Download module-iv --- combustion thermodynamic applied termodynamics

MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
VTU-NPTEL-NMEICT
Project Progress Report
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The Project on Development of Remaining Three Quadrants to NPTEL
Phase-I under grant in aid NMEICT, MHRD, New Delhi
Subject Matter Expert Details
web 
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Type of the
Course
Applied Thermodynamics
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Course Name:
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SME Name :
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Dr.A.R.ANWAR KHAN
Prof & H.O.D
Dept of Mechanical Engineering
VT
Module
II
DEPARTMENT OF MECHANICAL ENGINEERING,
GHOUSIA COLLEGE OF ENGINEERING,
RAMANARA -562159
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
CONTENTS
Sl.
No.
DISCRETION
1.
Quadrant -2
a. Animations.
b. Videos.
2.
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Quadrant -3
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c. Illustrations.
3.
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b. Open Contents
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a. Wikis.
Quadrant -4
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a. Problems.
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b. Assignments
c. Self Assigned Q & A.
d. Test your Skills.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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2014
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
MODULE-II
COMBUSTION THERMODYNAMICS
QUADRANT-2
Animations
1) http://mutuslab.cs.uwindsor.ca/schurko/animations/
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2) http://www.ohio.edu/mechanical/thermo/Applied/Chapt.7_11/Chapter11.html
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3) http://www.codecogs.com/library/engineering/thermodynamics/index.php
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4) http://www.slideshare.net/qiebti/ppt-application-of-second-law-thermodynamic
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5) http://yteach.com/index.php/resources/bond_enthalpy_dissociation_molecule_ch
arge_resonance_energy_t_page_11.html
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6) http://www.youtube.com/watch?v=q7ZKnnXz5R4
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7) http://www.awn.com/animationworld/combustion-2-action
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Videos
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1) http://www.youtube.com/watch?v=JK-K-QTSOqY
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2) http://www.youtube.com/playlist?list=PL1FF6C552B96BD57E
3) http://yteach.com/index.php/resources/bond_enthalpy_dissociation_molecule_charge_re
sonance_energy_t_page_11.html
4) http://www.youtube.com/watch?v=q7ZKnnXz5R4
5) http://www.tutorvista.com/chemistry/animations/combustion-animation
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
Illustrations
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
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or a gas turbine.
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The solid fuels are burned in beds or in pulversied from suspended in the air stream. The
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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
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are premixed or the fuel and air flow separately in to a burner or a furnace and simultaneously
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mix together as combustion proceeds.
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The Kg-mole or gram-mole is widely used in combustion calculations as a unit of weight.
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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.
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Consider a reaction
CH4 + 2O2
(12+4.032) + 64
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CO2 + 2H2O
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
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substance occupies the volume of 22.4m at NTP i.e., 1.013bar and 273K.
CH4 + 202
CO2 + 2H2O
1 volume of methane reacts with 2 volume of oxygen to form one volume of CO 2 and two
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
(ii)
C +O2
On mass basis (12) + (32)
CO2
(44)
1 Mol (C) + 1 Mol (O2)
1 Mol (CO2)
2C +O2
On mass basis (24) + (32)
2CO
(56)
Volume basis 2 Mol (C) + 1 Mol (O2)
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2C +O2
On mass basis (56) + (32)
2CO2
(88)
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(iii)
2Mol (CO)
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(i)
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volumes of H2O. Therefore in any reactions, the mass in confirmed but the no. of mol or
volumes may not be considered.
2 Mol (CO2)
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2 Mol (CO) + 1 Mol (O2)
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Combustion Stoichometry
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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
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stiochiometric coefficients, so that no excess of any constituent is present. In general a chemical
reaction may be written as
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aA + bB
cC + dD
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
The minimum amount of air required for the complete combustion of a fuel is known as
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
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.
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.
Or
1 Kg C
+ 4/3 Kg O2
7/3 kg CO
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+ 4/7 Kg O211/7 kg CO2
1 Kg CO
1K mol CO2
1K mol CO
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1K mol O2
½ K mol O2
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+
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On molal basis
1K mol C
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+ 32/24 Kg O256/24 kg CO
Similarly
1K mol C
1K mol C
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C
1 Kg C
2CO
2(28)
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Or
O2
+ (32)
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2C+
2(12)
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Consider a equation
C+
O2
CO2
(12) + (32)
(44)
Or
1
+ 8/3
11/3
Or
1 Kg C
+ 8/3 Kg O2 11/3 kg CO2
Similarly
+
½ K mol O2
1K mol CO2
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
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
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.
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Dew point of products:
The product of combustion containing water vapour are known as wet products. The water
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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
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before condensing, it is possible to find the saturation temp. corresponding to partial pressure
from the steam tables.
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Enthalpy of reaction
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Enthalpy of a reaction is defined as the difference between the enthalpy of the products at a
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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
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fuels. However there are so many different fuels and fuel mixtures that is not practical to list
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enthalpy of combustion values for all possible cases. Besides, the enthalpy of combustion is not
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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
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O2, N2, H2 and C a value of zero at standard reference state of 25 C 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
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
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
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change, H, has units kJ/mol and is defined:
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[C + H (fuel)] + [O2 + N2 (Air)] -> (Combustion Process) -> [CO2 + H2O + N2 (Heat)]
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where
(1)
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C = Carbon, H = Hydrogen, O = Oxygen, N = Nitrogen
H = H(products) – H(reactants)
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If energy is given off during a reaction, such as in the burning of a fuel, the products have less
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heat content than the reactants and H will have a negative value; the reaction is said to be
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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
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
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)
(4)
2CO (g) + O2 (g)
CO2 (g)
2CO2 (g)
H = –393.5 kJ
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C (s) + O2 (g)
H = –566.0 kJ
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(3)
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cannot be determined directly because carbon dioxide will also form. However, H can be
measured for:
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Multiplying equation (3) by 2 gives equation (5), and reversing equation (4) gives equation (6):
2C (s) + 2O2 (g)2CO2 (g)
H = –787.0 kJ
(6)
2CO2 (g)2CO (g) + O2 (g)
H = +566.0 kJ
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(5)
2C (s) + O2 (g)
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(2)
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Adding equations (5) and (6) gives the desired information:
2CO (g)
H = –221.0 kJ
For a reaction in which a compound is formed from the elements, the enthalpy change is called
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the heat of formation, Hf , 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:
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
2014
APPLIED THRMODYNAMICS
(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
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formation is per mole of compound, so Hf 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
you can calculate virtually any heat of reaction. The heat of a reaction is the sum of Hf
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for the products minus the sum of Hf values for the reactants. Expressed as a formula:
°
o
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Hrxn = Hf products- H f reactants

(10)
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values
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°
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Internal Energy of Combustion: It is defined as the difference between the internal energy
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of the products and the internal energy of the reactants when complete combustion occurs at
a given temperature and pressure. Uc = Up – UR
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p ne (hf + h – pv ) - R ni (hf + h – pv )
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=
(F/A) ideal
(F/A) actl
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comb =
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Combustion efficiency
It is defined as the ratio of ideal fuel-air to the actual fuel-air ratio
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
QUADRANT-3
Wikis
http://chemwiki.ucdavis.edu/Physical_Chemistry/Chemical_Equilibrium/Reversible_vs._I
rreversible_Reactions
http://www.slideshare.net/qiebti/ppt-application-of-second-law-thermodynamic
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Open Contents:
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Applied Thermodynamics by R. K. Rajput
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Applied Thermodynamics for Engineering Technologists by Eastop
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Applied Thermodynamics by B. K. Venkanna B. V. S
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Basic and Applied Thermodynamics by Nag
Applied Thermodynamics by D. S. Kumar
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A textbook of applied thermodynamics, steam and thermal ... by S. K. Kulshrestha
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Applied thermodynamics by Anthony Edward John Hayes
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
QUADRANT-4
Problems:
1) A coal sample gave the following analysis by weight, carbon 85 %, hydrogen 6%,
OXYGEN 6%, The Remainder Being Incombustible. Determine minimum weight of air
required per kg of coal for chemically correct composition.
Solution:
Element wt. (kg)
O 2 required (kg)
C=0.85
0.85*8/3=2.27
0.06*8=0.48
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H2=0.06
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O2=0.06
TOTAL O2=2.75
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Weight of O2 to be supplied = wt of O2 needed – wt of O2 already present in fuel
= 2.75-0.06=2.69kg
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Weight of air needed = 2.69*(100/23) =11.70kg
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C=0.848
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Element wt. (kg)
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2) The percentage composition of a sample of liquid fuel by weight is C=84.8%, H2=15.2%.
Calculate. i) the weight of air needed for the combustion of 1kgof fuel. ii) The volumetric
composition of the products of combustion if 155 excess air is supplied.
Solution:
H2=0.152
O2 required (kg)
Dry products (kg)
0.848*8/3=2.261
(0.848*11)/3 =3.109(CO 2)
0.152*8=1.216
Total O2 =3.477
Minimum air needed for combustion
=3.477*(100/23) = 15.11kg
Excess air supplied
= (15.11*15)/100 = 2.266kg
Weight of oxygen in excess air
= 2.266*(23/100) =0.521kg
Total air supplied
= minimum air + excess air
= 15.11+2.266 = 17.376kg
Weight of N2 in flue gases = 17.376*(77/100) = 13.38kg.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 12 of 29
MODULE-II --- Combustion Thermodynamics
2014
APPLIED THRMODYNAMICS
Volumetric composition of the product of combustion
Name of the gas
Weight (x)
Molecular
weight (y)
Proportional
volume(z)=x/y
CO2
O2
N2
3.109
0.521
13.38
44
32
28
0.0707
0.0163
0.4780
z=0.5650
Percentage
volume
=z/z*100
12.51%
2.89%
84.60%
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3) A single cylinder was supplied with a gas having the following % volumetric analysis
CO=5, CO2=10, H2=50, CH4=25, N2=10. The percentage volumetric analysis of dry gases
was CO2=8, O2=6 & N2=86.determine the air-fuel ratio by volume.
Solution:
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Combustion equations are :
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2H 2 + O 2 = 2H 2 O
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1 vol + 1/2 vol = 1 vol
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2CO + O 2 = 2CO 2
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CH 4 + 2O 2 = CO 2 + 2H 2 O
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1 vol + 1/2 vol = 1 vol
1 vol + 2 vol = 1 vol + 2 vol.
Vol (m 3)
O 2 needed (m 3)
0.05
0.025
CO 2
0.10
—
H2
0.50
0.25
CH 4
0.25
0.50
N2
0.10
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CO
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Gas
Total
1.0
Products (m 3)
CO 2
N2
0.05
—
0.10
—
—
0.25
—
—
—
0.10
—
0.775
0.4
Volume of air required = 0.775 × 100/21 = 3.69 m 3
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 13 of 29
0.10
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
Volume of nitrogen in the air = 3.69 × 79/100 = 2.92 m 3
Dry combustion products of 1 m 3 of gases (V) contain 0.4 m 3 of CO 2 + 0.1 m 3 of N 2
(as given in the table) + 2.92 m 3 of N 2
(From air supplied for complete combustion) = 3.42 m 3
Excess air supplied =
(O2*V) / (21-O2) = (6.0*3.42) / (21-6) =20.52/15= 1.37 m3
= 5.06
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Volume of fuel
= 5
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Air fuel ratio = Volume of air
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Total quantity of air supplied = 3.69 + 1.37 = 5.06 m3 .
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4) A sample of fuel has the following percentage composition : Carbon = 86 per cent ;
Hydrogen = 8 per cent ; Sulphur = 3 per cent ; Oxygen = 2 per cent ; Ash = 1 per cent. For an
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air-fuel ratio of 12 : 1, calculate :
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(i) Mixture strength as a percentage rich or weak.
(ii) Volumetric analysis of the dry products of combustion.
Solution.
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Element, wt. (kg)
C = 0.86
H 2 = 0.08
S = 0.03
O 2 = 0.02
O2 reqd. (kg)
0.86*(8/3) = 2.29
0.08*8 = 0.64
0.03*(1/1) = 0.03
Total O 2 = 2.96
Weight of oxygen to be supplied per kg of fuel = 2.96 – 0.02 = 294 kg.
Weight of minimum air required for complete combustion =(294 *100)/23= 12.78 kg
Hence “correct” fuel-air ratio = (1/1278): 1
But actual ratio is (1/12): 1.
(i) Mixture strength = (1278/12) × 100 = 106.5%
This shows that mixture is 6.5% rich. (Ans.)
Deficient amount of air = 12.78 – 12 = 0.78 kg
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 14 of 29
MODULE-II --- Combustion Thermodynamics
2014
APPLIED THRMODYNAMICS
Amount of air saved by burning 1 kg of C to CO instead of CO 2
= Oxygen saved ×100/23
=
*
=5.8 kg
Hence 0.78/58 = 0.134 kg of carbon burns to CO and as such 0.86 – 0.134 = 0.726 kg of carbon
burns to CO2 .
∴ CO formed = 0.134 ×7/3= 0.313 kg
CO 2 formed = 0.726 ×11/3= 2.662 kg
N 2 supplied = 12 × 0.77 = 9.24 kg
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SO 2 formed = 0.03 × 2 = 0.06 kg.
28
44
28
64
N
M
0.313
2.662
9.24
0.06
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CO
CO2
N2
SO2
Molecular weight
(y)
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Weight (x)
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C
Dry products
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ii) The percentage composition of dry flue gases is given as below:
Proportional
volume(z)=x/y
0.0112
0.0605
0.3300
0.0009
Σ z = 0.4026
Percentage volume
=z/z*100
2.78%
15.03%
81.97%
0.22%
5) A fuel (C 10 H 22) is burnt using an air-fuel ratio of 13: 1 by weight. Determine the complete
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volumetric analysis of the products of combustion, assuming that the whole Amount of
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hydrogen burns to form water vapour and there is neither any free oxygen nor any free carbon.
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The carbon burns to CO 2 and CO.Air contains 77% of nitrogen and 23% of oxygen by weight.
Solution:
Combustion equation is
2C 10 H 22 + 31O 2 = 20CO 2 + 22H 2 O
2 × 142 + 31 × 32 = 20 × 44 + 22 × 18
or
284 + 992 = 880 + 396
∴ Air required for complete combustion= (992 *100)/ (284/ 23) = 15.2 kg/kg of fuel
Air actually supplied = 13 kg/kg of fuel
∴ Deficiency of air = 15.2 – 13 = 2.2 kg/kg of fuel
Also 1 kg of C requires(4/3)*(100/23)= 5.8 kg of less air to burn to CO instead of CO2 .
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
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MODULE-II --- Combustion Thermodynamics
2014
APPLIED THRMODYNAMICS
Hence 22/58 = 0.379 kg C is burnt to CO;
– 0.379 = 0.466 kg of C is burnt to CO2.
And
Weight of CO2 formed = 0.466 ×11/3= 1.708 kg
Weight of CO formed = 0.379 ×7/3= 0.884 kg
Weight of H 2 O formed = (22/142)× 9 = 1.394 kg
Weight of N 2 from air = 13 × 0.77 = 10.01 kg.
44
28
18
28
0.0388
0.0316
0.0774
0.3575
Σ z = 0.5053
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1.708
0.884
1.394
10.01
Proportional
volume(z)=x/y
Percentage volume
=z/z*100
7.678%
6.254%
15.317%
70.750%
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C
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CO2
CO
H2O
N2
Molecular weight
(y)
t
Weight (x)
Pr
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Products of
combustion
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-N
PT
EL
-
N
M
6) The following results were obtained in a trial on a boiler fitted with economizer :
CO 2
CO
O2
N2
Analysis of gas entering the
8.3
0
11.4
80.3
economizer
Analysis of gas leaving
7.9
0
11.5
80.6
the economizer
(i)
Determine the air leakage into the economizer if the carbon content of the fuel
is 80 percent.
(ii)
(ii) Determine the reduction in temperature of the gas due to air leakage if
atmospheric temperature is 20°C and flue gas temperature is 410°C. Ash
collected from ash pan is 15 per cent by weight of the fuel fired.Take: Cp for
air = 1.005 kJ/kg K and c p for flue gas = 1.05 kJ/kg K.
Solution:
(i) Air supplied
=
Air supplied on the basis of conditions at entry to the economizer = (80.3 *80)/33(8.3+0)
= 23.45 kg
Air applied on the basis of conditions at exit = (80.6*80)/33(7.9+0) = 24.73 kg
∴ Air leakage = 24.73 – 23.45 = 1.28 kg of air per kg of fuel. (Ans)
For each kg of fuel burnt, the ash collected is 15% i.e., 0.15 kg.
∴ Weight of fuel passing up the chimney = 1 – 0.15 = 0.85 kg
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 16 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
∴ Total weight of products
= Weight of air supplied per kg of fuel+ Weight of fuel passing through chimney per kg
of fuel
= 23.45 + 0.85 = 24.3 kg
Heat in flue gases per kg of coal
= Weight of flue gases × Specific heat × Temperature rise above 0°C
= 24.3 × 1.05 × (410 – 0) = 10461 kJ
Heat in leakage air = Weight of leakage air × Specific heat × Temperature rise of air above 0°C
je
c
t
= 1.28 × 1.005 × (20 – 0) = 25.73 kJ.
specific heats, but sharing a common temperature t.
EI
C
(1.005 × 1.28 + 24.3 × 1.05) t = 10461 + 25.73
T
For heat balance :
Pr
o
We can still consider, in the mixture, the gas and the air as separate and having their own
26.8 t = 10486.73
N
M
∴ t = 391.3°C
PT
EL
-
∴ Fall in temperature as a result of the air leakage into the economizer
= 410 – 391.3 = 18.7°C.
U
-N
7) The chemical formula for alcohol is C 2 H 6 O. Calculate the stoichiometric air/fuel ratio by
mass and the percentage composition of the products of combustion per kg of C 2H 6O.
VT
Solution:. Chemical equation for complete combustion of given fuel can be written as
follows :
C2H 6O + 3O 2 → 2CO 2 + 3H 2O
(1 × 46) + (3 × 32) = (2 × 44) + (3 × 18)
For complete combustion of 1 kg of C2H 6O, oxygen required
=
(3×32)/46= 2.087 kg of oxygen
= 2.087 ×(100/23) = 9.074 kg of air
A : F ratio for complete combustion = 9.074 : 1. Ans.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 17 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
Also 46 kg of fuel produces products of combustion = 88 + 54 = 142 kg
∴ 1kg of fuel produces(142/46)= 3.087 kg of products of combustion (i.e CO and H 2O)
Hence CO2 produced by fuel = 88/46 × 100 = 1.913 or 191.3%. (Ans)
H 2 O produced by fuel =54/46× 100 = 1.174 or 117.4%. (Ans)
8) Calculate the amount of theoretical air required for the combustion of 1 kg of acetylene (C 2
H2 ) to CO 2 and H 2O.
je
c
t
Solution.
For combustion of acetylene (C 2 H 2 ) the stoichiometric equation is written as
Pr
o
C 2 H 2 + x O 2 → a CO 2 + b H 2O …(i)
Balancing the carbon atoms on both sides of the combustion eqn. (i), we get
T
i.e. a = 2
EI
C
2C = a C
Now balancing hydrogen atoms on both sides, we get
M
2H = 2bH
N
b=1
PT
EL
-
Thus, eqn. (i) becomes
C 2 H 2 + x O 2 → 2 CO 2 + 1 H 2O Now, balancing oxygen atoms in the above equation
2x = 2 × 2 + 1 = 5 i.e., x = 2.5
-N
Hence, the final combustion eqn. (i) is
VT
U
C 2 H 2 + 2.5 O 2 → 2 CO 2 + 1 H 2O...(ii)
Thus, for combustion of C 2 H 2 in air, we get
C 2 H 2 + 2.5O 2 + 2.5(79/21) N 2 → 2CO 2 + H 2O + 2.5(79/21)N 2
On a mass basis, this becomes
(2 × 12 + 2 × 1) C 2H 2 + 2.5 (2 × 16) O 2 + 2.5(79/21) (2 × 14) N 2
→ 2 (12 + 2 × 16) CO 2 + (1 × 2 + 1 × 16) H 2O + 2.5(79/21)(2 × 14) N 2
26 kg C 2 H 2 + 80 kg O 2 + 263.3 N 2 → 88 kg CO 2 + 18 kg H2O + 263.3 kg N2
or 1 kg C 2 H 2 + 3.076 kg O 2 + 10.12 kg N 2 → 3.38 kg CO 2 + 0.69 kg H 2O + 10.12 kg N 2
i.e., Amount of air = 3.076 + 10.12 = 13.196 kg of air per kg of C 2H 2 Hence amount of
theoretical air required for combustion of 1 kg acetylene = 13.196 kg.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 18 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
9) One kg of octane (C
H
8
18)
2014
is burned with 200% theoretical air. Assuming complete
combustion determine :
(i) Air-fuel ratio
(ii) Dew point of the products at a total pressure 100 kPa.
Solution:
The equation for the combustion of C 8 H 18 with theoretical air is
C 8 H 18 + 12.5 O 2 + 12.5(79/21) N2 → 8CO 2 + 9H 2O + 12.5(79/21) N2
For 200% theoretical air the combustion equation would be
Pr
o
→ 8CO 2 + 9H 2O + (1) (12.5) O 2 + (2) (12.5) (79/21) N 2
je
c
t
C 8 H 18 + (2) (12.5) O 2 + (2) (12.5) (79/21) N 2
Mass of fuel = (1) (8 × 12 + 1 × 18) = 114 kg/mole
T
Mass of air = (2) (12.5) (1+(79/21))*28.97 = 3448.8 kg/mole of fuel
EI
C
(i) Air-fuel ratio:
M
Air-fuel ratio, A/F = Mass of air /Mass of fuel=3448.8/114= 30.25
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EL
-
(ii) Dew point of the products, tdp:
N
i.e., A/F = 30.25. (Ans)
Total number of moles of products = 8 + 9 + 12.5 + (2) (12.5) (79/21)= 123.5 moles/mole fuel
Mole fraction of H 2O = 9/123.5 = 0.0728
-N
Partial pressure of H 2O = 100 × 0.0728 = 7.28 kPa
VT
point temperature.
U
The saturation temperature corresponding to this pressure is 39.7°C which is also the dew-
10) One kg of ethane (C
Hence tdp = 39.7°C.
2
H 6) is burned with 90% of theoretical air. Assuming complete
combustion of hydrogen in the fuel determine the volumetric analysis of the dry products of
combustion.
Solution:
The complete combustion equation for C 2 H 6 is written as:
C 2 H 6 + 3.5O 2 → 2CO 2 + 3H 2 O
The combustion equation for C 2 H 6 for 90% theoretical air is written as:
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 19 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
C 2 H 6 + (0.9) (3.5) O 2 + (0.9) (3.5) (79/21) N 2 → a CO 2 + b CO + 3H 2O +
(0.9) (3.5) (79/21) N 2
By balancing carbon atoms on both the sides, we get
2=a+b
...(i)
By balancing oxygen atoms on both the sides, we get
(0.9) (3.5) (2) = 2a + b + 3
...(ii)
Substituting the value of b (= 2 – a) from eqn. (i) in eqn. (ii), we get
(0.9) (3.5) (2) = 2a + 2 – a + 3
6.3 = a + 5
je
c
t
∴ a = 1.3
Pr
o
b = 2 – a = 2 – 1.3 = 0.7
Thus the combustion equation becomes:
EI
C
T
C 2 H 6 + (0.9) (3.5) O 2 + (0.9) (3.5) (79/21) N 2 → 1.3CO 2 + 0.7CO + 3H 2O +
(0.9) (3.5) (79/21) N 2
M
Total number of moles of dry products of combustion
N
= 1.3 + 0.7 + (0.9) (3.5) (79/21)
PT
EL
-
= 1.3 + 0.7 + 11.85 = 13.85 moles/mole of fuel
Volumetric analysis of dry products of combustion is as follows :
CO 2 = 13/ 13.85 × 100 = 9.38% (Ans)
-N
CO = 07/ 13.85 × 100 = 5.05%. (Ans)
VT
U
N 2 =11.85/13.85× 100 = 85.56% (Ans)
11) Methane (CH 4 ) is burned with atmospheric air. The analysis of the products on a ‘dry’
basis is as follows : CO 2 = 10.00%, O 2 = 2.37%, CO = 0.53%, N 2 = 87.10%.
(i) Determine the combustion equation ; (ii) Calculate the air-fuel ratio ;
(iii) Percent theoretical air.
Solution:
(i) Combustion equation :
From the analysis of the products, the following equation can be written, keeping in mind that
this analysis is on a dry basis.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 20 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
x CH 4 + y O 2 + z N 2 → 10.0 CO 2 + 0.53 CO + 2.37 O 2 + a H 2O + 87.1 N 2
To determine all the unknown co-efficients let us find balance for each of the elements.
Nitrogen balance : z = 87.1
Since all the nitrogen comes from the air, z/y=79/21; y = 87.1/ (79/21)= 23.16
Carbon balance : x = 10.00 + 0.53 = 10.53
Hydrogen balance : a = 2x = 2 × 10.53 = 21.06
Oxygen balance. All the unknown co-efficients have been solved for, and in this case the
oxygen balance provides a check on the accuracy. Thus, y can also be determined by an oxygen
balance
je
c
t
y = 10.00 +053/2+ 2.37 + (21.06/2) = 23.16
Pr
o
Substituting these values for x, y, z and a, we have,
10.53 CH 4 + 23.16 O 2 + 87.1 N 2 → 10.0 CO 2 + 0.53 CO + 2.37 O 2 + 21.06 H 2 O + 87.1 N 2
T
Dividing both sides by 10.53, we get the combustion equation per mole of fuel,
M
EI
C
CH 4 + 2.2 O 2 + 8.27 N 2 → 0.95 CO 2 + 0.05 CO + 2H 2O + 0.225 O 2 + 8.27 N 2
2.2 + 8.27 = 10.47 moles air/mole fuel
PT
EL
-
N
(ii) Air-fuel ratio A/F : The air-fuel ratio on a mole basis is
The air-fuel ratio on a mass basis is found by introducing the molecular weights
A/F = (10.47 28.97)/ (12+ (1* 4))= 18.96 kg air/kg fuel. (Ans)
-N
The theoretical air-fuel ratio is found by writing the combustion equation for theoretical air,
A/F theor. =
(iii)
VT
U
CH 4 + 2O 2 + 2 (79/21) N 2 → CO 2 + 2H 2 O + (2) (79/21) N 2
= 17.24 kg air/kg fuel. (Ans)
Percent theoretical air:
Per cent theoretical air = 18.96/17.24× 100 = 110%.
12) The following analysis relate to coal gas : H 2 = 50.4%, CO = 17 %, CH 4 = 20 %, C 4 H 8 =
2 %, O 2 = 0.4 %, N 2 = 6.2 %, CO 2 = 4 %.
(i) Calculate the stoichiometric A/F ratio.
(ii) Find also the wet and dry analyses of the products of combustion if the actual mixture
Is 30 per cent weak?
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 21 of 29
MODULE-II --- Combustion Thermodynamics
2014
APPLIED THRMODYNAMICS
Solution:
The example is solved by a tabular method ; a specimen calculation is given below :
For CH 4 :
CH 4 + 2O 2 → CO 2 + 2H 2 O
i.e., 1 mole CH 4 + 2 moles O 2 → 1 mole CO 2 + 2 moles H 2 O
There are 0.2 moles of CH 4 per mole of the coal gas, hence
0.2 moles CH 4 + 0.2 × 2 moles O 2
2
+ 0.2 × 2 moles H 2 O
∴ O 2 required for the CH 4 in the coal gas = 0.4 moles per mole of coal gas.
(i)
2
3
2H 2 + O2 → 2H 2O
2CO + O2 → 2CO2
CH 4 + 2O2 → CO 2 +2H 2O
C 4 H 8 + 6O2 → 4CO2 +4H 2O
—
—
—
O2
moles/mole
fuel
4
0.252
0.085
0.400
0.120
-0.004
—
—
5
—
0.17
0.20
0.08
—
—
0.04
6
0.504
—
0.40
0.08
—
—
—
0.853
0.49
0.984
Pr
o
je
c
Combustion equation
M
EI
C
T
0.504
0.17
0.20
0.02
0.004
0.062
0.04
N
1
H 2O
CO
CH 4
C4H8
O2
N2
CO 2
Moles/mole
fuel
PT
EL
-
Product
t
The oxygen in the fuel (0.004 moles) is included in column 4 as a negative quantity.
Products
CO 2
H2 O
Stoichiometric A/F ratio :
-N
Air required = 0.853/0.21 = 4.06 moles/mole of fuel
U
(where air is assumed to contain 21% O 2 by volume)
VT
∴ Stoichiometric A/F ratio = 4.06/1 by volume. (Ans)
(ii) Wet and dry analyses of the products of combustion if the actual mixture is 30% weak :
Actual A/F ratio with 30% weak mixture
= 4.06 + (30/100) × 4.06 = 1.3 × 4.06 = 5.278/1
Associated N2 = 0.79 × 5.278 = 4.17 moles/mole fuel
Excess oxygen = 0.21 × 5.278 – 0.853 = 0.255 moles
Total moles of N 2 in products = 4.17 + 0.062 = 4.232 moles/mole fuel.
Analysis by volume of wet and dry products:
Product
CO 2
Moles/mole fuel
0.490
% by vol. (dry)
9.97
% by vol. (wet)
8.31
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 22 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
—
5.19
84.84
100
H 2O
0.984
O2
0.255
N2
4.170
Total wet = 5.899– H 2O = 0.984
Total dry = 4.915
2014
16.68
4.32
70.69
100
Frequently asked Questions.
1) Write a short note on ‘excess air’.
2) What do you mean by stoichiometric air-fuel (A/F) ratio ?
je
c
t
3) What is enthalpy of formation (ΔHf) = ?
Pr
o
4 What is the difference between higher heating value (HHV) and lower heating value (LHV) of
EI
C
5) What is ‘adiabatic flame temperature’ ?
T
the fuel ?
6) The following is the analysis of a supply of coal gas :
N
M
H2 = 49.4 per cent ; CO = 18 per cent ; CH4 = 20 per cent ; C4H8 = 2 per cent ; O2 = 0.4 per
PT
EL
-
cent ; N2 = 6.2 per cent ; CO2 = 4 per cent. (i) Calculate the stoichiometric A/F ratio.
(ii) Find also the wet and dry analyses of the products of combustion if the actual mixture is 20
per cent weak.
-N
7) ) Methane (CH 4 ) is burned with atmospheric air. The analysis of the products on a ‘dry’
U
basis is as follows : CO 2 = 10.00%, O 2 = 2.37%, CO = 0.53%, N 2 = 87.10%.
VT
(i) Determine the combustion equation ; (ii) Calculate the air-fuel ratio ;
(iii) Percent theoretical air.
8) A fuel (C 10 H 22) is burnt using an air-fuel ratio of 13: 1 by weight. Determine the complete
volumetric analysis of the products of combustion, assuming that the whole Amount of
hydrogen burns to form water vapour and there is neither any free oxygen nor any free carbon.
The carbon burns to CO 2 and CO.Air contains 77% of nitrogen and 23% of oxygen by weight.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 23 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
Assignments:
1) Write a short note on ‘excess air’.
2) What do you mean by stoichiometric air-fuel (A/F) ratio ?
3) What is enthalpy of formation (ΔHf) = ?
4 What is the difference between higher heating value (HHV) and lower heating value (LHV) of
the fuel ?
5) What is ‘adiabatic flame temperature’ ?
je
c
t
6) During a trial in a boiler, the dry flue gas analysis by volume was obtained as CO2 = 13%, CO
Pr
o
= 0.3%,O2 = 6%, N2 = 80.7%. The coal analysis by weight was reported as C = 62.4%, H2 =
4.2%, O2 = 4.5%, moisture= 15% and ash 13.9%. Estimate : (a) Theoretical air required to burn
T
1 kg of coal. (b) Weight of air actually supplied per kg of coal. (c) The amount of excess air
EI
C
supplied per kg of coal burnt. [Ans. 8.5 kg ; 11.5 kg ; 3 kg]
N
PT
EL
-
remainder—incombustible matter.
M
7) In a boiler trial, the analysis of the coal used is as follows : C = 20%, H2 = 4.5% , O2 = 7.5%,
The dry flue gas has the following composition by volume : CO2 = 8.5%, CO = 1.2%, N2 =
80.3%, O2 = 10%. Determine : (i) Minimum weight of air required per kg of coal.
-N
(ii) Percentage excess air. [Ans. (i) 3.56 kg, (ii) 63.2%]
U
8) Calculate the stoichiometric air-fuel ratio for the combustion of a sample of dry anthracite of
VT
the following composition by mass : C = 90 per cent ; H2 = 3 per cent ; N2 = 1 per cent ; Sulphur
= 0.5 per cent ; ash = 3 per cent. If 20 per cent excess air is supplied determine :
(i) Air-fuel ratio (ii) Wet analysis of the products of combustion by volume.
[Ans. 11.25/1 (i) 13.5/1 ; (ii) CO2 = 16.3%, H2O = 0.03%, SO2 = 3.51%, N2 = 80.3%]
9) The percentage composition of a fuel by weight is as follows :
Carbon = 89.3 per cent ; Hydrogen = 5 per cent ; Oxygen = 4.2 per cent ; Nitrogen = 1.5 per cent
and the remainder ash. Determine the stoichiometric air-fuel ratio by mass. If 30 per cent excess
air is supplied, find the percentage composition of dry flue gases by volume.
[Ans. 11.74 ; CO2, = 14.3%, O2 = 4.9%, N2 = 80.8%]
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 24 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
Self Answered Question & Answer
1) Find the stoichiometric air-fuel ratio for the combustion of ethyl alcohol (C2H6O), in a petrol
engine. Calculate the air-fuel ratios for the extreme mixture strengths of 90% and 120%.
Determine also the wet and dry analyses by volume of the exhaust gas for each mixture strength.
[Ans. 8.96/1 ; 9.95/1 ; 7.47/1, Wet analysis : CO2 = 11.2%, H2O = 16.8%, O2 = 1.85%,
N2 = 70.2%, Dry analysis : CO2 = 13.45%, O2 = 2.22%, N2 = 84.4%
Wet analysis : CO2 = 6.94%, CO = 6.94%, H2 = 20.8%, N2 = 65.3%
je
c
t
Dry analysis : CO2 = 8.7%, CO = 8.7%, N2 = 82.5%]
Pr
o
2) The following is the analysis of a supply of coal gas :
H2 = 49.4 per cent ; CO = 18 per cent ; CH4 = 20 per cent ; C4H8 = 2 per cent ; O2 = 0.4 per
T
cent ; N2 = 6.2 per cent ; CO2 = 4 per cent. (i) Calculate the stoichiometric A/F ratio.
EI
C
(ii) Find also the wet and dry analyses of the products of combustion if the actual mixture is 20
per cent weak.
N
M
[Ans. (i) 4.06/1 by volume ; (ii) Wet analysis : CO2 = 9.0%, H2O = 17.5%, O2 = 3.08%,
PT
EL
-
N2 = 70.4%. Dry analysis : CO2 = 10.9%, O2 = 3.72%, N2 = 85.4%]
3) Calculate the stoichiometric air-fuel ratio for the combustion of a sample of dry anthracite of
the following composition by mass : C = 90 per cent ; H2 = 3 per cent ; N2 = 1 per cent ; Sulphur
-N
= 0.5 per cent ; ash = 3 per cent. If 20 per cent excess air is supplied determine :
VT
U
(i) Air-fuel ratio (ii) Wet analysis of the products of combustion by volume.
[Ans. 11.25/1 (i) 13.5/1 ; (ii) CO2 = 16.3%, H2O = 0.03%, SO2 = 3.51%, N2 = 80.3%]
4) The percentage composition of a sample of fuel was found to be C = 85%. H2 = 9%, S = 3%,
O2 = 1.5%, Ash = 1.5%. For an air-fuel ratio of 12 : 1. Calculate : (i) The mixture strength as a
percentage rich or weak. (ii) The volumetric analysis of the dry products of combustion.
[Ans. 8.83% rich ; CO2 = 12.16%, CO = 2.54%, N2 = 84.83%, SO2 = 0.47%]
5) A fuel has the following composition by weight : Carbon = 86% ; hydrogen = 11.75% and
oxygen = 2.25%. Calculate the theoretical air supply per kg of fuel, and the weight of products of
combustion per kg of fuel.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 25 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
[(Ans.) 13.98 kg ; 4.21 kg]
6) For a hydrocarbon fuel having 84.8% carbon, 15.2% hydrogen by mass determine the
stoichiometric air-fuel ratio. Also obtain the volumetric composition of combustion products
with 15% excess air being supplied.
[(Ans.) 15.11, 12.51% CO2, 2.89% O2, 84.60% N2]
7) During the combustion of coal having 22% carbon, 3.5% hydrogen, 6.5% oxygen and
remaining ash by mass, the dry flue gas has volumetric composition of 8% CO2, 1.5% CO,
80.5% N2 and 10% O2. Calculate the minimum air required per kg of coal and % excess air.
je
c
t
Consider air to have 23% and 77% of oxygen and nitrogen respectively by mass.
Pr
o
[(Ans.) 3.48 kg air, 43.3% excess air]
T
8) C10 H22 is burnt with air fuel ratio of 13 by mass. Combustion is such that complete of
EI
C
hydrogen burns into water and there is no free oxygen and no free carbon. Also the carbon
M
dioxide and carbon monoxide are present in combustion products. Determine volumetric analysis
N
of combustion products.
PT
EL
-
[(Ans.) 7.68% CO2, 6.25% CO, 15.32% H2O, 70.75% N2]
9) A coal sample has 66% C, 5.9% H2, 1% S, 19.9% O2, 1.5% N2, 5.6% ash by mass and
calorific value of 29000 kJ/kg. When this coal is burnt with 30% excess air the temperature of
-N
flue gases leaving is 300_C and ambient temperature is 17_C. Considering combustion to be
U
complete and partial pressure of moisture to be 0.07 bar in flue gases and energy accompanied
VT
being 3075 kJ/kg of steam, determine, (i) the air-fuel ratio by mass.
(ii) the volumetric analysis of combustion products.
(iii) the heat carried away by flue gases as percentage of heat released by coal.
[(Ans.) 11.55, 19.37% O2, 4.25% H2O, 0.16% SO2, 4.91% O2, 71.31% N2, 17.36%]
10) Determine the net and gross calorific values per kg of mixture at constant pressure for
stoichiometric mixture of air and C6H6 (benzene) vapour at 25_C. Enthalpy of combustion for
C6H6 at 25_C is – 3169500 kJ/kmol and the water is present in vapour phase in the combustion
products.
[(Ans.) 2861 kJ/kg, 2981 kJ/kg]
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 26 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
Test Your Skills:
Choose the Correct Answer :
1. The smallest particle which can take part in a chemical change is called
(a) Atom (b) Molecule (c) electron (d) compound.
2. A chemical fuel is a substance which releases ...... on combustion.
(a) chemical energy (b) heat energy
je
c
t
(c) sound energy (d) magnetic energy.
3. The most important solid fuel is
Pr
o
(a) wood (b) charcoal
EI
C
T
(c) coal (d) all of the above.
4. For each mole of oxygen, number of moles of nitrogen required for complete combustion of
N
PT
EL
-
(a) 20/21 (b) 2/21 (c) 77/21 (d) 79/21.
M
carbon are
5. Modern practice is to use ...... excess air.
-N
(a) 5 to 10 per cent (b) 15 to 20 per cent
U
(c) 20 to 25 per cent (d) 25 to 50 per cent.
VT
6. Stoichiometric air-fuel ratio by mass for combustion of petrol is
(a) 5 (b) 10 (c) 12 (d) 15.05.
7. An analysis which includes the steam in the exhaust is called
(a) dry analysis (b) wet analysis
(c) dry and wet analysis (d) none of the above.
8. The Orsat apparatus gives
(a) volumetric analysis of the dry products of combustion
(b) gravimetric analysis of the dry products of combustion
(c) gravimetric analysis of products of combustion including H2O
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 27 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
2014
(d) volumetric analysis of products of combustion including H2O.
9. In the Orsat apparatus KOH solution is used to absorb
(a) carbon monoxide (b) carbon dioxide
(c) oxygen (d) none of the above.
10. Enthalpy of formation is defined as enthalpy of compounds at
(a) 25°C and 10 atmospheres (b) 25°C and 1 atmosphere
Pr
o
(a) solid (b) gaseous (c) solid and gaseous (d) none of the above.
je
c
11. Bomb calorimeter is used to find the calorific value of ...... fuels.
t
(c) 0°C and 1 atmosphere (d) 100°C and 1 atmosphere.
T
12. When the fuel is burned and the water appears in the vapour phase, the heating value of fuel
EI
C
is called
M
(a) enthalpy of formation (b) lower heating value
PT
EL
-
N
(c) higher heating value (d) none of the above.
13. Heat released in a reaction at constant pressure is called
(a) entropy change (b) enthalpy of reaction
VT
U
(e) all of the above.
-N
(c) internal energy of reaction (d) none of the above
14. When the fuel is burned and water is released in the liquid phase, the heating value of fuel is
called
(a) higher heating value (b) lower heating value
(c) enthalpy of formation (d) none of the above.
15. Choose the correct statement :
(a) Number of atoms of each constituent are not conserved in a chemical reaction.
(b) The mass of all the substances on one side of the equation may not be equal to the mass of all
the substances on the other side.
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 28 of 29
MODULE-II --- Combustion Thermodynamics
APPLIED THRMODYNAMICS
(c) The number of atoms of each constituent are conserved in a chemical reaction.
(d) The number of moles of the reactants in a chemical equation are equal to the number of
moles of the products.
ANSWERS
je
c
t
1. (a) 2. (b) 3. (c) 4. (d) 5. (d) 6. (d) 7. (b)
Pr
o
8. (a) 9. (b) 10. (b) 11. (a) 12. (b) 13. (b) 14. (a)
VT
U
-N
PT
EL
-
N
M
EI
C
T
15. (c).
Dr. A.R Anwar Khan, Prof & HOD, GHOUSIA COLLEGE OF ENGINERING, RAMANAGARA
Page 29 of 29
2014