# Mehdi B. Nik

Energetics Course: Advanced Topic

Fall, 2008

Properties of High Temperature Gases

Mehdi B. Nik

Department of Mechanical Engineering and Material Science

University of Pittsburgh

Energetics Course: Advanced Topics

Fall, 2008

Preview

Some Definition

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

High Temperature Gases Examples

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

s
km
U
11

K
T
283

s
m
c
338

Apollo command vehicle

Returning from moon

5
.
32

M
From the tables

K
T
s
58000

Energetics Course: Advanced Topics

Fall, 2008

Example

Energetics Course: Advanced Topics

Fall, 2008

The Difference

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Deviation from Calorically Perfect Gas

Energetics Course: Advanced Topics

Fall, 2008

Molecules Modes of Energy

el
vib
rot
trans

trans el
  
  
 
Molecules

Atoms

Energetics Course: Advanced Topics

Fall, 2008

Energy Levels

Energetics Course: Advanced Topics

Fall, 2008

Energy levels

Total zero point energy

0
,

rot
el
vib
trans
o
o
o
o
o

Molecule Energy above zero Point Energy

rot
rot
k
k

trans
trans
trans
O
j
j

vib vib vib
l l O
  
 
 
el el el
m m O
  
 
 
zero point energy of a molecule
measures above the zero point energy
, thus all equal to zero at T=0K
trans rot vib el
i j k l m O
     
 
    
Energetics Course: Advanced Topics

Fall, 2008

Energy state

Quantized total energy level

Angular momentum and Quantized orientation

zero point energy of a molecule
measures above the zero point energy
, thus all equal to zero at T=0K
trans rot vib el
i j k l m O
     
 
    

Energy State

Degeneracy

Energetics Course: Advanced Topics

Fall, 2008

Macro State

System consisting of a fixed number of Molecule

j
j
N
N
j
N

Macro State

Total Energy of the system

j j
j
E N

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

Fall, 2008

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.

j
j
j
j
j
N
g
g
N
W
!
!
1
!
1

Fermions: number of the molecules that can be in
any degenerate state is one.

j
j
j
j
j
N
N
g
g
W
!
!
Energetics Course: Advanced Topics

Fall, 2008

Boltzman Distribution

It is valid in high temperature limit

Q
e
g
N
N
kT
j
j
j

*

j
kT
j
j
e
g
Q

Consequently

V
T
Q
RT
e

ln
2
V
T
Q
RT
RT
h

ln
2
V
T
Q
NkT
N
Q
Nk
S

ln
1
ln
T
V
Q
NkT
p

ln
Energetics Course: Advanced Topics

Fall, 2008

Evaluate the Partition Function

j
kT
j
j
Q g e

el
vib
rot
trans

box

the
of
dimension

are

,
,
a
number

quantum

are

n
,
n
,
n

,
8
3
2
1
3
2
1
2
3
2
3
2
2
2
2
2
1
2
1
2
a
a
a
n
a
n
a
n
m
h
trans

inertia

of
moment

I
number

quantum

rotational

1
8
2
2
J
J
J
I
h
rot

frequency

al
vibration
l
fundamenta

number

quantum

al
vibration
2
1
v
n
n
hv
vib

this
like
it

keep

we
el

2
2 2 2
1 2 3
1 1 1
0
8
0
1
2
O
O
O
trans
rot
vib
h
m a a a
hv

 

   
 
 

Energy measured above the zero point

O
O
O
O
el
el
el
vib
vib
vib
rot
rot
rot
trans
trans
trans
nhv
J
J
I
h
a
n
a
n
a
n
m
h

1
8
8
2
2
2
3
2
3
2
2
2
2
2
1
2
1
2
trans rot vib el
    
   
j
kT
j
j
Q g e

trans rot vib el
j i J n l
    
   
Energetics Course: Advanced Topics

Fall, 2008

Evaluate the partition function

0
2
2
2
3
2
1
1
8
2
l
kT
l
el
kT
hv
vib
rot
trans
el
vib
rot
trans
l
e
g
Q
e
Q
h
IkT
Q
V
h
mkT
Q
Q
Q
Q
Q
Q

Energetics Course: Advanced Topics

Fall, 2008

Evaluation of thermodynamic properties

V
T
Q
RT
e

ln
2
RT
e
trans
2
3

RT
e
rot

RT
e
kT
hv
e
kT
hv
vib
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

internal energy per unit mass
Electronic energy,obtained directly from
measured above zero-point
translational energy
Spectroscopic measurments
energy(sensible energy)
3
2
3
2
el
el
v v
v
e RT e
e
e
c c R
T
 

 
   
 
 
 
T

For molecules

Rotational energy
sensible energy
Electronic energy
translational energy
Vibrational energy
2
2
3
2
1
3
2
1
el
hv
kT
hv
kT
el
v v
hv
v
kT
hv
kT
e RT RT RT e
e
hv
e
e
e
kT
c c R R R
T T
e
   

 
     
 
 
 

Energetics Course: Advanced Topics

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

,,
AB A B
N N N
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

Reactant Products
AB A B
 
Change in zero- Zero-point energy Zero-po
int energy
point energy of products of reactants
     
 
     
     

'''
A B AB
O O O O
   
   
Energetics Course: Advanced Topics

Fall, 2008

Constraints

Total energy E is constant

'
'
'
A
B
AB
A A A A A
j j j j O
j j
B B B B B
j j j j O
j j
AB AB AB AB AB
j j j j O
j j
A B AB
E N N
E N N
E N N
E E E E const
  
  
  
  
  
  
   
 
 
 

Total number of A particles, free and combined are constants

A AB
j j A
j j
B AB
j j B
j j
N N N const
N N N const
  
  
 
 
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

A
j
B
j
AB
j
A
kT
j
A A
j
A
B
kT
j
B B
j
B
AB
kT
j
AB AB
j
AB
g e
N N
Q
g e
N N
Q
g e
N N
Q

Law of mass action

O
A B A B
kT
AB AB
N N Q Q
N Q

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

i i
pV N kT

Using Partial pressure

A B
A B
AB
AB
p p
N N V
N p kT

Combine it with law of mass action

O
A B
A B
kT
AB
AB
p p
kT Q Q
e
p V Q

Energetics Course: Advanced Topics

Fall, 2008

Equilibrium constants

Q is proportional to volume V, V’s cancel

O
A B
A B
kT
AB
AB
p p
kT Q Q
e
p V Q

A B
AB
p p
f T
p

This function of temperature is equilibrium constant

A B
P
AB
p p
K T
p

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

1 1 2 2 3 3 4 4 5 5
v A v A v A v A v A
  

5
4
4 5
3
1 2
1 2 3
v
v
A A
p
v
v v
A A A
p p
K T
p p p

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

2
2
2
2
O O
N N
N O NO
N O NO e
 

 
Energetics Course: Advanced Topics

Fall, 2008

Qualitative discussion

At high temperature, we will have a mixture of

2 2
,,,,,,
O N 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

2 2
i
-The partial pressures . For air, we hav
e ,,,,,,
-Concentration, the number of moles of s
pecies i per unit volume of the mixture,
X
-The mole-mass ratios, the number of mol
es of i
i O O N N NO
NO e
p p p p p p p p
 
i
per unit mass of mixture,
-The mole fractions. the number of moles
of species i per unit mole of mixture,
-The Mass fraction, the mass of species
i per unit mass of mixture,
i
i
X
c

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

2 2
I
O O N N NO
NO e
p p p p p p p p
 
      

Note: Dalton’s law holds only for perfect gases, gases
wherein intermolecular forces are negligible

Energetics Course: Advanced Topics

Fall, 2008

Problem, continue

2
2
2
2
2
2,
2
2,
,
,
2 II
2

V
O
p O
O
N
p N
N
NO
p NO
N O
e
NO
p NO
N O
p
O O K T
p
p
N N K T III
p
p
N O NO K T IV
p p
p p
N O NO e K T
p p

 
 
 
  
   

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

2
2
Avogadro Number
Avogadro Number
2
2
A O O NO O
NO
A N N NO N
NO
N N
N N
   
   

   
   

Also

or
i i i i
v
p v RT p
RT
 
 

Dividing the equations

2
2
2
.2
= =.25 VI
2.8
O O NO
NO
O
N N NO N
NO
p p p p
N
p p p p N

  

  

Electric charge must be conserved

VII
NO e NO e
p p
 
   
  
Energetics Course: Advanced Topics

Fall, 2008

Problem, summary

2 2
2
2
2
2
2
,
2
,
,
I

II

O O N N NO
NO e
O
p O
O
N
p N
N
NO
p NO
N O
p p p p p p p p
p
K T
p
p
K T III
p
p
K T
p p
 
      

2
2
,

V
2
.25 VI
2
e
NO
p NO
N O
O O NO
NO
N N NO
NO
NO e NO e
IV
p p
K T
p p
p p p p
p p p p
p p
 

   

  

  
  

VII

Seven nonlinear equation, seven unknown

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.

2
2
2
2
O O
N N
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

Example, M can be either
O2 or O

2
2
O M O M
  

Empirical result have shown

2
2
d O
k O M
dt

[O]: number of moles of O per unit volume of the mixture

k: reaction rate constant

Forward Reactions

2
2
f
k
O M O M
  

2
2
f
d O
k O M
dt

2
2
b
d O
k O M
dt
 

Reverse Reactions

2
2
b
k
O M O M
  

Net Rate

2
2
f
b
k
k
O M O M

 


2
2
2 2
f b
d O
k O M k O M
dt
 
Energetics Course: Advanced Topics

Fall, 2008

Chemical Rate Equation

If the system reaches chemical equilibrium

2
2
*
*
2
*
*
2
0
Define
equilibrium constant based on concentar
ion
f b
c
c
d O O
k k
dt
O
O
K
O
K
  

Also

1
c p
K K
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

2
2
O M O M
  

Non elementary Reactions

2 2 2
2 2
H O H O
 
2
2
2
2
2 2
2
2
H H
O O
H O OH O
O H OH H
OH H H O H

  
  
  
Energetics Course: Advanced Topics

Fall, 2008

Thank you