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) 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)
STRUCTURE OF CnHm
Complete Combustion: Occurs when all the combustible elements has been fully oxidized.
C + O 2 → CO 2
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.
C + 12 O 2 → CO
12 + 16 → 28
3+ 4 → 7
THE COMBUSTION CHEMISTRY
A. Oxidation of Carbon
C + O 2 → CO2
12 + 32 → 44
3 + 8 → 11
kgO 2 8
=
kgC
3
B. Oxidation of Hydrogen
H 2 + 12 O 2 → H 2O
2 + 16 → 18
1+ 8 → 9
kgO 2 8
=
kgH 2 1
C. Oxidation of Sulfur
S + O 2 → SO 2
32 + 32 → 64
1+1 → 2
kgO 2 1
=
kgS 1
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
Nitrogen
Oxygen
Argon
Carbon Dioxide
Water Vapor
Other Particles
% By Volume
78.09
20.95
0.93
0.04
1 (0.4% over the entire atmosphere)
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%
Moles of N 2 79
=
= 3.76
Mole of O2
21
COMBUSTION WITH AIR and Theoretical air requirement
Fuel + Air → Pr oducts
A) Combustion of Carbon with air
C + O 2 + 3.76 N 2 → CO2 + 3.76 N 2
12 + 32 + 3.76(28) → 44 + 3.76(28)
kg of air
= 11.44
kg of C
B) Combustion of Hydrogen with Air
H 2 + 12 O 2 + 12 (3.76) N 2 → H 2O + 12 (3.76) N 2
2 + 16 + 12 (3.76)(28) → 18 + 12 (3.76)(28)
kg of air
= 34.32
kg of H
C) Combustion of Sulfur with air
S + O 2 + (3.76) N 2 → SO2 + (3.76) N 2
32 + 32 + (3.76)(28) → 64 + (3.76)(28)
kg of air
= 4.29
kg of H
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
kg of air
A
=
 
 F T heoretical Kg of Fuel
EXCESS AIR
It is an amount of air in excess of the theoretical air required to influence complete combustion. With excess air, O 2 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
C n H m + aO 2 + a (3.76) N 2 → bCO2 + cH 2O + a (3.76) N 2
By Carbon balance
1(n) = b
b=n
By Hydrogen balance
1(m) = 2c
c = 0.5m
By Oxygen balance
2a = 2b + c
a = b + 0.5c
a = n + 0.25m
therefore for the combustion of one CnHm with 100% TA
a = n + 0.25m
b=n
c = 0.5m
32a + a (3.28)28 137.28a
A
=
=
 
12n + m
12n + m
 F T heoretical
137.28(n + 0.25m)
A
=
 
12n + m
 F T heoretical
B) With excess air
C n H m + (1 + e)aO 2 + (1 + e)a (3.76) N 2 → bCO2 + cH 2 O + dO 2 + (1 + e)a (3.76) N 2
By Oxygen balance
2(1 + e)a = 2b + c + 2d
2d = 2(1 + e)a - 2b - c
d = (1 + e)a - b - 0.5c
d = (1 + e)(n + 0.25m) − n − 0.25m
d = n + 0.25m + en + e(0.25m) − n − 0.25m
d = en + e(0.25m)
d = e(n + 0.25m )
137.28(1 + e)(n + 0.25m) kg of air
A
=
→ Actual A/F Ratio
 
F
12n + m
kg of C n H m
  Actual
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
ER =
(F A) actual
(F A)Stoichiometric
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
Percentage Excess Air =
(A F)
( F)
− A
(A F)
Actual
T heoretical
T heoretical
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.
O 
kg of air
A

  = 11.44C + 34.32 H 2 − 2  + 4.29S
8 
kg of fuel
 F 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.
100 % = %FC + %Volatiles + %Ash + %Moisture
ORSAT ANALYSIS
Orsat Analysis gives the volumetric or molal analysis of the products of combustion or exhaust gases on a Dry Basis.
100 % = %CO 2 + %O 2 + % N 2 + %CO + % NO x + %CH + ...
MASS FLOW RATE OF FLUE GAS
a) Without considering Ash loss:
A 
mgas = mFuel  + 1
F

b) Considering Ash loss
A

mgas = mFuel  + 1 − Ash Loss
F

where ash loss in decimal
•
Combustion equation with CO in the products due to incomplete combustion (100% theoretical air)
CnHm + aO2 + a (3.76 ) N 2 → bCO 2 + cH 2O + dCO + a (3.76 ) N 2
•
Combustion equation with CO in the products due to incomplete combustion (with excess air)
CnHm + (1 + e)aO2 + (1 + e)a (3.76 ) N 2 → bCO 2 + cH 2O + dCO + fO 2 + (1 + e)a (3.76 ) N 2
•
Typical Real-World Engine Combustion Process:
Fuel (CnHm) + Air (O 2 and N 2 ) → CO2 + H 2O + O 2 + N 2 + CH(VOC's) + CO + NOx
CH(VOC's) - Volatile Organic Compounds
CO - Carbon Monoxide
NOx - Nitrogen Oxides
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 widereaching, 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 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)
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.
haze that reduces visibility.
• PM is the main source of
• 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
fireplaces)
• Combustion of fossil fuels and wood (motor vehicles, woodstoves and
• 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.
plants and trees, leading to reduced yields.
•
Ozone
can
damage
• 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)
• 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
C n H m + aO2 + a (3.76) N 2 → bCO2 + cH 2O + a (3.76) N 2
where
a = n + 0.25m
b=n
c = 0.5m
b) With excess air e
C n H m + (1 + e)aO2 + (1 + e)a (3.76) N 2 → bCO2 + cH 2O + dO2 + (1 + e)a (3.76) N 2
where
d = e(n + 0.25m)
c)
With exhaust pollutants (emission gases)
CnHm + aO 2 + a(3.76)N2 ) → bCO2 + cH 2 O + dO 2 + fN 2 + gCH + hCO + iNOx
CH(VOC's) - Volatile Organic Compounds
CO - Carbon Monoxide
NOx - Nitrogen Oxides
Note : The above values of a, b, c and d in terms of n and m may not apply to actual combustion process
Combustion of Solid Fuels
a)
Combustion with 100% theoretical air
aC + bH 2 + cO2 + dN 2 + fS + gH 2O + xO 2 + x (3.76 ) N 2 → hCO 2 + iH 2O + jSO 2 + kN 2
b) Combustion with Excess air e
aC + bH 2 + cO 2 + dN 2 + fS + gH 2 O + (1 + e) xO 2 + (1 + e) x (3.76) N 2 → hCO 2 + iH 2 O + jSO 2 + LO 2 + mN 2
c)
Combustion with exhaust pollutants (emission gases)
aC + bH 2 + cO2 + dN 2 + fS + gH 2O + (1 + e)xO 2 + (1 + e)x(3.76) N 2 → hCO 2 + iH 2O + jSO 2 + LO 2 + mN 2 +
nCO + oCH + pNOx
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
C + O 2 → CO 2
12 + 32 → 44
3 + 8 → 11
kg of CO 2 11
=
kg of C
3
Kg of H2O per kg of H formed
H2 +
1
2 O2
→ H 2O
2 + 16 → 18
1+ 8 → 9
kg of H 2 O 9
=
kg of H
1
Kg of SO2 per kg of S formed
S + O 2 → SO 2
32 + 32 → 64
1+1 → 2
kg of SO 2 2
=
kg of S
1
Total Mass of Products
m Pr oducts = ΣniMi
m Pr oducts = n CO 2 M CO 2 + n H 2O M H 2O + n O 2 M O 2 + n SO 2 M SO 2 + n N 2 M N 2 + n CO M CO + n CH M CH + n NOx M NOx
Total Moles of Products
n Pr oducts = Σni
n Pr oducts = n CO 2 + n H 2O + n O2 + nSO 2 + n N 2 + n CO + n CH + n NOx
Dew Point Temperature
t dp = ( tsat ) Saturation temperatu re corresponding the partial pressure of H 2 O (PH 2O )
PH 2O = (P)
n H 2O
n Products
P = total pressure of product
Moles of Dry Flue Gas (The H2O is not included in the analysis)
n Dry Flue Gas = nCO2 + nO2 + nSO 2 + n N2 + nCO + nCH + n NOx
% of CO2 in the dry flue gas
n CO 2
n CO 2
x 100% =
x 100%
n Dry Flue Gas
n CO 2 + n O2 + nSO 2 + n N 2 + n CO + n CH + n NOx
Total Mass of Fuel
a.
For Hydrocarbon of Hydrocarbon Mixture
m Fuel = Σn F M F
b.
For Solid Fuels
mFuel = ΣniMi
mFuel = n CMC + n H 2MH 2 + n O2MO2 + n N 2M N 2 + nSMS + n H 2OMH 2O
Mass Flow Rate of Products (Known Fuel flow rate)
kg
m
kg
of Products = Products x Fuel Flow Rate
hr
mFuel
hr
Volume of Products at the product Pressure and Temperature (m 3)
VPr oducts =
n Pr oducts (R )T 3
m
P
Volume flow rate of Products at the product Pressure and Temperature (m 3/hr)
m3
V
m3
of Products = Products x Fuel Flow Rate
hr
mFuel
hr
Molecular Weight of Products
M=
n CO 2 M CO 2 + n H 2O M H 2O + n O 2 M O 2 + n SO 2 MSO 2 + n CO M CO + n CH M CH + n NOx M NOx + n N 2 M N 2 + ...n i M
n Pr oducts
Gas Constant of Products
8.3143 KJ
M kg - K
Σ(niMi )Ri
R = ΣxiRi =
Σ(niMi )
R=
Specific Heat of Products
Cp = Σx iCpi
Cv = Σx iC vi
R = Cp − C v
k=
Cp
Cv
Rk
KJ
k − 1 kg - K
R
KJ
Cv =
k − 1 kg - K
Cp =
Example No. 1: In the figure below, Determine
a. Percent excess air
b. Volumetric Analysis of Products
c. Orsat Analysis
mp = ma + mF
m p = 4000 + 200 = 4,200
kg
hr
4000
A
=
= 20
 
 F  Actual 200
A
A
= (1 + e ) 
 
F
  Actual
 F  T heoretical
A
F Actual − 1
e=
A
F T heoretical
CH 4 + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + a (3.76) N 2
( )
( )
137.27(n + 0.25m)
A
=
= 17.16
 
12n + m
 F  T heoretical
e = 0.17 = 17%
Combustion with EA; e = 0.17
CH 4 + (1.17)aO 2 + (1.17)a (3.76) N 2 → bCO2 + cH 2 O + dO 2 + (1.17)a (3.76) N 2
a = n + 0.25m = 2.34
b = n =1
c = 0.5m = 2
d = e(n + 0.25m) = 0.34
CH 4 + 2.34O 2 + 8.8N 2 → 1CO2 + 2H 2 O + 0.34O 2 + 8.8N 2
n Pr oducts = 1 + +2 + 0.34 + 8.8 = 12.14
Pr oduct Analysis
y CO 2 = 8.24%; y H 2 O = 16.48%; y O 2 = 2.8%; y N 2 = 72.48%
n Dry Pr oducts = 1 + 0.34 + 8.8 = 10.14
Orsat Analysis
y CO 2 = 9.86%
y O 2 = 3.35%
y N 2 = 86.78%
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
C8H18 + aO2 + a (3.76) N2 → bCO2 + cH2O + a (3.76) N2
C8H18 + 12.5O2 + 47N2 → 8CO2 + 9H2O + 47N2
Combustion with e = 20%
C8H18 + (1.20)12.5O2 + (1.20)47N2 → 8CO2 + 9H2O + dO2 + (1.20)47N2
C8H18 + 15O2 + 56.4N2 → 8CO2 + 9H2O + 2.5O2 + 56.4N2 → Actual Combustion eq.
a. Actual Air – Fuel Ratio
15(32) + (56.4)(28)
A
=
= 18.06
 
F
12(8) + 1(18)
 ACT UAL
b. the percentage CO2 by volume in the dry exhaust gases
n d − moles of dry exhaust gas
n d = 8 + 2.5 + 56.4 = 66.9
8
x100% = 11.95%
66.9
c. kg of water vapor formed per kg of fuel
y CO 2 =
n − moles of exhaust gas
n d = 8 + 9 + 2.5 + 56.4 = 75.9
kg H 2O = 9(18) = 162 kg
kg H 2O
162
162
kg
=
=
= 1.42
kg C8H18 12(8) + 1(18) 114
kg
d. volume of dry exhaust gas per kg of fuel if T = 290 K and P = 101.33 KPa
n d − moles of dry exhaust gas
n d = 8 + 2.5 + 56.4 = 66.9
PV = n RT
66.9(8.3143)(290) 3
Vd =
m
101.33
Vd = 1,591.9 m3
1,591.9
m3
= 14
kgC8H18
114
kgC8H18
e. the partial pressure of the water vapor in the exhaust
n − moles of exhaust gas
Vd
=
n = 8 + 9 + 2.5 + 56.4 = 75.9
9
y H 2O =
x100% = 11.86%
75.9
P
y H 2O = H 2O
P
PH 2O = 0.1186(101.33) = 12.015 KPa
f. the dew point temperature of the products
DPT = saturation temperatu re corresponding PH 2O
PH 2O = 12.015 KPa
From Steam Table
DPT = 46.467C
if the mixture is cooled below DPT, condensati on of H 2O 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: CO 2 = 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
aC12H 26 + bO 2 + b(3.76)N2 → 12.8CO2 + cH 2O + 0.2CO + 3.5O2 + 83.5N2
dividing the equation by a
C12H 26 + 20.5O2 + 77.08N2 → 11.816CO2 + 13H 2 O + 0.185CO + 3.23O2 + 77.08N2
20.5(32) + (77.08)(28)
A
=
= 16.56
 
12(12) + 1(26)
 F  Actual
Theoretica l Air - Fuel Ratio
137.28(n + 0.25m)
A
=
= 14.94
 
F
12n + m
  T heroretical
A
F Actual − 1 = 0.108 = 10.8%
e=
A
F T heoreical
( )
( )
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 :CO 2 = 11.07 % ; CO = 1.23 % ; O2 = 3.72
% ; N2 = 83.97 %. Also find percentage excess air.
12n
%C =
12n + m
12n
0.842 =
12n + m
12
12n + m =
n = 14.25n → eq.1
0.842
m
%H =
12n + m
m
0.158 =
12n + m
1
12n + m =
m = 6.33m → eq.2
0.158
14.25n = 6.33m
m = 2.25n → eq.3
For petrol Fuel the formula is
C n H 2n + 2
m = 2n + 2
2.25n = 2n + 2
n =8
m = 18
137.28(n + 0.25m)
A
=
= 14.57
 
12n + m
 F  T heoretical
Combustion Equation (Basis 100 moles of dry flue gas)
C8 H18 + aO 2 + bN 2 → 11.07CO2 + cH 2 O + 1.23CO + 3.72O 2 + 83.97 N 2
b = 83.97
a (3.76) = b
a = 22.33
c = 13.84
32a + 28b
A
=
= 17.49
 
F
  Actual 12n + m
17.49
e=
− 1 = 0.162 = 16.2%
14.57
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%
Combustion equation w ith 100 TA
64.91C + 25.13H2 + 3.14O2 + 0.45N2 + 0.79S + 5.58H2O → Fuel
aO 2 + a(3.76)N2 → Air
bCO2 + cH 2O + dSO2 + eN 2 → Products
64.91C + 25.13H2 + 3.14O2 + 0.45N2 + 0.79S + 5.58H2O → Fuel
75.12O2 + 282.46 N 2 → Air
64.91CO2 + 30.71H 2O + 0.79SO 2 + 282.91N 2 → Products
Combustion equation w ith 25% E.A.
64.91C + 25.13H2 + 3.14O2 + 0.45N2 + 0.79S + 5.58H2O → Fuel
(1.25)75.12O 2 + (1.25)282.46 N 2 → Air
64.91CO2 + 30.71H 2O + 0.79SO 2 + fO 2 + gN 2 → Products
2(3.14) + (5.58) + (1.25)75.12(2) = 2(64.91) + 30.71 + 2(0.79) + 2f
f = 18.78
2(0.45) + 2(1.25)282.46 = 2g
g = 353.52
64.91C + 25.13H2 + 3.14O2 + 0.45N2 + 0.79S + 5.58H2O → Fuel
93.9O2 + 353.07 N 2 → Air
64.91CO2 + 30.71H 2O + 0.79SO 2 + 18.78O 2 + 353.52 N 2 → Products
32(93.9) + 28(353.07)
A
=
= 12.07
 
 F Actual 12(64.91) + 2(25.13) + 32(3.14) + 28(0.45) + 32(0.79) + 18(5.58)
yi =
ni
n
n Pr oducts = 64.91 + 30.71 + 0.79 + 18.78 + 353.52 = 468.71
64.91
x100% = 13.85%
468.71
30.71
y H 2O =
x100% = 6.55%
468.71
0.79
ySO 2 =
x100% = 0.17%
468.71
18.78
yO2 =
x100% = 4.01%
468.71
353.52
y N2 =
x100% = 75.42%
468.71
y CO 2 =
M = ΣyiMi
64.91(44) + 30.71(18) + 0.79(64) + 18.78(32) + 353.52(28)
kg
M=
= 29.78
468.71
kgm
8.3143
KJ
R=
= 0.279
M
kg - K
PV = n RT
n RT
P
(64.91)(8.3143)(300 + 273)
VCO 2 =
= 3061.91 m3
101
VCO 2
3061.91
m3
=
= 2.87
kg of Fuel 12(64.91) + 2(25.13) + 32(3.14) + 28(0.45) + 32(0.79) + 18(5.58)
kg of Fuel
(0.79)(8.3143)(300 + 273)
VSO 2 =
= 37.04 m 3
101
VSO 2
37.04
m3
=
= 0.03
kg of Fuel 12(64.91) + 2(25.13) + 32(3.14) + 28(0.45) + 32(0.79) + 18(5.58)
kg of Fuel
V=
Example No. 6 (Alcohol)
Calculate the theoretical Oxygen/fuel ratio and Air/fuel ratio on a mass basis for the combustion of ethanol, C2H5OH.
C 2 H 5OH + aO2 + a (3.76) N 2 → bCO2 + cH 2O + a (3.76) N 2
2=b
6 = 2c
c=3
1 + 2a = 2b + c
a =3
Combustion Equation
C 2 H 5OH + 3O 2 + 11.28N 2 → 2CO2 + 3H 2O + 11.28N 2
A 3(32) + 11.28(28)
=
= 8.95
F
46
O2
32(3)
kg
=
= 2.09
Fuel
46
kg
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
Combustion with 100% TA
60CH 4 + 30CO + 10O 2 + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + O 2 + a (3.76) N 2
a = 125
b = 90
c = 120
Combustion with 305 EA
60CH 4 + 30CO + 10O 2 + (1.30)aO 2 + (1.30)a (3.76) N 2 → 90CO2 + 120H 2 O + dO 2 + (1.30)a (3.76) N 2
d = 37.5
60CH 4 + 30CO + 10O 2 + 162.5O 2 + 611N 2 → 90CO2 + 120H 2 O + 37.5O 2 + 611N 2
A
162.5(32) + 611(28)
=
= 11.52
F 60(16) + 30(28) + 10(32)
n Products = 90 + 120 + 37.5 + 611 = 858.5
ni
x100%
n
Volumetric Analysis of Products
y CO2 = 10.48%
yi =
y H 2O = 13.98%
y O2 = 4.37%
y N2 = 71.17%
Pi
P
PH20 = 0.1398(101.325) = 14.16 KPa
DPT = 52.8C
ORSAT ANALYSIS
n Dry Gas = 858.5 − 120 = 738.5
yi =
y CO2 = 12.2%
y O2 = 5.1%
y N2 = 82.7%
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.
3CH4 + 14H 2 + 50.9 N 2 + 0.6O2 + 27CO + 4.5CO2 + 32.38O2 + 121.73N 2 →
34.5CO2 + 20H 2O + 6.48O2 + 172.63N 2
A
kg
= 1.8
F
kg
A
mol
= 1.54
F
mol
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 CO 2 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
Combustion with 100% TA and CO emission
C3H 8 + aO 2 + a(3.76)N2 → bCO2 + cH 2 O + dCO + a(3.76)N2
d = 0.015(3) = 0.045
b = 3 − 0.045 = 2.955
1(8) = 2c
c=4
2a = 2b + c + d
a = 4.978
C3H 8 + 4.978O 2 + 18.715N 2 → 2.955CO2 + 4H 2 O + dCO + 18.715N 2
Combustion with 20% EA
C3H 8 + (1.20)4.978O 2 + (1.20)18.715N 2 → 2.955CO2 + 4H 2 O + 0.045CO + fO 2 + (1.20)18.715N 2
By oxygen balance
f = 0.996
COMBUSTION EQUATION
C3H 8 + 5.973O 2 + 22.458N 2 → 2.955CO2 + 4H 2 O + 0.045CO + 0.996O 2 + 22.458N 2
Actual Air - Fuel Ratio
5.973(32) + (22.458)( 28)
A
=
= 18.64
 
F
12(3) + 1(8)
  Actual
kg
Mass of air = 18.64(0.005)(3600) = 335.444
hr
Moles of Products = 2.955 + 4 + 0.045 + 0.996 + 22.458 = 30.45 moles
M = ΣyiMi
ni
yi =
n
2.955(44) + 4(18) + 0.045(28) + 0.996(32) + 22.458(28)
M = ΣyiMi =
30.45
kg
M = 28.37
kgm
8.3143
KJ
R=
= 0.293
28.37
kg - K
PV = n RT
m3
n RT  m Fuel  30.45(8.3143)(900)(0.005)
m3

 =
of Products =
= 0.26
sec
P  12(3) + 1(8) 
101(44)
sec
m3
n RT  m Fuel  0.045(8.3143)(900)(0.005)(3600)
m3

 =
of CO =
= 1.4
hr
P  12(3) + 1(8) 
101(44)
sec
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
Combustion with 100% theoretical air
With 100% TA
C12H 26 + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + a(3.76)N 2
Carbon balance
b=n
Hydrogen balance
m = 0.5m
Oxygen balance
a = n + 0.25m
With 25% EA
C12H 26 + (1 + e)aO 2 + (1 + e)a (3.76) N 2 → bCO2 + cH 2 O + dO 2 + (1 + e)a(3.76)N 2
d = e(n + 0.25m)
COMBUSTION EQUATION
C12H 26 + 23.125O 2 + 86.950N 2 → 12CO2 + 13H 2 O + 4.625O 2 + 86.950N 2
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
total
28
86.95
0.746
2434.6
0.728
20.884
74.6
72.8
116.575
1.00
3344.6
1.00
28.691
100.0
100.0
The combustion equation
C12H 26 + 23 .125 O 2 + 86 .95 N 2 → 12CO 2 + 13H 2 O + 4.625O 2 + 86 .95 N 2
The theoretical A/F ratio
137.28(n + 0.25m)
kg of air
A
=
= 14.94
 
12n + m
kg of C12H 26
 F T heoretical
The actual A/F ratio
kg of air
A
A
= (1 + e)  = 18.674
 
F
F
kg
of C12H 26
  Actual
 t
The Volumetric and gravimetric Analysis
Gases
yi(%)
xi(%)
CO2
10.3
15.8
H2O
11.2
7.0
O2
4.0
4.4
N2
total
74.6
72.8
100.0
100.0
Molecular weight and Gas constant
kg
M = ΣyiMi = 28.691
kg m
R
= ΣxiRi
M
8.3143
KJ
R=
= 0.2898
28.691
kg - K
R=
Kg of CO2 per kg of fuel
kg of CO2
528
=
= 3.106
kg of C12H 26 12(12) + 26
% C and % H in the fuel
12n
x 100%
12n + m
12(12)
%C =
x 100 = 84.7%
12(12) + 26
m
%H =
x 100%
12n + m
26
%H =
x 100 = 15.3%
12(12) + 26
%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.%
C
H2
O2
N2
S
H2O
Ash
78
5
8
0
1
2
6
100
94%
Ashless
U.A.
Ashless
83.0
5.3
8.5
0.0
1.1
2.1
100
Mi
xi/Mi
12
2
32
28
32
18
-
0.069
0.027
0.003
0.0003
0.0012
0.0999
Molal
Analysis
69.21
26.62
2.66
0.33
1.18
100
Combustion with 100% TA
Basis 100 moles of fuel
69.21C + 26.62H 2 + 2.66O 2 + 0.33S + 1.18H 2 O + xO 2 + x (3.76) N 2 → Air + Fuel
aCO2 + bH2O + cSO2 + dN2 → Pr oducts
Carbon Balance
69.21 = a
a = 69.21
Hydrogen Balance
26.62(2) + 1.18(2) = 2b
b = 27.8
Sulfur Balance
0.33 = c
Oxygen Balance
2.66(2) + 1.18(2) + 2x = 2a + b + 2c
x = 104.24
Nitrogen Balance
d = 301.50
Combustion with EA(0.30)
69.21C + 26.62H 2 + 2.66O 2 + 0.33S + 1.18H 2 O + (1.30) xO 2 + (1.30) x (3.76) N 2 → Air + Fuel
aCO2 + bH 2 O + cSO2 + eO 2 + fN 2 → Pr oducts
By oxygen blance
e = 24.06
Nitrogen balance
f = 391.94
104.24(32) + 391.94(28)
A
=
= 14.30
 
 F  Actual 69.21(12) + 26.62(2) + 2.66(32) + 0.33(32) + 1.18(18)
kg
m air = 14.30(500) = 7,149.24
hr
 kg Flue Gas  69.21(44) + 27.8(18) + 0.33(64) + 24.06(32) + 391.94(28)

 =
= 15.3
69.21(12) + 26.62(2) + 2.66(32) + 0.33(32) + 1.18(18)
 kg Fuel 
m Pr oducts = 15.3(500) = 7,649.24
kg
hr
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 = 900C
d. the Orsat analysis of the products
Combustion with 100% TA
Basis :100 moles of fuel
50C7 H16 + 50C8 H18 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + 15CO + 10NO + fN 2
Carbon Balance
50(7) + 50(8) = b + 15
b = 735
Hydrogen Balance
50(16) + 50(18) = 2c
c = 850
Oxygen Balance
2a = 2(735) + 850 + 15 + 10
a = 1172.50
Combustion with excess air (30%)
50C7 H16 + 50C8 H18 + (1.30)aO 2 + (1.30)a(3.76)N 2 → bCO2 + cH 2 O + 15CO + 10NO + gO 2 + hN 2
By oxygen balance
g = 351.75
by Nitrogen balance
h = 5726.18
Combustion Equation
50C7 H16 + 50C8 H18 + 1524.25O 2 + 5731.18N 2 → 735CO2 + 850H 2 O + 15CO + 10NO + 351.75O 2 + 5726.18N 2
1524.25(32) + 5731.18(28)
A
= 19.56
 Actual =
F
50(100) + 50(114)
 
nPollu tan t = 735 + 15 + 10 = 760 moles
n RT 760(8.3143)(900 + 273)
=
= 73,386.46 m 3 of exhaust pollutant
P
101
m 3 of exhaust pollutant
73,386.46
=
= 6.86
kg of Fuel
50(100) + 50(114)
n Dry Flue Gas = 735 + 15 + 10 + 351.75 + 5726.18 = 6,837.93
V=
Orsat Analysis
735
y CO2 =
x100% = 10.75%
6,837.93
15
y CO =
x100% = 0.22%
6,837.93
10
y NO =
x100% = 0.15%
6,837.93
351.75
yO2 =
x100% = 5.14%
6,837.93
5726.18
y N2 =
x100% = 83.74%
6,837.93
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.
C n H m + aO 2 + a (3.76) N 2 → 10.5CO2 + bH 2 O + 5.3O 2 + 84.2 N 2
By carbon balance
1(n) = 10.5
n = 10.5
By nitr0gen balance
a(3.76)2 = 84.2(2)
84.2
a=
= 22.394
3.76
Oxygen balance
2(22.394) = 2(10.5) + b + 2(5.3)
b = 13.19
By hydrogen balance
1(m) = 2b
m = 26.38
Combustion Equation
C10.5 H 26.38 + 22.394O 2 + 84.2 N 2 → 10.5CO2 + 13.19H 2 O + 5.3O 2 + 84.2 N 2
12(10.5)
= 82.69%
12(10.5) + 1(26.38)
1(26.38)
%H =
= 17.31%
12(10.5) + 1(26.38)
%C =
22.394(32) + 84.2(28)
A
=
= 20.18
 
12(10.5) + 1(26.38)
 F  Actual
137.2810.5 + 0.25(26.38)
A
=
= 15.4%
 
12(10.5) + 1(26.38)
 F  T heoretical
% Theoretica l air = (100 + 15.4) = 115.4%
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.
A 137.28(1.20)(7 + 4)
kg of air
=
= 20.592
F
(84 + 4)
kg fuel
m p = m F (20.592) + m F
m p = 50(21.592) = 1079.6 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.
Basis 100 moles of fuel
Combustion with 100% TA
80C12H 26 + 20C14H 30 + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + a (3.76) N 2
Carbon Balance
80(12) + 20(14) = b
b = 1,240
Hydrogen balsnce
80(26) + 20(30) = 2c
c = 1,340
Oxygen balance
2a = 2b + c
a = 1,910
Combustion with EA (e = 0.30)
80C12H 26 + 20C14H 30 + (1.30)aO 2 + (1.30)a (3.76) N 2 → bCO2 + cH 2 O + dO 2 + (1.30)a (3.76) N 2
By oxygen balance
d = 573
Combustion Equation
80C12H 26 + 20C14H 30 + 2,483O 2 + 9,336.08N 2 → 1,240CO2 + 1,340H 2 O + 573O 2 + 9,336.08N 2
n Pr oducts = 1,240 + 1,340 + 573 + 9,336.08 = 12,489.08
PH 2 O
P
=
n H 2O
n Pr oducts
1,340
= 10.84 KPa
12,489.08
at 10.84 KPa = 47.417C
PH 2 O = 101
DPT = t sat
Problem Natural Gas
A natural gas fuel showed the following percentages by volume: C 2H6 = 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 16C and a pressure of
101.6 KPa.
Combustion with 100% theoretic al air
9C 2 H 6 + 90CH4 + 0.2CO2 + 0.8N 2 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + dN 2
Carbon Balance
2(9) + 90 + 0.2 = b
b = 108.2
Hydrogen Balance
9(6) + 90(4) = 2c
c = 207
Oxygen Balance
2(0.2) + 2a = 2b + c
2(108.2) + 207 - 2(0.2)
a=
2
a = 211.5
Nitrogen Balance
2(0.8) + 2(a)(3.76) = 2d
2(0.8) + 2(211.5)(3.76)
d=
2
d = 796.04
Combustion with excess air
9C2 H 6 + 90CH4 + 0.2CO2 + 0.8N2 + (1.20)aO2 + (1.20)a(3.76)N2 → bCO2 + cH 2O + eO2 + fN 2
By oxygen balance
2(0.2) + 2(1.20)(211.5) = 2(108.2) + 207 + 2e
2(0.2) + 2(1.20)(211.5) − 2(108.2) − 207
e=
2
e = 42.3
By Nitrogen balance
2(0.8) + 2(1.20)(211.5)(3.76) = 2f
2(0.8) + 2(1.20)(211.5)(3.76)
f=
2
f = 955.088
Combustion Equation
9C2 H 6 + 90CH4 + 0.2CO2 + 0.8N2 + 253.8O2 + 954.288N2 → 108.2CO2 + 207H 2O + 42.3O2 + 955.088N2
Mass of Fuel
9(30) + 90(16) + 0.2(44) + 0.8(28) = 1741.2 kg
Mass of air
253.8(32) + 954.288(28) = 34841.664 kg
34841.664
kg of air
A
=
= 20.01
 
1741.2
kg of fuel
 F actual
A
A
= (1 + e) 
 
 F actual
 F  theoretical
20.01
kg of air
A
=
= 16.68
 
kg of fuel
 F  theoretical 1.20
Moles of Products
108.2 + 207 + 42.3 + 955.088 = 1312.588
108.2(44) + 207(18) + 42.3(32) + 955.088(28)
yiMi = M =
1312.588
kg
M = 27.9
kg mol

R=
8.3143
KJ
= 0.298
27.9
kg - K
PVa = n a RT
(253.8 + 954.288)(8.3143)(16 + 273)
= 28571.2 m 3
101.6
PVF = n F RT
Va =
(9 + 90 + .2 + 0.8)(8.143)(16 + 273)
= 2365 m 3
101.6
Va 28571.2
=
= 12.08
VF
2365
VF =
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
C12.8H 26.47 + 22.367O 2 + 84.1N 2 → 12.5CO2 + 13.234H 2O + 0.3CO + 3.1O2 + 84.1N 2
22.367(32) + 84.1(28)
kg of air
A
= 17.05
  =
12(12.8) + 26.47
kg of fuel
 F a
Combustion with 100% theoretical air
n = 12.8 ; m = 26.47
C12.8 H 26.47 + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + a (3.76) N 2
12.8 = b
26.47 = 2c
c = 13.235
2a = 2b + c
a = 19.4175
kg of air
 A  19.4175(32) + 19.4175(3.76)( 28
= 14.8
  =
12(12.8) + 26.47
kg of C12H 26
 F t
A
A
  = (1 + e) 
 F a
 F t
e = 0.152 = 15.2 %
12n
12(12.8)
=
= 85.3%
12n + m 12(12.8) + 26.47
m
26.47
%H =
=
= 14.7%
12n + m 12(12.8) + 26.47
%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.
ENTHALPY OF FORMATION
The “Enthalpy of Formation of a compound is the enthalpy at the Arbitrary Reference State (t = 25C and P = 1 Atm).
1 mole C
25C, 1 atm
Combustion
Chamber
1 mole CO2
25C, 1 atm
1 mole O2
25C, 1 atm
Q
Let;
HR – total enthalpy of Reactants
HP – total enthalpy of products
From 1st Law
Q + HR = HP
or
Q + Σ ni hi = Σ ni hi
R
P
but the enthalpy of all the reactants is Zero (for they are all elements)
Q = H P = -393,757 KJ
therefore
(h f ) CO 2 = −393,757
KJ
→ Enthalpy of formation of CO2
kg m
Note : Negative sign is due to the reaction' s being " Exothermic "
FIRST LAW ANALYSIS FOR STEADY STATE REACTING SYSTEM
W
1 mole C
25C, 1 atm
Combustion
Chamber
1 mole CO2
25C, 1 atm
1 mole O2
25C, 1 atm
By energy balance
Q
Q + Σ ni hi = W + Σ ni hi
R
P
In most cases, neither the reactants nor the products are at the reference state (t = 25C 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
(h − h 298 )
where
 − denotes that the pressure is 1 Atmosphere
General Equation for a Steady State, Steady flow reaction process



Q + Σ n i h f + (h − h298) i = W + Σ n j h f + (h − h298)
R
P

j
ADIABATIC FLAME TEMPERATURE
1 mole C
25C, 1 atm
Combustion
Chamber
1 mole CO2
25C, 1 atm
1 mole O2
25C, 1 atm
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)
HR = HP
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.
h RP = H P − H R

KJ
kg mol


h RP = Σ n j h f + (h − h298) − Σn i h f + (h − h298)
P

i
KJ
kg mol
(H P − H R ) KJ
M
kg
where
M - molecular weight
h RP =
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
O 
KJ

HHV = 33,820C + 144,212 H2 − 2  + 9,304S
8 
kg

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
HHV
A
kg of air/kg of fuel
  =
 F  t 3,041
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
141.5
131.5+ API
140
S=
130+ Be
S=
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
O 
KJ

HHV = 33,820C + 144,212 H2 − 2  + 9,304S
8 
kg

Converting the given U.A. to an ashless basis
C = 55% = 0.55
H 2 = 5.8% = 0.058
O 2 = 39% = 0.39
N 2 = 0.103% = 0.00103
S = 0.103% = 0.00103
O 
kg of air

A
  = 11.44C + 34.32 H 2 − 2  + 4.29S
F
8
kg
of fuel
 t


A
  = 6.6
 F t
O 
KJ

HHV = 33,820C + 144,212 H 2 − 2  + 9,304S
8 
kg

KJ
HHV = 19,889.8
kg
SAMPLE PROBLEM NO. 2
A small gas turbine uses C8H18 (liquid) for fuel and 400% theoretical air. The air and fuel enters at 25C 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.



Q + Σ n i h f + (h − h 298) i = W + Σ n j h f + (h − h 298 )
R
P
Combustion with 100% TA
C8H18 + 12 .5O 2 + 47 N 2 → 8CO 2 + 9H 2O + 47 N 2
Combustion with 400% theoretical air or e = 300%
C8H18 + 50 O 2 + 188 N 2 → 8CO 2 + 9H 2O + 37 .5O 2 + 188 N 2

j



Q + Σ n i h f + (h − h298) i = W + Σ n j h f + (h − h298)
R
W=
P

j
1000 
kg 
KJ
114
 = 456,000
0.25 
kgm 
kgm
Q + HR = 456,000 + HP
KJ
Q = −52,776
kg mFuel
Q = −463
KJ
kg Fuel
SAMPLE PROBLEM NO. 3
A diesel engine used gaseous Dodecane (C12H26) as fuel. The air and fuel enters the engine at 25C 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.
Products
Air and Fuel
W
Engine
Q = -232,000 KJ/kgmol



Q + Σ n i h f + (h − h298) i = W + Σ n j h f + (h − h298)
R
P

j
Combustion with 100% theoretical air
C12H26 + 18.5O2 + 69.56N2 → 12CO2 + 13H2O + 69.56N2
Combustion with 400% theoretical air or e = 300%
C12H26 + 37O 2 + 139.12N2 → 12CO2 + 13H2O + 18.5O 2 + 139.12N2
Q + HR = W + HP
W = Q − (HP − HR )
W = −232,000 − (−5,880,284.58) =
KJ
W = 5,648,284.58
kg mol Fuel
 1 
W = 5,648,284.58
 = 1568.97 KW
 3600 
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 25C and the products of combustion leaves the apparatus at 500 K. Determine the higher heating value of the fuel
considering;(assume hf(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
Combustion with 100% TA
C10H 22 + 15.5O 2 + 58.28N 2 → 10CO2 + 11H 2 O + 58.28N 2
Combustion with EA (e = 0.25)
C10H 22 + 19.375O 2 + 72.85N 2 → 10CO2 + 11H 2 O + 3.875O 2 + 72.85N 2
Incomplete Combustion
Combustion with 125% theoretical air and CO in the products
C10H 22 + 19 .063 O 2 + 71 .675 N 2 → 9.5CO 2 + 0.5CO + 11H 2O + 3.813 O 2 + 71 .675 N 2
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):
Dt =
1000H( air −  gas )
 w ater
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):
v = 2ghgas
where:
v - theoretical velocity of the flue gas in the chimney, m/sec
hg - draft in meters of flue gas
h gas =
D t ( w ater )
m of Gas
1000( gas )
ACTUAL VELOCITY OF FLUE GAS(v'):
v' = kv
where: K - velocity coefficient, whose value ranges from 0.30 to 0.50
DIAMETER OF CHIMNEY(D):
D=
4Q gas
v'
where:
Qg - volume flow rate of flue gas, m3/sec
VOLUME FLOW RATE OF FLUE GAS
Q gas
mgas m3
=
 gas sec
mg = mass flow rate of flue gas, kg/sec
MASS FLOW RATE OF FLUE GAS:
a) Without considering Ash loss:
A

m gas = mFuel  + 1
F

b) Considering Ash loss
A

mgas = mFuel  + 1 − Ash Loss
F

Where:
Ash loss in decimal
VOLUME FLOW RATE OF PRODUCTS OF COMBUSTION (Obtained from the balanced combustion equation)
PV = nRT
n + nH2O + n SO 2 + n O 2 + nN2 + n CO + n CH + nNO + nNO2 + nNO3
n = CO2
kg Fuel
nRT 
kg  m 3
V=
 mFuel

P 
sec  sec
Q gas = V in m 3/ sec
MOLECULAR WEIGHT and GAS CONSTANT OF PRODUCTS OF COMBUSTION
Mass of Products
Total Moles of Products
nCO2MCO2 + nH2OMH2O + nSO2MSO2 + nO2MO2 + nN2MN2 + nCOMCO + nCHMCH + nNOMNO + nNO2MNO2 + nNO3MNO3
M=
nPr oducts
M=
R=
R KJ
M kg − K
Problem No. 1
A power plant at an elevation 0f 600 m has a chimney with an actual draft of 19 mm of H 2O and frictional losses of 15%.
Flue gases at the rate of 5 kg/sec enters the chimney at 180C and leaves the chimney at 150C, 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 150C. 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
M = ΣyiMi
M Flue Gas = 12.7(44) + 2.2(18) + 0.1(64) + 6.8(32) + 77.9(28) + 0.30(28)
M Flue Gas = 30.12
R Flue Gas =
8.3143
KJ
= 0.276
30.12
kg - K
At 600 m elevation and standard atmosphere
P600 m = 94.44 KPa
T600 m = 290.1 K
ρ air =
P
kg
= 1.134 3
RT
m
 180 + 150 
=
 + 273 = 438K
2


94.44
kg
ρ Gas =
= 0.781 3
(0.276)( 438)
m
D a = D t − 0.15D t
TGas
ave
19
= 22.353 mm H 2 O
(1 − 0.15)
1000H(ρ air − ρ Gas )
Dt =
mm H 2 O
ρ water
Dt =
ρ water = 1000
H = 63.32 m
kg
m3
4Q gas
D=
πv'
v' = kv
v = 2gh gas
h gas =
D t (ρ water )
m of Gas
1000(ρ gas )
h gas = 28.62 m of flue gas
v = 23.70
m
sec
v' = 0.40(23.70) = 9.48
Qgas =
5
ρ gas
= 6.4
m
sec
m3
sec
D = 1.5 m
Problem No. 2
A steam power plant using high grade coal having the Ultimate analysis as follows, C = 81% ; H 2 = 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 650K 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 25C. For a
coal consumption of 1500 kg/hour as fired, determine the required height and diameter of the smokestack.
Ultimate Analysis
C = 81%
H 2 = 12%
O 2 = 1.2%
N 2 = 1.6%
S = 2.4%
H 2 O = 1.7%
A = 0.1%
Comp.
C
H2
O2
N2
S
H2O
Ash
Denominator
UA
81
12
1.2
1.6
2.4
1.7
0.1
(100-0.10
Ashless(xi)
81.1
12.01
1.201
1.602
2.402
1.702
100
Mi
12
2
32
28
32
18
-
Xi/Mi
0.068
0.06
0.0004
0.0006
0.0008
0.0009
0.1303
Yi(Molal)
51.87
46.1
0.29
0.44
0.58
0.73
100
Combustion with 100%TA
Basis :100 moles of Fuel
51.87C + 46.1H 2 + 0.29O 2 + 0.44 N 2 + 0.58S + 0.73H 2 O + xO 2 + x (3.76) N 2 → aCO2 + bH 2 O + cSO 2 + dN 2
51.87C + 46.1H 2 + 0.29O 2 + 0.44 N 2 + 0.58S + 0.73H 2 O + 75.21O 2 + 282.78N 2 → 51.87CO2 + 46.83H 2 O + 0.58SO 2 + 283.22 N 2
Combustion with EA(Excess air = 0.30)
51.87C + 46.1H 2 + 0.29O 2 + 0.44 N 2 + 0.58S + 0.73H 2 O + 1.30(75.21)O 2 + 1.30(282.78) N 2 → 51.87CO2 + 46.83H 2 O + 0.58SO 2 +
eO 2 + fN 2
51.87C + 46.1H 2 + 0.29O 2 + 0.44 N 2 + 0.58S + 0.73H 2 O + 97.77O 2 + 367.61N 2 → 51.87CO2 + 46.83H 2 O + 0.58SO 2 +
22.56O 2 + 368.05N 2
n Pr oducts = 489.89
n Pr oducts (8.3143)(650)
= 24,288.86m 3
109
Mass of Fuel = 767.63 kg
VPr oducts =
m 3 of Products
= 31.64
kg of Fuel
VF Pr oducts = 31.64(1500) = 47,462.07
VF Pr oducts = 13.18
m 3 of Products
hr
m 3 of Products
sec
Da = Dt - DL mm of H2O
27 = D t − 0.11D t
Dt =
27
= 30.34 mm H 2 O
1 − 0.11
h gas =
D t (ρ water )
m of Gas
1000(ρ gas )
ρ gas =
P
R g Tg
M g = ΣyiMi =
Σmi
ng
51.87(44) + 46.83(18) + 0.58(64) + 22.56(32) + 368.05(28)
= 28.96
51.87 + 46.83 + 0.58 + 22.56 + 368.05
Rg = 0.287
P
kg
ρ gas =
= 0.58 3
R g Tg
m
Mg =
101
kg
= 1.18 3
0.287(25 + 273)
m
30.34(1000)
=
m of Gas
1000(0.58)
= 51.92 m of gas
ρ air =
h gas
h gas
v = 2gh g = 31.92
m
sec
v' = kv = 0.35(31.92) = 11.17
Dt =
H=
1000H(ρ air − ρ Gas )
mm H 2 O
ρ water
D t ρ water
= 50.93 m
1000(ρ air − ρ Gas )
Qg =
D=
m
sec
πD 2 ( v' )
4
4Q g
π ( v' )
= 1.23 m
Problem no. 3
The actual velocity of a gas entering in a chimney is 8 m/sec. the gas temperature is 25C 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.
Fuel : C 7 H16
e : 20%
kg
hr
Combustion with 100% TA
C 7 H16 + aO 2 + a(3.76)N2 → bCO2 + cH 2 O + a(3.76)N2
m Fuel : 50
a = n + 0.25m = 11
b=n=7
c = 0.5m = 8
Combustion with EA (e = 0.20)
C 7 H16 + (1.20)aO 2 + (1.20)a(3.76)N2 → bCO2 + cH 2 O + dO 2 + (1.20)a(3.76)N2
d = e(n + 0.25m) = 2.2
Combustion Equation
C 7 H16 + 13O 2 + 49.632N 2 → 7CO2 + 8H 2 O + 2.2O 2 + 49.632N 2
Mass of Products (m products)
m products = 7(44) + 8(18) + 2.2(32) + 49.632(28) = 1,912.10 kg
Mass of Products per kg of fuel
m products 1,912.10 1,912.10
kg Products
=
=
= 19.12
m Fuel
12(7) + 16
100
kgFuel
M Pr oducts − mass flow rate of products
M Products = 19.12(50) = 956
kg
hr
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 16C and a pressure of 101.6
KPa.
Combustion with 100% TA (Basis :100 moles of fuel)
9C 2 H 6 + 90CH4 + 0.2CO2 + 0.8N 2 + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + dN 2
Carbon Balance
2(9) + 90 + 0.2 = b
b = 108.2
Hydrogen Balance
6(9) + 4(90) = 2c
c = 207
Oxygen Balance
0.2(2) + 2a) = 2b + c
a = 211.5
Nitrogen Balance
2(0.8) + a(3.76)2 = 2d
d = 796.04
9C 2 H 6 + 90CH4 + 0.2CO2 + 0.8N 2 + 211.5O 2 + 795.24 N 2 → 108.2CO2 + 207H 2 O + 796.04 N 2
n Air = 211.5+ = 795.24 = 1,006.74 Moles
n Gas = 100
PV = n RT
n RT
P
At the same temperaru te and pressure, the mole ratio is equal to volumetri c ratio
V=
m 3 of Air
1,006.74
=
= 10.06
100
m 3 of fuel
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
Combustion with 100% TA (basis; 100 moles of fuel)
80C12H 26 + 20 C14H 30 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + a(3.76)N 2
80(12) + 20(14) = b
b = 1240
80(26) + 20(30) = 2c
c = 1340
2a = 2b + c
a = 1910
Combustion with EA (e = 0.30)
80C12H 26 + 20 C14H 30 + (1.30)aO 2 + (1.30)a(3.76)N2 → bCO2 + cH 2 O + dO 2 + (1.30)a(3.76)N2
1.30a (2) = b(2) + c + 2d
d = 573
Combustion Equation
80C12H 26 + 20 C14H 30 + 2483O 2 + 9336.08N 2 → 1240CO2 + 1340H 2 O + 573O 2 + 9336.08N 2
2483(32) + 9336.08(28)
A
= 19.41
 Actual =
80(170) + 20(198)
F
n Pr oducts = 1240 + 1340 + 573 + 9336.08 = 12,489.08
PH2O
1340
=
P
12,489.08
PH2O
1340
=
101 12,489.08
PH2O = 10.84 KPa
DPT = tsat at 10.84 KPa
DPT = 47.417C
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.
For Fuel A
A
  = 11.44(0.885) + 34.32(0.115) = 14.07
F
For Fuel B
A
  = (1.06)(14.07) = 14.915
F
C + H =1
C = (1 − H)
A
  = 11.44(1 − H) + 34.32H = 14.915
F
14.915 − 11.44 = (34.32 − 11.44)H
H = 0.1519 = 15.19%
C = (1 − .1519) = 0.8481 = 84.81%
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 25C. Determine
a. the actual air-fuel Ratio
b. the m3 of CO2 formed per kg of fuel if the product temp. is 400C 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
Combustion with 130%TA
C12H 26 + (1.30)aO 2 + (1.30)a(3.76)N 2 → bCO2 + cH 2 O + dO 2 + (1.30)a(3.76)N 2
b=n
b = 12
a = n + 0.25m = 18.5
c = 0.5m = 13
d = e(n + 0.25m) = 5.55
C12H 26 + 24.05O 2 + 90.428N 2 → 12CO2 + 13H 2 O + 5.55O 2 + 90.428N 2
(1.30)137.28(n + 0.25m)
A
=
= 19.42
 
F
12m + m
  Actual
12(8.3143)(400 + 273)
VCO 2 =
= 662.7 m 3
101.325
VCO 2
662.7
=
= 3.9
m Fuel 12(12) + 26
n Pr oducts = 12 + 13 + 5.55 + 90.428 = 120.98
M=
m Pr oducts 12(44) + 13(18) + 5.55(32) + 90.428(28)
kg
=
= 28.7
n Pr oducts
120.98
kg mol
8.3143
KJ
= 0.29
28.7
kg - K
m Dry flue gas = 12(44) + 5.55(32) + 90.428(28) = 104.09
R=
M Dry Flue Gas =
12 + 5.55 + 90.428
kg
= 29.82
104.09
kg mol
Combustion with CO and with 100%TA
C12 H 26 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + 1.5CO + a(3.76)N 2
12 = b + 1.5
b = 12 - 1.5 = 10.5
26 = 2c
c = 13
2a = 2b + c + 1.5
c 1.5
a = b+ +
= 17.75
2 2
Combustion with CO and with EA (e = 0.30)
C12 H 26 + (1.30)aO 2 + (1.30)a(3.76)N 2 → bCO2 + cH 2 O + 1.5CO + dO 2 + (1.30)a(3.76)N 2
By oxygen balance
d = 5.325
C12 H 26 + 23.075O 2 + 86.762N 2 → 10.5CO 2 + 13H 2 O + 1.5CO + 5.325O 2 + 86.762N 2
n Pr oducts = 10.5 + 13 + 1.5 + 5.325 + 86.762 = 117.09
M=
10.5(44) + 13(18) + 1.5(28) + 5.325(32) + 86.762(28)
kg
= 28.51
117.09
kg mol
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.
H 2 + 1 O 2 → H 2O
2
C + O 2 → CO 2
2
+
16
→
18
12 + 32 → 44
3 + 8 → 11
kgO 2 8
= = 2.67
kgC
3
kgAir
= 11.44
kgC
1+ 8 → 9
kgO 2 8
= =8
KgH 1
kgAir
= 34.32
KgH
Problem 2: Calculate the theoretical Oxygen--fuel ratio and Air--fuel ratio on a mass basis for the combustion of ethanol,
C2H5OH.
Combustion with 100% TA
C 2 H 5OH + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + a (3.76) N 2
2=b
5 + 1 = 2c
c=3
1 + 2a = 2b + c
2(2) + 3 − 1
a=
=3
2
C 2 H 5OH + 3O 2 + 11.28N 2 → 2CO2 + 3H 2 O + 11.28N 2
kgOxygen
3(32)
96
=
=
kgFuel
124 + 5 + 16 + 1 46
kgAir
3(32) + 11.28(28)
=
=
kgFuel 124 + 5 + 16 + 1
Problem 3: Determine the molal analysis of the products of combustion when octane C8H18 is burned with 100% excess air.
Combustion with EA (e = 100%)
C8 H18 + (1 + 1)aO 2 + (1 + 1)a (3.76) N 2 → bCO2 + cH 2 O + dO 2 + (1 + 1)a (3.76) N 2
a = n + 0.25m
b=n
c = 0,5m
d = e(n + 0.25m)
C8 H18 + 24.5O 2 + 92.12 N 2 → bCO2 + cH 2 O + dO 2 + (1 + 1)a (3.76) N 2
C8 H18 + 25O 2 + 94 N 2 → 8CO2 + 9H 2 O + 12.5O 2 + 94 N 2
n Pr oducts = 8 + 9 + 12.5 + 94 = 123.5
8
y CO 2 =
x100% = 6.48%
123.5
9
y H 2O =
x100% = 7.29%
123.5
12.5
y O2 =
x100% = 10.12%
123.5
94
y N2 =
x100% = 76.11%
123.5
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?
Combustion with EA (e = 50%)
C10H 22 + 23.25O 2 + 87.42 N 2 → 10CO2 + 11H 2 O + 7.75O 2 + 87.42N 2
a = n + 0.25m
b=n
c = 0,5m
d = e(n + 0.25m)
Problem 5: A sample of pine bark has the following ultimate analysis, percent by mass: C = 53.4% ; H 2 = 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.
Fuel
UA
Ashless
C
H2
O2
N2
S
H20
Ash
Total
53.4
5.6
37.9
0.1
0.1
0
2.9
100
55
5.8
39
0.103
0.103
0
0
100
O 

A
= 11.44C + 34.32 H 2 − 2  + 4.29S = 6.6
 
8 
 F  T heoretical

A
A
= (1 + e) 
= 8.58
 
F
  Actual
 F  T heoretical
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 25C (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
Combustion with 100% TA
Basis :100 moles of fuel
15C 2 H 6 + 85CH4 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + 2CO + 1NO + dN 2
15C 2 H 6 + 85CH4 + 222O 2 + 834.72N 2 → 113CO2 + 215H 2 O + 2CO + 1NO + 834.22 N 2
Combustion with EXCESS AIR(e = 0.25)
Basis :100 moles of fuel
15C 2 H 6 + 85CH4 + (1.25)222O 2 + (1.25)834.72N 2 → 113CO2 + 215H 2 O + 2CO + 1NO + fO 2 + gN 2
15C 2 H 6 + 85CH4 + 227.5O 2 + 1,043.4 N 2 → 113CO2 + 215H 2 O + 2CO + 1NO + 55.5O 2 + 1,042.9 N 2
227.5(32) + 1,043.4(28)
A
=
= 21.05
 
15(30) + 85(16)
 F  Actual
Problem No. 2
Calculate the theoretical air – fuel ratio of the Hydrocarbon Fuel in the table below.
Fuel
C8H18
C7H16
C4H10
C9H20
C11H24
Theoretical
(By Mass)
15.05
15.10
15.38
15.02
14.96
A/F
Ratio
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 25C 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
hf of C7H16 = -187,900 KJ/kgm
For HHV
For LHV
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; CO 2 = 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 C 8H18. 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 C 12H26. 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 C 16H32.
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
2 b
3 a
4 d
5 a
6
7
8
9
10
a
b
c
d
d
11
12
13
14
15
a
b
c
d
a
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 H 2O and assume 10%
losses and average flue gas temperature of 350C .
Diameter
Generator Output = 1,500 KW
Height
Fuel and air
Diesel Engine
At 800 m above sea level
P = 92.21 KPa
T = 287.8 K
P
kg
ρ air =
= 1.12 3
RT
m
Combustion of C16H 34 with 50% excess air
C16H 34 + 36.75O 2 + 138.18N 2 → 16CO2 + 17H 2 O + 12.25O 2 + 138.18N 2
16(44) + 17(18) + 12.25(32) + 138.18(28)
kg
= 28.74
16 + 17 + 12.25 + 138.18
kgm
8.3143
KJ
R Pr oducts =
= 0.289
M
kg - K
183.43(8.3143)(350 + 273)
VPr oducts =
= 10,304.005 m 3
92.21
3
10,304.005
(0.30)(1500) hr = 5.7 m
Q Pr oducts =
12(16) + 34
3600 sec
sec
M Pr oducts =
92.21
kg
= 0.51 3
(0.289)(350 + 273)
m
D a = D t − 0.10D t = 25 mm water
ρ Pr oducts =
D t = 28.09 mm
28.09(1000)
= 54.85 m of flue gas
1000(0.51)
m
v = 2gh g = 32.80
sec
1000H ρ air - ρ gas
Dt =
mm water
ρ water
hg =
(
)
H = 46.05 m
May 08, 2019
BLOWERS
Qg =
π 2
D (kv)
4
D=
4(5.7)
= 0.7 m
π(0.40)(32.80)
Blower is a machine used to compressed air or gas by centrifugal force to a final pressure not exceeding 241 KPa gage.
Usually blower has no cooling system or it is not water cooled.
COMPRESSION OF GASES
The design of blower is usually based upon either an adiabatic or isothermal compression.
A. Isothermal Compression (PV = C) or (T1 = T2 = T)
W (Work/Power
of Blower
P
P2 T2
P1 T1
P2
m
m
P1V1 = P2 V2
1
V
P
W = P1V1 ln 2
P1
V1 = Q
WBlower = P1Q ln
W = −  VdP
PV = C
P1
T1 = T2
2
P2
P1
P1Q = mRT
W = QγH
H − Isothermal Head
ρg
1000
P
ρ=
RT
Pg
γ=
(1000)RT
γ=


P2
P1g
H
= Q

P1
 (1000)RT1 
(1000)RT1  P2 
 ln  meters
H=
 P 
g
1 

P1Q ln
B. Isentropic Compression (PVk = C; 0r S = C)
P1V1k = P2 V2 k
T2  P2
=
T1  P1
Q=0



k −1
k
W = Q − Δh − ΔKE − ΔPE

W = − VdP −ΔKE − ΔPE
For a compressor work is done
on the system; let - W = Wc
Wc − Ideal compressio n work/Po wer in blower
− W = WC = Δh − Q + ΔKE + ΔPE
k −1


kP1V1  P2  k

 
Wc =
− 1 + ΔKE + ΔPE



k − 1  P1 




If ΔKE & ΔPE are negligible

− W = WC = VdP + ΔKE + ΔPE

k −1




kP1V1  P2 k

 
VdP =
− 1 KW



k − 1  P1 




k −1




kP1V1  P2 k

 
Wc =
− 1



k − 1  P1 




P1V1 = mRT1
v 2 2 − v12
KW
2(1000)
mg (z 2 − z1 )
ΔPE =
KW
1000
ΔKE =
Q=0
k −1


kP1V1  P2  k
  − 1 = QγH
Wc =

k − 1  P1 




V1 = Q
γ=
P1g
(1000)RT1




P1g
k −1



kP1Q  P2 k
 


EFFICIENCY
A. Adiabatic Efficiency
ek =
Isentropic Work
x 100%
Brake or Shaft Work
B.Isothermal efficiency
eIso =
Isothermal Work
x 100%
Brake or Shaft Work
C. Brake Power
2πTN
KW
60,000
T - Brake Torque inN - m
N - n0. of RPM
BP =
RATIO OF THE ADIABATIC TEMPERATURE RISE TO THE ACTUAL TEMPERATURE RISE
K −1


K


P
2

T1  
− 1
 P1 



Y=
'
T2 − T1
(
)
RELATIONSHIP FOR CORRECTING PERFORMANCE CURVES
1. Volume Flow
QB N B
=
QA N A
2. Weight Flow
mB  N B   P1B   T1A 



=
mA  N A   P1A   T1B 
3. Pressure Ratio
K −1


K


P
2
 
− 1
 P1 


 B  N B
=
K
−
1


K
 NA
 P2 

−1
 P1 


A
P2
= rp (pressure ratio)
P1




2
 T1A

T
 1B




4. Head
H B N 2B
=
H A N 2A
5. Brake Power
BPB  NB 

=
BPA  N A 
3
 P1B

 P1A
 T1A 


T
 1B 
 P  K −1K

2
 
− 1
 P1 

BPB  P1B  Q B  
B


= 
BPA  P1A  Q A   P  K −1K

2
 
− 1
 P1 


A
1 - suction
2 - discharge
A - 1st condition
B - 2nd condition
R - gas constant, KJ/kg-K
P - absolute pressure in KPa
 - density, kg/m3
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
C
H2
O2
N2
S
H2O
Ash
U.A.
72
5
10
1.2
3.3
0.1
8.4
100
Ashless
78.6
5.46
10.92
1.31
3.6
0.11
100
Mi
12
2
32
28
32
18
-
Xi/Mi
0.066
0.027
0.003
0.0005
0.0011
0.0001
0.098
Molal Analysis
66.93
27.89
3.49
0.48
1.15
0.06
100
Combustion Equation w ith 50% Excess air
66.93C + 27.89H2 + 3.49O2 + 0.48N2 + 1.15S + 0.06H2O → Fuel
117.81O2 + 442.98N 2 → Air
66.93CO2 + 27.95H 2O + 1.15SO2 + 39.27O 2 + 443.46 N 2 → Pr oducts
M Pr oducts =
66.93(44) + 27.95(18) + 1.15(64) + 39.27(32) + 443.46(28)
= 29.71
66.93 + 27.95 + 1.15 + 39.27 + 443.46
8.3143
= 0.280
29.71
kg of Products
66.93(44) + 27.95(18) + 1.15(64) + 39.27(32) + 443.46(28)
=
= 16.83
kg of Fuel
66.93(12) + 27.89(2) + 3.49(32) + 0.48(28) + 1.15(32) + 0.06(18)
R=


hr
kg
 = 7.011
m Pr oducts = 16.83(1.5)(1000)
3600
sec)
sec


Products
CO2
H2O
SO2
O2
N2
n(Moles)
66.93
27.95
1.15
39.27
443.46
Cp = ΣxiCpi = 1.021
Cv = ΣxiCVi = 0.741
Cp
k=
= 1.378
Cv
k −1


kmRT1  P2  k

 
WBlower =
− 1
k − 1  P1 




P2 = 110
P1=10 0
T1 = 600 K
WBlower = 113.7 kw
WBlower
Brake Power
BP = 162.41 KW
MP  165 KW
e Blower =
Mi
44
18
64
32
28
nM(kg)
2,945.10
503.12
73.63
1,256.69
12,416.90
17,195.44
xi
17.13
2.93
0.43
7.31
72.21
100
Cp
0.845
1.866
0.624
0.918
1.041
Cv
0.656
1.404
0.494
0.658
0.744
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)
V
V
r = 1 = 4 →1
V2 V3
V1 = V4 and V2 = V3
V1 - volume at bottom dead center (BDC)
V2 – volume at top dead center (TDC)(clearance volume)
Displacement Volume (VD)
VD =V1 – V2 → Eq. 2
Percent Clearance (C)
C=
V2
VD
r=
1+ C
C
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 = Q A – QR
P,V and T Relations
At point 1 to 2 (S = C)
T2  V1
=
T1  V2



k −1
= (r )k −1
P
=  2
 P1



k −1
k
P2 = P1 (r ) k
V1
r
T2 = T1 (r ) k −1
V2 =
At point 2 to 3 (V = C)
T3 P3
=
T2 P2
Q A = mC v (T3 − T2 )
T3 =
QA
+ T2
mC v
P3 = P2
T3
T2
At point 3 to 4 (S = C)
k −1
P 
T3  V4 
 =  3 
=
T4  V3 
 P4 
P
P4 = k3
r
T3 = T4 (r ) k −1
k −1
k
= (r ) k −1
At point 4 to 1 (V = C)
T4 P4
=
T1 P1
Entropy Change
a) S during Heat Addition
T
S3 − S2 = mCv ln 3
T2
b) S during Heat Rejection
S1 − S4 = mCv
T1
T4
Thermal Efficiency
e=
W
x 100%
QA
e=
QA − QR
x100%
QA
 Q 
e = 1 − R  x 100%
 QA 
 (T − T ) 
e = 1 − 4 1  x 100%
 (T3 − T2 ) 

1 
e = 1 − k −1  x 100%
 (r ) 
Mean Effective Pressure
Pm =
W
KPa
VD
VD = V1 − V2
V2 =
V1
r
 1
VD = V1 1 − 
 r
mRT1  1 
VD =
1 − 
P1  r 
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.
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
V
V
r= 1 = 4
V2 V2
V1 V4

V2 V3
Cut-Off Ratio(rc)
V3
V2
Percent Clearance
V
C= 2
VD
rc =
r=
1+ C
C
Displacement Volume (VD)
VD =V1 – V2
Heat Added (QA)
At P = C ; Q = mCp(T)
QA = mCp(T3 – T2)
QA = mkCv(T3 – T2)
Heat Rejected (QR)
QR = mCv(T4 – T1)
Net Work (W)
W = Q
W = Q A – 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)
T2  V1 

=
T1  V2 
k −1
= (r )
k −1
P2 = P1r k
T2 = T1 (r ) k −1
At Point 2 to 3 (P = C)
T3 V3
=
= rc
T2 V2
Q A = mC p (T3 − T 2)
P2 = P1r k
P3 = P2
T3 =
QA
+ T2
mCp
T3 = T1 (r )k −1 (rc )
υ3 =
RT3
P3
rc =
υ3
υ2
P
=  2
 P1



k −1
k
At Point 3 to 4 (S = C)
T4  V3 

=
T3  V4 
υ1 = υ 4
k −1
P
=  4
 P3



k −1
k
P3 υ 3 k = P4 υ 4 k
υ
P4 = P3  3
 υ4
T4  P4 
= 
T3  P3 
P
T4 = T3  4
 P3
T4 = T1 (r )
T4 = T1
k

 = 423.4 KPa

k −1
k



k −1
k −1
k
V
(rc ) 3
 V4



k −1
V1k −1 V3 V3 k −1
V2 k −1 V2 V4 k −1
T4 = T1 (rc ) k
At Point 4 to 1 (V = C)
T4 P4
=
T1 P1
Entropy Change
a) S during Heat Addition
T
S3 − S2 = mCp ln 3
T2
b) S during Heat Rejection
T
S1 − S4 = mCv ln 1
T4
S1 - S4 = -( S3 – S2)
Thermal Efficiency
e=
W
x 100%
QA
e=
QA − QR
x 100%
QA
 Q 
e = 1 − R  x 100%
 QA 

(T − T ) 
e = 1 − 4 1  x 100%
k
(T3 − T2 ) 

Substituting Eq. 11, Eq. 13 and Eq. 14 to Eq. 22

1  r k − 1 
e = 1 − k −1  c
 x100%
 (r )  k (rc − 1) 
Mean Effective Pressure
W
KPa
VD
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
Pm =
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
V1
V2
V1 = V5 and V2 = V3
Cut-Off Ratio
r=
rc =
V4
V3
Pressure Ratio
P3
P2
P3 = P4
rp =
Percent Clearance
C=
V2
VD
1+ C
C
Displacement Volume
VD = V 1 – V2 m 3
VD =  1 -  2
r=
P, V, and T Relations
At point 1 to 2 (S = C)
T2  V1 

=
T1  V2 
k −1
= (r )
k −1
P 
=  2 
 P1 
k −1
k
T2 = T1 (r ) k −1
At point 2 to 3 (V = C)
T3 P3
=
= rP
T2 P2
T3 = T1 (r) k −1 (rP )
At point 3 to 4 (P = C)
T4  V4 
 = rc
=
T3  V3 
T4 = T1 (r ) k −1 (rP )(rc )
At point 4 to 5 (S = C)
T5  V4 

=
T4  V5 
T5 = T1
k −1

V1k −1
V2
k −1

(r ) V4  V4k−1 
k −1 P
V2  V1

T5 = T1 (rc ) (rP )
k
At 5 to 1 (V = C)
T5 P5
=
T1 P1
Entropy change
a) At 2 to 3 (V = C)
S3 − S2 = mC V
T3
T2
b) At 3 to 4 (P = C)
T4
T3
S4 – S2 = (S3 – S2) + (S4 – S3)
S4 − S3 = mCp ln
c) At 5 to 1 (V = C)
S1 – S5 = mCvln S1 − S5 = mCv
T1
T5
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
e=
W
x100%
QA
e=
QA − QR
x100%
QA
 Q 
e = 1 − R  x 100%
 QA 


(T5 − T1 )
e = 1 −
 x 100%
 (T3 − T2 ) + k(T4 − T3 ) 
Substituting


rP rc k − 1
1 
 x100%
e = 1 − k −1 
 (r )  (rP − 1) + krP (rc − 1) 
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 25C. 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 = 298K; QA = 1400 KJ/kg
Solution:
a.

1 
e = 1 − k −1  x 100%
 (r ) 

1 
e = 1 − 1.4 −1  x 100%
 (8)

e = 56.5%
b)
W
VD
Pm =
W = QA − QR
e=
W
QA
W = 0.565(1400)
KJ
W = 791
kg
VD = υ1 − υ 2
r=
υ1
υ2
υ2 =
υ1
r
υ1
r
 1
VD = υ1 1 − 
 r
P1υ1 = RT1
VD = υ1 −
m3
 1
1 − r  = 0.748 kg


Pm = 1057 KPa
VD =
RT1
P1
c)
1+ c
c
rc = 1 + c
c(r − 1) = 1
1
c=
= 0.143 = 14.3%
r −1
r=
Example No. 2
An engine operates on the air standard Otto cycle. The cycle work is 900 KJ/kg, the maximum cycle temperature is 3000C
and the temperature at the end of isentropic compression is
600C. Determine the engine's compression ratio and the thermal efficiency.(r = 6.35; e =
52.26%)
Given: W = 900 KJ/kg; T3 = 3273K; T2 = 873K
QA = Cv(T3 – T2) = 0.7175(3273 – 873) = 1722 KJ/kg
e = W/QA
e = 52.26%
1
 1  k −1
r =

1− e
r = 6.36
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 27C. 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%)
Given
r = 20
rc = 3
P1 = 100KPa
T1 = 573K
Q A = C p (T3 − T2 )
Rk
= 1.0045
k −1
At 1 to 2 (S = C)
Cp =
T2 = T1 (r )k −1 = 994.3K
T3 = rc (T 2) = 2983K
Q A = C p (T3 − T2 ) = 1997.65
KJ
kg
P2 = P1 (r ) k = 6229KPa
P2 = P3
At 3 to 4
k −1
υ 
T4  P4  k
=  
=  3 
T3  P3 
 υ4 
RT3
υ3 =
= 0.14
P3
k −1
T 4 = 1442.45K
Q R = Cv(T4 − T1 )
R
= 0.7175
k −1
KJ
Q R = 820
kg
W = QA − QR = 1167.65
W
e=
x100% = 59%
QA
Cv =
υ 
=  3 
 υ1 
k −1
Given: r = 20; rc = 3
Solution:
V
r= 1
V2
rk =
; P1 = 100 KPa ; T1 = 300C
P2
P1
P2 = 100(20)1.4 = 6229 KPa
P2 = P3 = 6229 KPa
T2
= (r ) k −1
T1
T2 = 300(20)1.4 −1 = 994.3 K
T3 V3
=
= rc
T2 V2
T3 = 994.3(3) = 2983 K
υ3 =
RT3 0.287(2983)
=
= 0.14 m 3
P3
6229
υ1 = υ 4 =
RT1 0.287(300)
=
= 0.861 m 3
P1
100
T4  V3 

=
T3  V4 
k −1
1.4 −1
 0.14 
T4 = 2983

 0.861 
= 1442.45 K
QA = Cp(T3 – T2) = 1.0045(2983 – 994.3)
QA = 1997.65 KJ/kg
QR = Cv(T4 – T1) = 0.7175(1442.45 – 300)
QR = 820 KJ/kg
W = QA - QR = (1997.65 – 820)
W = 1177.65 KJ/kg
e=
W
x 100%
QA
1177.65
x 100%
1997.65
e = 59 %
k

1  rc − 1 
e = 1 −
k −1 

 ( r )  k ( rc − 1 ) 
e=

 ( 3 )1.4 − 1 
1

 x 100%
e = 1 −
1.4 −1 
 1.4( 3 − 1 ) 
 ( 20 )
e = 60.6 %
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, 316K. 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


rP rc k − 1
1 
 x100%
e = 1 − k −1 
 (r − 1) + krP (rc − 1) 

 (r )  P
e = 58.7%
1+ C
C
1
C=
= 10 %
r −1
r=
QA = Q23 + Q34
Q23 = mCv(T3 – T2)
T2 = T1 (r) k −1 = 825 K
P2 = P1 (r ) k = 2,784 KPa
T3 = T2 (rp ) = 1238 K
Q23 = 0.454(0.7175)(1238 – 825) = 135 KJ
Q34 = mCp(T4 – T3)
T
rc = 4 = 1.6
T3
T4 = 1.6T3
T4 = 1981K
Q34 = 0.454(1.0045(1981 – 1238) = 339 KJ
QA = Q23 + Q34 = 474 KJ
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
Intake
Compression
Power
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.
Two Stroke Cycle Engine
1. Intake-Compression stroke
2. Power-exhaust stroke
Intake-compression
Power-Exhaust
Intake port
Exhaust port
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.
ENGINE PERFORMANCE
1.
2.
HEAT SUPPLIED BY FUEL
KJ
Qs = m F (HV )
hr
where:
mf - fuel consumption in kg/hr
HV - heating value of fuel in KJ/kg
INDICATED POWER
P πLD 2 Nn'
IP = mi
KW
4(60)
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
N = (RPM ) → (For 2 Stroke - Single acting)
N = 2(RPM) → (For 2 Stroke - Double acting)
(RPM)
N=
→ (For 4 Stroke - Single acting)
2
N = (RPM ) → (For 4 Stroke - Double acting)
3.
BRAKE POWER
BP =
PmbπLD 2 Nn'
KW
4(60)
where:
Pmb - brake mean effective pressure in KPa
N = (RPM ) → (For 2 Stroke - Single acting)
N = 2(RPM) → (For 2 Stroke - Double acting)
(RPM)
N=
→ (For 4 Stroke - Single acting)
2
N = (RPM ) → (For 4 Stroke - Double acting)
BP =
2 πTN
KW
60,000
where:
T - brake torque in N-m
N = RPM
4. FRICTION POWER
FP = IP − BP
5. INDICATED MEAN EFFECTIVE PRESSURE
AS'
Pmi =
KPa
L'
where:
A - area of indicator card, m2
S - spring scale, KPa/m
L' - length of indicator card, m
6. BRAKE TORQUE
T = (P − Tare )R N - m
where:
P - Gross load on scales, N
Tare - tare weight, N
R - length of brake arm, m
7. PISTON SPEED
m
min
8. DISPLACEMENT VOLUME
πLD 2 Nn'
VD =
KPa
4(60)
IP
VD =
KPa
Pmi
BP
VD =
KPa
Pmb
PS = 2LN
Where
N = (RPM ) → (For 2 Stroke - Single acting)
N = 2(RPM) → (For 2 Stroke - Double acting)
(RPM)
N=
→ (For 4 Stroke - Single acting)
2
N = (RPM ) → (For 4 Stroke - Double acting)
9. SPECIFIC FUEL CONSUMPTION
a. Indicated specific fuel consumption
m
kg
mfi = f
IP KW - hr
b. Brake specific fuel consumption
m
kg
mfb = f
BP KW - hr
c. Combined specific fuel consumption
m
kg
mfc = f
GP KW - hr
where: GP - Generator output
10. HEAT RATE (HR): Heat supplied divided by the KW produced.
a. Indicated heat rate
Q
KJ
HRi = s
IP KW - hr
b. Brake heat rate
Q
KJ
HRb = s
BP KW - hr
c. Combined heat rate
Q
KJ
HRc = s
GP KW - hr
11. GENERATOR SPEED
120f
N=
RPM
n
Where: n - number of generator poles (usually divisible by 4)
12. MECHANICAL EFFICIENCY
BP
ηm =
x 100%
IP
13. GENERATOR EFFICIENCY
GP
ηg =
x 100%
BP
14. Indicated Thermal Efficiency
3600(IP)
ei =
x 100%
Qs
15. Brake Thermal Efficiency
3600(BP)
eb =
x 100%
Qs
16. Combined Thermal efficiency
3600(GP)
ec =
x 100%
Qs
17. Indicated Engine Efficiency
e
ηi = i x 100%
e
18. Brake Engine Efficiency
e
ηb = b x 100%
e
where:
e - cycle thermal efficiency
19. Volumetric Efficiency
ηv =
Actual volume of air entering,
Displacement Volume,
m3
sec
m3
sec x 100%
20. Correction Factor for Nonstandard condition
a. Considering temperature and pressure effect
 B  T 
Ph = Ps  S  h 


 Bh  Ts 
b. Considering temperature effect alone
 T 
Ph = Ps  h 
 T 
 s
c. Considering pressure effect alone
B 
Ph = Ps  S 
 Bh 
Note: From US Standard atmosphere:
83.312h 

Bh =  Bs −
 mm Hg
1000 

6.5h 

Th =  Ts  K
1000 

where: B - barometric pressure, mm Hg
T - absolute temperature, K
h - at elevation h condition
s - at sea level condition
21. ENGINE HEAT BALANCE: The total heat supplied to th engine was broken down into four heat items.
Q2
Q1
QS
engine
Q3
Q4
QS = Q1 + Q2 + Q3 + Q4
Q1 - heat converted to useful work
Q2 - heat lost to cooling water
Q3 - heat lost to exhaust gases
Q4 - heat lost due to friction, radiation and unaccounted for
Q1 = 3600(BP) KJ/hr
Q2 = mwCpw(two - twi) KJ/hr
Q3 = Qa + Qb KJ/hr
Qa = mgCpg(tg - ta) KJ/hr
Qb = mf(9H2)(2442.7) KJ/hr
Q4 = QS - (Q1 + Q2 + Q3) KJ/hr
H2 = 0.26 - 0.15S kgH/kgfuel
141.5
S=
131.5 + API
140
S=
130 + BeI
where:
Qa - sensible heat of products of combustion
Qb - heat required to evaporate and superheat moisture formed from the
combustion of hydrogen in the fuel
tg - temperature of flue gas, C
ta - temperature of air, C
H2 - amount of hydrogen in the fuel kg H/kg fuel
TERMS AND DEFINITIONS
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.
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.
Governor - is a device used to govern or control the speed of an engine under varying load conditions.
Purifier - a device used to purify fuel oil and lube oil.
Generator - a device used to convert mechanical energy.
Crank scavenging - is one that the crankcase is used as compressor.
Thermocouple - is made of rods of different metal that are welded together at one end.
Centrifuge - is the purification of oil for separation of water.
Unloader - is a device for automatically keeping pressure constant by controlling the suction valve.
Planimeter - is a measuring device that traces the area of actual P-V diagram.
Tachometer - measures the speed of the engine.
Engine indicator - traces the actual P-V diagram.
Dynamometer - measures the torque of the engine.
Supercharging - admittance into the cylinder of an air charge with density higher than that of the surrounding air.
Bridge Gauge - is an instrument used to find the radial position of crankshaft motor shaft.
Piston - is made of cast iron or aluminum alloy having a cylinder form.
Atomizer - is used to atomize the fuel into tiny spray which completely fill the furnace in the form of hollow cone.
Scavenging - is the process of cleaning the engine cylinder of exhaust gases by forcing through it a pressure of m
fresh air.
Flare back - is due the explosion of a maximum fuel oil vapor and air in the furnace.
Single acting engine - is one in which work is done on one side of the piston.
Double acting engine - is an engine in which work is done on both sides of the piston.
Triple-expansion engine - is a three-cylinder engine in which there are three stages of expansion.
The working pressure in power cylinder is from 50 psi to 500 psi.
The working temperature in the cylinder is from 800F to 1000F.
Air pressure used in air injection fuel system is from 600 psi to 1000 psi.
Effect of over lubricating a diesel engine is:
Carbonization of oil on valve seats and possible explosive mixture is produced.
The average compression ratio of diesel engine is from 14:1 to 16:1.
Three types of piston:
1. barrel type
2. trunk type
3. closed head type
Three types of cam follower:
1. flat type
2. pivot type
3. roller type
Methods of mechanically operated starting valve:
1. the poppet
2. the disc type
Three classes of fuel pump:
1. continuous pressure
2. constant stroke
c. variable stroke
Type of pump used in transferring oil from the storage to the service tanks:
1. rotary pump
2. plunger pump
3. piston pump
4. centrifugal pump
Valve that is found in the cylinder head of a 4-stroke cycle engine:
1. fuel valve
2. air starting valve
3. relief valve
4. test valve
5. intake valve
6. exhaust valve
Four common type of governors used on a diesel engine:
1. constant speed governor
2. variable speed governor
3. speed limiting governor
4. load limiting governor
Kinds of piston rings used in an internal combustion engines:
1. compression ring
2. oil ring
3. firing ring
4. oil scraper ring
Reasons of smoky engine:
1. overload
2. injection not working
3. choked exhaust pipe
4. fuel or water and leaky things
Methods of reversing diesel engines:
1. sliding camshaft
2. shifting roller
c. rotating camshaft
Arrangements of cylinders:
1. in-line
2. radial
3. opposed cylinder
4. V
5. opposed piston
Position of cylinders:
1. vertical
2. horizontal
3. inclined
Methods of starting:
1. manual, crank, rope, and kick
2. electric (battery)
3. compressed air
4. using another engine
Applications:
1. automotive
2. marine
3. industrial
4. stationary power
5. locomotive
6. aircraft
Types of internal combustion engine:
1. Gasoline engine
2. Diesel engine
3. Kerosene engine
4. Gas engine
5. Oil-diesel engine
Methods of ignition:
1. Spark
2. Heat of compression
Reasons for supercharging:
1. to reduce the weight to power ratio
2. to compensate the power loss due to high altitude
Types of superchargers:
1. engine-driven compressor
2. exhaust-driven compressor
3. separately-driven compressor
Auxiliary systems of a diesel engine:
1. Fuel system
a. fuel storage tank
b. fuel filter
c. transfer pump
d. day tank
e. fuel pump
2. Cooling system
a. cooling water pump
b. heat exchanger
c. surge tank
d. cooling tower
e. raw water pump
3. Lubricating system:
a. lub oil tank
b. lub oil pump
c. oil filter
d. oil cooler
e. lubricators
4. Intake and exhaust system
a. air filter
b. intake pipe
c. exhaust pipe
d. silencer
5. Starting system
a. air compressor
b. air storage tank
Advantages of diesel engine over other internal combustion engines:
1. low fuel cost
2. high efficiency
3. needs no large water supply
4. no long warm-up period
5. simple plant layout
Types of scavenging:
1. direct scavenging
2. loop scavenging
3. uniflow scavenging
Color of the smoke:
1. efficient combustion - light brown
baze
2. insufficient air - black smoke
3. excess air - white smoke
Causes of black smoke:
1. fuel valve open too long
2. too low compression pressure
3. carbon in exhaust pipe
4. overload on engine
Causes of white smoke:
1. one or more cylinders not getting
enough fuel
2. too low compression pressure
3. water inside the cylinder
ENGINE FOUNDATION
Functions of a Foundation:
Support the weight of the engine.
Maintain proper alignment with the machinery and
Absorb the vibration produced by unbalanced forces created by reciprocating revolving masses.
Materials:
Mixture: 1 : 2 : 4
1 part Cement
2 parts sand
4 parts broken stone or Gravel
For good firm soil, reinforced concrete foundations for large engines may use a leaner mixture down to
1:3:6
Soil Bearing Pressure:
The safe loads vary from about 4,890 kg/m 2 for alluvial soil or wet clay, to 19,560 kg/m2 for gravel, coarse sand and dry clay.
(12,225 kg/m2 is assumed to be safe load average).
In computation 2,406 kg/m2 may be used as weight of concrete.
Depth:
The foundation depth maybe taken as a good practical rule, to be 3.2 to 4.2 times the engine stroke; the lower factor for wellbalanced multi-cylinder engines and increased factor for engines with fewer cylinders or on less firm soil.
Weight:
The minimum weight required to absorb vibration could be expressed as a function of the reciprocating masses and the speed
of the engine. However, for practical purposes it is simpler to use the empirical formula.
Volume:
If the weight and speed of the engine is not known, the volume of concrete for the foundation may be estimated from the data
in the table.
Anchor Bolts:
To prevent pulling out of the bolts when the nuts are tightened, the length embedded in concrete should be equal to at least
thirty (30) times the bolt diameter. The upper ends are surrounded by a 50 mm or 75 mm sheet metal pipe, 460 mm to 610
mm long to permit them to be bent slightly to fit the holes of the bedplate.
FORMULA:
WF = e (WE) N
Where:
WF – weight of foundation in kgs
WE – weight of engine in kgs
e – empirical coefficient
N – engine speed in RPM
Values of e
Type of Engine
Single Acting
Single Acting
Single Acting
Single Acting
Single Acting
Single Acting
Single Acting
Double acting
Double acting
Cylinder Arrangement
Vertical
Vertical
Vertical
Vertical
Horizontal
Horizontal Duplex
Horizontal Twin Duplex
Horizontal
Horizontal Twin Tandem
No. of Cylinders
1
2
3
4, 6, 8
1
2
4
1, 2
4
e
0.15
0.14
0.12
0.11
0.25
0.24
0.23
0.32
0.20
Volume of Concrete Foundation (Cubic Feet per HP)
No. of Cylinders
1
2
3
High speed engine
4.0
2.5
2.0
Medium speed engine
5.0
3.1
2.5
Low speed engine
6.0
4.0
3.0
Note: 1 cubic meter (m3) = 35.315 ft3
1 Horsepower = 0.746 KW
4
1.7
2.1
2.6
5-8
1.5
1.9
2.3
General Requirements
1. All heavy machinery shall be supported on solid foundation of sufficient mass and base area to prevent or
minimize the transmission of objectionable vibration to the building and occupied space and to maintain the
supported machine at its proper elevation and alignment.
2. Foundation mass should be from 3 to 5 times the weight of the machinery it is supposed to support. If the
unbalanced inertial forces produced by the machine can be calculated, a mass of weight equal to 10 to 20 times
the forces should be used to dampen vibration.
For stability, the total combined engine, driven equipment and foundation center of gravity must be kept below
the foundation’s top.
3. the weight of the machine plus the weight of the foundation should be distributed over a sufficient soil area which
is large enough to cause a bearing stress within the safe bearing capacity of the soil with a factor of safety of 5.
4. Foundation should be isolated from floor slabs and building footings by at least 25 mm around its perimeter to
eliminate transmission of vibration. Fill opening with water tight-mastic.
When installing machinery above grade level of a building, additional stiffness must be provided on its structural
members of the building to dampen machine vibration.
5. Foundations are preferably built of concrete in the proportions of one (1) measure of Portland Cement to two
(2) measures of sand and four (4) measures of screened crushed stones. The machine should not be placed
on thefoundation until ten (10) days have elapsed or operated until another ten (10) days have passed.
6. Concrete foundation should have steel bar reinforcements placed both vertically and horizontally, to avoid
thermal cracking. Weight of reinforcing steel should be from ½ % to 1 % of the weight of the foundation.
7. Foundation bolts of specified size should be used and surrounded by a pipe sleeve with an inside diameter of
at least three (3) times the diameter of the anchor bolt and a length of at least eighteen (18) times the diameter
of the bolt. No foundation bolts shall be less than 12 mm diameter.
8. Machine should be leveled by driving wedges between the machine’s base and concrete foundation and with
The aid of a spirit level. Grout all spaces under the machine bed with a thin mixture of one (1) part cement and
one part sand. The level wedges should be removed after grout has thoroughly set and fill wedges holes with
grout.
SAMPLE PROBLEMS
1.
A 2 - stroke, 4 - cylinders, 38 cm x 53 cm diesel engine is guaranteed to deliver 522 KW at 300 rpm. The fuel rate is
0.26 kg/KW-hr. If the heating value of the fuel is 44,320 KJ/kg Calculate:
a) the brake thermal efficiency (32%)
b) the brake mean effective pressure (422 KPa)
c) suction displacement in m3/min-KW of shaft power (0.15)
d) heat supplied to cylinder per Liter of displacement
Given: 2-stroke, n’ = 4; D = 0.38 m; L = 0.53 m; BP = 522 KW; N = 300 RPM
mFB = 0.26 kg/KW-hr; HV = 44,320 KJ/kg
3600BP 3600BP
3600
a. eb =
=
=
= 31.2%
.
Qs
mF (HV) mFB (HV)
PmBπLD 2 Nn'
KW
4(60)
N = RPM (for 2 - stroke, single acting)
240(BP)
PmB =
= 434 KPa
πLD 2 Nn'
b. BP =
.
c. BP = PmB(VD )
VD 1(60)
m3
=
= 0.14
BP PmB min - KW
d.
2.
QS
m (HV) KJ mFB (BP)(HV)PmB mFB (HV)PmB
= F
=
=
KJ/L = 1.4 KJ/L
VD BP(3600) m3
BP(3600)
(3600)(1000)
PmB
Find the volume in Liters needed for a two weeks supply of 26API fuel oil to operate a 750 KW engine 70 % of the time
at full load, 10 % at 3/4 load and idle 20% of the time. Fuel rate is 0.25 kg/KW-hr at full load and 0.24 kg/KW-hr at 3/4
load. Temperature of oil is 21C.
T = 2weeks(7 days)(24 hours) = 336 hrs
m F = 336(0.70)0.25(750) + 0.10(336)(0.75)(750)(0.24) = 48,636 kg
141.5
= 0.898
131.5 + 26
S @ t = 0.898 − 0.0007(21 − 15.56) = 0.895
kg
kg
ρ = 895 3 = 0.895
L
m
48,636
VF =
= 54,342 Liters
0.895
For Fuel Tank Design
H = 1.25D
D = 0.75H
S=
Sample Problem (ICE & ENGINE HEAT BALANCE)
A single cylinder, single acting, 30.5 cm by 46 cm, 4-stroke cycle diesel engine for a 3 storey office building was tested for one
(1) hour and the following data were obtained:
• Fuel consumed
7.5 kg
• Total revolutions
12,120
• Jacket Water
674 kg
• Rise in temperature of
Cooling water
28C
• Area of Indicator Card
5.56 cm2
• Length of Indicator Card
7 cm
• Spring Scale
815 KPa/cm
• Torque
1,139 N-m
• Heating Value of fuel
44,200 KJ/kg
Find:
a. The IP, BP, and Mechanical efficiency
b. The indicated and brake specific fuel consumption
c. The %age of total heat supplied converted into brake power
d. The %age of total heat absorbed by jacket water
e. The %age of heat lost to friction, exhaust gases, radiation and etc.
Solution:
a.
2πTN 2π(1139)12,120
BP =
=
= 24 KW
60,000
60,000(60)
A' S 5.56(815)
Pmi =
=
= 647.34 KPa
L'
7
Pmi πLD 2 Nn'
4(60)
12120
N=
= 202 RPM
60
647.34π(0.46)(0.305) 2 (202)1
IP =
= 37 KW
4(60)2
BP
24
ηm =
x 100% =
= 65%
IP
37
IP =
b.
m F 7.5
kg
=
= 0.203
IP 37
KW - hr
m
7.5
kg
= F =
= 0.3125
BP 24
KW - hr
m Fi =
m FB
c.
d.
e.
Qs = mF(HV) = 7.5(44,200) = 331,500 KJ/hr
Q1 = %age of heat converted to work
24(3600)
Q1 =
x100% = 26 %
331,500
Qw = heat absorbed by cooling water
Qw = mw(CPW) (tWB – tWA)
Qw = 674(4.187)(28) = 79,017.064 KJ/hr
Q2 = %age of total heat absorb by cooling or jacket water
79,017.064
Q2 =
x 100% = 24 %
331,500
Q3 = %age of total heat lost to exhaust gas, friction, radiation and etc.
Q3 = 100 –(Q1 + Q2) = 100 – (26 + 24) = 50 %
Sample Problem (ICE & ENGINE HEAT BALANCE)
A diesel generator set with a generator capacity of 1500 KW has a combined specific fuel consumption of 0.28 kg/KW-hr. The
engine uses fuel oil at 28API with an air-fuel ratio of 15:1. Atmospheric air in the plant averages 35C and 1 atmosphere
pressure. Exhaust gas temperature is 450C and generator efficiency is 92%. Cooling water is available at 16C and leaves the
engine at 32C. If the heat loss due to incomplete combustion, friction , radiation and unaccounted for amounts to 9% of the heat
supplied by fuel, determine the four heat items in KJ/hr and the cooling water required by the engine in L/sec. (C pg = 1.026 KJ/kgC)
141.5
141.5
=
= 0.887
131.5 + ο API 131.5 + 28
H 2 = 0.26 − 0.15S = 0.26 − (0.15)(0.887) = 0.127 kgH 2 / kgf uel
S=
LHV = HHV − Q L
Q L = 2442.7(9H 2 ) = 2442.7(9)(0.127) = 2,790.42KJ / kg
HHV = 51,716 − 8793.8(S) 2
HHV = 51,716 − 8793.8(0.887) 2 = 44,797KJ / kg
LHV = HHV - Q L KJ/kg
LHV = 44797 - 2790.42 = 42,006.58KJ / kg
Q1 = 3600(BP) KJ/hr
Q2 = mwCpw(two - twi) KJ/hr
Q3 = Qa + Qb KJ/hr
Qa = mgCpg(tg - ta) KJ/hr
Qb = mf(9H2)(2442.7) KJ/hr
Q4 = QS - (Q1 + Q2 + Q3) KJ/hr
150
KJ
(3600) = 5,869,565.2
0.92
hr
Q3 = Qa + Qb
Q1 =
Q a = m g C pg ( t g − t a )
mg = ma + m F
ma
= 15
mF
kg
hr
kg
m a = 15(420) = 6300
hr
kg
m g = 6,720
hr
m F = 0.28(1500) = 420
KJ
hr
KJ
Q b = 420(9)(0.127)(2442.7) = 1,171,976.4
hr
KJ
Q3 = 4,0333,285.2
hr
KJ
Qs = 420(42,006.58) = 17,642,763.6
hr
KJ
Q 4 = 0.09(17,642,763.6) = 1,587,848.724
hr
Q a = 6720(1.026)(450 − 35) = 2,861,308.8
Qs = Q1 + Q 2 + Q3 + Q 4
Q 2 = Qs − (Q1 + Q3 + Q 4 ) = 6,152,064.476
KJ
hr
Q 2 = m w (4.187)(32 − 16)
m w = 91,833
kg
hr
at 16C : ρw = 999
kg
m3
91,833  hr  1000L 
L
mw =
of cooling water


 = 26
3
999  3600  m 
sec
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.
Fuel : C 7 H16
e : 20%
kg
hr
Combustion with 100% TA
C 7 H16 + aO 2 + a(3.76)N2 → bCO2 + cH 2 O + a(3.76)N2
m Fuel : 50
a = n + 0.25m = 11
b=n=7
c = 0.5m = 8
Combustion with EA (e = 0.20)
C 7 H16 + (1.20)aO 2 + (1.20)a(3.76)N2 → bCO2 + cH 2 O + dO 2 + (1.20)a(3.76)N2
d = e(n + 0.25m) = 2.2
Combustion Equation
C 7 H16 + 13O 2 + 49.632N 2 → 7CO2 + 8H 2 O + 2.2O 2 + 49.632N 2
Mass of Products (m products)
m products = 7(44) + 8(18) + 2.2(32) + 49.632(28) = 1,912.10 kg
Mass of Products per kg of fuel
m products 1,912.10 1,912.10
kg Products
=
=
= 19.12
m Fuel
12(7) + 16
100
kgFuel
M Pr oducts − mass flow rate of products
M Products = 19.12(50) = 956
kg
hr
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 16C and a pressure of 101.6 KPa.
Combustion with 100% TA (Basis :100 moles of fuel)
9C 2 H 6 + 90CH4 + 0.2CO2 + 0.8N 2 + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + dN 2
Carbon Balance
2(9) + 90 + 0.2 = b
b = 108.2
Hydrogen Balance
6(9) + 4(90) = 2c
c = 207
Oxygen Balance
0.2(2) + 2a) = 2b + c
a = 211.5
Nitrogen Balance
2(0.8) + a(3.76)2 = 2d
d = 796.04
9C 2 H 6 + 90CH4 + 0.2CO2 + 0.8N 2 + 211.5O 2 + 795.24 N 2 → 108.2CO2 + 207H 2 O + 796.04 N 2
n Air = 211.5+ = 795.24 = 1,006.74 Moles
n Gas = 100
PV = n RT
n RT
P
At the same temperaru te and pressure, the mole ratio is equal to volumetri c ratio
V=
m 3 of Air
1,006.74
=
= 10.06
3
100
m of fuel
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
Combustion with 100% TA (basis; 100 moles of fuel)
80C12H 26 + 20 C14H 30 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + a(3.76)N 2
80(12) + 20(14) = b
b = 1240
80(26) + 20(30) = 2c
c = 1340
2a = 2b + c
a = 1910
Combustion with EA (e = 0.30)
80C12H 26 + 20 C14H 30 + (1.30)aO 2 + (1.30)a(3.76)N2 → bCO2 + cH 2 O + dO 2 + (1.30)a(3.76)N2
1.30a (2) = b(2) + c + 2d
d = 573
Combustion Equation
80C12H 26 + 20 C14H 30 + 2483O 2 + 9336.08N 2 → 1240CO2 + 1340H 2 O + 573O 2 + 9336.08N 2
2483(32) + 9336.08(28)
A
= 19.41
 Actual =
80(170) + 20(198)
F
n Pr oducts = 1240 + 1340 + 573 + 9336.08 = 12,489.08
PH2O
1340
=
P
12,489.08
PH2O
1340
=
101 12,489.08
PH2O = 10.84 KPa
DPT = tsat at 10.84 KPa
DPT = 47.417C
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.
For Fuel A
A
  = 11.44(0.885) + 34.32(0.115) = 14.07
F
For Fuel B
A
  = (1.06)(14.07) = 14.915
F
C + H =1
C = (1 − H)
A
  = 11.44(1 − H) + 34.32H = 14.915
F
14.915 − 11.44 = (34.32 − 11.44)H
H = 0.1519 = 15.19%
C = (1 − .1519) = 0.8481 = 84.81%
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 25C. Determine
f.
the actual air-fuel Ratio
g. the m3 of CO2 formed per kg of fuel if the product temp. is 400C and a pressure of 1 atm.
h. The M and R of the Products
i.
The M and R of the dry flue gas
j.
If 1.5 moles of carbon oxidizes to CO, determine the molecular weight of the resulting products
Combustion with 130%TA
C12H 26 + (1.30)aO 2 + (1.30)a(3.76)N 2 → bCO2 + cH 2 O + dO 2 + (1.30)a(3.76)N 2
b=n
b = 12
a = n + 0.25m = 18.5
c = 0.5m = 13
d = e(n + 0.25m) = 5.55
C12H 26 + 24.05O 2 + 90.428N 2 → 12CO2 + 13H 2 O + 5.55O 2 + 90.428N 2
(1.30)137.28(n + 0.25m)
A
=
= 19.42
 
12m + m
 F  Actual
12(8.3143)(400 + 273)
VCO 2 =
= 662.7 m 3
101.325
VCO 2
662.7
=
= 3.9
m Fuel 12(12) + 26
n Pr oducts = 12 + 13 + 5.55 + 90.428 = 120.98
M=
m Pr oducts 12(44) + 13(18) + 5.55(32) + 90.428(28)
kg
=
= 28.7
n Pr oducts
120.98
kg mol
R=
8.3143
KJ
= 0.29
28.7
kg - K
m Dry flue gas = 12(44) + 5.55(32) + 90.428(28) = 104.09
M Dry Flue Gas =
12 + 5.55 + 90.428
kg
= 29.82
104.09
kg mol
Combustion with CO and with 100%TA
C12 H 26 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + 1.5CO + a(3.76)N 2
12 = b + 1.5
b = 12 - 1.5 = 10.5
26 = 2c
c = 13
2a = 2b + c + 1.5
c 1.5
a = b+ +
= 17.75
2 2
Combustion with CO and with EA (e = 0.30)
C12 H 26 + (1.30)aO 2 + (1.30)a(3.76)N 2 → bCO2 + cH 2 O + 1.5CO + dO 2 + (1.30)a(3.76)N 2
By oxygen balance
d = 5.325
C12 H 26 + 23.075O 2 + 86.762N 2 → 10.5CO 2 + 13H 2 O + 1.5CO + 5.325O 2 + 86.762N 2
n Pr oducts = 10.5 + 13 + 1.5 + 5.325 + 86.762 = 117.09
M=
10.5(44) + 13(18) + 1.5(28) + 5.325(32) + 86.762(28)
kg
= 28.51
117.09
kg mol
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.
H 2 + 1 O 2 → H 2O
2
C + O 2 → CO 2
2
+
16
→
18
12 + 32 → 44
3 + 8 → 11
kgO 2 8
= = 2.67
kgC
3
kgAir
= 11.44
kgC
1+ 8 → 9
kgO 2 8
= =8
KgH 1
kgAir
= 34.32
KgH
Problem 2: Calculate the theoretical Oxygen--fuel ratio and Air--fuel ratio on a mass basis for the combustion of ethanol,
C2H5OH.
Combustion with 100% TA
C 2 H 5OH + aO 2 + a (3.76) N 2 → bCO2 + cH 2 O + a (3.76) N 2
2=b
5 + 1 = 2c
c=3
1 + 2a = 2b + c
2(2) + 3 − 1
a=
=3
2
C 2 H 5OH + 3O 2 + 11.28N 2 → 2CO2 + 3H 2 O + 11.28N 2
kgOxygen
3(32)
96
=
=
= 2.09
kgFuel
124 + 5 + 16 + 1 46
kgAir
3(32) + 11.28(28)
=
= 8.95
kgFuel 124 + 5 + 16 + 1
Problem 3: Determine the molal analysis of the products of combustion when octane C 8H18 is burned with 100% excess air.
Combustion with EA (e = 100%)
C8 H18 + (1 + 1)aO 2 + (1 + 1)a (3.76) N 2 → bCO2 + cH 2 O + dO 2 + (1 + 1)a (3.76) N 2
a = n + 0.25m
b=n
c = 0,5m
d = e(n + 0.25m)
C8 H18 + 24.5O 2 + 92.12 N 2 → bCO2 + cH 2 O + dO 2 + (1 + 1)a (3.76) N 2
C8 H18 + 25O 2 + 94 N 2 → 8CO2 + 9H 2 O + 12.5O 2 + 94 N 2
n Pr oducts = 8 + 9 + 12.5 + 94 = 123.5
8
y CO 2 =
x100% = 6.48%
123.5
9
y H 2O =
x100% = 7.29%
123.5
12.5
y O2 =
x100% = 10.12%
123.5
94
y N2 =
x100% = 76.11%
123.5
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?
Combustion with EA (e = 50%)
C10H 22 + 23.25O 2 + 87.42 N 2 → 10CO2 + 11H 2 O + 7.75O 2 + 87.42N 2
a = n + 0.25m
b=n
c = 0,5m
d = e(n + 0.25m)
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 airfuel
ratio.
Fuel
C
H2
O2
N2
S
H20
Ash
Total
UA
53.4
5.6
37.9
0.1
0.1
0
2.9
100
Ashless
55
5.8
39
0.103
0.103
0
0
100
O 

A
= 11.44C + 34.32 H 2 − 2  + 4.29S = 6.6
 
8 
 F  T heoretical

A
A
= (1 + e) 
= 8.58
 
 F  Actual
 F  T heoretical
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% C 2H6 and 85% CH4 that
requires 25% excess air. Air and fuel enters the combustor at 25C (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
f. The actual combustion equation
g. The actual air – fuel ratio
h. The heating value of the fuel (HP – HR) KJ/kg
i. The mass flow rate of fuel in kg/min
j. 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
Combustion with 100% TA
Basis :100 moles of fuel
15C 2 H 6 + 85CH4 + aO 2 + a(3.76)N 2 → bCO2 + cH 2 O + 2CO + 1NO + dN 2
15C 2 H 6 + 85CH4 + 222O 2 + 834.72N 2 → 113CO2 + 215H 2 O + 2CO + 1NO + 834.22 N 2
Combustion with EXCESS AIR(e = 0.25)
Basis :100 moles of fuel
15C 2 H 6 + 85CH4 + (1.25)222O 2 + (1.25)834.72N 2 → 113CO2 + 215H 2 O + 2CO + 1NO + fO 2 + gN 2
15C 2 H 6 + 85CH4 + 227.5O 2 + 1,043.4 N 2 → 113CO2 + 215H 2 O + 2CO + 1NO + 55.5O 2 + 1,042.9 N 2
227.5(32) + 1,043.4(28)
A
=
= 21.05
 
15(30) + 85(16)
 F  Actual
Problem No. 2
Calculate the theoretical air – fuel ratio of the Hydrocarbon Fuel in the table below.
Fuel
C8H18
C7H16
C4H10
C9H20
C11H24
Theoretical
(By Mass)
15.05
15.10
15.38
15.02
14.96
A/F
Ratio
COMBUSTION ENGINEERING
Activity No. 4 (APRIL 30, 2019)
Special Quiz
Name ___________________________________
A European made diesel engine uses liquid Octadecane (C18H38) fuel for power generation. This fuel is burned with 30% excess
air requirement. Air and fuel enters the engine at 25C (298 K) and 101 KPa and products of combustion leaves at 1200 K. It is
found that the products contains 2 moles of CO and 2 moles of NO during the test. Determine
a. The actual combustion equation
b. The actual air – fuel ratio
c. The heating value of the fuel (HP – HR) KJ/kg if (hf)C18H38 = -505,400 KJ/kgmol
d. The volume flow rate of emission gases (CO2, CO and NO) in m3/hr if the fuel consumption is 250 kg/hr
Combustion with 100% TA
Basis :1 mole of fuel
C18 H 38 + aO 2 + a(3.76)N 2 → bCO 2 + cH 2 O + 2CO + 2NO + dN 2
By Balancing equation
C18 H 38 + 27.5O 2 + 103.4 N 2 → 16CO 2 + 19H 2 O + 2CO + 2NO + 102.4 N 2
Combustion with 30% EA
C18 H 38 + 37.75O 2 + 134.42 N 2 → 16CO 2 + 19H 2 O + 2CO + 2NO + 8.25O 2 + 133.42 N 2
37.75(32) + 134.42(28)
A
=
= 19.32
 
12(18) + 38
 F  Actual
n EG = 16 + 2 + 2 = 20
20(8.3143)(1200)
= 1,975.68 m 3 → Volume of emission gases
101
V EG
1,975.68
=
= 7.78
kg of Fuel
254
VEG =
VFEG = 7.78(250) = 0.031
m3
of Emission Gases
hr
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 25C 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
hf of C7H16 = -187,900 KJ/kgm
For HHV
For LHV
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; CO 2 = 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 C 8H18. 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 C 12H26. 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 C 16H32.
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
2 b
3 a
4 d
5 a
6
7
8
9
10
a
b
c
d
d
11
12
13
14
15
a
b
c
d
a
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 H 2O and assume 10% losses and average flue gas
temperature of 350C .
Diameter
Generator Output = 1,500 KW
Height
Fuel and air
Diesel Engine
At 800 m above sea level
P = 92.21 KPa
T = 287.8 K
P
kg
ρ air =
= 1.12 3
RT
m
Combustion of C16H 34 with 50% excess air
C16H 34 + 36.75O 2 + 138.18N 2 → 16CO2 + 17H 2 O + 12.25O 2 + 138.18N 2
16(44) + 17(18) + 12.25(32) + 138.18(28)
kg
= 28.74
16 + 17 + 12.25 + 138.18
kgm
8.3143
KJ
R Pr oducts =
= 0.289
M
kg - K
183.43(8.3143)(350 + 273)
VPr oducts =
= 10,304.005 m 3
92.21
3
10,304.005
(0.30)(1500) hr = 5.7 m
Q Pr oducts =
12(16) + 34
3600 sec
sec
M Pr oducts =
92.21
kg
= 0.51 3
(0.289)(350 + 273)
m
D a = D t − 0.10D t = 25 mm water
ρ Pr oducts =
D t = 28.09 mm
28.09(1000)
= 54.85 m of flue gas
1000(0.51)
m
v = 2gh g = 32.80
sec
1000H ρ air - ρ gas
Dt =
mm water
ρ water
hg =
(
H = 46.05 m
)
Qg =
π 2
D (kv)
4
D=
4(5.7)
= 0.7 m
π(0.40)(32.80)
SPECIAL EXAM
MIDTERM
Name _______________________________
A coal fired steam power plant consumes Ten Metric Tons of coal per day. The coal has an as – received Ultimate Analysis of
C = 75% ; H2 = 5 % ; O2 = 7.0 %; N2 = 1.5 %; S = 2.0 %; M = 2.5 % and Ash = 7 %. This coal is burned with 50% excess air in
CFB boiler. Determine
a. The Ultimate Analysis on an ash-less basis
b. The molal analysis on an ash-less basis
c. The combustion equation
d. The actual air-fuel ratio
e. The required height and diameter of the smoke stacks if, Da = 30 mm H2O; Dlosses = 12%; k = 0.40
Air and fuel temp. = 298 K; Flue gas temp. = 400C; P = 101 KPa
Fuel
U.A.
C
H2
O2
N2
S
H2O
Ash
75.00
5.00
7.00
1.50
2.00
2.50
7.00
100
U.A.
(Ashless)
80.65
5.38
7.53
1.61
2.15
2.69
100
Mi
Xi/Mi
12
2
32
28
32
18
-
0.07
0.03
0.002
0.0006
0.0007
0.0015
-
Molal
Analysis
67.76
27.10
2.37
0.58
0.68
1.51
100
Combust ion equation w ith 50% EA
67.76C + 27.10H2 + 2.37O2 + 0.58N2 + 0.68S + 1.51H2O → Fuel
119.43O2 + 449.05N 2 → Air
67.76CO2 + 28.61H 2O + 0.68SO2 + 39.81O 2 + 449.63N 2 → Pr oducts


0.0753 
A

= (1.50) 11.44(0.8065) + 34.32 0.0538 −
 
 + 4.29(0.0215)
8 
 F  Actual



A
= 16.26
 
 F  Actual
101
kg
ρ air =
= 1.18 3
0.287(298)
m
kg
ρ water = 1000 3
m
67.76(44) + 28.61(18) + 0.68(64) + 39.81(32) + 449.63(28)
M Pr oducts =
= 29.67
67.76 + 28.61 + 0.68 + 39.81 + 449.63
KJ
R Pr oducts = 0.280
kg - K
101
kg
ρ Gas =
= 0.54 3
0.280(400 + 273)
m
D a = D t − 0.12D t
Da
30
=
= 34.09 mm
1 − Losses 1 − 0.12
34.09(1000)
h gas =
= 63.65 m of gas
1000(0.54)
m
v = 2g(hg) = 35.34
sec
m
v' = kv = 0.40(35.34) = 14.14
sec
10Mton  1000kg  day  hr 
kg
m Fuel =



 = 0.116
day  Mton  24hrs  3600 sec 
sec
kg
m Air = 16.26(0.116) = 1.189
sec
kg
m FlueGas = m Air + m Fuel = 2.002
sec
2.002
m3
Q FlueGas =
= 3.74
0.54
sec
1000H(ρ Air − ρ Gas )
Dt =
mm WG
ρ Water
Dt =
H = 52.83 m
π
Q FlueGas = D 2 ( v' )
4
D = 0.58 Meters
COMBUSTION ENGINEERING
FINAL EXAM SERIES S1
Name ___________________________________
Problem No. 1
In an air standard Otto cycle, the compression ratio is 7 and the compression begins at 35C and 100 KPa. The maximum
temperature of the cycle is 1100C. Find (a) the temperature and the pressure at various points in the cycle, (b) the heat supplied
per kg of air, (c) work done per kg of air, (d) the cycle efficiency and (e) the MEP of the cycle.
Q A = mC v (T3 − T2 )
Cv =
R
KJ
= 0.7175
k −1
kg − K
KJ
kg
KJ
Q R = mC v (T4 − T1 ) = 231.32
kg
KJ
W = Q A − Q R = 271.78
kg
W
e=
x100% = 50.08%
QA
Q A = mC v (T3 − T2 ) = 503.82
r=7
P1 = 100KPa
T1 = 35 + 273 = 308K
T3 = 1100 + 273 = 1,373K
T2 = T1 (r ) k −1 = 670.8K
P2 = P1 (r ) k = 1,524.5KPa
P2 P3
=
T2 T3

1 
e = 1 − k −1  x100% = 50.08%
 (r ) 
P3 = 3,120.4KPa
Pm =
 1 
T4 = T3  k −1  = 630.42K
 (r ) 
k
1
P4 = P3   = 204.67KPa
r
W
VD
RT1  1 
m3
VD =
1 −  = 0.758
P1  r 
kg
Pm = 358.6 KPa
Problem No. 2
In a Diesel cycle, the compression ratio is 15. Compression begins at 100 KPa, 40C. The heat added is 1,675 KJ/kg. Find
(a) the maximum temperature in the cycle, (b) work done per kg of air (c) the cycle efficiency (d) the temperature at the end of the
isentropic expansion (e) the cut-off ratio and (f) the MEP of the cycle.
r = 15
P1 = 100 KPa
T1 = 40 + 273 = 313 K
KJ
QA = 1,675
kg
T 2 = T1(r )k −1 =
COMBUSTION ENGINEERING
FINAL EXAM SERIES S2
Name ___________________________________
Problem No. 1
A 40 KW blast furnace engine shows by test a gas consumption of 4.2 m 3/KW-hr. Heating value of gas is 3,350 KJ/m 3.
Mechanical efficiency of the gas engine is 86%, Calculate:
a) The brake thermal efficiency
b) Indicated thermal efficiency
c) Heat rate in KJ/KW-hr
d) If the heat rejected to cooling water is 28% of the heat generated in engine cylinder, determine the quantity of cooling
water in L/hr to be circulated if the allowable rise in temperature of the cooling water is 10C
Problem No. 2
A mechanical draft cooling tower of a Diesel Power Plant is required to cool 15 kg/sec of water at 65C entering temperature to a
temperature of 35C. Atmospheric air at P = 97 KPa; tdA =28C and twA = 22C enters the cooling tower and leaves at tdB = 60C
(Psat = 19.925 KPa) and RH = 90%. If make-up water is supplied at 35C (hf = 146.51 KJ/kg) , Calculate
a) The kg/sec of dry air in the tower
b)
c)
d)
e)
f)
h)
The make-up water flow rate in kg/hr
The m3/sec capacity of Force draft fan at the bottom of the tower
The actual cooling Range (ACR)
the cooling tower approach (CTA)
the theoretical cooling range (CTR)
The cooling tower Efficiency