ENERGY CONVERSION

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• ENERGY CONVERSION
• ES 832a
• Eric Savory
• www.eng.uwo.ca/people/esavory/es832.htm
• Lecture 6 Basics of combustion
• Department of Mechanical and Material Engineering
• University of Western Ontario

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Fundamental Concepts of Combustion Theory

• The first step in any conversion system is to
  determine the amount of energy which will be
  available for conversion. For combustion
  processes, this information is important for
  determining the necessary mass rates and the
  temperatures of the products entering a turbine.
• Recall that a turbine is normally designed for a
  specific point of operation. By far, combustion
  is still the most common form of energy
  generation.

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Combustion Theory

• Determine the products of combustion
• e.g 2H2 O2 ? 2H2O
• Determine the heat released by combustion and the
  temperature of the products.
• Energy balance internal energy and enthalpy
• Q UP0 UR0 ?U ?Ho if K.E. negligible.
• Determine the energy available for conversion.

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Combustion processes (1)

• EXOTHERMIC (produces heat) reactions where mass
  is conserved.
• Most combustion processes use oxygen as oxidant
  (to burn with).
• The substances used before the reaction are known
  as REACTANTS and those arising from the process
  are referred to PRODUCTS.
• The substance undergoing oxidation is known as a
  FUEL.
• STOICHIOMETRIC mixtures are where the air mass
  available is exactly that required for
  combustion.
• EXCESS air is the amount of air present above
  stoichiometric ratios.

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Combustion processes (2)
PRINCIPAL CONSERVATION LAWS 1. The number
of atoms is conserved. 2. Mass is
conserved.
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Relative atomic mass of some common elements
Element name Symbol Relative atomic mass
Oxygen O 16
Nitrogen N 14
Hydrogen H 1
Carbon C 12
Sulphur S 32
Defined so that the elements mass is scaled such
that Carbon-12 is exactly 12.
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Relative molecular mass of common substances
Name Symbol Relative molecular mass
Oxygen O2 32
Nitrogen N2 28
Hydrogen H2 2
Water / steam H2O 18
Carbon dioxide CO2 44
Carbon monoxide CO 28
Sulphur dioxide SO2 64
mass of one molecule of that substance,
relative to 1/12th the mass of one atom of
Carbon-12
Molar mass M relative molecular mass,
numerically, but has units of kg / kmol
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1. Conservation of atoms

• Substances consist of atoms in combinations
  called molecules. The atoms recombine during
  chemical reactions, but their number remains
  unchanged.
• Example Combustion of hydrogen
• element Hydrogen (H) molecule H2
• element Oxygen (O) molecule O2
• Water molecule H2O
• Reactants Products
• nH2 mO2 ? H2O n, m of molecules

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To balance the reaction the number of atoms
remain the same (mathematically) Hydrogen 2
n 2 ? n 1 Oxygen 2 m 1 ? m ½
Hence, H2 ½ O2 ? H2O This could also
be written 2H2 O2 ? 2H2O (or any multiple !!)
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2. Conservation of Mass

• From (1), mass must also be conserved.
• Each atom has a characteristic mass which is a
  property of the isotope. This is usually
  expressed in terms of the relative molecular mass
  (either g / mol or kg / kmol where a mol is NA
  6.022 x 1023 units)
• For an ideal gas, 1 mol at STP occupies a volume
  of 23.6 litres. (STP 15oC, 1 atm)
• e.g. Hydrogen H MH 1 1 kg / kmol
• Oxygen O MO 16 16 kg / kmol
• ? H2 2 MH MH2 2 kg / kmol
• ? O2 2 MO MO2 32 kg / kmol
• ? H2O 2 MH ½ MO2 18 kg / kmol
• Hence, 2H2 O2 ? 2 H2O
• It can be stated that 2 (MH2) MO2 ? 2
  MH2O
• 4 kg / kmol 32 kg / kmol ? 36 kg / kmol

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Example

• Calculate the mass of oxygen required to burn 1
  m3 of fuel propane at STP. How much air must be
  supplied (at STP)?
• Solution
• (a) Need mO2 required to burn 1 m3 C3H8
  (propane) at STP
• To accomplish this task we must first determine
  the relative amount of reactants and products to
  burn the propane. This requires setting up the
  chemical reaction.
• The next step is to determine the relative mass
  of each based on the ratios determined from the
  chemical reaction.
• Then express everything in terms of 1m3 C3H8
  (propane) at STP. This will require the use of
  the ideal gas law (or tabled density values) for
  STP.
• Finally, since air is used, calculate the other
  components based on the composition of air.
• (b) Set-up the primary and secondary
  relationships.

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• Calculate the relative amounts first in terms of
  mol
• We know the products will be CO2, H2O
• Note all products are gaseous.
• n C3H8 m O2 ? p CO2 q H2O
• for C 3n p H 8n 2q O 2m
  2p q
• Solving in terms of q q 4n, p 3n ¾ q
  m p q/2 5/4 q
• n q/4, m 5/4 q, p ¾ q
• Thus, for n 1 C3H8 5O2 ? 3CO2 4H2O

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• Calculate the relative mass ratios using the
  molar mass
• Obtain molar mass for each species involved in
  the chemical reaction
• MO2 32 kg / kmol MH2O 18 kg / kmol
• MC 12 kg / kmol MH2 2 kg / kmol
• MC3H8 3 MC 4 MH2 44 kg / kmol
• Hence, for every mol of C3H8 we need 5 mol of O2
• or mO2/mC3H8 5MO2/1MC3H8
• (5 x 32) kg/kmol / 44 kg/kmol 3.636 kgO2 /
  kgC3H8
• (from tables rC3H8 2.02 kg/m3)

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Note from gas constant (R) RC3H8 R / M
(8.314 J/mol-K) / (44 kg/kmol)
189 J/kg-K (recall R is universal gas
constant) ? P / RT (101,300 Pa) / (189
J/kg-K) (273 K) 1.97
kg/m3 which is very close to 2.02 kg/m3 mC3H8
? Vol 2.02 kg/m3 1 m3 2.02 kg mO2 3.636
kgO2/kgC3H8 2.02 kgC3H8 7.34 kg of
oxygen
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• Calculate the total amount of air
• The molar (i.e. volumetric) ratios of air
  composition are approximately N2 79 Argon
  (Ar) 1 and O2 20
• For every 1 m3 of propane, we have (from the
  chemical equation of part (a))
• 5m3 O2 5(0.79/0.20)m3 N2 5(0.01/0.20)m3 Ar
• 25 m3 of air.
• What if the ideal gas equation had been used?
• We would then predict 7.16 kg of oxygen but the
  volume of air would remain the same.

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SUMMARY Combustion processes are exothermic
reactions in which mass and energy are conserved.
Molecular ratios are used to determine volumetric
and mass ratios. Stoichiometric mixtures are
those for which the amount of air is exactly that
need for 100 combustion. Excess air is the
amount of air above Stoichiometric conditions.
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Internal energy and enthalpy of combustion

• A heat conversion system is designed to produce
  a desired power output. Typically, the engineer
  is provided with the desired output levels, the
  turbine thermodynamic and mechanical efficiencies
  and the operating point (inlet temperature and
  pressure). The task of the engineer then becomes
  to determine the amount of energy generation and
  the mass flow rate of products needed to achieve
  these conditions. To do so requires that the
  engineer know
• the amount of fuel required
• the amount air required to combust these fuels
  and the mass flow rate and temperature of the
  products of combustion.

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Objective

• 1. Calculate the rate of energy which is
  liberated during a combustion process
• 2. Determine the minimum mass flow rate of air
  required for combustion
• 3. Determine the temperature and flow rate of
  products.

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The heat released, Q, during a chemical reaction
is a property of the reactants and the
process. For a closed system (no-flow) one
obtains Q (UP2 - UP0) (UR0 - UR1)
?U0 And for an open system (steady-flow) Q
(HP2 - HP0) (HR0 - HR1) ?H0 ?K.E.
Subscripts P product R reactant 0, 1, 2
temperatures T0, T1 and T2 (note sign
convention Q gt 0 if supplied by surroundings,
thus ?U0 lt 0 ?H0 lt 0 ) ?U0 and ?H0 are the
internal energy and enthalpy of combustion,
respectively. ?K.E. is kinetic energy
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• The enthalpy of combustion is usually given in
  terms of kJ/kmol, ?h0, such that
• ?H0 n?h0 where n kmol of fuel
• ?h0
• Is a property of the reaction. It is given at a
  standard reference (25C and 1atm)
• It is assumed that reactants and products are at
  the same temperature and pressure
• If not tabulated, it can be calculated from the
  energy of formation, hf, of the individual
  chemical species at 25C and 1 atm.
• ?h0 ?nhf - ?nhf n kmol of
  species/kmol fuel
• Products Reactants

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NOTES (1) It is important to know the phase
(liquid or gas) for reactants, fuel and products
since the enthalpy of vaporization, hfg, must be
included if present. (2) By definition ?u0
?h0 - (?P? - ?P?)
Products Reactants v specific molar volume,
P pressure For most practical situations ?u0
?h0 and so ?U0 ?H0
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• Typically, the reactants are supplied at a
  temperature T1, the combustion conditions are
  given at a temperature T0 and products are found
  at T2.

Reactants
U or h
Products
UR1 hR1
1
KE (open System)
UR0 hR0
0
UP2 hP2
2
UP0 hP0
2
0
Temp. T
T1
T2
T0
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• Rigorously, the enthalpy (or internal energy) at
  the states 0, 1 and 2 should be obtained from the
  tabulated thermodynamic values. However, for most
  engineering design purposes a good approximation
  is obtained from
• or
• where Cv or Cp are evaluated at (T0T1)/2 for
  reactants and (T0T2)/2 for products.

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1
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Example

• A stoichiometric mixture of air and gaseous
  methane at 54oC and 2 bar is buried in a 0.1 m3
  rigid vessel. The temperature of the products is
  measured to be 1,529oC. Given that the internal
  energy of combustion ?Uo - 802,310 kJ/kmol at
  25oC, calculate the amount of heat rejected to
  the environment.
• Hence, information given
• To 25oC 298 K ?Uo -802,310 kJ/kmol
• T1 54oC 327 K P1 2 bar V 0.1 m3
• T2 1,529oC 1,802 K
• Air is provided Use 79 N2 , 21 O2

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Summary The total energy liberated during a
chemical reaction is based on a 1st Law of
Thermodynamics balance (for either a closed or an
open system). The heat released during the
reaction, ?H0, is a property of the reaction.
?H0 is generally tabulated for a specific
condition. Equations of state and the 1st Law of
thermodynamics can be used to calculate the heat
release to the actual state. It can be calculated
from the energy of formation of the chemical
species.