AME513-F12-lecture2x

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Oct 27, 2013 (3 years and 11 months ago)

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AME 513


Principles of Combustion


Lecture 2

Chemical thermodynamics I


1
st

Law

2

AME 513
-

Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Outline


Fuels
-

hydrocarbons, alternatives


Balancing chemical reactions


Stoichiometry


Lean & rich mixtures


Mass and mole fractions


Chemical thermodynamics


Why?


1st Law of Thermodynamics applied to a chemically reacting
system


Heating value of fuels


Flame temperature

3

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Fuels & air


Usually we employ hydrocarbon fuels,
alcohols or coal
burning
in air, though other possibilities include
H
2
, CO
, NH
3
,
CS
2
, H
2
S, etc
.


For rocket fuels that do not burn air, many possible
oxidizers exist
-

ASTE
470, 570 & 572 discuss these


Why air?


Because it

猠晲敥e o映捯ur獥s⡷敬l, no琠r敡el礠睨敮 祯u 瑨in欠o映
all the money
we’ve
spent to clean up air)


Air ≈ 0.21 O
2

+ 0.79 N
2

(1 mole of air) or 1 O
2

+ 3.77 N
2

(4.77
moles of air)


Note for air, the average molecular weight is




0.21*32 + 0.79*28 = 28.9 g/mole



thus the gas constant = (universal gas constant / mole. wt.)



= (8.314 J/
moleK
) / (0.0289 kg/mole) = 287 J/
kgK


Also ≈ 1% argon, up to a few % water vapor depending on the
relative humidity, trace amounts of other gases, but
we’ll
usually assume just O
2

and
N
2

4

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Hydrocarbons


Alkanes
-

single bonds between carbons
-

C
n
H
2n+2
, e.g. CH
4
, C
2
H
6










Olefins or alkenes
-

one or more double bonds between carbons







Alkynes
-

one or more triple bonds between carbons
-

higher
heating
value than alkanes or alkenes due to strained
(endothermic) bonds
than alkanes or
alkenes, also very reactive

5

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Hydrocarbons


Aromatics
-

one or more ring structures











Alcohols
-

contain one or more OH groups














6

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Biofuels


Alcohols
-

produced by fermentation of food crops (sugars or
starches) or cellulose (much more difficult, not an industrial
process yet)


Biodiesel
-

convert vegetable oil or animal fat (which have very
high viscosity) into alkyl esters (lower viscosity) through

瑲慮獥獴敲ifi捡瑩on


睩瑨 慬捯hol





















Methyl linoleate

Ethyl stearate

Generic ester structure (R =
any organic radical, e.g. C
2
H
5
)

Methanol + triglyceride Glycerol+ alkyl ester

Transesterification

process

7

AME 513
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Fall 2012
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Lecture 2
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Chemical thermodynamics 1

Practical fuels


All practical fuels are
BLENDS of hydrocarbons and
sometimes other compounds


What distinguishes
fuels?


Flash point
-

temperature
above which fuel vapor
pressure is flammable when
mixed with air


Distillation curve
-

temp.
range over which molecules
evaporate


Relative amounts of
paraffins

vs. olefins vs. aromatics vs.
alcohols


Amount of impurities, e.g.
sulfur


Structure of molecules
-

affects octane
number
(gasoline) or
cetane

number
(Diesel)

8

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Gasoline
-

typical composition

Benzene

Toluene

J
.
Burri

et al.,
Fuel,

Vol. 83, pp. 187
-

193 (2004)

9

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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Property

Jet
-
A

Diesel

Gasoline

Heating value
(MJ/kg)

43

43

43

Flash point
(˚C) (T at which vapor
makes flammable mixture in air)

38

70

-
43

Vapor pressure

(at 100˚F) (psi)

0.03

0.02

8

Freezing point
(˚C)

−40

-
38

-
40

Autoignition

temperature

(˚C) (T at
which fuel
-
air mixture will ignite
spontaneously without spark or flame)

210

240

260

Density

(at 15˚C) (kg/m
3
)

810

850

720

Practical fuels
-

properties


Values NOT unique because


Real fuels are a mixture of many molecules, composition varies


Different testing methods & definitions

10

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Practical fuels
-

properties

http://www.afdc.energy.gov/afdc/pdfs/fueltable.pdf


11

AME 513
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Fall 2012
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Lecture 2
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Chemical thermodynamics 1

Practical fuels


What
doesn’t

distinguish one fuel from another?


Energy content (except for fuels containing alcohols, which are
lower)


Examples


Gasoline
-

low
-
T distillation point, easy to vaporize, need high
octane number; reformulated gasoline contains alcohols


Diesel
-

high
-
T distillation point, hard to vaporize, need LOW
octane number for easy ignition once fuel is inject


Jet fuel
-

medium
-
T distillation point; need low freezing T since it
will be used at high
altitude
/ low T

12

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Stoichiometry


Balancing of chemical reactions with

歮o睮


⡡(獵m敤⤠produ捴c


Example: methane (CH
4
) in air (O
2
+ 3.77N
2
)

CH
4

+
a(
O
2
+ 3.77N
2
)


b
CO
2

+
c
H
2
O +
d
N
2


(how do we know this know this set is reasonable? From 2nd
Law, to be discussed later)

Conservation of C, H, O, N atoms:



n
CH4
(1) + n
O2
(0) + n
N2
(0) = n
CO2
(b) + n
H2O
(0) + n
N2
(0)



n
CH4
(4)
+ n
O2
(0) + n
N2
(0) = n
CO2
(0)
+ n
H2O
(2c)
+ n
N2
(0)



n
CH4
(0)
+ n
O2
(2a)
+ n
N2
(0) = n
CO2
(2b)
+ n
H2O
(c)
+ n
N2
(0)



n
CH4
(0)
+ n
O2
(0)
+ n
N2
(3.77*2a)
= n
CO2
(0)
+ n
H2O
(0) + n
N2
(2d)

Solve: a = 2, b = 1, c = 2, d = 7.54

CH
4

+
2(
O
2
+ 3.77N
2
)


1
CO
2

+
2
H
2
O +
7.54
N
2


or in general





13

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Stoichiometry


This
is a special case where there is just enough fuel to combine
with all of the air, leaving no excess fuel or O
2

unreacted; this is
called a
stoichiometric mixture


In general, mixtures will have excess air (
lean mixture
) or excess
fuel (
rich mixture
)


This development assumed air = O
2

+ 3.77 N
2
; for lower or higher %
O
2

in the atmosphere, the numbers would change accordingly


14

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Stoichiometry


Fuel
mass fraction (f
)





n
i

= number of moles of species
i
,
M
i

= molecular weight of species
i



For the specific case of
stoichiometric

methane
-
air (x = 1, y = 4),

f = 0.0550
; a lean/rich mixture would have lower/higher f


For stoichiometric mixtures, f is similar for most
hydrocarbons but depends on the C/H ratio = x/y, e.g.


f = 0.0550 for CH
4

(methane)
-

lowest possible C/H ratio


f = 0.0703 for C
6
H
6

(benzene) or C
2
H
2

(acetylene)
-

high C/H ratio


F
uel
mole fraction
X
f




which varies a lot depending on x and y (i.e. much smaller
for big molecules with large x and y)

15

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Stoichiometry


Fuel
-
to
-
air ratio (FAR)



and
air
-
to
-
fuel ratio (AFR) = 1/(FAR)


Note also f = FAR/(1+FAR)


Equivalence
ratio

(

)






< 1: lean mixture;


> ㄺ ri捨 mi硴ure


What if we assume more products, e.g.

CH
4

+ ?(O
2
+ 3.77N
2
)


? CO
2

+ ? H
2
O + ? N
2

+ ? CO


In this case we have 4 atom constraints (1 each for C, H, O, and N
atoms) but 5 unknowns (5 question marks)
-

how to solve?

Need
chemical equilibrium

(discussed later) to decide how much
C and O are in the form of CO
2

vs. CO

16

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Fuel properties



Fuel

Heating value, Q
R

(J/kg)

f at stoichiometric

Gasoline

43 x 10
6

0.0642

Methane

50 x 10
6

0.0550

Methanol

20 x 10
6

0.104

Ethanol

27 x 10
6

0.0915

Coal

34 x 10
6

0.0802

Paper

17 x 10
6

0.122

Fruit Loops

16 x 10
6

Probably about the same as paper

Hydrogen

120 x 10
6

0.0283

U
235

fission

83,140,000 x 10
6

1

Pu
239

fission

83,610,000 x 10
6

1

2
H +
3
H fusion

339,00,000 x 10
6

2
H :
3
H = 1 : 1

17

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

introduction


Besides needing to know how to balance chemical
reactions, we need to determine
how much internal energy
or enthalpy is released

by such reactions and what the
final
state

(temperature, pressure, mole fractions of each species)
will be


What is highest temperature flame? H
2

+ O
2

at


㴠1?†
乯peⰠ
T‽ 3079䬠at‱ atm⁦r reactants at 298K


P
robably the highest is diacetylnitrile + ozone

C
4
N
2

+ (4/3)O
3


4 CO + N
2


T

= 5516K at 1 atm for reactants at 298K


Why should it? The H
2

+ O
2

system has
much

more energy
release per unit mass of reactants, but still a much lower
flame temperature

18

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

introduction


The problem is that the products are
NOT

just H
2
O, that is, we
don

琠g整

H
2

+ (1/2)O
2


H
2
O


but rather

H
2

+ (1/2)O
2


0.706 H
2
O + 0.062 O
2
+ 0.184 H
2





+ 0.094 H + 0.129 OH + 0.040 O

i.e. the water
dissociates

into the other species


Dissociation does 2 things that reduce flame temp.



More moles of products to soak up energy (1.22 vs. 1.00)



Energy is required to break the H
-
O
-
H bonds to make the other species


Higher pressures will reduce dissociation
-

Le
Chatelier’s

principle
:


When a system at equilibrium is subjected to a stress, the system shifts
toward a new equilibrium condition in such as way as to reduce the
stress




(more pressure, less space, system responds by reducing number
of moles of gas to reduce pressure)



19

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

introduction


Actually, even if we somehow avoided dissociation, the H
2

-

O
2

flame would be only 4998K
-

still not have as high a flame
temp. as the weird C
4
N
2

flame


Why? H
2
O is a
triatomic

molecule
-

more degrees of freedom
(DOFs) (i.e. vibration, rotation) than diatomic gases; each
DOF adds to the molecule

s ab楬楴y to⁳tore energy


So why is the C
4
N
2

-

O
3

flame so hot?


CO and N
2

are diatomic gases
-

fewer DOFs


CO and N
2

are very stable even at 5500K
-

almost no dissociation


O
3

decomposes exothermically to (3/2)O
2

20

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

goals


Given an initial
state

of a mixture (temperature, pressure,
composition), and an assumed
process

(constant pressure,
volume, or entropy, usually), find the final
state

of the
mixture


Three common processes in engine analysis


Compression

»
Usually constant entropy (isentropic)

»
Low P / high V to high P / low V

»
Usually P or V ratio prescribed

»
Usually composition assumed

晲o穥n


-

i映i琠reac瑥d be景re
compression, you wouldn

琠ge琠any work ou琡


Combustion

»
Usually constant P or v assumed

»
Composition MUST change (obviously…)


Expansion

»
Opposite of compression

»
May assume frozen (no change during expansion) or equilibrium
composition (mixture shifts to new composition after expansion)


21

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

assumptions


Ideal gases
-

note many

晬慶ars


o映瑨攠id敡e g慳al慷


PV =
n

T


PV =
mRT


Pv

= RT


P =

RT


P = pressure (N/m
2
); V = volume (m
3
); n = number of moles of gas;


= universal gas constant (8.314 J/
moleK
); T = temperature (K)

m = mass of gas (kg); R = mass
-
specific gas constant =

/
M

M
= gas molecular weight (kg/mole); v = V/m = specific volume (m
3
/kg)



= 1/v = density (kg/m
3
)



Adiabatic


Kinetic and potential energy negligible


Mass is conserved


Combustion process is constant P or V (constant T or s
combustion
isn’t
very interesting!)


Compression/expansion is reversible & adiabatic


(


i獥s瑲pi挬


= 〩

22

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

1st Law


1st Law of thermodynamics (conservation of energy),
control mass: dE =


-


W


E = U + PE + KE = U + 0 + 0 = U




W = 偤V


Combine: dU + PdV = 0


Constant pressure
: add VdP = 0 term


dU + PdV + VdP = 0


d⡕+偖⤠= 〠


dH = 0


H
reactants

= H
products


Recall h


H/m ⡭ = m慳猩a 瑨u猠h
reactants

= h
products


Constant volume
: PdV = 0


dU + PdV + VdP = 0


d⡕⤠= 〠


U
reactants

= U
products
, thus u
reactants

= u
products


h = u + Pv, thus (h
-

Pv)
reactants

= (h
-

Pv)
products


Most property tables report h not u, so h
-

Pv form is useful


New twist: h or u must include BOTH thermal and chemical
contributions!


23

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

1st Law


Enthalpy of a mixture (sum of thermal and chemical terms)























24

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

1st Law


Note we can also write h as follows




















Use these boxed expressions for h & u with h = constant (for constant P
combustion) or u = constant (for constant V combustion)

25

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

1st Law


Examples of tabulated data on h(T)
-

h
298
,

h
f
, etc.


(double
-
click table to open Excel spreadsheet with all data for CO,
O, CO
2
, C, O
2
, H, OH, H
2
O, H
2
, N
2
, NO at 200K
-

6000K)

26


Example: what are h and u for a
CO
2
-
O
2
-
CO at 10
atm
, 2500K
with X
CO

= 0.0129, X
O2

= 0.3376, X
CO2

= 0.6495?



Pressure
doesn

t affect h or u but T does; from the tables:



AME 513
-

Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

1st Law

27

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermodynamics
-

1st Law


Final pressure (for constant volume combustion)




28

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

heating value


Constant
-
pressure energy conservation equation (no heat
transfer, no work transfer other than PdV work)






Denominator = m = constant, separate chemical and thermal terms:








Term on left
-
hand side is the negative of the total
thermal

enthalpy
change per unit mass of mixture; term on the right
-
hand side is
the
chemical
enthalpy change per unit mass of mixture

29

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

heating value


By definition, C
P



(∂h/∂T)
P



For an ideal gas, h = h(T) only, thus C
P

= dh/dT or dh = C
P
dT


If C
P

is constant, then for the
thermal enthalpy




h
2

-

h
1

= C
P
(T
2

-

T
1
) = mC
P
(T
2

-

T
1
) /m


For a combustion process in which all of the enthalpy release by
chemical reaction goes into thermal enthalpy (i.e. temperature
increase) in the gas, the term on the left
-
hand side of the boxed
equation on page 27 can be written as








where is the constant
-
pressure specific heat
averaged
(somehow) over all species and averaged between the product
and reactant temperatures

30

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

heating value


Term on right
-
hand side of boxed equation on page 27 can be re
-
written as







Last term is the chemical enthalpy change per unit mass of fuel;
define this as
-
Q
R
, where Q
R

is the
fuel’s
heating value






For

our stereotypical hydrocarbons, assuming CO
2
, H
2
O and N
2

as
the only combustion products, this can be written as

31

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

flame temperature


Now write the boxed equation on page 27 (conservation of energy
for combustion at constant pressure) once again:








We

癥v獨o睮 瑨慴a瑨攠l敦e
-
h慮d 獩d攠=


慮d 瑨攠right
-
h慮d 獩d攠=
-

R
; combining these we obtain




This is our simplest estimate of the
adiabatic flame temperature

(T
products
, usually we write this as T
ad
) based on an initial
temperature (T
reactants
, usually written as T

) thus








(constant pressure combustion,








T
-
averaged C
P
)

32

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

flame temperature


This analysis has assumed that there is enough O
2

to burn all the
fuel, which is true for lean mixtures only; in general we can write




where for lean mixtures, f
burnable

is just f (fuel mass fraction)
whereas for rich mixtures, with some algebra it can be shown that





thus in general we can write

33

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

flame temperature


For
constant
-
volume combustion

(instead of constant pressure),
everything is the same except u = const, not h = const, thus the term on
the left
-
hand side of the boxed equation on page 27 must be re
-
written as








The extra PV terms (= mRT for an ideal gas) adds an extra mR(T
products
-
T
reactants
) term, thus







which means that (again, T
products

= T
ad
; T
reactants

= T

)







(constant volume combustion, T
-
averaged C
P
)


which is the same as for constant
-
pressure combustion except for the C
v

instead of C
P

34

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Chemical thermo
-

flame temperature


The constant
-
volume adiabatic flame (product) temperature on the
previous page is only valid for lean or stoichiometric mixtures; as
with constant
-
pressure for rich mixtures we need to consider how
much fuel can be burned, leading to







Note that the ratio of adiabatic temperature rise due to
combustion for constant pressure vs. constant volume is





In practice, one can determine by working backwards from a
detailed analysis; for stoichiometric CH
4
-
air, f = 0.055, Q
R

= 50 x
10
6

J/kg, constant
-
pressure combustion, T
ad

= 2226K for T


=
300K, thus ≈ 1429 J/kg
-
K (for other stoichiometries or other
fuels will be moderately different)

35

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Example of heating value


Iso
-
octane/air mixture:


C
8
H
18

+ 12.5(O
2
+ 3.77N
2
)


8 CO
2

+ 9 H
2
O + 12.5*3.77 N
2




















36

AME 513
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Fall 2012
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Lecture 2
-

Chemical thermodynamics 1

Comments on heating value


Heating
values
usually computed assuming that due to
reaction
with air all C



2
,
H


H
2
O, N



N
2
, S



2
, etc.


If one assumes liquid water, the result is called the
higher
heating value
; if one (more
realistically)
assumes gaseous
water, the result is called the
lower heating value


Most
hydrocarbons
have similar Q
R

(4.0
-

4.5 x 10
7

J/kg)
since the same C
-
C and C
-
H bonds are being broken and
same C
-
O and H
-
O bonds are being made


Foods similar
-

on a
dry weight basis
, about same Q
R

for all


Fruit Loops™ and Shredded Wheat™ have same

heating
value


⠱㄰(正慬/


= ㄮ㘠砠㄰
7

J/kg) although Fruit Loops™
is
mostly sugar whereas
Shredded Wheat™ has none
(the above
does not constitute a commercial endorsement)


Fats slightly higher than starches or sugars


Foods with (non
-
digestible) fiber lower

37

AME 513
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Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Comments on heating value


Acetylene is higher (4.8 x 10
7

J/kg)
due to C
-
C triple bond


Methane is higher (5.0 x 10
7

J/kg)
due to high
H/C ratio


H
2

is MUCH higher (12.0 x 10
7

J/kg)
due to

heavy


䌠Ctoms


Alcohols are lower (2.0 x 10
7

J/kg for methanol, CH
3
OH)
due
to

use汥ss


O atoms
-

add mass buto⁥ntha汰y re汥ase

38

AME 513
-

Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Example of adiabatic flame temperature


Lean iso
-
octane/air mixture, equivalence ratio 0.8, initial temperature
300K, average C
P

= 1400 J/kgK, average C
v

= 1100 J/kgK:


Stoichiometric: C
8
H
18

+ 12.5(O
2
+ 3.77N
2
)


8 CO
2

+ 9 H
2
O + 12.5*3.77 N
2




















39

AME 513
-

Fall 2012
-

Lecture 2
-

Chemical thermodynamics 1

Summary
-

Lecture 2


Many fuels, e.g. hydrocarbons, when chemically reacted with oxygen
or other oxidizing agents, will release a large amount of enthalpy


This chemical energy or enthalpy is converted into thermal energy or
enthalpy, thus in a combustion process the product temperature is
much higher than the reactant temperature


Only 2 principles are required to compute flame temperatures


Conservation of each type of atom


Conversation of energy (sum of chemical + thermal)


… but the resulting equations required to account for changes in
composition and energy can look formidable


The key properties of a fuel are its heating value Q
R

and its
stoichiometric fuel mass fraction
f
stoichiometric


The key property of a fuel/air mixture is its equivalence ratio (

)


A simplified analysis leads to