Chapter 14: CHEMICAL REACTIONS
The combustion process is a chemical reaction whereby fuel is oxidized and
energy is released.
Fuel
CnHm
Combustion
Chamber
CO2
H2O
N2
Air
Reactants
TR, PR
Products
PP, TP
Qnet
Fuels are usually composed of some compound or mixture containing
carbon, C, and hydrogen, H2.
Examples of hydrocarbon fuels are
CH4
Methane
C8H18 Octane
Coal
Mixture of C, H2, S, O2, N2 and non-combustibles
Initially, we shall consider only those reactions that go to completion. The
components prior to the reaction are called reactants and the components
after the reaction are called products.
Reactants → Products
For example, all carbon is burned to carbon dioxide and all hydrogen is
converted into water.
C + O2 → CO2
1
H2 + O2 → H2 O
2
A complete combustion process is one where all carbon is burned to carbon
dioxide (CO2) and all hydrogen is converted into water (H2O).
Chapter 14 -1
Example 14-1
A complete combustion of octane in oxygen is represented by the balanced
combustion equation. The balanced combustion equation is obtained by
making sure we have the same number of atoms of each element on both
sides of the equation. That is, we make sure the mass is conserved.
C8 H18 + A O2 → B CO2 + D H2 O
Note we often can balance the C and H for complete combustion by
inspection.
C8 H18 + A O2 → 8 CO2 + 9 H2 O
The amount of oxygen is found from the oxygen balance.
O: A(2) = 8(2) + 9(1)
A = 12.5
C8 H18 + 12.5 O2 → 8 CO2 + 9 H2 O
Note: Mole numbers are not conserved, but we have conserved the mass on
a total basis as well as a specie basis.
The complete combustion process is also called the stoichiometric
combustion, and all coefficients are called the stoichiometric coefficients.
In most combustion processes, oxygen is supplied in the form of air rather
than pure oxygen.
Air is assumed to be 21 percent oxygen and 79 percent nitrogen on a
volume basis. For ideal gas mixtures, percent by volume is equal to percent
by moles. Thus, for each mole of oxygen in air, there exists 79/21 = 3.76
moles of nitrogen. Therefore, complete or theoretical combustion of octane
with air can be written as
Chapter 14 -2
C8 H18 + 12. 5 ( O2 + 3. 76 N 2 ) →
8 CO2 + 9 H2 O + 47 N 2
Air-Fuel Ratio
Since the total moles of a mixture are equal to the sum of moles of each
component, there are 12.5(1 + 3.76) = 59.5 moles of air required for each
mole of fuel for the complete combustion process.
Often complete combustion of the fuel will not occur unless there is an
excess of air present greater than just the theoretical air required for
complete combustion.
To determine the amount of excess air supplied for a combustion process,
let us define the air-fuel ratio AF as
kmol air
AF =
kmol fuel
Thus, for the above example, the theoretical air-fuel ratio is
kmol air
12.5(1 + 3.76)
= 59.5
AFth =
kmol fuel
1
On a mass basis, the theoretical air-fuel ratio is
Chapter 14 -3
kg air
kmol air
kmol air
AFth = 59.5
kmol fuel [8(12) + 18(1)] kg fuel
kmol fuel
kg air
.
= 1512
kg fuel
28.97
Percent Theoretical and Percent Excess Air
In most cases, more than theoretical air is supplied to ensure complete
combustion and to reduce or eliminate carbon monoxide (CO) from the
products of combustion. The amount of excess air is usually expressed as
percent theoretical air and percent excess air.
Percent theoretical air =
Percent excess air =
AFactual
100%
AFth
AFactual − AFth
100%
AFth
Show that these results may be expressed in terms of the moles of oxygen
only as
Percent theoretical air =
N O2 actual
N O2 th
Chapter 14 -4
100%
Percent excess air =
N O2 actual − N O2 th
N O2 th
100%
Write the combustion equation of octane with 120 percent theoretical air
(20 percent excess air).
C8 H18 + 12
. (12.5) ( O2 + 3.76 N 2 ) →
8 CO2 + 9 H2 O + (0.2)(12.5) O2 + 12
. (47) N 2
Note that (1)(12.5)O2 is required for complete combustion to produce 8
kmol of carbon dioxide and 9 kmol of water; therefore, (0.2)(12.5)O2 is
found as excess oxygen in the products.
C8 H18 + 12
. (12.5) (O2 + 3.76 N 2 ) →
8 CO2 + 9 H2 O + 2.5 O2 + 12
. (47) N 2
Second method to balance the equation for excess air (see the explanation
of this technique in the text):
Chapter 14 -5
C8 H18 + 12
. Ath (O2 + 3.76 N 2 ) →
8 CO2 + 9 H2 O + 0.2 Ath O2 + 12
. Ath (3.76) N 2
O:
12
. Ath (2) = 8(2) + 9(1) + 0.2 Ath (2)
Ath = 12.5
Incomplete Combustion with Known Percent Theoretical Air
Consider combustion of C8H18 with 120 % theoretical air where 80 % C in
the fuel goes into CO2.
C8 H18 + 1.2(12.5) (O2 + 3.76 N 2 ) →
0.8(8) CO2 + 0.2(8) CO + 9 H2 O + X O2 + 12
. ( 47) N 2
O balance gives
O:
12
. (12.5)(2) = 0.8(8)(2) + 0.2(8)(1) + 9(1) + X (2)
X = 3.3
Why is X > 2.5?
Then the balanced equation is
C8 H18 + 1.2(12.5) ( O2 + 3.76 N 2 ) →
6.4 CO2 + 16
. CO + 9 H2 O + 3.3 O2 + 12
. ( 47) N 2
Chapter 14 -6
Combustion Equation When Product Gas Analysis Is Known
Propane gas C3H8 is reacted with air such that the dry product gases are
11.5 percent CO2, 2.7 percent O2, and 0.7 percent CO by volume. What
percent theoretical air was supplied? What is the dew point temperature of
the products if the product pressure is 100 kPa?
We assume 100 kmol of dry product gases; then the percent by volume can
be interpreted to be mole numbers. But we do not know how much fuel and
air were supplied or water formed to get the 100 kmol of dry product gases.
X C3 H8 + A (O2 + 3.76 N 2 ) →
115
. CO2 + 0.7 CO + 2.7 O2 + B H2 O + A( 3.76) N 2
The unknown coefficients A, B, and X are found by conservation of mass
for each species.
C: X (3) = 115
. (1) + 0.7(1)
X = 4.07
H: X (8) = B (2)
. (2) + 0.7(1)
O: A(2) = 115
+ 2.7(2) + B (1)
N 2 : A(3.76) = 85.31
B = 16.28
A = 22.69
The balanced equation is
4.07 C3 H8 + 22.69 ( O2 + 3.76 N 2 ) →
115
. CO2 + 0.7 CO + 2.7 O2 + 16.28 H2 O + 85.31 N 2
Second method to find A:
Assume the remainder of the 100 kmol of dry product gases is N2.
Chapter 14 -7
kmol N 2 = 100 − (115
. + 0.7 + 2.7) = 851
.
Then A is
A=
851
.
= 22.65
3.76
( fairly good check )
These two methods don’t give the same results for A, but they are close.
What would be the units on the coefficients in the balanced combustion
equation?
Generally we should write the combustion equation per kmol of fuel. To
write the combustion equation per unit kmol of fuel, divide by 4.07:
C3 H8 + 5.57 ( O2 + 3.76 N 2 ) →
2.83 CO2 + 017
. CO + 0.66 O2 + 4.0 H2 O + 20.96 N 2
The actual air-fuel ratio is
kg air
kmol air
kg fuel
1kmol fuel[3(12) + 8(1)]
kmol fuel
(5.57)(1 + 3.76) kmol air 28.97
AFactual =
kg air
= 17.45
kg fuel
The theoretical combustion equation is
Chapter 14 -8
C3 H8 + 5 ( O2 + 3.76 N 2 ) →
3 CO2 + 4.0 H2 O + 18.80 N 2
The theoretical air-fuel ratio is
kg air
kmol air
AFth =
kg fuel
1kmol fuel[3(12) + 8(1)]
kmol fuel
kg air
= 15.66
kg fuel
(5)(1 + 3.76) kmol air 28.97
The percent theoretical air is
AFactual
Percent theoretical air =
100%
AFth
17.45
=
100 = 111%
15.66
or
Percent theoretical air =
=
The percent excess air is
Chapter 14 -9
N O2 actual
N O2 th
100%
557
.
100 = 111%
5
AFactual − AFth
100%
AFth
17.45 − 15.66
=
100 = 11%
15.66
Percent excess air =
Dew Point Temperature
The dew point temperature for the product gases is the temperature at which
the water in the product gases would begin to condense when the products
are cooled at constant pressure. The dew point temperature is equal to the
saturation temperature of the water at its partial pressure in the products.
Tdp = Tsat at Pv = yv Pproducts
N water
yv =
∑ Ne
products
Chapter 14 -10
4
= 01398
.
2.83 + 017
. + 0.66 + 4 + 20.96
Pv = yv Pproducts = 01398
.
(100 kPa )
yv =
= 13.98 kPa
Tdp = Tsat at 13.98 kPa
= 52.44 o C
What would happen if the product gases are cooled to 100oC or to 30oC?
Example 14-11
An unknown hydrocarbon fuel, CXHY is reacted with air such that the dry
product gases are 12.1 percent CO2, 3.8 percent O2, and 0.9 percent CO by
volume. What is the average makeup of the fuel?
We assume 100 kmol (do you have to always assume 100 kmol?) of dry
product gases; then the percent by volume can be interpreted to be mole
numbers. We do not know how much air was supplied or water formed to
get the 100 kmol of dry product gases, but we assume 1 kmol of unknown
fuel.
C X HY + A (O2 + 3.76 N 2 ) →
12.1 CO2 + 0.9 CO + 3.8 O2 + B H2 O + D N 2
The five unknown coefficients A, B, D, X, and Y are found by conservation
of mass for each species, C, H, O, and N plus one other equation. Here we
use the subtraction method for the nitrogen to generate the fifth independent
equation for the unknowns.
Chapter 14 -11
C X HY + A (O2 + 3.76 N 2 ) →
12.1 CO2 + 0.9 CO + 3.8 O2 + B H2 O + D N 2
The unknown coefficients A, B, D, X, and Y are found by conservation of
mass for each species. Here we assume the remainder of the dry product
gases is nitrogen.
N 2 : D = 100 − (12.1 + 0.9 + 3.8) = 83.2
D
83.2
=
= 22.13
3.76 3.76
O: A(2) = (12.1)(2) + (0.9)(1) + (38
. )(2) + B(1)
B = 1154
.
O2 : A =
C: 1( X ) = 12.1(1) + (0.9)(1)
X = 13.0
H: 1(Y ) = B (2)
Y = 23.08
The balanced equation is
C13 H23.08 + 22.13 ( O2 + 3.76 N 2 ) →
12.1 CO2 + 0.9 CO + 38
. O2 + 1154
. H2 O + 83.2 N 2
Enthalpy of Formation
When a compound is formed from its elements (e.g., methane, CH4, from C
and H2), heat transfer occurs. When heat is given off, the reaction is called
exothermic. When heat is required, the reaction is called endothermic.
Consider the following.
C
Reaction chamber
(Back yard compost pile)
Methane
CH4
2H2
Reactants
TR = 298 K
PR = 1 atm
Qnet
Chapter 14 -12
Products
TP = 298 K
PP = 1 atm
The reaction equation is
C + 2 H2 → CH4
The conservation of energy for a steady-flow combustion process is
Ein = Eout
Qnet + H Reactants = H Products
Qnet = H Products − H Reactants
Qnet = ∑ N e he − ∑ N i hi
Products
Reactants
Qnet = 1hCH4 − (1hC + 2hH2 )
A common reference state for the enthalpies of all reacting components is
established as
The enthalpy of the elements or their stable
compounds is defined to be ZERO at 25oC (298 K)
and 1 atm (or 0.1 MPa).
Qnet = 1hCH4 − (1(0) + 2(0))
= hCH4
Chapter 14 -13
o
This heat transfer is called the enthalpy of formation for methane, h f .
The superscript (o) implies the 1 atm pressure value and the subscript (f)
o
implies 25oC data, h f is given in Table A-26.
During the formation of methane from the elements at 298 K, 0.1 MPa, heat
is given off (an exothermic reaction) such that
Qnet = h fo CH = −74,850
4
kJ
kmolCH4
o
The enthalpy of formation h f is tabulated for typical compounds. The
enthalpy of formation of the elements in their stable form is taken as zero.
The enthalpy of formation of the elements found naturally as diatomic
elements, such as nitrogen, oxygen, and hydrogen, is defined to be zero.
The enthalpies of formation for several combustion components are given
in the following table.
Substance
Air
Oxygen
Nitrogen
Carbon dioxide
Carbon monoxide
Water (vapor)
Water (liquid)
Methane
Acetylene
Ethane
Propane
Butane
Octane (vapor)
Dodecane
Formula
M
h fo
kJ/kmol
28.97
0
O2
32
0
N2
28
0
CO2
44
-393,520
CO
28
-110,530
H2Ovap
18
-241,820
H2Oliq
18
-285,830
CH4
16
-74,850
C2H2
26
+226,730
C2H6
30
-84,680
C3H8
44
-103,850
C4H10
58
-126,150
C8H18
114 -208,450
C12H26 170 -291,010
Chapter 14 -14
The enthalpies are calculated relative to a common base or reference called
the enthalpy of formation. The enthalpy of formation is the heat transfer
required to form the compound from its elements at 25oC (77 °F) or 298 K
(537 R), 1 atm. The enthalpy at any other temperature is given as
h = h fo + (hT − h o )
Here the term h o is the enthalpy of any component at 298 K. The enthalpies
at the temperatures T and 298 K can be found in Tables A-18 through A-25.
If tables are not available, the enthalpy difference due to the temperature
difference can be calculated from
(hT − h ) =
o
z
T
298 K
CP dT ′ =CP ,ave (T − 298)
The net heat transfer to the reacting system is
Qnet = H P − H R
=
∑ N [h
e
o
f
+ (hT − h o )]e −
Products
∑ N [h
i
o
f
+ (hT − h o )]i
Reactants
In an actual combustion process, is the value of Qnet positive or negative?
Chapter 14 -15
Example 14-3
Butane gas C4H10 is burned in theoretical air as shown below. Find the net
heat transfer per kmol of fuel.
Fuel
C4H10
CO2
H2O
N2
Combustion
Chamber
Theoretical
air
Reactants
TR = 298 K
PR = 1 atm
Products
TP = 1000 K
PP = 1 atm
Qnet
Balanced combustion equation:
C4 H10 + 6.5 (O2 + 3.76 N 2 ) →
4 CO2 + 5 H2 O + 24.44 N 2
The steady-flow heat transfer is
Qnet = H P − H R
=
∑ N [h
e
o
f
+ (hT − h o )]e −
Products
∑ N [h
i
o
f
+ (hT − h o )]i
Reactants
Reactants: TR = 298 K
Comp
C4H10
O2
N2
Ni
kmol/kmol
fuel
h fo
hT
kJ/kmol
kJ/kmol
1
6.5
24.44
-126,150
0
0
-8,682
8,669
Chapter 14 -16
ho
kJ/kmol
N i [h fo + (hT − h o )]i
-8,682
8,669
kJ/kmol fuel
-126,150
0
0
HR =
∑ N [h
i
o
f
+ (hT − h o )]i
Reactants
= −126,150
kJ
kmol C4 H10
Products: TP = 1000 K
Comp
CO2
H2O
N2
Ne
kmol/kmol
fuel
h fo
hT
kJ/kmol
kJ/kmol
4
5
24.44
-393,520
-241,820
0
HP =
ho
42,769
35,882
30,129
∑ N [h
e
o
f
kJ/kmol
N e [h fo + (hT − h o )]e
9,364
9,904
8,669
kJ/kmol fuel
-1,440,460
-1,079,210
+524,482
+ (hT − h o )]e
Products
= −1,995,188
kJ
kmol C4 H10
Qnet = H P − H R
= −1,869,038
kJ
kmol C4 H10
Chapter 14 -17
Adiabatic Flame Temperature
The temperature the products have when a combustion process takes place
adiabatically is called the adiabatic flame temperature.
Example 14-4
Liquid octane C8H18(liq) is burned with 400 percent theoretical air. Find
the adiabatic flame temperature when the reactants enter at 298 K, 0.1 Mpa,
and the products leave at 0.1MPa.
Fuel
C8H18
CO2
O2
H2O
N2
Combustion
Chamber
400%
Th. air
Reactants
TR = 298 K
PR = 0.1 MPa
Products
TP = ?
PP = 0.1MPa
Qnet = 0
The combustion equation is
C8 H18 + 4(12.5) ( O2 + 3.76 N 2 ) →
8 CO2 + 37.5 O2 + 9 H2 O + 188 N 2
The steady-flow heat transfer is
Qnet = H P − H R
=
∑ N [h
e
o
f
+ (hT − h o )]e −
Products
∑ N [h
i
Reactants
= 0 ( Adiabatic Combustion)
Chapter 14 -18
o
f
+ (hT − h o )]i
Thus, HP = HR for adiabatic combustion. We need to solve this equation for
TP.
Since the temperature of the reactants is 298 K, ( hT − h o )i = 0,
HR =
∑Nh
i
Reactants
o
f i
= 1( −249,950) + 4(12.5)(0) + 4(12.5)(3.76)(0)
= −249,950
kJ
kmol C4 H10
Since the products are at the adiabatic flame temperature, TP > 298 K
∑
HP =
Products
N e [h fo + (hTP − h o )]e
= 8(−393,520 + hTP − 9364)CO2
+ 9(−241,820 + hTP − 9904) H 2O
+ 37.5(0 + hTP − 8682)O2
+ 188(0 + hTP − 8669) N2
= (−7, 443,845 + 8hTP , CO2 + 9hTP , H 2O
+ 37.5hTP , O2 + 188hTP , N2 )
kJ
kmol C4 H10
Thus, setting HP = HR yields
∑N h
Pr oducts
e TP , e
= 8hTP , CO2 + 9hTP , H2 O + 37.5hTP , O2 + 188hTP , N 2
= 7,193,895
Chapter 14 -19
To estimate TP, assume all products behave like N2 and estimate the
adiabatic flame temperature from the nitrogen data, Table A-18.
242.5hTP , N 2 = 7,193,895
hTP , N 2 = 29,6655
.
kJ
kmol N 2
Tp ≅ 985 K
Because of the tri-atomic CO2 and H2O, the actual temperature will be
somewhat less than 985 K. Try TP = 960 K and 970K.
CO2
H2O
O2
N2
Ne
8
9
37.5
188
∑N h
Produts
e TP , e
h960 K
h970 K
40,607
34,274
29,991
28,826
7,177,572
41,145
34,653
30,345
29,151
7,259,362
Interpolation gives: TP = 962 K.
Example 14-5
Liquid octane C8H18(liq) is burned with excess air. The adiabatic flame
temperature is 960 K when the reactants enter at 298 K, 0.1 Mpa, and the
products leave at 0.1MPa. What percent excess air is supplied?
Fuel
C8H18(liq)
Combustion
Chamber
Excess
air
Reactants
TR = 298 K
PR = 0.1 MPa
Qnet = 0
Chapter 14 -20
CO2
O2
H2O
N2
Products
TP = 960 K
PP = 0.1 MPa
Let A be the excess air; then combustion equation is
C8 H18 + (1 + A)(12.5) ( O2 + 3.76 N 2 ) →
8 CO2 + 12.5 A O2 + 9 H 2 O + (1 + A)(12.5)( 3.76) N 2
The steady-flow heat transfer is
Qnet = H P − H R
=
∑ N [h
e
o
f
+ (hT − h o )]e −
Products
∑ N [h
i
o
f
+ (hT − h o )]i
Reactants
= 0 ( Adiabatic combustion)
Here, since the temperatures are known, the values of hT are known. The
product gas mole numbers are unknown but are functions of the amount of
excess air, A. The energy balance can be solved for A.
P
A=3
Thus, 300 percent excess, or 400 percent theoretical, air is supplied.
Example 14-6
Tabulate the adiabatic flame temperature as a function of excess air for the
complete combustion of C3H8 when the fuel enters the steady-flow reaction
chamber at 298 K and the air enters at 400 K.
The combustion equation is
C3 H8 + (1 + A)(5) ( O2 + 3.76 N 2 ) →
3 CO2 + 5 A O2 + 4 H 2 O + (1 + A)(5)( 3.76) N 2
Chapter 14 -21
where A is the value of excess air in decimal form.
The steady-flow heat transfer is
Qnet = H P − H R
=
∑ N [h
e
o
f
+ (hT − h o )]e −
Products
∑ N [h
i
o
f
+ (hT − h o )]i
Reactants
= 0 ( Adiabatic combustion)
Percent Excess Air
0
20
50
100
217
Adiabatic Flame Temp. K
2459.3
2191.9
1902.5
1587.1
1200
Enthalpy of Reaction and Enthalpy of Combustion
When the products and reactants are at the same temperature, the enthalpy
of reaction hR, is the difference in their enthalpies. When the combustion is
assumed to be complete with theoretical air supplied the enthalpy of
reaction is called the enthalpy of combustion hC. The enthalpy of
combustion can be calculated at any value of the temperature, but it is
usually determined at 25oC or 298 K. See Table A-27 for the enthalpy of
combustion at 25oC.
Chapter 14 -22
hC = H P − H R when TP = TR = 25o C = 298 K
=
∑N h
e
Products
o
f e
−
∑Nh
i
Reactants
o
f i
Heating Value
The heating value, HV, of a fuel is the absolute value of the enthalpy of
combustion or just the negative of the enthalpy of combustion.
HV = hC
The lower heating value, LHV, is the heating value when water appears as
a gas in the products.
LHV = hC = − hC with H2 Ogas in products
The lower heating value is often used as the amount of energy per kmol of
fuel supplied to the gas turbine engine.
The higher heating value, HHV, is the heating value when water appears
as a liquid in the products.
HHV = hC = − hC with H2 Oliquid in products
The higher heating value is often used as the amount of energy per kmol of
fuel supplied to the steam power cycle.
The higher and lower heating values are related by the amount of water
formed during the combustion process and the enthalpy of vaporization of
water at the temperature.
Chapter 14 -23
HHV = LHV + N H2O h fg H2 O
Example 14-7
The enthalpy of combustion of gaseous C8H18 at 25oC with liquid water in
the products is -5,512,200 kJ/kmol (see Table A-27). Find the lower
heating value of liquid octane.
LHVC8 H18 gas = HHVC8 H18 gas − N H2O h fg H2O
= 5,512,200
kJ
kmol H2 O
kJ
−9
(44,010)
kmol C8 H18
kmol C8 H18
kmol H2 O
= 5,116,110
kJ
kmol C8 H18
LHVC8 H18 liq = LHVC8 H18 gas − h fg C8 H18
= (5,116,110 − 41,460)
= 5,074,650
kJ
kmol C8 H18
kJ
kmol C8 H18 liq
Can you explain why LHVliq < LHVgas?
Chapter 14 -24
Closed System Analysis
Example 14-8
A mixture of 1 kmol C8H18 gas and 200 percent excess air at 25oC, 1 atm, is
burned completely in a closed system (a bomb) and is cooled to 1200 K.
Find the heat transfer from the system and the system final pressure.
Qnet
1 kmol C8H18 gas
200% excess air
T1 = 25oC
P1 = 1 atm
T2 = 1200 K
Wnet
Rigid
container
Apply the first law closed system:
Ein − Eout = ∆E
Qnet − Wnet = U P − U R
Wnet = 0 ( Rigid container )
b
Qnet = U P − U R = H P − ( PV ) P − H R − ( PV ) R
b
= H P − H R − ( PV ) P − ( PV ) R
g
Assume that the reactants and products are ideal gases; then
PV = NRu T
Chapter 14 -25
g
LM ∑
N
Qnet = H P − H R −
N e Ru Te −
Products
=
∑ N [h
e
o
f
∑ N i RuTi
Reactants
OP
Q
+ (hT − h o ) − Ru T ]e
Products
−
∑ N [h
i
o
f
+ (hT − h o ) − Ru T ]i
Reactants
The balanced combustion equation for 200 percent excess (300 percent
theoretical) air is
C8 H18 + ( 3)(12.5) ( O2 + 3.76 N 2 ) →
8 CO2 + 25 O2 + 9 H 2 O + 141 N 2
Qnet = 8( −393,520 + 53,848 − 9364 − 8.314(1200)) CO2
+ 9( −241,820 + 44,380 − 9904 − 8.314(1200)) H2 O
+ 25(0 + 38,447 − 8682 − 8.314(1200)) O2
+ 141(0 + 36,777 − 8669 − 8.314(1200)) N 2
− 1( −208,450 + h298 K − h o − 8.314(298)) C8 H18
− 37.5(0 + 8682 − 8682 − 8.314(298)) O2
− 141(0 + 8669 − 8669 − 8.314(298)) N 2
. ⋅ 106
= −112
kJ
kmol C8 H18
To find the final pressure, we assume that the reactants and the products are
ideal-gas mixtures.
Chapter 14 -26
PV
1 1 = N 1 Ru T1
PV
2 2 = N 2 Ru T2
where state 1 is the state of the mixture of the reactants before the
combustion process and state 2 is the state of the mixture of the products
after the combustion process takes place. Note that the total moles of
reactants are not equal to the total moles of products.
PV
N 2 Ru T2
2 2
=
PV
N 1 Ru T1
1 1
but V2 = V1.
P2 = P1
N 2 T2
N 1 T1
F
183 kmol products I F 1200 K I
= 1 atmG
H 179.5 kmol reactantsJK GH 298 K JK
= 4.11 atm
Chapter 14 -27
Second Law Analysis of Reacting Systems
Second law for the open system
Mass reservoir,
component i
Ti
Pi
Ni
Ne
Reacting
system
Mass reservoir,
component e
Te
Pe
Qk
Surroundings
Tk
The entropy balance relations developed in Chapter 6 are equally applicable
to both reacting and nonreacting systems provided that the entropies of
individual constituents are evaluated properly using a common basis.
Taking the positive direction of heat transfer to be to the system, the
entropy balance relation can be expressed for a steady-flow combustion
chamber as
Qk
∑ T + SReact − SProd + S gen ≥ ∆SCV
k
( kJ / k )
For an adiabatic, steady-flow process, the entropy balance relation reduces
to
S gen , adiabatic = S Prod − S React ≥ 0
The third law of thermodynamics states that the entropy of a pure crystalline substance at absolute zero temperature is zero. The third law
provides a common base for the entropy of all substances, and the entropy
values relative to this base are called the absolute entropy.
The ideal-gas tables list the absolute entropy values over a wide range of
temperatures but at a fixed pressure of Po = 1 atm. Absolute entropy values
at other pressures P for any temperature T are determined from
Chapter 14 -28
P
[ kJ / ( kmol ⋅ K )]
Po
For component i of an ideal-gas mixture, the absolute entropy can be
written as
s (T , P ) = s o (T , Po ) − Ru ln
si (T , Pi ) = sio (T , Po ) − Ru ln
yi Pm
Po
[ kJ / ( kmol ⋅ K )]
where Pi is the partial pressure, yi is the mole fraction of the component,
and Pm is the total pressure of the mixture in atmospheres.
Example 14-8
A mixture of ethane gas C2H6 and oxygen enters a combustion chamber at 1
atm, 25oC. The products leave at 1 atm, 900 K. Assuming complete
combustion, does the process violate the second law?
C2H6
CO2
Reacting
system
O2
TR = 25oC
PR = 1 atm
H2O
TP = 900 K
PP = 1 atm
Qsurr
Surroundings
Tsurr = 25oC
The balanced combustion equation is
C2 H6 + 35
. O2 → 2 CO2 + 3 H2 O
The mole fractions for the reactants and the products are
Chapter 14 -29
1
1
=
.
1 + 35
4.5
.
.
35
35
=
=
.
1 + 35
4.5
2
2
=
=
2+3 5
3
3
=
=
2+3 5
yC2 H6 =
yO2
yCO2
y H2 O
Now to calculate the individual component entropies.
For the reactant gases:
sC2 H6 (T , PC2 H6 ) = sCo2 H6 (T , Po ) − Ru ln
yC2 H6 Pm
Po
LM
F 1 1 AtmI OP
JJ P kJ
= M229.49 − 8.314 lnGG 4.5
MN
GH 1 Atm JK PQ kmol
C2 H6
= 242.0
kJ
kmolC2 H6 ⋅ K
Chapter 14 -30
⋅K
sO2 ( T , PO2 ) = sOo2 (T , Po ) − Ru ln
yO2 Pm
Po
LM
F 35. 1 AtmI OP
JJ P kJ
= M205.03 − 8.314 lnGG 4.5
MN
GH 1 Atm JK PQ kmol ⋅ K
O2
= 207.1
kJ
kmolO2 ⋅ K
∑N s
S React =
i i
Reactants
. (207.1)
= 1(242.0) + 35
= 966.9
kJ
kmolC2 H6 ⋅ K
For the product gases:
o
sCO2 (T , PCO2 ) = sCO
(T , Po ) − Ru ln
2
yCO2 Pm
Po
LM
F 2 1 AtmI OP
JJ P kJ
. − 8.314 lnGG 5
= M26356
MN
GH 1 Atm JK PQ kmol
CO2
.
= 2712
kJ
kmolCO2 ⋅ K
Chapter 14 -31
⋅K
sH2 O (T , PH2 O ) = sHo 2 O (T , Po ) − Ru ln
y H2 O Pm
Po
LM
F 3 1 AtmI OP
JJ P kJ
= M228.32 − 8.314 lnGG 5
MN
GH 1 Atm JK PQ kmol
H2 O
= 232.6
⋅K
kJ
kmolH2 O ⋅ K
∑N s
S Prod =
e e
Products
. ) + 3(232.6)
= 2(2712
= 1240.2
kJ
kmolC2 H6 ⋅ K
The entropy change for the combustion process is
S Prod − S React = (1240.2 − 966.9)
= 273.3
kJ
kmolC2 H6 ⋅ K
kJ
kmolC2 H6 ⋅ K
Now to find the entropy change due to heat transfer with the surroundings.
The steady-flow conservation of energy for the control volume is
Qnet sys = H P − H R
=
∑ N [h
e
o
f
+ (hT − h o )]e −
Products
∑ N [h
i
Reactants
Chapter 14 -32
o
f
+ (hT − h o )]i
Qnet sys = 2( −393,520 + 37,405 − 9364) CO2
+ 3( −241,820 + 31,828 − 9904) H2 O
− 1( −214,820 + h298 K − h o ) C2 H6
. (0 + 8682 − 8682) O2
− 35
.
= −1306
⋅ 106
kJ
kmol C2 H6
Qk Qnet sys
∑T = T
k
o
kJ
kmol C2 H6
=
(25 + 273) K
kJ
= −4,383
kmol C2 H6 ⋅ K
.
−1306
⋅ 106
The entropy generated by this combustion process is
Qk
≥0
Tk
kJ
= 273.3 − ( −4383)
kmol C2 H6 ⋅ K
kJ
= 4656.3
kmol C2 H6 ⋅ K
S gen = S Prod − S React − ∑
b
g
Since Sgen, or ∆Snet , is ≥ 0, the second law is not violated.
Chapter 14 -33