a) Thermal theory of explosion: the batch reactor model

a) Thermal theory of explosion: the batch reactor model

The following diagram shows the time evolution of temperature in an adiabatic batch reactor described by the following model:

 ¹  ^T

^{T T}

d e

d

 

 

 

    

 

with initial condition

  ^{0 0}

  ^.

0 20 40 60 80 100

0 500 1000 1500 2000 2500

Which one of these statements is true?

1.  T

and T

are constant, T

>T

>T

2.  T

and T

are constant, T

<T

<T

3. T

and T

are constant,  T

<  T

<  T

4. T

and T

are constant,  T

>  T

>  T

5. Nothing can be said Explain your answer.

T

T

T

t

b) Reactant and product mixtures

A combustor operates at an equivalence ratio of 0.31 with an air flowrate of 6.4 kg/s. The fuel is methane. Determine:

1. the fuel mass flowrate

2. the air-fuel ratio at which the engine operates 3. the mole fraction of oxygen in the product mixture

c) Adiabatic flame temperature

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

–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

, H

O, and N

.

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

+ T

)/2, where T

is guessed to be about 2100 K.

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

in the products,

estimate the adiabatic flame temperature for the system in Problem b).

--- Answers

a) The batch reactor model

ii.  T

and T

are constant, T

<T

<T

The “activation temperature” T

is proportional to the activation energy Ea in the Arrhenius expression. The higher the activation energy, the higher is the value of the temperature required to observe fast reaction.

b) Reactant and product mixtures

Given:

Equivalence ratio =0.31 air mass flowrate m 

 ^6.4 kg/s

molecular weight of air MW

= 28.85

molecular weight of fuel MW

= 1×12.01+4×1.008 = 16.04 find:

fuel mass flowrate m

operating air-fuel ratio (A/F)

oxygen mole fraction in the product mixture   x

By definition,  ^/ 

^4.76

fuel

A F a MW

 MW , where a = 2 is the stoichiometric coefficient of air for fuel combustion, in this case methane:

 

4 2 2 2 2 2

CH  2 O  3.76N  CO  2H O+7.52N Therefore  ^/  ^{2 4.76 28.85 17.12}

16.04

stoich

A F    . From the definition of  ,

   ^/  17.12

/ 55.24

0.31

stoich

A F  A F  

 The fuel flow rate is then

 ^/

 ^{6.4 kg/s 55.24 0.116 kg/s}

fuel

m m

 A F   



Finally we write the reaction

 

4 2 2 2 2 2 2

CH  a O  3.76N  CO  2H O  b O  3.76 N a

where a can be derived by  ^/  ^4.76

fuel

A F a MW

 MW , that is

 ^/  ^{55.24 16.04 6.45}

4.76 4.76 28.85

fuel air

a A F MW

 MW  

 We can write an atom balance of oxygen :

2 a    2 2 b  b a    2 b  4.45 and finally

 

4.45 0.14 1 2 3.76 7.45 3.76 6.45

O prod

x b

b a

  

    

c) Adiabatic flame temperature

c.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,

=2,

=7.52

N  N N

Properties:

Species Enthalpy of formation at 298 K

0 ,

h

(kJ/kmol)

Specific Heat at 1200 K

,

c

(kJ/kmol-K)

CH

–74831 (Table B.1, Turns’ textbook) –

CO

–393546 (Table A.2) 56.21 (Table A.2)

H

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

N

0 33.71 (Table A.7)

O

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/mol

H    

 

 

   

   

   

0

prod , ,

prod

CH

298

1 393546 56.21 298

2 241845 43.87 298

7.52 0 33.71 298 kJ/mol .

i f i p i ad

ad ad ad

H N h c T

T T T

 

    

      

      

      



Equating (first law) H

to H

and solving for T

yields

T

= 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

H   N h  T   N c

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

and the average heat capacity C

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

ad

p

T H

C

   .

Thus,

     

0

1 393546 2 241845 74831 802405 kJ/mol

H

          

CH4

1 56.21 2 43.87 7.52 33.71 397.4 kJ/mol K

C

      

then

 ⁸⁰²⁴⁰⁵ 

298 2317 K

397.4 T

 

   .

c.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>2 (2

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

0.31 2 6.45

0.31 0.31

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 4.45 b a    . We can now rewrite the species balance for   0.31 :

   

4 2 2 2 2 2 2

CH  6.45 O  3.76N  CO  2H O 4.45O   6.45 3.76 N

where in the products we find

2 2 2 2

CO

1,

=2,

=4.45,

=24.26

N  N N N

For simplicity we maintain the value of the temperature of 1200 K for the Specific Heats. In addition to the property values used in b.1, we need the Specific Heat at 1200 K for O

:

Species Enthalpy of formation at 298 K

0 ,

h

(kJ/kmol)

Specific Heat at 1200 K

,

c

(kJ/kmol-K)

O

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

       

react

1 74831 6.45 0 24.26 0

74831 kJ

H    

 

 

   

   

   

   

0

prod , ,

prod

298

1 393546 56.21 298

2 241845 43.87 298

4.45 0 35.59 298

24.26 0 33.71 298

i f i p i ad

ad ad ad

ad

H N h c T

T T T

T

 

    

      

      

      

      



Equating (first law) H

to H

and solving for T

yields

T

= 1014 K for   0.31

Finally, we note that the final temperature is much lower than the stoichiometric, hence we might

want to refine the computations by taking the average Specific Heats perhaps at 650 K rather than

1200 K.