3.13  Kinetics of the Change in Texture of the Potatoes During Cooking

Changes in texture of the potatoes during cooking were simulated with zero-order model (Harada, Tirtohusodo, and Paulus 1985):

E.3.13.1 d ( T J ) d t = - k t e x t u r e .

Where TJ stands for the texture judgment and k texture is the apparent rate constant. After integration we will have

E.3.13.2 [ T J ] = [ T J ] 0 - k t e x t u r e t ,

where [TJ]0 is the numerical value of the texture judgment at t = 0. MATLAB® code E.3.13 evaluates the apparent rate constant and compares the model with the experimental data.

Comparison of

Figure E.3.12.1   Comparison of Equation E.3.12.2 with the experimental data (shown in symbols). (From Aguilera, J. M., Oppermann, K., and Sanchez, F., Journal of Food Science, 52, 990–93, 1987.)

Figure E.3.12.2   Comparison of Equation E.3.12.5 with experimental data (shown in symbols). (From Aguilera, J. M., Oppermann, K., and Sanchez, F., Journal of Food Science, 52, 990–93, 1987.)

3

Arrhenius plots for the inactivation rate constant of PPO (upper line) and Hunterlab L value (o). (From Aguilera, J. M., Oppermann, K., and Sanchez, F.,

Figure E.3.12.3   Arrhenius plots for the inactivation rate constant of PPO (upper line) and Hunterlab L value (o). (From Aguilera, J. M., Oppermann, K., and Sanchez, F., Journal of Food Science, 52, 990–93, 1987.)

 

 

Figure E.3.13   Development of texture during cooking of potatoes. Texture judgment scores: 1 = very hard, 6 = optimal texture, 11 = pulpy. (From Harada, T., Tirtohusodo, H., and Paulus, K., Journal of Food Science, 50, 459–62, 472, 1985.) It should be noticed that the apparent similarity between the texture development and the first-order irreversible chemical reaction made it possible to develop an analogy model.

3.8  Metabolic Process Engineering

Metabolism is the set of a very large number of chemical reactions occurring in the cells to maintain life. Part of the metabolism, which consumes energy to maintain cellular activity and synthesizing chemicals, is named anabolism. Metabolic activity, which breaks down organic matter entering into the cells to supply raw material for energy metabolism and cellular synthesis is called catabolism.

Figure 3.6   Schematic description of the metabolic pathways important in food processing and preservation. The glycolytic pathway is anaerobic, whereas the TCA cycle and electron transport chain are aerobic pathways. Pyruvate is produced in the glycolytic pathway, than either converted into ethanol, lactic acid, and so on, or sent to the aerobic metabolic pathways.

Organisms store carbohydrates and fats to establish their own energy reserve. Bonds of the energy reserve chemicals are broken or made as part of the energy exchange and transformation with the environment. Starch is the most common energy reserve of the plant cells. Carbohydrates have a strong affinity for water, which makes the storage of large amounts of carbohydrates inefficient due to the large molecular weight of the solvated water–carbohydrate complex. In the animal cells, carbohydrates are converted into fat; the hydrophobic character of lipids makes them more compact for energy storage when compared to the hydrophilic carbohydrates. Triglycerides coming from the foods are split into glycerol and fatty acids in the intestine, carried with the blood to the adipose tissue, rebuilt there, and stored as fat. When the body requires extra energy, the hormone glucagon signals the breakdown of the triglycerides and the free fatty acids released by lipolysis enter into the bloodstream and circulate throughout the body. The fatty acids are broken down in mitochondria to generate Acetyl-CoA, which enters into the TCA cycle. A schematic description of the metabolic pathways important in food processing and preservation are summarized in Figure 3.6.

 

Example 3.14  Enthalpy of Formation and Bond Energy of Glucose

Metabolic pathways produce intermediary chemicals for the synthesis of the cellular components and provide energy. The apparent reaction of aerobic respiration is

  • C6H12O6(aq) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔHR = –2880 kJ.

Where ΔHR is the metabolic heat released as a result of respiration. Heat of formation of H2O and CO2 are given as

  • Cgraphite + O2(g) → CO2(g) ΔHf,CO2 = –417 kJ/mol,
  • H2(g) + ½O2(g) → H2O(l) ΔHf,H2O = –286 kJ/mol.

We may estimate the enthalpy of formation of glucose from the metabolic heat released during respiration as

  • ΔHR = 6 ΔHf,CO2 + 6 ΔHf,H2O – ΔHf,glucose – 6 ΔHf,O2

where ΔHf,O2 = 0 by definition.

  • ΔHf,glucose = (2880) + (6)(–417) + (6)(–286) = –1338 kJ/mol.

This is the enthalpy of formation of crystalline glucose. When glucose dissolves, OH groups make a hydrogen bond with the water and each hydrogen bond will have 20 kJ/mol of energy. Therefore the enthalpy of formation of dissolved glucose will be

  • ΔHf,glucose(aq) = ΔHf,glucose(s) + (5) ΔHf,OH bond,
  • ΔHf,glucose(aq) = –1338kJ/mol + (5)(20) = –1238 kJ/kg.

ΔHf,glucose calculated here is in very good agreement with ΔHf,glucose given in the literature as –1262 KJ/mol.

Glucose is the starting molecule of the energy metabolism. Energy demand of the cell is met by metabolizing glucose; that is, extracting the energy hidden in its bonds, through glycolytic pathway, citric acid cycle, and electron transport chain. Structure of glucose is given in the literature as depicted in Figure E.3.14

Solubility of glucose in water is very high due to the hydrogen bonds made between the OH groups and water. The total bond energy of the glucose molecule may be estimated by adding up the energies of all the bonds.

Bond

Bond Energy (kJ/mol)

Number of Bonds/Molecule

Total Bond Energy/mol (kJ/mol)

C–H

410

7

2870

C–O

330

7

2310

C–C

334

5

1670

O–H

456

5

2280

Hydrogen bonds

20

5

100

Total bond energy

9230

Glucose molecule.

Figure E.3.14   Glucose molecule.

Total bond energy of glucose implies that if the molecule should be divided into its atoms 9230 kJ/mol of energy may be released, implying an enthalpy change of –9230 kJ/mol and this is almost 7 times of ΔHf,glucose.

Metabolic activity in the fresh fruits and vegetable cells continue for some time after being detached from the plants. It alters the cellular composition in storage and reduces the quality. Enzymes that catalyze the metabolic reactions allow regulation of the rate of the metabolic activity by responding to the environmental conditions. Elevated carbon dioxide levels in storage environment forces the equilibrium at the CO2 producing steps of the metabolism to reestablish in favor of the lower metabolic activity. Carbon dioxide also dissolves in the cells and lowers the pH of the medium and the rate of the reactions. Lowering the temperature reduces the rates of the metabolic reactions as described by the Arrhenius expression 3.5, therefore the shelf life of some fruits and vegetables may be extended by refrigeration or freezing. Metabolic activity of the microorganism responds to the composition and the temperature of the storage environment in a similar way as those of the fruits and vegetables, therefore controlled atmosphere storage might also slow down the microbial decay. The best storage conditions varies for every fruit and vegetable with cultivar and stage of maturity. Controlled atmosphere storage of apples and kiwi fruit are among the most popular commercial practices, where the shelf life may be extended to about 7 months when appropriate temperature and gas compositions are provided (Gormley 1985; Raffo et al. 2009).

 

Example 3.15  Kinetics of Oxygen Consumption and Carbon Dioxide Generation in Controlled Atmosphere Storage

The rate-limiting step of the aerobic pathways of the energy metabolism may be described as

E.3.15.1 E n z y m e 1 + S + O 2 k 2 k 1 A C 1 k 3 E n z y m e 1 + C O 2 .

In the CO2 rich controlled atmosphere, competitive and noncompetitive inhibition by CO2 may be described as

E.3.15.2 E n z y m e 1 + C O 2 k 5 k 4 A C 2 ,

and

E.3.15.3 AC 1 + C O 2 k 7 k 6 A C 3 .

The respiration, for example, oxygen consumption and rate expression, based on the reaction mechanism described in Equations E.3.15.1 through E.3.15.3 is (Hertog et al. 1998)

E.3.15.4 r o 2 = [ k 0 O 2 exp { - E a O 2 R T } ] C o 2 ( 1 + C C o 2 K M C O 2 ) + C o 2 ( 1 + C C o 2 K M u C O 2 ) .

The rate-limiting step of the anaerobic pathway of the energy metabolism (glycolysis) may be described as

E.3.15.5 E n z y m e 2 + S k g k 8 A C 4 k 10 E n z y m e 2 + C O 2 .

Competitive inhibitory effect of O2 on glycolysis may be described as

E.3.15.6 E n z y m e 2 + O 2 k 12 k 11 A C 5 .

The competitive inhibitory effect of the excessive CO2 of the controlled atmosphere on glycolysis may be described as

E.3.15.7 E n z y m e 2 + C O 2 k 14 k 13 A C 6 .

When we consider the mechanism described in Equations E.3.15.1 through E.3.9.5 and the Arrhenius type temperature dependence, carbon dioxide production rate during anaerobic glucose to ethanol conversion is stated as (Hertog et al. 1998)

E.3.15.8 r c o 2 = k 0 c o 2 exp { - E a C O 2 R T } K M C O 2 ( 1 + c o 2 K M O 2 + c c o 2 K M C O 2 ) + 1 .

Total carbon dioxide production rate r CO2 is described as

E.3.15.9 r C O 2 , t o t al = ( R Q ) ( r o 2 ) + r c o 2 ,

where RQ is the respiratory quotient (ratio between CO2 production and O2 consumption). MATLAB® code E.3.15 prepares a 3-D plot to describe the variation of rCO2,total, as a function of temperature, carbon dioxide, and oxygen concentrations.

 

Example 3.16  Logistic Model of Depletion of the Carbohydrate Reserve in Mushrooms During Storage

Mushrooms may accumulate mannitol reaching up to 60% of their dry weight. Carbohydrates or mannitol may be used in the energy metabolism as depicted in Figure 3.3. Their depletion rates due to metabolic activity may be described as

E.3.16 d c d t = - k c r respiration ( c ( t ) - c min c min ) .

Where c(t) describes the concentration of either mannitol, glucose, or glycogen in the plant tissue at time t. The term (c(t) – c min)/c min implies that depletion slows down as c(t) approaches c min.

 

Figure E.3.15.1   Model describing the influence of oxygen and carbon dioxide concentrations on the carbon dioxide production rates of apples in the reduced atmosphere at T = 273 K. (Constants of the model adapted from Hertog et al. 1998.)

Parameter c min may attain different values depending on the state of the overall metabolic activity. It should be noticed that Equation E.3.16 describes the apparent metabolic activity and is used with modeling purposes only. MATLAB® code E.3.16.a tests the validity of the model by comparing it with the experimental data.

The rate of the slowest reaction step involved in the metabolic activity determines the rate of the metabolism. As long as the rate-limiting step remains unchanged, the apparent rate of the metabolic

Figure E.3.15.2   Model describing the influence of oxygen and carbon dioxide concentrations on the carbon dioxide production rates of apples in the reduced atmosphere at T = 278 K. (Constants of the model were adapted from Hertog et al. 1998.)

 

Figure E.3.16.1   Comparison of the logistic glycogen and mannitol consumption models with the data. The increase observed in the mannitol concentration in the first phase of the storage is assumed to be caused by the release of mannitol from degrading complex molecules. (Adapted from Varoquaux, P., Gouble, B., Barron, C., and Yildiz, F., Postharvest Biology and Technology, 16, 51–61, 1991.)

Figure E.3.16.2   Comparison of the logistic respiration rate and glucose consumption models with the data. (Adapted from Varoquaux, P., Gouble, B., Barron, C., and Yildiz, F., Postharvest Biology and Technology, 16, 51–61, 1991.)

activity may be studied the same way as the kinetics of the enzyme catalyzed reactions. Unfortunately, the rate-limiting step may change with temperature so the metabolic activity may not be studied like that of a simple reaction over a large range of experiments. The Arrhenius plot of the respiration rates is given in Figure E.3.16.3. Details of the computations are given in MATLAB® code E.3.16.b.

Alcoholic drink production processes are distinguished with respect to the source of the fermentable sugars and the additional flavor and quality improvement operations. The final product is ethanol in the case of beer, wine, sake, vodka, tequila, and whiskey; it is lactic acid in yogurt and cheese-making processes. The raw material of the metabolic final product industries depends on the availability determined by the climate. Europe may be divided into widely used (but loosely defined) wine, beer, and vodka belts. Wine, champagne, brandy (cognac), and the other wine-like products are produced in grape-growing countries of both the southern and northern hemisphere. Fresh grapes may not store economically for a long time, therefore wine is produced only for a few months of the year. Raisins (dried grapes) are stable over a year and used for raki (Turkish), ouzo (Greek), or arak (Arabic) like clear, colorless, unsweetened, aniseed-flavored distilled alcoholic drinks. The raki belt lies southeast of the wine belt and includes the Eastern Mediterranean and Northern African countries, where powerful solar energy makes it possible to dry the grapes easily. Rum is a distilled alcoholic drink produced from sugarcane juice or molasses in the tropical and subtropical countries. Molasses is the final by-product of the sugar industry. The carbon sources of wine-like, raki-like products (fructose and glucose), and rum (sucrose, glucose, and fructose) are readily fermentable and enter into the metabolic pathways of the microorganisms easily. Starch from grains is the carbon source of the fermentation processes leading to beer, whiskey, vodka, and saké production. Beer and saké (the Japanese alcoholic drink saké is also called rice-wine) are not distilled alcoholic drinks, whereas vodka and whiskey are. Starch is not readily fermentable, therefore some preparatory stages involve into beer, whiskey, vodka, and saké production, where large starch chains are cut into smaller sugars, to make their use possible in the metabolic pathways (Figure 3.3). Although starch, originating from the grains (mostly barley), is hydrolyzed prior to fermentation in the beer-making process, starch hydrolysis and fermentation occur simultaneously in the saké production.

Figure E.3.16.3   Arrhenius plot of the respiration rates. (Adapted from Varoquaux, P., Gouble, B., Barron, C., and Yildiz, F., Postharvest Biology and Technology, 16, 51–61, 1991.)

 

Vodka is a distilled beverage made of grains (usually rye or wheat), potatoes or sugar beet molasses in the countries where cultivation of grapes is very difficult. Whiskey is also made of grains (barley, rye, and wheat), corn, or molasses.

Tequila is a distilled alcoholic drink made in Mexico by using the agave juice. When producing 100% agave tequila, the only carbohydrate source is inulin hydrolyzed from the plant tissue by cooking. Inulins are a group of plant polysaccharides formed by linking several simple sugars together. Most plants, which synthesize and store inulin, do not store other materials such as starch.

3.9  Microbial Kinetics

Predictive microbial models may be used to describe the behavior of microorganisms under different physical and chemical conditions, such as temperature, pH, and water activity. These models allow the prediction of microbial safety or shelf life of the products and facilitate development of the HACCP programs (Chapter 5; Whiting and Buchanan 1994). Microbial kinetics are mostly based on the analogy between the microbial processes and the chemical or enzyme catalyzed reactions. Proliferation of the microorganisms in a batch growth medium may include four ideal stages as explained in Table 3.3. Some of these growth phases may be prevented intentionally with an appropriate experimental design.

Table 3.3   Ideal Growth Phases of Microorganisms in Batch Medium

Specific growth rate μ is a constant in the exponential growth phase, which implies that the growth rate is proportional with the viable microbial population, where all the members have equal potential for growth. The specific growth rate may be regarded as the frequency of producing new microorganisms by the ones already present. When microbial proliferation occurs in a substrate limited medium, specific growth rate may be related with the substrate concentration:

3.34a μ = μ max C S c s + K ,

where μmax is the maximum attainable specific growth rate. Equation 3.34a is called the Monod equation. There are numerous variations of the Monod equation available in the literature (Mulchandani and Luong 1989). The most common empirical modifications of Equation 3.34a to include substrate and product inhibition are

3.34b μ = μ max C S c s + K + c s 2 K s ,

and

3.34c μ = μ max C S c s + K K p K p + c p ,

where Ks and Kp are constants, cp is the product concentration.

The logistic model is frequently used to simulate microbial growth when a microbial population inhibits its own growth via depletion of a limited nutrient, product accumulation, or unidentified reasons (Example 1.2):

3.35 d x d t = μ x ( 1 - x x max ) ,

where μ is the initial specific growth rate and x max is the maximum attainable value of x. The logistic equation is an empirical model, it simulates the data when microbial growth curve follows a sigmoidal path to attain the stationary phase. It is mostly based on experimental observations only. When x << x max, the term in parenthesis is almost one and neglected, then the equation simulates the exponential growth (i.e., Equation 3.31). When x is comparable with x max, the term in the parenthesis becomes important and simulates the inhibitory effect of overcrowding on microbial growth. When x = x max, the term in the parenthesis becomes zero, then the equation will predict no growth (i.e., stationary phase as described with Equation 3.35). The logistic equation may be integrated as

3.36 x = x 0 e μ t 1 x 0 x max ( 1 e μ t ) .

The exponential growth model simulated with Equation 3.31 may be modified after substituting:

3.37 μ = μ 0 + μ 1 x - μ 2 x 2 ,

to simulate the Allee effect, which represents a population with maximum specific growth rate at intermediate microbial concentrations when μ0, μ1, and μ2 are positive constants (Edelstein-Keshet 1988).

When μ of Equation 3.31 is a function of time such that

3.38 d μ d t = - α μ ,

we obtain the Gompertz model, which may also be used to simulate the sigmoidal behavior of the microbial growth curve (α = constant). The Gompertz model is usually expressed in three equivalent versions (Edelstein-Keshet 1988):

3.39a d x d t = μ x , d μ d t = - α μ ,
3.39b d x d t = ( λ e α t ) x ,

and

3.39c d x d t = ( κ ln x ) x ,

where λ and κ are constants.

Microbial products may be roughly categorized in four groups as depicted in Table 3.4.

Primary metabolites are produced by the microorganism for its own metabolic activity. Secondary metabolites are usually produced against the external factors; that is, production of antibiotics starts in the stationary phase to prevent consumption of the limited nutrients by the other microbial species. The yeast Saccharomyces cerevisiae may be regarded as a product itself when produced as an additive to achieve leavening in the bakery industry. Sugars are consumed in the energy metabolism, some microorganisms may not convert them into carbon dioxide, but follow a shorter path and excrete the metabolic end products such as ethanol or lactic acid. Product formation models relates the product formation rate

Table 3.4   Microbial Products

Product

Examples

Biomass

Baker’s yeast

Metabolic end products

Ethanol, lactic acid, carbon dioxide

Primary metabolites

Amino acids, enzymes, vitamins

Secondary metabolites

Antibiotics

to fermentation variables (i.e., growth rate, biomass, or substrate concentration, etc.). The Luedeking and Piret (1959) model is among the most popular product formation models of food processing interest:

3.40 d c P r d t = α x + β d x d t ,

where cpr is the product concentration, α and β are the constants. The term αx represents the product formation rate by the microorganisms regardless of their growth; βdx/dt represents the additional product formation rate during growth in proportion with the growth rate. This is an empirical equation, because it simply relates the experimental observations mostly without considerable theoretical basis. When growth-associated product formation rates are much greater than the nongrowth-associated product formation rates, Equation 3.40 may be written as

3.41 d c P r d t = β d x d t .

When nongrowth-associated product formation rates are much greater than the growth-associated product formation rates, Equation 3.40 becomes

3.42 d c P r d t = α x .

Structured and age distribution models relate cellular structure or age distribution to growth and product formation rates, but need more information for application, generally difficult to use and not widely employed in food research, therefore not considered here; an interested reader may refer to Bailey and Ollis (1986) for detailed discussion.

In a fermentation process a substrate (i.e., nutrient), is allocated to three basic uses:

3.43 ( total rate of substrate utilization ) = { rate of   substrate utilization for biomass synthesis } + { rate of   substrate utilization for maintenance } + { rate of   substrate utilization for product formation } .

Substrate consumption to keep the cells alive without growth or product formation is referred to as maintenance. Equation 3.43 may be expressed in mathematical terms as

3.44 - d c s d t = 1 Y x / s d x d t + k M x + 1 Y p / s d c P r d t ,

where Yx/s is the cell yield coefficient defined as grams of biomass produced per grams of substrate used, and Yp/s is the product yield coefficient (i.e., grams of product produced per grams of substrate used). The yield coefficients usually remain constant as long as the fermentation behavior remains unchanged, but there are also examples to the variable yield coefficients in the literature, which may indicate a shift in substrate preference, availability of oxygen, and so on during the course of the process (Özilgen, Ollis, and Ogrydziak 1988).

Microorganisms produce heat as a by-product of the energy metabolism that may be summarized with typical examples as

3.45 Glucose + 36 P i + 36 ADP respiration 6 CO 2 + 42 H 2 O + 36 ATP + Q ,
3.46 Glucose + 2 P i + 2 ADP anaerobic fermentation 2 Lactate + 2 H 2 O + 2 ATP + Q ,

where Pi and Q represent inorganic phosphate atoms and metabolic heat generation, respectively. The ADP and ATP are abbreviations for adenosine diphosphate and adenosine triphosphate, respectively. The ATP is the energy currency of the cell. The ATP to ADP conversion is coupled with the energy consuming metabolic reactions (i.e., biosynthesis, cellular transport, etc.) and make them thermodynamically feasible. The ADP is converted back to ATP through Equation 3.45 or Equation 3.46. Metabolic heat generation rate may be related with the growth rate as

3.47 [ metabolic   heat generation   rate per   unit  volume of   fermentor ] = 1 ϒ Δ d x d t ,

where Y Δ is the heat generation coefficient defined as grams of biomass production coupled with one unit (i.e., kJ) of heat evaluation.

Most food and beverage fermentations involve mixed culture of microorganisms. Typical examples may include yogurt cultures (Streptococcus thermophilus and Lactobacillus bulgaricus), wine (Saccharomyces cerevisiae and wild microbial species), and cheese production (mixed culture of various microorganisms). The general interaction of two microorganisms are summarized in Table 3.5.

Table 3.5   General Interaction Ways of Microbial Populations

Microorganism 2

+

0

+

mutualism

predation

commensalism

Microorganism 1

competition

amensalism

0

neutralism

Note: Symbols + , –, and 0 indicate positive, negative, and no effect on the related microbial population.

Table 3.5 indicates that when microbial species 1 and 2 benefit from an interaction it is called mutualism; when microbial species 1 benefits but microbial species 2 is not affected from an interaction it is called commensalism. Interactions shown in Table 3.5 depend on the culture conditions. An initially neutral relation may turn in competition during the course of fermentation with depletion of the limiting substrate, or two microbial populations may have mutualistic relation in one medium, but competitive relation in another medium.

 

Example 3.17  Mixed Culture Interactions Between P. Vulgaris and S. Cerevisiae

Proteus vulgaris prefers to utilize sodium citrate when both sodium citrate and glucose are available, also nicotinic acid is needed for its growth. Saccharomyces cerevisiae can utilize only glucose and produces nicotinic acid. By varying the concentrations of the medium components, various mixed culture interactions may be created as depicted in Table E.3.17.

The chemical reactor and fermentor design are based on similar principles. A CSTR is called a chemostat when there is only one limited substrate. Equation 3.57 may be applied to a chemostat for a population balance of the microorganisms to yield

3.48 F x 0 - F x + V R x = d ( V x ) d t ,

where x 0 is the concentration of the microorganisms in the input stream. Under steady state conditions d(Vx)/dt = 0, when the microorganisms are in exponential growth phase Rx = μ x , we may rearrange Equation 3.48 to obtain

3.49 x = D x 0 D - μ ,

where x = microbial concentration in or at the exit of the fermentor,

3.50 D = F V = dilution   rate = 1 τ = 1 residence   time   in   the   fermentor .

While working with sterile nutrients (i.e., x 0 = 0), Equation 3.49 will become

3.51 ( D μ ) x = 0.

Equation 3.51 implies that a nonzero population may be maintained under steady state conditions only when

3.52 D = μ .

Table E.3.17   Mixed Culture Interactions Between P. vulgaris and S.cerevisiae

Sodium Citrate

Glucose

Nicotinic Acid

Interaction

Limiting concentration

limiting concentration

not added

commensalism

Not added

limiting concentration

not added

commensalism plus competition

Not added

limiting concentration

sufficient amount (not limiting)

competition

Excess amount

limiting concentration

not added

mutualism

Limiting concentration

limiting concentration

sufficient amount (not limiting)

neutralism

Source: Tseng, M. M.-C., and Phillips, C. R., Biotechnology and Bioengineering, 23, 1639–51, 1981.

It should be noticed that even under these conditions, an output stream will have the same biomass concentration as in the chemostat (Bailey and Ollis 1986).

When nutrients are sterile and the biomass growth agrees with the Monod Equation 3.48, it will become

3.53 D = μ max c s K s + c s ,

where cs is the substrate concentration in the fermentor. Equation 3.53 may be rearranged as

3.54 c s = D K s μ max - D .

Biomass yield coefficient Y x/s may be calculated as

3.55 Y x / s = x - x 0 c S 0 - c S ,

where c S0 is the substrate concentration in the input stream. When we use sterile nutrients (x 0 = 0), Equations 3.54 and 3.55 may be combined to calculate the biomass concentration in (also at the exit of) the fermentor:

3.56 x = Y x / s ( c S 0 - D K s μ max - D ) .

Equation 3.51 implies that stable fermentation operation may be maintained only when c S0 > DKs maxD.

 

Example 3.18  Kinetics of Microbial Growth, Gas Production, and Dough Volume Increase During Leavening

Commercial baker’s yeast contains strains of Saccharomyces cerevisiae, lactic acid bacteria, and low numbers of contaminating microorganisms. The yeast utilizes the simple sugars derived from flour and produces carbon dioxide. Growth of the microorganisms in the dough was simulated with a logistic equation (Akdogan and Özilgen 1992):

3.35 d x d t = μ x ( 1 - x x max ) .

The Luedeking–Piret equation was used to model gas production (Akdogan and Özilgen 1992):

3.40 d G d t = α x + β d x d t .

A fraction of the gas produced by the baker’s yeast is retained in the dough and the remaining gas diffuses out. The retained gas increases the volume of the dough. The volume increase rates are expected to decrease as the volume of the dough gets larger and the walls of the gas cell get thinner due to stretching. The rate of the dough volume increase may be related to the gas production rate (Akdogan and Özilgen 1992):

E.3.18.1 d V d t = ϕ ( 1 - V V max ) d G d t ,

where V and V max are the volume and the maximum attainable volume of the dough, respectively. Parameter ϕ is the ratio of the initial dough volume increase to rate of the initial gas production rate. Comparison of the model with a typical set of experimental data is shown in Figure E.3.18 MATLAB® code E.3.18 shows the details of the computations.

Comparison of the model (

Figure E.3.18   Comparison of the model (Equations 3.35, 3.40, and E.3.19.1) with the experimental data of microbial growth, gas production and dough volume increase. (From Akdogan, H., and Özilgen, M., Enzyme and Microbial Technology, 14, 141–3, 1992.)

 

 

Example 3.19  Kinetics of Microbial Growth and Lactic Acid Production by Mixed Cultures of Streptococcus Thermophilus and Lactobacillus Bulgaricus in Yogurt Production

Lactic acid production by mixed cultures of Streptococcus thermophilus and Lactobacillus bulgaricus is the most important chemical process involved in yogurt production. Interaction of these microorganisms in milk is favorable for both of the species, but not obligatory. Both numbers of the individual microorganisms and the total amounts of lactic acid produced by mixed cultures of these microorganisms are considerably higher than those obtained with pure cultures of each microorganism. In the mixed cultures, stimulation of growth of S. thermophilus was attributed to the production of certain amino acids (i.e., glycine, histidine, valine, leucine, and isoleucine) by L. bulgaricus. The S. thermophilus stimulates the growth of L. bulgaricus by producing formic acid. The logistic equation was modified to simulate the mixed culture growth of the microorganisms as (Ozen and Özilgen 1992)

E.3.19.1 d x s d t = μ S x S ( 1 - τ s x s + τ L x L x S max + x L max ) ,
E.3.19.2 d x L d t = μ L x L ( 1 - τ s x s + τ L x L x S max + x L max ) ,

where subscript L and S refer to L. bulgaricus and S. thermophilus, respectively. Some of the microbial species died at the end of the fermentation process, the death rates were simulated as (Ozen and Özilgen 1992)

E.3.19.3 d x d t = - k x .

The Luedeking–Piret equation was modified to simulate lactic acid production by the mixed culture of the microorganisms as (Ozen and Özilgen 1992):

E.3.19.4 d c P r d t = α s x s + α L x L + β S d x s d t + β L d x L d t .

Equations E.3.19.1 through E.3.19.4 are solved with MATLAB® code E.3.19 and compared with the experimental data.

 

 

Example 3.20  Kinetics of Spontaneous Wine Production

The wine-making process may be analyzed in two phases: alcoholic fermentation and malolactic fermentation. During the alcoholic fermentation process, wine microorganisms consume the fermentable sugars and produce ethanol. A spontaneous wine production process is actually a mixed-culture and multi-product process, commenced by the natural microorganisms of the grapes. Natural grape microorganisms consists of various genera of molds, yeasts, and lactic acid bacteria. Generally wine yeast, Saccharomyces cerevisiae, is extremely low in population on the grapes. In the wine-making process it multiplies with a strong fermentative capacity, excludes most of the other microorganisms from the medium and eventually invades the raw grape juice.

In the alcoholic fermentation process, growth of the biomass may be simulated in three phases (Özilgen, Celik, and Bozoglu 1991):

3.31b i . Exponential  growth:   d x / d t = μ x
3.31c ii . Stationary   phase :   d x / d t = 0
3.31d iii . Death   phase :   d x / d t = k d x

Figure E.3.19.1   Variation of the concentrations of L. bulgaricus and S. thermophilus in cultivations with 20% NFDM (nonfat dry milk) medium. (From Ozen, S., and Özilgen, M., Journal of Chemical Technology and Biotechnology, 54, 57–61, 1992.)

Variation of the lactic acid concentration with time. (From Ozen, S., and Özilgen, M.,

Figure E.3.19.2   Variation of the lactic acid concentration with time. (From Ozen, S., and Özilgen, M., Journal of Chemical Technology and Biotechnology, 54, 57–61, 1992.)

Total biomass concentration, denoted by x, is actually a mixed culture and the microbial species contribute to x actually change with time. In spontaneous wine fermentations, as the process proceeds, alcohol-sensitive microorganisms are inhibited and the alcohol tolerant microorganisms dominate the culture. Alcohol-tolerant microorganisms are generally better alcohol producers, therefore parameter β of the Luedeking–Piret equation may be related with the ethanol concentration in the medium:

E.3.20.1 β = β 0 + β 1 c p r + β 2 c p r 2 ,

where β0, β1, and β2 are constants. After substituting Equation E.3.20.1 in Equation 3.55:

E.3.20.2 d c P r d t = α x + ( β 0 + β 1 c p r + β 2 c p r 2 ) d x d t .

Substrate utilization in the exponential growth phase was

E.3.20.3 d c s d t = 1 Y x / s d x d t + 1 Y p / s d c P r d t .

Lactic acid bacteria is the major species contributing to the total biomass concentration x in the malolactic fermentation phase, where lactic acid bacteria uses malic acid as a substrate to produce lactic acid. The logistic equation was employed to simulate microbial growth in the malolactic fermentation phase:

E.3.20.4 d x d t M = μ M x ( 1 - x x max ) ,

where tM = time after commencement of the malolactic fermentation, and μ M = specific growth rate of the malolactic fermentation.

Temperature changes in the fermentation vessel were modeled after making thermal energy balance:

3.10.4.3 - Σ i = 1 n U i A i ( T - T e nv ) + V Y Δ d x d t = d ( ρ c V T ) d t ,

where the term Σ i = 1 n U i A i ( T T env )

represents the energy loss from the fermentor surfaces, and the term (V/Y Δ)(dx/dt) represents the thermal energy generation coupled with microbial growth. This term becomes zero after the end of the exponential growth in the alcoholic fermentation. The term dcVT)/dt is thermal energy accumulation. Equation E.3.25.5 was rearranged after substituting Equation 3.44 for dx/dt:

E.3.20.6 d T d t = K 1 ( T e n v - T ) + K 2 e μ t ,

where k 1 = Σ i = 1 n ( U i A i / ρ c V )

and K 2 = (μx 0/Y ΔρcV). The numerical value of parameter K 2 was zero after the end of the exponential growth phase of the alcoholic fermentation. MATLAB® code E.3.20 compares the model with a typical set of experimental data in Figures E.3.20.1 and E.3.20.2.

 

 

Example 3.21  Kinetics of Aspergillus Oryzae Cultivations on Starch

Cultivation of Aspergillus oryzae on starch is described as a combination of two rate processes: starch hydrolysis and the uptake of some of the fermentable hydrolysis products for cellular activities, including growth, enzyme production, and maintenance. These processes are inter-related in a cyclic way and neither of them can be accomplished without the other (Figure E.3.21.1). The first link between these processes is starch hydrolyzing extracellular enzymes (i.e., glucoamylases and amylases). They are produced by the microorganism in the first process (i.e., with the cellular activities of the microorganism) and catalyze the second process (i.e., starch hydrolysis). The second link between these processes is the starch hydrolysis products. They are produced in the second process and consumed in the first process.

Figure E.3.20.1   Course of fermentation with slow fermented wine. Experimental data are shown in symbols. (From Özilgen, M., Celik, M., and Bozoglu, T. F., Enzyme and Microbial Technology, 13, 252–6, 1991.)

Comparison of the model and the experimental temperatures. (From Özilgen, M., Celik, M., and Bozoglu, T. F.,

Figure E.3.20.2   Comparison of the model and the experimental temperatures. (From Özilgen, M., Celik, M., and Bozoglu, T. F., Enzyme and Microbial Technology, 13, 252–6, 1991.)

Figure E.3.21.1   Schematic description of the rate processes involved in A. oryzae cultivation on starch.

Biomass production was simulated with the logistic equation (Bayindirli, Özilgen, and Ungan 1991):

3.34 d x d t = μ x ( 1 - x x max ) .

The total starch hydrolyzing enzyme production (Bayindirli, Özilgen, and Ungan 1991):

3.10.4.3 d c E d t = k M d x d t - k N x ,

where kM and kN are constants. The first term kM (dx/dt) indicates that the enzyme production rate was proportional to the growth rate of the microorganism. The second term –kNx shows that the microorganisms degrade the starch hydrolyzing enzyme in proportion with their own concentration. The A. oryzae produces extracellular proteases, which may be among the causes of the starch hydrolyzing enzyme degradation.

Starch hydrolyzing enzymes of A. oryzae are extracellular. If the microorganisms are separated from the broth during cultivation, the remaining enzymes will continue to degrade starch in the broth. When the total starch hydrolyzing enzyme activity is constant, the amount of starch degraded in time interval dt will be

3.10.4.3 d c S = k S d t ,

where kS is the steady state starch degradation rate. If an additional enzyme is produced within this time interval, additional amounts of enzyme will be degraded in proportion with enzyme production (Bayindirli, Özilgen, and Ungan 1991):

3.10.4.3 d c S = k S d t + k U d c E ,

where kU is a constant (i.e., grams of starch degraded per unit activity of enzyme produced). Equation E.3.26.3 may be rearranged (Bayindirli, Özilgen, and Ungan 1991):

3.10.4.3 d c S d t = k s + k U d c E d t .

The reducing sugar accumulation rates in the broth were (Bayindirli, Özilgen, and Ungan 1991)

E.3.21.5 dC R dt = k R ( dC S dt ) 1 Y x / R dx dt k R x ,

where kR is a proportionality constant; Y x/R is the biomass yield defined as grams of biomass produced per grams of starch used. Parameter kR is the maintenance coefficient, defined as grams of starch used per gram of biomass in one hour to maintain vital activities other than growth. The term kR (dcS /dt) is the reducing sugar production rate in proportion to starch degradation. The terms (1/Y x/R )(dx/dt) and kRx are the reducing sugar consumption rates for growth and maintenance, respectively.

Glucose constitutes only a fraction of the reducing sugars and its accumulation rates in the medium may be expressed similarly to the reducing sugar accumulation as

E.3.21.6 dC G dt = k G ( dC S dt ) 1 Y x / G dx dt k G x ,

where the model constants had similar meanings to those of Equation E.3.21.5. MATLAB® code E.3.21 was used to compare the model with the experimental data.

 

Figure E.3.21.2   Time course of A. oryzae growth and starch consumption. (Original data (shown with symbols) from Bayindirli, A., Özilgen, M., and Ungan, S., Biocatalysis, 5, 71–8, 1991.)

Variation of enzyme activity, reducing sugar, and glucose concentrations during the course of the growth of

Figure E.3.21.3   Variation of enzyme activity, reducing sugar, and glucose concentrations during the course of the growth of A. oryzae on starch. (Original data (shown with symbols) from Bayindirli, A., Özilgen, M., and Ungan, S., Biocatalysis, 5, 71–8, 1991.)

3.10  Kinetics of Microbial Death

Microbial death kinetics has a significant importance in food processing since it is among the major phenomena occurring during pasteurization and sterilization processes. Microbial death is generally described in analogy with unimolecular, irreversible, first-order rate expression:

3.33 x ( live microoganism ) k d x d ( dead microoganism ) dx dt = k d x .

An alternative expression for Equation 3.33 is

3.57a d ( log  x ) dt = 1 D T ,

with

3.58b D T = 2.303 k d .

The DT value is defined as the heating time at constant temperature T to reduce microbial population by one log cycle, or 10% of its initial value. The z value is defined as the temperature difference required to change the DT value by a factor of 10, or one log cycle:

3.57c d ( log  D T ) dT = 1 z .

Equation 3.54 may be rearranged and integrated as

3.57d D T = D T ref 10 ( T ref T ) / z .

The z value is related with the activation energy of the Arrhenius expression:

3.58 z = 2.303 RTT ref E a .

During thermal processing of the foods, microbial death and inactivation of the enzymes are always accompanied with a loss of nutrients due to thermal degradation, since they share the same medium. Although death of the microorganisms and destruction of the enzymes and toxins are desired, loss of the nutrients is not desired. The range of the kinetic constants describing the resistance of the food components to thermal processing are given in Table 3.6.

In most processes, spores, vegetative cells, and enzymes are destroyed while color, flavor, and vitamins are desired to survive. Among the constituents to be destroyed enzymes usually have the highest D 121 (Table 3.6) values, therefore enzyme inactivation is almost the most difficult task to achieve in thermal processing.

Table 3.6   Kinetic Constants Describing the Resistance of the Food Constituents to Thermal Processing

Constituent

z (°C)

E a (Joules/mol)

D 121 (min)

Vitamins

20-30

8 .4 × 104 - 12.5 × 104

100-1000

Color, texture, flavor

25-5

4 .2 × 104 - 12.5 × 104

5-500

Enzymes

12-56

5.0 × 104 - 41.8 × 104

1-10

Vegetative cells

4-7

41.8 × 104 - 50.2 × 104

0.002-0.02

Spores

7-12

22.2 × 104 - 34.7 × 104

0.1-5.0

Cooking value (overall quality estimation)

15-45

5.5 × 104 - 17.0 × 104

1.2-12.5

Source: Lund, D., Food Technology, 31, no. 2, 71–78, 1977; Hallström, B., Skjöldebrand, C., and Trägårdh, C., Heat Transfer & Food Products. Elsevier Applied Science Publishers, London, 1988.

 

Example 3.22  MATLAB® Code for Conversion of the D T and z Values into kd and Ea and the Kinetic Compensation Relations for Microbial Death

The relation between the DT value and the thermal death rate reaction rate constant kd is

3.58 D T = 2.303 k d .

The DT values reported by Rodrigo et al. (1993) for destruction of Clostridium sporogenes PA3679 spores in the mushroom extract at different pH values and temperatures is the data, D matrix, of the MATLAB code E.3.22.a, where the D values are converted into the thermal death rate constants kd .

MATLAB® code E.3.22.b converts the thermal death rate constants back into the DT values. Temperature effects on the death rate constants were given with the Arrhenius expression:

3.5 k = k 0  exp  { E a RT } .

Equation 3.5 may be linearized as

ln ( k d ) = ln ( k d 0 ) E a RT .

MATLAB® code E.3.22.c determines the parameters ln(k d0) and E a .

The compensation relation is

E.3.22.1 ln ( k d 0 ) = 3.29 + 3.03 × 10 4 E a .

MATLAB® code E.3.22.d determines the compensation relation by using the ln (k d0) and Ea values. MATLAB® code E.3.22.e converts Ea values to z values. MATLAB® code E.3.22.f converts the z values back into Ea values.

 

 

 

 

 

Example 3.23  Kinetics of the Death of Spores When Subjected to Dynamic Lethal Temperatures

Sapru et al. (1993) suggested that when a population of dormant spores are subjected to dynamic lethal temperatures, a fraction of the population die without activation and the other fraction are activated first and then die:

E.3.23.1 dormant spores  ( x ) k d dead spores ,

 

 

Figure E.3.22   Comparison of the experimental data with Equation E.3.22.1.

E.3.23.2 dormant spores  ( x ) k a activared spores ( x a ) ,
E.3.23.3 activared spores  ( x a ) k ad dead spores .

Equation E.3.23.1 refers to the fraction that die without activation; Equations E.3.23.2 and E.3.23.3 refer to the other fraction. The process may be described as

E.3.23.4 dx dt = ( k d + k a ) x ,
E.3.23.5 dx a dt = k a x k ad x a .

MATLAB® code E.3.23 tests the validity of the model.

 

Figure E.3.23   Comparison of the model with the experimental data obtained with Bacillus stearothermophilus spores subjected to dynamic lethal temperatures. (Adapted from Sapru, V., Smerage, G. H., Teixeira, A. A., and Lindsay, J. A., Journal of Food Science, 223–8, 1993.)

 

Example 3.24  Kinetics of Ultraviolet Inactivation Processes

The Series-event model (Severin, Suidan, and Engelbrecht 1983) may be used to describe the ultraviolet inactivation processes. An event is assumed to be a unit of damage. The event occurs in a stepwise fashion:

E.3.24.1 M 0 kl M 1 kl M 2 kl ....... M n 1 kl M n ,

where k = constant, I = total point ultraviolet intensity, and Mi = an organism that has reached event level i. The rate at which an organism passes from one event level to the next is first order with respect to the ultraviolet intensity and independent of the event level occupied by the organism. As long as an organism is exposed to the ultraviolet collects damage. An event threshold exists such that an organism that collects damage greater than the tolerable event threshold is deactivated. The threshold may vary depending on the species, strain, or culturing conditions; however, for a given set of conditions the threshold level is constant.

For a well-mixed, flat, thin-layered, closed batch reactor the rate at which the organisms pass through event level i is

E.3.24.2 R xi = klx i 1 klx i .

Equation E.3.24.2 may be incorporated in Equation 3.57 written for species at damage level i in a batch reactor as

E.3.24.3 dx i dt = klx i 1 klx i .

With i = 0 (x 0 = concentration of the undamaged microorganisms) Equation E.3.24.3 is rewritten as

E.3.24.4 dx 0 dt = klx 0 .

The solution to Equation E.3.24.3 is

E.3.24.5 x 0 = x initial   e klt .

Equation E.3.24.5 describes the variation of the concentration of the undamaged microorganisms with time, where x initial is the initial concentration of the microorganisms before being subjected to UV radiation.

Equation E.3.24.3 may be rewritten and solved sequentially from i = 0 to i = n–1 (i.e., when i = 1 we have)

E.3.24.6 dx 1 dt = klx 0 klx 1 .

After substituting Equation E.3.24.5 in Equation E.3.24.6 and rearranging

E.3.24.7 dx 1 dt + klx 1 = klx initial   e klt .

The solution to Equation E.3.24.7 is

E.3.24.8 x 1 = x initial   e klt klt .

The general expression for time varying density of organisms at any level i is

E.3.24.9 x i = x initial ( klt ) i i ! e klt .

The total fraction of the surviving organisms between any level 0 and n–1 is

E.3.24.10 x x initial = Σ i = 0 n 1 x i x initial = exp ( klt ) Σ i = 0 n 1 ( klt ) i i ! ,

where n = threshold number of the damaged sites required for inactivation. MATLAB® code E.3.24 simulates the inactivation processes through Equation E.3.24.10.

 

 

Example 3.25  Weibull Model of Microbial Inactivation

The Weibull distribution model is often used in the industry with mechanical or electrical devices to estimate the probability of the failure p(t) at a given time t (Weibull 1951):

E.3.25.1 p ( t ) = e [ ( t γ ) / δ ] n .

Parameters n, γ, and δ are referred to as the shape factor, location parameter, and the characteristic life, respectively. The Weibull distribution model has three adjustable parameters, so it is commonly used in the industry to model a surviving fraction of the commodities that may not be represented with the other models. The location factor γ = 0 if failures may start at t = 0, then the Weibull model becomes

E.3.25.2 p ( t ) = e ( t / δ ) n .

Figure E.3.24   Inactivation processes as described by Equation E.3.23.10. Kinetic parameters, which simulate the data very closely (From Severin, B. F., Suidan, M. T., and Engelbrecht, R. S., Water Research, 17, 1669–78, 1983.) for individual species as: Escherichia coli n = 9, k = 1.538 × 10–3 cm2/mW s; Candida parasilosis n = 15, k = 0.891 × 10−3 cm2/mW s; f2 bacterial virus n = 1, k = 0.0724 × 10−3 cm2/mW s.

Equation E.3.25.2 may be used to estimate the fraction of the surviving microorganisms x(t)/x 0 in a sterilization process as

E.3.25.3 ln ( x ( t ) x 0 ) = ( t δ ) n ,

where parameters δ and n varies with the experimental conditions. The differential form of Equation E.3.25.3 is

E.3.25.4 dx dt = k app x ,

with the time-dependent apparent rate constant:

E.3.25.5 k app = n t ( t δ ) n

(van Boekel 2002), after modeling 55 sets of thermal processing data from the literature with the Weibull Equation E.3.25.3, concluded that when n < 1 the upward concave survival curve implies that the cells have the ability to adapt thermal stress and become more resistant to temperature effects. When n > 1, the downward concave survival curve implies that the cells are accumulating the damage and becoming increasingly heat sensitive.

Buzrul et al. (2008) employed Equation E.3.25.3 for modeling microbial inactivation in high pressure processing, where parameter δ was analogous to parameter DT of thermal processing operations. A variation of δ with pressure is:

E.3.25.6 d ( log  δ ) dP = 1 Z p .

MATLAB® code E.3.25.a prepares Figure E.3.25.1 to depict a variation of parameter δ with pressure. MATLAB code E.3.25.b computes the fraction of the microorganisms surviving the process (Figure E.3.25.2).

 

Figure E.3.25.1   Variation of the characteristic life δ with pressure during high pressure inactivation of E. coli in whole milk. (Adapted from Buzrul, S., Alpas, H., Largeteau, A., and Demazeau, G., European Food Research and Technology, 227, 443–8, 2008.)

Gomez et al. (2005) employed Equation E.3.25.3 for modeling microbial inactivation by pulsed electric fields. MATLAB code E.3.25.c computes the fraction of the microorganisms surviving the process (Figure E.25.3).

3.11  Ideal Reactor Design

Vessels used to conduct chemical reactions are called reactors. There are three basic types of ideal reactors: Batch, Continuously Stirred Tank (CSTR), and Plug Flow (PF). A batch reactor is a closed vessel with no input and output streams. Reactants are charged into the reactor, left for a certain period to be well mixed, and then removed when the required conversion is achieved. A CSTR reactor is equipped with input and output streams. Due to the well mixing, every point in the reactor and just at the exit are expected to have the same concentration. Plug flow reactors are tubular flow reactors, there is no velocity gradient in radial direction, and composition of the fluid varies in the flow direction only.

 

Figure E.3.25.2   Comparison of the Weibull model with the experimental data obtained during high pressure inactivation of E. coli in whole milk. (Adapted from Buzrul, S., Alpas, H., Largeteau, A., and Demazeau, G., European Food Research and Technology, 227, 443–8, 2008.)

 

Figure E.3.25.3   Comparison of the Weibull model with the experimental data obtained during inactivation of Listeria monocytogenes by pulsed electric fields. (Adapted from Gomez, N., Garcia, D., Alvarez, I., Condon, S., and Raso, J., International Journal of Food Microbiology, 103, 199–206, 2005.)

Material for a specific chemical balance around batch or CSTR reactor requires

3.59a ( input rate into the reactor ) ( output rate from the reactor ) + ( generation rate in the reactor ) = ( accumulation rate in the reactor ) .

In a batch reactor where reaction Ak products is occurring, with the rate expression RA = − kcA we have

3.59b ( input rate into the reactor ) = ( output rate from the reactor ) ,
3.59c generation of  A  rate in the reactor = R A V ,
3.59d ( accumulation rate of  A  in the reactor ) = d ( C A V ) dt .

With a constant reactor volume V, after substituting Equations 3.59b through d into Equation 3.57 we will obtain

3.59e dc A R A = R A .

After rearranging Equation 3.59e, we may calculate the time required to achieve a certain final reactant concentration cAf after starting with the initial concentration c A0 as

3.60 t = c A 0 c Af dc A R A .
 

Example 3.26  Acid Hydrolysis of Lactose in a Batch Reactor

Large amounts of lactose are produced in the cheese industry. It is not a good food ingredient because of its low solubility and limited sweetness. It is a pollutant when disposed into the environment. Lactose hydrolysis is generally achieved with an enzymatic process. Resin catalyzed hydrolysis may be preferred over enzymatic processes in some applications because of the higher temperature and conversion rates, and lower pH that prevents microbial contamination. Chen and Zall (1983) have shown that resin catalyzed lactose hydrolysis may be expressed with an irreversible first-order apparent rate expression as

E.3.26.1 Lactose k  products .

The major products, glucose and galactose, have higher solubility and sweetness. An Arrhenius type of rate expression was suggested to express the temperature effects on the apparent rate constant k with k 0 = 2.54 × 1021 h–1 and Ea = 1.543 × 105 J/mole. A batch reactor will be used to convert a 10% (w/w) lactose solution to obtain c Af /c A0 = 0.60. Initial temperature of the reactor will be 95°C. Chen and Zall (1983) observed formation of browning products during their studies; therefore, a linearly decreasing temperature profile,

E.3.26.2 T = T 0 α t ,

will be used during the hydrolysis experiments to reduce the formation of the browning reactions (T 0 = 95°C, α = constant). Parameter α will be chosen such that the temperature of the reactor will be 75°C at the end of the process. What should the process time and parameter α be?

Solution: The reaction rate expression is

E.3.26.3 dC A dt = R A ,

where

E.3.26.4 R A = kC A = k 0 C A  exp { E a / RT } .

After combining Equation E.3.26.4 and Equation E.3.26.2 and substituting values of constants we will obtain

E.3.26.5 dC A dt = 2.54 × 10 21 C A  exp { 18559 T } ,

we may use Equation E.3.26.2 to calculate the derivative

E.3.26.6 dT dt α ,

after combining Equations E.3.26.5 and E.3.26.6 and rearranging we will obtain

E.3.26.7 C A 0 0.6 C A 0 dC A C A = 368 348 2.54 × 10 21 a exp { 18559 T } dT .

We may define the integral as

/ = 368 348  exp { 18559 T } dT ,

and use the Simpson’s method to calculate it with ΔT = 2 K, N = (368–348)ΔT = 10:

/ = 2 3 { f 0 + 4 ( f 1 + f 3 + f 5 + f 7 + f 9 ) + 2 ( f 2 + f 4 + f 6 + f 8 ) + f 10 } ,

where

f 0 = exp { 18559 368 } = 1.25 × 10 22 , f 1 = exp { 18559 366 } = 9.5 × 10 23 , f 2 = exp { 18559 364 } = 7.19 × 10 23 , f 3 = exp { 18559 362 } = 5.42 × 10 23 .

After substituting the numbers in Equation E.3.26.7, we will calculate that α = 4.15, therefore the temperature profile should be T = 368–4.15 t. Since the final temperature is required to be 348 K the reaction time should be 4.8 hours. MATLAB® code E.3.26 shows the details of the computations.

In a CSTR (Figure 3.7) with input and output flow rates of F, where reaction A k

products is occurring (rate expression RA = – kcA ) we have

3.61a input rate of  A  into the reactor = Fc A 0 ,
3.61b output rate of the reactor = Fc A ,
3.61c generation of  A  rate in the reactor = R A V ,
3.61d accumulation rate of  A  in the reactor = d ( c A V ) dt ,

 

Figure 3.7   A continuously stirred tank reactor (CSTR).

after substituting Equations 3.61a through d into Equation 3.57 we will obtain

3.62 Fc A 0 Fc A + R A V = d ( c A V ) dt .

With a constant reactor volume V, under steady state conditions; that is, d(cAV)/dt = 0, Equation 3.62 may be rearranged as

3.63 V F = c A 0 c A R A .

The ratio V/F = τ is called the residence time. It is the average time spent by the entering liquid in the reactor.

 

Example 3.27  Acid Hydrolysis of Lactose in a CSTR

The lactose hydrolysis process described in Example 3.26 will be conducted in two series CSTRs (Figure E.3.27). The first reactor has a volume of 1 m3 and operated at 95°C. The second reactor has a volume of 3 m3 and operated at 50°C. What should be the volumetric flow rate F (m3/h) to obtain c A2/c A0 = 0.60?

Solution: Under the steady state conditions, the lactose balance around the first reactor is performed similarly as Equation 3.62 to obtain

E.3.27.1 Fc A 0 Fc A 1 k 1 C A 1 V 1 = 0.

A similar balance is performed for the second reactor:

E.3.27.2 Fc A 1 Fc A 2 k 2 C A 2 V 2 = 0.

After rearranging Equation E.3.27.1 we will obtain

E.3.27.3 C A 1 = Fc A 0 F + k 1 V 1 .

Equation E.3.27.3 is substituted in Equation E.3.27.2 to obtain

E.3.27.4 F 2 = C A 0 F + k 1 V 1 Fc A 2 k 2 C A 2 V 2 = 0.
The lactose hydrolysis process.

Figure E.3.27   The lactose hydrolysis process.

All the terms of Equation E.3.27.4 will be divided in c A0 to obtain

E.3.27.5 F 2 = 1 F + k 1 V 1 C A 2 C A 0 F k 2 V 2 C A 2 C A 0 = 0.

The Arrhenius expression is

3.5 k = k 0  exp { E a RT } .

After substituting the numbers in Equation 3.5 we will calculate the reaction rate constant as 0.317 h−1 at 95°C and at 2.82 × 10–4 h–1 50°C. After substituting c A2/c A0 = 0.6 and values of the rate constants and the reactor volumes in Equation E.3.27.5 we will calculate F = 0.44 m3/h. Details of the computations are available in MATLAB® code E.3.27.

In a PF reactor where reaction Ak products is occurring, with the rate expression RA = – kcA , the material described in Equation 3.57 may be done around a volume element of a PF reactor as described in Figure 3.8.

Each term appearing in the material may be written as follows:

3.64a ( input rate of  A  into the volume element ) = [ Fc A ] z ,

 

Figure 3.8   The volume element in a PF reactor used for material balance.

3.64b ( output rate of  A  from the volume element ) = [ Fc A ] z + Δ z ,
3.64c ( generation rate of  A  in the volume element ) = R A Δ z S ,

where S = cross-sectional area of the PF reactor:

3.64d ( accumulation rate of  A  from the volume element ) = d ( c A Δ z S ) dt .

Under steady state conditions; that is, d(cA Δ zS)/dt = 0, after substituting Equations 3.64a through d into Equation 3.57 we will obtain

3.65 [ Fc A ] z [ Fc A ] z + Δ z + R A Δ zS = 0.

Equation 3.65 may be rearranged as

3.66 F [ c A ] z [ c A ] z + Δ z Δ z = R A S .

After taking the limit of the left-hand side of Equation 3.66 as lim Δ z → 0 we will obtain

3.67 F dc A dz = R A S .

We may calculate the reactor length (L) for required conversion after rearranging and integrating Equation 3.67:

3.68 L = F S c A 0 c Af dc A R A .
 

Example 3.28  Acid Hydrolysis of Lactose in a Plug Flow Reactor

The lactose hydrolysis process described in Example 3.26 will be conducted in a PF reactor. The temperature is 95°C at the inlet of the reactor and will decrease linearly as T = T 0 – β z along the reactor and will be 75°C at the exit. What should S/F ratio in (h/m) to obtain cAf /c A0 = 0.60 in a 2 m long reactor?

Solution: Under the steady state conditions, the lactose balance around a shell element of the reactor is performed similarly to Equation 3.67 to obtain

E.3.28.1 F dc A dz = kc A S .

The data implies that β = 10°C/m and the temperature profile along the reactor is

E.3.28.2 T = 95 10 z .

The reaction rate constant is expressed in terms of the Arrhenius expression as

E.3.28.3 k = 2.54 × 10 21 exp { 18559 273 + ( 95 10 z ) } .

After combining Equations E.3.28.1 and E.3.28.3 and rearranging we will obtain

E.3.28.4 1 2.54 × 10 21 c A 0 c Af dC A C A = S F 0 2  exp { 18859 368 10 z } dz .

The integral appearing on the right-hand side needs to be evaluated numerically, similarly as that of Example 3.26. After substituting the numbers in Equation E.3.28.4, we will calculate S/F = 2.4 h/m. Details of the computations are given in MATLAB® code E.3.28.

 

 

Example 3.29  Acid Hydrolysis of Lactose With Immobilized β-Galactosidase in CSTR and Plug Flow Reactors

  • Kinetic constants for lactose hydrolysis by β-galactosidase from Escherichia coli were reported as KM = 1.9 moles/m3 and v max = 6.55 × 10–6 moles/(min mg enzyme) at 20°C and pH = 7.6 (Whitaker 1994). If this enzyme should be immobilized on a nonporous support to obtain 10 mg enzyme/cm3 of the reactor with η = 0.8, what should be the volumetric input flow rate of 50 moles/m3 lactose solution to a 0.7 m3 CSTR to achieve c exit/c 0 = 0.30? Solution: It is required that c A0 = 50 moles/m3 and cA = 15 moles/m3. The apparent reaction rate is
    E.3.29.1 R App = h C A V max K M + C A .
    After substituting the numbers, we will calculate that RApp = –4.65 × 10–6 moles/min mg enzyme. Since we have 10 mg enzyme/cm3 of a reactor, the apparent rate may be expressed in terms of the reactor volume as RApp = –46.5 moles/min m3. The lactose balance around the reactor under steady state conditions requires
    E.3.29.2 Fc A 0 Fc A + R App V = 0.
    Equation E.3.29.2 will be rearranged as
    E.3.29.3 F = V R App C A 0 C A .
    After substituting the numbers in Equation E.3.29.3 we will obtain F = 0.93 m3/min.
  • The same enzyme immobilized support is filled into a packed bed PF reactor (reactor diameter = 10 cm). What should be the reactor length to achieve the same conversion as in section (a)? Solution: The lactose balance around the reactor under steady state conditions requires
    3.68 L = F S C A 0 C Af dC A R App .
    We have already calculated that F = 0.93 m3/min, it is also known that c A0 = 50 moles/m3 and cA = 15 moles/m3 and the cross-sectional area of the reactor is S = π(reactor radius)2 = 7.85 × 10–3 m2. The apparent reaction rate is RApp = –52.4 cA /1.9 + cA moles/min m3. After substituting the numbers in Equation 3.68 we will obtain
    L = 0.93 m 3 / min 7.85 × 10 3 m 2 50 15 1 52.4 C A 1.9 + C A dc A = 84.3  m .
    The volume of the reactor = (L)(S) = 0.66 m3 and the smaller reactor volume required for the same conversion in a CSTR.
  • If the immobilized enzyme loses 10% of its activity after 24 hours of operation with reaction:
    E E i active enzyme inactive enzyme
    what will be the cAf /c 0 ratio after 3 days of operation with both CSTR and PF reactors? Solution: We have already described the enzyme inactivation rate as
    3.20 dc E dt = k d C E .

After integrating and rearranging Equation 3.20 we will have

E.3.29.4 k d = 1 t ln ( C E C E 0 ) .

Since cE /c E0 = 0.90 when t = 24 hours we may use Equation E.3.29.4 to calculate kd = 4.4 × 10–3 h−1. Equation E.3.29.4 may be rearranged to calculate the fraction of the remaining enzyme activity after 3 days (72 hours) of operation as

C E C E 0 = exp ( k d t ) = 0.72

The initial maximum enzyme activity v max was defined as

3.9 V max = k 3 C E 0 .

The remaining maximum enzyme activity after 3 days is

V max 3 days = ( k 3 C E 0 ) ( C E / C E 0 ) = ( 6.55 × 10 6 ) ( 0.72 ) = 4.72 × 10 6

moles/(min mg enzyme) since we have initially 10 mg enzyme immobilized/cm3 of reactor V max 3 days = 47.2 moles/min m3.

  • The cAf /c 0 ratio after 3 days of operation with CSTR We will substitute V max 3 days in Equation E.3.18.1 for v max to calculate RApp . Under these conditions Equation E.3.18.2 will be rewritten as
    E.3.29.5 Fc A 0 Fc A η V max 3 days C A K M + C A V = 0.
    After substituting the numbers, we will calculate c = 23.7 moles/m3 from Equation E.3.29.5, implying that cAf /c 0 = 0.474.
  • The cAf /c 0 ratio after 3 days of operation with PF reactor We will substitute V max 3 days in Equation E.3.18.1 for v max to calculate RApp . Under these conditions Equation 3.68 will be rewritten as
    E.3.29.6 84.3   m = 0.93   m 3 / min 7.85 × 10 3   m 2 50 C A   f 1 37.7 C A 1.9 + C A dc A .
    Equation E.3.29.6 will be simplified as
    E.3.29.7 16.7 = ln ( C Af ) + 0.54 C Af .
    After solving Equation E.3.29.7 we will obtain cAf = 23.8 moles/m3, implying that c Af /c A0 = 0.474. The computations are also carried out by MATLAB® code E.3.29.

 

 

Example 3.30  Reactors-in-Series Model for Digestion in Stomach and Intestines

The Eyriés (2003) model simulates stomach and intestines as a combination of ideal reactors as depicted in Figure E.3.30.1. Taking a caplet of medication may be regarded as an impulse passing through the stomach and the intestine. The degradation rate of the drug may be simulated as

E.3.30.1 R digestion = kc .

The medication balance around the first CSTR is

E.3.30.2 Fc 0 δ ( t ) Fc 1 ( t ) kc 1 ( t ) V 1 = d [ C 1 ( t ) V 1 ] dt .

Where V 1 is the volume of CSTR1; the medication concentrations are depicted with c 0(t) before entering and with c 1(t) after leaving CSTR1. After substituting τ = V 1/F = constant Equation E.3.30.2 is rearranged as

E.3.30.3 τ dc 1 ( t ) dt = C 0 δ ( t ) C 1 ( t ) k τ C 1 ( t ) .

The Laplace transform of Equation E.3.30.3 is

E.3.30.4 SC 1 ( S ) C 1 ( 0 ) = C 0 τ ( k + 1 τ ) C 1 ( S ) ,
Stomach (two CSTRS in series) and intestines (a plug flow reactor) as described by the Eyriés model. (From Eyriés, P., A Distributed-Parameter Modeling Approach for Performance Monitoring of Oral Drug Delivery Systems, Master of Science Thesis, Chemical Engineering Department, Worcester Polytechnic Institute, Worcester, Massachusetts, 2003.) The model considers the stomach as a nonideal CSTR, which behaves in-between an ideal CSTR and a plug flow reactor. The nonideality is described by replacing the single CSTR with a cascade of ideal CSTRs. The intestine is modeled as a single cylindrical plug flow reactor surrounded with a wall of constant width.

Figure E.3.30.1   Stomach (two CSTRS in series) and intestines (a plug flow reactor) as described by the Eyriés model. (From Eyriés, P., A Distributed-Parameter Modeling Approach for Performance Monitoring of Oral Drug Delivery Systems, Master of Science Thesis, Chemical Engineering Department, Worcester Polytechnic Institute, Worcester, Massachusetts, 2003.) The model considers the stomach as a nonideal CSTR, which behaves in-between an ideal CSTR and a plug flow reactor. The nonideality is described by replacing the single CSTR with a cascade of ideal CSTRs. The intestine is modeled as a single cylindrical plug flow reactor surrounded with a wall of constant width.

since c 1(0) = 0 we will calculate c 1(s) as

E.3.30.5 C 1 ( S ) = C 0 / τ 1 S + ( 1 + k / τ 1 ) .

The back transform of Equation E.3.30.3 is (Table 2.20)

E.3.30.6 C 1 ( t ) = C 0 τ exp { ( k + 1 τ ) t } .

When τ1 = τ2 concentration of the medication at the exit of the second CSTR may be calculated with a similar procedure as

E.3.30.7 C 2 ( t ) = C 0 t τ 2 exp { ( k + 1 τ ) t } .

The exit concentration of the nutrient from the cascade of j CSTRs may be expressed as

E.3.30.8 C j ( t ) = C 0 t j 1 ( j 1 ) ! τ j exp { ( k + 1 τ ) t } .

MATLAB® code E.3.30.a carries out the computations to determine the variation of the normalized drug concentration, c 2(t)/c 0, with time by using the “ two CSTRs in series” stomach model with k = 1.66 × 10–4 s–1 and τ = 900 s.

 

Figure E.3.30.2   Variation of the normalized drug concentration, c 2(t)/c 0, with time at the exit of the stomach, which is simulated as a cascade of two CSTRs.

The medication is ingested in the intestine while transported by convection through the lumen. The equation of continuity describes it as

2.13 C t + ( 1 ( rN r ) r + 1 r N θ θ + N x x ) = R .

Mass transport does not occur in r and θ directions in a plug flow reactor, implying that Nr = Nq = 0. We also have

E.3.30.9 N drug = x drug ( N drug + N lumen fluid ) D drug dc drug dx .

In a PF reactor, mass transfer with diffusion is negligible when we compare it with that of convection and x drug(N drug + N lumen fluid) = vc drug, where v is the constant flow velocity in the intestine through the lumen. After substituting Equation E.3.30.1 in Equation 2.13 we will have

E.3.30.10 C ( t , x ) t = V C ( t , x ) x kc ( t , x ) .

The drug concentration at the inlet of the intestine (PF reactor) is the same as that of the jth CSTR of the stomach, implying a BC for Equation E.3.30.10 as

E.3.30.10a BC  c ( 0 , t ) = C j ( t )
E.3.30.10b lC  c ( x , 0 ) = 0.

The Laplace transform of Equation E.3.30.7 is

E.3.30.11 SC ( S ) C ( 0 ) = V dc ( S ) dx kc ( S ) .

When j = 2 and we use transformation L{ tn e –α t } = n!/(s + α) n + 1 (Table 2.19) boundary condition of Equation E.3.30.8 will be c 0(s) = c 2(s) = c 02/[ s + (k + 1/τ)]2. Equation E.3.30.8 may be rearranged as ∫ dc(s)/c(s) = − ∫ k + s/v dx + κ, where κ is the integration constant. Upon integration we will have ln(c(s)) = −(k + s/v) x + κ or c(s) = κ1 exp (− k + s/vx) where κ1 = exp(κ). When x = 0 c 0(s) = c 2(s) = c 02/[ s + (k + 1/τ)]2 = κ1, then we will have

E.3.30.12 C ( S ) = C 0 / τ 2 [ s + ( k + 1 / τ ) ] 2 exp { k + s V x } = ( C 0 τ 2 e kx / v ) e ( x / v ) s [ s + ( k + 1 / τ ) ] 2 .

We will use the “ delayed nth power frequency shift” back transformation formula

L 1 { e β s ( s + α ) n + 1 } = ( t β ) n n ! e α ( t β )

to obtain

E.3.30.13 C ( t ) = ( t + x v ) C 0 τ 2 exp { kx v } exp { ( k + 1 τ ) ( t + x v ) } .

It should be noticed that [E.3.30.13] predicts c(t) = c 2(t) when x = 0. MATLAB® code E.3.30.b computes variation of the normalized drug concentration with time and position along the intestine.

 

Figure E.3.30.3   Variation of the normalized drug concentration, c(t)/c 0, with time (seconds) and dimensionless position along the intestine. Arrival of the drug to the inlet of the intestine (where the dimensionless position is zero) is described with a very nice peak as a function of time. Depletion of the drug along the intestine (as a function of time) is also seen clearly.

 

Example 3.31  A Model for Pasteurization With Microwaves in a Tubular Flow Reactor

In most microwave ovens, the microwaves change their polarization 2450 times in 1 second. Polar molecules try to align themselves according to the polarity of the microwave field. Their rapid movement causes tremendous intermolecular collisions, which heats up the medium that may be used to pasteurize the liquid foods. A section of an experimental microwave PF pasteurization reactor is shown in Figure E.3.31.1.

At a constant temperature, the thermal death rate of the microorganisms is

3.31d dx dt = k d x .

Temperature effects on the death rate constant kd may be expressed with the Arrhenius equation:

3.5 k d = k d 0  exp { E a RT } .

Heating of a stagnant fluid via microwave absorption may be expressed as

E.3.31.1 p = ω ɛ 0 E 2 κ ,

where p = absorbed microwave power, ω = angular frequency, ε0 = dielectric constant of free space, E = electric field intensity coupled by matched load, and κ = relative dielectric loss factor.

Figure E.3.31.1   Shell element of a pasteurization reactor.

The constant κ is the overall measure of the ability of the material to respond to the microwave field. Temperature effects on κ may be expressed as

E.3.31.2 κ = ϕ  exp ( β T ) ,

where β = constant. Heating of a stagnant liquid may be expressed after combining Equations E.3.31.1 and E.3.31.2 as

E.3.31.3 p = α  exp ( β T ) ,

where α = ωε0 E 2ϕ = constant. Thermal energy balance around an infinitely small shell element of the pasteurization reactor as depicted in Figure E.3.31.1 requires

E.3.31.4 ( input rate into the shell element ) ( output rate from the shell element ) + ( generation rate in the shell element ) = ( accumulation rate in the shell element ) .

After assuming no radial temperature distribution, we may convert Equation E.3.31.4 in mathematical terms as (Özilgen and Özilgen 1991)

E.3.31.5 π D r 2 v ρ c 4 [ T - T r e f ] z - π D r 2 v ρ c 4 [ T - T r e f ] z + Δ z - π D r Δ Z U ( T - T e n v ) + π D r 2 Δ z α e - π T 4 = π D r 2 Δ z ρ c 4 d T d T ,

where Dr is the diameter of the reactor, v, c, and ρ are the velocity, specific heat, and density of the liquid. The term ( π D r 2 v ρ c / 2 ) [ T T ref ] z

is the enthalpy of the liquid entering into the shell element at distance z from the entrance of the heating section of the reactor. It should be noticed that the enthalpy is calculated with respect to an arbitrarily chosen reference temperature T ref. The term ( π D r 2 v ρ   c / 2 ) [ T T ref ] z + Δ z is the enthalpy of the liquid leaving the shell element. The length of the shell was Δ z as depicted in Figure E.3.31.2. The term π Dr Δ z(TT env) describes the enthalpy loss from the shell element to the environment with convection. The heating rate of the liquid in the shell element with microwaves was described as π D r 2 Δ z α   e π   T / 4 , unsteady state thermal energy accumulation in the shell element was described as ( π D r 2 Δ z α   e π   T ) ( dT / dt ) .

Comparison of the model and the experimentally determined temperatures along the flow reactor. (Adapted from Özilgen, S., and Özilgen, M.,

Figure E.3.31.2   Comparison of the model and the experimentally determined temperatures along the flow reactor. (Adapted from Özilgen, S., and Özilgen, M., Enzyme and Microbial Technology, 13, 419–23, 1991.)

Under steady state conditions the unsteady state thermal energy accumulation term becomes zero and the equation was rearranged as

E.3.31.6 dT dz = K 1 exp ( β T ) K 2 ( T T env ) ,

where K 1 = α/vρ c and K 2 = U/Drvρ c. Solutions of Equation E.3.31.6 as given in the MATLAB® code E.3.31 simulates the variation of the temperature profiles along the reactor.

Although temperature varies drastically with distance, the flow reactor may be considered as a combination of 1 cm long small segments, and the temperature of any of these segments may be considered constant. The fraction of biomass surviving each segment may be calculated as

E.3.31.7 x i x 0 = exp ( k di V ) ,

where x 0 and xi are the viable cell counts at the entrance and exit of the ith segment, respectively. Parameter kdi is the death rate constant calculated with Equation 3.5 at the average temperature of the ith segment Ti . Equation E.3.31.7 is solved in the MATLAB® code E.3.31, where parameter v was the average velocity of the medium in the sterilization reactor. Parameter T 0 was the temperature of the medium before entering the microwave reactor, the ratio xi /x 0 was the viable biomass fraction surviving the whole pasteurization process. The numerical values of xi /x 0 were plotted against the reactor length in the microwave oven in Figure E.3.31.3.

 

Figure E.3.31.3   Comparison of the model and the experimentally determined surviving microbial fractions along the flow reactor. (Adapted from Özilgen, S., and Özilgen, M., Enzyme and Microbial Technology, 13, 419–23, 1991.)

Questions for Discussion and Problems

  • Metabolic Process Engineering and Microbial Kinetics
    • B.1. The following data were adapted from Cruz (2003), during the course of fermentation of the agave wort by a native yeast strain during a tequila production process:

       

      Complete the MATLAB® code to develop biomass, ethanol, isoamyl plus isobutyl alcohol production, and reducing sugars consumption models. You will need to evaluate the constants of the model yourself and compare the models with the data.
    • B.2. The following models were used by Wang et al. (2004) while studying the kinetics growth of apple wine yeast Saccharomyces cerevisiae on fructose:
      3.34 Microbial growth : dx dt = μ x ( 1 x x max ) ,
      Q.3.B.2.1 Ethanol production : dP dt = Y p / x dx d ( t Δ t ) .
      Where Δ t was introduced to describe delay of ethanol production to cell growth
      Q.3.B.2.2 Sugar consumption: dS dt = 1 Y s / x dx dt + mx ,
      where sugar is assumed to be consumed for biomass production and maintenance. The initial values were x 0 = 0.5 g/L and s 0 = 90 g/L and the coefficients of the model were μ = 0.08 g/L, x max = 7.23 g/L, Y p/x = 5.79 g/g, Δ t = 14.6 h, Y s/x = 0.19 g/g, and m = 0.07 h–1. Write a MATLAB® code to simulate the process.
  • Microbial Death Kinetics
    • B.1. The surviving number of microorganisms were counted after processing a food for the following given times at 110°C and 121°C are

      t (min) at 110°C 0 5 10 15 20
      x (cfu/ml) 60,000 21,000 7400 2650 930

      t (min) at 121°C 0 1 2 3 4
      x (cfu/ml) 60,000 4100 290 19 2
      • Determine the death rate constant at 110°C and 121°C.
      • Convert these death rate constants into D values.
      • The death rate constant of the microorganism was calculated at different temperatures as

        T (°C) 110 113 115 118 121 124
        kd (min−1) 0.208 0.397 0.743 1.44 2.62 5.01
        Determine the activation energy and frequency factor of the Arrhenius expression for the death rate constant.
      • Calculate the z value from the activation energy at 110°C and 124°C (T ref = 121°C). What is the average z value?
  • Rector Design
    • C.1. Solve the Example 5.30 with 3 and 4 CSTRs in series. Notice: τ = 600 s when there are 3 CSTRs and τ = 450 s when there are 4 CSTRs. Discuss your results by employing graphs when possible.

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