thermodynamics

1. ‫الرحيم‬ ‫الرحمن‬ ‫هللا‬ ‫بسم‬
”
‫قليال‬ ‫اال‬ ‫العلم‬ ‫من‬ ‫أوتيتم‬ ‫وما‬
“
‫العظيم‬ ‫اللة‬ ‫صدق‬
Course contents:
1. POWER AND REFRIGERATION
CYCLES
2. AIR VAPOR MIXTURE AND
PSYCHROMETRY
3. FUEL AND COMBUSTION

2. Vapor power cycles
• A gas power cycle considers air as the working fluid
→ single phase
• A vapor power cycle considers steam as the working
fluid → might be two phases
• The use of steam as working fluid
- Low cost
- Availability
- High enthalpy of vaporization
• Carnot cycle
- Give maximum thermal efficiency
- But not be suitable to use for analyzing the efficiency of
vapor power cycle

3. Objectives
• Analyze vapor power cycles in which the
working fluid is alternately vaporized and
condensed.
• Analyze power generation coupled with process
heating, called cogeneration.
• Investigate methods to modify the basic
Rankine vapor power cycle to increase the cycle
thermal efficiency.
• Analyze the reheat and regenerative vapor
power cycles.

4. Carnot vapor cycle

5. The working fluid, steam (water), undergoes a
thermodynamic cycle from 1-2-3-4-1.

6. Carnot Cycle Analysis:
• The larger the TH the larger the ηth
Carnot.
• The smaller the TL the larger the
ηth . Carnot
• To increase the thermal efficiency
in any power cycle, we try to
increase the maximum
temperature at which heat is
added.

7. Reasons why the Carnot cycle is
not used!
• Pumping process 1-2 requires the pumping of a
mixture of saturated liquid and saturated vapor
at state 1 and the delivery of a saturated liquid at
state 2.
• To superheat the steam to take advantage of a
higher temperature elaborate controls are
required to keep Th constant while the steam
• expands and does work.

8. The following is the T-s diagram and schematic for a
Simple Rankine cycle. All process are assumed to
be internally reversible.

9. • Ideal Rankine Cycle Processes
• 1-2 Isentropic compression in pump
• 2-3 Constant pressure heat addition in boiler
• 3-4 Isentropic expansion in turbine
• 4-1 Constant pressure heat rejection in condenser

10. • Assume steady-flow
• - Neglecting ΔPE & ΔKE
• - Assume adiabatic and reversible processes
• Since the pumping process involves an
• incompressible liquid, state 2 is in the
• compressed liquid region
• dh = T ds + v dP
• Since the ideal pumping process 1-2 is
• isentropic, ds = 0.




 m
m
m 2
1




 m
m
m 2
1

12. • Boiler
• To find the heat supplied in the boiler, assume
Conservation of mass and energy for steady flow
• Neglect ΔPE & ΔKE
• No work is done on the steam in the boiler
• The heat transfer per unit mass :

13. Turbine
• To find turbine work assume Conservation of
mass and energy for steady flow.
• The process is adiabatic and reversible
• Neglect ΔPE & ΔKE
• T he turbine work per unit mass
• wT = h3-h4 =
T
W

T
W

T
W


14. • The net work done by the cycle :
• Net power
• net work done wnet = wT − wp kJ/kg
• The thermal efficiency (ηth)
• It can be seen that efficiency can be
increased by increasing heat input or
increasing turbine work how?

15. • Condenser
• Process 4 - 1 occurs in a condenser, which is a type of heat
exchanger. As the working fluid condenses it releases thermal
energy, which is rejected to the surroundings by heat transfer. In
the ideal case , the working fluid leaves the condenser as a
saturated liquid.
• Many power plants are located next to large body of waters which
the plants use as a thermal energy sink. Due to concern about the
large amounts of thermal energy added to these bodies of water
by large power plants, termed thermal pollution, cooling towers
are now often used to reject at least part of this waste thermal
energy to the air.

16. • The amount of cooling water used for cooling the
condenser cooling water can be obtained from the
energy conservation equation as
• Frequently Asked Question: Is the heat transfer
process to the surroundings from states 4 to 1
really necessary, as is this not just wasting energy?
•

18. • Example

22. • Operating pressures of
boilers have gradually
increased over the years
from about 2.7 MPa (400
psia) in 1922 to over 30
MPa (4500 psia) today,
generating enough steam to
produce a net power output
of 1000 MW or more in a
large power plant. Today
many modern steam power
plants operate at
supercritical pressures (P
22.06 MPa) and have
thermal efficiencies of about
40 percent for fossil-fuel
plants and 34 percent for
nuclear plants.

23. • Example

26. • Ideal Reheat Rankine Cycle

27. • Ideal Reheat Rankine Cycle

29. • Rankine Cycle with Reheat

30. Example

32. • Ideal Regenerative Rankine Cycle

34. • Cycle with an open feed-water heater

36. • Cycle with a closed feed-water heater
with pump to boiler pressure

37. Example

38. • Example
• A steam power plant operates on an ideal regenerative Rankine cycle.
Steam enters the turbine at 6 MPa 450OC and is condensed in the
condenser at 20 kPa. Steam is extracted from the turbine at 0.4 MPa to
heat the feedwater in a closed feedwater heater. Water leaves the heater at
the condensation temperature of the extracted steam and that the extracted
steam leaves the heater as a saturated liquid and is pumped to the line
carrying the feedwater. Determine (a) the net work output per kilogram of
steam flowing through the boiler and (b) the thermal
• efficiency of the cycle.

39. Deviation from Actual Cycles

41. • Second law analysis of vapor power cycles

43. Example

48. Example

49. Example

50. The Brayton Cycle with Regeneration, Intercooling,
& Reheating
• The early gas turbines built in the 1940’s
and even 1950’s had simple-cycle
efficiencies of about 17 percent .
• The efforts to improve the cycle efficiency
concentrated in three areas:
• 1. Increasing the turbine inlet (or firing)
temperatures.
• 2. Increasing the efficiencies of turbo-
machinery components.
• 3. Add modifications to the basic cycle.

53. Brayton cycle components:
• Compressor :
• axial-flow compressor and the centrifugal – or
radial-flow compressor .
• Compressors contain a row of rotating blades
followed by a row of stationary (stator) blades.
• All work done on the working fluid is done by
the rotating rows, the stators converting the
fluid kinetic energy to pressure and directing
the fluid into the next rotor.

54. Combustion/Combustor:
• The function of the combustion chamber is to
accept the air from the compressor and to deliver
it to the turbine at the required temperature,
ideally with no loss of pressure.
• It is a direct-fired air heater in which fuel is
burned with less than one-third of the air after
which the combustion products are then mixed
with the remaining air.
• Fuel is commonly gaseous (natural gas, metane,
butane, etc) or liquid (kerosene, light diesel
oil, heavy residual oil, etc)

55. Turbine:
• Gas turbines move relatively large quantities of air
through the cycle at very high velocities.
• Air-standard assumptions:
Assumptions that the compression and expansion
processes are adiabatic (insulated) and reversible
(isentropic), that there is no pressure drop during the
heat addition process, and that the pressure leaving the
turbine is equal to the pressure entering the
compressor.
When the changes in kinetic and potential
energies are neglected, the energy balance for
a steady-flow process can be express, on a unit-
mass basis, as :

56. T-s and P-v Diagrams of an Ideal Brayton Cycle
• The ideal cycle that the working
fluid undergoes in the closed
loop is the Brayton cycle, which
is made up of four internally
reversible processes:
– 1-2 Isentropic compression (in a
compressor)
– 2-3 Constant pressure heat
addition
– 3-4 Isentropic expansion (in a
turbine)
– 4-1 Constant pressure heat
rejection

57. • Analysis of the Brayton cycle gas turbine engine
differs from that of the Otto and Diesel cycles in
two important respects:
1. Because the gas turbine is a steady flow open system,
the energy balances on the components are written in
terms of enthalpies rather than internal energies; and
2. While the Otto and Diesel cycles are most
conveniently described in terms of volume ratios (rV,
rc and re), the defining parameter for the Brayton cycle
is the pressure ratio across the compressor:
• Using the usual Air Standard cycle assumptions,
we can write the energy balances for each of the
components.
• When the changes in kinetic and potential energies are
neglected, the energy balance for a steady-flow
process can be expressed, on a unit–mass basis, as

58. Then the thermal efficiency of the ideal Brayton cycle under
the cold-air standard assumptions becomes

59. • Substituting these equations into the thermal efficiency relation and
simplifying give
• Efficiency under cold air assumption can be found as:
= 1-1/rP
(k-1)/k
• where rP is the pressure ratio and k the specific heat ratio.
• rP = P2 /P1
• The highest temperature in the cycle occurs at the
end of the combustion process (state 3), and it is
limited by the maximum temperature that the turbine
blades can withstand. This also puts limit on the
pressure ratios that can be used in the cycle.
• Back work ratio: Some of the turbine work output
(usually more than half) is used to drive the
compressor. The ratio of the compressor work to
the turbine work is called the back work ratio.

61. Development of Gas Turbines
• The gas turbine has experienced phenomenal progress and
growth since its first successful development in the 1930s.
The early gas turbines built in the 1940s and even 1950s had
simple cycle efficiencies of about 17 percent because of the
low compressor and turbine efficiencies and low turbine inlet
temperatures due to metallurgical limitations of those times.
• The efforts to improve the cycle efficiency
concentrated in three areas:
1. Increasing the turbine inlet (or firing) temperatures. (425
0C to 1425 0C and more today)
• - new materials
• -modern technique of cooling (coating blades with ceramic or cooling
them by air from compressor or steam).

62. 2. Increasing the efficiencies of turbo-machinery
components.
3. Adding modifications to the basic cycle:
- regeneration
- reheating
- inter-cooling
• Deviation of Actual Gas-Turbine Cycles from
Idealized Ones:
• The actual cycle differs from the ideal cycle due to:
• 1- pressure drop during heat addition and heat
rejection.
• 2- the actual work from turbine is less and from
compressor is more due to irreversibilites.

63. The deviation of actual
compressor and turbine
behavior from the idealized
isentropic behavior can be
accurately accounted
for by utilizing the
isentropic efficiencies of
the turbine and
compressor as

64. 1- THE BRAYTON CYCLE WIT REGENERATION
The high-pressure air leaving the compressor can be heated by
transferring heat to it from the hot exhaust gases in a counter-
flow heat exchanger, which is also known as a regenerator
or a recuperator.

65. • Assuming the regenerator to be well insulated and any changes in kinetic and
potential energies to be negligible, the actual and maximum heat transfers
from the exhaust gases to the air can be expressed as
The extent to which a regenerator approaches an ideal regenerator is called
the effectiveness ` ε ‘ and is defined as
The effectiveness of most regenerators used in practice is below 0.85.

66. • Under the cold-air-standard assumptions, the thermal efficiency of an ideal
Brayton cycle with regeneration is
Therefore, the thermal efficiency of an ideal Brayton cycle with regeneration depends
on the ratio of the minimum to maximum temperatures as well as the pressure ratio.

67. • THE BRAYTON CYCLE WITH REHEATING
In real turbines the reheaters are combustion chambers and the number of
combustors in series is limited by the supply of oxygen in the working fluid,
which decreases after each combustion process.

69. • THE BRAYTON CYCLE WITH INTERCOOLING
The net power output of a gas turbine engine can be increased not
only by increasing the power produced by the turbine, but also by
decreasing the work consumed by the compressor.

71. • THE BRAYTON CYCLE WITH REGENERATION,
INTERCOOLING AND REHEATING:

72. The Four Stroke Otto Cycle

73. • Consider the automotive spark-ignition power
cycle.
Processes
• Intake stroke
• Compression stroke
• Power (expansion) stroke
• Exhaust stroke
• Often the ignition and combustion process begins
before the completion of the compression stroke.
The number of crank angle degrees before the
piston reaches TDC on the number one piston at
which the spark occurs is called the engine timing

74. Otto Cycle P-V & T-s Diagrams
Pressure-Volume Temperature-Entropy
The air-standard Otto cycle is the ideal cycle that approximates the spark-
ignition combustion engine.
Process Description
1-2 Isentropic compression
2-3 Constant volume heat addition
3-4 Isentropic expansion
4-1 Constant volume heat rejection

75. The compression ratio r of an engine is the ratio of the maximum
volume to the minimum volume formed in the cylinder.
The mean effective pressure (MEP) is a fictitious pressure that, if it
operated on the piston during the entire power stroke, would produce
the same amount of net work as that produced during the actual cycle.

76. • Thermal Efficiency:
– For a constant volume heat addition (and rejection)
process;
– Assuming constant specific heat:
Otto Cycle Derivation

77. • For an isentropic compression (and expansion)
process:
• where: γ = Cp/Cv
• Then, by transposing,
Otto Cycle Derivation
Leading to

78. The compression ratio (rv) is a volume ratio and is
equal to the expansion ratio in an otto cycle
engine.
• Compression Ratio
where Compression ratio is defined as
Otto Cycle Derivation

79. • Then by substitution,
The air standard thermal efficiency of the Otto cycle
then becomes:
Otto Cycle Derivation

80. • Summarizing
where
and then
Isentropic
behavior
Otto Cycle Derivation

81. • Heat addition (Q) is accomplished through fuel
combustion
• Q = Lower Heat Value (LHV) BTU/lb, kJ/kg
Otto Cycle Derivation
also

83. • Determine the temperatures and pressures at each point in
the Otto Cycle.
Compression Ratio = 9.5:1
T1 temperature = 25oC = 298oK
P1 pressure = 100 kPa
Otto Cycle P & T Prediction

84. Diesel Cycle P-V & T-s Diagrams
The air-standard Diesel cycle is the ideal cycle that approximates the Diesel
combustion engine
Process Description
1-2 Isentropic compression
2-3 Constant pressure heat addition
3-4 Isentropic expansion
4-1 Constant volume heat rejection

85. Diesel Cycle Derivation
• Thermal Efficiency (Diesel):
For a constant pressure heat
addition process;
For a constant volume heat
rejection process;
Assuming constant specific heat:
where: γ = Cp/Cv

86. PV
T
PV
T
V V
T
T
P
P
4 4
4
1 1
1
4 1
4
1
4
1
 

where
where rC is called the cutoff ratio, defined as V3 /V2 .. Recall
processes 1-2 and 3-4 are isentropic, so
PV PV PV PV
k k k k
1 1 2 2 4 4 3 3
 
and
Since V4 = V1 and P3 = P2, we divide the second equation by
the first equation and obtain
Therefore,

87. th Diesel
c
k
c
k
c
k
c
k
T T T
T T T
k
T
T
r
r
r
r
k r
,
( / )
( / )
( )
( )
 


 


 



1
1 1
1
1
1 1
1
1
1 1
1
1 4 1
2 3 2
1
2
1

88. • Determine the temperatures and pressures at each point in
the Diesel Cycle.
Compression Ratio = 20:1
T1 temperature = 25oC = 298oK
P1 pressure = 100 kPa
Diesel Cycle P & T Prediction

89. Otto-Diesel Cycle Comparison

90. Dual Cycle P-V Diagrams:

91. • Dual Cycle Thermal Efficiency
Dual Cycle Efficiency
where: γ = Cp/Cv

92. REFRIGERATORS AND HEAT PUMPS
• The transfer of heat from a low-
temperature region to a high-
temperature one requires special
devices called refrigerators.
• Another device that transfers heat
from a low-temperature medium to a
high-temperature one is the heat
pump.
• The performance of refrigerators and
heat pumps is expressed in terms of
the coefficient of performance
(COP), defined as

93. This relation implies that COPHP 1 since COPR is a positive
quantity. That is, a heat pump functions, at worst, as a resistance
heater, supplying as much energy to the house as it consumes.
The cooling capacity of a refrigeration system:
the rate of heat removal from the refrigerated space—is often
expressed in terms of tons of refrigeration (TR).
(TR) is defined as the capacity of a refrigeration system that can
freeze 1 ton (2000 lbm) of liquid water at 0°C (32°F) into ice at
0°C in 24 hr.
1 TR = (m. L )/t
One ton of refrigeration is equivalent to 211 kJ/min or 200
Btu/min or 3.5167 kw.

94. THE REVERSED CARNOT CYCLE

95. • The coefficients of performance of Carnot refrigerators and
heat pumps are expressed in terms of temperatures as
The reversed Carnot cycle is the most efficient refrigeration
cycle operating between two specified temperature levels.
Therefore, processes 1-2 and 3-4 can be approached closely in
actual evaporators and condensers. However, processes 2-3
and 4-1 cannot be approximated closely in practice.
Why?

96. THE IDEAL VAPOR-COMPRESSION REFRIGERATION CYCLE
• The vapor-compression refrigeration cycle is the most widely
used cycle for refrigerators, air-conditioning systems, and heat
pumps. It consists of four processes:
• 1-2 Isentropic compression in a compressor
• 2-3 Constant-pressure heat rejection in a condenser
• 3-4 Throttling in an expansion device
• 4-1 Constant-pressure heat absorption in an evaporator

97. Assuming steady flow processes, neglect change in kinetic and potential energry,
then the steady flow energy equation on a unit–mass basis reduces to

98. • The COPs of refrigerators and heat pumps operating on the
vapor-compression refrigeration cycle can be expressed as
How can get h2 and h4?
EXAMPLE
A refrigerator uses refrigerant-134a as the working fluid and operates
on an ideal vapor-compression refrigeration cycle between 0.14 and 0.8
MPa. If the mass flow rate of the refrigerant is 0.05 kg/s, determine (a)
the rate of heat removal from the refrigerated space and the power input
to the compressor, (b) the rate of heat rejection to the environment, and
(c) the COP of the refrigerator.

99. • Improvement of the vapor compression cycle:
• 1- Increasing or decreasing the condenser pressure .
• 2- Increasing or decreasing the evaporator pressure.
• 3- subcooling the refrigerant liquid out from the condenser.
• 4- superheating the refrigerant vapor out from the evaporator.
• 5- subcooling and superheating with using heat exchanger.

100. COPR= qL/wC= (h1-h6)/(h3-h2)
= (h2-h6’)/(h3-h2)
Where h2-h1 = h4-h5

101. Azeotropic mixtures:
R 500: Mixture of R 12 (73.8 %) and R 152a (26.2%) R11, R12 (CFC)
R 502: Mixture of R 22 (48.8 %) and R 115 (51.2%) R22 (HCFC)
R503: Mixture of R 23 (40.1 %) and R 13 (59.9%) R134a (HFC)
R507A: Mixture of R 125 (50%) and R 143a (50%)
Zeotropic mixtures:
R404A : Mixture of R 125 (44%), R 143a (52%) and R 134a (4%)
R407A : Mixture of R 32 (20%), R 125 (40%) and R 134a (40%)
R407B : Mixture of R 32 (10%), R 125 (70%) and R 134a (20%)
R410A : Mixture of R 32 (50%) and R 125 (50%)

102. Gas Vapor Mixtures and Air-Conditioning
Definitions
• Dry Air Atmospheric Air
or moist air
Air with zero water A mixture of dry air and water
vapor vapor content
• The water vapor concentration in atmospheric air can
strongly affect human comfort.
• For air-conditioning applications, air pressure is
nominally 1 atmosphere and air temperature varies in
the range −10 T 50 0C. Under these conditions, dry air
and atmospheric air are treated as ideal gases.

104. • With the partial pressure for dry air and water vapor, (1) can be
modified to:
p = ρa 286.9 T + ρwv 455 T
Where ρa = 0.0035 pa / T
ρwv = 0.0022 pwv/ T
Dry air is more dense that humid air!

107. Degree of saturation μ: The degree of saturation is the ratio of the
humidity ratio ω to the humidity ratio of a saturated mixture ωs at the same
temperature and pressure. μ = ω /ωs

109. • Specific Enthalpy of Dry Air - Sensible Heat (ha):
• ha = cP,a .t kJ/kg
• cP,a = 1.006 (kJ/kg 0C) for air temp. (-100 to 100 0C)
= 0.240 (Btu/lb 0F)
• Specific Enthalpy of Water Vapor - Latent Heat (hwv):
• hwv = cP,wv t + hfg (at 0 0C) kJ/kg
• cP,wv = 1.84 (kJ/kg 0C)
= 0.444 (Btu/lb 0F)
• hfg = 2502 kJ/kg
= 1061 (Btu/lb)
• h = (1.006 ) t + ω [(1.84 ) t + (2502 )] kJ/kg
• h = (0.240 ) t + ω [(0.444 ) t + (1061)] Btu/lb

110. φ

111. Common properties in the charts includes
• dry-bulb temperature (td.b) - wet-bulb temperature (tw.b)
• relative humidity (φ) - humidity ratio (ω) - specific volume (v)
• dew point temperature (td.P) - enthalpy (h)
Dry Bulb Temperature – Td.b.
The air temperature is indicated by a thermometer not affected by
the moisture of the air.
Wet Bulb Temperature – Twb
It is the temperature indicated by a
moistened thermometer bulb
exposed to the air flow.
Dew Point Temperature – Tdp
The temperature at which water vapor
starts to condense out of the air,

112. • If td.P ~ tair, then φ is high.
• If moisture condensates on a cold bottle from the refrigerator ,
then ………………?
Most common air condition processes are
1. Sensible heating – sensible cooling
2. mixing air
3. cooling and dehumidifying air
4. humidifying air by adding steam and/or water
Sensible heating (Process A-B):
The supplied heat – Qh,s
Qh,s = ma(hB – hA) kJ/kg
QL = 0 and ωA= ωB

113. • Mixing Air of different Conditions
The heat balance for the mixture can be expressed as
mA hA + mC hC = (mA + mC ) hB
The moisture balance for the mixture can be expressed as:
mA ωA + mC ωC = (mA + mC ) ωB

114. • Sensible cooling (Process O-A)
• The heat transfer rate during this
process is given by:
Q c,s= ma (h O-h A)
= ma c P,a (t O-t A) kJ/kg
Cooling and dehumidification (Process O-C):

115. • When moist air is cooled below its dew-point by bringing it in
contact with a cold surface as shown in Fig.28.3, some of the
water vapor in the air condenses and leaves the air stream as
liquid, as a result both the temperature and humidity ratio of air
decreases as shown.
• By applying mass balance for the water:
• ma ωa =ma ωc + mw
• By applying mass balance for the water:
• Qtot= QC,s + QC,L = ma (hw –hc) + ma(ho-hw)
• = ma cP,a (to – tc) + mw hfg (ωo – ωc)
• Qtot = ma (ho – hc)

116. HEAT PUMP SYSTEMS

117. FUELS AND COMBUSTION
• TYPE OF FUELS
solid, liquid, and gaseous
Solid Fuels:
Biomass: woody – nonwoody
1-Woody: (soft woods and hard woods)
Soft woods: evergreen trees with needle – made up of vertical
oriented, hollow, tubular fibers from 2 to 7 cm long.
Hard woods: broad-leafed trees that shed their leaves. Have
shorter fibers and more porous. More denser than soft woods.

118. • Charcoal: made by heating wood in the absence of air.
• It can be pulverized and made into briquettes (using
starch as binder ).
• Dry wood consists of cellulose, hemi-cellulose, lignin,
resins and ash forming minerals.
2- Non-Woody Biomass: include agriculture residue such as
bagasse, straws, stalks, and pits.
3- Peat :
Fromed form decaying of woody plants, reeds, sedegs, and
mosses in watery bogs. It is dark-brown in color and
fibrous charater.
It has low % of carbon, high moisture 30% and more, not
suitable for producing power in power plants.

119. 4- Coal:
Coal classification
1. Anthracite (semi) (oldest type): hard – little
volatile – almost no moisture – high carbon. ( rare)
2. Lignite (Youngest type): soft – has high volatile and
moisture – low fixed carbon
3- Bituminous (sub and semi): lie in between lignite
and anthracite – the most type used in steam power
plant.
Physical and chemical properties of coal:
• Physical properties of coal include the heating
value, moisture content, volatile matter and
ash.

120. • The chemical properties of coal refer to the various elemental
chemical constituents such as carbon, hydrogen, oxygen,
nitrogen, and sulphur.
• GCV ranged from 1300 – 6200 and more (kcal/kg).
Analysis of Coal:
1. Proximate analysis: determine only fixed carbon, volatile matter,
moisture, and ash percentage.
2. Ultimate analysis: determine all components of coal, solid or gas.
The role of each constituent in heating value of fuel:
• Fixed carbon is the solid fuel left in the furnace after
volatile matter is distilled off. also contains some
hydrogen, oxygen, sulphur and nitrogen not driven off
with the gases. It gives a rough estimate of the heating
value of coal.

121. • Volatile matter: combustible as ( H2, HC, CH4,CO) –
incomustible as (CO2,N2) – 20 to 35% .
1. Proportionately increases flame length, and helps in
easier ignition of coal
2. Sets minimum limit on the furnace height and volume
3. Influences secondary air requirement and distribution aspects
4. Influences secondary oil support
Ash content:
Ash is an impurity that will not burn. Typical range is 5% to 40%.
• Reduces handling and burning capacity
• Increases handling costs
• Affects combustion efficiency and boiler efficiency
• Causes clinkering and slagging

122. Moisture content:
It decreases the heat content / kg of coal. Typical range is 0.5 to 10%.
• Affects clinkering and slagging tendencies
• Corrodes chimney and other equipment such as
air heaters and economizers
• Limits exit flue gas temperature
Refuse Solid Fuel:
Include municipal solid waste and commercial, institutional,
industrial, and agriculture waste. Recently, these waste
materials has been used to produce steam. The
combustible materials can be separated, and called
Refuse-Derived Fuel (RDF).

123. • Liquid Fuels:
• The products come out from the refining and
distillation of petroleum crude oil such as naptha,
gasoline, kerosene, gas oil, heavy fuel oil, ..etc.
• The various properties of liquid fuels are:
1. Density: measured by hydrometer
2. Specific Gravity: The density of fuel, relative to water.
3. Viscosity: a measure of internal resistance of fluid to flow.
• Viscosity α (1/T).
• It influences the degree of pre- heating required for handling,
storage and satisfactory atomization.
4- Flash Point: The lowest temperature at which the fuel can
be heated.

124. • Pour Point: The lowest temperature at which it will pour or
flow when cooled under prescribed conditions.
• Specific Heat: Light oils have a low specific heat, whereas
heavier oils have a higher specific heat.
• Calorific Value:The measurement of heat or energy produced.
Sulphur:

125. • Ash Content: These salts may be compounds of sodium,
vanadium, calcium, magnesium, silicon, iron, aluminum, nickel,
etc.
• Carbon Residue: the tendency of oil to deposit a
carbonaceous solid residue on a hot surface, such as a burner or
injection nozzle, when its vaporizable constituents evaporate.
• Water Content: cause damage to the inside surfaces
of the furnace during combustion especially if it
contains dissolved salts. It can also cause spluttering of
the flame at the burner tip, possibly extinguishing the
flame, reducing the flame temperature or lengthening
the flame.

126. • Types of gaseous fuel
Fuels naturally found in nature:
• - Natural gas Methane is the main constituent of natural gas and
accounting for about 95% of the total volume. Other components are: Ethane,
Propane, Butane, Pentane, Nitrogen, Carbon Dioxide, and traces of others. - -
- Methane from coal mines
Fuel gases made from solid fuel
• - Gases derived from coal
• - Gases derived from waste and biomass
• - From other industrial processes (blast furnace gas)
Gases made from petroleum
• - Liquefied Petroleum gas (LPG)
•Gases from oil gasification
•Gases from some fermentation process
•Gaseous fuels use are liquefied petroleum gases (LPG), Natural
gas, producer gas, blast furnace gas, coke oven gas etc.

127. • Refinery gases
• Gases from oil gasification
• Gases from some fermentation process
Gaseous fuels in common use are liquefied petroleum gases (LPG),
Natural gas, producer gas, blast furnace gas, coke oven gas etc.
• Properties of gaseous fuels:
• Fuel should be compared based on the net calorific value.

129. OXIDANT:
• Usually air ( the oxygen in the air )
• Where air is unavailable the oxidant has to be carried as well as
the fuel - space vehicles, rockets etc;.
MIXING:
• SOLIDS :- pulverizing to powder or small lumps
• LIQUIDS :- spray nozzles, atomizers, vaporizers,
• carburetors, 'burners'.
• GASES :- mixing valves, chambers, burners; usually
need precautions to avoid explosion and flash back .

130. IGNITION
• Simply mixing methane and air will not cause it
to burn. The molecules need to reach a certain
threshold energy level before the combustion
process will proceed. This may be provided
initially by another flame, a spark or hot surface.
The combustion process itself (if sustained ) then
continues the ignition process.
COMBUSTION
• The chemical dissociation of the fuel and it's
recombination with oxygen. Energy and Mass
are conserved.

133. Calorific Values of Fuel:
• The heat released by burning 1 kg of fuel or 1 m3 of
gas is called the calorific value.

134. Calculation of stochiometric air needed for combustion of furnace
oil. The first step is to determine the composition of the furnace oil.
Typical specifications of furnace oil from lab analysis is given below:

137. • Calculation of theoretical CO2 content in the flue gases:
Calculation of constituents of flue gas with excess air
Now we know the theoretical air requirements and the theoretical
CO2 content of flue gases.
The next step is to measure the actual CO2 percentage in the flue
gases. In the calculation
below it is assumed that the measured %CO2 in the flue gas is 10%.

139. • Calculation of theoretical CO2% in dry flue gas by
volume
For optimum combustion of fuel oil the CO2 or O2 in flue gases should be
maintained as follows:
CO2 = 14.5–15 %
O2 = 2–3 %

144. Example
• A fuel that is 80% CH4 ( methane ) and 20% C2H6 ( ethane ).
• We could choose any quantity of fuel but for convenience we
• shall use 1 kmole
( 0.8 CH4 + 0.2 C2 H6 ) + Y( 0.79 N2 + 0.21 O2 ) →
• (the reactants)
• a CO2 + b H2 O + c O2 + d N2
• (the products of complete combustion)
• The number of atoms on the L.H.S. of the equation must exactly
balance the number on the R.H.S.

145. • Solving equations (i) to (iv) simultaneously :-
• from (i) a = 1.2
• from (iii) b = 2.2
• from (ii) c = 0.21 Y - 2.3
• from (iv) d = 0.79 Y
• If we need to calculate the Stoichiometric Air-Fuel ratio,
then there no excess air and O2 in the product must equal zero,
so c = 0
How would we convert it to a mass ratio ?
10.95 kmol air = 10.95 x 29 kg/kmol = 317.62 kg
1 kmol fuel = 0.8 x 16 + 0.2 x 30 = 18.80 kg
Therefore the AFR (by mass) is 317.62:18.80 or 16.89:1

146. • (i) If the AFR is LESS than stoichiometric complete
• combustion cannot occur i.e.: we have an excess of
fuel, or a rich mixture, leading to unburnt or
incompletely burnt fuel. CO will be present. This
clearly is undesirable as we will not be releasing the
maximum available heat energy from the fuel.
• (ii) If the AFR is GREATER than stoichiometric we
can obtain complete combustion but we may be
supplying too much air thus reducing the temperature
of the combustion products*and therefore reducing the
heat transfer from them.

147. • Excess air percentage can be obtained as
% excess air = {AFexcess –AFStoich.}/AFstoich . 100%
For Hydrocarbon fuels (CaHb):
On volume bases
CaHb + (a+b/4) O2 = (a) CO2 + (b/2) H2O2
On mass bases
(12a+b) CaHb + 32(a+b/4) O2 = (44a) CO2 + (9b) H2O
If the air is used instead of O2
CaHb + (100/21 ) (a+b/4) of air = (a) CO2 + (b/2) H2O +
(79/21)(a+b/4) N2 on volume basis

148. (12a+b)CaHb + 32(100/23.3 ) (a+b/4) of air = (44a) CO2
+ (9b) H2O + 32(76.6/23.3)(a+b/4) N2 on mass basis
AFR = 32(100/23.3 ) (a+b/4) / (12a+b)
If excess air is used
CaHb + (1+X)(100/21 ) (a+b/4) of air = (a) CO2 + (b/2)
H2O + (1+X)(79/21)(a+b/4) N2 + X(a+b/4) O2
on volume basis
(12a+b)CaHb + 32(100/23.3 ) (a+b/4)(1+X) of air =
(44a) CO2 + (9b) H2O + 32(76.6/23.3)(a+b/4)(1+X) N2 +
32 X(a+b/4) O2 on mass basis

149. Then AFR with excess air for complete combustion is
AFR = 32(1+X)(100/23.3 ) (a+b/4) / (12a+b)
Where X = excess air factor = % excess air/100