Section1, Lecture 2

Equation of State, Review of Thermodynamics,

Isentropic Processes
work performed by piston on system
with heat addition

p A
v c

dx

q
Ff

A
p c
v- dv

Ff

Anderson: Chapter 1 pp. 19-39 1

 Review
• Key Concepts:

i) High Speed flows often seem “counter-intuitive” when

Compared with low speed flows

ii) Flow regimes

Subsonic - All flow everywhere on the aircraft less than local speed of sound.
Transonic - Some flow is subsonic and some is supersonic.
Supersonic - All flow everywhere on the aircraft is supersonic.
Hypersonic - Fluid flows that are Much Higher than sonic velocity

iii) Mach number - ratio of true airspeed to local speed of sound

1
iv) Mach Angle … angle of shock wave generated by “point object” 
sin M 
v) Sonic Velocity in a gas  p 
c  
   s0
2
 Equation of State for a Perfect Gas
• Relationship Between pressure, temperature, and density
derived empirically in Modern form by John Dalton

• Theoretically derived by Ludwig Boltzmann using statistical Thermodynamics

• In perfect gas … intermolecular (van der Waals) forces are neglected

p
p v = n Ru T v
T
QuickTime™ and a •p- pressure acting on gas
TIFF (LZW) decompressor
n
are needed to see this picture.
•v- volume of gas in system
•n- Number of moles of gas in system
• Ru - Universal gas constant
•T- Temperature of gas
John Dalton
1-mole --> 6.02 x 1023
Avagadro's number
3
 Equation of State for a Perfect Gas (cont’d)
• Re organizing the equation of state
M n Ru Ru
p RuT  T  T  RgT
V M M /n Mw
•p- pressure acting on gas
•v- volume of gas in system
•n- Number of moles of gas in system
• Ru - Universal gas constant
•T- Temperature of gas
• Mw- Molecular weight of gas
• Rg - Gas Specific Constant
•M- Mass of gas contained in volume

• Useful working form for Gas Dynamics

R R
p  u T  RgT  >Rg  u
 Mw Mw 4
 Equation of State for a Perfect Gas (cont’d)
• Numerical Values for Universal Gas Constant
Ru = 1545.40 ft-lbf/R-(lbm-mole)
Ru = 49722.01 ft-lbf/R-(slug-mole)
Ru = 8314.4126 J/K-(kg-mole)
• Molecular of various gases
• Gas Specific constant is
Universal constant divided
by the average molecular
weight of the gas (steam)

5
 Equation of State for a Perfect Gas (concluded)
• Molecular weight of Air
Average molecular weight of the gases in the atmosphere.
Air on earth at sea level is a mixture of approximately 78% nitrogen,
21%oxygen, with the remaining one percent a mix of argon,
carbon dioxide, neon, helium and other rare gases,

~ 28.96443 kg/kg-mole

• Numerical Values for Air Specific Gas Constant

Rg = 53.355 ft-lbf/R-(lbm)

Rg = 1716.658 ft-lbf/R-(slug)
Rg = 287.056 J/K-(kg)

6
 Specific Heats, Internal Energy, and Enthalpy
(cont’d)

• e is the internal energy per unit mass

The total kinetic and potential energy associated with the motions and relative
positions of the molecules a gas, solid, or liquid. An increase in internal
energy results in a rise in temperature or a change in phase

• h is the enthalpy per unit mass

.
The measure of the heat content of a chemical or physical system;
equivalent to the sum of the internal energy of the system plus the
product of its volume multiplied by the pressure exerted on it by its
Surroundings … i.e.
Symbols e --> u often
h=e+pv Used interchangeably
In scientific literature
h is the specific enthalpy, (SI - joules/kg)
e is the specific internal energy, (SI - joules.kg)
P is the pressure of the system, (SI - pascals)
v is the specific volume, (SI cubic metres/kg) = 1/

7
 Specific Heats, Internal Energy, and Enthalpy
(cont’d)

• Specific heat definitions:

- Constant Volume:
 de   dh  d
cv 
 dT  v
cp   
 dt  p dT
e  pv  
- Constant Pressure:
de d
cp  dh 


dT dT
RgT cv  Rg
 dT  p

Universal relationship cp cv  Rg

applies to all perfect gases
8
 First Law of Thermodynamics
work performed by piston on system
with heat addition
• Change in internal
p
Energy = heat added
v
A
c
+ work performed+
Energy dissipated
dx

q de dq  dwrev  dwirrev

Ff 1 1
dwrev  F dx  pAc dx
m m
p
A
c  A dx 
v- dv  p  c   pdv
 m 
Ff
1
 dwrev  Ffric dx
m
9
 First Law of Thermodynamics (continued)
• Recall from earlier lecture
1) Adiabatic Process… no heat loss or addition
2) Reversible Process … no dissipative phenomena occur
3) Reversible adiabatic process is referred to as isentropic

work performed by piston on system

with heat addition
dq =0 (adiabatic process)
p

1
A
v c

de  pdv  Ffric dx

dx

q m
Ff Ffric = 0 (isentropic process)
A
c

de  pdv
p
v- dv

Ff

10
 First Law of Thermodynamics (concluded)
• In terms of Enthalpy
- reversible process

de dq  pdv
h e  pv >dh de  vdp  pdv
dh
 dq  pdv  vdp  pdv dq  vdp
dh dq  vdp
- isentropic process (reversible, adiabatic)

dh vdp
11
 Second Law of Thermodynamics (cont’d)
•Thermodynamic entropy, s, is a measure of the amount of energy
in a physical system that cannot be used to do work.

• It is a relative measure of the disorder present in a system.

(SI unit of entropy is J/K)

dq
ds   dsirrev
T
Reversible heat addition Effect of dissipative processes

• Calculate change in entropy:

First Law: de dq  pdv  dwirrev

Second Law: Tds dq  Tdsirrev

12
 Second Law of Thermodynamics (cont’d)
For a reversible process
de dq  pdv
Tds de  pdv
Tds dq
Writing e in terms of enthalpy (h = e + p v )

dh de  pdv  vdp and

Second Law: Tds de  pdv

First Law: dh de  pdv  vdp
dh Tds  pdv  pdv  vdp 
dh Tds  vdp 13
 Second Law of Thermodynamics (cont’d)
Solving for ds and integrating between state 1 and state 2
dh vdp dh vdp 
s2  s1 1 
2
ds   
T T T T 

 dh  2  c p dT vdp 
cp  s2  s1 1  
But from definition  dT  p T T 
 

From the equation of state

1
p  RgT  RgT
v
RgT
v
p
14
 Second Law of Thermodynamics (concluded)
• Substituting equation of state into entropy integral
 RgT 
 dp 
2  c p dT vdp  2 c p dT p
v
RgT s2  s1        
p 1
 T T  1
 T T 
 
2  c p dT dp 
1  T  Rg p  Valid for thermally perfect gas

• For a thermally perfect gas, cp ~ constant

 T2   p2 
s2  s1 c p ln 2    Rg ln   Valid for adiabatic
 T1   p1  process

15
 Isentropic Process Relationships
• For an isentropic process, (adiabatic, reversible) dq = 0

dq
ds   dsirrev = 0
T
cp Rg
 T2   p2   T2   p2 
and c p ln 2    Rg ln      
 T1   p1   T1   p1 
Solving for p2 in terms of T2
cp cp
p2  T2  Rg  T2  c p  cv
   
p1  T1   T1 
16
 Isentropic Process Relationships (cont’d)
cp
• Defining a ratio of specific heats, 
cv

p2  T2   1 More “later”
 
p1  T1 
• Applying the equation of state
T2 p2 1

T1 p1 2

• and 1
 1 

 p2   1  p2   1  2   1  p2   2 
       >   
p
 1   p1   1   p1   1 
17
 Isentropic Process Relationships (concluded)
• Returning for the expression for sonic velocity

V
M
 RgT
Then for a calorically perfect gas
 
p 2p2   2  2  p1   
   p
 pk    p k 
p 1p1   1  1 
    2
1 2

  p 1 1
    k   1
 k  

 p

 RgT  >c   RgT !
 s

18
 Thermodynamic properties of a mixed gas
• Molecular weight:

j n j M w j j --- constituent gases

Mw  
j n j
p
ideal gas law  pV nRuT  n  V
RuT
@constant temperature and pressure  n j Vj “Dalton’s law”

j V j M w j  V  Vj
Mw    M j
w j

V frac j M w j  V frac j 
V
j V j j  V

k
k

 j k
k

19
 Thermodynamic properties of a mixed gas
(cond’t)
• Example: air

j

M w  V frac M w j j


Mwair =
2140.78084 + 2160.209476 + 400.00934 + 12 + 216  0.000314
+ …. = 28.96443 kg/kg-mole
8314.4126
--> Rg = = 287.056 J/K-(kg)
28.9644
20
 Thermodynamic properties of a mixed gas
(cond’t)

• Specific heat

j n j c 'j
p

  c '  c
_
cp ' V ' Molar specific heat
j n j
frac j p j p J
( )
j kg-moloK


_
cp ' 1
 c '  mass specific heat
_
cp 
Mw

Mw
 V frac j p j (
J
)
j kgoK


cp ' j
 1

_

cp j  cp 
Mw
 V Mw c frac j j pj

Mw j

Same holds for cv

21
 Thermodynamic properties of a mixed gas
(cond’t)
• Specific heat example air
Gas Molar Property cvair =
……………………..cv(J/kg-mol0K)
Ar 12500
206100.78084 + 217000.209476 + 125000.00934 + 282000.000314
He 12500
28.9541
CO 20700
H2 20400
HCl 21400 + …. = 717.64 J/kg-K
N2 20610
NO
O2
20900
21700

c p air  Rg  cv air  
Cl2 24800 287.056 + 717.64  J/kg-oK 1004.696 J/kg-oK
CO2 28200

N2O 28500

Data at 15°C and 1 atmosphere. 

 = cp/cv=1007.696/717.64 = 1.400

22
 Ratio of Specific Heats
 varies as a function of temperature and drops off significantly
• 
at higher temperatures (molecular spin mode)

 effects become significant for hypersonic flow

• 

• For Most
Of this Properties
class…we For air
will use
 1.40
• Careful …… for
rockets (where
Combustion occurs)
Invariably…  1.40
23
 Ratio of Specific Heats (concluded)
cp

•  key parameter for compressible flow analysis
cv
Approximate Specific Heat Ratio for Various
Gases, at moderate temperatures

QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.

Water (steam) 1.2855

24
 Thermodynamics Summary
Ru
• Equation of State: p  RgT  >Rg 
 Mw
- Ru = 8314.4126 J/K-(kg-mole)

- Rg (air) = 287.056 J/K-(kg-mole)


• Relationship of Rg to specific heats,  

cp cv  Rg
• Internal Energy and Enthalpy
 de   dh 
h = e + Pv cv  cp 
 dT  v  dT  p

25
 Thermodynamics Summary (cont’d)
• First Law of Thermodynamics, reversible process

de dq  pdv dh dq  vdp

• First Law of Thermodynamics, isentropic process
(adiabatic, reversible)

de  pdv dh vdp

26
 Thermodynamics Summary (cont’d)
• Second Law of Thermodynamics, reversible process

 T2   p2 
s2  s1 c p ln 2    Rg ln  
 T1   p1 

• Second Law of Thermodynamics, isentropic process

(adiabatic, reversible) ------> s2 - s1 = 0

 
p2 T   1  p2   2 
 2     
p1  T1   p1   1 

27
 Thermodynamics Summary (cont’d)
control
volume p p + dp
C C- d v A
c

 + d

• But from conservation of mass and momentum ..

sonic velocity is


 p2   2     p  1

     p2   p  
 1 1  2   p k    k 
p
 1  1 

p 1  p
  k     RgT  c   RgT
  
28
 Thermodynamics Summary (cont’d)
• Speed of Sound for calorically Perfect gas

c   RgT

• Mathematic definition of Mach Number

V
M
 RgT

29
 Thermodynamics Summary (concluded)
• For a mixed gas

j V j M w j  V 
Mw    M j
w j

V frac j M w j 
j V j j  V

k
k

 j

j n j c 'j
p

  c '  c
_
cp ' V ' Molar specific heat
j n j
frac j p j p J
( )
j kg-moloK

_
cp ' 1
 
_
cp 
M

M
 V frac j M w j c p j  mass specific heat J
( o )
 w w j kg K

Same holds for cv

30
Supersonic Flow Field Examples

• Mach Angle • Which aircraft

1 Is moving faster?

sin M 

31
 SR-71 Near Field Shock Wave Patterns
c   RgT

Speed of sound across each successive shock wave is

Higher (temperature increases) … wave catch up and
Reinforce each other
32
Effect of Local Speed of Sound on Shock
Wave Propagation

Speed of sound across each successive shock wave is

Higher (temperature increases) … wave catch up and
Reinforce each other
33
 Sonic Boom N-wave
By time boom reaches ground, shocks
Have coalesced into familiar sounding
N-waves “boom-boom” … entirely a result
of local sonic velocity changing with temperature

34
Speed of Sound Can Tell You a Lot

• What is the difference?

35
 Homework, Section 1
• A sample return Probe is being sent on a 1-year mission
from Earth to Mars Via Venus Using “aero-gravity” assist
( both both gravity And aerodynamics at Venus used to turn the
corner to Mars)

• The aero-assist maneuver at Venus is performed at

An altitude of 110 km above the Surface at a peak
Atmospheric velocity of 13.09 km/sec

• At 110 km altitude, the ambient temperature is 147.63k

• Assume that the Venutian atmospheric cp At -100 C

Composition by volume is {97% CO2, 3% N2}
CO2 ~ 0.845kJ/kg °K

• Calculate the Probe Mach number at the N2 ~ 0.995kJ/kg °K

Venus aero-assist interface
Note Units!
36
 Homework, Section 1 (cont’d)
• Show that for a reversible process

 T2   1 
s2  s1 cv ln    Rg ln  
 T1   2 

• and that for a reversible, adiabatic process

 1
 T2   2 
   
 T1   1 
37
 Homework, Section 1 (cont’d)
• Show that for an ideal gas the following
 hold


cp  Rg
 1
1
cv  Rg
 1

38