Fuels and Combustion
Fuels are substance primarily composed of hydrocarbons which upon combustion release
heat energy.
Combustion is the rapid chemical union with oxygen of an element whose exothermic heat of
reaction is sufficiently great and whose rate of reaction is fast enough that useful quantities of
heat are liberated at elevated temperature.
Classification of fuels:
According to nature of source:
1. Natural or Naturally occurring fuels.
2. Synthetic or manufactured and industrial by products.
According to Physical Structured:
1. Solids fuels
Coal, coke, peat, briquettes, wood, charcoal, bagasse, and a variety of waste
products including coconut shells and rice husks.
2. Liquids Fuels
Petroleum and its derivatives, synthetic liquid fuels manufactured from natural gas
and coal, shale oil, coal-by- products and alcohol.
3.) Gaseous Fuels
Natural gas or methane, liquefied petroleum gas (LPG) primarily composed of
butane and propane which are stored and delivered as liquids under pressure but are
consumed in the gaseous state.
Fuel and Lubricant Properties:
Specific Gravity SG
A dimensionless parameter, it is the ratio of the mass of a unit volume of fuel
to the mass of the same volume of a standard substance at a specified
temperature.
In Mathematical expression,
SG =
density of liquid fuel
density of water
or
SG =
density of gaseous fuel
density of air
Specific gravity of a fuel normally measured at 15.60C affects the spray penetration as
the fuel is injected into the cylinders aside from being a measure of fuel heat content.
Correction factor as applied to specific gravity at a certain temperature expressed in
degrees Celsius:
SGt = SG15.60C - 0.0007(t - 15.6), SI / metric
SGt= SG9600F - 0.0004(t - 60), English
American Petroleum Institute Gravity Unit, 0API
The accepted standard by the petroleum and oil industry, it was drawn up to correct
the values measured by incorrectly calibrated hydrometers.
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In mathematical expression:
141.5
API = SG 0 – 131.5
15.6 C
o
Baume Gravity Unit, 0Baume
Another standard commonly associated with brine, expressed mathematically as:
140
o
Baume = SG 0 - 130
15.6 C
Viscosity
Property of a fluid by virtue of which it offers resistance to shear and angular deformation
Considerably varies temperature, i.e. as temperature increase, viscosity for liquids
decrease and for gases increase.
Viscosity indicates how readily the oil atomize and it will affect the injection pump, it affects
the fuel spray governing the atomization and penetrating qualities of the spray.
 Viscosimetry
Laboratory instruments used to measure viscosity are of the efflux type which consists
of a tank fitted with a small orifice; most common instrument used is the Saybolt
viscosimeter, consisting of a temperature bath surrounding a standard oil tube:
The time in seconds for 60 cc of oil at 37.80C to flow through the tube is know as the
Universal Viscosity (abbreviated SUV) of the fluid, or seconds Saybolt Universal
(Abbreviated SSU).
Generally, there are two types of Saybolt viscosimeters:
Saybolt Universal Viscosimeter
Used for light lubricating oils having an efflux time greater than 32 seconds,
V centistokes = 0.308 (SSU – 26)
Viscosity for most diesel fuels varies from 30 to 45 SSU at 37.80C.
Furol Viscosimeter
Used for very viscous fluids, as heavy oils above 250 SSU; similar to the Saybolt
Universal, except that it has a larger diameter tube resulting in an efflux time
approximately one- tenth
that of the Saybolt Universal Viscosimeter.
Viscosity index is in terms of Saybolt Furol Viscosity
(Abbreviated SFV), or seconds Saybolt Furol ( Abbreviated SSF ).
62 SSF = 600 SSU
Units of Measurement
Absolute Viscosity, 
Expressed as the ratio of shear intensity to velocity gradient, in metric system,
1 gram mass
1 poise = 1 cm x 1 sec
2
Numerically equal to
1dyne x 1 sec
, or 0.1 Pa-sec
1cm2
in English system,
1 reyn =
1lb force x 1sec
1 in2
the poise being a large unit for ordinary use, centipoises (0.01 poise) is generally used;
centipose is almost equal to the viscosity of water at 200C which is 1.005 centipoises.
Kinematic Viscosity, v
Expressed as the ratio of absolute Viscosity to the fluid density as measured indirectly
by the viscosimeter;
In mathematical expression:
v=
absolute Viscosity μ
fluid density ρ
In metric system,
1 stoke = 1 cm2 / sec
measurement is limited to a Newtonian fluid, a fluid in which the velocity gradient is directly
proportional to the shear stress:
the stroke is also a large unit for ordinary use, centistokes (0.01 stoke) is generally preferred.
Volatility
It is the ability of a liquid fuel to change into vapor which is manifested in the
temperature range at which various portions of the fuel are vaporized.
High- speed engines require larger fractions of low boiling point compounds as
compared to low – speed engines since the latter needs more time for combustion.
American Society for Testing and Materials (ASTM) Distillation Test:
A measured sample of oil is slowly heated; as the various boiling points of its
compounds are reached, they then distilled, allowed to condense and measured.
After complete distillation, data obtained are used to construct a distillation curve
should be smooth for good balance of compounds. Average range between initial
boiling point
and endpoint is 1630C to 37100C.
Carbon Residue
Also referred to as conradson carbon, it is the carbonaceous residue which remains
after
destructive distillation, expressed in percentage by weight of the original sample.
Flash and Fire Point
Flash point is the temperature at which a substance will give off vapor that will burn
momentarily.
Fire point is the temperature which the vapor will continue to burn after being ignited.
Flash and fire points- serve as safely precaution for handling and storing fuels, the
lower the flash point, the greater the hazard.
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measuring instrument used are the Cleveland open tester (open – type) and the
pensky- martens closed tester ( closed type)
Cloud and Pour Point
Cloud point- is the temperature at which the hydrocarbon components if a liquid fuel
become insoluble; this means that the paraffin wax and other solid substance in the
lubricating oil begin to crystallize and separate from the lubricating oil.
Pour point – is the minimum temperature at which a certain amount will become
insoluble to prevent the flow under specified conditions.
Dropping Point
It is the temperature at which grease will pass from a semi- solid to a fluid state;
ASTM method D- 566 is the procedure used in determining dropping point.
Grease, semi- fluid oil, primarily consists of soap materials and mineral oil with an
alkali; common types of soap used are sodium, lime, mixed base (soda lime) lithium,
lead, aluminum, barium or strontium.
Aniline Point
This is an indication of the paraffinity of a compression – ignition engine fuel;
technically, it is the temperature at which equal volumes of the fuel and Aniline are
just miscible.
Composition of Fuels:
Liquid and gaseous fuels are usually mixtures of hydrocarbons which may be
classified as follows:
Paraffins, CnH2n+2
Saturated hydrocarbons, very stable in character
Olefins, CnH2n
unsaturated hydrocarbons, characterized by the presence of a double bond
between carbon atoms.
Diolefins, CnH2n-2
less unsaturated than olefins, characterized by the presence of two double
bonds.
Analysis of Fuel Composition
Ultimate Analysis
The straightforward chemical analysis of the material in terms of the
component elements; common elements specified are carbon, hydrogen, oxygen,
nitrogen, and
sulfur.
Proximate Analysis
The determination of the moisture, ash, volatile matter (CO, H2, CH4, C2H6
and
H2S), coke and fixed carbon.
Volumetric Analysis
For gaseous mixture of fuel
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Heating Value of fuels, HV
or calorific value, it is the amount of heat evolved when a unit amount of the fuel is
burned completely and the combustion products cooled to a base or reference temperature.
Classifications
1. Higher Heating Value (ASME), HHV
or gross calorific Value (ASTM), it is the amount of heat evolved during a fuel’s
complete combustion where the water leaves in the liquid state.
2. Lower Heating Value (ASME), LHV
or Net Calorific Value (ASTM) it is the amount of heat evolved during a fuel’s
complete combustion where the water leaves in the gaseous or vapor state.
LHV = HHV - 9H2(2442)
where: 9H2 = lb. of water formed per lb. or cu. ft. of fuel burned
2442 = latent heat of vaporization of 1kg.water
also,
h2 =26 -15 (SG), percent by weight
Methods of determining heating value:
A fuel’s content can either be determined experimentally or calculated based on its
composition.
1.) Laboratory experimentation
a.) bomb calorimeter for solid and liquids fuels
b.) gas calorimeter for gaseous fuels
2.) Calculation using empirical or derived formulas
note: units expressed in kJ/ kg
A Dulong’s formula for solid fuels of known ultimate analysis
HHV = 33,820C + 144,212 (H – 0/8) + 9304S
A ASME formula for petroleum Products
HHV = 41,130 + 139.6 (0API)
A Bureau of Standard formula
HHV= 51,716 - 8,793.8(SG)2
Fuel Foundation Processes:
1. fractional Distillation the primary method of crude oil refining
2. thermal cracking changing heavy oil into gasoline by means of high pressure, high
temperature and longer exposure time
3. Catalytic Cracking Subjects oil to high pressure and high temperature in the presence
of a catalyst; permits an accurate control of the compounds formed and produces a
gasoline of higher octane number that he one produced in thermal cracking
4. Hydrogenation Process of catalytic cracking in a hydrogen atmosphere; obtained are
more saturated products than those from cracking process alone
5. Isomerization Process by which the atoms of carbon and hydrogen in normal
hydrocarbons are re- arranged to produce a more complex structure of higher antiknock value
6. Polymerization Makes use of high pressure,high temperature and a catalyst to combine
light and volatile gases into gasoline
7. alkylation process of combining an isoparaffin usually iso- butane, with an olefin
usually butane or propane, to form a larger isoparaffin molecule, usually iso- octane or
iso- heptane, having a very high octane number
5
8. reforming used to obtain fuels with substantially higher than 100 octane number
currently used to process about forty percent of motor gasoline
9. Hydrodesulfurization Process of adding hydrogen to unsaturated hydrocarbons and
reducing the sulfur content of the resulting fuel oil
Stoichiometry and Energetic of combustion
The following may be use in analyzing combustion processes:
1. Composition by weight
76.8 percent nitrogen, 23.2 percent oxygen
thus, for 100 lb of air
76.8/ 23.2= 3.3 lb of nitrogen / lb of oxygen
2. Composition by volume
79.0 percent nitrogen, 21.0 percent oxygen
thus, for 100 moles of air
79.0 /21.0 = 3.76 moles of nitrogen/ mole of oxygen
3. Molecular weight, MW
note: Units expressed in kg/kmol
air = 28.97
H2 = 2
O2 = 32
N2 = 28
C = 12
S =32
Typical combustion Reaction
Fuel + air
→
products of combustion or flue gases
1.) Theoretical air – fuel ratio, (A/F)t
exact amount of air needed to furnish oxygen for complete theoretical
combustion (stoichiometric condition)
Combustion of solid fuel with known ultimate analysis, kg air / kg fuel
(A/F)t = 11.5C + 34.5 (H - 0/8) + 4.3S
2.) Actual air – fuel ratio (A/F)a
determined by the presence of excess air which is defined as the amount of air
supplied over and above the theoretical air
in mathematical expression
Percent excess air =
(AF/)a – (AF)t
(AF)t
Classification of combustion
1. combustion reaction with chemically correct or stoichiometric condition
generally chemical formula of the fuel is CnHm
CnHm+ (n+0.25m)O2 + 3.76(n + 0.25m)N2 → nCO2 + 0.5mH2O + 3.76(n+0.25m)N2
2. combustion reaction with greater, amount of theoretical air, having fuel –
lean mixture
3. combustion reaction with lesser amount of theoretical air, or having a fuel – rich
mixture
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Equivalence ratio for a given mass of air
actual mass of fuel in combustion
(A/F)t
Φ = stoichiometric mass of fuel used = (A/F)
a
Note:
Φ = 1, for stoichiometric mixture
Φ < 1, for fuel lean, mixture;
Φ > 1, for fuel – rich mixture
Sample Problems
1. A cylindrical oil tank 2.4m diameter x 6m long is filled to the neck with fuel oil which is
checked at 21° Bé at 31 °C. Estimate the kJ heating value stored in this tank.
SOLUTION:
2.4 m
6 m
Baume =
140
− 130
SG15.6°
;
21 =
140
− 130
SG15.6°
SG15.6°C = 0.927
SGt= SG15.6°C – 0.0007 ( t – 15.6°C )
@t = 31°C
SG31° = 0.927 – 0.0007 (31-15.6)
SG31°C = 0.916
HHV = 51716 – 8793.8(SG)2
HHV = 51716– 8793.8(0.916)2
HHV = 44333.96 kJ/kg
Mass of oil
mf = vf ρf
0.916 = ρf / 1000
ρf = 916 kg/m3
vf = volume of fuel = d2h = (2.4)2(6) = 27.14 m3
So, mf = (27.14)(916) = 24, 860.24 kg
Finally,
Qh=mfHHV = (24, 860.24)(44333.96)
Qh = 1.102 x 109 kJ
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2. A diesel engine consumes 1 bbl of industrial fuel of 25 °API of 80.6 °F in a single-day
operation. Determine the amount of heat liberated by the fuel as a result of combustion.
SOLUTION:
Bbl =159 liter = 0.159 m3
t = 80.6 °F = 27 °C
141.5
141.5
°API =
− 131.5 ; 25 =
− 131.5
SG15.6°
SG15.6°
SG15.6°C= 0.904
@t = 27°C
SG27°C = SG15.6 - (0.0007)(T-15.6)
SG27°C =0.904 - (0.0007)(27-15.6)
SG27°C = 0.89602
HHV = 41130 + 139.6 °API = 44620 kJ/kg
ρf = (1kg/li) (0.89602) = 0.89602 kg/li
mf = Vf ρf
mf = (159 liter) (0.89602 kg/li)
mf =142.47 kg
Qh= mf x HHV
Qh=(142.47 kg/24 hr) x (44620 kJ/kg)
Qh=264,875.475 kJ/hr
3. A horizontal cylindrical fuel oil storage tank 2.4 m dia. x 5 m long is gauged at 825.5 mm oil
depth. A delivery of oil is made which raises the level to 2165 mm, measured after the
temperature of the contents had attained ground temperature, i.e., 7.2oC. This oil is purchased
for $ 3.31 per bbl at 15.6oC/15.6oC. The delivery is billed at $ 380.50. Is this correct to the
nearest dollar?
SOLUTION:
Before
2400mm
825.5mm
5000mm
cos
θ 0.3745
=
2
1.2
θ = 143 .63 o
A
sec tor
=
1 2
θ
2r
1
π
A sec tor = (1.2) 2 (143.630 )(
)
2
1800
A sec tor = 1.805 m2
8
1
1
θ
θ
bh = (r sin )(2)(r cos )
2
2
2
2
1 2
= r sin θ
2
1
= (1.2) 2 (sin 143.630 )
2
A t = 0.427m2
At =
A =A
1
sector
- At
= 1.805 - 0.427
A
1
= 1.378 m2
After: oil raise to 2165mm,
2165 mm
A
w
2400mm
= A sec tor - A t
θ
0.965
=
2
1.2
θ = 72.94 0
1
1
Aw = 2 r 2 θ - 2 r 2 sin θ
1
π
= (1.2) 2 [ (72.94 0 x
) - sin 72.940 ]
2
1800
sin ce : cos
A
w
A
2
= 0.2283 m2
= Acircle - Aw
= π(1.2) 2 - 0.2283 m2
= 4.3 m2
V= volume of oil deliver
= (A2-A1) L
= (4.3-1.378) (6)
= 17.532 m3= 17,532 Li
= 110.26 bbL
$380.50
110.26 bbl
Pr ice = $3.45 bbl
No, because it exceed $3.31 per barrel
Pr ice =
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4. A coal sample has an ultimate analysis 0f 80.0% C, 4.0% O2, 4.5 % H2, 1.7 % N , 1.5 % S, and
8.3% ash. Determine the heating value and mass air/fuel ratio for complete combustion with
100% theoretical air.
SOLUTION:
HHV = 33820C + 144212(
9304(0.015)
HHV = 32964.04 kJ/kg
) + 9304S = 33820(0.80) + 144212(
) +
Theoretical Air/Fuel ratio
A
O
0.04
= 11.5C + 34.5( H − ) + 4.3S = 11.5(0.80) + 34.5(0.045 −
) + 4.3(0.015)
F
8
8
A
kg − air
= 10.64
F
kg − fuel
5. A fuel with a mass analysis of 83% C, 12% H2 , and 5% O2 is burned with 100% theoretical air.
Determine:
The complete combustion equation.
The molal analysis of the products with and without water vapour considered.
SOLUTION:
Xi
0.83
0.12
0.05
C
H2
O2
Yi =
Mi
12
2
32
Xi/Mi
0.0691
0.06
0.00156
0.13066
Yi
0.53
0.46
0.01
Xi
Mi
Xi

Mi
0.53C + 0.46H2 + 0.01O2 + aO2 + 3.76aN2 = bCO2 + cH2O + 3.76aN2
Carbon balance
0.53 = b
Hydrogen Balance
0.46 = c
Oxygen balance
0.01(2) + 2a = 2b + c
0.01(2) + 2a = 2(0.53) + 0.46
a = 0.75
Complete combustion equation
0.53C + 0.46H2 +0.001O2 + 0.75O2 + 3.76(0.75)N2 = 0.53CO2 + 0.46H2O + 0.75(3.76)N2
b.) Molal analysis
with vapor
np = 0.53 + 0.46 + 0.75(3.76) = 3.81
yCO2 = 0.53 / 3.81 = 0.14
yH2O = 0.46 /3.81 = 0.12
yN2 = (0.75)(3.76) / 3.81 = 0.74
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Without vapour
np = 0.53 + (0.75)(3.76) = 3.35
yCO2 = 0.53 / 3.35 = 0.16
yN2 = (0.75)(3.76)/3.35 = 0.84
6. A fuel oil is burned with 50% excess air, and the combustion characteristics of the fuel oil are
similar to C12 H26. Determine the complete combustion equation, volumetric analysis of the
products of combustion, and determine the dew point for the products if the pressure is 101
kPa.
SOLUTION:
CnHm + (1+e)(n + 0.25m)O2 + (1 + e)(n + 0.25m)(3.76)N2 = nCO2 + 0.5mH2O + (1 +
e)(3.76)(n + 0.25m)N2 + e (n + 0.25m)O2
Complete combustion equation
C12H26 + (1.5)(12 + 6.5)O2 + (1.5)(12 + 6.5)(3.76)N2 = 12CO2 +13H2O + 1.5(3.76)(12 +
6.5)N2 + 0.5(12 + 6.5)O2
C12H26 + 27.75O2 + 104.34N2 = 12CO2 + 13H2O + 104.34N2 + 9.25O2
The total mole of the product
np = 12 + 13 + 104.34 + 9.25 = 138.59 mole
YCO2 = 12/138.59 = 0.087
YH2O = 13/138.59 = 0.094
YN2 = 104.34/138.59 = 0.753
YO2 = 9.25 / 138.59 = 0.067
Partial water vapor pressure
PH2O/P = YH2O
; PH2O/101 = 0.094
PH2O = 9.49 kPa
Tdp = dew point = tsat @ 9.49kPa = 45°C
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