MCG5191 Assignment 1: Combustion of Methane and NO Production

This document presents a comprehensive solution to a mechanical engineering assignment (MCG5191) focusing on combustion, thermodynamics, and fluid dynamics. The assignment includes calculating the adiabatic flame temperature for methane combustion under stoichiometric and excess air co...Read More

Chemistry
Problem 1
Consider the following global reaction for the combustion of methane in air (21% O2 and
79 % N2 by volume):
CH4 + 2O2 → CO2 + 2H2O
Species hi;ref [kJ/mol]
O2 0
N2 0
CO2 -393.5
H2O -241.8
CH4 -74.87
Reference enthalpy of selected species at Tref = 298.16 K.
a) Calculate the adiabatic flame temperature for the complete combustion of a
stoichiometric mixture at initial temperature of 300K and constant pressure. Take
the mixture heat capacity at constant pressure to be cp = 1.4 kJ/ (kg K).
Solution
Adiabatic flame temperature is obtained by reacting given two gases, at a certain pressure to
expect the end temperature.
Enthalpy changes is given by
h2 −h1=−q 1= ( hf
o ) unit mass
Final temperature level.
hf −h2=q 1
The specific heat as constant
C p avg (T F −T 2)=q 1
Overall process
∆ h1−∆ h2 =∆ hadiabatic =0
Temperature change is given by,
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T f −T2=q1/C p avg=hf
o/C p avg
CH4 + 2O2 → CO2 + 2H2O
(h¿ ¿ f )¿ CH4 → ¿ hf ¿CO2 + 2(h¿ ¿ f )¿ H2O +2(h¿ ¿ f )¿ O2
Hreactant = -74.87+2(0) = -74.87 Kj/ mol
Hproduct =∑ni [∆H0f,i +Cp,I ( Tad - 298.16) = -393.5+2(-241.8) +1.4 (Tad - 298.16)
Hreactant =Hproduct
-74.87 Kj/ mol = -393.5 +2(-241.8) +1.4 (Tad - 298.16)
-74.87 = -393.5 – 483.6+1.4 (Tad - 298.16)
-74.87= -877.1+ 1.4 (Tad - 298.16)
802.23/1.4 =1.4/1.4(Tad - 298.16)
573.02 =(Tad - 298.16)
=871.71K
b) Recalculate the adiabatic flame temperature if there is now an excess of air such
that only half of the O2 molecules are needed to fully consume the methane
molecules present in the reactants.
Equation for the reaction
CH4 +2 O2 → CO2 + 2H2O
T f −T2=q1/C p avg=hf
o/C p avg
the amount of leftover moles of the excess reagent is given by,
nexcessreagent
o −nlimiting reagent
o × molar ratio of excess reagent
molar ratio of limiting reagent =nexcess reagent
To react one mole of methane it requires 2moles of oxygen. Air is 50% excess so oxygen
supplied is 1.5 times the required moles which is 2 times which equals to 3. Number of moles of
nitrogen supplied as the air contains 79% of nitrogen and 21 % of oxygen is
79 moles of N2
21 moles of O2
X 3 moles of O2 =11.28 moles N 2
Number of components of i in equilibrium is given as;
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n i = nio + ∑ vi where ,
vi is the stoimetric coefficient of i
nio is the initial number of moles
CH4 O2 N2 CO2 H2O
nio 1 3 11.28 0 0
Vi -1 -2 0 1 2
ni 0 1 11.8 1 2
hi;ref [kJ/mol] -74.87 0 0 -393.5 -241.8
CH4 + 2O2 → CO2 + 2H2O
(h¿ ¿ f )¿ CH4 → ¿ hf ¿CO2 + 2(h¿ ¿ f )¿ H2O +2(h¿ ¿ f )¿ O2
Hreactant = 0(-74.87)+2(0) = 0 Kj/ mol
Hproduct =∑ni [∆H0f,i +Cp,I ( Tad - 298.16) = -393.5+2(-241.8) +1.4 (Tad - 298.16)
Hreactant =Hproduct
O Kj/ mol = -393.5 +2(-241.8) +1.4 (Tad - 298.16)
0 = -393.5 – 483.6+1.4 (Tad - 298.16)
0 = -877.1+ 1.4 (Tad - 298.16)
877.1/1.4 =1.4/1.4(Tad - 298.16)
626.5 = (Tad - 298.16)
Tad = 924.5K
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c) N2 can react to form NO. We will consider the global thermal NO production
reaction presented in class:
N2 + O2 = 2NO
Assume that reaction 2 is in equilibrium. Recall that Kp (T) = exp (-∆G oR=RT) At 298 K,
the Gibbs free energy of reaction is∆G oR=173 kJ/mol. In order to obtain the Gibbs free
energy of reaction at other temperatures, it is necessary to know the temperature
dependence of ∆HOR and ∆SOR . The temperature dependence of enthalpy and entropy of
individual species at reference pressure are commonly provided in polynhomial forms in
thermochemical databases such as the NIST Chemistry WebBook. Here, we take the
shortcut version and take the following fit the equilibrium constant:
Kp (T) = 17.38T0.0247 exp (-21719/T)
Find an expression for the concentration of NO as a function of temperature. Evaluate this
concentration for the two temperatures identified in a) and b). State your assumptions if
any is required and comment on the results.
The equilibrium mole fraction of NO in air at T = 298.16 K and T = 300 K by assuming that the
mole fractions of oxygen (XO2 = 0.21) and nitrogen (XN2 = 0.79) remain unchanged.
The equilibrium constant of the reaction N2+ O2= 2 NO
Kp (T) = 17.38T0.0247 exp (-21719/T)
For the partial pressure of NO one has
p NO =(pN2p O2Kp ) 1/2
Neglecting the consumption of N2 and O2 as a first approximation, their partial pressures in air
may be approximated as
pN2 =0.79p pO2 =0.21p
Given the following;
Kp (T) = 17.38T0.0247 exp (-21719/T)
∆G oR=173 kJ/mol
Kp (T) = exp (-∆G oR=RT)
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∆GO = -RT in Kp (T)
-∆GO/RT = in Kp (T)
Problem 2
Consider the simplified mechanism for H2-O2 ignition discussed in class:
1) H + O2 → OH + O
2) O + H2 → OH + H
3) OH + H2 →H2O + H
4) H + O2 +M→ HO2 +M
5) H2 + O2→ H + HO2
associated with rate constants k1, k2, k3, k4 and k5, respectively.
a) Assuming k2 and k3 are very large, we can put O and OH in quasi-steady state.
Show that the reduced mechanism will be
1) 3H2 + O2 →2H + 2H2O
2) H + O2 +M →HO2 +M
3) H2 + O2 →H + HO2
with rates of progress
qI = k1 [O2] [H]
qII = k4 [O2] [H] [M]
qIII = k5 [O2] [H2]
Solution
Hydrogen peroxide combustion consists many elementary reactions that involves reacting
species. Hydrogen-oxygen reactions includes;
H + O2 ⇌OH + O
H2 + O ⇌ OH + H
H2 + OH ⇌ H2O + H
H2O + O⇌ 2OH
the first three reactions are sufficient to describe radical chain branching in that neglecting the
fourth reaction results in an erroneous balance between O and OH in very lean deflagrations.
The combined equation becomes;
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;
H + O2 + M → HO2 + M
H + OH + M ⇌ H2O + M
H + H + M⇌ H2 + M
O + O + M⇌ O2 + M
H + O + M ⇌OH + M
O + OH + M ⇌HO2 + M
The above reactions is most important under all conditions as they improves accuracy of
calculated deflagration velocities in stoichiometric and rich mixtures. When the equations is
combined the reactions are included as they approach equilibrium. Hydrogen peroxide
consuming reactions includes.
HO2 + H → 2OH
HO2 + H⇌ H2 + O2
HO2 + OH →H2O + O2
HO2 + H⇌ H2O + O
HO2 + O⇌ OH + O2
Hydroperoxil radicals, created through this process are consumed in the reaction being only
significant in fuel environments in which hydroxyl ions appears more concentrated than that
hydrogen ions. The above reactions can be reduced and to identify the reduced equations
includes;
C˙H2O + C˙O + C˙OH + 2C˙H2O2 = 2ωI
C˙O2 + C˙HO2 = - ωI
C˙H2 – C OH - 2C˙O + C˙HO2 - 2C˙H2O2 = ωII - 3ωI
C˙H + C˙OH + 2C˙O - C˙HO2 + 2C˙H2O2 = 2ωI - 2ωI
The concentrations at steady state of the species are smaller than those of production rates
indicating that with the approximations introduced, the equations reduces further to the two
overall as shown below.
3H2 + O2⇌ 2H2O + 2H
H + H + M ⇌H2 + M
The overall rates above are reduced equations that can be expressed in terms of different
alternative sets of overall reactions, the resulting formulations are all equivalent.
The results above is the two-step mechanism which is not sufficiently to describe flames fails
and auto ignition processes and therefore needs to be improved. With both H and HO2 out of
steady state, the overall reactions of the resulting reduced mechanism final becomes
3H2 + O2 →2H + 2H2O
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H + O2 +M →HO2 +M
H2 + O2 →H + HO2
with rates of progress as given below
qI = k1 [O2] [H]
qII = k4 [O2] [H] [M]
qIII = k5 [O2] [H2]
b) In class, we identified that the cross-over temperature (temperature that
delimitates \short" and \long" or \impractical" ignition delays) is the root of
K 4 ( Tc ) p
K 1 ( Tc ) RTc -2=0
Find the root numerically to determine the cross-over temperature for pressures of
p = 1 bar and p = 0.1 bar. The rate constants are of the form
kk(T) = AkTβk exp(Eak/RT) Take A1 = 2 x 108 m3mol-1s-1, β1 = 0, Ea=70.3 kJmol-1, A4 =
2.3 x106 m6mol-2s-1K0.8, β4=-0.8, Ea4=0.
The cross-over temperature equation is;
T c= √b2−ac
2 π kB sinϑ
And the cross-over temperature where are points is;
T c= √ H x (2 Ds−Hx )
2 π k B
When the rate constants are of the form;
k k ( T )= Ak T β k
exp (−E a , k
RT )
k k ( T )= Ak T β k
exp (−E a , k
RT )= ( 2.3−2 ) T0.8 exp 0 /RT
k k ( T )=¿ Ak T β k
exp (−E a , k
RT )= ( 2.3−2 ) T0.8 exp 0 /RT
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c) Using Cantera (see the Cantera installation section), generate an ignition delay [ms]
vs. 1000/temperature [K-1] semi-log plot for a stoichiometric hydrogen-air mixture
(log scale for the ordinates).Your initial temperature should range from approx. 750
K to approx. 1500 K. Consider two pressures: p = 1 bar and p = 0.1 bar. Use a
constant pressure homogenous reactor for your simulations. Comment on your
results and discuss how it links to your results in part b).
Coding
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The result of simulation
Problem 3.
Show that, for any fluid quantity Q,
ρ DQ
Dt = ∂ ( ρQ )
∂ t + ∆ ( ρQμ )
In a volume v bound by surface s that is fixed in space, mass inside is given by the derivative;
∫v ρdQ
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Rate of mass flux out of Q = ∫ V ρv . dS=∫ V ∇ .(ρq)dQ
We then use green’s formula to convert to quantity integral. The integral ∇ .(ρq) in the
Cartesian coordinates x = (x, y, z), q = (u, v,w) as;
∇ . ( ρv )= δ ( ρu )
δx + δ ( ρv )
δy + δ ( ρw )
δz
This shows that the gradient in the flow field are required for non-zero net flux
δρ
δt +∇ . ( ρQu ) =0
Therefore, by fluid quantity Q, is given
ρ DQ
Dt = ∂ ( ρQ )
∂ t + ∆ ( ρQμ ) =0
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