**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. * T**a** and T*0 are constant, ^{}^{T}^{max 1}>^{}^{T}^{max 2}>^{}^{T}^{max 3}
2. * T**a** and T*0 are constant, *T*^{max 1}

*<**T*^{max 2}

*<**T*^{max 3}
3. *T*^{max}*and T*0* are constant, T**a1 **>T**a2**>T**a3*

4. *T*^{max}*and T*0* are constant, T**a1 **<T**a2**<T**a3*

5. Nothing can be said Explain your answer.

*T*1

*T*2

*T*3

*t*

**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

**c) Adiabatic flame temperature**

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 (T*i** + T**ad**)/2, where T**ad* is guessed to be about 2100 K.

---
**Answers**

**a) The batch reactor model**

Statement 4. (*T*^{max}*and T*0* are constant, T**a1 **<T**a2**<T**a3* ) is true. The “activation

*temperature” T**a **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 *m** _{air}* 6.5 kg/s

*molecular weight of air MW**air *= 28.85

*molecular weight of fuel MW**fuel *= 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

###

*x*

*O*2

*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 4.76}^{28.85} ^{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/s^{2}

/ 68.48

*air*
*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} ^{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 :

2*a* 2 2 *b* *b a* 2 *b*6.00
and finally

###

26.00 1

1.54 10 1 2 3.76 7.45 3.76 8.00

*O* *prod*

*x* *b*

*b* *a*

**c) Adiabatic flame temperature**

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)

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* *H* *N h*

4

react

CH

1 74831 2 0 7.52 0 74831 kJ/mol

*H*

_{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*

*ad*
*ad*
*ad*

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

*T*
*T*
*T*

Equating (first law) *H*^{react}to ^{H}^{prod}and solving for *T yields*^{ad}

*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 ^{}^{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*

; ^{p}^{prod} ^{i p i}^{,}
*C* *N c*
then we have

0

298 ^{R}

*ad*

*p*

*T* *H*

*C*

. Thus,

_{4}

0

1 393546 2 241845 74831 802405 kJ/molCH

*H**R*

CH4

1 56.21 2 43.87 7.52 33.71 397.4 kJ/mol K

*C**p*

then

^{802405}

298 2317 K

397.4
*T**ad*

.