Transcript Filter Density Function for Subgrid Scale Modeling of Turbulent

Properties of High Temperature Gases
Mehdi B. Nik
Department of Mechanical Engineering and Material Science
University of Pittsburgh
Energetics Course: Advanced Topic
Fall, 2008
Preview
High Temperature Gases Examples
• Combustion
Exothermic chemical reaction between fuel and oxidant accompanied by
production of the heat
• Explosion
A large volume of gas is liberated in combustion besides the production of heat
The sudden evolution of large quantities of gas creates excessive pressure
Some Definition
•
•
•
e=e(T,v),h=h(T,p)
Thermally perfect gas:
e = e(T) => Cv = Cv(T) , h = h(T) => Cp = Cp(T)
Calorically Perfect Gas
Cv = cte, Cp = cte
Energetics Course: Advanced Topics
Fall, 2008
Properties of High Temperature Gases
• Equilibrium Gases
• Nonequilibrium Gases
•Behind the normal shock, the molecules does not
have enough time to reach to the equilibrium
condition
Energetics Course: Advanced Topics
Fall, 2008
One example
Apollo command vehicle
Returning from moon
U   11
km
s
T  283K
c  338
m
s
M  32.5
From the tables
Ts  58000 K
Energetics Course: Advanced Topics
Fall, 2008
Example
Energetics Course: Advanced Topics
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The Difference
Energetics Course: Advanced Topics
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Deviation from Calorically Perfect Gas
Energetics Course: Advanced Topics
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Molecules Modes of Energy
Molecules
   rot
   vib
   el
    trans
Atoms
   el
    trans
Energetics Course: Advanced Topics
Fall, 2008
Energy Levels
Energetics Course: Advanced Topics
Fall, 2008
Energy levels
•
Total zero point energy
•
Molecule Energy above zero Point Energy
j
trans
 o   o
trans
  ovib   oel ,  orot  0
  jtrans   O trans
 k   k
rot
rot
 l   l   O
vib
vib
vib
 m   m   O
el
el
 i   j
trans
el
  krot   lvib   mel 
measures above the zero point energy
, thus all equal to zero at T=0K
Energetics Course: Advanced Topics
 O
zero point energy of a molecule
Fall, 2008
Energy state
•
Quantized total energy level
 i   j
trans
  krot   lvib   mel 
measures above the zero point energy
, thus all equal to zero at T=0K
•
Angular momentum and Quantized orientation
•
Energy State
•
Degeneracy
Energetics Course: Advanced Topics
 O
zero point energy of a molecule
Fall, 2008
Macro State
•
System consisting of a fixed number of Molecule
N Nj
j
•
Macro State
Nj
•
Total Energy of the system
E    j N j
j
Energetics Course: Advanced Topics
Fall, 2008
Thermodynamic Equilibrium
•
•
Over a period of time, one particular macro state will
occur more frequently than any other. This particular
macro state is called the most probable macro state. It is
the macro state which occurs when the system is in
thermodynamics equilibrium.
The main problem in statistical thermodynamics is, given
a system with a fixed number of identical particles and
fixed energy, find the most probable macro state.
Energetics Course: Advanced Topics
Fall, 2008
Microstate
Energetics Course: Advanced Topics
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Most probable macro state
•
•
In any given system of molecules,
the microstates are constantly
changing, due to the molecular
collisions. One assumption is
that, each of these microstate of
the system occurs with equal
probability.
Based on this assumption, the
most probable macro state is that
macro state which has the
maximum number of microstates.
Energetics Course: Advanced Topics
Fall, 2008
Bosons and Fermions
•
Boson: number of the molecules that can be in any
degenerate state is unlimited.

N
W 
g
j
•
 g j  1!
j
j
 1! N j !
Fermions: number of the molecules that can be in
any degenerate state is one.
W 
j
g
g j!
j
 N j ! N j
Energetics Course: Advanced Topics
Fall, 2008
Boltzman Distribution
•
It is valid in high temperature limit
N N
*
j
g je

j
kT
Q
Q   g je

j
kT
j
•
Consequently
  ln Q 
e  RT 2 


T

V
 Q 
  ln Q 
S  Nk  ln  1  NkT

N

T



V
  ln Q 
h  RT  RT 2 

 T V
  ln Q 
p  NkT 

 V T
Energetics Course: Advanced Topics
Fall, 2008
Evaluate the Partition Function
Q   g je

j
kT

 trans

O
j
h2  1
1
1 
 2  2  2 0
8m  a1 a2 a3 
   rot
   vib
   el
    trans
 0
 rot
h 2  n12 n22 n32 
   ,

 trans 
8m  a12 a22 a32 
n1 , n 2 , n 3 are quantum number
  hv
 vib
a1 , a2 , a3 are dimension of the box
 
 rot
h2
J  J  1
8 I
J rotational quantum number
I moment of inertia
2
1

2

n vibration al quantum number
v fundamenta l vibration al frequency
  hv n  
 vib
 el we keep it like this
O
O
1
2
Energy measured above the zero point
   transO 
 trans   trans
h2
   rot 
 rot   rot
O
   vib
 vib   vib
O
 el   el   el
h 2  n12 n22 n32 
   
8m  a12 a22 a32 
8 2 I
 nhv
J  J  1
O
   trans   rot   vib   el
Q   g je

j
kT
j
 j  i
trans
Energetics Course: Advanced Topics
  J rot   nvib   lel
Fall, 2008
Evaluate the partition function
Q  QtransQrotQvibQel
3
2
 2mkT 
Qtrans  
 V
2
 h

8 2 IkT
Qrot 
h2
1
Qvib 
hv

1  e kT

Qel   g l e

l
kT
l 0
Energetics Course: Advanced Topics
Fall, 2008
Evaluation of thermodynamic properties
  ln Q 
e  RT 2 

 T V
etrans 
3
RT
2
erot  RT
hv
kT
evib 
e
hv
kT
RT
1
Theorem of equipartition energy (kinetic theory): each thermal degree of
freedom of molecule contribute RT/2 to the energy per unit mass
•It is not valid for vibrational motion of a diatomic molecule.
Energetics Course: Advanced Topics
Fall, 2008
In summary
•
For atoms
3
RT
2

e
internal energy per unit mass
measured above zero-point
energy(sensible energy)

translational energy
eel
Electronic energy,obtained directly from
Spectroscopic measurments
eel
3
 e 
cv  

c

R

v

2
T
 T v
•
For molecules
e
sensible energy

3
RT
2


RT
Rotational energy
translational energy
3
 e 
cv  

c

RR
v

2
 T v
hv
kT RT 
eel
e kT  1
Electronic energy
hv
Vibrational energy

hv
e
Energetics Course: Advanced Topics
kT
hv
kT

2
e
hv

1
kT
2
R
eel
T
Fall, 2008
Specific heat
Energetics Course: Advanced Topics
Fall, 2008
Some remarks
•
•
•
•
Energy and specific heat are temperature dependent.
this is the case for thermally perfect non reacting case.
Because we assumed intermolecular forces are
negligible.
Constant gamma will not be valid beyond 600K
As T->infinity, Cv->7/2R and it become constant. Which
is not valid. Long before that, the gas will dissociate and
ionize.
We can calculate the internal energy above the zero
point. At zero point we can not calculate it.
Energetics Course: Advanced Topics
Fall, 2008
The Equilibrium Constant
•
•
•
•
•
Up to now, was just single chemical species
High Temperature gases are mixture of several species
Lets consider a mixture of three arbitrary chemical species A,B
and AB
Assumption
– Mixture is confined in a given volume at given pressure and
temperature
– The system has existed long enough for the composition to become
fixed ( the reaction is taking place an equal number of times to both
the right and left)
This is the case of chemical equilibrium
AB
A B
Energetics Course: Advanced Topics
Fall, 2008
Equilibrium Constant
N AB , N A , N B number of AB,A,B particles in the mixture at chemical equilibrium
AB,A,B particles each they have their own set of energy levels, populations and degeneracy
Energetics Course: Advanced Topics
Fall, 2008
Energy Ladder
•
We don’t know the absolute values of the zero point
energies, but we know they are not equal. Therefore we
show it with different height in the ladder.
Energetics Course: Advanced Topics
Fall, 2008
Change in zero point energy
AB  A  B
Reactant
Products
Change in zero-  Zero-point energy   Zero-point energy 
 point energy   of products
  of reactants


 
 


 O    
'A
O
Energetics Course: Advanced Topics
'B
O

' AB
O
Fall, 2008
Constraints
•
Total energy E is constant




E   N    N jA  jA   OA
A 'A
j j
A
j
j
E   N    N Bj  Bj   OB
B
B
j
'B
j
j
E
AB
j
N 
AB
j
j
' AB
j

  N jAB  jAB   OAB

j
E  E A  E B  E AB  const
•
Total number of A particles, free and combined are constants
A
AB
N

N
 j  j  N A  const
j
j
B
AB
N

N
 j  j  N B  const
j
j
Energetics Course: Advanced Topics
Fall, 2008
Properties of the system at chemical equilibrium
•
We must find the most probable macro state of the system, the
same way we did before
N N
A
j
N N
B
j
N
•
•
AB
j
A
g Aj e

 Aj
kT
QA
B
N
g Bj e

 Bj
kT
QB
AB
g AB
j e

 AB
j
kT
Q AB

Law of mass action
A B
 O Q Q
N AN B
kT

AB
N
Q AB
It relate the amount of different species to the change in
zero point energy and the ratio of partition functions
Energetics Course: Advanced Topics
Fall, 2008
Partial pressure
•
Perfect gas equation for the mixture
pV  NkT
•
Partial pressure
pV
 Ni kT
i
•
Using Partial pressure
N A N B p A pB V

AB
N
p AB kT
•

Combine it with law of mass action
Energetics Course: Advanced Topics
p A pB kT  kTO Q AQ B

e
p AB
V
Q AB
Fall, 2008
Equilibrium constants

p A pB kT  kTO Q AQ B

e
p AB
V
Q AB
•
Q is proportional to volume V, V’s cancel
p A pB
 f T 
p AB
•
This function of temperature is equilibrium constant
p A pB
 K P T 
p AB
•
Equilibrium constant is the ratio of partial pressures of the
products of the reaction to the partial pressures of the reactants
Energetics Course: Advanced Topics
Fall, 2008
Generalizing the idea
•
Considering the general chemical equation
v1 A1  v2 A2  v3 A3
v4 A4  v5 A5
p  p 

T  
p  p  p 
v5
v4
Kp
A4
v1
A1
•
•
•
A5
v2
A2
v3
A3
In summary, we calculated the equilibrium Constant
We showed it is a function of temperature
You can use partition functions to calculate Equilibrium
constant as well
Energetics Course: Advanced Topics
Fall, 2008
Qualitative discussion
•
•
•
•
•
Air at normal room temperature and pressure
79% N2, 20% O2, 1% traces of Ar,He,CO2,H2O by
volume
Ignoring these traces, we will have N2, O2
If we heat the air to a high temperature 2500K<T<9000K
Chemical reaction s will occur among N2 and O2
O2
2O
N2
2N
N O
NO
N O
NO   e 
Energetics Course: Advanced Topics
Fall, 2008
Qualitative discussion
•
At high temperature, we will have a mixture of
O2 , N 2 , O, N , NO, NO  , e
•
•
If the air is brought to a given T and p, then left for a period
of time until the above reactions are occurring an equal
amount in both the forward and reverse direction, we
approach the condition of chemical equilibrium
For air in chemical equilibrium at a given p and T, the
species are present in specific, fixed amounts, which are
unique functions of p and T
Energetics Course: Advanced Topics
Fall, 2008
Equilibrium Composition
•
Several different ways of specifying the composition of a
gas mixture
-The partial pressures pi . For air, we have pO2 , pO , pN2 , pN , pNO , pNO , pe
-Concentration, the number of moles of species i per unit volume of the mixture,  Xi 
-The mole-mass ratios, the number of moles of i per unit mass of mixture,i
-The mole fractions. the number of moles of species i per unit mole of mixture, X i
-The Mass fraction, the mass of species i per unit mass of mixture, ci
•
If we know each of these, we can find the other ones
•
We are going to work with partial pressures.
Energetics Course: Advanced Topics
Fall, 2008
Problem
•
•
A high temperature gas at given T and p, assume the 7
species are present. Find the partial pressures?
Dalton’s Law: the total pressure of the mixture is the
sum of the partial pressures
p  pO2  pO  pN2  pN  pNO  pNO  pe
•
I
Note: Dalton’s law holds only for perfect gases, gases
wherein intermolecular forces are negligible
Energetics Course: Advanced Topics
Fall, 2008
Problem, continue
p 
2O  O
O2
 K p ,O2 T 
 II 
 K p , N 2 T 
 III 
pNO
 K p , NO T 
pN pO
 IV 
pNO pe
V
pO2
2N 
N2
N O
N O
•
2
 pN 
2
pN2
NO 


NO  e 
pN pO
 K p , NO  T 
Kp can be found from statistical mechanics or from
JANAF tables
Energetics Course: Advanced Topics
Fall, 2008
Problem, continue
•
Indestructibility of matter
– The number of O nuclei, both in free and combined state, must remain
constant
– The number of N nuclei, both in free and combined state, must remain
constant
Avogadro Number
NA
Avogadro Number
NA
 2
O2
 O   NO   NO  NO

 2
N2
  N   NO   NO  N N

v
RT
2 pO2  pO  pNO  pNO
•
Also
•
Dividing the equations
•
Electric charge must be conserved
pi v  i RT
or i  pi
2 pN2  pN  pNO  pNO
NO  e  pNO  pe

Energetics Course: Advanced Topics




NO .2
= =.25
N N .8
 VI 
 VII 
Fall, 2008
Problem, summary
•
Seven nonlinear equation, seven unknown
p  pO2  pO  pN2  pN  pNO  pNO  pe
 pO 
2
 K p ,O2 T 
 II 
 K p , N 2 T 
 III 
pNO
 K p , NO T 
pN pO
 IV 
pNO pe
V
pO2
 pN 
2
pN2
pN pO
 K p , NO T 
2 pO2  pO  pNO  pNO
2 pN2  pN  pNO  pNO
 NO  e  pNO  pe

•
I


 .25

 VI 
 VII 
Providing P,T, we should be able to calculate partial pressures of
different species
Energetics Course: Advanced Topics
Fall, 2008
Remark on chemical species
•
The proper choice of the type of species in the mixture is a
matter of experience and common sense. If there is a any
doubt, it is always safe to assume all possible combinations
of the atoms and molecules as potential species. If any of
choices turn out to be trace species, the results of the
calculation will state so.
O2
2O
N2
2N
N O
N O
NO   e 
Energetics Course: Advanced Topics
Fall, 2008
Air Composition
Energetics Course: Advanced Topics
Fall, 2008
Chemical Rate Equations
•
•
•
•
•
Oxygen at p=1atm and T=3000K is partially dissociated
Increase the temperature to 4000K
In 4000K, O increases and O2 decreases
It takes time for the molecule to adjust to a new
equilibrium condition
During this period, chemical reaction taking place at a
definite net rate.
Energetics Course: Advanced Topics
Fall, 2008
Chemical Rate Equation
•
•
•
•
•
•
•
O2  M  2O  M
Example, M can be either
O2 or O
d O 
Empirical result have shown
dt
 2k O2  M 
[O]: number of moles of O per unit volume of the mixture
k: reaction rate constant
Forward Reactions
f
O2  M 
 2O  M
Reverse Reactions
kb
O2  M 
 2O  M
Net Rate
k
dt
d O 
dt
 2k f O2  M 
 2kb O   M 
2
f

 2O  M
O2  M 

k
kb
d O 
dt
Energetics Course: Advanced Topics
d O 
 2k f O2  M   2kb O   M 
2
Fall, 2008
Chemical Rate Equation
•
If the system reaches chemical equilibrium
d O 
dt
O 
*2
 0  k f  kb
O2 
*
O 
Define K c 
*
O2 
*2
K c equilibrium constant based on concentarion
•
•
Also
1
Kc 
Kp
RT
In practice, Kf is found from experiment and Kb comes
from relation for Kc
Energetics Course: Advanced Topics
Fall, 2008
Reactions
•
Elementary Reactions is one that takes place in a single step
O2  M  2O  M
•
Non elementary Reactions
2H 2  O2  2H 2O
H 2  2H
O2  2O
H  O2  OH  O
O  H 2  OH  H
OH  H 2  H 2O  H
Energetics Course: Advanced Topics
Fall, 2008
Thank you
Energetics Course: Advanced Topics
Fall, 2008