Download - Lecture Chemical Reaction
2
Ideal Gas Model
The ideal gas equation of state is:
TRnTM
RmmRTPV
where is the Universal Gas Constant (8.314 kJ/kmol K), is the molecular
weight and n is the number of moles.
R M
dTTcTh p )()(
dTTcTu v )()(
Specific enthalpy (units: kJ/kg)
Specific entropy (units: kJ/kg K) )ln()(),( o
o PPRTsTPs
(Po = 1 bar, so entropy at Po)
Specific internal energy (units: kJ/kg)
3
Ideal Gas Model for Mixtures
n
iimm
1
The mass m of a mixture is equal to the sum of the mass of n components
where are mass specific values (units: kJ/kg)
The mixture internal energy U and enthalpy H (units: kJ) is:
n
iii
n
iii hmmhHummuU
11
ii hu and
The mass fraction, xi, of any given species is defined as:
n
ii
ii x
m
mx
1
1 and
n
iii
n
i
iin
iii
n
i
ii hxm
hm
m
Hhux
m
um
m
Uu
1111
The mixture specific internal energy u and enthalpy h is:
4
The mixture internal energy U and enthalpy H (units: kJ) is:
n
iii
n
iii hnHunU
11
where are molar specific values (units: kJ/kmol) ii hu and
n
iinn
1
The total number of moles in the mixture is:
Ideal Gas Model for Mixtures
The mole fraction, yi, of any given species is defined as:
n
ii
ii y
n
ny
1
1 and
n
iii
n
iii hyhuyu
11
The mixture molar specific internal energy and enthalpy (units kJ/ kmol) is:
5
Ideal Gas Model for Mixtures
The partial pressure of a component, Pi, in the mixture (units: kPa) is:
PyPP
P
RTPV
RTVP
n
ny ii
i
i
ii or
Mass specific entropy (kJ/kg K) molar specific mixture entropy (kJ/mol K) :
o
n
iii
o
ii PPRPPRsxs ln)/ln(1
o
n
iii
o
ii PPRyRsys lnln1
The mixture molecular weight, M, is given by:
i
n
ii
n
i
ii
n
ii
Myn
Mn
n
m
n
mM
11
1
6
Composition of Standard Dry Air
Air is a mixture of gases including oxygen (O2), nitrogen(N2), argon (Ar),
carbon dioxide (CO2), water vapour (H20)….
For combustion dry air is taken to be composed of 21% O2 and 79% N2
by volume (yO2=0.21, yN2=0.79).
76.321.0
79.0
2
2
2
2
2
2 O
N
O
tot
tot
N
O
N
y
y
n
n
n
n
n
n
For every mole of O2 there are 3.76 moles of N2.
The amount of water in moist air at temperature T is specified by the
specific humidity (w or the relative humidity (F) defined as follows:
10 )(
22 FFTP
P
m
m
sat
OH
air
OHw
kg/kmol 84.28)28(79.0)32(21.0
22221
NNOOi
n
iiair MyMyMyMMolecular weight of air is
7
Combustion Stoichiometry
If sufficient oxygen is available, a hydrocarbon fuel can be completely
oxidized, the carbon is converted to carbon dioxide (CO2) and the hydrogen
is converted to water (H2O).
Elements cannot be created or destroyed, so
C balance: 3 = b → b= 3
H balance: 8 = 2c → c= 4
O balance: 2a = 2b + c → a= 5
OcHbCOaOHC 22283
# of moles species
The overall chemical equation for the complete combustion of one mole of
propane (C3H8) with oxygen is:
OHCOOHC 22283 435
Thus the stoichiometric reaction is:
8
Combustion Stoichiometry
Air contains molecular nitrogen N2, if products are at a “low” temperature
the nitrogen is not significantly affected by the reaction, it is considered inert.
222224
76.32
)76.3(4
NOHCONOHC
C balance: = b → b = H balance: = 2c → c = /2
O balance: 2a = 2b + c → a = b + c/2 → a = + /4
N balance: 2(3.76)a = 2d → d = 3.76a/2 → d = 3.76( + /4)
Example: The stoichiometric reaction of octane (C8H18) = 8 and = 18
22222188 4798)76.3(5.12 NOHCONOHC
22222 )76.3( dNOcHbCONOaHC
The complete reaction of a general hydrocarbon CH with air is:
9
Combustion Stoichiometry
Note above equation only applies to stoichiometric mixture
The stoichiometric mass based air/fuel ratio for CH fuel is:
HC
NO
fuelii
airii
fuel
airs
MM
MM
Mn
Mn
m
mFA
22 476.3
4/
112
)2876.332(4
1
)/(
1/
s
sAF
FA
Substituting the respective molecular weights and dividing top and bottom
by one gets the following expression that only depends on the ratio of the
number of hydrogen atoms to hydrogen atoms (/) in the fuel.
Example: For octane (C8H18), / = 2.25 → (A/F)s = 15.1
(gasoline (A/F)s≈ 14.6
10
Fuel Lean Mixture
Fuel-air mixtures with more than stoichiometric air (excess air) can burn
With excess air you have fuel lean combustion
a fuel lean mixture has excess air, so g > 1
2222222
)76.3)(4
( eOdNOHCONOHC
g
At low combustion temperatures, the extra air appears in the products as
O2 and N2:
Above reaction equation has two unknowns (d, e) and we have two
atom balance equations (O, N) so can solve for the unknowns
11
Fuel-air mixtures with less than stoichiometric air (excess fuel) can burn.
With less than stoichiometric air you have fuel rich combustion, there is
insufficient oxygen to oxidize all the C and H in the fuel to CO2 and H2O.
2222222
)76.3)(4
( fHeCOdNOHCONOHC
g
Fuel Rich Mixture
Above reaction equation has three unknowns (d, e, f) and we only have
two atom balance equations (O, N) so cannot solve for the unknowns
unless additional information about the products is given.
a fuel rich mixture has insufficient air → g < 1
Get incomplete combustion where carbon monoxide (CO) and molecular
hydrogen (H2) also appear in the products.
12
The equivalence ratio, f, is commonly used to indicate if a mixture is
stoichiometric, fuel lean, or fuel rich.
s
mixture
mixture
s
AF
AF
FA
FA
/
/
/
/f
Off-Stoichiometric Mixtures
stoichiometric f = 1
fuel lean f < 1
fuel rich f > 1
Products)76.3(4
22
NOHC
Stoichiometric mixture:
Off-stoichiometric mixture:
Products)76.3(4
122
NOHC
f
fg
1
13
Off-stoichiometric Conditions
Example: Consider a reaction of octane with 10% excess air, what is f?
Stoichiometric : 22222188 4798)76.3(5.12 NOHCONOHC
10% excess air is: 222222188 98)76.3)(5.12(1.1 bNaOOHCONOHC
O balance: 1.1(12.5)(2) = 16 + 9 + 2a → a = 1.25,
N balance: 1.1(12.5)(3.76)(2) = 2b → b = 51.7
Other terminology used to describe how much air is used in combustion:
110% stoichiometric air = 110% theoretical air = 10% excess air
1.1 )76.3)(4
( 2283 g
g NOHC → mixture is fuel lean
91.01/)76.4)(5.12(1.1
1/)76.4(5.12
/
/
mixture
s
FA
FAf 1.1
91.
11
f
14
First Law Analysis for Reacting System
Consider a constant pressure process in which nf moles of fuel react with
na moles of air to produce np moles of product:
PnAnFn paf
Applying First Law with state 1 being the reactants at P1, T1 and state 2
being products at P2, T2:
Reactants Products
Reactants Products
Reaction
State 1 State 2
Q
W
15
RRii
Ppii
RP
ThnThnQ
HH
HH
VPUVPU
)()(
)()(
12
111222
First Law Analysis for Reacting System
)()( 121221 VVPUUQ
WUQ
16
Enthalpy of Reaction
Consider the case where the final temperature of the products is the same as
the initial temperature of the reactants (e.g., calorimeter is used to measure Q).
Reaction Q
W
P1=P2=Po
T1=T2=To
The heat released under this situation is referred to as the enthalpy of
reaction, HR ,
fuel of kmolor kgper kJ :units )()(
)()(
Roii
Poii
RRii
Ppii
R
ThnThn
ThnThn
HQ
To
17
The maximum amount of energy is released from a fuel when reacted with a
stoichiometric amount of air and all the hydrogen and carbon contained in the
fuel is converted to CO2 and H2O
Heat of Combustion
222224
76.32
)76.3(4
NOHCONOHC
HR = HP – HR < 0 (exothermic)
HR = HP – HR > 0 (endothermic)
This maximum energy is commonly referred to as the heat of combustion, or
the heating value, given per unit mass of fuel or air
18
Heat of Combustion
There are two possible values for the heat of combustion depending on
whether the water in the products is taken to be saturated liquid or vapour.
Tp
T
S
hg hf
From steam tables:
hfg = hg – hf > 0
The term higher heat of combustion and heating value (HHV) is used
when the water in the products is taken to be in the liquid state (hH20 = hf )
The term lower heat of combustion and heating value (LHV) is used
when the water in the products is taken to be in the vapour state (hH20 = hg )
19
0
h(kJ/kg fuel)
T
Products with H2O (g)
Reactants
Products with H2O (l)
hfg,H2O per kg fuel
hlow
hhigh
Heat of Combustion, graphical
298K
21
Fuel A/F Energy
density
(MJ/L fuel)
Specific
energy
(MJ/kg air)
Gasoline*
14.6 32 2.9
Butanol 11.2 29.2 3.2
Ethanol 9.0 19.6 3.0
Methanol
Hydrogen
6.5
34.3
16
**
3.1
4.1
* Diesel about 10% more energy per volume than gasoline
** liquefied hydrogen has roughly ¼ the energy density of gasoline
22
Heat of Formation
Consider the following reactions taking place at atmospheric pressure and
with TP = TR = 298K
222
2222
/ 000,394 )()()(
/ 000,286 )()()(2/1
COkmolkJQgCOgOsC
OHkmolkJQlOHgHgO
Values for standard heat of formation for different species are tabulated
In these reactions H2O and CO2 are formed from their elements in their
natural state at standard temperature and pressure (STP) 1 atm and 298K.
ofh
kmolkJQh
kmolkJQh
o
f, CO
o
Of, H
/ 000,394
/ 000,286
2
2
Reactions of this type are called formation reactions and the corresponding
measured heat release Q is referred to as the standard heat of formation
( ) so:
24
Enthalpy Scale for a Reacting System
By international convention, the enthalpy of every element in its natural state
(e.g., O2(g), N2(g), H2(g), C(s)) at STP has been set to zero
0)298,1( o
fhKatmhi.e,
Consider the following identity:
)]298,1(),([)298,1(),( KatmhTPhKatmhTPh
)]298,1(),([ ),( , KatmhTPhhTPh ii
o
ifi
sensible enthalpy chemical enthalpy 298K T
K ip dTc298 ,
Therefore, the enthalpy of the i’th component in a mixture is:
25
-14000
-5000
-9000
298 2800
H2O
CO2
Temperature K
Enthalpy (kJ/kg)
The data is also found in the JANNAF tables provided at course web site
,
o
ifh
27
Adiabatic Flame Temperature
Consider the following adiabatic constant pressure process:
For a given reaction where the ni’s are known for both the reactants and the
products, Ta can be calculated explicitly.
Fuel
Air Products
(Ta )
Reactants
(T1)
)()(
0)()(
1
Rii
Paii
RRii
Ppii
ThnThn
ThnThnQ
For a constant pressure process, the final products temperature, Ta, is known
as the adiabatic flame temperature (AFT).
28
Adiabatic Flame Temperature
R
iiP
aii ThnThn )()( 1
P R
o
ifi
o
ifiR
T
ipiP
T
ipi hnhndTcndTcn ia
,,298 ,298 ,OR
R
ii
o
ifiP
iai
o
ifi KhThhnKhThhn )298()()298()( 1,,
o
RHSensible heat of reactants
(equal to 0 if T1 = 298K)
Sensible heat of products
P R
o
ifi
o
ifiR
iiiP
iaii hnhnKhThnKhThn ,,1 )298()()298()(
29
o
HCf
o
OHf
o
COf
NaNOHaOHCOaCO
hhh
KhThKhThKhTh
18822
222222
,,, 98
)298()(47)298()(9)298()(8
22222188 4798)76.3(5.12)( NOHCONOlHC
Consider constant pressure complete combustion of stoichiometric liquid
butane-air initially at 298K and 1 atm
Look up enthalpy of formation values and iterate Ta → 2410K
Adiabatic Flame Temperature, example
P R
o
ifi
o
ifiR
iiiiP
aii hnhnKhThnKhThn ,,1 )298()()}298()({
022 ,, o
Nf
o
Of hhNote:
31
Now consider octane air with 10% excess air
Look up values and iteration gives Ta = 2262K
Adiabatic Flame Temperature, example
222222188 7.5125.198)76.3)(5.12(1.1 NOOHCONOHC
o
HCf
o
OHf
o
COfNaN
OaOOHaOHCOaCO
hhhKhTh
KhThKhThKhTh
1882222
222222
,,, 98 )298()(7.51
)298()(.251 )298()(9)298()(8
HR same as for f =1
Excess air adds 6 moles of diatomic molecules (O2 and N2) into the
products that does not contribute to heat release just soaks it up.
33
Constant Volume AFT
Reaction Q
W
Consider the case where the piston is fixed and the cylinder is perfectly
insulated so the process is adiabatic (Q = 0)
)()(
0)()(
1
Rii
Paii
RRii
Ppii
TunTun
TunTunQ
Note h = u + pv = u + RT, so
T) R)(Th(nTRThnR
iiP
aii 1))((
34
p Riai
P R
o
ifi
o
ifiR
iiiP
iaii
TRnTRn
hnhnKhThnKhThn
1
,,1
)298()()298()(
Extra term compared to constant pressure AFT ( term > 0)
Constant Volume AFT
R
iii
o
ifiP
iiai
o
ifi TRKhThhnTRKhThhn )298()()298()( 1,,
The AFT for a constant volume process is larger than for a constant
pressure process.
The AFT is lower for constant pressure process since there is Pdv work
done
35
i
a
R
p
i
CV
R
p
R
p
R
p
p
pp
R
RR
PR
T
T
n
n
P
P
T
T
n
n
P
P
P
TRn
P
TRn
VV
Constant Volume Combustion Pressure
Assuming ideal gas behaviour:
For large hydrocarbons like octane the mole ratio term is close to one
22222188 4798)76.3(5.12)( NOHCONOlHC
06.15.60
64
)76.4(5.121
4798
R
p
n
n
36
Stoichiometric octane-air (HR= 47.9 MJ/kg-fuel):
4798)76.3(5.12 22222188 NOHCONOHC
1.8604.706.1298
2266
5.60
64
i
a
R
p
i
CV
T
T
n
n
P
P
Stoichiometric hydrogen-air (HR= 141.6 MJ/kg-fuel):
88.11)76.3(5.0 22222 NOHNOH
8.60.885.0298
2383
38.3
88.2
i
a
R
p
i
CV
T
T
n
n
P
P
Engine Fuel Comparison
Stoichiometric ethanol-air (HR= 29.7 MJ/kg-fuel):
2222262 2.1332)76.3(5.3 NOHCONOOHC
5.737.702.1298
2197
7.17
2.18
i
a
R
p
i
CV
T
T
n
n
P
P