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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
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.
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 = ∆U0 = ∆Ho if K.E. negligible.
[∆U0 is the internal energy of combustion and is always
–ve since heat is transferred to the surroundings]
• Determine the energy available for conversion.
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.
Combustion processes (2)
PRINCIPAL CONSERVATION LAWS:
1. The number of atoms is conserved.
2. Mass is conserved.
Relative atomic mass of some common elements
Element name
Symbol
Oxygen
O
Relative
atomic mass
16
Nitrogen
N
14
Hydrogen
H
1
Carbon
C
12
Sulphur
S
32
Defined so that the element’s mass is scaled such that
Carbon-12 is exactly 12.
Relative molecular mass of common substances
Name
Oxygen
Nitrogen
Hydrogen
Water / steam
Carbon dioxide
Carbon monoxide
Sulphur dioxide
Symbol
O2
N2
H2
H2O
CO2
CO
SO2
Relative
molecular mass
32
28
2
18
44
28
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
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
To balance the reaction the number of atoms
remain the same (mathematically):
Hydrogen
2n=2→n=1
Oxygen
2m=1→m=½
Hence,
H2 + ½ O2 → H2O
This could also be written 2H2 + O2 → 2H2O
(or any multiple !!)
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 MH2 + 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
Example
• Calculate the mass of oxygen required to burn 1 m3 of fuel
propane at 1 bar, 273 K. How much air must be supplied (at
these conditions)?
• Solution:
(a) Need mO2 required to burn 1 m3 C3H8 (propane)
– 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 1 bar, 273 K. This will require the use of the ideal gas
law (or tabled density values) for these conditions.
– Finally, since air is used, calculate the other
components based on the composition of air.
(b) Set-up the primary and secondary relationships.
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;
Solving in terms of q:
O: 2m = 2p + 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
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 , at 1 bar, 273 K)
[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)
r = 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 = r Vol = 2.02 kg * 1 m3 = 2.02 kg
mO2 = 3.636 kgO2/kgC3H8 * 2.02 kgC3H8
= 7.34 kg of oxygen
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.
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.
Excess air is usually supplied in practice to ensure
complete combustion.
A deficiency of air when combusting hydrocarbons
 Hydrogen is complete but C gives CO2 + CO.
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
(i) the amount of fuel required;
(ii) the amount air required to combust these fuels and
the mass flow rate and temperature of the products
of combustion.
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 the products.
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 > 0 if supplied by
surroundings, thus ∆U0 < 0; ∆H0 < 0 )
∆U0 and ∆H0 are the internal energy and enthalpy of
combustion, respectively. ∆K.E. is kinetic energy
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 (25°C 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 25°C
and 1 atm.
∆h0 = ∑n*hf - ∑n*hf ; n = kmol of species/kmol fuel
Products Reactants
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 - (∑Pν
∑Pν)
Products Reactants
v = specific molar volume, P = pressure
For most practical situations: ∆u0 ≈ ∆h0 and so
∆U0 ≈ ∆H0
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
UR0 hR0
UP2 hP2
UP0 hP0
1
0
2
0
T0 T1
KE (open
System)
2
T2
Temp. T
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:
H P02  H R10  P mCP T2  T0   R mCP T2  T0   H fg
or
U P02  U R10  P mCV T2  T0   R mCV T1  T0   U fg
where: Cv or Cp are evaluated at (T0+T1)/2 for
reactants and (T0+T2)/2 for products.
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
Solution
We want to find
Q = Uo +
 (U0 – U1)
+
R
 (U2 – U0)
P
To do this we need the mass m of each component
and Cv for each at the average temperature (from
tables).
For m we need to find the total number of moles, N:
Recall: n = mol/m3 = P1 / R T1
= 2 bar / [(8.314 J/mol-k) * (327 K)]
= 2 x 105 / (8.314 * 327) = 73.57 mol/m3
Hence, N = 0.1 * 73.57 = 7.357 mol
This consists of CH4 (methane), O2, N2 and so we
now need to know how much O2 we have (for
stoichiometric conditions):
CH4 + 2 O2 → CO2 + 2 H2O
For 1 mol of CH4 we have 2 mol of O2 and also
0.79/0.21 * 2 of N2 (because air is provided)
NCH4 * (1 + 2 + 0.79/0.21 * 2) = 7.357 mol
Hence, NCH4 = 0.699 mol
So,
NO2 = 2 NCH4 = 1.398 mol
NH20 = 1.398 mol
NN2 = 5.529 mol
NCO2 = 0.699 mol
The heat given off by the reaction at 25oC is
Uo = - 802,310 kJ/kmol * 0.699 mol * 10-3 kmol/mol
= - 560.8 kJ
Next we need to determine the internal energy
change required to bring the reactants to the
“tabulated” reaction temperature and then the
products back up to the measured exit
temperature. From tables we find
Units: kJ/kmol-K
Reactants Cv [at (25+54)/2]
CH4
27.8
O2
21.1
N2
20.8
Products Cv [at (25+1529)/2]
CO2
46.6
H2O
33.5
N2
24.6
Hence, for the reactants (R) and products (P):
 (U0 – U1) = NCH4 * CvCH4 * (T0 – T1) +
R
N02 * Cv02 * (T0 – T1) + NN2 * CvN2 * (T0 – T1)
= [ NCH4CvCH4 + N02Cv02 + NN2CvN2 ] * (T0 – T1)
= [0.699 * 27.8 + 1.398 * 21.1 + 5.529 * 20.8]
* (25 – 54) * 10-3 = - 4.75 kJ
 (U2 – U0) = NC02 * CvC02 * (T2 – T0) +
P
NH20 * CvH20 * (T2 – T0) + NN2 * CvN2 * (T2 – T0)
= [ NC02CvC02 + NH20CvH20 + NN2CvN2 ] * (T0 – T0)
= [0.699 * 46.6 + 1.398 * 33.5 + 5.529 * 24.6]
* (1529 – 25) * 10-3 = 324.3 kJ
So the overall balance is
Q = Uo +  (U0 – U1) +  (U2 – U0)
R
P
= -560.8 kJ – 4.75 kJ + 324.3 kJ  - 241 kJ
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.