This presentation had been prepared for the aircraft propulsion class to my undergraduate and graduate students at Kasetsart University and Chulalongkorn University - Bangkok, Thailand.

1. Aeropropulsion
Unit
Thermodynamics
Review of Fundamentals
Assist. Prof. Anurak Atthasit, Ph.D.
2005 - 2010
International School of Engineering, Chulalongkorn University
Regular Program and International Double Degree Program, Kasetsart University

2. Aeropropulsion
Unit
2A. ATTHASITKasetsart University
Topics
1. Brief Review of Thermodynamics
1. Notion of Calorically Perfect
2. Isentropic Process and Isentropic Flow
3. Stagnation State
2. Quasi-One-Dimensional Flow and
Compressible Flow
1. Speed of Sound & Mach Number
2. Area-Mach Number Relationship
3. Sonic Throat

3. Aeropropulsion
Unit
3A. ATTHASITKasetsart University
Perfect Gas and Kinetic Theory
of Gases
Assumptions for enable using the kinetic theory of gases are:
1. Intermolecular forces between the molecules is negligibly small.
2. The volume of molecules that occupy a space is negligibly small and ignored.
Perfect Gas Law may be derived from the kinetic theory of gas:
P RT
R
R
M
Where the gas constant (R)
M is the molecular weight
8314
.
J
R
kmol K

Gas is Air

4. Aeropropulsion
Unit
4A. ATTHASITKasetsart University
Specific Heat for a Perfect Gas
and Calorically Perfect Gas
Specific heats at constant pressure and volume:
p
v
dh C dT
de C dT


In general
( )
( )
p p
v v
C C T
C C T


There is often a simplifying assumption of constant specific heats, which
is a valid approximation to gas behavior in a narrow temperature range
0
0
1500
p p
v v
C Const
C Const



5. Aeropropulsion
Unit
5A. ATTHASITKasetsart University
First Law of Thermodynamics
Statement of conservation of energy for a system of fixed mass m.
q de w  
δ = “path dependent”
2
1 2
1
2
2 1
1
rev
q q
de e e
w pdv



 



The difference in the convention for positive
heat and work interaction with the system
explains the opposite sides of equation where the
two energy exchange terms are located in this
equation.
(1st Law pt. of view) There is no distinction
between heat and mechanical work exchange
with the system.
Flow work on fluid
created by the
pressure forces at
reversible

6. Aeropropulsion
Unit
6A. ATTHASITKasetsart University
Second Law of Thermodynamics
1. A statement of impossibility of a heat engine exchanging heat with a
single reservoir and producing mechanical work continuously.
Necessary having a second reservoir at a lower temperature where
heat is rejected to by the heat engine.
2. All mechanical work may be converted into system energy whereas
not all heat transfer to a system may be converted into system energy
continuously.
Clausius Inequality:
Tds q
(2nd Law pt. of view) Distinguishing between heat and work

7. Aeropropulsion
Unit
7A. ATTHASITKasetsart University
1st and 2nd Law combination:
Gibbs Equation
The pressure forces within the fluid perform reversible work, and the
viscous stresses account for dissipated energy of the system (into heat)
Tds de pdv 
1. It looks as if we
have substituted the
reversible forms of
heat and work into
the first law to
obtain the Gibbs
equation
Enthalpy: combination 2 forms of fluid energy; internal
energy (thermal energy) and flow work (pressure
energy)
h e pv
dh de pdv vdp
 
  
Tds dh vdp 

8. Aeropropulsion
Unit
8A. ATTHASITKasetsart University
2nd Law of Thermodynamics:
Entropy Change in Process
Tds dh vdp 
From Gibbs Equation:
p vdh C dT de RdT C dT RdT    where
2
2
2 1
11
ln
p p
p
dT v dT dP
ds C dP C R
T T T P
PdT
s s s C R
T P
   
    
Thereis often a simplifyingassumptionof constantspecificheats, which
is a valid approximation to gas behavior in a narrow temperaturerange
0
0
1500
p p
v v
C Const
C Const


2 2
2 1
1 1
ln ln
T P
s s Cp R
T P
  
2
2
2 1
11
ln
T
Cp
T
  
Tabulated thermodynamic function 

9. Aeropropulsion
Unit
9A. ATTHASITKasetsart University
Specific Heat and The Ratio
Think about p vdh C dT de RdT C dT RdT    
1
1
, ,
1 1
p v
p v
p v
p v
v v
C C R
C C
R R
C C R
C R C R
C C


 
 
 

   
 
One can obtain
The ratio of specific heats is related to the
degrees of freedom of the gas molecules, n, via:
2n
n


 Diatomic gas at
‘normal’ temperature
5 2
1.4
5


 
• Diatomic gas at
normal Temp.: 5
degrees of freedom
are; 3 translation
motion, 2 rotational
• 600K<Temp.<2000
K, vibration mode
activated, n=6,
γ=1.33
• >2000K, n=7,
γ=9/7=1.29
0
0
1500
?hot 
?cold 

10. Aeropropulsion
Unit
10A. ATTHASITKasetsart University
Specific Heat and The Ratio
Specific Heat at Constant Pressure (Cp) in kJ/kg.K and Specific Heat Ratio 

11. Aeropropulsion
Unit
11A. ATTHASITKasetsart University
y = 0,958e0,000x
R² = 0,993
y = -3E-11x3 + 1E-07x2 + 7E-05x + 0,978
R² = 0,996
0,9
0,95
1
1,05
1,1
1,15
1,2
1,25
1,3
1,35
1,4
0 500 1000 1500 2000 2500
1atm
10 atm
Expon. (1atm)
Poly.(1atm)
Specific Heat
Specific Heat at Constant Pressure (Cp) in kJ/kg.K

12. Aeropropulsion
Unit
12A. ATTHASITKasetsart University
Specific Heat Ratio
Specific Heat Ratio 
y = -6E-14x4 + 3E-10x3 - 4E-07x2 + 0,000x + 1,386
R² = 0,999
1,2
1,25
1,3
1,35
1,4
1,45
0 500 1000 1500 2000 2500
1 atm
10 atm
Poly.(1 atm)

13. Aeropropulsion
Unit
13A. ATTHASITKasetsart University
• To analyze the
performance of Ideal
Engine, one can assume
every components are
working under the
isentropic process
EXCEPT ‘Combustor’
Isentropic Process and Isentropic
Flow for a Calorically Perfect
ds
1
2 2 2
1 1 1
PC
R
p
P T TdT dP
C R
T P P T T

    
       
   
From Gibbs Equation:
Using the Perfect Gas Law:
2 2
1 1
P
P



 
  
 
0
0
1500

14. Aeropropulsion
Unit
14A. ATTHASITKasetsart University
Exercise1: Isentropic Flow for a
Calorically Perfect
2 2
1 1
P
P



 
  
 
From the Isentropic Flow for Perfect Gas:
Show that
dp P
d

 


15. Aeropropulsion
Unit
15A. ATTHASITKasetsart University
Stagnation State
We define the stagnation state of a gas as the state reached in
decelerating a flow to rest reversibly and adiabatically and without any
external work
2
2
t
V
h h 
(The total energy of the gas does not change in the
deceleration process)
Calorically Perfect Gas
2 2
2 2
t t
P
V V
h h T T
C
    
2th
3th
1th

16. Aeropropulsion
Unit
16A. ATTHASITKasetsart University
Need a Break?
Thrust reversal, also called
reverse thrust, is the
temporary diversion of an
aircraft engine's exhaust or
changing of propeller pitch
so that the thrust produced is
directed forward, rather than
aft.

17. Aeropropulsion
Unit
17A. ATTHASITKasetsart University
CONSERVATION PRINCIPLES

18. Aeropropulsion
Unit
18A. ATTHASITKasetsart University
Conservation Principles for
Systems and Control Volumes
( )
0
system V t
dm d
dV
dt dt
 
Continuity equation:
Leibnitz’ rule of differentiating such integrals is the bridge
between the system and the control volume approach:
0 0( ) ( ) ( )
ˆ
V t V t S t
d
dV dV V ndS
dt t

 

  
  
. .
. . . .
( )C V
C V C V
dV dV m
t t t


  
 
   
Unit vector normal
to the control
surface and
pointing outward
Unsteady accumulation of mass within the
control volume (vanishes for steady flows) Net crossing of the
mass through the
control surface per
unit time

19. Aeropropulsion
Unit
19A. ATTHASITKasetsart University
Conservation of Mass for a
Control Volume
( )
0
system V t
dm d
dV
dt dt
 
Continuity equation:
0 0( ) ( ) ( )
ˆ
V t V t S t
d
dV dV V ndS
dt t

 

  
  Where
. .
. .
( )C V
C V
dV m
t t
 

 
. .
ˆ ˆ ˆ out in
C S Outlets Inlets
V ndS V ndS V ndS m m          
There is no mass crossing the
boundaries of a control volume
at any other sections besides
the “inlets” and “outlets”
ˆ ( )V n  
ˆ ( )V n  
. .( ) 0C V out in
d
m m m
dt
  
We note that the difference between a
positive outgoing mass and a negative
incoming mass appears as a mass
accumulation or depletion in the C.V.

20. Aeropropulsion
Unit
20A. ATTHASITKasetsart University
Momentum Equation: Steady
Flow
( )
net
d mV
F
dt

The time rate of change of linear momentum of an object of
fixed mass to the net external forces that act on the object
0 0( ) ( ) ( )
( )
ˆ( )
V t V t S t
d V
VdV dV V V n dS
dt t

 

  
  
The unsteady momentum within the control
volume (vanishes identically for a steady flow)
Net flux of momentum
in and out of the C.S.
( ) ( )out in netmV mV F 
• External Forces exerted on the fluid at its boundaries
(ex. Drag forces)
• Body forces (volume forces) such as gravitational forces
,
,
,
( ) ( )
( ) ( )
( ) ( )
X out X in net X
Y out Y in net Y
Z out Z in net Z
mV mV F
mV mV F
mV mV F
 
 
 
(Vector form)
3 spatial directions:
Cartesian coordinates

21. Aeropropulsion
Unit
21A. ATTHASITKasetsart University
Exercise 2: Application of Momentum Equation
and Isentropic Flow on Wave Propagation
2 1
u
P
T
s

P dP
T dT
d
s
 



du
“Particle” mass
motion behind the
wave
Plane acoustic wave
propagation in a gas at rest
Gas at rest
Acoustic wave
(stationary)
Complete the relative
properties when flow is
seen by an observer
fixed at the wave
?1 ?2
?3 ?4

22. Aeropropulsion
Unit
22A. ATTHASITKasetsart University
Exercise 3: Application of Momentum Equation
and Isentropic Flow on Wave Propagation
Acoustic wave
(stationary)
P dP
T dT
d
s
 



P
T
s

2 1
uu du
๑. Showing, by applying steady conservation
principles to a control volume, the continuity to
have the following relation
( )d du ud
du ud
  
 
 
 
๒. Neglecting the second order derivative and under the constant mass flow rate at
a unit flow area “u”, show that the momentum equation simplifies to:
udu dP 
And simplified by
๓. From ๑. & ๒., proof that the square of the acoustic wave propagation in terms
of pressure and density changes that occur as a result of wave propagation in a
medium at rest is

23. Aeropropulsion
Unit
23A. ATTHASITKasetsart University
Exercise 3: Application of Momentum Equation
and Isentropic Flow on Wave Propagation
Acoustic wave
(stationary)
P dP
T dT
d
s
 



P
T
s

2 1
uu du
๓. From ๑. & ๒., proof that the square of the
acoustic wave propagation in terms of pressure
and density changes that occur as a result of
wave propagation in a medium at rest is
2 dP
u
d

๔. Defining the speed of sound in a gas is “a”, show that
the speed of sound is:
a RT

24. Aeropropulsion
Unit
24A. ATTHASITKasetsart University
Exercise 4: Stagnation State and Local
Mach Number of a Calorically Perfect Gas
1. Based on the stagnation enthalpy, proof that the
total to stagnation temperature ratio can be derived as:
21
1
2
tT
M
T
  
  
 
2. What is the assumption to have the relation
of the total to stagnation pressure ratio as:
1
21
1
2
tP
M
P

   
   
  
3. Show that the density ratio of a calorically
perfect gas in isentropic process is:
1
1
21
1
2
t
M
 

  
   
   2tT
3tT
1tT

25. Aeropropulsion
Unit
25A. ATTHASITKasetsart University
Law of Conservation of Energy
for a Control Volume
Rate of energy transfer (Apply by the1st Law of thermodynamics)
dE
Q W
dt
 
• Internal Energy, Kinetic Energy, Potential Energy
• Changing
potential
energy in gas
turbine
engines or
most other
aerodynamic
applications is
negligibly
small and
often ignored
2 2 2
( ) . . . .
ˆ( )
2 2 2V t C V C S
dE d V V V
e dV e dV e V n dS
dt dt t
  
      
            
      
  
Rate of energy in Leibnitz' rule of integration
The unsteady energy within the control volume
(vanishes identically for a steady flow)
Net flux of fluid
power in and out
of the C.S.

26. Aeropropulsion
Unit
26A. ATTHASITKasetsart University
Law of Conservation of Energy
for a Control Volume
Rate of mechanical energy transfer due to pressure on the boundary
• The pressure forces act normal to the boundary and point inward (opposite to ň)
Rate of energy transfer (Apply by the1st Law of thermodynamics)
dE
Q W
dt
 
• Pressure Energy Transfer on the boundary
• Energy transfer by Shear Forces
• Shaft Power (P_s)
• Viscous Shear Stresses (W_viscous-shear)
Pressure Energy Transfer on the boundary
. .
ˆ
C S
pV ndS 
Rate of work
done by the
surroundings
on the C.V.
. .
ˆ
C S
pV ndS
Rate of work
done by the
C.V. on the
surroundings

27. Aeropropulsion
Unit
27A. ATTHASITKasetsart University
Law of Conservation of Energy
for a Control Volume
Rate of energy transfer (Apply by the1st Law of thermodynamics)
dE
Q W
dt
 
2 2 2
( ) . . . .
ˆ( )
2 2 2V t C V C S
dE d V V V
e dV e dV e V n dS
dt dt t
  
      
            
      
  
_
. .
ˆ s viscous shear
C S
W pV ndS W   
Rearranging:
2 2
_
. . . .
ˆ( )
2 2
s viscous shear
C V C S
V p V
Q e dV e V n dS W
t
 

    
           
     
 
Define: internal energy and flow work as enthalpy
2
_
. . . .
ˆ( )
2
t s viscous shear
C V C S
V
Q e dV h V n dS W
t
 
  
       
   
 

28. Aeropropulsion
Unit
28A. ATTHASITKasetsart University
Law of Conservation of Energy for a
Control Volume: The Simplification
2
_
. . . .
ˆ( )
2
t s viscous shear
C V C S
V
Q e dV h V n dS W
t
 
  
       
   
 
In steady flows: a time derivative vanishes
Adiabatic flows: the rate of heat transfer through
the walls of the control volume vanishes.
In the absence of shaft work: (as inlets
and nozzles) Shaft Power vanishes
Viscous shear stresses order: this term is small compared
with the net energy flow in the fluid, hence neglected.
( ) ( )t out t in smh mh Q   Ex: Multiple inlets, uniform flow

29. Aeropropulsion
Unit
29A. ATTHASITKasetsart University
Exercise 5: Conservation of
Energy for a Control Volume
1
2
42.0Q MW
A control volume with heat transfer rate and
mechanical power (i.e., shaft power) exchange
specified at its boundaries. There is a single inlet
and a single outlet where mass crosses the
boundary in a steady and uniform flow.
1
1
1
1
50 /
25
150 /
1004 / .
1.4
p
m kg s
T C
u m s
C J kg K


 



2
1
2
400 /
1243 / .
1.3
p
u m s
C J kg K




100s kW  
Calculate the exit
total and static
temperature (note
that the gas is not
calorically perfect)
Tt2=927K
T2=863K

30. Aeropropulsion
Unit
30A. ATTHASITKasetsart University
Quasi-One-Dimensional Flow
An invicid fluid model
accurately estimates the
pressure, temperature,
and flow development
along the duct axis.
Primary flow
Direction
Nozzle and Diffuser “Inviscid
Assumption” For high Reynolds number
flows where the boundary layer is thin
compared with lateral dimensions of a
duct and assuming an attached boundary
layer.
Taylor series approximation
of the slab of inviscid fluid
P dP
T dT
d
A dA
u du
 





P
T
A
u

dx
2
dP
P 

31. Aeropropulsion
Unit
31A. ATTHASITKasetsart University
Exercise 6: Continuity of Quasi-
One-Dimensional Flow
P dP
T dT
d
A dA
u du
 





P
T
A
u

dx
2
dP
P 
1. Derive the continuity equation to this thin slice of the
flow.
2. Simplify the continuity equation from 1. and show that
this can be in this form:
0
dA d du
A u


  
3. Take the natural logarithm of both sides of
“uA=constant” and take the differential of both sides,
see the result.
4. Make conclusion from 2. and 3.
• It is called logarithmic derivative of
the continuity equation
0
dA d du
A u


  

32. Aeropropulsion
Unit
32A. ATTHASITKasetsart University
Exercise 7: Momentum and Energy
Equation of Quasi-One-Dimensional Flow
P dP
T dT
d
A dA
u du
 





P
T
A
u

dx
2
dP
P 
1. Balance the momentum equation in the streamwise
direction of flow on the slab system.
2. Simplify the momentum equation from 1. and show that
this can be in this form:
udu dP  
3. What is the energy equation for an adiabatic flow with
no shaft work of the slab system?
4. Show that the equation of state for a perfect gas may be
written in logarithmic derivative form as:
dP d dT
P T


 

33. Aeropropulsion
Unit
33A. ATTHASITKasetsart University
Governing Equations for an
Isentropic Flow Through a Duct
These governing
equations shall
apply to
compressible flow
problems of interest
to propulsion
2
0
0t
dA d du
A u
udu dP
dh dh udu
dP d dT
P T
a
P







  
 
  
 


34. Aeropropulsion
Unit
34A. ATTHASITKasetsart University
Isentropic Flow Through a Duct 
Area-Mach Number Relationship
2
0
0t
dA d du
A u
udu dP
dh dh udu
dP d dT
P T
a
P







  
 
  
 

Let us divide the momentum
equation by d, to get
2dP
udu a
d d

 

  
Which yields
2
d udu
a


 
2
2
0
( 1)
udu du dA
a u A
du dA
M
u A
   
  
Substitute in the continuity equation,

35. Aeropropulsion
Unit
35A. ATTHASITKasetsart University
Isentropic Flow Through a Duct
 Sonic Throat
*
2
1
*
2
1
1
*
*
2
*
1
2( 1)
2
*
1
2
1
1
2
1
2
1
1
2
1
2
1
1
2
1
1
1 2
1
2
T
T
M
P
P
M
P
P
T
M
T
P
m AV AM
R T
M
A
A M






















 
  
 
 
 
 
       
 
 
  
       
 
  
   
  
 
 
 

36. Aeropropulsion
Unit
36A. ATTHASITKasetsart University
Conclusion
*
2
1
*
2
1
1
*
*
2
*
1
2( 1)
2
*
1
2
1
1
2
1
2
1
1
2
1
2
1
1
2
1
1
1 2
1
2
T
T
M
P
P
M
P
P
T
M
T
P
m AV AM
R T
M
A
A M






















 
  
 
 
 
 
       
 
 
  
       
 
  
   
  
 
 
 
2
0
0t
dA d du
A u
udu dP
dh dh udu
dP d dT
P T
a
P







  
 
  
 

P dP
T dT
d
A dA
u du
 





P
T
A
u

dx
2
dP
P 
Cabin Crew!
Prepare for
take-off!

37. Aeropropulsion
Unit
37A. ATTHASITKasetsart University
Cowling Damage
• Emirates Airbus Rolls
Royce engine Cowling
Went to pieces....No
Details On Why.
• Happened 18 Oct. 2006
En Route From BHX
(Birmingham Int'l
Airport) to DXB (Dubai
Int'l Airport)
• This Happened On
Approach Over Water To
Land.