ME470 Chemical Reactions Hw Solutions
Inst: Shoeleh Di Julio
Chapter 15, Solution 14.
Propane is burned with 75 percent excess air during a combustion process. The AF ratio is to be
determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis The combustion equation in this case can be written as
C 3 H 8  1.75a th O 2  3.76N 2  
 3CO 2  4H 2 O  0.75a th O 2  (1.75  3.76)a th N 2
where ath is the stoichiometric coefficient for air. We have
automatically accounted for the 75% excess air by using the
C3H8
factor 1.75ath instead of ath for air. The stoichiometric amount of
oxygen (athO2) will be used to oxidize the fuel, and the remaining
Air
excess amount (0.75athO2) will appear in the products as free
oxygen. The coefficient ath is determined from the O2 balance,
75% excess
O2 balance:
175
. ath  3  2  0.75ath
Substituting,
C3H 8  8.75 O 2  3.76 N 2


Products
ath  5

 3CO 2  4H 2O  3.75O 2  32.9 N 2
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
AF 
mair
8.75  4.76 kmol 29 kg/kmol 

 27.5 kg air/kg fuel
mfuel 3 kmol 12 kg/kmol   4 kmol 2 kg/kmol 
Chapter 15, Solution 15.
Acetylene is burned with the stoichiometric amount of air during a combustion process. The AF ratio is to
be determined on a mass and on a mole basis.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis This is a theoretical combustion process since C2H2 is burned completely with stoichiometric
amount of air. The stoichiometric combustion equation of C2H2 is
C 2 H 2  a th O 2  3.76N 2  
 2CO 2  H 2 O  3.76 a th N 2
O2 balance:
ath  2  0.5


ath  2.5
Substituting,
C 2 H 2  2.5O 2  3.76N 2  
 2CO 2  H 2 O  9.4N 2
C2H2
Products
1
0
0
%
The air-fuel ratio is determined by taking the ratio of the mass of the air to the mass of the fuel,
t
mair

2.5  4.76 kmol 29 kg/kmol 
h  13.3 kg air/kg fuel
AF 

mfuel 2 kmol 12 kg/kmol   1 kmol 2 kg/kmole
o
On a mole basis, the air-fuel ratio is expressed as the ratio of rthe mole numbers of the air to the mole
numbers of the fuel,
e
t
i
c
a
l
a
AFmole basis 
N air
(2.5  4.76) kmol

 11.9 kmol air/kmol fuel
N fuel
1 kmol fuel
Chapter 15, Solution 18.
Propylene is burned with 50 percent excess air during a combustion process. The AF ratio and the
temperature at which the water vapor in the products will start condensing are to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, O2, and N2 only. 3
Combustion gases are ideal gases.
Properties The molar masses of C, H2, and air are 12 kg/kmol,
Products
C3H6
2 kg/kmol, and 29 kg/kmol, respectively (Table A-1).
Analysis (a) The combustion equation in this case can be
5
written as
0
C 3 H 6  1.5 a th O 2  3.76N 2  
 3CO 2  3H 2 O  0.5 a th O 2% (1.5  3.76) a th N 2
where ath is the stoichiometric coefficient for air. It is determined from
e
x
O2 balance: 15
. ath  3  15
.  0.5ath

 ath  4.5
c
Substituting, C 3 H 6  6.75 O 2  3.76N 2  3CO 2  3H 2 O  2e.25O 2  25.38N 2
The air-fuel ratio is determined by taking the ratio of the mass of the airs to the mass of the fuel,
s
m
6.75  4.76 kmol 29 kg/kmol 
AF  air 
 22.2 kg air/kg fuel
mfuel 3 kmol 12 kg/kmol   3 kmol 2 kg/kmol 
a
i temperature of the water vapor in
(b) The dew-point temperature of a gas-vapor mixture is the saturation
r
the product gases corresponding to its partial pressure. That is,
Thus,
 Nv 
 3 kmol 
P
105 kPa   9.367 kPa
Pv  

 N prod  prod  33.63 kmol 


Tdp  Tsat@9.367 kPa  44.5C
Chapter 15, Solution 23.
Gasoline is burned steadily with air in a jet engine. The AF ratio is given. The percentage of excess air used
is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis The theoretical combustion equation in this
case can be written as
C8 H18  a th O 2  3.76N 2

 8CO 2  9H 2O  3.76a th N 2
where ath is the stoichiometric coefficient for air. It is determined from
O2 balance:
ath  8  4.5


ath  12.5
Gasoline
(C8H18)
Jet engine
Products
Air
The air-fuel ratio for the theoretical reaction is determined by taking the ratio of the mass of the air to the
mass of the fuel for,
AFth 
m air,th
m fuel

12.5  4.76 kmol 29 kg/kmol 
 15 .14 kg air /kg fuel
8 kmol 12 kg/kmol   9 kmol 2 kg/kmol 
Then the percent theoretical air used can be determined from
Percent theoretical air 
AFact
AFth

18 kg air/kg fuel
 119%
15.14 kg air/kg fuel
Chapter 15, Solution 24.
Ethane is burned with air steadily. The mass flow rates of ethane and air are given. The percentage of
excess air used is to be determined.
Assumptions 1 Combustion is complete. 2 The combustion products contain CO2, H2O, and N2 only.
Properties The molar masses of C, H2, and air are 12 kg/kmol, 2 kg/kmol, and 29 kg/kmol, respectively
(Table A-1).
Analysis The theoretical combustion equation in this case can be written as
C 2 H 6  a th O 2  3.76N 2

 2CO 2  3H 2 O  3.76a th N 2
C2H6
where ath is the stoichiometric coefficient for air. It is determined from
O2 balance:
ath  2  15
.


ath  35
.
The air-fuel ratio for the theoretical reaction is determined by
taking the ratio of the mass of the air to the mass of the fuel for,
AFth 
m air,th
m fuel

Combustion
chamber
Products
Air
3.5  4.76 kmol 29 kg/kmol 
 16.1 kg air /kg fuel
2 kmol 12 kg/kmol   3 kmol 2 kg/kmol 
The actual air-fuel ratio used is
AFact 
m air 176 kg/h

 22 kg air/kg fuel
m fuel
8 kg/h
Then the percent theoretical air used can be determined from
Percent theoretical air 
AFact
AFth

22 kg air/kg fuel
 137 %
16.1 kg air/kg fuel
Thus the excess air used during this process is 37%.
Chapter 15, Solution 51.
Liquid propane is burned with 150 percent excess air during a steady-flow combustion process. The mass
flow rate of air and the rate of heat transfer from the combustion chamber are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air and combustion gases are ideal gases. 3 Kinetic and
potential energies are negligible. 4 Combustion is complete.
Properties The molar masses of propane and air are 44 kg/kmol and 29 kg/kmol, respectively (Table A-1).
Analysis The fuel is burned completely with excess air, and thus the products will contain only CO 2, H2O,
N2, and some free O2. Considering 1 kmol of C3H8, the combustion equation can be written as
C 3 H 8    2.5 a th O 2  3.76N 2  
 3CO 2  4H 2 O  1.5 a th O 2  2.5 3.76 a th N 2
where ath is the stoichiometric coefficient and is
determined from the O2 balance,

Q
2.5ath  3  2  1.5ath
Thus,


ath  5
C 3 H 8    12.5 O 2  3.76N 2  
 3CO 2  4H 2 O  7.5O 2  47N 2
C3H8
25C
Air
12C
Combustion
chamber
Products
1200 K
P = 1 atm
(a) The air-fuel ratio for this combustion process is
m
12.5  4.76 kmol 29 kg/kmol 
AF  air 
 39.22 kg air/kg fuel
mfuel 3 kmol 12 kg/kmol   4 kmol 2 kg/kmol 
 air  AFm
 fuel   39.22 kg air/kg fuel1.2 kg fuel/min   47.1 kg air/min
Thus,
m
(b) The heat transfer for this combustion process is determined from the energy balance
Ein  Eout  Esystem applied on the combustion chamber with W = 0. It reduces to
 Qout 
 N h
P

f
h h
   N h
P
R

f
 h h

R
Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,
Substance
C3H8 ()
O2
N2
H2O (g)
CO2
hf
kJ/kmol
-118,910
0
0
-241,820
-393,520
h 285 K
h 298 K
h1200 K
kJ/kmol
--8296.5
8286.5
-----
kJ/kmol
--8682
8669
9904
9364
kJ/kmol
--38,447
36,777
44,380
53,848
The h f of liquid propane is obtained by adding h fg of propane at 25C to h f of gas propane.
Substituting,
Qout  3393 ,520  53,848  9364   4241,820  44,380  9904   7.50  38,447  8682 
 47 0  36,777  8669   1 118 ,910  h298  h298   12 .50  8296 .5  8682 
 47 0  8286 .5  8669 
 190 ,464 kJ /kmol C 3 H 8
or
Qout  190 ,464 kJ /kmol C 3 H 8
Then the rate of heat transfer for a mass flow rate of 1.2 kg/min for the propane becomes
 1.2 kg/min 
 m 
190,464 kJ/kmol   5194 kJ/min
Q out  N Qout   Qout  
N
 44 kg/kmol 
Chapter 15, Solution 64.
A mixture of benzene gas and 30 percent excess air contained in a constant-volume tank is ignited. The
heat transfer from the combustion chamber is to be determined.
Assumptions 1 Both the reactants and products are ideal gases. 2 Combustion is complete.
Analysis The theoretical combustion equation of C6H6
with stoichiometric amount of air is
C6H6 g   ath O2  3.76N 2  
 6CO 2  3H 2O  3.76 ath N2
Q
C6H6+ Air
25C, 1 atm
where ath is the stoichiometric coefficient and is determined from the O2 balance,
ath  6  1.5  7.5
1000 K
Then the actual combustion equation with 30% excess air becomes
C6H6 g   9.75 O2  3.76N 2  
 5.52CO 2  0.48CO  3H 2O  2.49O 2  36.66N 2
The heat transfer for this constant volume combustion process is determined from the energy balance
Ein  Eout  Esystem applied on the combustion chamber with W = 0. It reduces to
 Qout 
 N h
P

f
 h  h   Pv
   N h
 h  h   Pv

f
R
P

R
Since both the reactants and the products behave as ideal gases, all the internal energy and enthalpies
depend on temperature only, and the Pv terms in this equation can be replaced by RuT.
It yields
 Qout 
 N h
P

f
 h1000 K  h298 K  Ru T
   N h
R
P

f
 Ru T

R
since the reactants are at the standard reference temperature of 25C. From the tables,
hf
kJ/kmol
82,930
0
0
-241,820
-110,530
-393,520
Substance
C6H6 (g)
O2
N2
H2O (g)
CO
CO2
h 298 K
h1000 K
kJ/kmol
--8682
8669
9904
8669
9364
kJ/kmol
--31,389
30,129
35,882
30,355
42,769
Thus,
Qout  5.52 393 ,520  42 ,769  9364  8.314  1000 
 0.48  110 ,530  30 ,355  8669  8.314  1000 
 3 241,820  35,882  9904  8.314  1000 
 2.49 0  31,389  8682  8.314  1000 
 36 .66 0  30,129  8669  8.314  1000 
 182,930  8.314  298   9.75 4.76  8.314  298 
 2,200, 433 kJ
or
Qout  2,200,433 kJ
Chapter 15, Solution 70.
Hydrogen is burned with 20 percent excess air during a steady-flow combustion process. The exit
temperature of product gases is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air and combustion gases are ideal gases. 3 Kinetic and
potential energies are negligible. 4 There are no work interactions. 5 The combustion chamber is adiabatic.
Analysis Adiabatic flame temperature is the temperature at which the products leave the combustion
chamber under adiabatic conditions (Q = 0) with no work interactions (W = 0). Under steady-flow
conditions the energy balance Ein  Eout  Esystem applied on the combustion chamber reduces to
 N h

H2
The combustion equation of H2 with 20% excess air is
7C
P

f
h h
   N h
P
R

f
h h
R
H 2  0.6 O 2  3.76N 2  
 H 2 O  0.1O 2  2.256N
From the tables,
2
Air
20% excess air
7C
Combustion
chamber
Products
TP
hf
kJ/kmol
0
0
0
-241,820
Substance
H2
O2
N2
H2O (g)
Thus,
h 280 K
h 298 K
kJ/kmol
7945
8150
8141
9296
kJ/kmol
8468
8682
8669
9904
1 241,820  hH O  9904   0.10  hO  8682   2.256 0  hN  8669 
 10  7945  8468   0.60  8150  8682   2.256 0  8141  8669 
2
It yields
2
2
hH 2O  0.1hO 2  2.256 h N 2  270 ,116 kJ
The adiabatic flame temperature is obtained from a trial and error solution. A first guess is obtained by
dividing the right-hand side of the equation by the total number of moles, which yields 270,116/(1 + 0.1 +
2.256) = 80,488 kJ/kmol. This enthalpy value corresponds to about 2400 K for N 2. Noting that the majority
of the moles are N2, TP will be close to 2400 K, but somewhat under it because of the higher specific heat
of H2O.
At 2300 K:
hH 2O  0.1hO 2  2.256 h N 2  198,199   0.179 ,316   2.256 75,676 
 276 ,856 kJ Higher tha n 270,116 kJ 
At 2250 K:
hH 2O  0.1hO 2  2.256 h N 2  195,562   0.177 ,397   2.256 73,856 
 269 ,921 kJ Lower than 270,116 kJ 
By interpolation,
TP = 2251.4 K
Chapter 15, Solution 73.
Acetylene gas is burned with 30 percent excess air during a steady-flow combustion process. The exit
temperature of product gases is to be determined.
Assumptions 1 Steady operating conditions exist. 2 Air and combustion gases are ideal gases. 3 Kinetic and
potential energies are negligible. 4 There are no work interactions.
Analysis The fuel is burned completely with the excess air, and thus the products will contain only CO2,
H2O, N2, and some free O2. Considering 1 kmol of C2H2, the combustion equation can be written as
C2H 2  1.3 ath O2  3.76N 2  
 2CO 2  H 2O  0.3 athO2  1.3 3.76 ath N 2
where ath is the stoichiometric coefficient and is determined
from the O2 balance,
C2H2
1.3ath  2  0.5  0.3ath

 ath  2.5
25C
Thus,
C2H2  3.25 O2  3.76N 2  
 2CO 2  H2O  0.75O 2  12.22N 2
Under steady-flow conditions the energy balance Ein  Eout  Esystem
applied on the combustion chamber with W = 0 reduces to
 Qout 
N P h f  h  h  P 
N R h f  h  h 
 
  
Air
75,000 kJ/kmol
Combustion
chamber
30% excess air
27C

R
Assuming the air and the combustion products to be ideal gases, we have h = h(T). From the tables,
Products
TP
hf
kJ/kmol
226,730
0
0
-241,820
-393,520
Substance
C2H2
O2
N2
H2O (g)
CO2
Thus,

h 298 K
h 300 K
kJ/kmol
--8682
8669
9904
9364
kJ/kmol
--8736
8723
-----
 
 8682   12 .22 0  hN
 75,000  2  393 ,520  hCO 2  9364  1  241,820  hH 2O  9904



 0.75  0  hO2
 8669  1226 ,730 
2
 3.25 0  8736  8682   12 .22 0  8723  8669 
It yields
2hCO2  hH2O  0.75hO2  12.22hN 2  1,321184
,
kJ
The temperature of the product gases is obtained from a trial and error solution. A first guess is obtained
by dividing the right-hand side of the equation by the total number of moles, which yields 1,321,184/(2 + 1
+ 0.75 + 12.22) = 82,729 kJ/kmol. This enthalpy value corresponds to about 2500 K for N 2. Noting that
the majority of the moles are N2, TP will be close to 2500 K, but somewhat under it because of the higher
specific heats of CO2 and H2O.
At 2350 K:
2hCO 2  hH 2O  0.75 hO 2  12 .22 h N 2  2122 ,091   1100 ,846   0.75 81,243   12 .22 77 ,496 
 1,352 ,961 kJ Higher tha n 1,321,184 kJ 
At 2300 K:
2hCO 2  hH 2O  0.75 hO 2  12 .22 h N 2  2119 ,035   198,199   0.75 79 ,316   12 .22 75,676 
 1,320 ,517 kJ Lower than 1,321,184 kJ 
By interpolation,
TP = 2301 K