# Create a comprehensive master’s level Combustion study guide (conceptual) for final exam

Part of my final exam in combustion involves answering 25-30 conceptual questions. My instructor will take these questions primarily from material covered in the 8 sections attached. These slides come from Turns’ “Combustion Concepts and Applications third edition,” (Chs. 2, 4-6, 8, 16) so having this textbook would be a plus. I need someone to make a comprehensive conceptual study guide using the material from these eight sections. That is, it should focus on the concepts involved. Additionally, I would like that person to come up with some possible exam concept questions/answers for each section. Questions that could be asked on a final exam and answered in a sentence or two. I have combined some of the sections in order to upload all of them, but there are 8 total.
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Thermo-dynamics of Nonreacting Gas Mixtures
Black and Hartley, Ch. 11
Turns, Ch. 1
Denbigh, Ch. 1, 2, and 3
Non-reacting gas mixtures
 Many practical systems are composed of more than one substance




Air and water vapor
Methane and air
Air
Liquid water and water vapor
 Multiple phases of a single substance is usually covered in
 It is impractical to tabulate properties for every conceivable mixture
 Alternative is to develop procedures for predicting mixture properties
from properties of each component
 In our case, we focus on gases
Mole Fraction
Mass Fraction
Assume there are N total species in a nonreacting
gas mixture and ni be the number of moles of
Let Mi be the molecular weight of species i
then Mi ni is the mass of species i
mi = Mi ni
species i
The total mass of the mixture is
The total number of moles is
N
mtot = ? mi
N
ntot = ? ni
i =1
i =1
The mole fraction is defined as
ni
=
ntot
?i
=
ni
N
?n
i
i =1
The mass fraction is defined as
m
Mi ni
Yi = i =
N
mtot
? Mi ni
i =1
and
and
N
N
??
i =1
i
?Y
=1
i =1
The mixture molecular weight Mmix is defined as the
total mass of the mixture divided by the total number
of moles
N
=
Mmix
?M n
i
i
=
ntot
i =1
N
?? M
i =1
i
i
i
=1
and
Yi
=
Mi ni
Mi
?i
=
Mmix ntot
Mmix
Partial and component pressure
N
?i
=
Starting with ?
N
Y
=
?
=i 1=i 1
i
Yi ? i
=
1 and using
we obtain
Mmix =
1
N
Yi
?
i =1 M i
Species and mixture gas constant
Ri =
R
Mi
Rmix
R
=
Mmix
species gas constant
mixture gas constant
Mi
Mmix
Two widely used methods for predicting p-v-T behavior are
1. Dalton’s law for the mixture pressure
The total pressure of a mixture is equal to the sum of the
component pressure of each species, i.e., the pressure each
species would exert if they existed seperately at the same
temperature and volume as the mixture.
pmix
=
N
p where p
=
?
i =1
i
i
nRTmix
Vmix
2. Amagat’s law of additive volumes
The total volume occupied by a mixture of gases is equal
to the sum of the component volumes, i.e., the volume
occupied by each component if they existed separately
at the same temperature and pressure as the mixture.
N
Vm = ?Vi
i =1
p – v -T behavior of real and ideal gas mixtures
Dalton’s law
The partial pressure is defined as ? i pm .
The component pressure is found from
pi
Zi ,D ni RT
=
or pi v i RT
Vm
Therefore, the component pressure and the partial
pressure are the same for an ideal gas
where Zi ,D is the Dalton’s law compressibility factor
of species i at the same temperature and volume
as the mixture.
Zi ,D = Zi (T ,Vm , ni )
The average compressibility factor for the mixture is
For real gas mixtures, the partial pressure and the
Z
component pressure differ by i ,D
Zm,D
= ?=
Partial pressure
pi
i pm
Zi ,D
Zm,D
N
Zm,D = ? ? i Zi ,D
i =1
For ideal gases, Zi ,D = 1
?=
ni RT
i nmix RT
Therefore, the component pressure becomes
Zi ,D ni
Z
pi
=
pm ? i i ,D pm
Zm,D nmix
Zm,D
=
pi
=
Z=
1. Therefore,
For an ideal gas Z
i ,D
m,D
R
Yi nmix Mmix
T
Mi
pi
=
= Yi ?mix RiT
Vm
pi = ? i pm
Vm
Vm
or
Amagat’s law
Same as Dalton’s rule except the compressibility factor
The component volume is
is based on mixture molar specific volume rather than
component molar specific volume
Vi =
Zi ,A ni RT
pm
Zi ,A is the compressibility factor evaluted at the mixture
temperature and pressure.
pi =
Zi ,B ni RT
Vm
Zi ,B = Zi (T ,Vm , Nm )
Zi ,A = Zi (T , pm , ni )
The mixture compressibility factor is found from
N
Zm, A = ? ? i Zi , A
i =1
Kaye’s rule
An approximate method for finding the
compressibility factor for a gas mixture.
Define a psedo-critical pressure and temperature
Amagat’s law is expected to produce more accurate results
at high pressure than Dalton’s law [Black and Hartley].
N
Tc’ = ? ? iTc ,i
i =1
N
pc’ = ? ? i pc ,i
i =1
Zi ,A ni
Zi , A
Vi
=
=
?i
Vm Zm,A nmix Zm,A
Zm,K = Z (TR’ , pR’ )
For ideal gases, the component volume fraction equals the
mole fraction
TR’ =
p
T
and pR’ =
pc’
Tc’
Ideal gas mixtures
Ideal gas mixture
 How do we define an ideal gas mixture
 Each component gas and the mixture behaves as an ideal gas
 The heat of mixing and the internal energy change of mixing is zero
 This means that the partial molar enthalpy and internal energy are equal to their molar
specific values
 The Gibbs free energy and the entropy of mixing are usually nonzero
 Two such gases separated by a semi-permeable membrane are in equilibrium
when their partial pressures are equal
 Each components chemical potential obeys
0
0
µ=
µ
+
RT
ln
p
and
µ
i
i
i
i is a function of T only
Specific internal energy, enthalpy, and entropy
Specific properties are obtained using mass or molar weighted
averages, as follows
N
smix (T , p ) = ? Yi si (T , pi )
Enthalpy
i =1
N
mi hi
?
=1
hmix i=
=
mmix
N
smix (T , p ) = ? Yi si (T , pi )
N
?Y h
i
i =1
i
N
hmix
=
?n h
i
i
=
nmix
i =1
N
?? h
i =1
i
i
Internal energy
N
umix
=
?m u
i
i
=
mmix
i =1
Entropy
? p ?
si si0 (T , pref ) – Ri ln ? i ?
=
? pref ?
? p ?
si si0 (T , pref ) – R ln ? i ?
=
? pref ?
where, pref = 1 atm
N
?Y u
i
i =1
i
N
ni ui
?
=1
umix i=
=
nmix
i =1
si0 denotes the entropy of species i at the reference
N
?? u
i =1
Note: Entropy depends on the species partial pressure.
i
i
Note: The enthalpy and internal energy do not depend
on pressure for ideal gases.
pressure and that only depends on temperature.
pref = 1 atm
Extensive properties
specific heats
mass basis
Umix m=
=
mix umix
N
N
N
m u ? m=
Y u ?U
=
?
i i
=i 1=i 1
Hmix m=
=
mix hmix
N
mix i i
=i 1
N
N
m h ? m=
Y h ?H
=
?
i i
=i 1=i 1
Smix m=
=
mix smix
mass basis
i
N
mix i i
=i 1
N
i
N
m s ? m=
Y s ?S
=
?
i i
=i 1=i 1
mix i i
=i 1
i
Umix nmix
umix
=
=
=
? n (h
N
i =1
=
i
i
N
n u ?n
=
?
N
? u = ? Ui
i i
mix i i
i =1
i 1=i 1
=
– pi v i
N
)
N
? ??n h ?? – RT ? [n ] for an ideal gas
i i
mix
i 1=i 1
Hmix nmix
hmix
=
=
N
i
N
Smix nmix
smix
=
=
N
? h ?H
n h ?n=
=
?
i i
=i 1=i 1
N
N
mix i i
=i 1
N
? s ?S
n s ?n=
=
?
i i
=i 1=i 1
mix i i
=i 1
N
N
? ?hmix ?
? ?hi ?
c p,m ?=
Y
Yi c p,i
=
=
?
?
i
?
?
?
? ?T ? p i 1=
? ?T ? p i 1
=
molar basis
molar basis
N
N
N
? ?umix ?
? ?ui ?
cv ,m ?=
Yi cv ,i
=
=
?
?
i
?
?Y
?
T
T
?
?
?
?v i 1=
?
?v i 1
=
i
i
N
N
? ?umix ?
? ?ui ?
cv ,m ?=
? i cv ,i
=
?
?
i
?
? ?=
?
T
T
?
?
?
?v i 1=
?
?v i 1
=
N
N
? ?hmix ?
? ?hi ?
c p,m=
=
=?
? i c p,i
?
? ?i
?
?
?
T
?
T
?
=
?
? p i 1=
?
?p i 1
Note: Technically, the intensive properties above are
partial molar properties. However, one property of ideal
solutions and gases is that the partial molar property is
equal to the intensive property of the pure species, i.e.,
the species in the mixture do not affect the properties of
other species in the mixture.
Thermodynamics of
Chemical Reactions
Introduction
 Previous thermodynamics studies focused on situations where either
 Only one species, or component, was present, or
 The composition of the mixture was fixed as the system proceeded between
equilibrium states
 Many situations, e.g., combustion, involve systems whose
composition is not fixed between starting and ending states
 These problems are the focus of this section
Mass conservation
 In chemically reacting systems, the components that exist prior to
reaction are called reactants and the components that exist after
reaction are called products
 The total product and reactant mass is conserved
 Species are not conserved. They may be created or destroyed during
chemical reaction
 However, individual atoms are conserved during reaction
Mass conservation
1
O2 ? H2O
2
Total mass balance:
H2 +
( mol )
(
)
g
1 mol ) 18 g
(18=
mol ) (
mreactants =2 g
mproducts
=
Atom balances:
H:
2 ·1 = 2 ·1
O:
2·1 =
1
2
(1 mol ) +
16 g
?1
?
mol
18 g
?=
mol ?? 2
?
Conservation of atomic nuclei
and stoichiometry
Consider at system composed of different constituent species
such as hydrogen, oxygen, carbon, nitrogen, etc.
Assuming no chemical reactions, the amounts of each constituent
remains constant as the state of the system changes. However,
chemical reactions may cause the composition of the system to
As chemical reactions occur, the amounts of these constituents
change over time.
change and new species appear and disappear.
For example, consider a container with CO2 ( g ) initially at room
Assume there are N constituents. Let’s denote each constituent
by a common symbol Ai ( p ) , where i = 1, 2, ?, N and p indicates
whether the species is a solid ( s ) , liquid ( l ) , or gas ( g ) . We may
choose to denote species by its chemical symbol, e.g., CO2 , H2O,
temperature. If the temperature is raised to 1000 K , some of the
CO2 will decompose (dissociate) into CO and O2 . Lowering the
temperature back to room temperature will restore the system
to its original condition (state).
etc.
Therefore, we see that the amount of a constituent may change
For example, consider a system containing solid carbon, gaseous
oxygen, and gaseous carbon dioxide. We may denote the species
as
O2 ( g ) ? A1 ( g )
C ( s ) ? A2 ( s )
CO2 ( g ) ? A3 ( g )
The amount of each species is denoted by ni , where i = 1, 2, ?, N
The mol fraction of each species is denoted by ? i =
N
require that
??
i =1
i
=1
ni
ntot
and also
as the state of the system changes. However, the number of atomic
nuclei is conserved regardless of any chemical reactions that occur.
Ex: Suppose we have one mol of CH4 with a stoichiometric amount
of air in a closed container at STP. The mixture is allowed to react
to form CO2 , H2O, and N2 .
CH4 + nair ??( 0.20 ) O2 + ( 0.80 ) N2 ?? ? nco2CO2 + nh2o H2O + nn2 N2
Since the number of atomic nuclei must be conserved,
C: 1=
1 · nco2 ? nco2 =
1
H : 4 =·
2 nh2o ? nh2o =
2
O : nair · 0.20 · 2 = 2 · nco2 + 1 · nh2o ? nair
2 · nco2 + 1 · nh2o
=
= 10
0.20 · 2
N : nair · 0.80 · 2 = 2 · nn2 ? nn2 = 8
Because this is a stoichiometrically balanced equation,
we use the ? i notation, rather than the ni , to denote stoichiometric
coefficients.
This chemical equation says nothing about the amount of
constituents before, during, or after reaction. It only indicates the
inter-relationship between constituent formation and destruction.
In other words, the stoichiometric equation specifies the mechanism
of the reaction.
The chemical equation is stoichiometrically balanced
because it expresses conservation of atomic nuclei
CH4 + 10 ??( 0.20 ) O2 + ( 0.80 ) N2 ?? ? CO2 + 2H2O + 8N2
The stoichiometric chemical equation may also be written as
For every 1 mol of CH4 destroyed, 10 mol of air are destroyed and
? 1 A1 ( g ) + ? 2 A2 ( g ) + ? 3 A3 ( g ) + ? 4 A4 ( g ) + ? 5 A5 ( g ) + ? 6 A6 ( g ) =
0
1 mol of CO2 , 2 mols of H2O, and 8 mols of N2 are formed.
?
???
?
– nch 4 CH4 ( g )
?
???
?
– no 2 O2 ( g )
?
???
?
– nn 2 N2
?
???
?
nco 2 CO2 ( g )
?1 =
-1 ? 2 =
-2
?3 =
-8
=
? 4 1=
?5 2 =
?6 8
or
N
?? A ( p ) = 0
i =1
i
i
?
???
?
nh 2 o H2 O ( g )
?
???
?
nn 2 N2 ( g )
H2 + CO + 2O2 ? xCO2 + yH2O + zO2
Atom balance
H:
2 = 2y
O:
1 + 2 · 2 = 2 x + 1y + 2z
C:
1=x
=
, x 1=
,z 1
y 1=
so, the stoichiometric equation is given by
H2 + CO + 2O2 ? CO2 + H2O + O2
Reactant mass balance
(
2g
mol
)
(1 mol ) +
(
28 g
mol
)
Product mass balance
(
44 g
mol
)
(1 mol ) +
(
18 g
mol
(1 mol ) +
)
(
(1 mol ) +
32 g
(
mol
32 g
94 g
) (2 mol ) =
mol
94 g
) (1 mol ) =
Theoretical combustion process
Theoretical combustion processes
 Combustion is a general term that refers to the chemical reaction of a
fuel and oxidizer resulting in a release of energy
 Fuel is any combustible material and may include solids, liquids, or
gases
 Most of our discussion will focus on hydrocarbon fuels such as
 Gases: methane, butane, propane, etc.
 Liquids: alcohols, kerosene, gasoline, diesel fuel, jet A, AvGas, JP8, etc.
 Solids: coal
 Gasoline, diesel, and aviation fuels are blends of liquid hydrocarbon fuels
distinguished by the range of the distillation column included in the fuel
 Gasoline approximated by octane C8H18
 Diesel approximated by dodecane C12H26
Theoretical combustion processes
 For fuel to burn, an oxidizer is required
 Air is a common oxidizer that is roughly 79% nitrogen and 21% oxygen
 Oxygen
0.21O2 + 0.79N2 =
1 mol of air
 Nitrous oxide N2O
O2 + 3.76N2 =
4.76 mol of air
 There are a few fuels and oxidizers that will decompose exothermically by
themselves
 Acetylene C2H2
 Nitrous oxide N2O
 Complete combustion means that there is enough oxidizer to oxidize the
fuel completely to H2O and CO2. There is no leftover fuel, CO, OH, etc.
 Stoichiometric proportions are defined as the minimum amount of
oxidizer required for complete combustion ? theoretical air
Theoretical combustion processes
 Incomplete combustion refers to the case where there is insufficient
oxidizer to completely oxidize the fuel
 CO and OH are indicators of incomplete combustion
 Hydrogen usually wins the battle with carbon for the available oxygen due to
its high kinetic rate
 This usually means that all of the hydrogen is reacted to form water but
 CO is left over rather than being converted completely to CO2
Excess air
Excess air is the amount of air in excess of the amount required
for complete combustion. 50% excess air means that there is 50%
more air than theoretical air.
Recall the stoichiometrically balanced chemical equation
CH4 + 10 ??( 0.20 ) O2 + ( 0.80 ) N2 ?? ? CO2 + 2H2O + 8N2
The stoichiometric amount of air required for complete combustion
is referred to as theoretical air (in this case nair,theoretical = 10 )
=
nair ,excess
nactual – ntheoretical
=
× 100% Excess air
ntheoretical
Example
Find the excess air for the following combustion process
CH4 + 20 ??( 0.20 ) O2 + ( 0.80 ) N2 ??
nair ,excess =
20 – 10
× 100%= 100%
10
Equivalence ratio
The equivalence ratio f is the ratio of the stoichiometric to actual
AFR or the ratio of the actual to stoichiometric FAR.
Air – fuel ratio
The molar air to fuel ratio AFR is
AFRmol =
nair
nfuel
The mass air to fuel ratio AFR is
AFR
=
mol
mair
nair Mair
M
=
= AFRmol air
mfuel nfuel Mfuel
Mfuel
Fuel – air ratio
The molar air to fuel ratio AFR is
FARmol
f
=
n
= fuel
nair
The mass air to fuel ratio AFR is
m
FARmass = fuel
mair
( AFR )st ( FAR )act
=
( AFR )act ( FAR )st
f is the same whether AFR or FAR is on a mass or mol basis
Example
Find f for the following reaction
CH4 + 20 ??( 0.20 ) O2 + ( 0.80 ) N2 ??
f
=
( AFRmol )st
=
( AFRmol )act
10
= 0.5
20
Stoichiometry
 There must be sufficient air (21%
 The molar stoichiometric air-fuel
O2 : 79% N2) to react with the fuel
ratio for methane is 9.52 (by mass
its 17.14)
 A stoichiometric reaction is one in
which there is no fuel or oxidizer
 If there is twice as much air as is
remaining
required, the molar air-fuel ratio is
19.04 (34.28 by mass)
 Lean refers to the case where
there is excess oxidizer
 Recall, the equivalence ratio f is
 Rich to the case where there is
( f / a )act ( a / f )stoich
insufficient oxidizer
=
f =
( f / a )stoich ( a / f )act
 The stoichiometric quantity of
oxidizer is determined by
balancing the reaction equation
 Note that f is same by mass or
mole
Cx H y + a ( 0.21O2 + 0.79 N 2 ) ? xCO2 +
a=
x+ y
y
H 2O + 0.79aN 2
2
4 where a is the stoichiometric quantity of air
0.21
Stoichiometry
 The chemical equation can also
be written in terms of ??
assuming we keep constant the
stoichiometric quantity of air
 The combustion efficiency ?c is
the fraction of fuel actually
burned
 Note: other definitions of
combustion efficiency exist out
in the real world, so make sure
you know what the other
 The equations below are for
complete combustion, not
equilibrium
y
H 2O + 0.79aN 2 + ?
2
(1 – ?cf ) 0.21aO2 + (1 – ?c ) f Cx H y
f Cx H y + a ( 0.21O2 + 0.79 N 2 ) ? f?c x CO2 + f?c
x+ y
4 where a is the stoichiometric quanity of air
0.21
?c is the combustion efficiency, and
a=
f is the equivalence ratio
Actual combustion processes
 In reality, even when there is more than
enough oxygen around to completely
oxidize the available fuel and the fuel and
oxidizer are well mixed, CO, OH and other
species are present
 Usually little or no H2 is present in the
products due to the affinity of hydrogen to
form water
 Degree of mixing affects composition of
products
 Composition has to be measured
 Gas chromatograph
 Gas analyzer
 IR Absorption
 UV emission
 Chemical cells
Equivalence ratio from gas analyzer (dry) measurements
fCx H y + a ( 0.21O2 + 0.79N2 ) ? f?c x CO2 + f?c
y
H2O + 0.79aN2 + ?
2
(1 – ?cf ) 0.21aO2 + (1 – ?c ) fCx H y
No2= (1 – ?cf ) 0.21a
Ntot =(1 – f ) 0.21a + f x + 0.79a
(1 – ?cf ) 0.21a
No2
X
, assuming ?c
=
=
o2
Ntot (1 – f ) 0.21a + f x + 0.79a
??(1 – f ) 0.21a + f x + 0.79a ?? X=
(0.21a – 0.21af )
o2
=1
X o2 0.21a – 0.21af X o2 + f xX o2 + 0.79aX o2 = 0.21a – 0.21af
0.21af – 0.21af X o2 + f xX o2 = 0.21a – X o2 0.21a – 0.79aX o2
f ( 0.21a – 0.21aX 02 + xX 02 ) = 0.21a – X o2 a
f ??0.21a (1 – X 02 ) + xX 02 ??= a ( 0.21 – X 02 )
f
x+y
a ( 0.21 – X 02 )
4
, a
=
0.21a (1 – X 02 ) + xX 02
0.21
Find dry O2 from wet O2 measurements
no2
no2
=
ntot ,wet ntot ,dry + nh2o
ntot ndry nh2o
=
+
no2 no2
no2
ndry ntot nh2o
=

no2 no2 no2
? o2,wet
no2
1
= ?=
=
o2 ,dry
ntot nh2o 1 – ? h2o,wet
ndry

no2 no2
Dry air reference for gas analysis
measurements
 NOx is regulatory focus
 In the US, the EPA has instituted
standards depending on engines
input energy and intended use
 Standards are referenced to 15%
oxygen, dry
 Most gas analyzers measure and
report dry mole fractions
 The dry reference is used to allow
measurements to be comparable
across different data
( NOx )15%O 2 =
( 20.9 – 15) ( NOx,msd )
( 20.9 – O
2, msd
)
Enthalpy of formation
Enthalpy of formation
Energy associated with breaking or forming of
chemical bonds
Enthalpy
We may think of enthalpy as having parts consisting
of sensible, phase changes, and chemical
Sensible
The difference between the enthalpy of 1 mol of a compound
at the standard reference state and the enthalpy of the stable
Amount of heat added to cause a change in temperature
elements that form the compound at the std. ref. state.
N
?h = 1· hi – ? n j h j0
?
j =1
compound
?
???
?
T
hs (T ) – hs (Tref ) =
? cpdT
0
f ,i
Tref
elements
Latent heats
N
Amount of heat that must be added to change the phase of
j =1
j
0
j
= 0 by definition
hi0
of formation of the compound at the SRS, i.e., ?hf0,i =
hif ? fusion
?hf0,i < 0 for exothermic reactions hig ? sublimation ?hf0,i > 0 for endothermic reactions
1 mole of
compound
@ STP
Qin
?n h
The enthalpy of a compound at the SRS is equal to the enthalpy
a substance constant p,T
hfg ? vaporization
Elements
@ STP
0
Formation reaction
Heat interaction that must occur in forming 1 mol of
product (compound) at the SRS from it elements in their
naturally occurring state at the SRS.
Example: Write the formation reaction for H2O ( …
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Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.

### Free-revision policy

Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.

Your email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.

### Fair-cooperation guarantee

By sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.

## Calculate the price of your order

550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
\$26
The price is based on these factors: