CHAPTER
Fuels and Combustion
3
3.1 Review of Concepts Related to Combustion :
(a)Fuel :
Fuel can be defined as the source of heat energy which is released in a reactive system by chemical
or nuclear reactions.
(b)Classification of fuels :
The fuels are classified as solid, liquid or gaseous fuels.
Automotive engines generally use liquid fuels i.e. petrol and diesel fuels.
3.1.1 Chemical Reactions :
The main constituents of a combustible matter are carbon and hydrogen. To promote the
combustion of a fuel it reacts with O2 of air to form oxides of reacting elements.
The basic combustion equations can be written as :
(a)2H2 + O22H2O
2 moles of H2 + 1 mole of O22 moles of water vapour
i.e. 2 volumes of H2 + 1 volume of O22 volume of H2O
In terms of mass :
4 kg of H2 + 32 kg of O2 36 kg of H2O
or
1 kg of H2 + 8 kg of O29 kg of H2O ...(3.1.1)
(b) C + O2 CO2
1 mole of C + 1 mole of O21 mole of CO2
12 kg of C + 32 kg of O2 44 kg of CO2
(c) 2C + O22 CO
2 mole of C + 1 mole of O2 2 moles of CO
24 kg of C + 32 kg of O256 kg of CO
(d) S + O2SO2
1 mole of S + 1 mole of O2 1 mole of SO2
32 kg of S + 32 kg of O2 64 kg of SO2
i.e. 1 kg of S + 1 kg of O2 2 kg of SO2...(3.1.4)
(e) CH4 (Methane) + 2O2CO2 + 2H2O
1 mole of CH4 + 2 mole O2 1 mole of CO2 + 2 mole of H2O
(1  12 + 4  1 = 16 kg) of CH4 + 64 kg of O244 kg of CO2 + 36 kg of H2O
Above equations are helpful in calculating the amount of required to burn the fuel.
Instead of supplying O2, in most cases the fuel is burnt in presence of air consisting of
O2 and N2 mainly with small traces of other inert gases. With the known composition of air, the
mass of air can be determined for combustion of unit quantity of fuel.
3.1.2 Composition of Air :
Approximate composition of air is as shown in Table 3.1.1.
Table 3.1.1 : Composition of air
3.1.3 Mass Fraction :
It is defined as the ratio of mass of a constitutent of a mixture or a compound to the total
mass of mixture or a compound.
3.1.4 Mole Fraction :
The ratio of moles of a constitutent gas to the total moles of mixture of gases is called
the mole fraction.
3.1.5 Stoichiometric Air and Stoichiometric Mixture :
The theoretical amount of air required for complete combustion of unit quantity
of fuel is called Stoichiometric air.


A mixture of theoretical air and fuel is called Stoichiometric or Chemically correct
mixture.
3.1.6 Excess Air :

In practice the combustion of fuel is never complete due to non-homogeneity of
mixture.

In order to ensure complete combustion of fuel, usually, actual air supplied is more than
the theoretical air required for complete combustion of fuel. Then,
Excess air = Actual air – Stoichiometric (theoretical) air ...(3.1.8)
The magnitude of excess air supplied depends on the homogeneity of mixture,
turbulence and maximum pressure and temperature to be attained in combustion process.
3.1.7 Rich Mixture, Lean Mixture and Mixture Strength (Equivalence Ratio) :


If actual A.F. ratio is more than stoichiometric A.F. ratio, the mixture is said to lean or
weak mixture.

If actual A.F. ratio is less than stoichometric A.F. ratio, the mixture is called rich
mixture.
Mixture strength or equivalence ratio is defined as the ratio of stoichiometric A.F.
ratio to actual A.F. ratio. Therefore,

3.2 Determination of Minimum Air Required Per kg of Solid or Liquid Fuel for Complete
Combustion :
Let us consider 1 kg of fuel, the ultimate analysis of which shows that carbon is ‘C’ kg,
hydrogen is ‘H’ kg, sulphur is ‘S’ kg and oxygen is ‘O’ kg. The amount of oxygen required
can be computed with the help of combustion.
Equations (3.1.1) to (3.1.5) as follows :
Similarly,
H kg of hydrogen requires (8 H) kg of O2
S kg of sulphur requires (S) kg of O2
 Total oxygen required = C + 8H + S kg
(O) kg of oxygen is already present in the fuel.
Therefore, minimum oxygen required for complete combustion
 Minimum air required becomes,
Since air contains 23% of oxygen by mass in air.
Ex. 3.2.1 :Determine the air-fuel ratio and the theoretical amount of air required by mass for complete combustion
of a fuel containing 85% carbon, 8% hydrogen, 3% oxygen, 1% sulphur and remaining is ash. If 40%
of excess air is used, what volume of air at 27 C and 1.05 bar pressure, does this represent per kg of

fuel ?
Soln. :
Given :C = 85%, H2 = 8 %, O2 = 3 %, S = 1%
Actual air supplied = 40% in excess of theoretical air, T = 27ºC = 300 K, p = 1.05 bar
Minimum air required per kg of fuel,
...According to Equation (3.2.2)
= 12.55 kg/kg of fuel...Ans.

Air fuel ratio by mass = 12.55 : 1...Ans.
Since, air supplied is 40% in excess.
The actual amount of air supplied = 1.4  12.55 i.e. 17.57 kg.
It’s volume occupied can be calculated as follows (by using gas equation) :
pV = mRT
 (1.05  105)V = 17.57  287  (273 + 27)
 V= 14.407 m3...Ans.
3.3 Determination of Flue Gas Analysis by Mass and by Volume :

The main constituents of the products of combustion (flue gas) of a fuel are CO, CO2,
H2O (water vapour), SO2 and nitrogen of the air supplied.

If an excess air is supplied, oxygen will also be present as one of the constituent in the
flue gases.

The mass of each constituent of the flue gas can be determined with the help of
combustion Equations (3.1.1) to (3.1.5) and each can be expressed as percentage of the total
flue gas. It would represent the mass analysis of the wet flue gases.
The flue gas analysis by volume can be determined with the help of Avogadro’s law
from the mass analysis by dividing the percentage mass analysis of each constituent by their
respective molecular mass.

In case of dry flue gas analysis the constituents water vapour (H2O) and sulphur
dioxide (SO2) are not considered since these are condensed at N.T.P. conditions.

(a)Procedure of conversion of mass analysis of flue gases into volumetric analysis :
The mass analysis of flue gases can be converted into volumetric analysis by using
Avogadro’s law. The procedure is :
Step 1 :Divide the percentage of mass of each constituent by their respective molecular mass. This
will represent the proportionate volume of each constituent in the mixture of flue gases, since,
Step 2 :Add all the volumes so obtained.
Step 3 :Divide the volume of each constituent by total volume and express it as percentage. It gives
the percentage volumetric analysis of flue gases.
(b)Procedure of conversion of volumetric analysis into mass analysis or gravimetric
analysis :
therefore,
mass= Number of moles  molecular mass
Using this concept, the volumetric analysis can be converted into mass analysis by
following method :
Step 1 :Multiply the percentage volume of each constituent by their respective molecular mass. It
gives the proportionate mass of each constituent in the flue gases.
Step 2 :Add the proportionate mass of all the constituents to get total mass.
Step 3 :Divide the mass of each constituent by total mass and express it as percentage.
This gives the mass analysis of flue gases.
Ex. 3.3.1 :The percentage analysis by mass of a solid fuel is as follows :
C = 87%, H2 = 3%, O2 = 3%, N2 = 1%, S = 1% and the remainder is ash. If 50% excess air
is supplied. Find :
(a)The theoretical amount of air required for complete combustion of fuel.
(b)The volumetric analysis of wet flue gases and the mass of flue gas per kg of fuel.
Soln. : Refer Equations (3.1.1) to (3.1.5).
(a)Mass of air required for complete combustion :
(b) N2 present in theoretical air :
Since 50% of excess air is supplied, following will be the masses of each constituent in
the products of combustion.
CO2 = 3.19 kg; SO2 = 0.02 kg; H2O = 0.27 kg
O2 from 50% excess air = 0.5  (2.57 – 0.03) = 1.27 kg
N2 in flue gases = N2 in air + N2 in fuel
 Total mass of flue gases = Mass of (CO2 + SO2 + H2O + O2 + N2)
= 3.19 + 0.02 + 0.27 + 1.27 + 12.76
= 17.51 kg/kg of fuel ...Ans.
The mass analysis of the wet flue gases becomes
Total = 100.00%
The mass analysis of flue gases can be converted into volumetric analysis as follows :
Ex. 3.3.2 :A fuel has the following percentage composition by mass :
C = 85%, H2 = 15%
Determine :
(i)The stoichiometric mass of air required for complete combustion of fuel.
(ii)The percentage composition by mass of dry products of combustion.
Air contains 23.2% oxygen by mass.
Soln. :
Given :
(i)To find stoichiometric mass of air :
Chemical reactions are :
Consider 1 kg of fuel. Therefore,
mass of carbon in fuel = 0.85 kg
mass of H2 in fuel = 0.15 kg
From Equation (i)
From Equation (ii)
1 kg of H2 requires 8 kg of O2 and produces 9 kg of H2O .
0.15 kg of H2 requires 8  0.15= 1.2 kg of O2
and it produce 9  0.15 = 1.35 kg of H2O
Hence, stoichiometric O2 = 2.267 + 1.2 = 3.467 kg
 Stoichiometric air required
(ii)To find percentage composite by mass :
Mass of constituents in dry flue gases are :
CO2 = 3.117 kg

Total mass of dry flue gases = 3.117 + 11.477 = 14.594 kg.
Mass analysis is
Ex. 3.3.3 :The petrol used in an SI engine is assumed to have a chemical formula C H . Determine : (a) the
7
16
stoichiometric A : F ratio and (b) If 50% excess air is supplied then find the volumetric composition of
dry exhaust products. Air contains 23% of O and 77% of N by Mass. (GTU - May 2011, 7 Marks)
2
2
Soln. :
Numerical :Fuel C7 H16 ;
1 kg of Air contains : O2 = 0.23 kg, N2 = 0.77 kg
(a)Stoichiometric A : F. ratio :
Complete Chemical reaction of fuel C7 H16 with air can be written as :
C7 H16 + 11 O2 + 11  3.76 N2
 7 CO2 + 8 H2 O + 11  3.76 N2
i.e. C7 H16 + 11 O2 + 41.36 N2 = 7 CO2 + 8 H2 O + 41.36 N2…(i)
 Stoichiometric A : F ratio,
(b)Volumetric composition of dry exhaust gases if 50% excess air is supplied.
Chemical reaction equation (i) with 50% excess air can be written as :
C7 H16 + 11  1.5 O2 + 41.36  1.5 N2 = 7 CO2 + 8 H2 O + 11 0.5O2 + 41.36  1.5 N2
i.e.C7
H16 + 16.5 O2 + 62.04 N2 = 7 CO2 + 8 H2 O + 5.5 O2 + 62.04 N2
The no. of moles of dry products of combustion are
…Ans.
Ex. 3.3.4 :Petrol (by mass 85% carbon and 15% hydrogen) is burned with 14 times its mass of air, which is
insufficient for complete combustion. Assume that all the hydrogen is burned, no carbon is deposited
and that there is no free oxygen in the exhaust, find the mass of each of the gases in the exhaust per
kg of fuel.
Soln. :
Given :
Let C1 mass of carbon is burned to CO2 and remainder i.e. (0.85 – C1) kg of C is burned
to CO.
The chemical reactions can be written as :
C + O2 CO2
12 kg of C + 32 kg of O2 44 kg of CO2
12 kg of C + 16 kg of O2  28 kg of CO
2 kg of H2 + 16 kg of O2 18 kg of H2O
1 kg of H2 + 8 kg of O2  9 kg of H2O
0.15 kg of H2 + (8  0.15) kg of O2  (9  0.15) kg of H2O ...(iii)
 Total O2 required
= (5.797 C1 + 1.449) kg...(iv)
 Mass of gases in exhaust from Equations (i) to (iv) are :
H2O = 9  0.5 = 1.35 kg...Ans.
Ex. 3.3.5 :The percentage analysis of gaseous fuel by volume is given as follows :
CO = 8%, CO = 22%, O = 4%, H = 30% and N = 36%.
2
2
2
2
Determine the minimum volume of air required for complete combustion of
1 m of gas and calculate the percentage composition by volume of the dry products of
3
combustion.
If 1.4 m of air is supplied per m of gas, what will be the percentage by volume of CO in
3
3
the dry products of combustion.
Soln. :
Given : The volume of constituents per m3 of fuel gas is,
CO2 = 0.08 m3, CO = 0.22 m3, O2 = 0.04 m3 , H2= 0.3 m3 and N2 = 0.36 m3
2
The chemical reaction and the volumes of O2 required for combustion with the volumes
of products are given in the table below.
Minimum volume of air required,
= 1.0476 m3/m3 of fuel gas…Ans.
Products of combustion per m3 of fuel gas are :
CO = 0.3 m3, H O = 0.3 m3
2
2
= 1.1876 m3
 Total volume of products of combustion is
= 0.3 + 0.3 + 1.1876 = 1.7876 m3 /m3 of fuel gas.
If 1.4 m3 of air is supplied then,
Volume of CO2, H2O and N2 in products are,
CO2 = 0.3 m3,
H2O = 0.3 m3
Total volume of products of combustion =0.3 + 0.3 + 1.466 =2.066 m3
3.4 Necessity of Exhaust Gas Analysis :

During the discussion in preceding sections, the calculation of air-fuel ratio and the
products of combustion were based on the premise that complete information is available on
the reactants entering into the combustion process.

It was also assumed that all the carbon will burn to CO2 in the presence of excess air.

However, it is a common experience based on experimental measurements that the
products of combustion carry carbon monoxide and hydrocarbons which may be due to
incomplete combustion of fuel or dissociation of gases at high temperatures even if the
excess air has been used.

Also, in many practical situations and applications, it is difficult to obtain the actual
air-fuel ratios.

To overcome these problems, fairly accurate air-fuel ratio can be determined if the
exhaust gas analysis is determined experimentally.

There is a further need to control the air-fuel ratio to strictly maintain the air-pollution
emissions set by the state and central government authorities.

Therefore, the exhaust gas analysis is necessary for the following reasons :
(a)To control exhaust emissions like CO and HC
(b)To determine emissions fairly accurate air-fuel ratios.
(c)To ensure complete combustion of fuel as far as possible.
(d)To control the maximum temperatures of combustion products by varying air-fuel ratio.
(e)Efficient operation of combustion chambers and work out the heat energy balance in I.C.
engines, boilers etc.
The exhaust gas analysis can be carried out with the help of Orsat’s apparatus or using
other equipments like Non-Despersive Infra-Red analysis (NDIR), Flame Ionization Detector
(FID), flue gas analyser in boilers etc.
3.5 Flue Gas Analysis (By Orsat Apparatus) :
A simple and convenient apparatus used for analysis of dry flue gases by volume is called
as Orsat Apparatus shown in Fig. 3.5.1. It consists of a graduated measuring bottle also called
as eudiometer, an aspirating bottle, three double reagent pipettes to absorb CO2, O2 and CO.
Eudiometer is connected to aspirating a bottle by means of rubber tube.

The first reagent pipette next to eudiometer contains the KOH solution
(33% KOH + 67% water by weight) to absorb CO2. The second pipette contains alkaline
solution of pyrogallic acid (5 gm of pyrogallic acid in 15 cc of water + 33 gm of KOH in
67 gm of water) to absorb O2 and third consists of acidic solution of cuprous chloride
(5 cc of CuO in 100 cc of HCl) to absorb CO.

Each pipette is made of two pipettes joined together by a glass tube at the bottom as
shown in Fig. 3.5.1. Front row of pipettes are connected to the front flue gas sample tube and
rear row of pipettes are connected to another tube which runs parallel to flue gas sample
tube. Each front pipette is provided with number of glass tubes inside to increase the wetted
surface area, hence, it accelerates the absorbing action of the solution.
Fig. 3.5.1 : Orsat Apparatus

Firstly, the existing air or the gas is expelled from the eudiometer by raising the
aspirating bottle and by keeping three way cock open to atmosphere. The cocks a, b, c and
three way cock are closed. The aspirating bottle is lowered so that the eudiometer reads zero
on graduated scale. The three way cock is open and the sample of flue gas is drawn in the
measuring bottle. The flue gas is expelled by raising the aspirating bottle and the fresh flue
gas measuring 100 cc is again drawn in by lowering the aspirating bottle. Now the three way
cock is kept in the closed position.
Now the cock ‘A’ is opened and the sample of flue gas is forced into the first reagent
pipette containing KOH by raising the aspirating bottle. Here the CO2 component of the flue
gas is absorbed. Aspirating bottle is moved up and down several times to ensure the complete
absorption of CO2 by KOH solution in reagent pipette. The sample of gas is returned to
eudiometer and cock ‘A’ is closed by keeping the reagent to its original level of solution.
The eudiometer reading is taken and the difference of two readings gives the percentage of
CO2 by volume in the gas sample.

The same procedure is repeated with the pipettes ‘B’ and ‘C’ to find the percentage
analysis of O2 and CO respectively in the flue gas.


When the percentage of CO2, O2 and CO have been determined, the remainder of the
gas sample is assumed to be nitrogen, N2 .

The gases must be absorbed in the order of CO2 and O2 and CO since pyrogallic acid
solution will absorb O2 as well as CO2 and cuprous chloride solution absorbs CO2, O2 and
CO.

It should be noted that the Orsat apparatus gives the analysis of dry flue gases since the
water is condensed at atmospheric conditions.

Additional pipettes may be provided for determination of other hydrocarbons and H2,
present due to incomplete combustion of fuel.
3.6 To Determine the Air Supplied From Volumetric Analysis of Dry Flue Gases :
(a)First Method :
The quantity of air supplied to the combustion system cannot be readily determined by
direct measurements. Indirectly the air-fuel ratio may be determined if the volumetric analysis
of the dry flue gases is known (approximately).
Given :
Let C% be the mass of carbon in the fuel and the analysis of dry flue gas by volume is
given as :
CO2 = C1 %; CO = C2 %; O2 = O% and N2 = N%.
The relative masses of the flue gases will be as follows,
Since 44 kg of CO2 is obtained from 12 kg of carbon
 44 C1 kg of CO2 would be obtained from
Similarly, 28 C2 kg of CO is produced by,
 Total mass of carbon in flue gases = 12 C1 + 12 C2 = 12 (C1 + C2) kg
It follows that,
Neglecting the mass of nitrogen in fuel, the above ratio will also represent the ratio of
N2 in air supplied per kg of fuel. Therefore,
Therefore, approximately the ratio of excess air to the total air is the ratio of N2 in excess
air to the total air, i.e.
Note :Equation (3.6.2) gives the excess air without taking into account the burning of C to CO.
(b)Second Method :
In case the mass analysis of the fuel and the volumetric composition of dry flue gas is
given, following procedure can be adopted to determine the air supplied per kg of fuel :
Step 1 :Determine the stoichiometric air required per kg of fuel from the given mass analysis of the
fuel.
Step 2 :If O2 appears in the volumetric analysis of dry flue gases, it shows that the air supplied is in
excess.
Step 3 :Convert the volumetric analysis of flue gases into mass analysis.
Step 4 :Consider 1 kg of dry flue gas by mass and determine the mass of carbon associated per kg of
dry flue gas (DFG) by the method explained above in method (a).
Step 5 :Determine the mass of dry flue gases per kg of fuel, as follows :
Step 6 :Calculate the mass of water vapour formed per kg of fuel from the known mass of hydrogen
in fuel (if given).
Step 7 :If ma is the mass of air supplied per kg of fuel, it can be calculated as follows :
ma + mass of fuel = mass of DFG + mass of water vapour
...(3.6.3)
Alternatively, mass of air can also be calculated from the mass analysis of dry flue gases as follows
:
(Mass
ma 0.77
of
N2 in DFG/kg
...(3.6.4)
of
DFG)  (Mass
of
DFG/kg
Step 8 :Mass of excess air supplied = Mass of air supplied – Stoichiometric air
of
fuel)
=
...(3.6.5)
This method has been explained with the help of Examples 3.6.2 and 3.6.3.
Ex.
3.6.1
:In
a
boiler
trail
the
composition
of
dry
flue
gases
by
volume
is,
CO =
2
11.5%,
CO = 2.5%, O2 = 5% and N2 = 81% (by difference). If the fuel used has 80% of carbon by mass,
determine the air-fuel ratio and the excess air supplied per kg of fuel.
Soln. :
Given : C1 = 11.5,
C2 = 2.5, O = 5
and N = 81, C = 80.
To find air-fuel ratio :
Using Equation (3.6.1),
Air-fuel ratio = 14.026 : 1...Ans.
To find air-fuel ratio :
Ex. 3.6.2 :The constituents of coal by mass are,
C = 80%, H = 6%, O = 7% and ash = 7%
2
2
Orsat's analysis of dry products of combustion by volume is as follows :
CO = 10%, CO = 1.6%, O = 8% and N = 80.4%
2
2
2
Determine the following :
(a)Mass of dry flue gases in kg/kg of fuel.
(b)Air supplied per kg of fuel.
(c)Stoichiometric air supplied per kg of fuel and percentage of excess air supplied.
(d)Heat carried away by products of combustion if flue gas exit temperature is 400 C. C for
pg
dry
flue
gases
=
1.026
kJ/kg
K
and
C for
ps
steam
= 2.1 kJ/kg K.
Air temperature = 30C.
Soln. :
The volumetric analysis of dry flue gases can be converted into mass analysis as
follows :
(a)To find mass of dry flue gases in kg/kg of fuel :

Since 44 kg of CO2 contains 12 kg of carbon and 28 kg of CO contains 12 kg of
carbon, it follows that the total carbon associated per kg of dry flue gases (DFG) will
be,
= 0.0401 + 0.0064 = 0.0465 kg/kg of DFG
Fuel contains 0.8 kg of carbon per kg of fuel. Therefore,
= 17.2 kg …Ans.
(b)Air supplied per kg of fuel :

Fuel also contains 0.06 kg of H2 per kg of fuel which will produce,

From the law of conservation of mass,

Mass of fuel + Mass of air supplied, ma per kg of fuel
= mass of DFG/kg of fuel + mass of H2O


(0.8 kg of C + 0.06 kg of H2 + 0.07 kg of O2) + ma = 17.2 + 0.54
ma = 16.81 kg/kg of fuel…Ans.
Alternative method :
(Mass of N2 in DFG/kg of DFG)  (Mass of DFG/kg of fuel) = ma 0.77
 0.752407  17.2 = ma  0.77
 Mass of air supplied, ma = 16.8 kg/kg of fuel …Ans.
Alternatively,
Mass of air supplied/kg of fuel can also be calculated with the help of Equation (3.6.1)
as follows :
(c)Stoichiometric air supplied/kg of fuel :
Percentage of excess air supplied
(d)Heat carried away by flue gases :
= Heat carried away by (DFG + water vapour)
= mg  Cpg (t – t0) + ms  Cps (t – t0)
= 17.2  1.026 (400 – 30) + 0.54  2.1  (400 – 30)
= 6949 kJ …Ans.
Ex. 3.6.3 :A fuel oil has the mass composition as :
C = 85%, H = 13%, O = 2%
2
2
The dry exhaust gases has the following volumetric percentage composition.
CO = 9%, CO = 1.5%, O = 7%, N = 82.5%
2
2
2
Determine the mass of air supplied per kg of fuel and the percentage excess air supplied.
Soln. :
Theoretical O2 required/kg of fuel can be calculated as follows :
 Total O2 air required = 2.267 + 1.040 – 0.020 = 3.287 kg/kg of fuel
Volumetric analysis of dry flue gases can be converted into mass analysis as follows :
= 0.042394 kg/kg of dry flue gas
Mass of C in fuel = 0.85 kg/kg of fuel
Mass of excess O2 per kg of DFG
…(By allowing for unburnt CO)
Mass of air supplied = Stoichiometric air + excess air
= 14.291 + 5.867
= 20.158 kg/kg of fuel…Ans.
= 41.05 %…Ans.
3.7 To Determine A.F. Ratio with the Help of Exhaust Gas Analysis using Carbon Hydrogen Balance Method :

In case of gas turbines, I.C. engines, jet propulsion etc., the combustion may take place
with excess or insufficient air while the time available for combustion is very short being
high speed machines.

In such cases, the combustion is never complete and the products of combustion in
exhaust may have some amount of methane, hydrogen, carbon monoxide and other
hydrocarbon. This may be due to :
(i)Intermediate reactions during combustion process.
(ii)Dissociation of gases at high temperatures.
(iii)Re-association of gases at the time of exhaust when temperatures are low.

The exhaust gas analysis i.e. the analysis of products of combustions can be carried out
by having more absorbing reagent pipettes in case of Orsat's apparatus or by special
techniques like by infrared gas analyser with recorder.

The A.F. ratio can be determined from the known exhaust gas analysis by special
methods called as Carbon-hydrogen balance method or by other methods. This method is
fairly accurate. This method can be applied in the following cases :
1.When the chemical formulae of the hydrocarbon fuel is known (Refer Ex. 3.7.1).
2.When the chemical formulae of the hydrocarbon fuel is unknown, say, it is Cx Hy .
Procedure to be followed :
Step 1 :Write the chemical reaction with percentage of exhaust gases written on right side.
Step 2 :Add some known H2O moles, say, z H2O since the Orsat apparatus gives the analysis of dryflue gases only.
Step 3 :Balance the moles of C and H2 on both sides.
Following examples will illustrate the C – H2 balance method for determining the A.F.
ratios.
Ex. 3.7.1 :A fuel oil C H is burnt in air at atmospheric pressure. The Orsat analysis of the products of combustion
12
26
yields :
CO2 = 12.8%, O2 = 3.5%, CO = 0.2%, N2 = 83.5%
Determine the air-fuel ratio, the stoichiometric air required and combustion equation.
Soln.. :
In this problem, we shall employ other methods.
To find air-fuel ratio :
Let ‘x’ moles of C12 H26 are burnt with ‘y’ moles of air consisting of y moles of O2 and
3.76 y moles of N2 which produces the products of combustion by volume as given. Assume
that ‘z’ moles of water vapour are condensed during measurement of exhaust analysis.
The combustion equation can be written as :
x C12 H26 + yO2 + 3.76 yN212.8 CO2 + 3.5 O2 + 0.2 CO + 83.5 N2 + zH2O
By carbon balance,
By N2 balance,
3.76 y = 83.5; y = 22.2
By H2 balance,
Therefore, the combustion equation can be written as,
C12 H26 + 22.2 O2 + 3.76  22.2 N212.8 CO2 +3.5 O2 + 0.2 CO + 83.5 N2 + 14.08 H2O
Based on 1 mole of fuel burnt, the equation becomes,
C12 H26 + 20.492 O2 + 20.492  3.76 N211.815 CO2 + 3.23 O2 + 0.1846 CO + 77.077 N2 +
12.996 H2O …Ans.
= 16.548 : 1 …Ans.
To find stoichoimetric air :
Stoichiometric combustion equation will be,
C12 H26 + 18.5 O2 + 18.5  3.76 N2 12 CO2 + 13 H2O + 18.5 3.76 N2
= 14.939 …Ans.
Ex. 3.7.2 :Determine the air-fuel ratio by mass used in an engine if the products of combustion have the following
volumetric analysis :
CO = 6%, O = 0.2%, CO = 13.4%, H = 8%
2
2
2
CH = 0.5% and N = 71.9% (by difference)
4
2
Soln. :
In this problem, the combustion of fuel is not known. Let the fuel used is a hydrocarbon
of chemical formulae, Cx Hy. The combustion equation can be written as,
Since 71.9% N2 present must be due to air supplied, then, 71.9/3.76 moles will be of
O2 associated with air.
Carbon balance :
x = 6 + 13.4 + 0.5 = 19.9
Hydrogen balance :
So as to determine the number of moles of H2O, we can apply theoxygen balance as
follows :
 z = 12.4; (It represents the number of moles of H2O formed)
Now balancing the hydrogen atoms on both sides we get,
y = 8  2 + 0.5  4 + 12.44  2 = 42.88
The combustion equation becomes,
+ 71.9 N2 + 12.4H2O
Ex. 3.7.3 :The products of combustion of hydrocarbon burnt in air are analysed by an orsat apparatus. It gave the
following
percentage
composition,
CO =
2
9%,
CO = 1%, O = 8%, N = 82%.
2
2
Determine the air-fuel ratio from the analysis given above and composition of fuel on mass basis. Also, calculate the
percentage of excess air supplied.
Soln. :
Given :Percentage composition of,
CO2 = 9%, CO = 1%, O2 = 8%, N2 = 82%.
To find air fuel ratio :
Let the fuel be Cx Hy. The combustion equation is,
By carbon balance,
x= 9 + 1 = 10
By oxygen balance,
or,
z = 8.61
Using hydrogen balance,
y = 2z = 2  8.61 = 17.22
Therefore, the combustion equation is,
To find composition of fuel on mass basis :
To find excess air supplied :
= 14.504 kg/kg of fuel …Ans.
= 50.42% …Ans.
Ex. 3.7.4 :An unknown hydrocarbon fuel has the following Orsat analysis :
CO = 12.5%, CO = 0.3%, O = 3.1%, N = 84.1%
2
2
Determine :
(i) Air-fuel ratio
(ii) Fuel composition on mass basic
(iii) Percentage of excess air.
Soln. :
Combustion equation is,
By carbon balance,
x = 12.5 + 0.3 = 12.8
By oxygen balance,
By hydrogen atoms balance,
y = 2z = 2  13.234 = 26.468
Therefore the combustion equation is,
2
C12.8 H26.468 + 84.1 N2 + 22.367 O2  12.5 CO2 + 0.3 CO + 3.1 O2 + 84.1 N2 + 13.234 H2O
(i)Air-fuel ratio :
(ii)Fuel composition on mass basis :
(iii)Percentage of excess air :
Stoichiometric air required
Ex. 3.7.5 :A petrol engine uses Octane (C H ) as fuel. The measurements of fuel and air shows the air-fuel ratio as
8
18
17.025 : 1. Calculate the stoichiometric air-fuel ratio, mixture strength and volumetric analysis of
products of combustion.
Soln. :
The combustion equation can be written as :
C8H18 + 12.5 O2 + 12.5  3.76 N2 8 CO2 + 9 H2O + 12.5 3.76 N2…(i)
= 15.053 : 1 …Ans.
Actual air-fuel ratio = 17.025 : 1 (given)
= 0.8842 or 88.42% …Ans.
Since the mixture strength is less than 100%, it shows that the mixture is weak i.e. air
supplied is in excess.
Actual combustion Equation (i) can now be modified as :
C8H18 + 1.131  12.5 O2 + 12.5  1.131  3.76 N2  8 CO2+ 9 H2O + 0.131  12.5 O2 +
12.5  1.131  3.76 N2
i.e. the products of combustion are :
Ex. 3.7.6 :The analysis of dry exhaust from an internal combustion engine gave 12% CO , 2% CO, 4% CH , 1% H ,
2
4
2
4.5% O and the remainder nitrogen. Calculate proportion by mass of carbon to hydrogen in fuel
2
assuming it to be a pure hydrocarbon.
Soln. :
Given : Products of combustion :
CO2 = 12% CO = 2% CH4 = 4% H2 = 1% O2 = 4.5% and remainder N2
 N2 = 100 – (12 + 2 + 4 + 1 + 4.5) = 76.5%
Let the hydrocarbon fuel be Cx Hy. The chemical reaction can be written as by assuming
hydrogen produces z moles of H2O vapour.
Cx Hy + 20.346 O2 + 76.5 N2  12 CO2 + 2 CO + 4 CH4 + 1H2 + 4.5 O2 + 76.5 N2 + z H2O
By carbon balance :
x = 12 + 2 + 4 = 18 …(i)
By hydrogen (H) balance :
y = 4  4 + 1  2 + 2z = 18 + 2z…(ii)
By O2 balance :
or, z = 5.69 …(iii)
Substituting the value of z from Equation (iii) in Equation (ii)
y = 18 + 2  5.69 = 29.4
 Combustion equation becomes :
C18 H29.4 + 20.346 O2 + 76.5 N2  12 CO2 + 2CO + 4 CH4 + 1 H2 + 4.5 O2 + 76.5 N2
+ 5.69 H2O
3.8 Calorific Value or Heating Value of Fuels :
If a unit quantity of fuel is burnt in presence of oxygen, the heat energy is liberated. The
heat energy released is measured with the help of calorimeters by cooling the products of
.combustion upto S.T.P.(Standard temperature and pressure taken as 25C and 1
atmospheric pressure).
Definition of calorific value of fuel :
Therefore the calorific value of a fuel can be defined as the heat energy released during
complete combustion of unit quantity of fuel when the products of combustion are cooled back
at S.T.P. Its unit are kJ/kg or kJ / kgmole (for solid and liquid fuels) or kJ / m3 (for gaseous fuels).
The fuels containing hydrogen as one of the constituent will produce water vapour during
combustion process. During the calorimetric measurements when the products of combustion
are cooled back to their standard temperature, the enthalpy of vapourisation of steam would be
given off to the water jacket of the calorimeter. The calorific value so obtained is
called higher or gross calorific value (H.C.V.) of fuels.
Therefore, higher calorific value (HCV) is defined as the amount of heat energy
released due to complete combustion of unit quantity of fuel when the products of combustion
are cooled back to STP and water vapour is condensed.
Whereas, the lower calorific value (L.C.V.) of the fuels is a fictitious quantity of heat that
would be obtained due to combustion of unit quantity of fuel if the water vapour formed in the
products of combustion are cooled back to 25C at atmospheric pressure and still remains in
gaseous state.
The relation between higher and lower calorific value of fuels is as follows :
– Lower calorific value (L.C.V.) = – Higher calorific value (H.C.V.) –
mw  L
...(3.8.1)
where, mw = Mass of water vapour formed per kg of fuel
L = Latent heat of vaporisation and part of sensible heat
L = 2445 kJ/kg of water vapour at constant pressure
L = 2308 kJ/kg of water vapour at constant volume
The calorific value of some of the fuels are given in table in kJ/kg mole.
The calorific value of fuel per kg will be,
Note :In actual combustion problems e.g. in case of boilers, internal gas turbines, combustor of gas turbines,
furnaces etc., the gases are never cooled upto the atmospheric temperature. It implies that the minimum
heat energy released during the combustion process corresponds to lower heating value of fuels. For
this reason in all thermal designs, we always take L.C.V. of fuels rather than H.C.V. of fuels. Such an
assumption ensures that we can obtain the expected power output and thermal efficiency of the engine.
3.9 Calorific Values of Fuels at Constant Pressure and at Constant Volume :

A combustion process can be carried in many ways, however, the combustion processes
at constant pressure and at constant volume are of importance to us.

If a combustion process is carried out at constant pressure, the heat energy released
when the products of combustion are cooled back to their standard conditions is called
as heating value of fuel at constant pressure, Qp.

Similarly, if the combustion process is carried out at constant volume, the heat energy
released represents the heating value of fuel at constant volume, Qv.
3.9.1 Relationship between Qp and Qv :
Applying 1st law equation to combustion process, neglecting changes in K.E. and P.E.’s.
dQ – dW = dU...(i)
For a constant volume process :
d Qv – 0 = dU
or,
Qv = U = Up – Um...(ii)
where, Qv = heating value of fuel at constant volume
Up = total internal energy of products of combustion
Um = total internal energy of reactants or mixture before combustion.
Here, we have introduced the term total internal energy (U) so as to distinguish from
the sensible internal energy ( Us ) which is the function of temperature alone. The total
internal energy is the sum of sensible internal energy and the chemical energy (  ).
Therefore,
U = Us + ...(iii)
Using Equation (iii), the Equation (ii) can be rewritten as :
Qv = ( Usp + p ) – ( Usm + m )
But, p = 0 since the chemical energy of products has already been released because it
is assumed that the complete combustion of fuel has taken place.
 Qv = Usp – ( Usm + m )...(iv)
For a constant pressure combustion process :
From Ist law Equation (i)
dQp – p  dV = dU
Qp –  p  dV = Up – Um
or,
Qp – p ( Vp – Vm ) = Usp – ( Usm + m )...(v)
Qp = ( Usp + p  Vp ) – ( Usm + pVm + m ) = Hsp – ( Hsm + m )...(vi)
Also,
Subtracting Equation (iv) from Equation (v) we get,
Qp – Qv = p  Vp – p  Vm...(vii)
pV = nRT
But,
where,
n = number of moles,
R = universal gas constant = 8.3143 kJ / kgmole K
 p  Vp = np  RT;p  Vm = nm  RT
Substituting the above quantities in Equation (vii) and on solving we get,
Qp – Qv = ( p – m )RT...(3.9.1)
It should be noted that Qp and Qv both represent the higher heating of fuels in the
Equation (3.9.1) above.
Note :1.Q and Q are always negative since the heat is transferred from the system to the surroundings in an
p
v
exothermic reaction.
2.Q and Q both represent higher heating values of fuels in Equation (3.9.1)
p
v
since the water vapour ( H O ) formed is condensed at S.T.P. from gaseous state. Hence, H O gives away its
2
2
latent heat of vaporisation and part of it’s sensible heat.
3.If H O at S.T.P. is assumed to be gas, the Q and Q will represent their lower heating value or lower calorific
2
p
v
value of fuel (LCV). It represents a fictitious quantity. However, LCV is used in design calculations for
safe design of systems.
4.Relationship between H.C.V. and L.C.V. are as follows (in kJ/kgmol).
– ( HCV ) = – ( LCV ) + m  2308 ( in kJ /kg
V
V
w
mole
– ( H.C.V ) = – ( L.C.V ) + m  2445 ( in kJ /kg
p
p
w
where, m is the mass of vapour formed per kg
w
) ...(3.9.2)
mole
mole
) ...(3.9.3)
of fuel.
Table 3.9.1 gives the H.C.V. and L.C.V. of some of the fuels as enthalpy of reaction.
Table 3.9.1 : Enthalpy of reactions for various fuels
3.10 Experimental Determination of Calorific Value of Fuels :
The apparatus used for determining the calorific value of fuels is known as calorimeter.
The basic principle used in determining the calorific value of fuels is that the known quantity
of fuel is burned and the heat energy liberated is transferred to a medium of known mass and
specific heat and the rise in temperature of medium is measured. Though there are various types
of calorimeters available. We shall only discuss the Bomb, Boy’s and Junkers gas calorimeters
used for determining the calorific value of fuels.
Bomb calorimeter :
This calorimeter is used to determine the calorific value of fuel of solid and liquid fuels.
Its schematic diagrams shown in Fig. 3.10.1(a) and the electrical circuit for firing the fuse (A)
is shown in Fig. 3.10.1(b).
Fig. 3.10.1 : Bomb calorimeter
Description :
The calorimeter E consists of a thick walled bomb G made of stainless steel. It has a
capacity of 650 c.c. and it is designed to withstand high pressure upto 200 atmospheres. The
bomb at the bottom is screwed at the base cap J. The bomb with its base cap rests on supports
K. The bomb is connected to a non-return oxygen valve C and it has a release valve B. The
bomb also has a stirring device (not shown in Fig. 3.10.1) usually driven by a small motor.
The crucible made of silicon or quartz is carried on support ring S which can slide on
insulated pillars P. Both pillars have a slot through which the fuse wire A can be attached to
the leads of the electrical circuit. The calorimeter has a lid H and a sensitive thermometer D
which is used to record the temperatures of known quantity of water (2500 c.c. approx.) and
calorimeter during the experiment.
Working :
Before the start of experiment the crucible F and the calorimeter is weighed. A known
quantity of water is filled in the calorimeter. A pillet of known quantity of solid fuel say coal
is formed and 1 gm of this pillet is kept into the crucible. The bomb cover is screwed in position
on the calorimeter after making the electrical connections and installation of stirring device and
the thermometer. The oxygen is admitted slowly to the bomb by opening the oxygen valve until
a pressure of 25 atmospheres approximately is reached. The stirring is continued throughout
the experiment and the temperature readings are taken every minute. After 5 minutes when the
equilibrium sets in, the fuel is ignited by closing the switch of the electrical circuit. The
temperature of the water rises quickly due to heat energy released by the combustion of fuel
and the temperature readings are noted at regular interval of time of
1 minute.
When the temperature reaches to its maximum value, it is observed that the temperature
again starts falling due to heat transfer losses to the surroundings. When the temperature fall
shows a steady rate, the temperature readings are made for a period of not less than 5 minutes.
The rate of cooling is determined which helps in applying the correction factor in order to
determine the actual rise in temperature of water and calorimeter. At the end of experiment the
release valve is opened so that the pressure inside the bomb reduces to atmospheric pressure.
In case the calorific value of liquid fuels is determined by this calorimeter, the volatile
fuels are kept in gelatine capsules while the other fuels can be kept directly in the crucible.
Calculations :
Let, mf = Amount of fuel burnt, ;C = Calorific value of fuel
mc = Mass of calorimeter,
;mw = Mass of water
Sc = Specific heat of calorimeter, ;Sw = Specific heat of water ≃ 4.19 kJ/kg K
t = Actual temperature rise of water and calorimeter
The product ( mc  Sc ) is called water equivalent of calorimeter.
Heat given by fuel = Heat absorbed by calorimeter and water
mf  C = mc  Sc  t + mw  Sw  t...(3.10.1)
With the help of Equation (3.10.1) the calorific value of the fuel can be calculated.
Equation (3.10.1) is modified for determination of C.V. of liquid fuels as follows :
m  c1 + mf  c = mc  sc  t + mw  sw  t...(3.10.2(A))
where, m = mass of capsule and C1 is its calorific value.
3.10.1 Cooling Correction :
The cooling correction can be computed with the help of Newton’s law of cooling.
According to this the law of heat loss dQ is the function of the temperature difference between
the hot body and the surroundings.
Alternately, the cooling correction factor as recommended by Regnault Pfaundler is
given as,
where, N = number of minutes during the period the temperature reaches to its maximum
value from the point of firing (chief period).
v1 = rate of fall of temperature per minute during the pre-firing period.
v2 = rate of fall of temperature per minute after reaching maximum temperature.
t1 = average temperature during pre-firing period.
t2 = average temperature during chief period i.e. from the point of firing to the time it reaches
to its maximum temperature.
3.11 Boy’s Gas Calorimeter :

Fig. 3.11.1 shows the sketch of a Boy’s gas calorimeter used for determination of
calorific value of gaseous fuels. The main parts of the calorimeter are burner, chimney and
the cooling coils.

The gas is supplied whose calorific value is to be determined. A gas meter is attached
to measure the rate of volume of gas supplied. The fuel gas is burned and the products of
combustion rise in the chimney. The heat liberated due to combustion of fuel is transferred
to cooling water.

The inlet of the water is first to the outer cooling coils and then it returns upwards
through the inner coils as shown in Fig. 3.11.1. The water vapour formed gets condensed in
the outer portion of the chimney and the condensate is collected and weighed. Three
thermometers are provided to measure the temperature of cooling water at inlet and outlet
and also to measure the temperature of products of combustion at exit. It is ensured that these
flue gases are cooled upto atmospheric temperature.
Fig. 3.11.1 : Boy’s gas calorimeter
Calculations :
Let, Vg = Volume of gas supplied, ;pg = Pressure of gas supplied
Tg = Temperature of gas supplied, ;mw = Mass of water circulated
t1 = Temperature of cooling water at inlet, t2 = Temperature of cooling water at outlet
C = Calorific value of gaseous fuel in kJ/m3,;
Sw = Specific heat of water ≃ 4.19 kJ / kgK
V = Volume of gas supplied at S.T.P.
Where, p = 1.013 barT = Standard temperature = 25C or 298 K
Energy balance :
Heat given by the fuel = Heat absorbed by cooling water
V  C = mw  Sw  ( t2 – t1 )...(3.11.1)
From the Equation (3.11.1) the calorific value of fuel, C at S.T.P. can be calculated.
3.11.1 Junker Gas Calorimeter : (MU - May 2016)
A Junker’s gas calorimeter is shown in Fig. 3.11.2. It is used to determine the C.V. of
gaseous fuels.
Fig. 3.11.2 : Junker Gas calorimeter
The gas whose C. V. is to be determined us supplied to the combustion chamber where
it is burnt by a gas burner. The supply of gas is metered by the gas flow meter and its pressure
is measured by the manometer attached to the pressure regulator. A water jacket surrounds the
combustion chamber through which the cold water flows. It absorbs the heat released by
combustion of gases and the hot water flows from the top as shown. The gases are cooled upto
the room temperature as for as possible so that the entire heat of combustion of gases is
absorbed by the circulating water. The condensate of gases is collected in a pot as shown.
Ti and To refers to the inlet and outlet temperature of cooling water and Tg the exit
temperature of the gases of exit as measured by the thermometers.
The calorific value of gases can be determine as :
hfg = Latent heat of water at atmo. pressure
Volume flow rate of gases is converted at S.T.P i.e. at 25C and 760 mm of Hg.
Ex. 3.11.1 : During Bomb calorimeter test on diesel oil, the following data were recorded :
Room temperature = 25C
Weight of crucible = 8.231 gm
Weight of crucible and diesel oil = 8.803 gm
Weight of calorimeter vessel = 1.05 kg
Weight of water and calorimeter vessel = 3.5 kg
Water equivalent of calorimeter = 0.56 kg
Rise in temperature of water and calorimeter = 2.35C
Cooling correction = 0.02C
Find the HCV and LCV when mass of condensate is 0.32 gm.
The partial pressure of water vapour is 8 kPa.
Soln. :
Given : T0 = 25C
Weight of crucible, mcr = 8.231 gm; Weight of crucible and oil = 8.803 gm
 Weight of oil, mo = 8.803 – 8.231 = 0.572 gm
Weight of calorimeter vessel, mc = 1.05 kg
Weight of water and calorimeter, (mw + mc ) = 3.5 kg
 mw = 3.5 – 1.05 = 2.45 kg
Water equivalent of calorimeter, Mc = mc  Cpc = 0.56 kg
(T) = 2.35C ; Cooling correction = 0.02C
 Actual temperature rise of calorimeter and water,
T = T + Cooling correction = 2.35 + 0.02 = 2.37C
Mass of condensate, mH2O = 0.32 gm
Partial pressure of water vapour, ps = 8 kPa = 0.08bar
(i)H.C.V. of fuel, C :
Heat given by fuel = Heat absorbed by water and calorimeter
mo  C = (mw  Cpw + mc  Cpc) T
(0.572  10– 3)  C = (2.45  4.187 + 0.56)  2.37
C = 44823.5 kJ/kg of fuel...Ans.
(ii)LCV of fuel, C1 :
0.572  10– 3 kg of oil forms the condensate = 0.32  10– 3 kg
From steam tables at ps = 0.08 bar we get :
hg = 2577.1 kJ/kg
Sensible heat of water of condensate at (T0 + T)C, i.e. at (25 + 2.37)C = 27.37C,
We get,
hf = Cpw  27.37 = 4.187  27.37 = 114.6 kJ/kg
Enthalpy given by vapour, h = hg – hf = 2577.1 – 114.6 = 2462.5 kJ/kg
 LCV of fuel = HCV of fuel – mw  h
= 44823.5 – 0.5594  2462.5
= 43446 kJ/kg...Ans.
Ex. 3.11.2 :The following data was obtained during experimental determination of calorific value of fuel by Bomb
calorimeter.
Mass of coal = 0.78 gm
Mass of fuse wire = 0.032 gm
Calorific value of fuse wire = 7 kJ/gm
Mass of water in calorimeter = 2 kg
Water equivalent of calorimeter = 0.4 kg
Rise in temperature of calorimeter water = 3.2C
Cooling correction = 0.01C
Determine HCV and LCV of coal at NTP conditions. Given the coal contains 90% of carbon and 5% of hydrogen.
Soln. :
Mass of coal, mf = 0.78 gm = 0.78  10– 3 kg
Mass of fuse wire, mfw = 0.032 gm and (C  V1) = 7 kJ/gm = 7000 kJ/kg
Mass of water in calorimeter, mW = 2 kg,
Water equivalent of calorimeter, (mc  Cpc) = 0.4 kg
 T = 3.2 C ; Cooling correction = 0.01C
C = 90% and H2 = 5%
Corrected temperature rise of water and cooling water,
T = T + Cooling correction = 3.2 + 0.01 = 3.21C
(i)Higher calorific value of coal, HCV
Heat given by fuel and fuse = Heat absorbed by water and calorimeter
mf  HCV + mfW  C  V = (mW  CPW + mC  Cpc ) T
(0.78  10–3) HCV + (0.032  10–3)  7000 = (2  4.187 + 0.4) 3.21
HCV = 35821.2 kJ /kg ...Ans.
(ii) LCV of coal
Amount of water vapour formed /kg of fuel can be calculated as follows :
(mH2 = 0.05 kg /kg of fuel)
2 kg 16 kg 18 kg
1 kg 8 kg 9 kg
LCV = HCV – mH2O  L = 35821.2 – 0.45  2445 = 34721 kJ/kg...Ans.
Ex. 3.11.3 :Following were the test results obtained during the experimentation on Bomb Calorimeter :
Mass of oil = 0.6 gm
Mass of water = 1500 gm
Temperature rise of cooling water = 2.95C
Cooling correction factor = 0.05C
Mass of fuse wire = 0.02 gm
Calorific value of fuse wire = 1600 kJ/kg
Water equivalent of calorimeter ( m S ) =3 kJ/K
c
c
Find the calorific value of the fuel.
Soln. : Actual temperature rise of cooling water,
t = Measured temperature rise + Cooling correction factor
= 2.95 + 0.05 = 3C
Let ‘C’ be the calorific value of fuel.
Heat given by (fuel + fuse wire) =Heat absorbed by (calorimeter + cooling water)
or, Mass of oil  C + Mass of fuse wire  C.V. of fuse wire
= ( mc  Sc )  t + mw  Sw  t
0.6  10– 3  C + 0.02  10– 3  1600 = 3  3 + (1500  10– 3 ) 4.19  3
or,
0.6  10– 3 C + 0.032 = 27.855
or,
C = 46371.7 kJ/kg ...Ans.
Ex. 3.11.4 :During bomb calorimeter test on diesel oil according to BSS specifications, the following data were
recorded :
Room temperature 25C ; Weight of crucible = 8.116 gm ;
Weight of crucible and oil =8.702 gm ; Weight of can = 1.051 kg
Weight of can and water = 3.492 kg ; Water equivalent of can = 0.559 kg
Rise in temperature of can and water = 2.305C ;
Find the higher calorific value of fuel.
Soln. :
 Weight of water, mw = Weight of water and can – weight of can
= 3.492 – 1.051 = 2.441 kg
Water equivalent of can = mc  sc = 0.559 kg
Rise in temperature of can and water, t = 2.305C
C be the calorific value of oil
Heat given by fuel = Heat absorbed by (can + water)
moil  C = mc sc  t + mw  sw + t
 C = 42400.5 kJ/kg
Ex. 3.11.5 :The following results are obtained when sample of gas is tested by gas calorimeter :
Gas burnt in calorimeter = 0.08 m
3
Pressure of gas supply = 5.2 cm of water
Barometer = 75.5 cm of Hg
Temperature of gas = 13C
Weight of water heated by gas = 28 kg
Temperature of water at inlet = 10C
Temperature of water of outlet = 23.5C
Steam condensed = 0.06 kg
Find HCV per m of gas at 15C and barometric pressure of 76 cm of Hg.
3
Soln. :
Volume of gas, Vg= 0.08 m3,
Pressure of gas, pg = Barometric pressure + Pressure above atmospheric pressure
= 75.5 mm of Hg + 5.2 cm of water
Temperature of gas, Tg = 13C = 286 K
Mass of water, mw = 28 kg
 Tw = Two – Twi = 23.5 – 10 = 13.5C
Steam condensed, ms = 0.06 kg
Volume of gas at T = 15C = 288 K can be calculated as follows :
V = 0.080393 m3
Let C.V. be the HCV of fuel gas.
Heat given by fuel gas = Heat absorbed by cooling water
V  C.V. = mw  Cpw  Tw
0.080393  C.V. = 28  4.187  13.5
C.V. = 19687 kJ/m3 of fuel gas…Ans.
Ex. 3.11.6 :The following observations were made during a test on coal gas :
Volume of gas used = 0.06 m
3
Mass of cooling water circulated = 9.8 kg
Mass of condensed steam collected = 0.009 kg
Rise in temperature of cooling water = 6.3 C
Pressure of gas tested above atmosphere = 45 mm of water
Temperature of gas tested = 14C
Barometric pressure = 750 mm of Hg
Calculate the higher and lower calorific values at N.T.P.
Soln. :
Volume of gas, Vg = 0.06 m3 ; Tg = 14C = 287 K
Pressure of gas, Pg =Barometric pressure + Press. of gas above atmosphere
N.T.P. means, p = 1.013 bar and T = 0C = 273 K
Let V be the volume of gas supplied at N.T.P. Then,
Cooling water :
mw = 9.8 kg ;
(t)w = 6.3C
Let ‘C’ be the higher calorific value of gas.
Heat given by fuel gas = Heat absorbed by cooling water
V  C = mw  Cpw  (t)w
0.05657  C = 9.8  4.187  6.3
C (H.C.V) = 4569.6 kJ/m3 of gas at NTP...Ans.
Amount of steam condensed is 0.009 kg when the fuel gas burnt is 0.05657 m3.
Therefore,
L.C.V = H.C.V – m  L = 4569.6 – 0.1591  2445 = 4180.6 kJ/m …Ans.
3
s

In a reactive system involving chemical reactions, the composition of substances at the
end of combustion process are no longer the same as that in the beginning of the process.

Therefore, in order to determine the energy before and after chemical reaction, it is
necessary to determine the energy of different substances before and after chemical reaction
with reference to a certain standard state so that no ambiguity exists.
Standard reference state :

Standard reference state is taken as, temperature, T0 = 25C or 298K and pressure, p0 =
1 atmosphere

Thus, with the above datum, enthalpy values can be assigned to compounds for study
of reactive systems.
Example of formation of CO2 :
Fig. 3.12.1

Consider the formation of CO2 from its elements carbon and O2. (Refer Fig. 3.12.1)

It is assumed that the reactants (C and O2) and the product (CO2) both enter and leave
the combustion chamber at standard reference state of p0 = 1 atmosphere and T0 = 25 C
The reaction being exothermic, an amount of heat energy equal to chemical energy of
fuel carbon as Q = 393520 kJ/kgmole will be transferred from the system to the surroundings.


This energy was defined as heat of reaction or calorific value of fuel. Accordingly,
HP – HR = – 393520 kJ/kgmole of carbon... (i)
Enthalpy of formation of natural or stable elements :
Enthalpy datum for all naturally occuring stable elements are assigned zero
enthalpy.


For example, the stable form of gases oxygen, hydrogen and nitrogen are respectively
O2, H2 and N2 while the natural form of carbon is graphite. All these elements are assigned
zero enthalpy of formation.
Enthalpy of formation of CO2:

Referring to Equation (i) above, since the enthalpy of formation of reactants C and O2 is
zero and the product consists of 1 kgmole of CO2 at the same reference standard state, it follows
that the enthalpy of formation of CO2 is – 393520 kJ/kgmole i.e.

Negative value indicates that heat energy is released in forming the compound from its
stable elements i.e. reaction is exothermic.

A positive value will indicate that heat energy is absorbed in formation of compound
from its stable elements. It also represents that reaction is endothermic.
3.12.1 Specific Enthalpy of a Compound at (p, T) :

For determining the specific enthalpy of a compound at any other state (p, T) which is
different than the standard state (p0, T0), it can be written as follows :
Where,
Enthalpy change.
3.12.2 Enthalpy of Formation Table :
Table 3.12.1, represents the enthalpy of formation, Gibbs formation and absolute entropy
of various substance at 25 C and 1 atmosphere pressure.
Table 3.12.1: Enthalpy of formation, Gibbs function and absolute entropy at
25 C and 1 atmosphere
3.12.3 Application of Enthalpy of Formation :

Enthalpy of formation can be used to calculate the enthalpy of reaction for various
compounds.
For example: consider the formation of methane. Its chemical reaction is :
CH4 + 2 O2CO2 + 2 H2O... (i)

Enthalpy of reaction of any compound can be calculated with the help of following
equation,
= – 802330 kJ/ kgmole
Therefore, enthalpy of reaction of CH4 is (– 802330) kJ/ kgmole.
Ex. 3.12.1 : Calculate the enthalpy of reaction of methyl alcohol (C H OH) at 25 C with the help of enthalpy of
2
5
formation table.
Soln. : The chemical reaction is,
C2 H5 OH + 3O22 CO2 + 3 H2O (g)
From Equation (3.12.3), we have,
= – 1277220 kJ/ kgmole...Ans.
Ex. 3.12.2 :A hydrocarbon fuel (C H ) has enthalpy of combustion – 4856920 kJ/kg mole. Find its value of enthalpy
7
16
of formation. Take the values of enthalpy of formation of CO and H O respectively as – 393791 and
2
2
136 288 kJ/kg mole.
Soln. :
The chemical reaction of C7 H16 with O2 can be written as :
C7 H16 + 11 O27CO2 + 8 H20
 Enthalpy formation of C7 H16 i.e.
3.13 Application of First Law of Thermodynamics to Reactive Systems

First law of thermodynamics is applicable both to reactive and non-reactive systems.
However, analysis of reactive system, in addition, involves the chemical energy of fuel. The
reactants and the products enter and leave at different temperatures.

We shall carry out first, the first law analysis of steady flow system and then for the
closed system.
3.13.1 First Law Analysis of Steady Flow Systems :

Enthalpy of any substance at any other state (p, T) was defined by Equation (3.12.2) as
follows :

Consider any hypothetical chemical reaction as follows for a steady flow process shown
in Fig. 3.13.1.
n1  A + n2  Bn3 C + n4 D

n1 and n2 are the moles of reactants A and B respectively and n3 and n4 are the moles of
products C and D respectively.
Fig. 3.13.1 : Steady flow reactive system

General, steady flow energy equation (S.F.E.E) with suffix P for products and suffix R
for reactants can be written as :
QC – WC = HP – HR [If,  K.E. = 0,  P. E. = 0] ...(3.13.2A)
If the changes in K.E. and P.E. are neglected, Equation (3.13.1) reduces to :

In the above equation ni represents the number of moles of products and reactants per
mole of fuel supplied. Reactants (R) refer the homogeneous mixture of fuel and air and
products (P) refer to products of combustion.
3.13.2 Heat of Reaction and Heat of Combustion :

Sensible enthalpy values for various reactants and products at various temperatures
can be read with the help of tables given at Table 3.14.1 and values of internal energy and
entropy from Table 3.13.1.
3.13.3 First Law Analysis of Closed System :

First law equation for a closed system can be written as :
Q – W =  E...(3.13.3)
Where, Total energy, E = Internal energy, U + K.E. + P.E. ...(3.13.4)

In case the changes in K.E. and P.E. are neglected, the equation for a chemical reaction
can be rewritten as follows :
Q – W =  U = UP – UR... (3.13.5)
Or, Since, U = H – p  V, Equation (3.13.4) reduces to :

In the Equation (3.13.7) ni represents the moles of various constituents of ideal gases.
Since the volume of solids and liquids is assumed to be negligible compared to gaseous volume, it
implies that ni = 0 for solids and liquids constituents.

Heat transfer Q represents the heat of combustion of a closed system.
Internal energy values of gases at different temperatures are given in Table 3.13.1.
Ex. 3.13.1 :What do you understand by standard heat of reaction and heat of formation? Carbon reacts with oxygen
to form carbon dioxide in a steady flow chamber. Reactants and products are at 25C and 1atm. Find
the energy involved and type of reaction. Assume enthalpy of formation of CO gas as (–) 393520
2
kJ/kmol.
Soln. :
Theory : Refer Sections 3.9.1 and 3.12
Numerical : Refer Fig. P. 3.13.1
p0 = 1 atm.,
T0 = 25C = 298 K
Chemical reaction is :
On neglecting  K.E. and  P.E., First law equation can be written as:
Qc – 0 = – 393520 –  0 + 0 
Qc = – 393520 kJ/kmole.…Ans.
Negative sign shows that the reaction is exothermic since heat is rejected from the
combustion chamber.
Ex. 3.13.2 :Liquid octane C8 H18 at 25C is used as a fuel. Air used is 150% of theoretical air and is supplied at 25 C.
Assume a complete combustion and the product leaves the combustion chamber at 1500 K. Calculate
the heat transfer per kg mole of fuel. Use the following data :
Soln. :
Refer Fig. P. 3.13.2.
Fig. P. 3.13.2
The combustion equation of C8 H18 with theoretical air can be written as :
C8 H18 (l) + 12.5 O2 + 12.5 × 3.76 N2  8CO2 + 9 H2 O + 12.5 × 3.76 N2
When air used is 150%, above equation can be rewritten as :
C8 H18 + 1.5 × 12.5 O2 + 1.5 × 12.5 × 3.76 N2
 8CO2 + 9 H2 O + 0.5 × 12.5 O2 + 1.5 × 12.5 × 3.76 N2
i.e. C8 H18 + 18.75 O2 + 70.5 N2  8CO2 + 9 H2 O + 6.25 O2 + 70.5 N2
From Ist law of thermodynamics (since Wc = 0) we have,
= 8 (– 393.5 + 71.075 – 9.36) + 9 (– 241.8 + 57.99 – 9.9)
+ 6.25 (0 + 49.29 – 8.68) + 70.5 (0 + 47.07 – 8.67)
= – 2654.28 – 1732.32 + 253.8125 + 2707.2 = – 1425.5875 MJ/kgmole
Energy of reactants hT = h298
= 1 (– 250) + 18.75 × zero + 70.5 × zero = – 250…(iii)
…(ii)
Substituting the values in equation (i),
Qc = – 1425.5875 – ( – 250) = – 1175.5875 kJ/kgmol of fuel…Ans.
3.14 Adiabatic Flame Temperature

In case of chemical reactions, the heat energy released due to combustion is used
internally to raise the temperature of products of combustion and partially it is transferred to
the surroundings.

In case the combustion chamber is completely insulated (i.e. no heat transfer is
permitted between the system and surroundings), then the chemical energy released during
combustion will be utilised to raise the temperature of products of combustion alone. In such
a situation, the combustion process is said to be adiabatic and temperature attained by the
products is called the adiabatic flame temperature.
Definition :
Adiabatic flame temperature or theoretical flame temperature is defined as the
theoretical temperature attained by the products of combustion in an adiabatic process
assuming complete combustion.

In actual practice, the actual flame temperatures are less than adiabatic flame
temperature for the following reasons.

1.There is inevitable heat transfer to surroundings since the system cannot be made perfectly
insulated.
2.Combustion is never complete.
3.At high temperatures the gases are not stable, therefore, these may dissociate and absorb
energy internally from the system, thereby, reducing the flame temperatures.
3.14.1 Equation for Adiabatic Flame Temperature :

Consider a perfectly insulated combustion chamber. The reactants are supplied at p0,
T1 and the products leave at p0 and T2.

Where T2 represents the adiabatic flame temperature.
Fig. 3.14.1

is :
From steady flow energy Equation (3.13.2A), with negligible change in K.E. and P.E.
QC – WC = HP – HR...(i)
But, QC = 0 (Combustion chamber is perfectly insulated)
WC = 0 (Since no work transfer)
Equation (i) reduces to
HP = HR...(3.14.1)

With the help of Equation (3.14.2) the adiabatic flame temperature T2 can be calculated
with the help of chemical reaction equation, enthalpy of formation table and enthalpy of
gases table.

Adiabatic flame temperatures gives an important data since it represents the maximum
temperature that can be attained in a combustion chamber. It helps in selecting the materials
for combustion chamber and design of combustion chamber from thermal stresses
considerations.

The adiabatic flame temperatures can be reduced by increasing the air-fuel ratio, i.e.
either by increasing the rate of air supplied or by reducing the rate of fuel supplied to
combustion chamber.

Adiabatic flame temperature is affected by :
(a)Heat losses from the combustion chamber.
(b)Dissociation of gases.
(c)Incomplete combustion of fuel since mixture of fuel and air is never homogeneous.
(d)Adiabatic flame temperature is maximum with stoichiometric air. It is lower if excess air is
supplied since the total heat of reaction of fuel is utilized by more number of moles of
products of combustion. The adiabatic flame temperature is also lower with deficient air,
it is due to the fact that heat of reaction is not released since the O2 available is not
sufficient to burn the entire fuel.
Ex. 3.14.1 :Determine the adiabatic flame temperature which would be obtained from complete combustion of liquid
octane (C H ) if the fuel and air temperature is 25C.
8
18
Heats of formation are as follows :
Soln. :
The stoichiometric chemical reaction of C8 H18 with air is :
C8 H18 (l) + 12.5 O2 + (12.5  3.76 = 47) N2  8 CO2 + 9 H2 O + 47 N2
Given : T1 = 25C = 298 K, hence,
= – 249950 + 0 + 0 = – 249950 kJ/kgmole...(i)
Substituting the values in Equation from Equations (i) and (iii) we get :
In Equation (iv) T2 is unknown we have to use hit and trail method to determine T2 with
the help of enthalpy of gases values given in Table P. 3.14.1.
From Table 3.14.1. This enthalpy of N2 corresponds to 2600 K. Since the products also
have CO2 and H2O whose specific enthalpy is more than the enthalpy of nitrogen, actual
temperature will be lower than T2. We tabulate the values for LHS of Equation (iv) at
T2 = 2300 K and 2400 K as follows :
 Approximate adiabatic temperature (by using linear law),
= 2389.3 K ...Ans.
Ex. 3.14.2 :Liquid Octane at 30C is burned with 400% theoretical air at 30C in a steady flow process. Determine
the adiabatic flame temperature.
Soln. :
Stoichiometric chemical reaction for Octane (C8 H18) fuel is
C8 H18 (l) + 12.5 O2 + 12.5  3.76 N2  8 CO2 + 9 H2 O + 12.5 3.76 N2
With 400% theoretical air it becomes
C8 H18 (l) + 4  12.5 O2 + 4  12.5  3.76 N2  8 CO2 + 9 H2 O + 3 12.5 O2 + 4  12.5  3.76 N2
i.e. C8 H18 (l) + 50 O2 + 188 N2  8 CO2 + 9 H2 O + 37.5 O2 + 188 N2
Given: T1 = 30C = 303 K, Heats of formation from Table 3.12.1 are :
For adiabatic combustion process :
Using the gas tables, we can calculate the energies as follows :
Hence,
= – 216330 kJ/kgmole of C8 H18... (ii)
From Equations (i), (ii) and (iii) we get,
= 7434966 – 216330 = 7218636 kJ... (iv)
 n = 8 + 9 + 37.5 + 188 = 242.5
i
Since N2 is maximum, T2 for above enthalpy corresponds to 1000 K approximately. We
can compute the table of enthalpy values as follows :
 T2 lies between 900 K and 1000 K. By interpolation :
 Adiabatic flame temperature is 964.6 K ...Ans.
Summary
Fuel is defined as the source of heat energy which is released in a reactive system by
chemical or nuclear reactions. Fuels may be classified as solid, liquid or gaseous
fuels which may be eithernaturally occurring or prepared fuels.

The main constituents of fuel are C, H2, compounds of C and H2, S, N2, O2, moisture,
ash and other incombustible matters.


Composition of air
By mass : N2 = 77%, O2 = 23%
By volume : N2 = 79%, O2 = 21%

Mixture strength or Equivalence ratio

Depending upon whether the actual A.F. ratio is more or less than stoichiometric A.F.
ratio, mixture is called lean or rich mixturesrespectively.

Minimum air required for complete combustion of fuel is given by

The mass analysis can be converted into volumetric analysis by using Avogadro's law
i.e. by dividing the mass analysis of gases by their respective molecular masses in order to
obtain their proportionate volumes.
Exhaust gas analysis is necessary to control exhaust emissions, determine A.F. ratio,
ensure complete combustion of fuel, control maximum temperatures in combustion
chambers and their efficient operation.


Exhaust gas analysis is done by Orsat's apparatus.
The air-fuel ratio can be determined by using C – H2 balance method in case the
volumetric analysis of dry products of combustion is known. It is also called as
mole method.

Calorific value of a fuel is defined as the amount of heat energy released during
complete combustion of unit quantity of fuel when the products of combustion are cooled
back to S.T.P. It representsH.C.V. since water vapour is condensed at S.T.P.

While the L.C.V. of fuel represents the fictitious quantity of heat that would be
obtained due to combustion of unit quantity of fuel if the water vapour formed in the products
are cooled back to S.T.P. (upto 25C at atomospheric pressure) and the water vapour still
remains in gaseous state.

– LCV = – H.C.V – mW  L
L = 2445 kJ/kg of water vapour at constant pressure
L = 2308 kJ/kg of water vapour at constant volume
mW = mass of water vapour formed per kg of fuel.

Calorific fuel of fuel can be determined experimentally by :
1.Bom’b calorimeter (For solid and liquid fuels)
2.Boys and Junker’s gas calorimeters for gaseous fuels

Enthalpy of a compound at (p, T)
Enthalpy of formation can be used to determined the enthalpy of reaction of
various compounds.

[According to Equation(3.12.1)]

Application of first law to reactive systems.
(a)For steady flow system :
If changes in K.E. and P.E. are neglected :
QC – WC = HP HR
If work transfer is zero, QC is called the heat of reaction or heat of combustion.
(b)Closed system :
Adiabatic flame temperature is defined as the theoretical temperature attained by the
products of combustion in an adiabatic process assuming complete combustion.

HP = HR
Exercise
[ Note : For answers refer the section numbers indicated in bracket. ]
Theory :
Q. 1Explain why excess air is used for burning of fuel ? [Section 3.1.6]
Q. 2For what purpose an Orsat apparatus is used ? Discuss its working with the help of a neat sketch. [Section
3.5]
Q. 3Define terms
(i) Excess air [Section 3.1.6]
(ii)Stoichiometric air-fuel analysis. [Section 3.1.5]
Q. 4Write short note on Orsat apparatus for exhaust gas analysis. [Section 3.5]
Q. 5Differentiate between mass fraction and mole fraction. [Sections 3.1.3 and 3.1.4]
Q. 6Define mixture strength or equivalence ratio. How the concept of mixture strength related to rich and weak
mixtures of fuel and air. [Section 3.1.7]
Q. 7Explain the method of converting mass analysis of products of combustion into volumetric analysis and
vice-versa. [Section 3.3]
Q. 8Explain the concept of (C – H2) balance or mole method for determining the approximate A.F.
ratios. [Section 3.7]
Q. 9What is the purpose of carrying out exhaust analysis ? Is it possible to determine A.F. ratio from exhaust
gas analysis ? If so, explain how ?[Sections 3.4, 3.6 and 3.7]
Q. 10Define and differentiate between internal energy of reaction and enthalpy of reaction.[Section 3.9]
Q. 11Prove that :
Q – Q = ( n – n ) RT [Section 3.9.1]
p
v
p
R
Q. 12Discuss the method for determining the calorific value of solid and liquid fuels. [Section 3.10]
Q. 13Name the apparatus used for measurement of C.V. of gaseous fuels and discuss its working with the help
of neat sketch. [Section 3.11]
Q. 14Explain the application of heat of formation of compounds. [Section 3.12.3]
Q. 15Write the S.F.E.E. and reduce it for a reactive system. [Section 3.13.1]
Q. 16Define adiabatic flame temperature and explain. [Section 3.14]
Numerical :
Q. 1The ultimate analysis of a solid fuel by mass is, C = 86%, H = 12%, O = 1% and
2
2
S = 1%. Determine the theoretical amount of air required for complete combustion at NTP and its
volume
occupied.
0.773 m /kg.
3
Ans. :14.1 kg; 10.9 m3
Assume,
specific
volume
of
air
at
NTP
to
be
Q. 2In problem, determine the percentage composition of products of combustion by mass.
Ans. :CO2 = 20.8%, H2O = 7.15%, SO2 = 0.13%, N2 = 71.92%
Q. 3A fuel consists of carbon and hydrogen only and the quantity of theoretical air required of its combustion
per kg is 15 kg. Determine the composition of fuel.
Ans. :C = 85.3%, H2 = 14.7%
Q. 4The percentage analysis of the fuel used in a boiler by mass is as follows : C = 90%, H = 3.5%, O = 3%,
2
2
N = 1%, S = 1% and the remainder is ash. Find :
2
(a)Theoretical amount of air required and the mass analysis of dry products of combustion.
(b)If 50% excess air is supplied, determine the volumetric analysis of dry products of
combustion.
Ans. :(a) 11.583 kg/kg of fuel; CO2 = 27.00%; N2 = 73%,
(b) CO2 = 18.31%, O2 = 7.4% and N2 = 74.29%
Q. 5The mass analysis of coal is as follows :
C = 77.2%, H = 5.2%, S = 1.25%, N = 1.5%, O = 8.65% and the remainder is ash. Determine
2
2
2
the amount of theoretical air required for complete combustion and the volumetric analysis of
products of combustion with 25% excess air.
Ans. :10.33 kg/kg of fuel; CO2 = 13.8%, O2 = 4.1%, H2O = 5.6%, SO2 = 0.1% and N2 = 76.4%
Q. 6A gaseous fuel has the following composition by volume :
CO = 6%, CH = 29.8%, H = 56.2% and N = 4.8% and O = 1.2%.
2
4
2
2
2
Determine the theoretical air required for complete combustion and the analysis of dry products
of combustion with 50% excess air.
Ans. : 4.26 m3 of air/m3 of fuel; CO2 = 6.38%, O2 = 7.56% and N2 = 86.06%
Q. 7A liquid fuel has 86% carbon and the remainder is hydrogen and inert material. This fuel is supplied to an
internal combustion engine which consumes 12 kg of fuel/hr. If the exhaust from the engine shows 10%
of CO by mass, find the total mass of exhaust leaving the engine per hour.
2
Ans. :258 kg/hr
Q. 8A boiler uses coal which has the mass composition as : C = 84%, H = 3% and remainder as ash. The orsat
2
apparatus
gave
the
volumetric
analysis
as
follows
:
CO = 11.5%, O = 8.4% and N = 80.1%. Calculate the air-fuel ratio and the mass of excess air supplied.
2
2
2
Ans. :17.8 : 1; 7.0 kg
Q.
9The
volumetric
analysis
of
dry
exhaust
gases
as
determine
by
orsat
apparatus
is,
CO = 10%, CO = 3.5%, O = 8% and N = 78.5%.
2
2
2
The fuel used has the mass composition as : C = 80%, H = 6%, O = 7% and the remainder
2
2
being ash. Determine by weight the air supplied per kg of fuel and the excess air supplied.
Ans. :14.095 kg/kg of fuel; 3.037 kg/kg of fuel
Q. 10 The volumetric analysis of flue gases obtained from the combustion of an unknown hydrocarbon fuel is :
CO = 12.1%, CO = 0.5%, O = 3.2% and N = 84.2%.
2
2
2
Determine the excess air supplied in percentage of theoretical air.
Ans. :15%
Q. 11A fuel water gas has the composition as :
CO = 6%, H = 48%, O = 0.5%, CH = 2%, CO = 38% and N = 5.5%. The dry flue gas analysis
2
2
2
4
2
showed CO = 15.5%, O = 4.75% and CO = 0.2%.
2
2
Find the percentage of excess air supplied and the volume of flue gases.
Ans. :29.1%, 5.94 m3/m3 of water gas
Q. 12Determine the air fuel ratio by mass of an engine whose products of combustion show,
CO = 8%, O = 1%, CO = 13%, H = 8%, CH = 1% and the remainder is N by volume on dry basis by
2
2
2
4
2
carbon hydrogen balance method.
Ans. :8.528 kg of air/kg of fuel
Q. 13An engine uses C H as the fuel and the volumetric exhaust gas analysis shows the following :
8
18
CO = 10%, CO = 7%, H = 3.4%, CH = 0.2% and N = 79.4% (by difference).
2
2
4
2
Calculate the air-fuel ratio by carbon hydrogen balance method.
Ans. :11.967 : 1
Q. 14Compute the heating value of methane at constant pressure if the heating value at constant volume is (–
886038) kJ/kg
mole
at 25C and water vapour formed is in the liquid state.
Ans. : – 898994 kJ/kgmole
Q. 15L.H.V. of propane (C H ) at constant pressure and 25C is (– 2045475) kJ/kg . Find the H.H.V at constant
3
8
mole
pressure and at constant volume.
Ans. : Qp = – 221515 kJ/kgmole, Qv = – 2047906 kJ/kgmole
Q. 16Calculate the enthalpy of reaction of n-butane (C H ) using the concept of heats of formation.
4
10
Ans. : – 2657080 kJ/kgmole
Q.
17Calculate
the
Table 3.12.1.
enthalpy
of
reaction
of
CH with
4
the
help
of
heats
of
formation
Ans. : – 802330 kJ/kgmole
Q. 18Calculate the enthalpy of reaction of ethane (C H ) with the help of heats of formation
2
6
Table 3.12.1.
Ans. : – 1427860 kJ/kgmole
Q. 19Methane (CH ) gas initially at 500 K is burnt with 100% excess air at 500 K both at
4
1 atmosphere pressure. The temperature of products of combustion is 1500 K. Find the heat transfer
during the combustion process.
Ans. : – 10559.7 kJ/kgmole
Q. 20In a steady flow combustion chamber, the reactants liquid propane (C H ) and 100% excess air flow into
3
8
the combustion chamber. The reactants are at 1 atmosphere and 25C. The exit temperature of
products of combustion is 1600 K.
For a mass flow rate of 0.05 kg/min of propane, find the air-fuel ratio and heat transfer during
the combustion process.
Given :
Ans. : A.F. ratio = 31.2:1, QC = + 29.867 kW
3.15 University Questions and Answers
May 2011
Q. 1The petrol used in an SI engine is assumed to have a chemical formula C H . Determine : (a) the
7
16
stoichiometric A : F ratio and (b) If 50% excess air is supplied then find the volumetric composition of
dry exhaust products. Air contains 23% of O and 77% of N by Mass. (Ex. 3.3.3)(7 Marks)
2
2
May 2016
Q. 2Explain construction and working of Junker’s gas calorimeter with neat sketch.(Section 3.11.1) (7 Marks)
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