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

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a) Thermal theory of explosion: the batch reactor model

The following diagram:

0 20 40 60 80 100

0 500 1000 1500 2000 2500

shows the time evolution of temperature in an adiabatic batch reactor described by the model

0

0

a

a

E RT

E

r RT

V

dC k e C

dt

dT H

k e C dt c

 

 



with initial conditions

 0 0;  0 0

C C T T . If we refer to the following quantities:

0

max r ; a a

V

H C E

T T

c R



which one of these statements is true?

1. Ta and T0 are constant, Tmax 1>Tmax 2>Tmax 3 2. Ta and T0 are constant, Tmax 1

<Tmax 2

<Tmax 3 3. Tmaxand T0 are constant, Ta1 >Ta2>Ta3

4. Tmaxand T0 are constant, Ta1 <Ta2<Ta3

T1

T2

T3

t

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b) Reactant and product mixtures

A combustor operates at an equivalence ratio of 0.25 with an air flowrate of 6.5 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

1. Estimate the constant-pressure adiabatic flame temperature for the combustion of a stoichiometric CH4–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 CO2, H2O, and N2.

 The enthalpy of the product mixture is estimated using constant specific heats evaluated at 1200 K  (Ti + Tad)/2, where Tad is guessed to be about 2100 K.

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a) The batch reactor model

Statement 4. (Tmaxand T0 are constant, Ta1 <Ta2<Ta3 ) is true. The “activation

temperature” Ta 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 =0.25 air mass flowrate mair 6.5 kg/s

molecular weight of air MWair = 28.85

molecular weight of fuel MWfuel = 1×12.01+4×1.008 = 16.04 find:

fuel mass flowrate mfuel operating air-fuel ratio (A/F)

oxygen mole fraction in the product mixture

###  

xO2 prod

By definition,

/ stoich 4.76 air

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 4.7628.85 17.12

16.04

stoich

A F  

. From the definition of  ,

  / 17.12

/ 68.48

0.25

stoich

A F A F

The fuel mass flow rate is then

6.5 kg/s 9.49 10 kg/s2

/ 68.48

air fuel

m m

A F

Finally we write the reaction

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4 2 2 2 2 2 2

CH a O 3.76N CO 2H ObO 3.76 Na

where a can be derived by

/ 4.76 air

fuel

A F a MW

MW

, that is

/ 68.48 16.04 8.00

4.76 4.76 28.85

fuel air

a A F MW

MW

We can write an atom balance of oxygen :

2a  2 2 b b a  2 b6.00 and finally

###  

2

6.00 1

1.54 10 1 2 3.76 7.45 3.76 8.00

O prod

x b

b a

  

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 ,

hf i

(kJ/kmol)

Specific Heat at 1200 K

,

cp i

(kJ/kmol-K)

CH4 –74831 (Table B.1, Turns’ textbook)

CO2 –393546 (Table A.2) 56.21 (Table A.2)

H2O –241845 (Table A.6) 43.87 (Table A.6)

N2 0 33.71 (Table A.7)

O2 0

By applying the first law of thermodynamics:

react prod

react prod

= =

i i i i

H N h HN h

      

4

react

CH

1 74831 2 0 7.52 0 74831 kJ/mol

H

 

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 

 

4

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

H N h c T

T T T





 

Equating (first law) Hreactto Hprodand solving for T yieldsad

T = 2317 K for ad  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 HR0 and the average heat capacity Cp

of the product mixture, respectively as:

0 0 0

, ,

prod react

R i f i i f i

H N h N h

; p prod i p i, C N c then we have

0

298 R

p

T H

C



. Thus,

  4

0

1 393546 2 241845 74831 802405 kJ/molCH

HR

         

CH4

1 56.21 2 43.87 7.52 33.71 397.4 kJ/mol K

Cp    

then

802405

298 2317 K

.

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