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UNIT-1 COMBUSTION THERMODYNAMICS
1.1 Introduction
All conventional fossil fuels, whether, solid, liquid or gaseous, contain basically carbon and
hydrogen which invariably react with the oxygen in the air forming carbon dioxide, carbon
monoxide or water vapour. The heat energy released as a result of combustion can be utilized for
heating purposes or for generation of high pressure steam in a boiler or as power from an engine or
a gas turbine.
The solid fuels are burned in beds or in pulversied from suspended in the air stream. The
liquid fuels are burned either by vaporising and mixing with air before ignition, when they behave
like a gaseous fuel. The gaseous fuels are either burned in burners when the fuel and air are
premixed or the fuel and air flow separately in to a burner or a furnace and simultaneously mix
together as combustion proceeds.
The Kg-mole or gram-mole is widely used in combustion calculations as a unit of weight.

The molecular weight of any substance in kg represents one kilogram mole or 1K mole. 1Kmol of
hydrogen has a mass of 2.016Kg and 1Kmol of carbon has a mass of 12Kg.
Consider a reaction
CH4 + 2O2 CO2 + 2H2O 1.1
(12+4.032) + 64 44 + 2 (2.032 + 16)
16.032kg of methane reacts with 64Kg of oxygen to form 44kg of carbon dioxide and
36.032kg of water. We can also simply state that 1Kmol of methane reacts with 2Kmol of oxygen
to form 1Kmol of carbon dioxide and 2K mol of water, this has advantage of permitting easy
conversion between the mass and volumetric quantities for the gaseous fuel and the product of
combustion. If the gases are considered ideal then according to Avogadro hypothesis, all gases
contain the same number of molecules per unit volume. It implies that 1K mole of any gaseous
substance occupies the volume of 22.4m3 at NTP i.e., 1.013bar and 273K.
CH4 + 202 CO2 + 2H2O 1.2
1 volume of methane reacts with 2 volume of oxygen to form one volume of CO2 and two volumes
of H2O. Therefore in any reactions, the mass in confirmed but the no. of mol or volumes may not
be considered.
(i) C +O2 CO2

On mass basis (12) + (32) (44)
1 Mol (C) + 1 Mol (O2) 1 Mol (CO2)
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(ii) 2C +O2 2CO

On mass basis (24) + (32) (56)
Volume basis 2 Mol (C) + 1 Mol (O2) 2Mol (CO)
(iii) 2C +O2 2CO2

On mass basis (56) + (32) (88)
2 Mol (CO) + 1 Mol (O2) 2 Mol (CO2)
1.2 Combustion Stoichometry
A balanced chemical equation for complete Combustion of the reactions with no excess air
in the product is known as a stiochiometric equation. A stiochiometric mixture of the reactants is
one in which the molar proportions of the reactants are exactly as given by the stiochiometric
coefficients, so that no excess of any constituent is present. In general a chemical reaction may be
written as
aA + bB - cC + dD 1.3
Where the reactants A and B react to form the products C and D. The small letters a, b, c and d are
known as the stiochiometric coefficients.
For the combustion of any fuel the most common oxidizer is air which is a mixture of 21%
O2 and 79% N2 (on volume basis). One mol of oxygen is accompanied by 79/21 (3.76) mol of
Nitrogen. The Chemical equation for the stiochiometric combustion of carbon with air is written
as
C + O2 + 3.76N2 CO2 + 3.76N2 1.4
The minimum amount of air required for the complete combustion of a fuel is known as
theoretical air. However in practice it is difficult to achieve complete combustion with theoretical
air. Therefore fuel requires some excess air for different application and may vary from 5% ~ 20%
and in gas turbine it may go up to 400% of theoretical quantity.
1.3 Theoretical air required for complete combustion.
If the fuel composition is known, the requirement of oxygen or air can be calculated either by mass
balance or by mole method.
Consider a equation
C + O2 CO2 1.5
(12) + (32) (44)
Or 1 + 8/3 11/3
Or 1 Kg C + 8/3 Kg O2 11/3 kg CO2
Similarly
2C + O2 2CO 1.6
2(12) + (32) 2(28)
Or 1 Kg C + 32/24 Kg O2 56/24 kg CO
Or 1 Kg C + 4/3 Kg O2 7/3 kg CO
Similarly
1 Kg CO + 4/7 Kg O2 11/7 kg CO2 1.7
On molal basis
1K mol C + 1K mol O2 1K mol CO2

1K mol C + K mol O2 1K mol CO
1K mol C + K mol O2 1K mol CO2
1.4 Conversion of Gravimetric analysis to volumetric basis and vice versa
If the composition of fuel is given on gravimetric (or weight) basis it can be converted to
volumetric (or mole) basis as follows. Divide the weight of each constituents of the mixture by its
molecular weight. This will give the relative volume (or mole) of each constituents. Add all the
relative volumes of the constituents then,
Indivisual (relative) volume of the constituents X 100

Total volume of all the constituents
will give the %age by volume of each constituents in the fuel.
If the volumetric composition of a fuel is given, it can be converted to gravimetric (or weight) basis
as follows. Multiply the indivisual volume of each constituent by its molecular weight. This will
give relative weight of each constituent. Add all the relative weights of the constituents then
Indivisual weight of the constituents X 100

Total(reative)weights of the constituents
will give the %age by weight of each constituent in the fuel.
1.4.1 Calculation of the minimum amount of air for a fuel of known composition.
Example 1
Calculate the minimum volume of air required to burn one Kg of coal having the following
composition by weight
C = 72.4%, H2 5.3%, N2 = 1.81, O2 = 8.5%, moisture 7.2%

S = 0.9% and ash 3.9%
On weight basis:
Taking 1kg coal as basis weight of oxygen required to burn 1kg of coal
C + O2 CO2
0.724 x 32/12 = 1.93 kg

0.53x 16/2 = 0.424 kg
0.009x32/32 = 0.009 kg
Total O2 = 2.363 kg per kg of coal
But 0.085kg O2 is available in coal, therefore O2 required

= 2.363 0.085 = 2.278kg per Kg of coal.
Air contains 23% of oxygen by weight. Therefore the weight of the air supplied is
2.278x 100/23 = 9.9 kg per kg of coal

Density of air required at NTP
P v = mRT
P= m/v RT = RT,
= Molecular weight
Volume
= P/RT = 1.013x105/287x273 = 1.29 kg/m3
Therefore volume of air required = 9.9(kg)/1.29(kg) = 7.67 m3
On mole basis
Consider 100kg of coal
C = 72.4/12 = 6.03K mol, O2 = 8.5/32 = 0.265K mol

H2 = 5.3/2 = 2.65K mol, H2O = 7.2/18 = 0.4K mol
N2 = 1.8/28 = 0.064K mol, S = 0.9/32 = 0.028K mol
1K mol C + 1K mol O2 1Km CO2
Therefore 6.03 K mol of carbon requires

6.03 K mol of oxygen
1 K mol H2 + K mol O2 1K mol H2O

H2 - 2.65 x = 1.325K mol
S- 0.028x1 = 0.028
Total O2 required 6.03 + 1.325 + 0.028 = 7.383
The oxygen present in coal 0.265K mol

Net O2 required = 7.383 0.265 = 7.118K mol
Air required
7.118x100/21 = 33.89K mol / 100kg of coal = 0.3389K mol / 1kg coal
Volume of air supplied

0.3389K mol/kg X 22.4m3 = 7.59m3/kg of coal
Example 2
Calculate the volumetric analysis of the flue gases when coal burns with 20% excess air from the
previous calculation the actual air required 33.89K mol/100kg coal. Therefore the actual air is
33.89 x 120/100 = 40.67K mol/ 100 kg coal
The amount of N2 associated with this
40.67 x 79/100 = 32.13K mol
The amount of O2 present 40.67 x 21/100 = 8.54K mol
The actual amount of O2 required was 7.118K mol excess O2 will appear in exhaust gas = 8.54
7.118 = 1.422K mol.
Therefore:
CO2 = 6.03K mol
SO2 = 0.028K mol
N2 = 32.13K mol (air) + 0.064 (fuel)
= 32.194K mol
O2 = 1.422K mol os excess oxygen.
Therefore the Total volume = (6.03 + 0.028 + 32.194 + 1.422)
= 39.674K mol
The volumetric composition of the gas
CO2 = (6.03/39.674) x 100 = 15.12%

SO2 = (0.028/39.674) x 100 = 0.07%
N2 = (32.13/39.674) x 100 = 81.15%
O2 = (1.422/39.674) x 100 = 3.58%
1.5 Calculation of the composition of fuel and excess air supplied from the exhaust gas
analysis:

Some times the composition of fuel is unknown and it becomes necessary to judge whether the
amount of air supplied is sufficient or not, or excess. This can be obtained by analyzing the sample
of exhaust gases.
Example 3
The composition of dry flue gases obtained by burning a liquid fuel containing only hydrogen and
carbon is CO2 10.7%, O2 5.1%, N2 84.2%. Calculate the composition of fuel by weight and excess
air used.
Solution: consider 100K mol of dry flue gases. They will contain 10.7K mol of O2 (from CO2) +
5.1K mole of (as max. oxygen) = 15.8K mol
Using nitrogen balance the actual air used 84.2 x 100/79 = 106.58K mol of dry flue gases and
oxygen in the air supplied 106.58 x 21/100 = 22.38K mol. Therefore the amount of O2 present in
the water produced by the combustion of H2 is 22.38 15.8 = 6.58K mol O2. We know that 1 K
mole of H2 combines with K mol O2 to produce water. Therefore the amount of hydrogen
present is 6.58x2 = 13.16K mol/100K mol of dry flue gases, and the carbon present is 12X10.7 =
128.4kg/100K mol of dry flue gas. Therefore the composition of fuel (by weight) is 128.4kgC
and 26.32Kg H2 on the %age basis.
C = (128.4/(128.4+26.32) x 100 = 82.99%
H = (26.32/(128.4+26.32) x 100 = 17.01%
Excess air supplied
The amount of O2 required to burn 10.7K mol C is 10.7K mol and to burn 13.16K mol H2 is 13.16
X = 6.58
Total O2 required = 10.7 + 6.58 = 17.28K mol/100K mol of dry flue gases
%age of excess air = (22.38 17.28)/(17.28) x 100 = 29.5%
1.6 Dew point of products:

The product of combustion containing water vapour are known as wet products. The water vapour
present in combustion product is cooled down to a point of condensation the vapour turn in to
liquid and volume will be reduced. Knowing the partial pressure exerted by the water before
condensing, it is possible to find the saturation temp. corresponding to partial pressure from the
steam tables.

1.7 Flue gas analysis

A fuel has the following %age volumetric analysis
H2 = 48 CH4 : 26, CO2 : 11, CO : 5, N2 = 10
The %age volumetric analysis of the dry exhaust gases in CO2:8.8, O2: 5.5, N2: 85.7
Determine the air/fuel ratio by volume if air contains 21% O2 by volume
Solution: the chemical equation for the reaction of 100 moles of fuel gas with air may be written
as
48H2 + 26CH4 + 11CO2 + 5CO + 10N2 + x (O2 + 3.76N2) aCO2 + bH2O + cO2 + dN2
Carbon balance (C) 26 + 11 + 5 = a = H2

H2 48 + 52 = b = 100
O2 11 + 2.5 + x = a + c (i)
N2 10 + 3.76x = d (ii)
Solving (i) and (ii) we, have
From (i) 11 + 2.5 + x = 100/2 + a + c

13.5 + x = 50 + a + c
Adding 10 + 3.76x = d
23.5 + 4.76x = 50 + a + c + d
Or a + c + d = 4.76x 26.5
% CO2 by volume in dry gas
(a/a+c+d) x 100 = 8.8
or (42/4.76x 26.5) = 0.088
4.76x = 503.77
AF = Total mol air = 503.77 = 5.038%
Total mol fuel 100
Example 4
A blast furnace gas has the following volumetric analysis H2 CO-24%, CH4 2%, CO2-6%, O2-3%
and N2-56%
Determine the Ultimate gravimetric analysis

Given volumetric analysis, H2 9%, CO-24%, CH4 2%, CO2-6%, O2-3% and N2-56%

Solution: The volumetric analysis may be converted into mass or granite metric analysis by
completing the table as follows:
Constituent Volume in Molecular Proportional Mass in kg per kg of % by mass =
1m3 of mass (b) mass the gas (d)=(c)/ (d)x100
flue gas (c)=(a)x(b)
(a)
CO 0.24 28 6.72 6.72/18.48 = 0.36 36%
CH4 0.02 16 0.32 0.32/18.48 = 0.0173 1.73%
CO2 0.06 44 2.64 264/18.48 = 0.142 14.2%
O2 0.03 32 0.96 0.96/18.48 = 0.0519 5.19%
N2 0.56 14 7.84 7.84/18.48 = 0.42 42%
c = 18.48 (d) = 1 100
The volumetric analysis of flue gas components becomes

CO-0.36, CH4 0.0173, CO2- 0.142, O2-0.0519 and N2-0.42
Example 5
Determine the fuel gas analysis and air fuel ratio by weight when fuel oil with 84.9% carbon,
11.4% hydrogen, 3.2% sulphur, 0.4% oxygen and 0.1% ash by weight is burnt with 20% excess
air, assume complete combustion.
Solution: Consider 1kg of fuel
Oxygen required / Kg of fuel
For burning of 1kg C - 0.849 x32/12

For burning of 1kg H - 0.114 x16/2
For burning of 1kg S - 0.032 x32/32
Total O2 required is 3.208 kg.
Amount of O2 contained in the fuel = 0.004Kg
Net O2 supplied / kg of fuel = 3.208 0.004
= 3.204 kg O2
Net air supplied = 3.204x100/23 = 13.93 kg/kg of fuel
When 20% excess air supplied
Total air supplied = 13.93 x 1.2 = 16.716 kg/kg of fuel.

N2 actually supplied = 16.716 x 77/100 = 12.871 kg/kg of fuel
O2 actually supplied = 16.716 x 23/100 = 3.845 kg/kg of fuel
Total free O2 in fuel gas = 3.845 6.204
= 0.641 kg/kg of fuel
Total free N2 in fuel gas = 12.87 kg/kg of fuel
Flue gas analysis:
C converted to CO2 = 0.849x44/12 = 3.113 kg CO2

H converted to H2O = 0.114x18/2 = 1.026 kg H2O
S converted to SO2 = 0.032x64/32 = 0.064 kg SO2
Flue gas / kg of fuel:
= 3.113 + 1.26 + 0.064 + 0.641 + 12.871

CO2 H2O SO2 O2 N2
= 17.715kg.
Therefore:
CO2 = (3.113/17.715)x100 = 17.573%
SO2 = (0.064/17.715)x100 = 0.36%
O2 = (0.641/17.715)x100 = 3.618%
H2O = (1.026/17.715)x100 = 5.79%
N2 = (12.871/17.715)x100 = 72.656%
Air fuel mixture ratio is = 16.716 : 1
Example 6
A blast furnace gas has the following volumetric analysis.
H2 = 9%, CO = 24%, CH4 = 2%, CO2 = 6%, O2 = 3% and N2 = 56 %
Determine the ultimate gravimetric analysis.
Solution:
Total H2 in the blast furnace gas.

% volumetric analysis = 9H2 + 2H4
Proportional mass = % volumetric analysis X mol. Mass of element
= (9x2) + (2x4) = 18 + 8 = 26 kg.
Total C in the blast furnace gas.

% of volumetric analysis = 24C + 2C + 6C
Proportional mass = (24+2+6) x 12
= 384 kg
Total O2 in the blast furnace gas
% of volumetric analysis = 24xO + 6O2 + 3O2

Proportional mass = (24+16) x 9 (32)
= 672 kg
Total N2 in the blast furnace gas
% of volumetric analysis = 56 N2
Proportional mass of N2 = 56 x 28 = 1568 Kg.
Total weight of blast furnace gas:

= 384kg C + 26kg H2 + 672kg O2 + 1568kg N2
= 2650kgs
Gravimetric %age compositon:
C = (384/2650)x100 = 14.49%
H2 = (26/2650)x100 = 0.98%
O2 = (672/2650)x100 = 25.36%
N2 = (1568/2650)x100 = 59.17%
Example 7
The analysis of coal used in a boiler trial is as follows. 82% carbon, 6% hydrogen, 4% oxygen, 2%
moisture and 8% ash. Determine the theoretical air required for complete combustion of 1kg of
coal. If the actual air supplied is 18kg per kg of coal the hydrogen is completely burned & 80%
carbon burned to CO2 ,the reminder is CO, Determine the volumetric analysis of the dry products
of combustion.

Solution: For complete combustion.
O2 required is
For carbon - 0.82 = 2.186 kg of O2
For hydrogen - 0.006 = 0.48 kg of O2
Total O2 required = 2.666kg.
Net O2 supplied = Total O2 required O2 present in the fuel
= 2.66 0.004
= 2.662 kg/kg of coal
Theoretical minimum air required for complete combustion [C burns to CO2 totally]
Air supplied = 2.626x100/23 = 11.417 kg/kg of coal
Flue gas analysis:

But actually only 80% carbon is burns to CO2
CO2 = 0.8 x 0.82 x44/12 = 2.405kg of CO2
20% carbon is burnt to CO

CO = 0.2 x 0.82 x 28/12 = 0.383 kg of CO
O2 actually required for 80% carbon burnt to CO2

= 0.8 x 0.82 x 3232/12 = 1.749 kg of O2
O2 actually required for 20% carbon burnt to CO
= 0.2 x 0.82 x 16/12 = 0.219 kg of O2
O2 required by Hydrogen:
= 0.06 x 8 = 0.48 kg of O2.
H2O produced = 0.06 x 9 = 0.54 kg of H2O

But actual air supplied = 18kg
Actually O2 supplied = 18 x23/100 = 4.14 kg of O2
Free O2 in the flue gas = 4.14 + 0.04 1.749 0.219 0.48
= 1.732 kg of O2/kg of coal
N2 in the flue gas = 18x77/100 = 13.86 kg/kg of coal
Volumetric analysis of the dry products of combustion.

CO2 = (2.405/44)x 100 = 0.0546m3/K. mol
CO = (0.383/28) x 100 = 0.0137 m3/K.mol
O2 = (1.732/32)x 100 = 0.0541m3/K.mol
N2 = (13.86/28)x 100 = 0.495m3/K.mol
In % of volume:
CO2 = (0.0546/0.6174)x 100 = 8.84%
CO = (0.0137/0.6174)x 100 = 2.22%
O2 = (0.0541/0.6174)x 100 = 8.76%
N2 = (0.495/0.6174)x 100 = 80.70%
1.8 Enthalpy of reaction

Enthalpy of a reaction is defined as the difference between the enthalpy of the products at a
specified state and the enthalpy of the reactants at the same state for a complete reaction. For
combustion process, the enthalpy of a reaction is usually referred to as the enthalpy of
combustion it is obviously a very useful property for analyzing the combustion processes of fuels.
However there are so many different fuels and fuel mixtures that is not practical to list enthalpy of
combustion values for all possible cases. Besides, the enthalpy of combustion is not of much use
when the combustion is incomplete. Therefore a more practical approach would be have a more
fundamentally property to represent the chemical energy of an element or compound at some
reference state. This property is the enthalpy of formation which can be viewed as the enthalpy
of a substance at a specified state due to its chemical composition. To establish a starting point it is
assigned the enthalpy of formation for all stable elements such as O2, N2, H2 and C a value of zero
at standard reference state of 25oC and 1 atm. For all stable compounds.
In a chemical reaction bonds are broken in the reactants and new bonds formed in the products.
Energy is required to break bonds and energy is released when bonds are formed. The energy
associated with a chemical reaction depends on the number and type of bonds broken and/or formed.
Every chemical species has a certain amount of "heat content," or enthalpy, H, which cannot be
measured. However, differences in enthalpy can be measured. The net energy change for a reaction
performed at constant pressure is the enthalpy change for the reaction. This enthalpy change, H ,
has units kJ/mol and is defined:

[C + H (fuel)] + [O2 + N2 (Air)] -> (Combustion Process) -> [CO2 + H2O + N2 (Heat)]
where
C = Carbon, H = Hydrogen, O = Oxygen, N = Nitrogen
(1) H = H(products) H(reactants)
If energy is given off during a reaction, such as in the burning of a fuel, the products have less heat
content than the reactants and H will have a negative value; the reaction is said to be
exothermic. If energy is consumed during a reaction, H will have a positive value; the reaction
is said to be endothermic.
The enthalpy change for a chemical change is independent of the method or path by which the
change is carried out as long as the initial and final substances are brought to the same temperature.
This observation, known as HESS'S LAW, has important practical utility.
Thermochemical equations may be treated as algebraic equations: they may be written in the
reverse direction with a change in the sign of H even though the reverse reaction may not
actually occur; they may be added and subtracted algebraically; the equation and associated
H value may be multiplied or divided by factors. Hess's Law allows the calculation of enthalpy
changes that would be difficult or impossible to determine directly, i.e. by experiment.
The enthalpy change for the reaction:
(2) 2C (s) + O2 (g) 2CO (g)
cannot be determined directly because carbon dioxide will also form. However, H can be
measured for:
(3) C (s) + O2 (g) CO2 (g) H = 393.5 kJ
(4) 2CO (g) + O2 (g) 2CO2 (g) H = 566.0 kJ
Multiplying equation (3) by 2 gives equation (5), and reversing equation (4) gives equation (6):
(5) 2C (s) + 2O2 (g) 2CO2 (g) H = 787.0 kJ

(6) 2CO2 (g) 2CO (g) + O2 (g) H = +566.0 kJ
Adding equations (5) and (6) gives the desired information:
(2) 2C (s) + O2 (g) 2CO (g) H = 221.0 kJ
For a reaction in which a compound is formed from the elements, the enthalpy change is called
the heat of formation, Hf o, for the compound. The superscript "o" indicates standard conditions
of one atmosphere pressure. Equation (2) and (3) are such reactions. Some others:
(7) S (s) + O2 (g) SO2 (g) H = 296.9 kJ
(8) Mg (s) + Cl2 (g) MgCl2 (s) H = 641.8 kJ
In reactions (2), (3), (7), and (8) H for the reaction is Hf o for the compound. For the reaction:
(9) 2S (s) + 3O2 (g) 2SO3 (g) H = 790.4 kJ
the heat of reaction is associated with the formation of two moles of SO3. But heat of formation
is per mole of compound, so Hfo for SO3 is half of 790.4, or 395.2 kJ.
Extensive listings of heats of formation are available in handbooks. With these values of H of ,
you can calculate virtually any heat of reaction. The heat of a reaction is the sum of H f o values
for the products minus the sum of Hfo values for the reactants. Expressed as a formula:
(10) Hrxn = H f products- H f reactants

o o
Heats of formation for several compounds are given below. Note that the phase of the compound

is important when choosing a Hf

for a free element is zero.
o o
STANDARD HEATS OF FORMATION, H f , kJ/mole, at 25 C
AgCl (s) 127.1 Ca(OH)2 (s) 986.1 K3PO4 (aq) 2002.9

AgNO3 (aq) 100.7 Ca(OH)2 (aq) 1002.9 K2SO4 (aq) 1409.2
AlCl3 (s) 695.4 HCl (g) 92.3 MgCl2 (aq) 797.1
AlCl3 (aq) 1027.2 HCl (aq) 167.4 Mg(NO3)2 (aq) 875.1
Al(OH)3 (s) 1272.8 H2O (g) 241.8 NaCl (aq) 407.1
Al2(SO4)3 (aq) 3753.5 H2O (l) 285.8 NaHCO3 (s) 947.7
BaCl2 (aq) 873.2 H3PO4 (aq) 1294.1 NaNO3 (aq) 446.2
Ba(NO3)2 (aq) 951.4 H2SO4 (l) 814.0 NaOH (aq) 469.4
BaSO4 (s) 1473.2 H2SO4 (aq) 888.0 Na2SO4 (aq) 1387.0
CaCl2 (aq) 877.8 KOH (aq) 481.2 ZnCl2 (aq) 487.4
EXAMPLE: Using H
f data calculate the heat of reaction for:
o
(11) AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq)
H = [Hf AgCl (s) + Hf NaNO3 (aq)] [H

f AgNO3 (aq) + H
f NaCl (aq)]
o o o o
= [(127.0) + (446.2)] [(100.7) + (407.1)]
= [573.2] [507.8] = 573.2 + 507.8 = 65.4 kJ
EXAMPLE Using Hf data calculate the heat of reaction for

o f
(12) 2 AgNO3 (aq) + MgCl2 (aq) 2 AgCl (s) + Mg(NO3)2 (aq)
H = [2 Hf AgCl (s) + Hf Mg(NO3)2 (aq)] [2 f AgNO3 (aq) + H MgCl2

o o H o o (aq)]
= [2(127.0) + (875.1)] [2(100.7) + (797.1)]
= [1129.1] [998.5] = 1129.1 + 998.5 = 130.6 kJ

APPLIED THERMODYNAMICS
Note: the values off H o

chemical equation.
are multiplied by the stiochiometric coefficients from the balanced
(13) 2C (s) + O2 (g) 2CO (g) H = 221.0 kJ

(14) C (s) + O2 (g) CO2 (g) H = 393.5 kJ
(15) 2H2 (g) + O2 2H2O (g) H = 483.6 kJ
(16) 2H2 (g) + O2 2H2O (l) H = 571.8 kJ
(17) CaO (s) + CO2 (g) CaCO3 (s) H = 178.1 kJ

(18) 2Ca (s) + O2 (g) 2CaO (g) H = 1271.0 kJ
Using Hess' Law with appropriate equations from (13)-(18), above, calculate H for each
of the following reactions:
1) H2O (l) H2O (g)

2) C (s) + H2O (g) CO (g) + H2 (g)
3) Ca (s) + H2O (g) CaO (s) + H2 (g)
4) CO (g) + 1/2 O2 (g) CO2 (g)
5) 2Ca (s) + 2C (s) + 3O2 (g) 2CaCO3 (s)
Using heats of formation values from page T-56 calculate H for each of the following
reactions:
6) 2Al (s) + 3Cl2 (g) 2AlCl3 (s)

7) 2Al (s) + 3H2SO4 (aq) Al2(SO4)3 (aq) + 3H2 (g)
8) 2Al (s) + 3ZnCl2 (aq) 2AlCl3 (aq) + 3Zn (s)
9) 3BaCl2 (aq) + Al2(SO4)3 (aq) 3BaSO4 (s) + 2AlCl3 (aq)
10) Na2SO4 (aq) + BaCl2 (aq) BaSO4 (s) + 2NaCl (aq)
11) BaCl2 (aq) + 2AgNO3 (aq) 2AgCl (s) + Ba(NO3)2 (aq)
12) Ca(OH)2 (aq) + 2HCl (aq) CaCl2 (aq) + 2H2O (l)
13) 2Al(OH)3 (s) + 3H2SO4 (aq) Al2(SO4)3 (aq) + 6H2O (l)
14) AlCl3 (aq) + 3NaOH (aq) Al(OH)3 (s) + 3NaCl (aq)
15) 3KOH (aq) + H3PO4 (aq) K3PO4 (aq) + 3H2O (l)
Dr G PUNDARIKA, BMSCE Page 1

Answers to Problems
1) +44.1 kJ 9) 76.9 kJ
2) +131.3 kJ 10) 19.2 kJ
3) 393.7 kJ 11) 130.8
4) 283.0 kJ 12) 111.7
5) 2414.2 kJ 13) 258.7
6) 1390.8 kJ 14) 58.7 kJ
7) 1089.5 kJ 15) 122.6
8) 592.2 kJ
1.9 Internal Energy of Combustion: It is defined as the difference between the internal
energy of the products and the internal energy of the reactants when complete
combustion occurs at a given temperature and pressure.
Uc = Up UR
= p ne (hf + h pv ) - R ni (hf + h pv )
1.10 Combustion efficiency

It is defined as the ratio of ideal fuel-air to the actual fuel-air ratio
comb = (F/A) ideal

(F/A) actl
Example 8
Consider the following reaction, which occurs in a steady state, steady flow processes.
CH4 + 2O2 CO2 + 2H2O (l)
The reactants and products are each at total pressure of 0.1Mpa and 25oC. Determine the
heat transfer for per K mol of fuel entering the combustion chamber.
Solution : using the values of enthalpy of formation
Q = hf = p nehf - R nihf
R nihf = (hf) CH4 = -74873KJ

p nehf = (hf) CO2 + 2 (hf) H2O (l)
= - 393522+2 (-2852830) = - 965182KJ
Therefore Q = - 965182 (-74873) = - 890309KJ

Example 9
A small gas turbine uses C8H18 as fuel and 400% theoretical air. The air and fuel
enters at 25oC and the products of combustion leaves at 900K. The output of the engine
and the fuel consumption are measured and it is found that the specific fuel consumption
is 0.25kg/Sec of fuel per MW out put. Determine the heat transfer from the engine per K
mol of fuel. Assume complete combustion
Solution:
The combustion equation is
C8H18 + 4 (12.5)O2 + 4 (12.5) ( 3.76)N2 - 8CO2 + 9H2O + 37.5O2 + 188N2
By first law
Q + R ni (hf + h); = W + p ne (hf + h)
R ni (hf + h) = (hf) C8H18 = 250105KJ/K mol fuel at 25oC
Considering the products
p ne (hf + h) = nCO2 (hf + h) CO2+ nH2O (hf + h) H2O + nO2 (h)O2 + nN2 (h)N2
hf of O2, N2 = O h = Enthalpy of formation from 298oK to 900K
Therefore p ne (hf + h) = 8 (- 393522+288030)+9(-241826+21892)

+37.5(19249)+188(18222) = -755769KJ/K mol fuel.
W = 1000(KW) X 114 Kg = 456920KJ/K mol
0.25 K mol
Therefore Q = -755769+456920- (-250105) = -48744KJ/K mol fuel

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UNIT-1 COMBUSTION THERMODYNAMICS

The Kg-mole or gram-mole is widely used in combustion calculations as a unit of weight.

(i) C +O2 CO2

(ii) 2C +O2 2CO

(iii) 2C +O2 2CO2

2 Mol (CO) + 1 Mol (O2) 2 Mol (CO2)

1.2 Combustion Stoichometry

C + O2 + 3.76N2 CO2 + 3.76N2 1.4

1 Kg CO + 4/7 Kg O2 11/7 kg CO2 1.7

1K mol C + 1K mol O2 1K mol CO2

1.4 Conversion of Gravimetric analysis to volumetric basis and vice versa

Indivisual (relative) volume of the constituents X 100

will give the %age by volume of each constituents in the fuel.

Indivisual weight of the constituents X 100

will give the %age by weight of each constituent in the fuel.

C = 72.4%, H2 5.3%, N2 = 1.81, O2 = 8.5%, moisture 7.2%

0.724 x 32/12 = 1.93 kg

But 0.085kg O2 is available in coal, therefore O2 required

2.278x 100/23 = 9.9 kg per kg of coal

C = 72.4/12 = 6.03K mol, O2 = 8.5/32 = 0.265K mol

N2 = 1.8/28 = 0.064K mol, S = 0.9/32 = 0.028K mol

1K mol C + 1K mol O2 1Km CO2

Therefore 6.03 K mol of carbon requires

1 K mol H2 + K mol O2 1K mol H2O

The oxygen present in coal 0.265K mol

Net O2 required = 7.383 0.265 = 7.118K mol

7.118x100/21 = 33.89K mol / 100kg of coal = 0.3389K mol / 1kg coal

Volume of air supplied

The amount of N2 associated with this

40.67 x 79/100 = 32.13K mol

The amount of O2 present 40.67 x 21/100 = 8.54K mol

The volumetric composition of the gas

CO2 = (6.03/39.674) x 100 = 15.12%

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Excess air supplied

1.6 Dew point of products:

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1.7 Flue gas analysis

Carbon balance (C) 26 + 11 + 5 = a = H2

From (i) 11 + 2.5 + x = 100/2 + a + c

Determine the Ultimate gravimetric analysis

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The volumetric analysis of flue gas components becomes

Solution: Consider 1kg of fuel

Oxygen required / Kg of fuel

For burning of 1kg C - 0.849 x32/12

When 20% excess air supplied

Total air supplied = 13.93 x 1.2 = 16.716 kg/kg of fuel.

Total free N2 in fuel gas = 12.87 kg/kg of fuel

Flue gas analysis:

C converted to CO2 = 0.849x44/12 = 3.113 kg CO2

Flue gas / kg of fuel:

= 3.113 + 1.26 + 0.064 + 0.641 + 12.871

CO2 = (3.113/17.715)x100 = 17.573%

SO2 = (0.064/17.715)x100 = 0.36%

H2O = (1.026/17.715)x100 = 5.79%