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```									COMBUSTION

COMBUSTION EQUATIONS

Let us begin our review of this particular variety of chemical-
reaction equations by considering the combustion of propane in
a pure oxygen environment. The chemical reaction is
represented by:

C3H8 + 5 O2 → 3 CO2 + 4 H2O

Note that the number of moles of the elements on the left-hand
side may not equal the number of moles on the right-hand side.
However, the number of atoms of an element must remain the
same before, after, and during a chemical reaction; this
demands that the mass of each element be conserved during
combustion.
In writing the equation we have demonstrated some knowledge
of the products of the reaction. Unless otherwise stated we will
assume complete combustion: the products of the
combustion of a hydrocarbon fuel will be H2O and CO2.
Incomplete combustion results in products that contain H2,
CO, C, and/or OH.
For a simple chemical reaction, such as the combustion of
propane, we can immediately write down a balanced chemical
equation. For more complex reactions the following systematic
method proves useful:
1. Set the number of moles of fuel equal to 1.
2. Balance CO2 with number of C from the fuel.
3. Balance H2O with H from the fuel.
4. Balance 02 from CO2 and H2O.
For the combustion of propane we assumed that the process
occurred in a pure oxygen environment. Actually, such a
combustion process would normally occur in air. For our
purposes we assume that air consists of 21% 02 and 79% N2 by
volume so that for each mole of 02 in a reaction we will have:
79        mol N 2
 3.76
21        mol O 2

Thus, on the (simplistic) assumption that N2 will not undergo
any chemical reaction, the chemical reaction of the combustion
of propane is replaced by:

C3H8 + 5(O2 + 3.76N2) → 3CO2 + 4H2O + 18.8N2
The minimum amount of air that supplies sufficient 02, for the
complete combustion of the fuel is called theoretical air or
stoichiometric air. When complete combustion is achieved
with theoretical air, the products contain no O2, as in the our
reaction. In practice, it is found that if complete combustion is to
occur, air must be supplied in an amount greater than
theoretical air. This is due to the chemical kinetics and
molecular activity of the reactants and products. Thus we often
speak in terms of percent theoretical air or percent excess
air, where:
% theoretical air = 100% + % excess air
Slightly insufficient air results in CO being formed; some
hydrocarbons may result from larger deficiencies.
The parameter that relates the amount of air used in a
combustion process is the air-fuel ratio (AF),which is the ratio
of the mass of air to the mass of fuel. The reciprocal is the fuel-
air ratio (FA). Thus:
m                                m fuel
AF  air                         FA 
m fuel                            mair

Again, considering propane combustion with theoretical air, the
air-fuel ratio is:
m      5  4.76  29         kg air
AF  air                 15.69
m fuel     1  44            kg fuel
where we have used the molecular weight of air as 29 kg/kmol
and that of propane as 44 kg/kmol. If, for the combustion of
propane, AF > 15.69, a lean mixture occurs; if AF < 15.69, a
rich mixture results.
The combustion of hydrocarbon fuels involves H2O in the
products of combustion. The calculation of the dew point of the
products is often of interest; it is the saturation temperature at
the partial pressure of the water vapour. If the temperature
drops below the dew point, the water vapour begins to
condense. The condensate usually contains corrosive
elements, and thus it is often important to ensure that the
temperature of the products does not fall below the dew point.

EXAMPLE 01
Butane is burned with dry air at an air-fuel ratio of 20. Calculate
( a ) the percent excess air, ( b )the volume percentage of CO2
in the products, and ( c ) the dew-point temperature of the
products.
The reaction equation for theoretical air is:
C4H10+ 6.5(O2 + 3.76N2) → 4CO2+ 5H2O + 24.44N2
( a ) The air-fuel ratio for theoretical air is:
m      6.5  4.76  29         kg air
AFth  air                    15.47
m fuel      1  58             kg fuel

This represents 100% theoretical air. The actual air-fuel ratio is
20. The excess air is then:
 AF  AFth              20  15.47 
% excess air   act         100%                100%  29.28%
   AFth                 15.47 

( b ) The reaction equation with 129.28% theoretical air is:

C4H10 + (6.5)(1.2928)(O2 + 3.76N2) → 4CO2 + 5H20 + 1.903O2 + 31.6N2

The volume percentage is obtained using the total moles in the
products of combustion. For CO2, we have:
 4 
% CO 2          100%  9.41%
 42.5 

( c ) To find the dew-point temperature of the products we need
the partial pressure of the water vapour. It is found using the
mole fraction to be:
 5 
pv  zH O  patm          100  11.76 kPa
2
 42.5 

where we have assumed an atmospheric pressure of 100 kPa.
Using Table C-2 (Properties of Saturated H2O – Pressure
Table), we find the dew-point temperature to be Td.p.= 49°C.

EXAMPLE 02
Butane is burned with 90% theoretical air. Calculate the volume
percentage of CO in the products and the air-fuel ratio. Assume
no hydrocarbons in the products.
For incomplete combustion we add CO to the products of
combustion. Using the reaction equation from Example 01, we
have:

C4H10 + (0.9)(6.5)(O2 + 3.76N2) → aCO2 + 5H2O + 22N2 + bCO
With atomic balances on the carbon and oxygen we find:
C:        4=a+b
O:     11.7 = 2a + 5 + b
a = 2.7 and b = 1.3
The volume percentage of CO is then:
1.3 
% CO     100%  4.19%
 31 

The air-fuel ratio is:
m    0.9  6.5  4.76  29         kg air
AF  air                          13.92
mfuel         1 58                 kgfuel
EXAMPLE 03
Butane is burned with dry air, and volumetric analysis of the
products on a dry basis (the water vapour is not measured)
gives: 11.0% CO2,1.0% CO, 3.5% O2, and 84.5% N2. Determine
the percent theoretical air.
The problem is solved assuming that there is 100 moles of dry
products. The chemical equation is:
aC4H10 + b(O2 + 3.76N2) → 11CO2 + 1CO + 3.5O2 + 84.5N2 + cH2O

We perform the following balances:
C:        4a = 11 + 1 :. a = 3
H:         10a = 2c :.      c = 15
0:         2b = 22 + 1 + 7 + c :.     b = 22.5
A balance on the nitrogen allows a check: 3.76b = 84.5, or b =
22.47. This is quite close, so the above values are acceptable.
Dividing through the chemical equation by the value of a so that
we have 1 mol fuel:
C4H10 + 7.5(O2 + 3.76N2) → 3.67CO2 + 0.33CO + 1.17O2 + 28.17N2 +
5H2O

Comparing this with the combustion equation of Example 01
using theoretical air, we find:
 7.5 
% theoretical air     100%  107.7%
 6.5 
EXAMPLE 04

Volumetric analysis of the products of combustion of an
unknown hydrocarbon, measured on a dry basis, gives 10.4%
CO2, 1.2% CO, 2.8% O2, and 85.6% N2. Determine the
composition of the hydrocarbon and the percent theoretical air.
The chemical equation for 100 mol dry products is:
CaHb + c(O2 + 3.76N2) → 10.4CO2 + 1.2CO + 2.8O2 + 85.6N2 + dH2O

Balancing each element:
C: a = 10.4 + 1.2 :. a = 11.6
N: 3.76c= 85.6 :. c = 22.8
0: 2c = 20.8 + 1.2 + 5.6 + d :. d = 18.9
H: b = 2d :. b = 37.9

The chemical formula for the fuel is C11.6H37.9. This could
represent a mixture of hydrocarbons, but it is not any species
listed in Appendix B, since the ratio of hydrogen atoms to
carbon atoms is 3.27 = 13/4.
To find the percent theoretical air we must have the chemical
equation using 100% theoretical air:
C11.6H37.9+ 21.08(O2 + 3.76N2) → 11.6CO2 + 18.95H2O + 79.26N2

Using the number of moles of air from the actual chemical
equation, we find:
 22.8 
% theoretical air           100%  108%
 21.08 
Solved Problems
12.1 Ethane (C,H,) is burned with dry air which contains 5 mol 0, for each mole of fuel.

a
Calculate (a) the percent of excess air, ( b ) the air-fuel ratio, and ( c ) the dew-point temperature.

The stoichiometric equation is C,H, + 3.5(0, + 3.76N2) -, 2C0, + 3H20 + 6.58N2. The required
combustion equation is
C,H, + 5(0, + 3.76N2) -, 2C0, + 3H20 + 1.50, + 18.8N,
( a ) There is excess air since the actual reaction uses 5 mol 0, rather than 3.5 mol. The percent of
excess air is

(
-,.:*5)(100%) = 42.9%
% excess air =
CHAP.121 COMBUSTION 279
( b ) The air-fuel ratio is a mass ratio. Mass is found by multiplying the number of moles by the
molecular weight:
(c) The dew-point temperature is found using the partial pressure of the water vapor in the
combustion products. Assuming atmospheric pressure of 100 kPa, we find
P,= yHZOPa=tm(&)(loo)                     = 1.86    kPa
Using the Table C-2,we interpolate and find Td.p.4=9°C.
12.2 A fuel mixture of 60% methane, 30% ethane, and 10% propane by volume is burned with
stoichiometric air. Calculate the volume flow rate of air required if the fuel mass is 12 lbm/h
assuming the air to be at 70°F and 14.7 psia.
The reaction equation, assuming 1 mol fuel, is
0.6CH4 + 0.3C2H, + O.lC,H,        +
a(0, + 3.76N2)4 bCO, + cH,O + dN,
We find a, b, c, and d by balancing the various elements as follows:
C: 0.6     + 0.6 + 0.3       =b   :. b = 1.5
H: 2.4       + 1.8 + 0.8 = 2c         :. c = 2.5
0: 2a     = 2b + c :. a = 2.75
N: (2X3.76a)= 2d :. d = 10.34
The air-fuel ratio is
AF = (0.6)(16) + (0.3)(30) + (0.1)(44)
(2.75)(4.76)(29) 379.6
23 =-= lbm air
16S -lbm fuel
and hai=r(AF)m,,, = (16.5)(12) = 198 lbm/h. To find the volume flow rate we need the air density.
It is
whence
(The volume flow rate is usually given in ft3/min (cfm).)
12.3 Butane (C,H,,) is burned with 20°C air at 70% relative humidity. The air-fuel ratio is 20.
Calculate the dew-point temperature of the products. Compare with Example 12.1.
The reaction equation using dry air (the water vapor in the air does not react, but simply tags
along, it will be included later) is
C,H,, + "(0,+ 3.76N2)+ 4C0, + 5H,O + 6 0 , + cN,
The air-fuel ratio of 20 allows us to calculate the constant a, using Mhe=, 58 kg/kmol, as follows:
A F = -mdryair
- (a)(4.76)(29) = 2o :. a = 8.403
mtiel ( 1) ( 5 8 )
We also find that b = 1.903 and c = 31.6.The partial pressure of the moisture in the 20°C air is
P,.= \$Pg= (0.7)(2.338) = 1.637kPa
280 COMBUSTION [CHAP.12
The ratio of the partial pressure to the total pressure (100 kPa) equals the mole ratio, so that

N, p, = N P = [ (8.403)(4.76) + Nu](      E) or Nu = 0.666 kmol/kmol fuel
We simply add N,. to each side of the reaction equation:
C,H,, + 8.403(02+ 3.76N2) + 0.666H20 4C0, + 5.666H20 + 1.9030, + 31.6N2
-+

The partial pressure of water vapor in the products is P, = = (100)(5.666/43.17) = 13.1 kPa.
From Table C-2 we find the dew-point temperature to be Td,p.=51 "C, which compares with 49 "C using
dry air as in Example 12.1. Obviously the moisture in the combustion air does not significantly
influence the products. Consequently, we usually neglect the moisture.
12.4 Methane is burned with dry air, and volumetric analysis of the products on a dry basis gives
10% CO,, 1% CO, 1.8% O,, and 87.2% N,. Calculate ( a )the air-fuel ratio, ( b )the percent
excess air, and ( c )the percentage of water vapor that condenses if the products are cooled to
30 "C.
Assume 100 mol dry products. The reaction equation is
uCH, + b ( 0 , + 3.76N2) 10C0, + CO
-+              +
1.80, + 87.2N2 + c H 2 0
A balance on the atomic masses provides the following:
c: a = 1 0 + 1 :. a = 11
H: 4a = 2 c :. c = 22
0: 2 b = 2 0 + 1 + 3 . 6 + ~ :. b = 23.3
Dividing the reaction equation by a so that we have 1 mol fuel:
CH, + 2.12(0, + 3.76N2) 0.909C0, + 0.091CO + 0.1640,
-+                                  +   7.93N2 + 2H,O
( a ) The air-fuel ratio is calculated from the reaction equation to be
( b ) The stoichiometric reaction is CH, + 2(0, + 3.76N2) -+ CO, + 2H,O + 7.52N2. This gives the
excess air as

% excess air =
~                ( 2*1:-   )(100%) = 6%
( c ) There are 2 mol water vapor in the combustion products before condensation. If N, represents
moles of water vapor that condense when the products reach 30 "C, then 2 - N, is the number of
water vapor moles and 11.09 - N, is the total number of moles in the combustion products at
30°C. We find N, as follows:
N, - p, 2 - N, - 4.246 .. N P 11.09 - N, - 1oO       N, = 1.597 mol H 2 0
The percentage of water vapor that condenses out is

% condensate =    ( -1*77) (         =      1 ~ ~)
79.     8%
12.5 An unknown hydrocarbon fuel combusts with dry air; the resulting products have the

a         following dry volumetric analysis: 12% CO,, 1.5% CO, 3% 0,, and 83.5% N,. Calculate the percent
excess air.
The reaction equation for 100 mol dry products is
C,H, +  c ( 0 , + 3.76N2) -+ 12C0, + 1.5CO + 30, + 83.5N2 + dH,O
CHAP. 121 COMBUSTION 281
A balance on each element provides the following:
C: a = 12 + 1.5 :. a = 13.5
N: 3.76~= 83.5 :. c = 22.2
0: 2c = 24 + 1.5 + 6 + d :. d = 12.9
H: b = 2d :. b = 25.8
The fuel mixture is represented by Cl13.5Hz.8F.or theoretical air with this fuel, we have
C13.5Hz.8+ 19.95(0, + 3.76N2) 13.5c0, + 12.9H20 + 75.0N2
-+

Comparing this with the actual equation above, we find

% excess air =   ( 22-21;\$.95)(100%)      = 11.3%
12.6 Carbon reacts with oxygen to forni carbon dioxide in a steady-flow chamber. Calculate the
energy involved and state the type of reaction. Assume the reactants and products are at
25°C and 1 atm.
The reaction equation is C + 0, CO,. The first law and Table B-6 give
-+

Q      = Hp - HR =   C N,.(%)i - N,(h"f)i
prod react
-393 520) - 0 - 0 = -393 520 kJ/kmol
= (1)(
The reaction is exothermic (negative Q).
12.7 Methane enters a steady-flow combustion chamber at 77°F and 1 atm with 80% excess air
which is at 800 "R and 1atm. Calculate the heat transfer if the products leave at 1600OR and
1 atm.
The reaction equation with 180% theoretical air and with the water in vapor form is
CH, + 3.6(0, + 3.76N2) CO, + 2H,O(g) + 1.60, + 13.54N2
-+

The first law, with zero work, provides the heat transfer:
Q = C N , ( % + h - h " ) i - CN,(%+h-h")
prod react

= (1)( -169,300   +
15,829 - 4030) + (2)( -104,040 + 13,494 - 4258) + (1.6)(11,832 - 3725)
+(13.54)(11,410 - 3730) - (-32,210) - (3.6)(5602 - 3725) - (13.54)(5564 - 3730)
= -229,500 Btu/lbmol fuel
12.8 Ethane at 25°C is burned in a steady-flow combustion chamber with 20% excess air at
127"C, but only 95% of the carbon is converted to CO,. If the products leave at 1200 K,
calculate the heat transfer. The pressure remains constant at 1 atm.
The stoichiometric reaction equation is
C,H,     +
3.5(0,, + 3.76N2) 2C0, + 3H,O-+       ll.28N2  +
With 120% theoretical air and the product CO, the reaction equation becomes
C,H, + 4.2(0, + 3.76N2:)-+ 1.9C0, + 0.1CO + 3H,O + 0.750, + 11.28N2
The first law with zero work is Q = HP - HR.The enthalpy of the products is [see (12.9)]
Hp = (1.9)( -393 520 + 53 850 - 9360) + (0.1)( -110530 + 37 100 - 8670)
+ (3)( -241 820 + 44 380             -.   9900) + (0.75)(38 450 - 8680) + (11.28)(36 780 - 8670)
= -1   049 000 kJ/kmol fuel
-- -
282 COMBUSTION [CHAP.12
The enthalpy of the reactants is
HR = -84 680 + (4.2)( 11710 - 8680) + (15.79)( 11640 - 8670) = -25 060 kJ/kmol fuel
Then Q = -1049 000 - ( -25 060) = -1024 000 kJ/kmol fuel.
12.9 A rigid volume contains 0.2 lbm of propane gas and 0.8 lbm of oxygen at 77°F and 30 psia.
The propane burns completely, and the final temperature, after a period of time, is observed
to be 1600OR.Calculate ( a ) the final pressure and ( b )the heat transfer.
The moles of propane and oxygen are Npropa=n e0.2/44 = 0.004545 lbmol and NoWgen= 0.8/32
= 0.025 lbmol. For each mole of propane there is 0.025/0.004545 = 5.5 mol 0,.The reaction equation
for complete combustion is then
C3H8 + 5.502 + 3 c 0 , + 4H,O(g) + 0.502
( a ) We use the ideal-gas law to predict the final pressure. Since the volume remains constant, we
have
v =          1N ETl N2RT2 (6.5)( 537) - (7.5)( 1600) :. - P2= 103.1 psia
PI
( b ) By (12.11), with
Q
p 2 30 p2
= 1.986     Btu/lbmol- OR,we have for each mole of propane:
=   E&(% + h           -   h"   - RT)i -          E&.(%+ h - h"      -
prod react

= (3)[ -169,300   + 15,830 - 4030 - (1.986)(1600)]
+ (4)[ -104,040 + 13,490 - 4260 - (1.986)(1600)]
+ (0.5) [11,830 - 3720 - (1.986)( 1600)]
-(l)[ -44,680 - (1.986)(537)] - (5.5)[( -1.986)(537)]
= -819,900 Btu/lbmol fuel
Thus Q = ( -819,900)(0.004545)= 3730 Btu.
12.10 Propane is burned in a steady-flow combustion chamber with 80% theoretical air, both at

Solved Problems
12.1 Ethane (C,H,) is burned with dry air which contains 5 mol 0, for each mole of fuel.

a
Calculate (a) the percent of excess air, ( b ) the air-fuel ratio, and ( c ) the dew-point temperature.

The stoichiometric equation is C,H, + 3.5(0, + 3.76N2) -, 2C0, + 3H20 + 6.58N2. The required
combustion equation is
C,H, + 5(0, + 3.76N2) -, 2C0, + 3H20 + 1.50, + 18.8N,
( a ) There is excess air since the actual reaction uses 5 mol 0, rather than 3.5 mol. The percent of
excess air is

% excess air =   ( -,.:*5)(100%)   = 42.9%

CHAP.121 COMBUSTION 279
( b ) The air-fuel ratio is a mass ratio. Mass is found by multiplying the number of moles by the
molecular weight:
(c) The dew-point temperature is found using the partial pressure of the water vapor in the
combustion products. Assuming atmospheric pressure of 100 kPa, we find
P,= yHZOPa=tm(&)(loo)                = 1.86 kPa
Using the Table C-2,we interpolate and find Td.p.4=9°C.
12.2 A fuel mixture of 60% methane, 30% ethane, and 10% propane by volume is burned with
stoichiometric air. Calculate the volume flow rate of air required if the fuel mass is 12 lbm/h
assuming the air to be at 70°F and 14.7 psia.
The reaction equation, assuming 1 mol fuel, is
0.6CH4 + 0.3C2H, + O.lC,H,         +
a(0, + 3.76N2)4 bCO, + cH,O + dN,
We find a, b, c, and d by balancing the various elements as follows:
C: 0.6     + 0.6 + 0.3        =b   :. b = 1.5
H: 2.4       + 1.8 + 0.8 = 2c          :. c = 2.5
0: 2a     = 2b + c :. a = 2.75
N: (2X3.76a)= 2d :. d = 10.34
The air-fuel ratio is
AF = (0.6)(16) + (0.3)(30) + (0.1)(44)
(2.75)(4.76)(29) 379.6
23 =-= lbm air
16S -lbm fuel
and hai=r(AF)m,,, = (16.5)(12) = 198 lbm/h. To find the volume flow rate we need the air density.
It is
whence
(The volume flow rate is usually given in ft3/min (cfm).)
12.3 Butane (C,H,,) is burned with 20°C air at 70% relative humidity. The air-fuel ratio is 20.
Calculate the dew-point temperature of the products. Compare with Example 12.1.
The reaction equation using dry air (the water vapor in the air does not react, but simply tags
along, it will be included later) is
C,H,, + "(0,+ 3.76N2)+ 4C0,              + 5H,O + 6 0 , + cN,
The air-fuel ratio of 20 allows us to calculate the constant a, using Mhe=, 58 kg/kmol, as follows:
A F = -mdryair
- (a)(4.76)(29) = 2o :. a = 8.403
mtiel ( 1) ( 5 8 )
We also find that b = 1.903 and c = 31.6.The partial pressure of the moisture in the 20°C air is
P,.= \$Pg= (0.7)(2.338) = 1.637kPa
280 COMBUSTION [CHAP.12
The ratio of the partial pressure to the total pressure (100 kPa) equals the mole ratio, so that

N, p, = N P = [ (8.403)(4.76) + Nu](    E)   or Nu = 0.666 kmol/kmol fuel
We simply add N,. to each side of the reaction equation:
C,H,, + 8.403(02+ 3.76N2) + 0.666H20 4C0, + 5.666H20 + 1.9030, + 31.6N2
-+

The partial pressure of water vapor in the products is P, = = (100)(5.666/43.17) = 13.1 kPa.
From Table C-2 we find the dew-point temperature to be Td,p.=51 "C, which compares with 49 "C using
dry air as in Example 12.1. Obviously the moisture in the combustion air does not significantly
influence the products. Consequently, we usually neglect the moisture.
12.4 Methane is burned with dry air, and volumetric analysis of the products on a dry basis gives
10% CO,, 1% CO, 1.8% O,, and 87.2% N,. Calculate ( a )the air-fuel ratio, ( b )the percent
excess air, and ( c )the percentage of water vapor that condenses if the products are cooled to
30 "C.
Assume 100 mol dry products. The reaction equation is
uCH, + b ( 0 , + 3.76N2) 10C0, + CO
-+              +
1.80, + 87.2N2 + c H 2 0
A balance on the atomic masses provides the following:
c: a = 1 0 + 1 :. a = 11
H: 4a = 2 c :. c = 22
0: 2 b = 2 0 + 1 + 3 . 6 + ~ :. b = 23.3
Dividing the reaction equation by a so that we have 1 mol fuel:
CH, + 2.12(0, + 3.76N2) 0.909C0, + 0.091CO + 0.1640,
-+                                 +    7.93N2 + 2H,O
( a ) The air-fuel ratio is calculated from the reaction equation to be
( b ) The stoichiometric reaction is CH, + 2(0, + 3.76N2) -+ CO, + 2H,O + 7.52N2. This gives the
excess air as

~   % excess air =   ( 2*1:- ) (100%)   = 6%

( c ) There are 2 mol water vapor in the combustion products before condensation. If N, represents
moles of water vapor that condense when the products reach 30 "C, then 2 - N, is the number of
water vapor moles and 11.09 - N, is the total number of moles in the combustion products at
30°C. We find N, as follows:
N, - p, 2 - N, - 4.246 .. N P 11.09 - N, - 1oON, = 1.597 mol H 2 0
The percentage of water vapor that condenses out is

% condensate =       ( -1*77) (         =      1 ~ ~)
79.   8%
12.5 An unknown hydrocarbon fuel combusts with dry air; the resulting products have the

a
excess air.
following dry volumetric analysis: 12% CO,, 1.5% CO, 3% 0,, and 83.5% N,. Calculate the percent

The reaction equation for 100 mol dry products is
C,H,     +
c ( 0 , + 3.76N2) -+ 12C0, + 1.5CO + 30, + 83.5N2 + dH,O
CHAP. 121 COMBUSTION 281
A balance on each element provides the following:
C: a = 12 + 1.5 :. a = 13.5
N: 3.76~= 83.5 :. c = 22.2
0: 2c = 24 + 1.5 + 6 + d :. d = 12.9
H: b = 2d :. b = 25.8
The fuel mixture is represented by Cl13.5Hz.8F.or theoretical air with this fuel, we have
C13.5Hz.8+ 19.95(0, + 3.76N2) 13.5c0, + 12.9H20 + 75.0N2
-+

Comparing this with the actual equation above, we find

% excess air =   ( 22-21;\$.95)(100%)    = 11.3%

SUPPLEMENTARY PROBLEMS
1. The following fuels combine with stoichiometric air: (a) C2H4,
(b) C,H,, (c) C4H10,(d) C5H12, (e) C8H18, and (f) CH,OH.
Provide the correct values for x , y, z in the reaction equation:
C,H, + w ( 0 , + 3.76N2) -+ xC0, + yH,O + zN,

Am. ( a )2,2,11.28 ( b )3,3,16.92 (c)4,5,24.44 ( d )5,6,30.08
(e)8,9,47 (f)1,2,5.64
12.13 Methane (CH,) is burned with stoichiometric air and the products are cooled to 20°C assuming
complete combustion at 100 kPa. Calculate (a) the air-fuel ratio, (6) the percentage of CO2 by weight
of the products, (c) the dew-point temperature of the products, and ( d )the percentage of water vapor
condensed.
Am. ( a ) 17.23 ( b )15.14% ( c )59°C ( d )89.8%

12.14 Repeat Prob. 12.13 for ethane (CZH6).

Ans. (a) 16.09 ( b )17.24% ( c ) 55.9"C ( d )87.9%

12.15 Repeat Prob. 12.13 for propane (C3H8).
Am. ( a ) 15.67 ( 6 ) 18.07% ( c ) 53.1"C ( d ) 87.0%
12.16 Repeat Prob. 12.13 for butane (C4H10).
Ans. ( a ) 15.45 ( 6 )18.52% ( c ) 53.9"C ( d ) 86.4%
12.17 Repeat Prob. 12.13 for octane (c4H18).
Am. ( a ) 16.80 ( 6 )15.92% ( c ) 57.9"C ( d ) 89.2%
12.18 Ethane (C,H,) undergoes complete combustion at 95 kPa with 180% theoretical air. Find ( a ) the
air-fuel ratio, (6) the percentage of CO, by volume in the products, and ( c ) and dew-point temperature.
Ans. ( a ) 28.96 ( 6 )6.35% ( c ) 43.8"C
12.19 Repeat Prob. 12.18 for propane (C3H8). Ans. ( a ) 28.21 ( b )6.69% ( c ) 42.5"C
12.20 Repeat Prob. 12.18 for butane (C4H10). Am. ( a ) 27.82 ( 6 )6.87% ( c ) 41.8"C
12.21 Repeat Prob. 12.18 for octane (C5Hl8). Ans. ( a ) 30.23 ( 6 ) 10.48% ( c ) 45.7"C
12.22 Calculate the mass flux of fuel required if the inlet air flow rate is 20 m3/min at 20°C and 100 kPa
using stoichiometric air with ( a ) methane (CH,), ( 6 )ethane (C,H,), ( c ) propane (C3H8), ( d ) butane
(C4H10), and ( e )octane (c5H18).
Am. ( a ) 1.38 kg/min ( 6 ) 1.478 kg/min ( c ) 1.518 kg/min ( d ) 1.539 kg/min
( e ) 1.415 kg/min
12.23 Propane (C3H8) undergoes complete combustion at 90 kPa and 20°C with 130% theoretical air.
Calculate the air-fuel ratio and the dew-point temperature if the relative humidity of the combustion air
is ( a ) 90%, ( 6 )80%, ( c ) 60%, and ( d ) 40%.
Ans. ( a ) 20.67, 50.5"C ( 6 )20.64, 50.2"C ( c ) 20.57, 49.5"C ( d ) 20.50, 48.9"C
12.24 An air-fuel ratio of 25 is used in an engine that burns octane (C8HI8). Find the percentage of excess air
required and the percentage of CO, by volume in the products. Am. 165.4%, 7.78%
12.25 Butane (C4Hlo) is burned with 50% excess air. If 5% of the carbon in the fuel is converted to CO,
calculate the air-fuel ratio and the dew-point of the products. Combustion takes place at 100 kPa.
Am. 23.18, 46.2"C
12.26 A fuel which is 60% ethane and 40% octane by volume undergoes complete combustion with 200%
theoretical air. Find ( a ) the air-fuel ratio, ( 6 )the percent by volume of N , in the products, and ( c ) the
dew-point temperature of the products if the pressure is 98 kPa.
Ans. ( a ) 30.8 (6) 76.0% ( c ) 40.3"C
12.27 One lbm of butane, 2 lbm of methane, and 2 lbm of octane undergo complete combustion with 20 lbm
of air. Calculate ( a ) the air-fuel ratio, (6) the percent excess air, and ( c ) the dew-point temperature of
the products if the combustion process occurs at 14.7 psia.
Ans. ( a ) 19.04 ( 6 ) 118.7% ( c ) 127°F
12.28 Each minute 1 kg of methane, 2 kg of butane, and 2 kg of octane undergo complete combustion with
stoichiometric 20°C air. Calculate the flow rate of air required if the process takes place at 100 kPa.
Ans. 65.92 m3/min
12.29 A volumetric analysis of the products of butane (C4H1o) on a dry basis yields 7.6% CO,, 8.2% O,,

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