Download Thermochemistry

Lecture 2
Thermochemistry
X.S. Bai
Thermochemistry
Design a power plant
X.S. Bai
Thermochemistry
When we study a combustion device, what do we want to know?
X.S. Bai
• 
• 
• 
• 
heat generated
power production
combustion efficiency
combustion control
• 
• 
• 
• 
Temperature
pressure
species concentrations
flow velocity
Thermochemistry
When we study a combustion device, what do we want to know?
X.S. Bai
• 
• 
• 
• 
heat generated
power production
combustion efficiency
combustion control
• 
• 
• 
• 
Temperature
pressure
species concentrations
flow velocity
Thermochemistry
Outlines
•  Multi-component mixture
–  definition of thermodynamic variables
•  Law of mass conservation
–  molecules are not conserved
–  elements are conserved
•  First Law of thermodynamics
–  energy is conserved
•  Second Law of thermodynamics
–  chemical reactions follows certain directions
X.S. Bai
Thermochemistry
Mixture in a combustion system
X.S. Bai
Thermochemistry
Mixture of a combustion gas (1)
•  Mixture of a combustion gases contains
–  fuel (contain C, H, O …)
•  fossil fuels (Coal, natural gas, gasoline, kerosene ..)
•  biomass fuel (wood chips, city waste …)
–  air (23.3% oxygen, 76.7% nitrogen in mass)
–  Products
•  CO2 (green house gas)
•  H2O
•  minor species (CO, soot, NOx …)
X.S. Bai
Thermochemistry
Mixture of combustion gas
A macro-scopic view
X.S. Bai
Thermochemistry
Mixture of combustion gas (2)
A micro-scopic view
Molecules in a gas mixture moves randomly at a speed of sound, the distance between
molecules is in the order of mean free path
Time 1
X.S. Bai
Time 2
Time 3
Thermochemistry
Molecule Units
•  A molecule consists of atom. An atom consists of nucleus and electrons. A
nucleus consists of protons and neutrons ...
–  1 proton weighs 1.6726 x 10-27 kg
–  1 neutron weighs about the same as 1 proton
–  1 electron weighs 0.9109 x 10-30 kg
Too light ?
–  1 mole molecules = 6.0221 x 1023 molecules
–  1 mole C weighs = 12 g
–  molecular weight [kg/kmole]
X.S. Bai
Thermochemistry
Mixture (3)
•  In a combustion system, usually there are a lot of species (different
molecules), say N. It is often useful to know mass percentage and mole
percentage …
–  mass fraction Yi=
mass of species i
total mass
–  mole fraction Xi=
number of moles of i
total number of moles
–  number of mole i =
mass of species i
molecule weight of i
Ex:
X.S. Bai
Species i =
all molecules of type i
1 kg CO = 1000/28=36 mole CO
Thermochemistry
Mixture (4)
•  Mole concentration, mole fraction and mass fraction can be converted if the
molecule weight of species is known.
X i MWi
Yi =
,
MWmix
Yi MM mix
Xi =
,
MWi
X.S. Bai
i
)1
#N Y &
= " X i MWi = %% " MWi ((
i
$
'
i=1
i=1
N
MWmix
Yi
Ci = ! MW
Thermochemistry
Mixture (5)
•  The concept of stoichiometry
–  stoichiometric air-fuel ratio
(A/F)stoic=(mair/mfuel)stoic
For a stoichiometric
methane/air and
propane/air system
calculate the above
quantities
–  stoichiometric fuel-air ratio
(F/A)stoic=1/(A/F)stoic
–  equivalence ratio φ=(A/F)stoic /
(A/F)=(F/A)/(F/A)stoic
–  percent stoichiometric air
=100%/φ
methane
17.16
1
Propane
15.60
1
A/F
φ
–  percent excess air = 100%*(1φ)/φ
X.S. Bai
Thermochemistry
Mixture (6)
- ideal gas
•  The concept of ideal gas
–  An ideal gas is referred to a gas that has no inter-molecular
forces and no volume
–  introduced in relating pressure and density (equation of state)
N
p / ! = nR u T,
Yi
1
n ="
=
MWmix
i=1 MWi
Xi
pi = X i p =
!R u T
MWmix
X.S. Bai
Thermochemistry
Mixture of combustion gas (7)
- macroscopic view
•  1 m3 air at standard condition contains about 30 mole molecules, that is
1.8*1025 molecules.
•  Tennekes & Lumley (A first course in turbulence)
–  The smallest flow scale is Kolmogrov scale
η ≈ 10 −4 m
–  the distance between molecules is the mean free path
ξ ≈ 10 −8 m
–  the ratio between these scale is
ξ / η ≈ 10 −4 ∝ Ma / Re1/ 4
•  So, it can be assumed that in combustion system, the gases can be treated
as continuum media.
X.S. Bai
Thermochemistry
Mass conservation in a combustion
system
X.S. Bai
Thermochemistry
Law of mass conservation
In the universe, matters can be transformed to different
forms, but the total mass of the matter involved in the
transformation is not changed
X.S. Bai
Thermochemistry
Mass conservation in a combustion system
•  Matter is comprised of molecules; molecule is comprised of atoms; atom is
comprised of nuclei and electrons, nucleus is comprised of protons and
neutrons…
H
H
H2
o : o
X.S. Bai
Thermochemistry
Mass conservation in a combustion system
•  Processes involving formation
and destruction of new
molecules are referred to as
chemical reactions
•  Combustion involves only
chemical reactions
–  molecules are not
conserved in combustion
–  atoms (elements) are
conserved in combustion !!!
X.S. Bai
Thermochemistry
Mass conservation in a combustion system
•  A chemical reaction can be denoted as …
–  ex. CO+0.5O2
CO2
–  this equation indicates 1 mole of CO react with 1/2 mole
of oxygen and form 1 mole of carbon dioxide
–  recall 1 mole of matter = 6.02*1023 molecules
•  In the above example none of the three molecules are
conserved, but the two atoms involved (C & O) are
conserved. The factor 0.5 is due to C & O conservation
during chemical reactions
•  In general
N
ν i' M i →
∑
i =1
X.S. Bai
N
ν i''M i
∑
i =1
Thermochemistry
Mass conservation in a combustion system
•  How to use Law of mass conservation to compute Yi
–  use mass conservation of N atoms to construct N
algebraic equations
–  For a C, H, O, N system, 4 algebraic relations are
obtained
YJ =
mJ
∑m
=
J
∑
i
MWJ
Yi nJi
MWi
Number of atom J
in molecule i
X.S. Bai
Thermochemistry
Mass conservation –
example: biomass combustion
•  10 kg of dry wood chips are supplied to a combustor
–  C (52% mass)
–  H (6% mass)
–  O (41% mass)
•  Air
•  Products
–  CO2, H2O
•  Find out:
–  Stoichiometric Air
–  CO2, H2O
X.S. Bai
Thermochemistry
Mass conservation – example biomass combustion
Air required: x kg
Total reactants: 10+x kg
Total products: 10+x kg
The final products are: CO2, H2O, N2
Conservation of C, H, N, O: 4 equations
4 unknowns: x, YCO2, YH2O, YN2
X.S. Bai
Thermochemistry
Mass conservation – example biomass combustion
•  Air required: 62.52 kg
•  Products
–  YCO2=0.26291
–  YH2O=0.07446
–  YN2=0.66123
X.S. Bai
Thermochemistry
Energy conservation in a combustion
system
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  First Law of thermodynamics says for a fixed mass
system
–  Energy can be converted from one form to another
–  change of total energy = heat added to the system work done by the system to the surroundings
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  Internal energy
–  a sum of all the microscopic form of energy, which are related
to molecular structure and degree of the molecular activity.
–  Thermal energy and chemical energy has to be taken into
account
•  Kinetic energy
–  Macroscopic form of energy related to the fluid motion
•  Potential energy
–  Macroscopic form of energy related to the fluid height
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  Kinetic energy
•  Potential energy
•  Energy transfer
–  heat transfer
–  work
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
Internal energy
a sum of all the microscopic form of
energy, which are related to molecular
structure and degree of the molecular
activity.
•  Some physical insight to
internal energy
–  thermal energy
•  sensible energy
•  latent energy
–  chemical energy
–  nuclear energy
X.S. Bai
Thermochemistry
Energy conservation in a
combustion system …
•  Some physical insight to
sensible energy
•  thermal energy
–  sensible energy
–  latent energy
X.S. Bai
Thermochemistry
Chemical bonds
Oxygen molecule
Water molecule
Nitrogen molecule
X.S. Bai
Thermochemistry
X.S. Bai
Thermochemistry
Electron configuration of Neon
X.S. Bai
Thermochemistry
Chemical bonds
X.S. Bai
Thermochemistry
Energy in combustion systems
•  Total energy of system per unit mass
=e + 1/2v2 + gz
–  e=specific internal energy
–  1/2v2 =specific kinetic energy, v=velocity
–  gz=specific potential energy
X.S. Bai
Thermochemistry
Energy in combustion systems
•  Energy conservation
–  Change of total energy = heat to the system – work done by
the system to the surroundings
1 2
!(e + v + gz) = !Q " pdv
2
–  for an adiabatic constant pressure system energy
conservation says …
1 2
e + p / ! + v + gz = const.
2
•  Specific enthalpy: h=e+p/ρ
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  The concept of enthalpy
–  has no direct physical meaning, just a combined property
–  very useful in combustion system
•  Zero enthalpy
–  Enthalpy of the element in their naturally occurring state and
at standard condition is zero
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  In combustion problems we assume gas mixtures are ideal
N
N
h = ! h i Yi ,
e=
i=1
h i = h 0i,f+
N
!e Y ,
i
i
i=1
p=
!p ,
i
pi = X i p
i=1
T
!c
pi
dT
Tref
•  The concept of enthalpy of formation: hi,f0
–  to take into account the chemical energy, define a reference
state (p=1atm, T=25C) and let enthalpy of formation for the
element in their naturally occurring state zero.
–  NASA has a dada base for most elements and compounds
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  The concept of zero
enthalpy
–  enthalpy is a relative
quantity in isothermal
flows
–  in combustion, one
must take into account
the chemical energy in
enthalpy !! And nuclear
energy is not taken into
account !!
Zero enthalpy: element at its naturally existing state at standard condition has zero enthalpy
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  The concept of enthalpy of combustion (heat of combustion)
X.S. Bai
Thermochemistry
Energy conservation in a combustion system …
•  Using the concept of enthalpy to calculate adiabatic flame temperature
(constant pressure system)
X.S. Bai
Thermochemistry
The second Law of thermodynamics
•  The second law of thermodynamics:
–  consider a fixed-volume, adiabatic reaction vessel (a
close system), the system develops towards a state at
which entropy reaches its maximum (or equivalently
Gibbs free energy reaches its minimum).
–  This state is a chemical equilibrium state
–  It is an ideal state
–  It is a important reference state
X.S. Bai
Thermochemistry
Chemical equilibrium
•  Second Law of
Thermodynamics
–  CO+0.5O2
–  CO2
CO2
ds>0
ds<0
CO+O2 ?
–  CO + H2O =CO2+H2
X.S. Bai
Thermochemistry
Calculation of equilibrium quantities
•  Element mass conservation
•  First Law of Thermodynamics
–  or equivalently given temperature
•  Second Law of Thermodynamics
–  specify a chemical equilibrium reaction
–  Maximal entropy or
–  minimal Gibbs free energy
•  NASA code CEC86 available upon request
X.S. Bai
Thermochemistry
Summary
•  Combustion is a process in which fuel oxidizes
•  chemical energy is released in the form of sensible
(thermal) energy, by breaking old bonds and forming new
bonds
•  First Law of thermodynamics deals with how the energy is
converted
•  Second Law of thermodynamics tells us the maximal
extend of this conversion
•  To know exactly how much energy is converted one needs
to learn the next chapter - chemical kinetics
X.S. Bai
Thermochemistry