**a) Thermal theory of explosion: the batch reactor model**

### 1. Derive an expression for the time at which the concentration of a reactant reduces to one fourth of its initial value for a batch reactor under isothermal conditions and a first order reaction kinetics. Does this depend on the initial concentration?

### 2. Is it possible to avoid, by cooling, a thermal explosion in a closed, fixed-volume vessel in which an exothermic reaction can occur? If yes, explain how; if not, explain why.

**b) Adiabatic flame temperature**

### 1. Estimate the constant-pressure adiabatic flame temperature for the combustion of a stoichiometric CH

4### –air mixture. The pressure is 1 bar and the initial reactant temperature is 298 K. Use the following assumptions:

### “Complete combustion”, i.e., the product mixture consists only of CO

2### , H

2### O, and N

2### .

### The enthalpy of the product mixture is estimated using constant specific heats evaluated at 1200 K (T

*i*

* + T*

*ad*

*)/2, where T*

*ad*

### is guessed to be about 2100 K.

### 2. Under the same assumptions, but obviously considering excess O

2### in the products,

### estimate the adiabatic flame temperature for a CH

4### –air mixture with equivalence

### ratio = 0.7.

### --- **Answers**

**a.1**

### The equation that describes the dynamics of a batch reactor under isothermal conditions and a first order reaction kinetics is:

*dC* *kC*

*dt*

*where k is constant. The solution is *

###

0### exp

*C t* *C* *kt*

### If we call

^{t}^{1}

_{4}

### the time at which the concentration

*C*

### reduces to half of its initial value

*C*0

### , we have:

##

^{1}

^{4}

##

^{1}4

##

^{1}4

^{ }

0

### 1 1

### exp ln ln 1

### 2 4

*C t* *kt* *kt*

*C*

###

^{}

^{0}

^{} ^{ln 4} ^{ } ^{kt}

^{kt}

^{1}

_{4}

### The time at which the concentration of a reactant reduces to half of its initial value is then:

###

14

*t*

### ln 4

###

*k*

**a.2**

### Under the assumptions made in the thermal theory of explosion written for a batch reactor model, it is possible to avoid explosion in a cooled reactor in which an exothermic reaction can occur.

### Indeed, for an exothermic reaction the change in enthalpy from reactants to products is NEGATIVE at constant temperature. Until the reactor is at the same temperature as the coolant, no energy can be exchanged with the environment, hence the reactor enthalpy must remain constant.

### The decrease in chemical potential energy due to the reaction must then be compensated by an increase in the temperature, that corresponds to the “sensible heat”. Temperature increases during reaction and the reaction rate constant increases as a function of the temperature. Thus, the reaction accelerates. However, the difference between the increased reactor temperature and the coolant temperature produces heat removal. If heat removal is efficient, the reactor temperature will remain relatively low and the process will not lead to thermal explosion. Towards the end of the process, due to lack of reactants, the reaction rate will decrease and the temperature will finally achieve the value of the temperature of the coolant medium.

### In conclusion, it is possible to avoid thermal explosion if the reactor is cooled efficiently.

**b.1**

### The mixture composition follows from balancing the chemical equation for a stoichiometric mixture of methane and air:

###

4 2 2 2 2 2

### CH 2 O 3.76N CO 2H O+7.52N where in the product mixture we find

2 2 2

CO

### 1,

H O### =2,

N### =7.52

*N*

###

*N*

*N*

### Properties:

### Species Enthalpy of formation at 298 K

0 ,

*h**f i*

### (kJ/kmol)

### Specific Heat at 1200 K

,

*c**p i*

### (kJ/kmol-K)

### CH

4### –74831 (Table B.1, Turns’ textbook) –

### CO

2### –393546 (Table A.2) 56.21 (Table A.2)

### H

2### O –241845 (Table A.6) 43.87 (Table A.6)

### N

2### 0 33.71 (Table A.7)

### O

2### 0 –

### By applying the first law of thermodynamics:

react prod

react prod

### = =

*i i* *i i*

*H* *N h* *H* *N h*

###

4

react

CH

### 1 74831 2 0 7.52 0 74831 kJ/kmol

*H*

###

###

###

###

###

_{4}

0

prod , ,

prod

CH

### 298

### 1 393546 56.21 298

### 2 241845 43.87 298

### 7.52 0 33.71 298 kJ/kmol .

*i* *f i* *p i* *ad*

*ad*
*ad*
*ad*

*H* *N h* *c* *T*

*T* *T* *T*

###

###

###

###

###

###

### Equating (first law)

*H*react

### to

*H*prod

### and solving for

*T*

*ad*

### yields

*T**ad*

### = 2317 K for 1 (stoichiometric conditions)

### Another way is:

0

react ,

react react

### (in our case)

*i i* *i f i*

*H* *N h* *N h*

###

0

prod , ,

prod prod

*i f i* *ad*

### 298

*i p i*

*H* *N h* *T* *N c*

### If we define, on a one-mole-of-fuel basis, the enthalpy of reaction

*H*

*R*

^{0}

### and the average heat capacity

*C*

*p*

### of the product mixture, respectively as:

0 0 0

, ,

prod react

*R* *i f i* *i f i*

*H* *N h* *N h*

### ;

^{,}

prod

*p* *i p i*

*C* *N c* then we have

0

### 298

^{R}*ad*

*p*

*T* *H*

*C*

### .

### Thus,

###

_{4}

0

### 1 393546 2 241845 103847 802405 kJ/kmol

CH*H**R*

###

CH4

### 1 56.21 2 43.87 7.52 33.71 397.4 kJ/kmol K

*C**p*

###

### then

### ^{802405}

### 298 2317 K

### 397.4 *T*

*ad*

###

### .

**b.2**

### In this case, the mixture composition follows from balancing the chemical equation for a *mixture of methane and air, with excess air. Suppose that we provide 4.76a moles of air, with a>5* (5 is the value for a stoichiometric mixture). We can write the species balance as:

###

4 2 2 2 2 2 2

### CH

*a*

### O 3.76N CO 2H O 3.76 N

*a*

###

*b*

### O

*where a and b must be computed from the knowledge of the equivalence ratio. From the definition* of the equivalence ratio we can compute the value of a:

###

###

###

*stoic*

*stoic*

*A F* *F A*

*A F* *F A*

###

### where

### ^{4.76}

### 1

*air* *stoic* *air*

*stoic*

*fuel* *stoic* *fuel*

*m* *a* *MW*

*A F* *m* *MW*

###

### ; ^{4.76}

### 1

*air* *air*

*fuel* *fuel*

*m* *a MW*

*A F* *m* *MW*

###

### hence

### 0.7 2 2.86

### 0.7 0.7

*stoic* *stoic*

*a* *a*

*a* *a*

###

### From the balance of oxygen atoms we find

### 2 1 2 2 1 2

*a*

###

*b*

### that is

### 2 0.86

*b a*

### . We can now rewrite the species balance for 0.7 :

###

4 2 2 2 2 2 2

### CH 2.86 O 3.76N CO 2H O 10.74N 0.86O

### where in the products we find

2 2 2 2

CO

### 1,

H O### =2,

N### =10.74,

O### =0.86

*N*

###

*N*

*N*

*N*

### In addition to the property values used in b.1, we need the Specific Heat at 1200 K for O

2### : Species Enthalpy of formation at 298 K

0 ,

*h**f i*

### (kJ/kmol)

### Specific Heat at 1200 K

,

*c**p i*

### (kJ/kmol-K)

### O

2### 0 35.593 (Table A.11)

### By applying the first law:

react prod

react prod

### = =

*i i* *i i*

*H* *N h* *H* *N h*

###

4

react

CH

### 1 74831 2.86 0 10.74 0 74831 kJ/kmol

*H*

###

###

###

###

###

###

_{4}

0

prod , ,

prod

CH

### 298

### 1 393546 56.21 298

### 2 241845 43.87 298

### 10.74 0 33.71 298

### 0.86 0 35.59 298 kJ/kmol

*i* *f i* *p i* *ad*

*ad*
*ad*
*ad*
*ad*

*H* *N h* *c* *T*

*T* *T* *T* *T*

###

###

###

###

###

###

###

### Equating (first law)

*H*react

### to

*H*prod

### and solving for

*T*

*ad*

### yields

*T**ad*