Me 312 module 1

MODULE 1
COMBUSTION ENGINEERING
By: Engr. YURI G. MELLIZA
Combustion
Combustion occurs when fossil fuels, such as natural gas, fuel oil, coal, or gasoline, react with oxygen in the air to produce
heat. The heat from burning fossil fuels is used for industrial processes, environmental heating or to expand gases in a cylinder
and push a piston. Boilers, furnaces, and engines are important users of fossil fuels.
Fossil fuels are hydrocarbons, meaning they are composed primarily of carbon and hydrogen. When fossil fuels are burned,
carbon dioxide (CO2) and water (H2O) are the principle chemical product, formed from the reactants carbon and hydrogen in
the fuel and oxygen (O2) in the air.
Fuel: A substance containing combustible elements which in rapid chemical union with oxygen produced combustion.
TYPES OF FUEL
1) Solid Fuels
Example:
a. coal
b. charcoal
c. coke
d. woods
2) Liquid Fuels (obtained by the distillation of petroleum)
Example:
a. Gasoline
b. kerosene
c. diesel
d. Fuel oil
e. alcohol (these are not true hydrocarbons, since it contains oxygen in the molecule)
3) Gaseous Fuels (a mixture of various constituent’s hydrocarbons, its combustion products do not
have sulfur components)
Example:
a. Natural Gas (example: methane, ethane, propane)
b. Coke oven gas -obtained as a byproduct of making coke
c. Blast furnace gas - a byproduct of melting iron ore
d. LPG
e. Producer Gas - fuel used for gas engines
4) Nuclear Fuels
Example:
a. Uranium
b. Plutonium
COMBUSTIBLE ELEMENTS
1. Carbon (C)
2. Hydrogen (H2)
3. Sulfur (S)
TYPES OF HYDROCARBONS
1) Paraffin - all ends in "ane"
Formula: CnH2n+2
Structure: Chain (saturated)
Example:
GAS
a. Methane(CH4)
b. Ethane (C2H6)
LPG
a. Propane (C3H8)
b. Butane (C4H10)
c. Pentane (C5H12)

GASOLINE
a. n-Heptane (C7H16)
b. Triptane (C7H16)
c. Iso- octane (C8H18)
FUEL OIL
a. Decane (C10H22)
b. Dodecane (C12H26)
c. Hexadecane (C16H34)
d. Octadecane (C18H38)
2) Olefins - ends in "ylene" or "ene"
Formula: CnH2n
Structure: Chain (unsaturated)
Example:
a. Propene (C3H6)
b. Butene (C4H8)
c. Hexene (C6H12)
d. Octene (C8H16)
3) DIOLEFIN - ends in "diene"
Formula: CnH2n-2
Structure: Chain (unsaturated)
Example:
a. Butadiene (C4H6)
b. Hexadiene (C6H10)
4) NAPHTHENE - named by adding the prefix "cyclo"
Formula: CnH2n
Structure: Ring (saturated)
Example:
a. Cyclopentane (C5H10)
b. Cyclohexane (C6H12)
5) AROMATICS - this hydrocarbon includes the;
A. Benzene Series (CnH2n-6)
B. Naphthalene Series (CnH2n-12)
Structure: Ring (unsaturated)
Example:
a. Benzene (C6H6)
b. Toluene (C7H8)
c. Xylene (C8H10)
6) ALCOHOLS - These are not true hydrocarbon, but sometimes used as fuel in an internal combustion engine. The
characteristic feature is that one of the hydrogen atom is replaced by an OH radical.
Example:
a. Methanol (CH4O or CH3OH)
b. Ethanol (C2H6O or C2H5OH)
Covalent Bond: A molecular bond that involves the sharing of electron pairs between atoms. These electron pairs are known
as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share
electrons, is known as covalent bonding.
Ring Structure (Heterocyclic Compounds): It is a cyclic compounds that has atoms of at least two different elements as
members of its ring(s). Heterocyclic chemistry is the branch of organic chemistry dealing with the synthesis, properties, and
applications of these heterocycles.
Chain Structure: A crystalline structure in which forces between atoms in one direction are greater than those in other
directions, so that the atoms are concentrated in chains.

Saturated Hydrocarbon – hydrocarbon that contain only single bonds between carbon atoms. They are the simplest class of
hydrocarbons. They are called saturated because each carbon atom is bonded to as many hydrogen atoms as possible. All the
carbon atoms are joined by a single bond.
Unsaturated Hydrocarbon – are hydrocarbons that have double or triple covalent bonds between adjacent carbon atoms.
Those with at least one carbon – to carbon double bond are called “alkenes” and those with at least one carbon – to – carbon
triple bond are called “alkynes”. (it has two or more carbon atoms joined by a double or triple bond)
Isomers – each of two or more compounds with the same formula but at different arrangement of atoms in the molecule and
different properties.( two hydrocarbons with the same number of carbon and hydrogen atoms, but at different structure)
STRUCTURE OF CnHm
Complete Combustion: Occurs when all the combustible elements has been fully oxidized.
2
2 CO
O
C →
+
Incomplete Combustion: Occurs when some of the combustible elements have not been fully oxidized and it may result from;
a. Insufficient oxygen
b. Poor mixing of fuel and oxygen
c. the temperature is too low to support combustion.
Result: Soot or black smoke that sometimes pours out from chimney or smokestack.

1
2
2
C O CO
+ →
THE COMBUSTION CHEMISTRY
A. Oxidation of Carbon
2 2
2
C O CO
Mole Basis
1 1 1
Mass Basis
1(12) 1(32) 1(44)
12 32 44
3 8 11
kgO 8
kgC 3
+ →
+ →
+ →
+ →
+ →
=
2
kg CO 11
kg C 3
=
2
kg C 3
kg CO 11
=
B. Oxidation of Hydrogen
1
2 2 2
2
1
2
1
2
2
2
H O H O
Mole Basis
1 1
Mass Basis
1(2) (32) 1(18)
1 8 9
kgO 8
kgH 1
+ →
+ →
+ →
+ →
=
2
2
kg H O 9
kg H 1
=

C. Oxidation of Sulfur
2 2
2
S O SO
Mole Basis
1 1 1
Mass Basis
1(32) 1(32) 1(64)
1 1 2
kgO 1
kgS 1
+ →
+ →
+ →
+ →
=
2
kg SO 2
kg S 1
=
Combustion of Carbon with “Incomplete Combustion”
1
C O2 CO
2
Mole Basis
1
1 1
2
Mass Basis
1
1(12) (32) 1(28)
2
12 16 28
3 4 7
kg C 3
kg CO 7
+ →
+ →
+ →
+ →
+ →
=

Actual Composition of Air
• Oxygen. The most important gas in the composition is oxygen. ...
• Nitrogen. To balance out oxygen, there is Nitrogen. ...
• Argon. ...
• Carbon dioxide. ...
• Water vapor. ...
• Other Particles
Gases % By Volume
Nitrogen 78.09
Oxygen 20.95
Argon 0.93
Carbon Dioxide 0.04
Water Vapor 1 (0.4% over the entire atmosphere)
Other Particles
In theoretical Combustion, the composition of atmospheric air is considered as a mixture of O2 and N2 only.
a) Volumetric or Molal analysis
O2 = 21%
N2 = 79%
b) Gravimetric Analysis
O2 = 23.3%
N2 = 76.7%
Molal or Volumetric ratio of N2 to O2 in air
76
.
3
21
79
O
of
Mole
N
of
Moles
2
2
=
=
COMBUSTION WITH AIR and Theoretical air requirement
Fuel Air Products
+ →
A) Combustion of Carbon with air
2 2 2 2
C + O + 3.76N CO + 3.76N
Mole Basis
1+ (1+ 3.76) 1+ 3.76
Mass Basis
1(12) +1(32) + 3.76(28) 1(44) + 3.76(28)
kg Air 32 + 3.76(28)
= = 11.44
kg C 12
→
→
→

B) Combustion of Hydrogen with Air
( ) ( )
( ) ( )
( ) ( )
1 1 1
2 2 2
2 2 2 2 2
1 1 1
2 2 2
1 1 1
2 2 2
2
H O 3.76 N H O 3.76 N
Mole Basis
1 3.76 1 3.76
Mass Basis
1(2) (32) 3.76 (28) 1(18) 3.76 (28)
2 16 (3.76)14 18 3.76(14)
kg Air 16 3.76(14)
34.32
kg H 2
+ + → +
+ + → +
+ + → +
+ + → +
+
= =
C. Combustion of Sulfur with air
( ) ( )
2 2 2 2
S + O + 3.76 N SO + 3.76 N
Mole Basis
1+1+ 3.76 1+ 3.76
Mass Basis
1(32) +1(32) + 3.76(28) 1(64) + 3.76(28)
kg Air 32 + 3.76(28)
= = 4.29
kg S 32
→
→
→
Minimum air requirement for each kg of combustible
element
Combustible
Elements
Air kg
Carbon 11.44 Kg
Hydrogen 34.32 Kg
Sulfur 4.29 kg

t
A Mass of Air
=
F Mass of Fuel
 
 
 
( )
A t
Actual Air - Fuel Ratio
A A
= 1+ e
F F
Where
e - excess air in decimal
   
   
   
THEORETICAL AIR
The minimum amount of air that supplies sufficient oxygen for the complete combustion of all the carbon, hydrogen, and sulfur
present in the fuel is called the theoretical amount of air. (With theoretical air, no O2 is found in the product)
EXCESS AIR
It is an amount of air more than the theoretical air required to influence complete combustion. With excess air, O2 is present in
the products. Excess air is usually expressed as a percentage of the theoretical air. But in actual combustion, although there is
an amount of excess air, the presence of CO and other emission gases in the products cannot be avoided.
Example: 25% excess air is the same as 125% theoretical air.
COMBUSTION OF HYDROCARBON FUEL(CnHm)
Combustion of CnHm with 100% theoretical air (No emission gases and with complete combustion)
n m 2 2 2 2 2
a. With100% TA
C H + aO + a(3.76)N bCO + cH O + a(3.76)N
a = n + 0.25m
b = n
c = 0.5m
→
n m 2 2 2 2 2 2
b. With excess air e
C H + (1+ e)aO + (1+ e)a(3.76)N bCO + cH O + dO + (1+ e)a(3.76)N
a = n + 0.25m
b = n
c = 0.5m
d = e(n + 0.25m)
→
Note: The values of a,b,c, and d above in terms of n and m is applicable only for the combustion of one type of hydrocarbon.

n m
Theoretical
n m
Actual
A 137.28(n + 0.25m) kg Air
=
F 12n + m kg C H
A 137.28(n + 0.25m) kg Air
= (1+ e)
F 12n + m kg C H
 
 
 
   
   
   
EQUIVALENCE RATIO
( )
( )
( )
( )
Actual Theoretical
Stoichiometric Actual
F A
A F
ER
F A
A F
= =
Stoichiometric (or chemically correct) mixture of air and fuel, is one that
contains just sufficient oxygen for complete combustion of the fuel
A “Weak Mixture”, is one which has an excess of air
A “Rich Mixture”, is one which has a deficiency of air
( ) ( )
( ) l
Theoretica
l
Theoretica
Actual
F
A
F
A
F
A
Air
Excess
Percentage
−
=

DEW POINT TEMPERATURE
The Dew Point Temperature (tdp) is the saturation temperature corresponding the partial pressure of the water vapor in the
mixture (products of combustion).
ULTIMATE ANALYSIS
Ultimate Analysis gives the amount of C, H2, O2, N2, S and moisture in percentages by mass, sometimes the percentage amount
of Ash is given.
Air – Fuel Ratio with known Ultimate Analysis
fuel
of
kg
air
of
kg
S
29
.
4
8
O
H
32
.
34
C
44
.
11
F
A 2
2
t
+






−
+
=






where: C, H, O and S are in decimals obtained from the Ultimate Analysis (on an ashless basis)
PROXIMATE ANALYSIS
Proximate Analysis gives the percentage amount of Fixed Carbon, Volatiles, Ash and Moisture.
Moisture
%
Ash
%
Volatiles
%
FC
%
%
100 +
+
+
=
ORSAT ANALYSIS
Orsat Analysis gives the volumetric or molal analysis of the products of combustion or exhaust gases on a Dry Basis.
...
CH
%
NO
%
CO
%
N
%
O
%
CO
%
%
100 x
2
2
2 +
+
+
+
+
+
=

MASS FLOW RATE OF FLUE GAS
a) Without considering Ash loss:
gas Fuel
A
m m 1
F
 
= +
 
 
b) Considering Ash loss
gas Fuel
A
m m 1- Ash Loss
F
 
= +
 
 
where ash loss in decimal
• Combustion equation with CO in the products due to incomplete combustion (100% theoretical air)
2
2
2
2
2 N
)
76
.
3
(
a
dCO
O
cH
bCO
N
)
76
.
3
(
a
aO
CnHm +
+
+
→
+
+
• Combustion equation with CO in the products due to incomplete combustion (with excess air)
2
2
2
2
2
2 N
)
76
.
3
(
a
)
e
1
(
fO
dCO
O
cH
bCO
N
)
76
.
3
(
a
)
e
1
(
aO
)
e
1
(
CnHm +
+
+
+
+
→
+
+
+
+
• Typical Real-World Engine Combustion Process:
Oxides
Nitrogen
-
NO
Monoxide
Carbon
-
CO
Compounds
Organic
Volatile
-
s)
CH(VOC'
NOx
CO
s)
CH(VOC'
N
O
O
H
CO
)
N
and
(O
Air
(CnHm)
Fuel
x
2
2
2
2
2
2 +
+
+
+
+
+
→
+
EMISSIONS
Emissions are any kind of substance released into the air from natural or human sources — flows of gases, liquid droplets or
solid particles. Not all emissions become air pollutants, but many do, causing significant health and environmental problems.
The amount of air pollutants in an area depends on the number
and size of emission sources, along with the weather and lay of the land.
The main sources of emissions are:
❖ Point Sources
Point sources are stationary industrial facilities such as pulp and paper mills and factories that burn fossil fuels. They operate
under ministry authorization (a regulation, permit, approval, or code of conduct), or under an air-discharge permit issued by
Philippine Govt.
❖ Area Sources
Area sources are stationary sources that are not normally required to obtain a discharge permit from the ministry. They include
prescribed burning, residential wood use, light industry, and other residential, commercial, and institutional sources. Emissions
from most of these area sources individually are small compared to point sources but can be significant when considered
collectively.

❖ Mobile Sources
Mobile sources include motor vehicles mainly involved in the transportation of people and goods (e.g., passenger cars, trucks,
and motorcycles), aircraft, marine vessels, trains, off-road vehicles, and small off-road engines (e.g., agricultural, lawn/garden,
construction and recreational equipment).
❖ Natural Sources
Natural sources of emissions occur in nature without the influence of human beings, such as wildfires, plants, wildlife, and
marine aerosol.
Pollutants:
Air pollutants are any gas, liquid or solid substance that have been emitted into the atmosphere and are in high enough
concentrations to be considered harmful to the environment, or human, animal and plant health.
Pollutants emitted directly into the air are called "primary pollutants." "Secondary" pollutants" are formed in the air when they
react with other pollutants. Ground-level ozone is an example of a secondary pollutant that forms when nitrogen oxides (NOx)
and volatile organic compounds (VOCs) react in the presence of sunlight.
We come in contact with many kinds of air pollutants every day. Depending on the type and amount emitted, these pollutants
may affect air quality at the local, regional, and/or global scale. For example, smoke from woodstoves or backyard burning,
and motor vehicle exhaust are pollutant mixtures that affect air quality in our neighborhoods and communities, and inside our
homes. Smoke from forest fires or ground-level ozone can cover an entire region. Long-lasting pollutants can contribute to
serious global problems, such as ozone depletion and climate change.
An air pollutant can become dangerous to our health when we are exposed to it for a long time, and also when we breathe in a
large amount of it. Health effects can last for a short while (e.g., coughing) or become a long-term problem (e.g., lung and heart
disease, cancer). Pollution can also cause death. The young, the elderly and those with pre-existing heart or lung disease are
the most sensitive to the effects of air pollution.
Common Pollutants
Air pollutants can be visible (e.g., the brownish-yellow colour of smog) or invisible. Besides affecting human health and the
environment, air pollutants can also hamper our ability to see very far (visibility).
Air pollution can have local and regional impacts — such as ground-level ozone and wood smoke. It can also have wide-
reaching, global effects — such as climate change and depletion of the ozone layer.
Health effects from local air pollution can last for a short while (e.g., coughing) or become a long-term problem (e.g., lung and
heart disease, cancer). Pollution can also cause death. An air pollutant can become dangerous to our health when we are exposed
to it for a long time, as well as when we breathe in a large amount of it.
The Most Significant Air Pollutants
The air pollutants that pose the most serious local threat to our health are particulate matter and ground-level ozone — the key
ingredients of smog. They mainly affect the lowest part of the atmosphere, which holds the air we breathe. Particulate matter
is a significant problem in rural areas, as well, due to wood burning.
Particulate Matter (PM)
Particulate matter refers to tiny solid or liquid particles that float in the air. Some particles are large or dark enough to be seen
as smoke, soot or dust. Others are so small that they can only be detected with a powerful, electron microscope. PM occurs in
two forms: primary and secondary.
• Primary PM is emitted directly into the atmosphere by wood burning (e.g., in wood stoves, open burning, wood stoves)
and fossil fuel burning (e.g., in motor vehicles, oil/gas furnaces and industry). Primary PM also includes pollen, spores
and road dust.
• Secondary PM is formed in the atmosphere through chemical reactions involving nitrogen dioxide, sulphur dioxide,
volatile organic compounds and ammonia.

We measure particulate matter in microns (micrometres). One micron is a millionth of a metre. Particulate matter between 10
and 2.5 microns in diameter or less is called PM10. That’s about seven times smaller than the width of a human hair. It is
invisible to the naked eye and small enough to inhaled into our nose and throat.
Particulate matter that’s 2.5 microns and less is called PM2.5. This is the particulate matter of greatest concern because it can
travel deep into the lungs and become lodged there, causing heart and lung disease, and premature death. Fine particles that
comprise PM2.5 are also efficient at scattering light, resulting in a degradation in visibility.
Ground-Level Ozone (O3)
Ground-level ozone is formed by the reaction of two types of chemicals — volatile organic compounds and nitrogen oxide —
in the presence of sunshine and warm temperatures. When the air is still (stagnant), the ozone will build up.
Ground-level ozone usually occurs in the warmer months of the year. Ground-level ozone collects over urban areas that produce
large amounts of VOCs and NOx. Rural areas can be affected, too, though. That’s because the ozone can travel up to several
hundred kilometres away, carried by the wind.
Low concentrations of ground-level ozone can irritate the eyes, nose and throat. Ozone can also irritate the lung airways, and
make them red and swollen (inflammation). People with lung problems are most at risk, but even healthy people who are active
outdoors can be affected when ozone levels are high.
EXHAUST POLLUTANTS
HYDROCARBONS (HC): Hydrocarbon emissions result when fuel molecules in the engine do not burn or burn only partially.
Hydrocarbons react in the presence of nitrogen oxides and sunlight to form ground-level ozone, a major component of smog.
Ozone can irritate the eyes, damage lungs, and aggravate respiratory problems. It is our most widespread urban air pollution
problem. Some kinds of exhaust hydrocarbons are also toxic, with the potential to cause cancer.
NITROGEN OXIDES (NOx): Under the high pressure and high temperature conditions in an engine, nitrogen and oxygen
atoms in the air we breathe react to form various nitrogen oxides, collectively known as NOx. Nitrogen oxides, like
hydrocarbons, are precursors to the formation of ozone. They also contribute to the formation of acid rain.
CARBON MONOXIDE (CO): Carbon monoxide is a product of incomplete combustion and occurs when carbon in the fuel
is partially oxidized rather than fully oxidized to carbon dioxide. Carbon monoxide reduces the flow of oxygen in the
bloodstream and is particularly dangerous to persons with heart disease.
CARBON DIOXIDE (CO2): Carbon dioxide does not directly impair human health, but it is considered a “greenhouse gas”.
In other words, as it accumulates in the atmosphere, it is believed to trap the earth’s heat and contribute to the potential for
climate change.
Evaporative Emissions
HYDROCARBONS: Hydrocarbons also escape into the air through fuel evaporation. With today’s efficient exhaust emission
controls and today’s clean burning gasoline formulations, evaporative losses can account for much of the total hydrocarbon
pollution from current model cars on hot days when ozone levels are highest. Evaporative emissions occur from fuel.
OTHER KINDS OF AIR POLLUTANTS
There are many more air pollutants than particulate matter and ground-level ozone. They are usually grouped into four
categories, as shown in the table below.
Pollutant Category Types of Pollutants
Common Air Contaminants(CACs)
(also known as "criteria air contaminants") particulate matter (PM), sulphur oxides (SOx), nitrogen oxides (NOx), volatile
organic compounds (VOCs), carbon monoxide (CO) and ammonia (NH3).
Ground-level ozone (O3) is often included with CACs because it is a byproduct of CAC interactions.

Persistent Organic Pollutants(POPs)
e.g., dioxins and furans
Heavy Metals
e.g., mercury
Air Toxics
e.g., benzene, polycyclic aromatic hydrocarbons(PAHs)
Table of Common Pollutants
Not included here are the pollutants that influence the larger atmosphere, causing global environmental problems: stratospheric
ozone depletion and global climate change.
MAIN (COMMON) POLLUTANTS
Pollutant Description and Sources Health Impact Environment
Particulate Matter (PM)
Dust, soot, and tiny bits of solid material. PM10 — Particles smaller than 10µm (microns) in diameter.
Far too small to see — 1/8th the width of a human hair. • Road dust; road construction
• Mixing and applying fertilizers/ pesticides
• Forest fires
• Coarse particles irritate the nose and throat, but do not normally penetrate deep into the lungs. • PM is the main source of
haze that reduces visibility.
• It takes hours to days for PM10 to settle out of the air.
• Because they are so small, PM2.5 stays in the air much longer than PM10, taking days to weeks to be removed.
• PM can make lakes and other sensitive areas more acidic, causing changes to the nutrient balance and harming aquatic life.
PM2.5–Particles smaller than 2.5µm in diameter • Combustion of fossil fuels and wood (motor vehicles, woodstoves and
fireplaces)
• Industrial activity
• Garbage incineration
• Agricultural burning • Fine particles are small enough to make their way deep into the lungs. They are associated with all
sorts of health problems — from a runny nose and coughing, to bronchitis, asthma, emphysema, pneumonia, heart disease, and
even premature death.
• PM2.5 is the worst public health problem from air pollution in the province. (Research indicates the number of hospital visits
increases on days with increased PM levels).
Ground level Ozone (O3)
Bluish gas with a pungent odour • At ground level, ozone is formed by chemical reactions between volatile organic
compounds (VOCs) and nitrogen dioxide (NO2) in the presence of sunlight.

• VOCs and NO2 are released by burning coal, gasoline, and other fuels, and naturally by plants and trees.
• Exposure for 6-7 hours, even at low concentrations, significantly reduces lung function and causes respiratory
inflammation in healthy people during periods of moderate exercise. Can be accompanied by symptoms such as chest pain,
coughing, nausea, and pulmonary congestion. Impacts on individuals with pre-existing heart or respiratory conditions can be
very serious.
• Ozone exposure can contribute to asthma, and reduced resistance to colds and other infections. • Ozone can damage plants
and trees, leading to reduced yields.
• Leads to lung and respiratory damage in animals.
• Ozone can also be good: the ozone layer above the earth (the stratosphere) protects us from harmful ultraviolet rays.
Other Pollutants • sulfur dioxide (SO2)
• carbon monoxide (CO)
• nitrogen dioxide (NO2)
• total reduced sulphur (TRS)
• volatile organic compounds (VOCs)
• persistent organic pollutants (POPs)
• lead (Pb)
• polycyclic aromatic hydrocarbons (PAHs)
• dioxins and furans
Most of these pollutants come from combustion and industrial processes or the evaporation of paints and common chemical
products. • The health impacts of these pollutants are varied.
• Sulphur dioxide (SO2), for example, can transform in the atmosphere to sulphuric acid, a major component of acid rain.
• Carbon monoxide is fatal at high concentrations, and causes illness at lower concentrations.
• Dioxins and furans are among the most toxic chemicals in the world. • While some of these pollutants have local impact on
the environment (e.g., lead) or are relatively short lived (NO2) some are long lived (POPs) and can travel the world on wind
currents in the upper atmosphere.

Combustion of Hydrocarbon Fuel with emission gases
a) With 100% theoretical air
0.5m
c
n
b
0.25m
n
a
where
N
)
76
.
3
(
a
O
cH
bCO
N
)
76
.
3
(
a
aO
H
C 2
2
2
2
2
m
n
=
=
+
=
+
+
→
+
+
b) With excess air e
0.25m)
n
(
e
d
where
N
)
76
.
3
(
a
)
e
1
(
dO
O
cH
bCO
N
)
76
.
3
(
a
)
e
1
(
aO
)
e
1
(
H
C 2
2
2
2
2
2
m
n
+
=
+
+
+
+
→
+
+
+
+
c) With exhaust pollutants (emission gases)
process
combustion
actual
apply to
not
may
m
and
n
of
in terms
d
and
c
b,
a,
of
values
above
The
:
Note
Oxides
Nitrogen
-
NOx
Monoxide
Carbon
-
CO
Compounds
Organic
Volatile
-
s)
CH(VOC'
NOx
i
CO
h
CH
g
N
f
O
d
O
H
c
CO
b
)
a(3.76)N
O
a
CnHm 2
2
2
2
2
2 +
+
+
+
+
+
→
+
+
Combustion of Solid Fuels
a) Combustion with 100% theoretical air
2
2
2
2
2
2
2
2
2
2 kN
jSO
O
iH
hCO
N
)
76
.
3
(
x
xO
O
gH
fS
dN
cO
bH
aC +
+
+
→
+
+
+
+
+
+
+
b) Combustion with Excess air e
mN
LO
jSO
O
iH
hCO
N
)
76
.
3
(
x
)
e
1
(
xO
)
e
1
(
O
gH
fS
dN
cO
bH
aC 2
2
2
2
2
2
2
2
2
2
2 +
+
+
+
→
+
+
+
+
+
+
+
+
+
c) Combustion with exhaust pollutants (emission gases)
x
2
2
2
2
2
2
2
2
2
2
2
pNO
oCH
nCO
mN
LO
jSO
O
iH
hCO
N
)
76
.
3
(
x
)
e
1
(
xO
)
e
1
(
O
gH
fS
dN
cO
bH
aC
+
+
+
+
+
+
+
→
+
+
+
+
+
+
+
+
+
Note: In balancing combustion equation for Solid fuels, convert the Ultimate Analysis of Coal to Molal or volumetric analysis,
then reduced to and Ashless basis
Example: Reduction of Ultimate coal analysis to Molal ashless analysis

Kg of CO2 per kg of C formed
3
11
C
of
kg
CO
of
kg
11
8
3
44
32
12
CO
2
O
C
2
2
=
→
+
→
+
→
+
Kg of H2O per kg of H formed
1
9
H
of
kg
O
H
of
kg
9
8
1
18
16
2
O
H
2
O
H
2
2
2
1
2
=
→
+
→
+
→
+
Kg of SO2 per kg of S formed
1
2
S
of
kg
SO
of
kg
2
1
1
64
32
32
SO
2
O
S
2
2
=
→
+
→
+
→
+
Total Mass of Products
NOx
NOx
CH
CH
CO
CO
2
N
2
N
2
SO
2
SO
2
O
2
O
O
2
H
O
2
H
2
CO
2
CO
oducts
Pr
oducts
Pr
M
n
M
n
M
n
M
n
M
n
M
n
M
n
M
n
m
niMi
Σ
m
+
+
+
+
+
+
+
=
=
Total Moles of Products
NOx
CH
CO
2
N
2
SO
2
O
O
2
H
2
CO
oducts
Pr
oducts
Pr
n
n
n
n
n
n
n
n
n
ni
Σ
n
+
+
+
+
+
+
+
=
=
Dew Point Temperature
product
of
pressure
total
P
n
n
)
P
(
P
)
(P
O
H
of
pressure
partial
the
ing
correspond
re
temperatu
Saturation
)
tsat
(
t
oducts
Pr
O
2
H
O
H
O
H
2
dp
2
2
=
=
=
Moles of Dry Flue Gas (The H2O is not included in the analysis)
NOx
CH
CO
2
N
2
SO
2
O
2
CO
Gas
Flue
Dry n
n
n
n
n
n
n
n +
+
+
+
+
+
=
% of CO2 in the dry flue gas
100%
x
n
n
n
n
n
n
n
n
100%
x
n
n
NOx
CH
CO
2
N
2
SO
2
O
2
CO
2
CO
Gas
Flue
Dry
2
CO
+
+
+
+
+
+
=

Total Mass of Fuel
a. For Hydrocarbon of Hydrocarbon Mixture
F
F
Fuel M
n
Σ
m =
b. For Solid Fuels
O
2
H
O
2
H
S
S
2
N
2
N
2
O
2
O
2
H
2
H
C
C
Fuel
Fuel
M
n
M
n
M
n
M
n
M
n
M
n
m
niMi
Σ
m
+
+
+
+
+
=
=
Mass Flow Rate of Products (Known Fuel flow rate)
hr
kg
Rate
Flow
Fuel
x
m
m
Products
of
hr
kg
Fuel
Products
=
Volume of Products at the product Pressure and Temperature (m3
)
3
oducts
Pr
oducts
Pr m
P
T
)
R
(
n
V =
Volume flow rate of Products at the product Pressure and Temperature (m3
/hr)
hr
m
Rate
Flow
Fuel
x
m
V
Products
of
hr
m 3
Fuel
Products
3
=
Molecular Weight of Products
oducts
Pr
i
2
N
2
N
NOx
NOx
CH
CH
CO
CO
2
SO
2
SO
2
O
2
O
O
2
H
O
2
H
2
CO
2
CO
n
M
n
...
M
n
M
n
M
n
M
n
M
n
M
n
M
n
M
n
M
+
+
+
+
+
+
+
+
=
Gas Constant of Products
)
niMi
(
Σ
Ri
)
niMi
(
Σ
xiRi
Σ
R
K
-
kg
KJ
M
3143
.
8
R
=
=
=
Specific Heat of Products
K
-
kg
KJ
1
k
R
C
K
-
kg
KJ
1
k
Rk
C
C
C
k
C
C
R
C
x
Σ
C
C
x
Σ
C
v
p
v
p
v
p
vi
i
v
pi
i
p
−
=
−
=
=
−
=
=
=

For Coal Fuel
1. Ultimate Analysis (%ages by mass)
100% = %C + %H2 + %O2 + %N2 + %S + %Moisture + %Ash
2. Proximate Analysis (%ages by mass)
100% = %Fixed Carbon + %Volatiles + %Moisture + %Ash
3. Orsat Analysis (%ages by volume)
100% = %CO2 + %CO + %O2 + %SO2 + %N2 (Mostly SO2 is not given)
4. Reduction of UA to an Ashless basis
2
2
2
2
2
2
2
D = 100 - %Ash
Ashless Ultimate Analysis
%C
C = x100%
D
%H
H = x100%
D
%O
O = x100%
D
%N
N = x100%
D
%H
S = x100%
D
%M
M = x100%
D
5. Mass of Products (Known Orsat Analysis and Ultimate Analysis on an ashless basis)
Flue Gas 2 2 2 2
kg Flue Gas
M = 44(%CO ) + 28(%CO) + 32(%O ) + 28(%N ) + %M + 9(%H ) + 2(%S)
kg Coal
6. Amount of Carbon in Products
Carbon
Carbon 2
Coal
Carbon
Carbon 2
Coal
Carbon
Carbon 2
Coal
kg
12 12
M = 44(%CO ) + 28(%CO)
44 28 kg
kg
M = 12(%CO ) +12(%CO)
kg
kg
M = 12 (%CO ) + (%CO)
kg
 
 

7. Unburned Carbon in the Refuse
u r
r
r
C
u
Coal
Refuse
r
Coal
r
C = W - Ash
A
W =
1- C
kg
C - unburn carbon in the refuse,
kg
kg
W - Amount of refuse,
kg
C - kg of combustible per kg of refuse
8. Carbon Actually Burned
C
a u
Coal
a r
kg
C = C - C
kg
C = C - W + Ash

SAMPLE PROBLEMS
Example No. 1: In the figure below, Determine
a. Percent excess air
b. Volumetric Analysis of Products
c. Orsat Analysis
( )
( )
( )
2
2
2
2
2
2
4
2
2
2
2
2
2
4
l
Theoretica
2
2
2
2
2
4
l
Theoretica
Actual
l
Theoretica
Actual
Actual
p
F
a
p
N
8
.
8
O
34
.
0
O
H
2
CO
1
N
8
.
8
O
34
.
2
CH
34
.
0
)
m
25
.
0
n
(
e
d
2
m
5
.
0
c
1
n
b
34
.
2
m
25
.
0
n
a
N
)
76
.
3
(
a
)
17
.
1
(
dO
O
cH
bCO
N
)
76
.
3
(
a
)
17
.
1
(
aO
)
17
.
1
(
CH
0.17
e
EA;
with
Combustion
%
17
17
.
0
e
16
.
17
m
n
12
)
m
25
.
0
n
(
27
.
137
F
A
N
)
76
.
3
(
a
O
cH
bCO
N
)
76
.
3
(
a
aO
CH
1
F
A
F
A
e
F
A
e
1
F
A
20
200
4000
F
A
hr
kg
200
,
4
200
4000
m
m
m
m
+
+
+
→
+
+
=
+
=
=
=
=
=
=
+
=
+
+
+
→
+
+
=
=
=
=
+
+
=






+
+
→
+
+
−
=






+
=






=
=






=
+
=
+
=
%
78
.
86
y
%
35
.
3
y
%
86
.
9
y
Analysis
Orsat
14
.
10
8
.
8
34
.
0
1
n
%
48
.
72
y
%;
8
.
2
y
%;
48
.
16
y
%;
24
.
8
y
Analysis
oduct
Pr
14
.
12
8
.
8
34
.
0
2
1
n
2
2
2
2
2
2
2
N
O
CO
oducts
Pr
Dry
N
O
O
H
CO
oducts
Pr
=
=
=
=
+
+
=
=
=
=
=
=
+
+
+
+
=

Example No. 2 (Combustion of Gasoline)
Typical gasoline C8H18 is burned with 20% excess air by weight. Find
a. the air-fuel ratio
b. the percentage CO2 by volume in the dry exhaust gases
c. kg of water vapor formed per kg of fuel
d. volume of dry exhaust gas per kg of fuel if T = 290 K and P = 101.33 KPa
e. the partial pressure of the water vapor in the exhaust
f. the dew point temperature of the products
Fuel: C8H18
Excess air: e = 20%
Product Temperature = 290 K
Product Pressure = 101.33 KPa
Combustion with 100% theoretical air
2
2
2
2
2
18
8
2
2
2
2
2
18
8
N
47
O
H
9
CO
8
N
47
O
5
.
12
H
C
N
)
76
.
3
(
a
O
cH
bCO
N
)
76
.
3
(
a
aO
H
C
+
+
→
+
+
+
+
→
+
+
Combustion with e = 20%
eq.
Combustion
Actual
N
4
.
56
O
5
.
2
O
H
9
CO
8
N
4
.
56
O
15
H
C
N
47
)
20
.
1
(
dO
O
H
9
CO
8
N
47
)
20
.
1
(
O
5
.
12
)
20
.
1
(
H
C
2
2
2
2
2
2
18
8
2
2
2
2
2
2
18
8
→
+
+
+
→
+
+
+
+
+
→
+
+
a. Actual Air – Fuel Ratio
06
.
18
)
18
(
1
)
8
(
12
)
28
)(
4
.
56
(
)
32
(
15
F
A
ACTUAL
=
+
+
=






b. the percentage CO2 by volume in the dry exhaust gases
%
95
.
11
%
100
x
9
.
66
8
y
9
.
66
4
.
56
5
.
2
8
n
gas
exhaust
dry
of
moles
n
2
CO
d
d
=
=
=
+
+
=
−
c. kg of water vapor formed per kg of fuel
kg
kg
42
.
1
114
162
1(18)
12(8)
162
kg
kg
kg
162
)
18
(
9
kg
9
.
75
4
.
56
5
.
2
9
8
n
gas
exhaust
of
moles
n
18
H
8
C
O
2
H
O
2
H
d
=
=
+
=
=
=
=
+
+
+
=
−
d. volume of dry exhaust gas per kg of fuel if T = 290 K and P = 101.33 KPa
18
H
8
C
3
18
H
8
C
d
3
d
3
d
d
d
kg
m
14
114
1,591.9
kg
V
m
1,591.9
V
m
33
.
101
)
290
)(
3143
.
8
(
9
.
66
V
T
R
n
PV
9
.
66
4
.
56
5
.
2
8
n
gas
exhaust
dry
of
moles
n
=
=
=
=
=
=
+
+
=
−

e. the partial pressure of the water vapor in the exhaust
KPa
015
.
12
)
33
.
101
(
1186
.
0
P
P
P
y
%
86
.
11
%
100
x
9
.
75
9
y
9
.
75
4
.
56
5
.
2
9
8
n
gas
exhaust
of
moles
n
O
2
H
O
2
H
O
2
H
O
2
H
=
=
=
=
=
=
+
+
+
=
−
f. the dew point temperature of the products
H2O
H2O
2
DPT saturation temperature corresponding P
P 12.015 KPa
From Steam Table
DPT 49.467 C
if the mixture is cooled below DPT, condensation of H O in the
mixture will occur
=
=
= 
Example No. 3 (Known Orsat analysis and Fuel)
A fuel oil C12H26 is used in an internal combustion engine and the Orsat analysis are as follows: CO2 = 12.8% ; O2 = 3.5%; CO
= 0.2% and N2 = 83.5%. Determine the actual air-fuel ratio and the percent excess air.
Solution:
(Basis 100 moles of dry flue gas and a moles of fuel)
aC12H26 + bO2 + b(3.76)N2 → 12.8CO2 + cH2O + 0.2CO + 3.5O2 + 83.5N2
By C balance
12a = 12.8 + 0.2
a = 1.0833
By N2 Balance
b(3.76) = 83.5
b = 22.207
By H balance
26a = 2c
c = 26(1.0833)/2
c = 14.083
( )
( ) %
8
.
10
108
.
0
1
F
A
F
A
e
94
.
14
m
n
12
)
m
25
.
0
n
(
28
.
137
F
A
Ratio
Fuel
-
Air
l
Theoretica
56
.
16
)
26
(
1
)
12
(
12
)
28
)(
08
.
77
(
)
32
(
5
.
20
F
A
77.08N
+
3.23O
+
0.185CO
+
O
13H
+
CO
816
.
1
1
77.08N
+
20.5O
+
H
C
a
by
equation
the
dividing
83.5N
+
3.5O
+
0.2CO
+
O
cH
+
12.8CO
b(3.76)N
+
bO
+
H
aC
Theoreical
Actual
al
Theroretic
Actual
2
2
2
2
2
2
26
12
2
2
2
2
2
2
26
12
=
=
−
=
=
+
+
=






=
+
+
=






→
→

Example No. 4 (Gasoline with Orsat analysis)
The following is the ultimate analysis of a sample of petrol by weight : C = 84.2 % ; H = 15.8 %. Calculate the ratio of air to
petrol consumption by weight if the volumetric analysis of the dry exhaust gas is :CO2 = 11.07 % ; CO = 1.23 % ; O2 = 3.72 %
; N2 = 83.97 %. Also find percentage excess air.
18
m
8
n
2
n
2
n
25
.
2
2
n
2
m
H
C
is
formula
the
Fuel
petrol
For
3
.
eq
n
25
.
2
m
m
33
.
6
n
25
.
14
2
.
eq
m
33
.
6
m
158
.
0
1
m
n
12
m
n
12
m
158
.
0
m
n
12
m
H
%
1
.
eq
n
25
.
14
n
842
.
0
12
m
n
12
m
n
12
n
12
842
.
0
m
n
12
n
12
C
%
2
2n
n
=
=
+
=
+
=
→
=
=
→
=
=
+
+
=
+
=
→
=
=
+
+
=
+
=
+
%
2
.
16
162
.
0
1
57
.
14
49
.
17
e
49
.
17
m
n
12
b
28
a
32
F
A
84
.
13
c
33
.
22
a
b
)
76
.
3
(
a
97
.
83
b
N
97
.
83
O
72
.
3
CO
23
.
1
O
cH
11.07CO
bN
aO
H
C
gas)
flue
dry
of
moles
100
(Basis
Equation
Combustion
57
.
14
m
n
12
)
m
25
.
0
n
(
28
.
137
F
A
Actual
2
2
2
2
2
2
18
8
l
Theoretica
=
=
−
=
=
+
+
=






=
=
=
=
+
+
+
+
→
+
+
=
+
+
=







Example No. 5 (Coal Fuel)
The following data were obtained from a boiler test: Ultimate analysis of coal as fired is; C = 62%, H2 = 4%, O2 = 8%, N2 = 1
%, S = 2%, H2O = 8% and Ash = 15%. Excess air is 25% for complete combustion. Fuel and air temperature and pressure are,
25C and 101 KPa, respectively. Flue gas temperature is 300C and P = 101 KPa. Determine
a. Ultimate analysis on an ashless basis
b. Molal analysis of fuel on an ashless basis
c. Combustion equation
d. Actual air – fuel ratio in kg/kg
e. Volumetric Analysis of Products
f. Molecular Weight and Gas Constant of Products
g. Cubic meter of CO2 per kg of fuel burnt
h. Cubic meter of SO2 per kg of fuel burnt
Ashless U.A.
C =72.9% ; H2 = 4.7% ; O2 = 9.4% ; N2 = 1.2% ; S = 2.4% ; M = 9.4%
Molal analysis on an ashless basis
C =64.91% ; H2 = 25.13% ; O2 = 3.14% ; N2 = 0.45% ; S = 0.79% ; H2O = 5.58%
Products
N
52
.
353
O
78
.
18
SO
79
.
0
O
H
71
.
0
3
64.91CO
Air
N
07
.
353
93.9O
Fuel
O
5.58H
0.79S
0.45N
3.14O
25.13H
64.91C
52
.
353
g
g
2
46
.
82
2
)
25
.
1
(
2
2(0.45)
18.78
f
2f
2(0.79)
30.71
2(64.91)
2(2)
(1.25)75.1
(5.58)
2(3.14)
Products
N
g
fO
SO
79
.
0
O
H
71
.
0
3
64.91CO
Air
N
46
.
82
2
)
25
.
1
(
2O
(1.25)75.1
Fuel
O
5.58H
0.79S
0.45N
3.14O
25.13H
64.91C
E.A.
25%
ith
equation w
Combustion
Products
N
91
.
82
2
SO
79
.
0
O
H
71
.
0
3
64.91CO
Air
N
46
.
82
2
75.12O
Fuel
O
5.58H
0.79S
0.45N
3.14O
25.13H
64.91C
Products
eN
dSO
O
cH
bCO
Air
a(3.76)N
aO
Fuel
O
5.58H
0.79S
0.45N
3.14O
25.13H
64.91C
TA
100
ith
equation w
Combustion
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
→
+
+
+
+
→
+
→
+
+
+
+
+
=
=
+
=
+
+
+
=
+
+
→
+
+
+
+
→
+
→
+
+
+
+
+
→
+
+
+
→
+
→
+
+
+
+
+
→
+
+
+
→
+
→
+
+
+
+
+
07
.
12
)
58
.
5
(
18
)
79
.
0
(
32
)
45
.
0
(
28
)
14
.
3
(
32
)
13
.
25
(
2
)
91
.
64
(
12
)
07
.
353
(
28
)
9
.
93
(
32
F
A
Actual
=
+
+
+
+
+
+
=







%
42
.
75
%
100
x
71
.
468
353.52
y
%
01
.
4
%
100
x
71
.
468
18.78
y
%
17
.
0
%
100
x
71
.
468
0.79
y
%
55
.
6
%
100
x
71
.
468
30.71
y
%
85
.
13
%
100
x
71
.
468
64.91
y
71
.
468
52
.
353
78
.
18
79
.
0
71
.
0
3
64.91
n
n
ni
yi
2
N
2
O
2
SO
O
2
H
2
CO
oducts
Pr
=
=
=
=
=
=
=
=
=
=
=
+
+
+
+
=
=
K
-
kg
KJ
279
.
0
M
3143
.
8
R
kgm
kg
78
.
29
71
.
468
)
28
(
52
.
353
)
32
(
78
.
18
)
64
(
79
.
0
)
18
(
71
.
0
3
64.91(44)
M
yiMi
Σ
M
=
=
=
+
+
+
+
=
=
Fuel
of
kg
m
03
.
0
)
58
.
5
(
18
)
79
.
0
(
32
)
45
.
0
(
28
)
14
.
3
(
32
)
13
.
25
(
2
)
91
.
64
(
12
04
.
37
Fuel
of
kg
V
m
04
.
37
101
)
273
300
)(
3143
.
8
)(
79
.
0
(
V
Fuel
of
kg
m
87
.
2
)
58
.
5
(
18
)
79
.
0
(
32
)
45
.
0
(
28
)
14
.
3
(
32
)
13
.
25
(
2
)
91
.
64
(
12
91
.
3061
Fuel
of
kg
V
m
91
.
3061
101
)
273
300
)(
3143
.
8
)(
91
.
64
(
V
P
T
R
n
V
T
R
n
PV
3
2
SO
3
2
SO
3
2
CO
3
2
CO
=
+
+
+
+
+
=
=
+
=
=
+
+
+
+
+
=
=
+
=
=
=
Example No. 6 (Alcohol)
Calculate the theoretical Oxygen/fuel ratio and Air/fuel ratio on a mass basis for the combustion of ethanol, C2H5OH.
kg
kg
09
.
2
46
)
3
(
32
Fuel
O
95
.
8
46
)
28
(
28
.
11
)
32
(
3
F
A
N
28
.
11
O
H
3
CO
2
N
28
.
11
O
3
OH
H
C
Equation
Combustion
3
a
c
b
2
a
2
1
3
c
c
2
6
b
2
N
)
76
.
3
(
a
O
cH
bCO
N
)
76
.
3
(
a
aO
OH
H
C
2
2
2
2
2
2
5
2
2
2
2
2
2
5
2
=
=
=
+
=
+
+
→
+
+
=
+
=
+
=
=
=
+
+
→
+
+

Example No. 7 (Gaseous Fuel Mixture)
A gaseous fuel mixture has the following volumetric analysis, CH4 = 60% ; CO = 30% and O2 = 10% If this fuel is burned with
30% excess air by volume, determine
a. The combustion equation
b. The actual fuel ratio
c. The Orsat analysis
d. The dew point temperature (assume P = 101.325 KPa)
Combustion Equation
%
7
.
82
y
%
1
.
5
y
%
2
.
12
y
5
.
738
120
5
.
858
n
ANALYSIS
ORSAT
C
8
.
52
DPT
KPa
16
.
14
)
325
.
101
(
1398
.
0
20
PH
P
Pi
yi
%
17
.
71
y
%
37
.
4
y
%
98
.
13
y
%
48
.
10
y
Products
of
Analysis
Volumetric
%
100
x
n
ni
yi
5
.
858
611
5
.
37
120
90
n
52
.
11
)
32
(
10
)
28
(
30
)
16
(
60
)
28
(
611
)
32
(
5
.
162
F
A
N
611
O
5
.
37
O
H
120
CO
90
N
611
O
5
.
162
O
10
CO
30
CH
60
5
.
37
d
N
)
76
.
3
(
a
)
30
.
1
(
dO
O
H
120
CO
90
N
)
76
.
3
(
a
)
30
.
1
(
aO
)
30
.
1
(
O
10
CO
30
CH
60
EA
305
with
Combustion
120
c
90
b
125
a
N
)
76
.
3
(
a
O
O
cH
bCO
N
)
76
.
3
(
a
aO
O
10
CO
30
CH
60
TA
100%
with
Combustion
N2
O2
CO2
Gas
Dry
N2
O2
O
2
H
CO2
oducts
Pr
2
2
2
2
2
2
2
4
2
2
2
2
2
2
2
4
2
2
2
2
2
2
2
4
=
=
=
=
−
=

=
=
=
=
=
=
=
=
=
=
+
+
+
=
=
+
+
+
=
+
+
+
→
+
+
+
+
=
+
+
+
→
+
+
+
+
=
=
=
+
+
+
→
+
+
+
+

Example No. 8 (Producer’s Gas)
Producer gas from bituminous coal contains following molar analysis. CH4 = 3 %, H2 = 14.0%, N2 = 50.9%, O2 = 0.6%, CO
= 27.0% and CO2 = 4.5%. This is burned with 25% excess air, Calculate the air/fuel ratio on a volumetric basis and on a mass
basis.
mol
mol
1.54
F
A
kg
kg
8
.
1
F
A
N
63
.
172
O
48
.
6
O
H
20
CO
5
.
34
N
73
.
121
O
38
.
32
CO
5
.
4
CO
27
O
6
.
0
N
9
.
50
H
14
CH
3
2
2
2
2
2
2
2
2
2
2
4
=
=
+
+
+
→
+
+
+
+
+
+
+
Example No. 9 (Combustion with emission gas)
An combustor of a small scale industrial plant burns liquid Octane (C3 H8 ) at the rate of 0.005 kg/sec, and uses 20% excess
air. The air and fuel enters the engine at 25C and the combustion products leaves the engine at 900 K. It may be assumed that
90% of the carbon in the fuel burns to form CO2 and the remaining 1.5% burns to form CO.(P = 101 KPa) Determine
a. The actual air – fuel ratio
b. The kg/s of actual air per hour
c. The M and R of the products
d. The m3
/sec of products at the product temperature and pressure
e. The cubic meter of CO emission for 24 hrs operation
2
2
2
2
2
2
8
3
2
2
2
2
2
2
8
3
2
2
2
2
2
8
3
2
2
2
2
2
8
3
N
458
.
22
O
996
.
0
CO
045
.
0
O
4H
CO
955
.
2
N
458
.
22
O
973
.
5
H
C
EQUATION
COMBUSTION
996
.
0
f
balance
oxygen
By
N
715
.
8
1
)
20
.
1
(
fO
CO
045
.
0
O
4H
CO
955
.
2
N
715
.
8
1
)
0
(1.2
8O
(1.20)4.97
H
C
EA
20%
with
Combustion
N
715
.
8
1
dCO
O
4H
CO
955
.
2
N
715
.
8
1
O
978
.
4
H
C
978
.
4
a
d
c
b
2
a
2
4
c
c
2
)
8
(
1
955
.
2
045
.
0
3
b
045
.
0
)
3
(
015
.
0
d
a(3.76)N
dCO
O
cH
bCO
a(3.76)N
aO
H
C
emission
CO
and
TA
100%
with
Combustion
+
+
+
+
→
+
+
=
+
+
+
+
→
+
+
+
+
+
→
+
+
=
+
+
=
=
=
=
−
=
=
=
+
+
+
→
+
+

sec
m
4
.
1
)
44
(
101
)
3600
)(
005
.
0
)(
900
)(
3143
.
8
(
045
.
0
)
8
(
1
)
3
(
12
m
P
T
R
n
CO
of
hr
m
sec
m
26
.
0
)
44
(
101
)
005
.
0
)(
900
)(
3143
.
8
(
45
.
30
)
8
(
1
)
3
(
12
m
P
T
R
n
Products
of
sec
m
T
R
n
PV
K
-
kg
KJ
293
.
0
37
.
28
3143
.
8
R
kgm
kg
37
.
28
M
30.45
22.458(28)
0.996(32)
0.045(28)
4(18)
2.955(44)
yiMi
Σ
M
n
ni
yi
yiMi
Σ
M
moles
45
.
30
22.458
0.996
0.045
4
2.955
Products
of
Moles
hr
kg
335.444
5)(3600)
18.64(0.00
air
of
Mass
64
.
18
)
8
(
1
)
3
(
12
)
28
)(
458
.
22
(
)
32
(
973
.
5
F
A
Ratio
Fuel
-
Air
Actual
3
Fuel
3
3
Fuel
3
Actual
=
=








+
=
=
=








+
=
=
=
=
=
+
+
+
+
=
=
=
=
=
+
+
+
+
=
=
=
=
+
+
=






Example No. 10 (Hydrocarbon Fuel)
A hydrocarbon fuel represented by C12H26 is used as fuel in an IC engine and requires 25 % excess air for complete combustion.
Determine
a. The combustion equation
b. The theoretical air – fuel ratio
c. The actual air – fuel ratio
d. The volumetric and gravimetric analysis of the products
e. The molecular weight M and gas constant R of the products
f. The kg of CO2 formed per kg of fuel
g. % C and %H in the fuel
Solution
Fuel: C12H26
12 26 2 2 2 2 2
12 26 2 2 2 2 2 2
With 100% TA
C H aO a(3.76)N bCO cH O a(3.76)N
Carbon balance
b n
Hydrogen balance
c 0.5m
Oxygen balance
a n 0.25m
With 25% EA
C H (1 e)aO (1 e)a(3.76)N bCO cH O dO (1 e)a(3.76)N
d e(n 0.
+ + → + +
=
=
= +
+ + + + → + + + +
= +
12 26 2 2 2 2 2 2
25m)
COMBUSTION EQUATION
C H 23.125O 86.950N 12CO 13H O 4.625O 86.950N
+ + → + + +

Products
Gases Mi ni yi mi xi yiMi yi(%) xi(%)
CO2 44 12 0.103 528 0.158 4.529 10.3 15.8
H2O 18 13 0.112 234 0.070 2.007 11.2 7.0
O2 32 4.625 0.040 148 0.044 1.270 4.0 4.4
N2 28 86.95 0.746 2434.6 0.728 20.884 74.6 72.8
total 116.575 1.00 3344.6 1.00 28.691 100.0 100.0
The combustion equation
2
2
2
2
2
2
26
12 N
95
.
86
4.625O
O
H
3
1
12CO
N
95
.
86
O
125
.
23
H
C +
+
+
→
+
+
The theoretical A/F ratio
26
12
l
Theoretica H
C
of
kg
air
of
kg
94
.
14
m
n
12
)
m
25
.
0
n
(
28
.
137
F
A
=
+
+
=






The actual A/F ratio
26
12
t
Actual H
C
of
kg
air
of
kg
674
.
18
F
A
)
e
1
(
F
A
=






+
=






The Volumetric and gravimetric Analysis
Gases yi(%) xi(%)
CO2 10.3 15.8
H2O 11.2 7.0
O2 4.0 4.4
N2 74.6 72.8
total 100.0 100.0
Molecular weight and Gas constant
K
-
kg
KJ
2898
.
0
691
.
28
3143
.
8
R
xiRi
Σ
M
R
R
kg
kg
691
.
28
yiMi
Σ
M
m
=
=
=
=
=
=
Kg of CO2 per kg of fuel
106
.
3
26
)
12
(
12
528
H
C
of
kg
CO
of
kg
26
12
2
=
+
=
% C and % H in the fuel

%
3
.
15
100
x
26
)
12
(
12
26
H
%
100%
x
m
n
12
m
H
%
%
7
.
84
100
x
26
)
12
(
12
)
12
(
12
C
%
100%
x
m
n
12
n
12
C
%
=
+
=
+
=
=
+
=
+
=
Properties of Fuels and Lubricants
a) Viscosity - a measure of the resistance to flow that a lubricant offers when it is subjected to shear stress.
b) Absolute Viscosity - viscosity which is determined by direct measurement of shear resistance.
c) Kinematics Viscosity - the ratio of the absolute viscosity to the density
d) Viscosity Index - the rate at which viscosity changes with temperature.
e) Flash Point - the temperature at which the vapor above a volatile liquid forms a combustible mixture with air.
f) Fire Point - The temperature at which oil gives off vapor that burns continuously when ignited.
g) Pour Point - the lowest temperature at which a liquid will continue to flow
h) Dropping Point - the temperature at which grease melts.
i) Condradson Number(carbon residue) - the percentage amount by mass of the carbonaceous residue remaining
after destructive distillation.
j) Octane Number - a number that provides a measure of the ability of a fuel to resist knocking when it is burnt in a
gasoline engine. It is the percentage by volume of iso-octane in a blend with normal heptane that matches the
knocking behavior of the fuel.
k) Cetane Number - a number that provides a measure of the ignition characteristics of a diesel fuel when it is burnt in a
standard diesel engine. It is the percentage of cetane in the standard fuel.
END OF MODULE 1

Me 312 module 1

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Me 312 module 1