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 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.
0 20 40 60 80 100
0 500 1000 1500 2000 2500
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 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.
--- 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 Φ=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
2prod
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 Φ ,
( ) ( / ) 17.12
/ 68.48
0.25
stoich
A F = A F = = Φ
The fuel mass flow rate is then
( )
6.5 kg/s 2
9.49 10 kg/s / 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
(
2) 6.00 1.54 10 1
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
p i ,
c (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