Transcript lecture1 - Concordia University

Lecture #1
Review of Chemical Thermodynamics
Textbook: Introduction and Chapter 1
Combustion
MECH 6191
Department of Mechanical and Industrial Engineering
Concordia University
Objective
Fuel + oxidizer -> products of combustion + energy
reactants
products
Basic question:
How much energy is released?
1. Stoichiometry – First “approximation” of how much fuel reacts with how much
oxidizer to give how much product
2. Energy balance – concerns with energetics, first law of thermodynamics
Combustion stoichiometry
Combustion involves chemical changes:
ex.
H2 + 1/2 O2  H2O
In chemical reactions, the total mass of the system is conserved, but the
mole of chemical species are not!!
First basic task in combustion analysis:
Stoichiometry deals with the conservation of atomic species as the basic rule
in figuring out the number of moles of the different product species for a given
number of moles of reactants.
- What does it mean when a mixture is said to be in
stoichiometric condition?
Combustion stoichiometry
Stoichiometric: just enough O2 to consume the fuel for complete reaction, i.e.
right amount of O2 reacting with the fuel to form H2O and CO2.
Carbon goes to CO2
Hydrogen to H2O
complete combustion
Example: The overall chemical equation for the complete combustion of
one mole of propane (C3H8) with oxygen is:
C3 H8  aO2  bCO2  cH 2O
# of moles
species
Elements cannot be created or destroyed, so
C balance:
b= 3
H balance:
2c= 8  c= 4
O balance:
2b + c = 2a  a= 5
Thus the above reaction is:
C3 H8  5O2  3CO2  4H 2O
i.e. 16.7% (by mole) of propane
burns with 83.3% oxygen for
complete reaction
Some hydrocarbon fuels at stoichiometric condition
Fuel-air combustion
• In many combustion technology, the fuel is usually burned with air
• For stiochiometric mixture with air instead of with pure oxygen:
 For every molecule of O2 we need from air, we
bring in 3.76 moles of N2 as well.
 For stoichiometric analysis, N2 is usually inert (no
dissociation) and there is the same amount of mole of
N2 in the product as in the reactant.
Oxidation by air
• Except for chemical rockets, most of the mobile power plants burn fuels with air
(air-breathing engine device).
• Air is a mixture of gases including oxygen (O2), nitrogen(N2), argon
(Ar), carbon dioxide (CO2), water vapour (H2O)….
Composition of Standard Dry Air
• In combustion, since oxygen is the oxidizer in air, it is usually sufficiently
accurate to consider dry air composed of 21% O2 and 79% N2 by volume
(mole fraction).
nN 2
nN 2 ntot xN 2 0.79




 3.76
ntot nO2 xO2 0.21
nO2
• For every mole of O2 there are 3.76 moles of N2.
• The molecular weight of air is
n
M air   xi M i  xO2  M O2  xN 2  M N 2
i 1
 0.21(32)  0.79(28)  28.84 kg/kmol
Some hydrocarbon fuels combustion with
air at stoichiometric condition
Combustion Stoichiometry
The complete reaction of a general hydrocarbon CaHb with air is:
Ca H b  a(O2  3.76N2 )  bCO2  cH2O  dN2
C balance:
H balance:
O balance:
N balance:
a=b
b = 2c  c = b/2
2a = 2b + c  a = b + c/2  a = a + b/4
2(3.76)a = 2d  d = 3.76a/2  d = 3.76(a + b/4)
b
b
b


Ca H b  a  (O2  3.76 N 2 )  aCO2  H 2O  3.76a   N 2
4
2
4


The above equation defines the stoichiometric proportions of fuel and air.
Example: For propane (C3H8) a= 3 and b= 8
C3 H8  5(O2  3.76 N 2 )  3CO2  4H 2O  3.765N 2
Combustion Stoichiometry
The complete reaction of a general hydrocarbon CaHbON with air is:
Ca H b O N  a(O2  3.76N2 )  bCO2  cH2O  dN2
C balance:
H balance:
O balance:
N balance:
a=b
b = 2c  c = b/2
2a+ = 2b + c  a = a + b/4 – /2
2(3.76)a+ = 2d  d = /2 + 3.76(a + b/4 – /2)

b 
b
b  


Ca H b O N  a   (O2  3.76N 2 )  aCO2  H 2O    3.76a    N 2
4 2
2
4 2 


2
Mixture rules (Basic definition)
The total number of moles and the mole concentration in the mixture is:
n
n   ni
i 1
C
n
V
The mole fraction, xi, of any given species is defined as:
ni
xi 
n
n
and
x
i 1
i
1
The mixture internal energy U and enthalpy H (units: kJ) is:
n
U   ni ui
i 1
n
H   ni hi
i 1
where ui and hi are molar specific values (units: kJ/kmol)
The mixture molar specific internal energy and enthalpy (units kJ/ kmol) is:
n
u   xi ui
i 1
n
h   xi hi
i 1
Mixture rules (Basic definition)
The mixture can be expressed either by its mass (m) or mole number (n).
The mass m of a mixture is equal to the sum of the mass of n components
n
m   mi
i 1
m
And the total density is:  
V
The mass fraction, yi, of any given species is defined as:
m
yi  i
m
n
and
y
i 1
i
1
The mixture internal energy U and enthalpy H (units: kJ) is:
n
U  m u   mi ui
i 1
n
H  m h   mi hi
i 1
The mixture specific internal energy u and enthalpy h (units: kJ/kg) is:
n
n
mi
u   ui  yi ui
i 1 m
i 1
n
n
mi
h   hi  yi hi
i 1 m
i 1
Mixture rules (Basic definition)
Relationship:
yi 
M i ni

 M i ni
i 1, N
M i xi
 M i xi
xi 
i / M i

 i / M i
i 1, N
i 1, N
yi / M i
 yi / M i
i 1, N
where Mi = molar weight of species i
i / M i
xi 
 yi M / M i
i / M
M i ni M i xi
yi 

Mn
M
where the mixture molecular weight, M, given by:
n
m
M  
n
m
i 1
n
i
n
n
ni M i

  xi M i
n
i 1
i 1
Mixture rules (Basic definition)
H2 + 0.5 (O2 +3.76 N2) = H2 + 2.38 air
Mole fraction of H2 = nH2/ntotal = 1/(2.38+1) = 0.295
Mole fraction of Air = nair/ntotal = 2.38/(2.38+1) = 0.705
C8H18 + 12.5 (O2 +3.76 N2) = C8H18 + 59.5 air
Mole fraction of C8H18 = nC8H18/ntotal = 1/(59.5+1) = 0.01653
Mole fraction of Air = nair/ntotal = 59.5/(59.5+1) = 0.98347
For heavy hydrocarbons we need only about 1% fuel for stoichiometric combustion.
Mixture rules (Basic definition)
H2 + 0.5 (O2 +3.76 N2) = H2 + 2.38 air
weight percent of H2 = massH2/masstotal
= nH2*MwH2/(nH2*MwH2+nair*Mwair)
= 2/(2+2.38*29) = 2.816%
Mixture rules
For most of the combustion problems, the perfect gas equation of state
holds accurately.  Each specie behaves as if the others are not present.
Consider a system divided into different compartments with nA moles of
specie A in a compartment of volume VA, nB moles in a Volume VB, etc. All
the gases in the various compartments have the same pressure p and
temperature T. Applying the perfect gas law:
p
nA RT nB RT nc RT


 ...
VA
VB
VC
V  VA  VB  VC  ...
pA 
n A RT
V
pB 
n  nA  nB  nC  ...
nB RT
V
pC 
nC RT
V
n
p  p A  pB  pC  ...   pi
i 1
Partial pressure
Mixture rules
n A p A nB pB nC pC

,

,

, etc
n
p
n
p
n
p
piV
ni
RuT pi
xi  

pV
n
p
RuT
or
pi  xi p
The mole fraction of the various species is just the partial pressure ratios.
Important relationship used to prepare a mixture in practice.
cannot measure mole fraction experimentally
instead monitor the pressure during preparation of a
combustible mixture
Ideal Gas Model for Mixtures
Method of partial pressure
Let say I want to prepare a mixture with nA of species A and
nB mole of species B with total pressure of 1 atm
1. Calculate the corresponding partial pressure for each species
in the mixture
nA pA

n
p
nB p B

n
p
2. Evacuate your tank, fill up the tank with gas A until it reach the
pressure pA
3. Continue to fill the tank with gas species B until the final
desired total pressure p is reached.
Combustion Stoichiometry
The stoichiometric air/fuel ratio:
 A / F s
m
 air 
m fuel
 A / F s 
 n M 
 n M 
i
i air
i
i fuel
nair
n fuel
Mass-based
Mole-based
fuel/air ratio:
 A / F s 
1
( F / A) s
The fuel-air or air-fuel ratio refers to the ratio of the amount of fuel to the
amount of air (by moles or mass) or vice versa for air-fuel ratio.
Combustion Stoichiometry
The stoichiometric mass based air/fuel ratio for CaHb fuel is:
 A / F s 
 ni M i air
mair

m fuel  ni M i  fuel
a  b M  3.76a  b M

 O2

 N2
4
4





aM C  bM H
Substituting the respective molecular weights and dividing top and bottom
by a one gets the following expression that only depends on the ratio of the
number of hydrogen atoms to carbon atoms (b/a) in the fuel.
 A / F s
 b a  
1 
(32  3.76  28.16)
1
4 


( F / A) s
12.011 b a  1.008
Note above equation only applies to stoichiometric mixture
For typical petroleum based fuel: (A/F)stoich~14.2-15
By mass
Off-stoichiometric Mixture
• Fuel-air mixtures with more than stoichiometric air (excess air) can burn.
With excess air you have fuel lean combustion
- excess air (i.e. O2) in the product
- major products: CO2, H2O, O2, N2
- minor products: HC, CO, H2, NO
• With less than stoichiometric air you have fuel rich combustion.
- insufficient oxygen to oxidize all the C and H in the fuel to CO2 and H2O.
- Incomplete combustion where carbon monoxide (CO) and molecular
hydrogen (H2) also appear in the products.
- major products: CO2, H2O, CO, H2, N2
- minor products: HC, O2, NO
Off-Stoichiometric Mixtures
The equivalence ratio, f, is commonly used to indicate if a mixture is
stoichiometric, fuel lean, or fuel rich.


F / Amixture
A / F s
f

F / As
 A / F mixture
stoichiometric f = 1
fuel lean
f<1
fuel rich
f>1
Stoichiometric mixture:
b

Ca H b   a  (O2  3.76 N 2 )  Products
4

Off-stoichiometric mixture:
1
b
Ca H b  a  (O2  3.76 N 2 )  Products
f
4
Off-Stoichiometric Conditions
Other terminology used to describe how much air is used in combustion:
%theoretica
l air 
100%
f
%excessair  %theoretical air  100% 
100%(1  f )
f
e.g. 150% theoretical air = 50% excess air = lean mixture
85% theoretical air = 15% deficient in air = rich mixture
Example
Consider a reaction of octane with 10% excess air, what is the
equivalence ratio, f?
The stoichiometric reaction is:
C8 H18  12.5(O2  3.76N 2 )  8CO2  9H 2O  47 N 2
10% excess air is:
C8 H18  1.1(12.5)(O2  3.76 N 2 )  8CO2  9H 2O  aO2  bN2
16 + 9 + 2a = 1.1(12.5)(2)  a = 1.25,
b = 1.1(12.5)(3.76) = 51.7
 A / F s
12.5(4.76) / 1
f

 0.91
 A / F m ixture 1.1(12.5)(4.76) / 1
%excessair  %theoretical air  100% 
100%(1  f )
f
Summary
Objective of this first review lecture
- Be aware of different basic definition in thermochemistry
• Combustion theory
- Stoichiometry
> reaction and element balance
> definition: theoretical air, excess air,
mass-based A/F or F/A,
equivalence ratio
• Next class
- Energy balance and chemical equilibrium
Concordia University