1. COMBUSTION ENGINEERING
By: Engr. YURI G. MELLIZA
Fuel - a substance composed of chemical elements, which in rapid chemical union with oxygen produced combustion.
Combustion - is that rapid chemical union with oxygen of an element, whose exothermic heat of reaction is sufficiently great
and whose rate of reaction is sufficiently fast, whereby useful quantities of heat are liberated at elevated temperatures. It is the
burning or oxidation of the combustible elements.
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. diesoline
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. 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 ttractive 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)
3. STRUCTURE OF CnHm
A. Chain Structure (saturated)
B. Chain Structure (unsaturated)
C. Ring Structure (saturated)
4. Complete Combustion: Occurs when all the combustible elements has been fully oxidized.
22 COOC
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.
743
281612
COOC 22
1
THE COMBUSTION CHEMISTRY
A. Oxidation of Carbon
3
8
kgC
kgO
1183
443212
COOC
2
22
B. Oxidation of Hydrogen
1
8
kgH
kgO
981
18162
OHOH
2
2
222
1
2
C. Oxidation of Sulfur
1
1
kgS
kgO
211
643232
SOOS
2
22
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
5. In theoretical Combustion, the composition of atmospheric air is considered as;
a) Volumetric or Molal analysis
O2 , = 21%
N2 = 79%
b) Gravimetric Analysis
O2 = 23.3%
N2 = 76.7%
76.3
21
79
OofMole
NofMoles
2
2
COMBUSTION WITH AIR and Theoretical air requirement
oductsPrAirFuel
A) Combustion of Carbon with air
44.11
Cofkg
airofkg
)28(76.344)28(76.33212
N76.3CON76.3OC 2222
B) Combustion of Hydrogen with Air
32.34
Hofkg
airofkg
)28)(76.3(18)28)(76.3(162
N)76.3(OHN)76.3(OH
2
1
2
1
22
1
222
1
22
1
2
C) Combustion of Sulfur with air
29.4
Hofkg
airofkg
)28)(76.3(64)28)(76.3(3232
N)76.3(SON)76.3(OS 2222
6. 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
FuelofKg
airofkg
F
A
lTheoretica
EXCESS AIR
It is an amount of air in excess of 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)
A) Combustion of CnHm with 100% theoretical air
mn12
0.25m)n(28.137
F
A
mn12
a28.137
mn12
28)28.3(aa32
F
A
0.5mc
nb
0.25mna
TA100%withCnHmoneofcombustionfor thetherefore
0.25mna
0.5cba
c2b2a
balanceOxygenBy
0.5mc
2c1(m)
balanceHydrogenBy
nb
b1(n)
balanceCarbonBy
N)76.3(aOcHbCON)76.3(aaOHC
lTheoretica
lTheoretica
22222mn
B) With excess air
m25.0ned
)m25.0(eend
m25.0n)m25.0(eenm25.0nd
m25.0n)m25.0n)(e1(d
0.5c-b-e)a(1d
c-2b-e)a2(12d
2dc2be)a2(1
balanceOxygenBy
N)76.3(a)e1(dOOcHbCON)76.3(a)e1(aO)e1(HC 222222mn
7. RatioA/FActual
HCofkg
airofkg
mn12
)m25.0n)(e1(28.137
F
A
mnActual
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.
EQUIVALENCE RATIO
tricStoichiomeA
F
actualA
F
)(
)(
ER
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
lTheoretica
lTheoreticaActual
F
A
F
A
F
A
AirExcessPercentage
8. 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.
fuelofkg
airofkg
S29.4
8
O
H32.34C44.11
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 x222
MASS FLOW RATE OF FLUE GAS
a) Without considering Ash loss:
1
F
A
mm Fuelgas
b) Considering Ash loss
LossAsh1
F
A
mm Fuelgas
where ash loss in decimal
9. Combustion equation with CO in the products due to incomplete combustion (100% theoretical air)
22222 N)76.3(adCOOcHbCON)76.3(aaOCnHm
Combustion equation with CO in the products due to incomplete combustion (with excess air)
222222 N)76.3(a)e1(fOdCOOcHbCON)76.3(a)e1(aO)e1(CnHm
Typical Real-World Engine Combustion Process:
OxidesNitrogen-NO
MonoxideCarbon-CO
CompoundsOrganicVolatile-s)CH(VOC'
NOxCOs)CH(VOC'NOOHCO)Nand(OAir(CnHm)Fuel
x
222222
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.
10. 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.
11. 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 a majority 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)
12. 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 • sulphur dioxide (SO2)
• carbon monoxide (CO)
• nitrogen dioxide (NO2)
• total reduced sulphur (TRS)
• volatile organic compounds (VOCs)
• persistent organic pollutants (POPs)
13. • 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
a) With 100% theoretical air
0.5mc
nb
0.25mna
where
N)76.3(aOcHbCON)76.3(aaOHC 22222mn
b) With excess air e
0.25m)n(ed
where
N)76.3(a)e1(dOOcHbCON)76.3(a)e1(aO)e1(HC 222222mn
c) With exhaust pollutants (emission gases)
processcombustionactualapply tonotmaymandnofin termsdandcb,a,ofvaluesaboveThe:Note
OxidesNitrogen-NOx
MonoxideCarbon-CO
CompoundsOrganicVolatile-s)CH(VOC'
NOxiCOhCHgNfOdOHcCOb)a(3.76)NOaCnHm 222222
Combustion of Solid Fuels
a) Combustion with 100% theoretical air
2222222222 kNjSOOiHhCON)76.3(xxOOgHfSdNcObHaC
b) Combustion with Excess air e
mNLOjSOOiHhCON)76.3(x)e1(xO)e1(OgHfSdNcObHaC 22222222222
c) Combustion with exhaust pollutants (emission gases)
x
22222222222
pNOoCHnCO
mNLOjSOOiHhCON)76.3(x)e1(xO)e1(OgHfSdNcObHaC
14. 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
Cofkg
COofkg
1183
443212
CO2OC
2
2
Kg of H2O per kg of H formed
1
9
Hofkg
OHofkg
981
18162
OH2OH
2
22
1
2
Kg of SO2 per kg of S formed
1
2
Sofkg
SOofkg
211
643232
SO2OS
2
2
Total Mass of Products
NOxNOxCHCHCOCO2N2N2SO2SO2O2OO2HO2H2CO2COoductsPr
oductsPr
MnMnMnMnMnMnMnMnm
niMiΣm
Total Moles of Products
NOxCHCO2N2SO2OO2H2COoductsPr
oductsPr
nnnnnnnnn
niΣn
15. Dew Point Temperature
productofpressuretotalP
n
n
)P(P
)(POHofpressurepartialtheingcorrespondretemperatuSaturation)tsat(t
oductsPr
O2H
OH
OH2dp
2
2
Moles of Dry Flue Gas (The H2O is not included in the analysis)
NOxCHCO2N2SO2O2COGasFlueDry nnnnnnnn
% of CO2 in the dry flue gas
100%x
nnnnnnn
n
100%x
n
n
NOxCHCO2N2SO2O2CO
2CO
GasFlueDry
2CO
Total Mass of Fuel
a. For Hydrocarbon of Hydrocarbon Mixture
FFFuel MnΣm
b. For Solid Fuels
O2HO2HSS2N2N2O2O2H2HCCFuel
Fuel
MnMnMnMnMnMnm
niMiΣm
Mass Flow Rate of Products (Known Fuel flow rate)
hr
kg
RateFlowFuelx
m
m
Productsof
hr
kg
Fuel
Products
Volume of Products at the product Pressure and Temperature (m3
)
3oductsPr
oductsPr m
P
T)R(n
V
Volume flow rate of Products at the product Pressure and Temperature (m3
/hr)
hr
m
RateFlowFuelx
m
V
Productsof
hr
m 3
Fuel
Products
3
Molecular Weight of Products
oductsPr
i2N2NNOxNOxCHCHCOCO2SO2SO2O2OO2HO2H2CO2CO
n
Mn...MnMnMnMnMnMnMnMn
M
16. Gas Constant of Products
)niMi(Σ
Ri)niMi(Σ
xiRiΣR
K-kg
KJ
M
3143.8
R
Specific Heat of Products
K-kg
KJ
1k
R
C
K-kg
KJ
1k
Rk
C
C
C
k
CCR
CxΣC
CxΣC
v
p
v
p
vp
viiv
piip
Example No. 1: In the figure below, Determine
a. Percent excess air
b. Volumetric Analysis of Products
c. Orsat Analysis
17.
2222224
2222224
lTheoretica
222224
lTheoretica
Actual
lTheoreticaActual
Actual
p
Fap
N8.8O34.0OH2CO1N8.8O34.2CH
34.0)m25.0n(ed
2m5.0c
1nb
34.2m25.0na
N)76.3(a)17.1(dOOcHbCON)76.3(a)17.1(aO)17.1(CH
0.17eEA;withCombustion
%1717.0e
16.17
mn12
)m25.0n(27.137
F
A
N)76.3(aOcHbCON)76.3(aaOCH
1
F
A
F
A
e
F
A
e1
F
A
20
200
4000
F
A
hr
kg
200,42004000m
mmm
%78.86y
%35.3y
%86.9y
AnalysisOrsat
14.108.834.01n
%48.72y%;8.2y%;48.16y%;24.8y
AnalysisoductPr
14.128.834.021n
2
2
2
2222
N
O
CO
oductsPrDry
NOOHCO
oductsPr
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
22222188
22222188
N47OH9CO8N47O5.12HC
N)76.3(aOcHbCON)76.3(aaOHC
Combustion with e = 20%
18. eq.CombustionActualN4.56O5.2OH9CO8N4.56O15HC
N47)20.1(dOOH9CO8N47)20.1(O5.12)20.1(HC
222222188
222222188
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%100x
9.66
8
y
9.664.565.28n
gasexhaustdryofmolesn
2CO
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
kg162)18(9kg
9.754.565.298n
gasexhaustofmolesn
18H8C
O2H
O2H
d
d. volume of dry exhaust gas per kg of fuel if T = 290 K and P = 101.33 KPa
18H8C
3
18H8C
d
3
d
3
d
d
d
kg
m
14
114
1,591.9
kg
V
m1,591.9V
m
33.101
)290)(3143.8(9.66
V
TRnPV
9.664.565.28n
gasexhaustdryofmolesn
e. the partial pressure of the water vapor in the exhaust
KPa015.12)33.101(1186.0P
P
P
y
%86.11%100x
9.75
9
y
9.754.565.298n
gasexhaustofmolesn
O2H
O2H
O2H
O2H
f. the dew point temperature of the products
occurwillmixture
in theOHofoncondensatiDPT,belowcooledismixturetheif
C467.46DPT
TableSteamFrom
KPa015.12P
PingcorrespondretemperatusaturationDPT
2
O2H
O2H
19. 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
%8.10108.01
F
A
F
A
e
94.14
mn12
)m25.0n(28.137
F
A
RatioFuel-AirlTheoretica
56.16
)26(1)12(12
)28)(08.77()32(5.20
F
A
77.08N+3.23O+0.185CO+O13H+CO816.1177.08N+20.5O+HC
abyequationthedividing
83.5N+3.5O+0.2CO+OcH+12.8COb(3.76)N+bO+HaC
Theoreical
Actual
alTheroretic
Actual
2222222612
2222222612
20. 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.
18m
8n
2n2n25.2
2n2m
HC
isformulatheFuelpetrolFor
3.eqn25.2m
m33.6n25.14
2.eqm33.6m
158.0
1
mn12
mn12
m
158.0
mn12
m
H%
1.eqn25.14n
842.0
12
mn12
mn12
n12
842.0
mn12
n12
C%
22nn
%2.16162.01
57.14
49.17
e
49.17
mn12
b28a32
F
A
84.13c
33.22a
b)76.3(a
97.83b
N97.83O72.3CO23.1OcH11.07CObNaOHC
gas)fluedryofmoles100(BasisEquationCombustion
57.14
mn12
)m25.0n(28.137
F
A
Actual
222222188
lTheoretica
21. 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,
25C and 101 KPa, respectively. Flue gas temperature is 300C 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%
ProductsN52.353O78.18SO79.0OH71.0364.91CO
AirN07.35393.9O
FuelO5.58H0.79S0.45N3.14O25.13H64.91C
52.353g
g246.822)25.1(22(0.45)
18.78f
2f2(0.79)30.712(64.91)2(2)(1.25)75.1(5.58)2(3.14)
ProductsNgfOSO79.0OH71.0364.91CO
AirN46.822)25.1(2O(1.25)75.1
FuelO5.58H0.79S0.45N3.14O25.13H64.91C
E.A.25%ithequation wCombustion
ProductsN91.822SO79.0OH71.0364.91CO
AirN46.82275.12O
FuelO5.58H0.79S0.45N3.14O25.13H64.91C
ProductseNdSOOcHbCO
Aira(3.76)NaO
FuelO5.58H0.79S0.45N3.14O25.13H64.91C
TA100ithequation wCombustion
22222
22
2222
22222
22
2222
2222
22
2222
2222
22
2222
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
23. 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
%7.82y
%1.5y
%2.12y
5.7381205.858n
ANALYSISORSAT
C8.52DPT
KPa16.14)325.101(1398.020PH
P
Pi
yi
%17.71y
%37.4y
%98.13y
%48.10y
ProductsofAnalysisVolumetric
%100x
n
ni
yi
5.8586115.3712090n
52.11
)32(10)28(30)16(60
)28(611)32(5.162
F
A
N611O5.37OH120CO90N611O5.162O10CO30CH60
5.37d
N)76.3(a)30.1(dOOH120CO90N)76.3(a)30.1(aO)30.1(O10CO30CH60
EA305withCombustion
120c
90b
125a
N)76.3(aOOcHbCON)76.3(aaOO10CO30CH60
TA100%withCombustion
N2
O2
CO2
GasDry
N2
O2
O2H
CO2
oductsPr
22222224
22222224
22222224
24. 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
N63.172O48.6OH20CO5.34
N73.121O38.32CO5.4CO27O6.0N9.50H14CH3
2222
2222224
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 25C 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
Combustion with 100% theoretical air
22222283
22222283
2222283
2222283
N458.22O996.0CO045.0O4HCO955.2N458.22O973.5HC
EQUATIONCOMBUSTION
996.0f
balanceoxygenBy
N715.81)20.1(fOCO045.0O4HCO955.2N715.81)0(1.28O(1.20)4.97HC
EA20%withCombustion
N715.81dCOO4HCO955.2N715.81O978.4HC
978.4a
dcb2a2
4c
c2)8(1
955.2045.03b
045.0)3(015.0d
a(3.76)NdCOOcHbCOa(3.76)NaOHC
emissionCOandTA100%withCombustion
26. 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
2222222612 N95.864.625OOH3112CON95.86O125.23HC
The theoretical A/F ratio
2612lTheoretica HCofkg
airofkg
94.14
mn12
)m25.0n(28.137
F
A
The actual A/F ratio
2612tActual HCofkg
airofkg
674.18
F
A
)e1(
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
xiRiΣ
M
R
R
kg
kg
691.28yiMiΣM
m
Kg of CO2 per kg of fuel
106.3
26)12(12
528
HCofkg
COofkg
2612
2
% C and % H in the fuel
27. %3.15100x
26)12(12
26
H%
100%x
mn12
m
H%
%7.84100x
26)12(12
)12(12
C%
100%x
mn12
n12
C%
COMBUSTION ENGINERING
SAMPLE EXERCISES
Example No. 1 (Coal Fuel)
In a furnace. 500 kg of coal with an Ultimate analysis of C = 78 %, H2 =5 %, O2 = 8 %, S = 1%, M = 2% and an A = 6
% is fully consumed hourly with 30% excess air. How much air must be fed into the furnace, and how much flue gas is
produced in kg/hr.
Converting the UA on an ashless basis, then to molal ashless analysis
Fuel U.A.%
U.A.
Ashless
Mi xi/Mi
Molal
Analysis
C 78 83.0 12 0.069 69.21
H2 5 5.3 2 0.027 26.62
O2 8 8.5 32 0.003 2.66
N2 0 0.0 28 - -
S 1 1.1 32 0.0003 0.33
H2O 2 2.1 18 0.0012 1.18
Ash 6 - - - -
100 100 0.0999 100
Ashless 94%
29. Example No. 2(Hydrocarbon mixture with emission)
A fuel mixture with a molal analysis of 50% C7H16 and 50% C8 H18 is oxidized with 30% excess air. If 15 moles of the
carbon atoms oxidizes to CO and 10 of Nitrogen atoms goes to NO, Determine
a. the combustion equation
b. the actual air fuel ratio
c. the cubic meter of exhaust pollutants formed per kg of fuel at P = 101 KPa and t = 900C
d. the Orsat analysis of the products
18.5726h
balanceNitrogenby
75.351g
balanceoxygenBy
hNgO10NO15COOcHbCOa(3.76)N)30.1(aO)30.1(H50CH50C
(30%)airexcesswithCombustion
50.1172a
01158502(735)2a
BalanceOxygen
850c
2c50(18)50(16)
BalanceHydrogen
735b
15b50(8)50(7)
BalanceCarbon
fN10NO15COOcHbCOa(3.76)NaOH50CH50C
fuelofmoles100:Basis
TA100%withCombustion
222222188167
22222188167
%74.83%100x
93.837,6
18.5726
y
%14.5%100x
93.837,6
75.351
y
%15.0%100x
93.837,6
10
y
%22.0%100x
93.837,6
15
y
%75.10%100x
93.837,6
357
y
AnalysisOrsat
93.837,618.726575.3511015357n
86.6
)114(50)100(50
46.386,73
Fuelofkg
pollutantexhaustofm
pollutantexhaustofm46.386,73
101
)273900)(3143.8(760
P
TRn
V
moles7601015735ttannPollu
56.19
)114(50)100(50
)28(18.5731)32(25.1524
Actual
F
A
N18.7265O75.35110NO15COOH508CO357N18.5731O25.1524H50CH50C
EquationCombustion
2N
2O
NO
CO
CO2
GasFlueDry
3
3
222222188167
30. Example No.3 (Unknown Fuel with known Orsat analysis)
A gas turbine power plant receives an unknown type of hydrocarbon fuel. Some of the fuel is burned with air, yielding the
following Orsat analysis of the products of combustion, CO2 = 10.5%; O2 = 5.3% and N2 = 84.2%. Determine
a. the percentage, by mass, of carbon and hydrogen in the fuel
b. the percentage of theoretical air.
115.4%15.4)(100airlTheoretica%
%4.15
)38.26(1)5.10(12
)38.26(25.05.1028.137
F
A
18.20
)38.26(1)5.10(12
)28(2.84)32(394.22
F
A
%31.17
)38.26(1)5.10(12
)38.26(1
H%
%69.82
)38.26(1)5.10(12
)5.10(12
C%
N2.84O3.5OH19.132CO5.10N2.84O394.22HC
EquationCombustion
38.26m
b21(m)
balancehydrogenBy
19.13b
2(5.3)b2(10.5)2(22.394)
balanceOxygen
394.22
3.76
84.2
a
84.2(2)a(3.76)2
balancenitr0genBy
10.5n
10.51(n)
balancecarbonBy
N2.84O3.5ObH2CO5.10N)76.3(aaOHC
lTheoretica
Actual
2222238.265.10
22222mn
31. kg/hr6.1079)592.21(50m
m)592.20(mm
fuelkg
airofkg
592.20
)484(
)47)(20.1(28.137
F
A
p
FFp
Example No.4 (Hydrocarbon Fuel)
A fuel represented by C7H16 is oxidized with 20% excess air and the mass of fuel required for combustion is 50 kg/hr. Determine
the mass flow rate of the products in kg/hr.
Problem: (Hydrocarbon Mixture and DPT)
A fuel with a molal analysis of 80% C12H26 and 20% C14H30 is burned with 30% excess air. The flue gas is at atmospheric
pressure. Find the minimum exhaust temperature to avoid condensation if P = 101 KPa.
22222230142612
22222230142612
2222230142612
N08.336,9O573OH340,1CO240,1N08.336,9O483,2H20CH80C
EquationCombustion
573d
balanceoxygenBy
N)76.3(a)30.1(dOOcHbCON)76.3(a)30.1(aO)30.1(H20CH80C
0.30)(eEAwithCombustion
910,1a
c2b2a
balanceOxygen
340,1c
2c20(30)80(26)
balsnceHydrogen
240,1b
b20(14)80(12)
BalanceCarbon
N)76.3(aOcHbCON)76.3(aaOH20CH80C
TA100%withCombustion
fuelofmoles100Basis
C417.47KPa10.84attDPT
KPa84.10
08.489,12
340,1
101P
n
n
P
P
08.489,1208.336,9573340,1240,1n
sat
OH
oductsPr
OHOH
oductsPr
2
22
Problem Natural Gas
A natural gas fuel showed the following percentages by volume: C2H6 = 9%; CH4 = 90%; CO2 = 0.2 % and N2 = 0.8 %. If this
gas is used as fuel and is burned with 20% excess air for complete combustion, Determine
a) The Combustion Equation
b) The theoretical and actual air fuel ratio
c) The molecular weight and gas constant of the products
d) The volume of air required per m3
of natural gas if the gas and air are at temperature of 16C and a pressure of 101.6
KPa.
33. Sample Problem (Unknown Fuel – Known Orsat analysis)
An unknown hydrocarbon is used as fuel in a diesel engine, and after an emission test the orsat analysis shows,
CO2 = 12.5% ; CO = 0.3% ; O2 = 3.1% ; N2 = 84.1%.Determine
a. the actual air-fuel ratio
b. the percent excess air
c. the fuel analysis by mass
CnHm + aO2 + a(3.76)N2 12.5CO2 + bH2O + 0.3CO + 3.1O2 + 84.1N2
By Carbon balance
n = 12.5 + 0.3
n = 12.8
By Hydrogen balance
m = 2b eq. 1
By Oxygen balance
2a = 2(12.5) + b + 0.3 + 2(3.1) eq. 2
By Nitrogen balance
a(3.76) = 84.1
a = 22.367
substituting a to eq. 2
b = 13.234
substituting b to eq. 1
m = 26.47
fuelofkg
airofkg
05.17
47.26)8.12(12
)28(1.84)32(367.22
F
A
N1.842O1.3CO3.0OH234.13CO5.12N1.84O367.22HC
a
2222247.268.12
Combustion with 100% theoretical air
n = 12.8 ; m = 26.47
08.12
2365
28571.2
V
V
m2365
6.101
)27316)(143.8)(8.02.909(
V
TRnPV
m28571.2
6.101
)27316)(3143.8)(288.9548.253(
V
TRnPV
F
a
3
F
FF
3
a
aa
K-kg
KJ
0.298
9.27
3143.8
R
kg
kg
9.27M
1312.588
)955.088(2842.3(32)207(18)108.2(44)
MyiMi
1312.588955.08842.3207108.2
ProductsofMoles
mol
35. 1 mole C
1 mole O2
1 mole CO2
25C, 1 atm
25C, 1 atm
25C, 1 atm
Combustion
Chamber
Q
W
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.
ENTHALPY OF FORMATION
The “Enthalpy of Formation of a compound is the enthalpy at the Arbitrary Reference State (t = 25C and P = 1 Atm).
1 mole C
1 mole O2
1 mole CO2
25C, 1 atm
25C, 1 atm
25C, 1 atm
Combustion
Chamber
Q
Let;
HR – total enthalpy of Reactants
HP – total enthalpy of products
From 1st Law
"Exothermic"beingsreaction'thetodueissignNegative:Note
COofformationofEnthalpy
kg
KJ
757,393)h(
therefore
KJ-393,757HQ
elements)allare(for theyZeroisreactantstheallofenthalpythebut
hnΣhnΣQ
or
HHQ
2
m
COf
P
ii
P
ii
R
PR
2
FIRST LAW ANALYSIS FOR STEADY STATE REACTING SYSTEM
36. Atmosphere1ispressurethat thedenotes
where
)hh( 298
By energy balance
In most cases, neither the reactants nor the products are at the reference state (t = 25C and P = 1 Atm). In these case we must
account for the property change between the reference state and the actual state. The change in enthalpy between the reference
state and the actual state is
General Equation for a Steady State, Steady flow reaction process
ADIABATIC FLAME TEMPERATURE
1 mole C
1 mole O2
1 mole CO2
25C, 1 atm
25C, 1 atm
25C, 1 atm
Combustion
Chamber
If Q = 0 ; W = 0 ; KE = 0 and PE = 0, all the thermal energy would go into raising the temperature of the products of
combustion. When the combustion is complete, the maximum amount of chemical energy has been converted into thermal
energy and the temperature of the product is at its maximum. This maximum temperature is called the “ Adiabatic Flame
Temperature” (AFT)
PR HH
ENTHALPY OF COMBUSTION OR HEATING VALUE
It is the difference between the enthalpies of the products and the reactants at the same temperature T and Pressure P.
weightmolecular-M
where
kg
KJ
M
)HH(
h
kg
KJ
)hh(hnΣ)hh(hnΣh
kg
KJ
HHh
RP
RP
mol
i298fi298fj
P
RP
mol
RPRP
ii
P
ii
R
hnΣWhnΣQ
j298fj
P
i298fi
R
)hh(hnΣW)hh(hnΣQ
37. HEATING VALUE FORMULAS
Heating Value - is the energy released by fuel when it is completely burned and the products of combustion are cooled to the
original fuel temperature.
Higher Heating Value (HHV) - is the heating value obtained when the water in the products is liquid.
Lower Heating Value (LHV) - is the heating value obtained when the water in the products is vapor.
For Solid Fuels
kg
KJ
S304,9
8
O
H212,144C820,33HHV 2
2
where: C, H2, O2, and S are in decimals from the ultimate analysis on an ashless basis
For Coal and Oils with the absence of Ultimate Analysis
fuelofair/kgofkg
041,3
HHV
F
A
t
HHV = 31,405C + 141 647H KJ/kg
For Liquid Fuels
HHV = 43,385 + 93(Be - 10) KJ/kg
Be - degrees Baume
For Gasoline
HHV = 41,160 + 93 (API) KJ/kg
LHV = 38,639 + 93 (API) KJ/kg
For Kerosene
HHV = 41,943 + 93 (API) KJ/kg
LHV = 39,035 + 93 (API) KJKkg
For Fuel Oils
HHV = 41,130 + 139.6(API) KJ/kg
LHV = 38,105 + 139.6(API) KJ/kg
API - American Petroleum Institute
For Fuel Oils (From Bureau of Standard Formula)
HHV = 51,716 - 8793.8 (S)2 KJ/kg
LHV = HHV - QL KJ/kg
QL = 2,442.7(9H2) KJ/kg
H2 = 0.26 - 0.15(S) kg of H2/ kg of fuel
Be130
140
S
API131.5
141.5
S
38. j298fj
P
i298fi
R
)hh(hnΣW)hh(hnΣQ
Where:
S @ t = S - 0.0007(t-15.56)
S - specific gravity of fuel oil at 15.56 C
H2 - hydrogen content of fuel oil, Kg H/Kg fuel
QL - heat required to evaporate and superheat the water vapor formed by the combustion of hydrogen in the fuel,KJ/kg
S @ t - specific gravity of fuel oil at any temperature t
Oxygen Bomb Calorimeter - instrument used in measuring heating value of solid and liquid fuels.
Gas Calorimeter - instrument used for measuring heating value of gaseous fuels.
Sample Problem No. 1 (Coal Fuel/Heating Value)
A sample of coal fuel has the following ultimate analysis, percent by mass:H2 = 5.6% ; C = 53.4% ; S = 0.1% ; N2 = 0.1% ;
O2 = 37.9% and Ash = 2.9%. This coal will be used as a fuel by burning it with no excess air in a furnace. Determine
a) the air--fuel ratio on a mass basis
b) the molar analysis of products of combustion
c) the HHV of the fuel in KJ/kg
kg
KJ
S304,9
8
O
H212,144C820,33HHV 2
2
00103.0%103.0S
00103.0%103.0N
39.0%39O
058.0%8.5H
55.0%55C
basisashlessantogiven U.A.theConverting
2
2
2
6.6
F
A
fuelofkg
airofkg
S29.4
8
O
H32.34C44.11
F
A
t
2
2
t
kg
KJ
19,889.8HHV
kg
KJ
S304,9
8
O
H212,144C820,33HHV 2
2
SAMPLE PROBLEM NO. 2
A small gas turbine uses C8H18 (liquid) for fuel and 400% theoretical air. The air and fuel enters at 25C and the products of
combustion leave at 900 K. The output of the engine and the fuel consumption are measured and it is found that the specific
fuel consumption is 0.25 kg/sec of fuel per Megawatt output. Determine the heat transfer from the engine per kgmol of fuel.
Assume complete combustion.
Combustion with 100% TA
22222188 N47OH9CO8N47O5.12HC
Combustion with 400% theoretical air or e = 300%
222222188 N188O5.37OH9CO8N188O50HC
39. Air and Fuel Products
Q = -232,000 KJ/kgmol
WEngine
SAMPLE PROBLEM NO. 3
A diesel engine used gaseous Dodecane (C12H26) as fuel. The air and fuel enters the engine at 25C and 100% excess air is
required for combustion. The products of combustion leaves at 600 K and the heat loss from the engine is 232,000 KJ/kgm
fuel. Determine the work for a fuel rate of 1 kgm/hr of fuel.
Combustion with 100% theoretical air
222222612 N56.69OH13CO12N56.69O5.18HC
Fuel
mFuel
j298fj
P
i298fi
R
kg
KJ
463Q
kg
KJ
776,52Q
HP000,456HRQ
kgm
KJ
456,000
kgm
kg
114
25.0
1000
W
)hh(hnΣW)hh(hnΣQ
j298fj
P
i298fi
R
)hh(hnΣW)hh(hnΣQ
40. Combustion with 400% theoretical air or e = 300%
2222222612 N12.139O5.18OH13CO12N12.139O37HC
SAMPLE PROBLEM No. 4
A fuel oil represented by C10H22 is burned with 25% excess air in a laboratory to determine its heating value. Air and fuel enters
at 25C and the products of combustion leaves the apparatus at 500 K. Determine the higher heating value of the fuel
considering;(assume hf(C10H22) = -289,402 KJ/kgmol Fuel
a) Complete combustion with no CO in the products
b) Incomplete combustion with 5% of the carbon atoms oxidizes to CO
222222210 N28.58OH11CO10N28.58O5.15HC
TA100%withCombustion
2222222210 N85.72O875.3OH11CO10N85.72O375.19HC
0.25)(eEAwithCombustion
KW1568.97
3600
1
585,648,284.W
kg
KJ
585,648,284.W
)58.284,880,5(000,232W
)HRHP(QW
HPWHRQ
Fuelmol
42. CHIMNEY
By. ENGR. YURI G. MELLIZA
FUNCTIONS OF CHIMNEY
1) To dispose the exhaust gases at suitable height so that no pollution will occur in the vicinity.
2) To produce the necessary draft required for the flow of gases.
DRAFT - is the difference between the absolute gas pressure at any point in a gas flow passage,(furnace, chimney, air heater
and etc.) and the ambient atmospheric pressure. Draft is positive if Pa Pgas and is negative if Pa Pgas.
D - diameter of chimney, m
H - height of chimney, m
THEORETICAL DRAFT(Dt):
water
gasair
t
)(H1000
D
where:
Dt - theoretical draft in mm of water
H - height of chimney or smokestack in meters
a - density of air in kg/m3
g - density of flue gas in kg/m3
w - density of gage fluid (water) in kg/m3
ACTUAL DRAFT(Da):
Da = Dt - DL mm of H2O
where: DL - draft losses, and usually expressible in percentage of the theoretical draft.
THEORETICAL VELOCITY(v):
gasgh2v
where:
v - theoretical velocity of the flue gas in the chimney, m/sec
hg - draft in meters of flue gas
43. Gasofm
)(1000
)(D
h
gas
watert
gas
ACTUAL VELOCITY OF FLUE GAS(v'):
kv'v
where: K - velocity coefficient, whose value ranges from 0.30 to 0.50
DIAMETER OF CHIMNEY(D):
'v
Q4
D
gas
where:
Qg - volume flow rate of flue gas, m3
/sec
VOLUME FLOW RATE OF FLUE GAS
sec
mm
Q
3
gas
gas
gas
mg = mass flow rate of flue gas, kg/sec
MASS FLOW RATE OF FLUE GAS:
a) Without considering Ash loss:
1
F
A
mm Fuelgas
b) Considering Ash loss
LossAsh1
F
A
mm Fuelgas
Where:
Ash loss in decimal
VOLUME FLOW RATE OF PRODUCTS OF COMBUSTION (Obtained from the balanced combustion equation)
44. secminVQ
sec
m
sec
kg
m
P
TRn
V
kg
nnnnnnnnnn
n
TRnPV
3/
gas
3
Fuel
Fuel
3NO2NONOCHCO2N2O2SOO2H2CO
MOLECULAR WEIGHT and GAS CONSTANT OF PRODUCTS OF COMBUSTION
Kkg
KJ
M
R
R
n
MnMnMnMnMnMnMnMnMnMn
M
ProductsofMolesTotal
ProductsofMass
M
oductsPr
3NO3NO2NO2NONONOCHCHCOCO2N2N2O2O2SO2SOO2HO2H2CO2CO
Problem No. 1
A power plant at an elevation 0f 600 m has a chimney with an actual draft of 19 mm of H2O and frictional losses of 15%. Flue
gases at the rate of 5 kg/sec enters the chimney at 180C and leaves the chimney at 150C, with a volumetric analysis of
CO2 = 12.7%
H2O = 2.2%
SO2 = 0.1%
O2 = 6.8%
N2 =77.9%
CO = 0.30%
and leaves the chimney at 150C. Calculate the height and diameter of the chimney assuming velocity coefficient of k = 0.40.
(Ps = 760 mm Hg and Ts = 294 K)
ANSWER: H = 98.6 M ; D = 1.3 m
K-kg
KJ
276.0
30.12
8.3143
R
12.30M
0.30(28)77.9(28)6.8(32)0.1(64)2.2(18)12.7(44)M
yiMiΣM
GasFlue
GasFlue
GasFlue
46. Problem No. 2
A steam power plant using high grade coal having the Ultimate analysis as follows, C = 81% ; H2 = 12.0% ; O2 = 1.2%; N2 =
1.6% ; S = 2.4%; H2O = 1.7% and Ash = 0.1% requires 30% excess air for complete combustion. Exhaust gases in the
smokestack has an average temperature of 650K and pressure of 109 KPaa and produces and actual draft of 27 mm of water
with average draft losses of 11% and coefficient k in the stack of 0.35. Ambient air condition is 101 KPa and 25C. For a coal
consumption of 1500 kg/hour as fired, determine the required height and diameter of the smokestack.
0.1%A
1.7%OH
2.4%S
1.6%N
1.2%O
12%H
81%C
AnalysisUltimate
2
2
2
2
Comp. UA Ashless(xi) Mi Xi/Mi Yi(Molal)
C 81 81.1 12 0.068 51.87
H2 12 12.01 2 0.06 46.1
O2 1.2 1.201 32 0.0004 0.29
N2 1.6 1.602 28 0.0006 0.44
S 2.4 2.402 32 0.0008 0.58
H2O 1.7 1.702 18 0.0009 0.73
Ash 0.1 - - - -
Denominator (100-0.10 100 0.1303 100
sec
Productsofm
18.13V
hr
Productsofm
07.462,47)1500(64.31V
64.31
Fuelofkg
Productsofm
kg63.767FuelofMass
m86.288,24
109
)650)(3143.8(n
V
89.489n
N05.368O2.562
SO58.0OH83.46CO87.51N61.367O77.97OH73.0S58.0N44.0O29.0H1.4651.87C
fNeO
SO58.0OH83.46CO87.51N)78.282(30.1O)21.75(30.1OH73.0S58.0N44.0O29.0H1.4651.87C
0.30)airEA(ExcesswithCombustion
N22.283SO58.0OH83.46CO87.51N78.282O21.75OH73.0S58.0N44.0O29.0H1.4651.87C
dNcSOObHaCON)76.3(xxOOH73.0S58.0N44.0O29.0H1.4651.87C
Fuelofmoles100:Basis
100%TAwithCombustion
3
oductsPrF
3
oductsPrF
3
oductsPr
oductsPr
oductsPr
22
222222222
22
222222222
2222222222
2222222222
3
Da = Dt - DL mm of H2O
48. Problem no. 3
The actual velocity of a gas entering in a chimney is 8 m/sec. the gas temperature is 25C and pressure of 98 KPa with a gas
constant of 0.287 KJ.kg-K. Determine the chimney diameter if mass of gas is 50,000 kg/hr. (1.37 m)
COMBUSTION ENGINEERING
QUIZ NO. 1(April 15, 2019)
Name _______________________
Problem No. 1
A fuel represented by C7H16 is oxidized with 20% excess air and the mass of fuel required for combustion is 50 kg/hr. Determine
the mass flow rate of the products in kg/hr.
hr
kg
956)50(12.19M
productsofrateflowmassM
kgFuel
Productskg
12.19
100
1,912.10
1612(7)
1,912.10
m
m
fuelofkgperProductsofMass
kg1,912.10)28(632.49)32(2.28(18)7(44)m
)(mProductsofMass
N632.49O2.2O8H7CON632.49O13HC
EquationCombustion
2.2)m25.0n(ed
a(3.76)N)20.1(dOOcHbCOa(3.76)N)20.1(aO)20.1(HC
0.20)(eEAwithCombustion
8m5.0c
7nb
11m25.0na
a(3.76)NOcHbCOa(3.76)NaOHC
TA100%withCombustion
hr
kg
50:m
20%:e
HC:Fuel
Products
oductsPr
Fuel
products
products
products
222222167
222222167
22222167
Fuel
167
49. Problem No. 2
The analysis of the natural gas showed the following percentages by volume: C2H6 = 9%; CH4 = 90%; CO2 = 0.2 % and N2 =
0.8 %. Find the volume of air required per cu,m. of gas if the gas and air are at temperature of 16C and a pressure of 101.6
KPa.
04.796d
2da(3.76)22(0.8)
BalanceNitrogen
5.211a
c2b2a)0.2(2)
BalanceOxygen
207c
2c4(90)6(9)
BalanceHydrogen
2.108b
b0.2902(9)
BalanceCarbon
dNOcHbCON)76.3(aaO0.8N0.2CO90CHH9C
fuel)ofmoles100:(BasisTA100%withCombustion
2222222462
06.10
100
1,006.74
fuelofm
Airofm
ratiocvolumetritoequalisratiomolethepressure,andtetemperarusametheAt
P
TRn
V
TRnPV
100n
Moles1,006.7424.7955.211n
N04.796OH207CO2.108N24.795O5.2110.8N0.2CO90CHH9C
3
3
Gas
Air
2222222462
50. Problem No. 3
A fuel with a molal analysis of 80% C12H26 and 20% C14H30 is burned with 30% excess air. The flue gas is at P = 101 KPa.
Dtermine
a. the combustion equation
b. the actual air-fuel ratio
c. the partial of H2O and DPT in the products
C417.47DPT
KPa84.10attsatDPT
KPa84.10P
08.489,12
1340
101
P
08.489,12
1340
P
P
12,489.0808.933657334012401n
41.19
)198(20)170(80
)28(08.9336)32(2483
Actual
F
A
N08.9336O573OH3401CO2401N08.9336O2483HC20H80C
EquationCombustion
573d
d2c)2(b)2(a30.1
a(3.76)N)30.1(dOOcHbCOa(3.76)N)30.1(aO)30.1(HC20H80C
0.30)(eEAwithCombustion
1910a
cb2a2
1340c
c2)30(20)26(80
1240b
b)14(20)12(80
a(3.76)NOcHbCOa(3.76)NaOHC20H80C
fuel)ofmoles100(basis;TA100%withCombustion
H2O
H2O
H2O
oductsPr
22222230142612
22222230142612
2222230142612
COMBUSTION ENGINEERING
QUIZ NO. 2 (April 23, 2019)
Name ______________________________
Problem No. 1
The mass analysis of hydrocarbon fuel A is 88.5% Carbon and 11.5% Hydrogen. Another hydrocarbon fuel B requires 6%
more air than fuel A for complete combustion. Calculate the mass analysis of Fuel B.
51. %81.848481.0)1519.1(C
%19.151519.0H
H)44.1132.34(44.11915.14
915.14H32.34)H1(44.11
F
A
)H1(C
1HC
915.14)07.14)(06.1(
F
A
BFuelFor
07.14)115.0(32.34)885.0(44.11
F
A
AFuelFor
Problem No. 2
A diesel engine uses a hydrocarbon fuel represented by C12H26 and is burned with 30% excess air. The air and fuel is supplied
at 1 atm and 25C. Determine
a. the actual air-fuel Ratio
b. the m3
of CO2 formed per kg of fuel if the product temp. is 400C and a pressure of 1 atm.
c. The M and R of the Products
d. The M and R of the dry flue gas
e. If 1.5 moles of carbon oxidizes to CO, determine the molecular weight of the resulting products
9.3
26)12(12
7.662
m
V
m7.662
325.101
)273400)(3143.8(12
V
42.19
mm12
0.25m)(n28.137)30.1(
F
A
N428.90O55.5OH31CO21N428.90O05.24HC
55.50.25m)e(nd
130.5mc
18.50.25mna
12b
nb
a(3.76)N)30.1(dOOcHbCOa(3.76)N)30.1(aO)30.1(HC
130%TAwithCombustion
Fuel
2CO
3
2CO
Actual
2222222612
2222222612
K-kg
KJ
29.0
7.28
3143.8
R
kg
kg
7.28
98.120
)28(428.90)32(55.5)18(13)44(12
n
m
M
98.120428.9055.51312n
moloductsPr
oductsPr
oductsPr
53. COMBUSTION ENGINEERING
QUIZ NO. 3 (April 24, 2019)
Name ___________________________________
Problem 1: Calculate the theoretical oxygen and air required to burn 1 kgmol of carbon, and 1 kgmol of Hydrogen.
44.11
kgC
kgAir
67.2
3
8
kgC
kgO
1183
443212
COOC
2
22
32.34
KgH
kgAir
8
1
8
KgH
kgO
981
18162
OHO
2
1H
2
222
Problem 2: Calculate the theoretical Oxygen--fuel ratio and Air--fuel ratio on a mass basis for the combustion of ethanol,
C2H5OH.
1165241
)28(28.11)32(3
kgFuel
kgAir
46
96
1165241
)32(3
kgFuel
kgOxygen
N28.11OH3CO2N28.11O3OHHC
3
2
13)2(2
a
cb2a21
3c
c215
b2
N)76.3(aOcHbCON)76.3(aaOOHHC
TA100%withCombustion
2222252
2222252
Problem 3: Determine the molal analysis of the products of combustion when octane C8H18 is burned with 100% excess air.
222222188
222222188
222222188
N94O5.12OH9CO8N94O25HC
N)76.3(a)11(dOOcHbCON12.92O5.24HC
)m25.0n(ed
m5,0c
nb
m25.0na
N)76.3(a)11(dOOcHbCON)76.3(a)11(aO)11(HC
100%)(eEAwithCombustion
%11.76%100x
5.123
94
y
%12.10%100x
5.123
5.12
y
%29.7%100x
5.123
9
y
%48.6%100x
5.123
8
y
5.123945.1298oductsPrn
2N
2O
O2H
2CO
54. Problem 4: A certain fuel has the composition C10H22. If this fuel is burned with 50% excess air, what is the composition of the
products of combustion?
)m25.0n(ed
m5,0c
nb
m25.0na
N42.87O75.7OH11CO10N42.87O25.23HC
50%)(eEAwithCombustion
2222222210
Problem 5: A sample of pine bark has the following ultimate analysis, percent by mass: C = 53.4% ; H2 = 5.6% ; O2 = 37.9%
; N2 = 0.1% ; S = 0.1% ; Ash = 2.9%. This bark will be used as a fuel by burning it 30% excess air. Determine the actual
air-fuel
ratio.
58.8
F
A
)e1(
F
A
6.6S29.4
8
O
H32.34C44.11
F
A
lTheoreticaActual
2
2
lTheoretica
COMBUSTION ENGINEERING
Activity No. 4 (APRIL 29, 2019)
NAME __________________________________
Problem No. 1
A combustor receives 340 m3
/min of natural gas having a volumetric chemical composition of 15% C2H6 and 85% CH4 that
requires 25% excess air. Air and fuel enters the combustor at 25C (298 K)and 101 KPa and products of combustion leaves at
1100 K. Emission test gives that 2 moles of CO and 1 mole NO is found in the products. Determine
a. The actual combustion equation
b. The actual air – fuel ratio
c. The heating value of the fuel (HP – HR) KJ/kg
d. The mass flow rate of fuel in kg/min
e. The volume flow rate of emission gases (CO2, CO and NO) in m3
/hr
C2H6: M = 30; Cp = 1.7549 KJ/kg-K; Cv = 1.4782 KJ/kg-K
CH4: M = 16; Cp = 2.1377 KJ/kg-K; Cv = 1.6187 KJ/kg-K
N9.042,1O5.551NO2COOH152CO131N4.043,1O5.22785CHH15C
NgfO1NO2COOH152CO131N72.348)25.1(O222)25.1(85CHH15C
fuelofmoles100:Basis
0.25)AIR(eEXCESSwithCombustion
N22.3481NO2COOH152CO131N72.348O22285CHH15C
dN1NO2COOcHbCOa(3.76)NaO85CHH15C
fuelofmoles100:Basis
TA100%withCombustion
222222462
222222462
22222462
22222462
Fuel UA Ashless
C 53.4 55
H2 5.6 5.8
O2 37.9 39
N2 0.1 0.103
S 0.1 0.103
H20 0 0
Ash 2.9 0
Total 100 100
56. COMBUSTION ENGINEERING
Activity No. 5 (May 02, 2019)
NAME __________________________________
A hydrocarbon Fuel C7H16 (n-Heptane) was tested in a laboratory in order to determine its LHV and HHV. The fuel is burned
with 50% excess air requirement for complete combustion. Air and fuel enters the apparatus at 25C and 101 KPa, while the
products of combustion leaves the apparatus at 600 K. Determine
a. the combustion equation
b. the LHV in KJ/kg of fuel
c. the HHV in KJ/kg of fuel
hf of C7H16 = -187,900 KJ/kgm
For HHV
For LHV
57. COMBUSTION ENGINEERING
Activity No. 6 (May 03, 2019)
NAME __________________________________
Multiple Choice
Instruction: Write the letter, which corresponds to the correct answer in the table below.
1. A hydrocarbon fuel represented by C8H18 is burned with air. Its gas constant in KJ/kg-K is equal to;
a. 0.073 b. 0.286 c. 0.0826 d. 0.187
2. A gaseous mixture has the following volumetric analysis O2, 30%; CO2, 40% N2, 30%. Determine its molecular weight in
kg/kgmol.
a. 25.6 b. 35.6 c. 28.9 d. 30.2
3. The products of combustion of an automotive engine using a diesel fuel has an Orsat analysis as follows; CO2 = 12.5%; O2
= 3.5%;
CO = 0.5%; N2 = 83.5%. The gas constant of the dry flue gas is equal to;
a. 0.287 b. 1.0045 c. 0.1889 d. 1.007
4. A newly designed high speed European car uses high grade gasoline fuel represented by C8H18. If this fuel is burned with
30% excess
air, determine the actual air-fuel ratio.
a. 20.5 b. 10.87 c. 15.8 d. 19.6
5. A fuel represented by C7H16 is oxidized with 20% excess air and the mass of fuel required for combustion is 50 kg/hr.
Determine the
mass flow rate of the products in kg/hr.
a. 956.05 b. 906.05 c. 806.07 d. 856.07
6. There are 20 kg of flue gas formed per kg of fuel burned for the combustion of C12H26. What is the percent excess air.
a. 27.17 b. 16.56 c. 26.67 d. 8.21
7. For a certain ideal gas mixture, R = 0.270 KJ/kg-k and k = 1.3. Determine the mass of the mixture in kg for 10 moles of
the mixture.
a. 408 b. 308 c. 508 d. 208
8. Determine the molecular weight of a products if Cp = 1.1 KJ/kg-K and k = 1.3.
a. 28.75 b. 26.75 c. 32.75 d.39.75
9. Determine the mass ratio C/H of a hydrocarbon fuel with a chemical formua of C16H32.
a. 8 b. 10 c. 4 d. 6
10. The Dalton’s Law of partial pressure is called the:
a. Law of additive volume
b. Law of additive mass
c. Law of additive moles
d. Law of additive pressure
Problem: A Fuel Oil belonging to the paraffin family that is 85%C and 15%H is burned in an internal combustion engine
and requires 20% excess for complete combustion. Determine;
11) The fuel chemical formula
a. C17H36 b. C12H26 c. C10H22 d. C14H30
12) The actual air-fuel ratio
a. 14.25 b. 17.85 c. 19.26 d. 20.26
13) The kg of CO2 formed per kg of fuel
a. 2.118 b. 1.117 c. 3.117 d. 4.118
14) The partial pressure of H2O if the total pressure of the products is 101 KPa
a. 18.72 b. 10.93 c. 75.23 d. 11.54
15) The gas constant R of the products in KJ/kg-K
a. 0.2895 b. 0.3762 c. 0.2004 d. 0.3008
ANSWER TABLE
1 a 6 a 11 a
2 b 7 b 12 b
3 a 8 c 13 c
4 d 9 d 14 d
5 a 10 d 15 a
58. 3air
m
kg
12.1
RT
P
ρ
K8.287T
KPa21.92P
levelseaabovem800At
COMBUSTION ENGINEERING
MIDTERM EXAM
Name ______________________________
Problem
A medium size internal combustion engine power plant is situated at an altitude of 800 m above sea level. The engine is 4 –
stroke diesel engine using C16H34 with 50% excess air required for combustion. The plant produces 1.5 MW of electrical energy
with a specific fuel consumption of 0.30 kg/KW – hr. Sea level condition is P = 760 mm Hg and T = 293 K. Design the required
Height and diameter of the smoke stack in the plant with an actual draft of 25 mm of H2O and assume 10% losses and average
flue gas temperature of 350C .
Fuel and air
Generator Output = 1,500 KW
Height
Diameter
Diesel Engine
sec
m
7.5
sec3600
hr
)1500(30.0
3412(16)
10,304.005
Q
m10,304.005
21.92
)273350)(3143.8(43.183
V
K-kg
KJ
289.0
M
3143.8
R
kgm
kg
74.28
18.13825.121716
)28(18.138)32(25.12)18(17)44(16
M
N18.138O25.12OH17CO16N18.138O75.36HC
airexcess50%withHCofCombustion
3
oductsPr
3
oductsPr
oductsPr
oductsPr
2222223416
3416
m05.46H
watermm
ρ
ρ-ρ000H1
D
sec
m
80.322ghv
gasflueofm85.54
)51.0(1000
)28.09(1000
h
mm09.28D
watermm25D10.0DD
m
kg
51.0
)273350)(289.0(
21.92
ρ
water
gasair
t
g
g
t
tta
3oductsPr
63. T - absolute temperature, K
- specific weight, KN/m3
Q - capacity, m3
/sec
BP - brake power, KW
N - speed, RPM
W - work, KW
m - mass flow rate, kg/sec
H - head, m
Example (Blower)
A steam power plant uses coal as fuel having the ultimate analysis as follows: C = 72% ; H2 = 5%; O2 = 10%; N2= 1.2%; S =
3.3%; M = 0.1% & A = 8.4% This coal is burned with 50% excess air for complete combustion. Exhaust gases leaves the
boiler at 100 KPa and 600 K and is compressed in a blower isentropically to a pressure of 110 KPa producing the necessary
draft in the chimney. If the plant consumes 1.5 MTon per hr of coal, determine the motor power required by the blower for a
blower efficiency of 70%.
Fuel U.A. Ashless Mi Xi/Mi Molal Analysis
C 72 78.6 12 0.066 66.93
H2 5 5.46 2 0.027 27.89
O2 10 10.92 32 0.003 3.49
N2 1.2 1.31 28 0.0005 0.48
S 3.3 3.6 32 0.0011 1.15
H2O 0.1 0.11 18 0.0001 0.06
Ash 8.4 - - - -
100 100 0.098 100
sec
kg
011.7
sec)3600
hr
)1000)(5.1(83.16m
83.16
0.06(18)1.15(32)0.48(28)3.49(32)27.89(2)66.93(12)
)28(46.443)32(27.39)64(15.1)18(95.27)44(93.66
Fuelofkg
Productsofkg
280.0
71.29
3143.8
R
71.29
46.44327.3915.195.2793.66
)28(46.443)32(27.39)64(15.1)18(95.27)44(93.66
M
oductsPrN46.443O27.39SO15.1OH95.27CO93.66
AirN98.442117.81O
FuelO0.06H1.15S0.48N3.49O27.89H66.93C
airExcess50%ithEquation wCombustion
oductsPr
oductsPr
22222
22
2222
Products n(Moles) Mi nM(kg) xi Cp Cv
CO2 66.93 44 2,945.10 17.13 0.845 0.656
H2O 27.95 18 503.12 2.93 1.866 1.404
SO2 1.15 64 73.63 0.43 0.624 0.494
O2 39.27 32 1,256.69 7.31 0.918 0.658
N2 443.46 28 12,416.90 72.21 1.041 0.744
17,195.44 100
65. INTERNAL COMBUSTION ENGINE CYCLE
Air Standard Cycle
o Otto Cycle
o Diesel Cycle
o Dual Cycle
AIR STANDARD OTTO CYCLE
Otto, Nikolaus August
Born:June 10, 1832, Holzhausen, Nassau
Died: Jan. 26, 1891, Cologne
German engineer who developed the four-stroke internal-combustion engine, which offered the first practical alternative to
the steam engine as a power source.
Otto built his first gasoline-powered engine in 1861. Three years later he formed a partnership with the German industrialist
Eugen Langen, and together they developed an improved engine that won a gold medal at the Paris Exposition of 1867.
In 1876 Otto built an internal-combustion engine utilizing the four-stroke cycle (four strokes of the piston for each ignition).
The four-stroke cycle was patented in 1862 by the French engineer Alphonse Beau de Rochas, but since Otto was the first to
build an engine based upon this principle, it is commonly known as the Otto cycle. Because of its reliability, its efficiency,
and its relative quietness, Otto's engine was an immediate success. More than 30,000 of them were built during the next 10
years, but in 1886 Otto's patent was revoked when Beau de Rochas' earlier patent was brought to light.
Processes
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Volume Heat Addition (V = C)
3 to 4 – Isentropic Expansion (S = C)
4 to 1 – Constant volume Heat Rejection (V = C)
Compression Ratio (r)
1
V
V
V
V
r
3
4
2
1
V1 = V4 and V2 = V3
V1 - volume at bottom dead center (BDC)
V2 – volume at top dead center (TDC)(clearance volume)
66. Displacement Volume (VD)
VD =V1 – V2 Eq. 2
Percent Clearance (C)
D
2
V
V
C
C
C1
r
Heat Added (QA)
At V = C ;
Q = mCv(T)
QA = mCv(T3 – T2)
Heat Rejected (QR)
QR = mCv(T4 – T1)
Net Work (W)
W = Q
W = QA – QR
P,V and T Relations
At point 1 to 2 (S = C)
)r(TT
P
P
r
V
V
T
T
1k
12
k
1k
1
21k
1k
2
1
1
2
At point 2 to 3 (V = C)
2
3
2
3
P
P
T
T
At point 3 to 4 (S = C)
1k
43
1k
k
1k
4
3
1k
3
4
4
3
)r(TT
)r(
P
P
V
V
T
T
At point 4 to 1 (V = C)
1
4
1
4
P
P
T
T
Entropy Change
a) S during Heat Addition
2
3
23
T
T
lnmCvSS
b) S during Heat Rejection
4
1
41
T
T
mCvSS
67. Thermal Efficiency
100%x
Q
W
e
A
%100x
Q
QQ
e
A
RA
100%x
Q
Q
1e
A
R
100%x
)TT(
)TT(
1e
23
14
100%x
)r(
1
1e 1k
Mean Effective Pressure
KPa
V
W
P
D
m
where:
Pm – mean effective pressure, KPa
W – Net Work KJ, KJ/kg, KW
VD – Displacement Volume m3
, m3
/kg, m3
/sec
Displacement Volume
VD = V1 – V2 m3
Eq. 20
VD = 1 - 2 m3
/kg Eq.21
Note:
For Cold Air Standar: k = 1.4
For Hot Air Standard: k = 1.3
AIR STANDARD DIESEL CYCLE
Diesel, Rudolf (Christian Karl)
Born: March 18, 1858, Paris, France
Died: September 29, 1913, at sea in the English Channel
German thermal engineer who invented the internal-combustion engine that bears his name. He was also a distinguished
connoisseur of the arts, a linguist, and a social theorist.
Diesel, the son of German-born parents, grew up in Paris until the family was deported to England in 1870 following the
outbreak of the Franco-German War. From London Diesel was sent to Augsburg, his father's native town, to continue his
schooling. There and later at the Technische Hochschule (Technical High School) in Munich he established a brilliant
scholastic record in fields of engineering. At Munich he was a protégé of the refrigeration engineer Carl von Linde, whose
Paris firm he joined in 1880.
68. Diesel devoted much of his time to the self-imposed task of developing an internal combustion engine that would approach
the theoretical efficiency of the Carnot cycle. For a time he experimented with an expansion engine using ammonia. About
1890, in which year he moved to a new post with the Linde firm in Berlin, he conceived the idea for the diesel engine . He
obtained a German development patent in 1892 and the following year published a description of his engine under the title
Theorie und Konstruktion eines rationellen Wäremotors (Theory and Construction of a Rational Heat Motor). With support
from the Maschinenfabrik Augsburg and the Krupp firms, he produced a series of increasingly successful models,
culminating in his demonstration in 1897 of a 25-horsepower, four-stroke, single vertical cylinder compression engine. The
high efficiency of Diesel's engine, together with its comparative simplicity of design, made it an immediate commercial
success, and royalty fees brought great wealth to its inventor.
Diesel disappeared from the deck of the mail steamer Dresden en route to London and was assumed to have drowned.
Processes:
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Pressure Heat Addition (P = C)
3 to 4 – Isentropic Expansion (S = C)
4 to 1 – Constant Volume Heat Rejection (V = C)
Compression Ratio
2
4
2
1
V
V
V
V
r
3
4
2
1
V
V
V
V
Cut-Off Ratio(rc)
2
3
c
V
V
r
Percent Clearance
D
2
V
V
C
C
C1
r
Displacement Volume (VD)
VD =V1 – V2
Heat Added (QA)
At P = C ; Q = mCp(T)
QA = mCp(T3 – T2)
QA = mkCv(T3 – T2)
69. Heat Rejected (QR)
QR = mCv(T4 – T1)
Net Work (W)
W = Q
W = QA – QR
W = mkCv(T3 – T2) - mCv(T4 – T1)
W = mCv[k (T3 – T2) - (T4 – T1)]
P,V and T Relations
At point 1 to 2 (S = C)
11Eq.)r(TT
P
P
r
V
V
T
T
1k
12
k
1k
1
21k
1k
2
1
1
2
At point 2 to 3 (P = C)
c
2
3
2
3
r
V
V
T
T
)r()r(TT c
1k
13
At point 3 to 4 (S = C)
k
c14
1k
4
1k
3
2
3
1k
2
1k
1
14
1k
4
3
c
1k
14
k
1k
3
4
1k
4
3
3
4
)r(TT
V
V
V
V
V
V
TT
V
V
)r()r(TT
P
P
V
V
T
T
At point 4 to 1 (V = C)
1
4
1
4
P
P
T
T
Entropy Change
a) S during Heat Addition
2
3
23
T
T
lnmCpSS
b) S during Heat Rejection
4
1
41
T
T
lnmCvSS
S1 - S4 = -( S3 – S2)
Thermal Efficiency
AQ
W
e x 100%
A
RA
Q
QQ
e
x 100%
70. 100%x
Q
Q
1e
A
R
100%x
)TT(k
)TT(
1e
23
14
Substituting Eq. 11, Eq. 13 and Eq. 14 to Eq. 22
%100x
)1r(k
1r
)r(
1
1e
c
k
c
1k
Mean Effective Pressure
KPa
V
W
P
D
m
where:
Pm – mean effective pressure, KPa
W – Net Work KJ, KJ/kg, KW
VD – Displacement Volume m3
, m3
/kg, m3
/sec
Displacement Volume
VD = V1 – V2 m3
Eq. 25
VD = 1 - 2 m3
/kg Eq.26
AIR STANDARD DUAL CYCLE
Processes:
1 to 2 – Isentropic Compression (S = C)
2 to 3 – Constant Volume Heat Addition Q23 ( V = C)
3 to 4 – Constant Pressure Heat Addition Q34 (P = C)
4 to 5 – Isentropic Expansion (S = C)
5 to 1 - Constant Volume Heat Rejection (V = C)
Compression Ratio
2
1
V
V
r
V1 = V5 and V2 = V3
Cut-Off Ratio
3
4
c
V
V
r
71. Pressure Ratio
2
3
p
P
P
r
P3 = P4
Percent Clearance
D
2
V
V
C
C
C1
r
Displacement Volume
VD = V1 – V2 m3
VD = 1 - 2
P, V, and T Relations
At point 1 to 2 (S = C)
)r(TT
P
P
r
V
V
T
T
1k
12
k
1k
1
21k
1k
2
1
1
2
At point 2 to 3 (V = C)
P
2
3
2
3
r
P
P
T
T
)r()r(TT P
1k
13
At point 3 to 4 (P = C)
)r)(r()r(TT
r
V
V
T
T
cP
1k
14
c
3
4
3
4
At point 4 to 5 (S = C)
)r()r(TT
V
V
V
V
r
V
V
TT
V
V
T
T
P
k
c15
1k
1
1k
4
2
4
P1k
2
1k
1
15
1k
5
4
4
5
At 5 to 1 (V = C)
1
5
1
5
P
P
T
T
Entropy change
a) At 2 to 3 (V = C)
2
3
V23
T
T
mCSS
72. b) At 3 to 4 (P = C)
3
4
34
T
T
lnmCpSS
S4 – S2 = (S3 – S2) + (S4 – S3)
c) At 5 to 1 (V = C)
S1 – S5 = mCvln
5
1
51
T
T
mCvSS
Heat Added
QA = QA23 + QA34
QA23 = mCv(T3 - T2)
QA34 = mCp(T4 - T3) = mkCv(T4 - T3)
QA = mCv[(T3 - T2)+ k(T4 - T3)
Heat Rejected
QR = mCv(T5 - T1)
Net Work
W = (QA - QR)
W = mCv[(T3 - T2) + k(T4 - T3) - (T5 - T1)]
Thermal Efficiency
%100x
Q
W
e
A
%100x
Q
QQ
e
A
RA
100%x
Q
Q
1e
A
R
100%x
)TT(k)TT(
)TT(
1e
3423
15
Substituting
%100x
)1r(kr)1r(
1rr
)r(
1
1e
cPP
k
cP
1k
73. SAMPLE PROBLEMS
AIR STANDARD CYCLE
Example No. 1
An air standard Otto cycle has a compression ratio of 8 and has air conditions at the beginning of compression of 100 KPa and
25C. The heat added is 1400 KJ/kg. Determine:
a) the thermal efficiency (56.5%)
b) the mean effective pressure (1057 KPa)
c) the percent clearance c (14.3%)
Given:
r = 8; P1 = 100 KPa ; T1 = 298K; QA = 1400 KJ/kg
Solution:
56.5%e
100%x
)8(
1
1e
100%x
)r(
1
1e
.a
14.1
1k
75. Example No. 3
An air standard diesel cycle has a compression ratio of 20 and a cut-off ratio of 3. Inlet pressure and temperature are 100 KPa
and 27C. Determine:
a) the heat added in KJ/kg (1998 KJ/kg)
b) the net work in KJ/kg (1178 KJ/kg)
c) the thermal efficiency (61%)
78. Example No. 4
An ideal dual combustion cycle operates on 0.454 kg of air. At the beginning of compression air is at 97 KPa, 316K. Let rp =
1.5, rc= 1.6 and r = 11. Determine
a. the percent clearance (C = 10%)
b. the heat added, heat rejected and the net cycle work
c. the thermal efficiency
%7.58e
%100x
)1r(kr)1r(
1rr
)r(
1
1e
cPP
k
cP
1k
%10
1r
1
C
C
C1
r
QA = Q23 + Q34
Q23 = mCv(T3 – T2)
825)r(TT 1k
12
K
K1238)(rTT
KPa784,2)r(PP
p23
k
12
Q23 = 0.454(0.7175)(1238 – 825) = 135 KJ
Q34 = mCp(T4 – T3)
K1981T
T6.1T
6.1
T
T
r
4
34
3
4
c
Q34 = 0.454(1.0045(1981 – 1238) = 339 KJ
QA = Q23 + Q34 = 474 KJ
79. Internal Combustion Engine
By. ENGR. YURI G. MELLIZA
Diesel engine is a type of internal combustion engine that uses low grade fuel oil and which burns this fuel inside the cylinder
by heat of compression. It is used chiefly for heavy-duty work. Diesel engines drive huge freight trucks, large buses, tractors,
and heavy road-building equipment. They are also used to power submarines and ships, and the generators of electric-power
stations in small cities. Some motor cars are powered by diesel engines.
Gasoline engine - is a type of internal combustion engine, which uses high grade of oil. It uses electricity and spark plugs to
ignite the fuel in the engine's cylinders.
Kinds of diesel engines. There are two main types of diesel engines. They differ according to the number of piston strokes
required to complete a cycle of air compression, exhaust, and intake of fresh air. A stroke is an up or down movement of a
piston. These engines are (1) the four-stroke cycle engine and (2) the two-stroke cycle engine.
Four Stroke Cycle Engine
1. Intake
2. Compression
3. Power
4. Exhaust
In a four-stroke engine, each piston moves down, up, down, and up to complete a cycle. The first down stroke draws air into
the cylinder. The first upstroke compresses the air. The second down stroke is the power stroke. The second upstroke
exhausts the gases produced by combustion. A four-stroke engine requires exhaust and air-intake valves.
It completes one cycle in two revolutions of the crankshaft.
Intake Compression Power Exhaust
80. Two Stroke Cycle Engine
1. Intake-Compression stroke
2. Power-exhaust stroke
In a two-stroke engine, the exhaust and intake of fresh air occur through openings in the cylinder near the end of the down
stroke, or power stroke. The one upstroke is the compression stroke. A two-stroke engine does not need valves. These
engines have twice as many power strokes per cycle as four-stroke engines, and are used where high power is needed in a
small engine. It completes one cycle in one revolution of the crankshaft.
INTERNAL COMBUSTION ENGINE POWER PLANT
Two stroke cycle engine: An engine that completes one cycle in one revolution of the crankshaft.
Four stroke cycle engine: An engine that completes one cycle in two revolution of the crankshaft.
Intake-compression Power-Exhaust
Intake port
Exhaust port
81. ENGINE PERFORMANCE
1. HEAT SUPPLIED BY FUEL
hr
KJ
)HV(mQs F
where:
mf - fuel consumption in kg/hr
HV - heating value of fuel in KJ/kg
2. INDICATED POWER
KW
)60(4
'NnLDπP
IP
2
mi
where:
Pmi - indicated mean effective pressure in KPa
L - length of stroke in m
D - diameter of bore in m
n' - no. of cylinders
IP - indicated power in KW
acting)Double-Stroke4For()RPM(N
acting)Single-Stroke4For(
2
(RPM)
N
acting)Double-Stroke2For(2(RPM)N
acting)Single-Stroke2For()RPM(N
3. BRAKE POWER
KW
)60(4
'NnLDπP
BP
2
mb
where:
Pmb - brake mean effective pressure in KPa
acting)Double-Stroke4For()RPM(N
acting)Single-Stroke4For(
2
(RPM)
N
acting)Double-Stroke2For(2(RPM)N
acting)Single-Stroke2For()RPM(N
KW
000,60
TNπ2
BP
where:
T - brake torque in N-m
N = RPM
4. FRICTION POWER
BPIPFP
5. INDICATED MEAN EFFECTIVE PRESSURE
KPa
'L
'AS
Pmi
where:
A - area of indicator card, m2
S - spring scale, KPa/m
L' - length of indicator card, m
82. 6. BRAKE TORQUE
m-NR)TareP(T
where:
P - Gross load on scales, N
Tare - tare weight, N
R - length of brake arm, m
7. PISTON SPEED
min
m
LN2PS
8. DISPLACEMENT VOLUME
KPa
P
BP
V
KPa
P
IP
V
KPa
)60(4
'NnLDπ
V
mb
D
mi
D
2
D
Where
acting)Double-Stroke4For()RPM(N
acting)Single-Stroke4For(
2
(RPM)
N
acting)Double-Stroke2For(2(RPM)N
acting)Single-Stroke2For()RPM(N
9. SPECIFIC FUEL CONSUMPTION
a. Indicated specific fuel consumption
hr-KW
kg
IP
fm
fim
b. Brake specific fuel consumption
hr-KW
kg
BP
fm
fbm
c. Combined specific fuel consumption
hr-KW
kg
GP
fm
fcm
where: GP - Generator output
10. HEAT RATE (HR): Heat supplied divided by the KW produced.
a. Indicated heat rate
hr-KW
KJ
IP
sQ
HRi
b. Brake heat rate
hr-KW
KJ
BP
sQ
HRb
c. Combined heat rate
hr-KW
KJ
GP
sQ
HRc
11. GENERATOR SPEED
RPM
n
120f
N
83. Where: n - number of generator poles (usually divisible by 4)
12. MECHANICAL EFFICIENCY
100%x
IP
BP
mη
13. GENERATOR EFFICIENCY
100%x
BP
GP
gη
14. Indicated Thermal Efficiency
100%x
sQ
3600(IP)
ie
15. Brake Thermal Efficiency
100%x
sQ
3600(BP)
be
16. Combined Thermal efficiency
100%x
sQ
3600(GP)
ce
17. Indicated Engine Efficiency
100%x
e
ie
iη
18. Brake Engine Efficiency
100%x
e
be
bη
where:
e - cycle thermal efficiency
19. Volumetric Efficiency
100%x
sec
m
Volume,ntDisplaceme
sec
m
entering,airofvolumeActual
v 3
3
η
20. Correction Factor for Nonstandard condition
a. Considering temperature and pressure effect
sT
hT
hB
SB
sPhP
b. Considering temperature effect alone
sT
hT
sPhP
c. Considering pressure effect alone
hB
SB
sPhP
Note: From US Standard atmosphere:
K
1000
6.5h
-sThT
Hgmm
1000
83.312h
sBhB
where: B - barometric pressure, mm Hg
T - absolute temperature, K
