1) The document discusses the fundamentals of thermodynamics and mechanics, including concepts like momentum, work, energy, forces, equations of state, and the ideal gas law. 2) It describes microscopic and macroscopic views of matter, distinguishing between intensive and extensive state variables as well as homogeneous and heterogeneous systems. 3) The key concepts of thermodynamic equilibrium, temperature, and the relationships between pressure, volume, amount of substance and temperature defined by the ideal gas law are summarized.

1. A
B C
D
Application to
compressors
Thermodynamics
Basics
THERMO-
DYNAMICS

2. 2
Thermodynamic
March 2008
• The momentum characterizes a current state of the system
• The momentum is characterized by its magnitude and its direction  vector!
• For a mass point .......with m=const. 
• For a body with finite dimensions where is the velocity of the mass center
• The momentum can only be changed by external forces
I

v
m
I


*
 F
a
m



*
Mechanics


j
j
F
I
dt
d 

The motion of a body (mass point, rigid / deformable body) is governed in an
inertial reference system by
“equation of motion”: The rate of change of the momentum equals the sum of
the external forces acting on the system
S
v
m
I


*
 S
v


3. 3
Thermodynamic
March 2008
Forces
Forces act between bodies and occur in pairs (equal in magnitude and opposite in
direction.)
• Force also characterized by magnitude and direction
• Sum of forces = vector sum...parallelogram rule
• Forces may act over a distance (gravity,...) or via contact
• Forces give rise to changes of the momentum and/or the shape of a body
1
2
3
1
,
2
F

2
,
1
F

1
,
3
F

3
,
1
F

2
,
3
F

3
,
2
F

i
j
j
i F
F ,
,





4. 4
Thermodynamic
March 2008
Work & power
Def.: Work of a single force by an infinitesimal displacement of its point of
application at a body given by the scalar product
„Work equals the product of displacement of the point of application of the force and
component of force in direction of displacement“.....natural combination of motion
(deformation) and force.
Attention: A certain amount of work is always associated with a process
and never with a current state of a system...work is not a property! (in
contrast to so-called state variables such as mass, speed, momentum,
temperature, energy,....)
0

W
de 0

W
de

F

s
d


F

s
d

Power: v
F
dt
W
d
P e 



 W
s
m
kg
s
m
N
P 

 3
2
*
*
]
[

cos
*
* s
d
F
s
d
F
W
de






 J
s
m
kg
m
N
W 

 2
2
*
*
]
[

5. 5
Thermodynamic
March 2008
Power theorem
2
*
2
1
v
m
Ekin


Def.: Kinetic energy of a mass point ….scalar quantity
• For body with finite dimensions m
d
v
E
m
kin 
 *
2
1 2

• Special case: rigid body






 2
2 .
,
*
*
2
1
,

prop
rot
kin
v
m
trans
kin
kin E
E
E
s


Equation of motion 
„The rate of change of the kinetic energy of a system equals the power
of the internal and (!!) external forces“
)
(
)
( i
a
kin P
P
E
dt
d


)
(
2
1
)
(
2
1
1
,
2
,
i
a
kin
kin W
W
E
E 
 


First integral 
„The finite increase of the kinetic energy of a system equals the total work
of the external and internal forces in the time intervall t1t2 elapsed“

6. 6
Thermodynamic
March 2008
Work of internal forces
)
(t
ds

1.1.1
Examples of internal forces that do work:
• Frictional forces
  0
*
*
*
* )
(
)
(
)
(
2
)
(
1
,
2
)
(
1
)
(
2
,
1
)
(
)
(
1
,
2
)
(
2
,
1 






 t
t
t
t
t
t
n
n
n
e ds
F
ds
F
ds
F
ds
F
F
W
d
body 1
body 2
1
s
d

2
s
d


)
(
2
,
1
n
F

)
(
1
,
2
n
F

)
(
2
,
1
t
F

)
(
1
,
2
t
F

2
,
1
F

1
,
2
F

n
t
t
n
ds
s
c
W
d spring
e *
*


• Spring forces
0
l
s
s
c
Fspring *

ds

7. 7
Thermodynamic
March 2008
Potential energy
 Gravity (homogeneous amd parallel gravity field)
Work done in general dependent on what happens between 1 and 2
Exception: „Conservative“ forces
• Force just dependent on position and independent on how position was reached
• Work done by such a force no longer dependent on special path but just on starting and
end positions  potential energy
Def.: 2
1
1
,
2
, 


 W
E
E pot
pot
Examples:
 
1
2
1
,
2
, *
* z
z
g
m
E
E pot
pot 


 Hookean spring: s....elongation of spring
2
* 2
s
c
Epot 

8. 8
Thermodynamic
March 2008
Mechanical energy
i.e.
In a conservative mechanical system, the sum of kinetic and potential energy
remains constant, i.e. is conserved
1.1.1
Oscillation  conversion of potential into
kinetic energy and vice versa g
m *
In a conservative mechanical system we have
 
1
,
2
,
1
,
2
, pot
pot
kin
kin E
E
E
E 



  0




 pot
kin
pot
kin E
E
dt
d
const
E
E „First integral“ of
equation of motion
„Time-reversal invariance“
In a system with friction   0

 pot
kin E
E
dt
d
Example.: Pendulum
No time-reversal invariance
 arrow of time!

9. 9
Thermodynamic
March 2008
Summary mechanics
• The current motion-state of a system is characterized by its momentum
• The momentum of a system can just be changed by the action of external
forces
 momentum is a conserved quantity
• Work is a mechanical quantity
• Each mechanical system has a certain kinetic and potential energy. For
systems with friction / damping / plastic deformations (dissipative systems)
the mechanical energy is not conserved but decreases with time
• The variables temperature and heat are not defined mechanically.

10. 10
Thermodynamic
March 2008
Overview thermodynamics (TD)
Thermodynamics
• defines temperature,
• extends the energy (kinetic + potential + internal) so that energy is a
conserved quantitiy for all systems („first law of thermodynamics“...energy
cannot be destroyed or created,....),
• introduces heat as another way how systems can exchange energy with
each other,
• introduces the state variable entropy that imposes restrictions as to the
conversion of different energy forms into each other
(energy  available energy=exergy) and introduces an arrow of time,
• allows general statements as to physical-chemical relations for equilibrium.

11. 11
Thermodynamic
March 2008
Thermodynamic systems
Thermodynamic system:
Separated by system boundaries (material or virtual) from surroundings
(=everything except the system)
• Closed system....no transfer of matter across system boundaries
• Open system....transfer of matter across system boundaries
• Isolated system....no interaction with surroundings at all
Open system, system
boundaries fixed in space
open system,
deformable
system
boundaries

12. 12
Thermodynamic
March 2008
Homogeneous – heterogeneous systems
• Homogeneous system
chemical composition and physical properties the same throughout system, otherwise
• Heterogeneous system
• Phase.....homogeneous part of heterogeneous systems (solid, liquid, gaseous,...)
1.1.1
condensate
gas (pure substance, e.g. water vapour)
or mixture....homogeneous system
Two-phase mixture (gas/gas mixture
and condensate).....
heterogeneous system

13. 13
Thermodynamic
March 2008
State variables
State variable..... quantifiable, measurable macroscopic property of a system
• Intensive state variables
do not depend on the size of the system. Examples: pressure p,
temperature T
• Extensive state variables
depend on the size of the system and add when two systems are added at
equilibrium. Examples: volume V, mass m, energy U....denoted by capital
letters (except for the mass m)
• Specific state variables
Are referred to the mass of the system. Examples:
specific volume v=V/m=1/r,
specific heat capacity,....denoted by lowercase letters

14. 14
Thermodynamic
March 2008
Measures of amount of matter
Amount of matter („how much“) characterized by
• Mass m (unit kg)
• Particle number N
• amount of substance (unit kmol)
• The molar mass M [kg/kmol] is a constant characteristic for each substance and is
given (approx.) by the sum of the atomic masses. Example.: MC12
=12kg/kmol,
MH2
=2kg/kmol
• The number of particles per Kilomol is given by Avogadro´s constant
NA= 6.02214*1026 (kmol) –1
• Mass m and amount of substance n of a pure substance are related with each other via
m=n*M
• Molar mass and molecular mass mM (microsc. unit) are related with each other via
M=NA* mM

15. 15
Thermodynamic
March 2008
Thermodynamic equilibrium
A system whose state variables do not change when the system gets isolated
from the surroundings is in thermodynamic equilibrium.
Necessary conditions:
• Mechanic equilibrium
• Chemical equilibrium
• Same temperature throughout the system („thermal“ equilibrium)
Thermodynamic equilibrium is characterized by
• a tendency of all systems to establish it and
• a minimum amount of state variables required to characterize it.

16. 16
Thermodynamic
March 2008
• State of system determined by complete set of suitably chosen independent
state variables. All other state variables are dependent state variables.
• There are relations between dependent and independent state variables
called equations of state (equation, table,....). These state equations have
to be determined experimentally, are characteristic for each substance, and
mostly refer to the thermodynamic equilibrium state.
Important special case: „Simple“ systems
The state of a simple system is determined by two independent intensive
/specific state variables (pressure, temperature, specific volume=1/density,...) plus one
extensive state variable (mass, volume, amount of substance,...).
Examples:
• pure substances (=consisting of one chemical species)
• gas mixture
Equations of state

17. 17
Thermodynamic
March 2008
Thermal equation of state
The thermal equation of state describes the dependency of the temperature
(=intensive state variable) on the mechanical (intensive or specific) state
variables that uniquely determine the state of the system.
Example.: Ideal gas law
T
M
p
*
R

r
T
n
V
p *
*
* R

....absolute pressure [Pa]
p
....density [kg/m3]
r
....universal gas constant
 
K
kmol
J *
/
8314

R
....molar mass [kg/kmol]
M
....absolute Temperatur [K]
T
....special gas constant
M
R R


18. 18
Thermodynamic
March 2008
Ideal Gases
Example ideal gas law:
In a pressure reservoir (V=5l) there are 0.6 Mol CO2 at a temperature of 25°C.
What is the gas pressure?
Solution:
kMol
kg
M /
44
16
*
2
12
*
1 


kg
kMol
kg
kMol
m
M
n
m 0264
.
0
/
44
*
10
*
6
.
0
* 3



 
3
3
3
/
28
.
5
10
*
5
0264
.
0
m
kg
m
kg
V
m


 
r
Ideal gas law:
  bar
Pa
m
kg
K
kmol
kg
K
kmol
J
T
M
p 975
.
2
297458
28
.
5
*
25
15
.
273
*
44
*
8314
*
* 3




 r
R

19. 19
Thermodynamic
March 2008
Ideal Gases
ideal Gases.....
 are defined by ideal gas law
 may be used to define the absolute thermodynamic temperature
 all real gases behave approximately as ideal gases for not too high
pressures and temperatures  ideal gas law is asymptotic limiting law
for real gases

20. 20
Thermodynamic
March 2008
Microscopic point of view
Matter is composed of enormous amount of small particles (atoms,
molecules, ions,...) that have different kinetic and potential energies
and interact with each other.....statistic theories (probabilities!)
pressure...............sum of rate of change of momentum of all particles
„internal energy“...sum of kinetic plus potential energy of all particles
temperature..........proportional to mean kinetic energy of all particles
Parameters dependent on substance
- particle mass
- particle size
Parameters dependent on state
- density of particles
- mean particle distance d
 mean free path L
d
L

21. 21
Thermodynamic
March 2008
Microscopic point of view
Results of kinetic gas theory:
If
• size of particles negligible
• no interaction between particles (no potential energy)
then
• Ideal gas p*v=R*T
• internal energy proportional to absolute temperature

22. 22
Thermodynamic
March 2008
Microscopic point of view
state of aggregation  mobility of particles
Gas....swarm of mosquitos
- arbitrarily deformable
- strongly compressible, strong density changes
- no order
liquid....ant hill
- arbitrarily deformable
- weakly compressible, small density changes
- short range order, no long range order
solid....system of mass points in meshwork of springs
- weakly deformable
- weakly compressible, small density changes
- short range and long-range order

23. 23
Thermodynamic
March 2008
Pressure
Definitions
absolute pressure bara
absolute pressure psia
gauge pressure barg
gauge pressure psig
Units
1 Pa = 1 Pascal = 1 N/m²
1 bar = 100.000 Pa
1 psi = 6894 Pa
Standard conditions
atmospheric pressure
p0 = 1,0133 bara
1,013
bar
760
mm
Hg
14,696
psia
pressure above atmospheric pressure
Atmospheric pressure
absolute
pressure
perfect vacuum
absolute
pressure
vacuum
gauge
pressure
In thermodynamic calculations only the absolute
pressure may be used!

24. 24
Thermodynamic
March 2008
Definitions
absolute temperature T K
Celsius temperature J °C °C
Fahrenheit scale J °F °F
Units
1 K = Kelvin K
J °C = T - 273.15 °C
= (J °F - 32°F) / 1,8
J °F = 1,8 J °C + 32 °F
Standard conditions
T = 273.15 K = 0°C = 32°F
In thermodynamic calculations only the absolute
temperature may be used!
Temperature
Standard temp.: 0°C
273.15
K
temperature
in
°F
arbitrary temperature
absolute
temperature
in
K
absolute zero: 0 K
temperature
in
°C
32°F

25. 25
Thermodynamic
March 2008
Density
• Units
Density r kg/m³ “rho”
Standard density r0 kg/ms³ “rho-zero” = density at standard
condition p0 = 1.013 bar; J = 0°C
Specific gravity g r0 / r0 Luft density referred to density of air
• Examples
Air r0 = 1.29 kg/ mn³ M = 29 g/mol g = 1
Hydrogen r0 = 0.0897 kg/ mn³ M = 2 g/mol g = 0.0695
Methane r0 = 0.715 kg/ mn³ M = 16 g/mol g = 0.554
Nitrogen r0 = 1.25 kg/ mn³ M = 28 g/mol g = 0.967
Ammoniac r0 = 0.769 kg/ mn³ M = 17 g/mol g = 0.596

26. 26
Thermodynamic
March 2008
Gas mixtures
Under consideration: Mixture of n gases i=1,2,...n
• Pressure of mixture = sum of partial pressures ........„Dalton´s law“
1.1.1
Mixture of ideal gases:
• Volume of mixture = sum of partial volumes
• Partial pressure pi.... pressure gas i would have if it alone occupied the volume V at
the temperature of the mixture
• Partial volume Vi....volume that gas i would occupy at temperature of mixture
Definitions:

27. 27
Thermodynamic
March 2008
Ideal gas mixture
with Dalton´s law and it follows that
Ideal gas law for components T
n
V
p i
i *
*
* R

summed over all components T
n
V
p
n
i
i
n
i
i *
*
*
1
1
R

 


T
M
v
p *
*
R




n
i
i
i M
n
m
1
*
with the molar mass of the mixture 


n
i
i
i
M
n
n
M
1
*
Measures for concentration:
mole fraction
volumetric fraction
i=yi for ideal gas mixtures
n
n
y i
i /
:



n
i
i
i
i V
V
1
/
:


28. 28
Thermodynamic
March 2008
Ideal gas mixture
Example: Molar mass of dry air
Dry air consists of approx. 78% N2, 21%O2 und 1%Ar (volume fraction). What is
the molar mass of the air?
kmol
kg
kmol
kg
kmol
kg
kmol
kg
M
/
0
.
29
/
40
*
01
.
0
/
16
*
2
*
21
.
0
/
14
*
2
*
78
.
0





29. 29
Thermodynamic
March 2008
Real gas behaviour
Attempt to improve ideal gas law.....Van der Waals equation of state
  T
R
b
v
v
a
p *
*
2









a,b....positive constants characteristic of each substance
related to intermolecular attractive forces
related to particle size
• Van-der-Waals equation of state approaches ideal gas law for low densities
• Simplest equation of state that describes behaviour of pure substances
in gaseous and liquid state qualitatively correct
• Accuracy in general not sufficient

30. 30
Thermodynamic
March 2008
Real gas (compressibility) factor Z defined by
• describes deviations from ideal gas law
• Z=Z(p,T) has to be determined for each gas
„Law of corresponding states“: value of Z as a function of reduced pressure
and reduced temperature (i.e. referred to critical values) the same for all gases.
But: Only approximately valid  no physical law!
Nevertheless useful for practical purposes  Redlich-Kwong equation (ITKK)
Real gas (compressibility) factor
T
R
Z
v
p *
*
* 

31. 31
Thermodynamic
March 2008
Example for compressibility factor
Nitrogen
negative deviation from ideal gas behaviour
due to intermolecular forces (moderately high
pressures, for higher temperatures they do not
contribute as significantly)
positive deviation from ideal gas behaviour
at all temperatures for sufficiently high
pressures due to size of molecules

32. 32
Thermodynamic
March 2008
Two phase systems
Considered: Mixture of vapour and liquid of pure substance
Boiling liquid
T>Ts T=Ts T=Ts T=Ts
Superheated
vapour
Saturated
vapour
Mixture of vapour
and liquid
p=const=mK*g/AK
Experimental findings:
• At a certain pressure, equilibrium between vapour and liquid attainable only at
certain temperature („saturation temperature“)
• „liquid-vapour equilibrium line“ gives relationship between „saturation
temperature“ and „vapour pressure"
• above „critical pressure“ no phase transition gas-liquid

33. 33
Thermodynamic
March 2008
Wet vapour
1.1.1
• liquid phase nearly incompressible  saturated liquid line nearly vertical
• Specific volumes of saturated liquid and saturated vapour only dependend on pressure,
values obtained from „thermodynamic tables“
• Isotherms in wet vapour region horizontal
• For pressures above critical one pc no phase transition gas-liquid
v
liquid
gaseous
isotherms T=const
wet vapour
solid
solid/liquid
critical temperature
isotherm
saturated
vapour line
solid/gaseous
saturated
liquid line
p
0,5vc vc 2vc 10vc
critical point
triple line
solid
liquid
gaseous
solid-gaseous
ps
T
Tc
pc

34. 34
Thermodynamic
March 2008
p-v-T Diagram of pure substance
A solid
B liquid
C gaseous
D liquid-vapour
critical point
pk critical pressure
Tk critical temperature
vk critical specific volume
lg v
p
T
p = constant
T = constant
v = constant
pc
Tc
Vc
A
B C
D

35. 35
Thermodynamic
March 2008
Multi-component systems
Two independent intensive/specific state variables no longer determine state
 specification of concentration required
Important example: Wet air
General: mixture of non-condensing gases and condensable vapour
Special: Wet air.....mixture of dry air and water (steam/liquid)
Definitions:
Water content............ x=mw/mair x=xv+xliquid+xice=mv/mair+mliquid/mair+mice/mair
Relative humidity....... f=pv/ps pv partial pressure of steam,
ps=ps(T) saturation pressure of water
Saturation level.......... y=xv/xs xs=xs(T, p)....water content at saturation

36. 36
Thermodynamic
March 2008
Wet air
Derivation of relationship between vapour content and partial pressure:
 
    V
V
air
W
air
V
W
V
air
V
air
V
V
p
p
p
M
M
T
M
p
p
T
M
p
V
V
m
m
x





 *
*
/
*
*
/
*
*
*
:
R
R
r
r
Dalton: Partial pressure dry air pair + partial pressure
steam pV = total pressure p
Note: pVps(T) so that xV xs=MW/Mair*ps(T)/(p-ps(T))
Cooling down unsaturated wet air at p=const  ps decreases  xs decreases until xV=xs
.....“dew point“....further reduction in T then results in condensation of steam

37. 37
Thermodynamic
March 2008
• Whenever there is a change of state, the thermodynamic system
undergoes a thermodynamic process
• important limiting case: quasistatic processes (in contrast to nonstatic processes)
• change of state = series of thermodynamic equilibria
• tremendous simplification for analysis
• only approximately realizable
• a certain change of state may be realized by different processes
• not every process gives rise to a change of the state (steady processes)
Changes of state
p1 p2<p1
FG
Gas
state 1
state 2
Quasistatic expansion of a gas
in a cylinder with frictionlessly
guided piston of changing weight
Quasistatic expansion of gas in a throttle
with varying cross section (local
thermodynamic equilibrium)

38. 38
Thermodynamic
March 2008
Reversible / irreversible processes
• A reversible process is a process that, after it has taken place, can be
reversed so that it causes no changes in the system as well as the
surroundings. All other processes are irreversible.
• (virtually) all quasistatic processes are reversible
• nonstatic processes are irreversible

39. 39
Thermodynamic
March 2008
• Balance equations apply to extensive state variables
• An infinitesimally small change dZ of an extensive state variable z is
composed of two parts according
where deZ (external) is the infinitesimal supply of Z from the surroundings
and diZ (internal) stands for the infinitesimal production of Z inside the
system.
- mass balance dm=dem
- momentum balance
- energy balance dE=deE
- entropy balance dS=deS+diS
• Expressed in rates of change with time:
• Special cases:
• isolated system deZ=0 /
• steady system dZ=0 / dZ/dt=0
Balance equations
no internal production  conserved quantities
Z
d
Z
d
dZ i
e 

dt
Z
d
Z
dt
dZ i

 
rate of production
rate of supply
0

Z

I
d
I
d e




40. 40
Thermodynamic
March 2008
Energy balance….first law of TD
• The energy (extensive state variable) of a thermodynamic system is given
by
where U is the so-called internal energy
• the energy balance reads
where deW ( ) is the work (power) done on the system, deQ ( ) is the heat
(heat transfer rate) supplied to the system, and de
(m)E (energy flux) is the
energy that is supplied to the system via the entering mass („material supply of
energy“  open system)
Note: Heat and work are process quantities in contrast to state quantities such as
energy, mass, volume, pressure,.....heat is not a property of a system!
The subscript e stands for externally supplied
E
d
Q
d
W
d
dE m
e
e
e
)
(



)
(m
E
Q
W
dt
dE 

 


U
E
E
E pot
kin 


P
W,
 Q


41. 41
Thermodynamic
March 2008
• Closed system at rest
Example: Quasistatic, adiabatic (no heat transfer) expansion of gas
Work done on system....deW= - Fboundary*dz
Quasistatic change of state:
- mechanic equilibrium piston....Fboundary=FG
- pressure inside cylinder homogeneous.... Fboundary=p*A
 quasistatic work deW = - p*A*dz= - p*dV
1st law
Energy balance….special cases of interest
12
12
1
2 Q
W
U
U
Q
d
W
d
dU e
e 





state 2
FG
Ga
s
state 1
Fboundary
z
dz




2
1
1
2 *dV
p
U
U
p*dV
1
2
V2
V1
p2
p1
 W12 
Note:
• the work depends on the special path 12
• according to the 1st law, the work equals the change
in the internal energy where the latter just depends
on the states 1 and 2
• solution: further restriction of no heat transfer
leaves only one quasistatic path 
isentropic change of state

42. 42
Thermodynamic
March 2008
Energy balance….special cases of interest
• Open system, steady state
Mass balance:
Energy balance:
Dividing total work W into shaft work Ws and flow work deWf=p(m) v(m) dem (required to
move fluid through system boundaries) gives
with the specific enthalpy h:=u+p*v
 
 
)
(
,
)
(
,
)
(
)
(
,
)
(
,
)
(
*
0 m
out
pot
m
out
kin
m
out
m
in
pot
m
in
kin
m
in e
e
u
e
e
u
m
Q
W
dt
dE








 


m
m
m
m
m
dt
dm
out
in
out
in




 




 0
)]
:
*
(
:
*
[
*
0 )
(
,
)
(
,
)
(
)
(
)
(
)
(
)
(
,
)
(
,
)
(
)
(
)
(
)
( m
out
pot
m
out
kin
m
out
m
out
m
out
m
out
m
in
pot
m
in
kin
m
in
m
in
m
in
m
in
s
e
e
h
v
p
u
e
e
h
v
p
u
m
Q
W 












 

 


 

 




in
m

out
m

in
v
p
u )
,
,
(
out
v
p
u )
,
,
(
Q

s
W

in
z
out
z

43. 43
Thermodynamic
March 2008
Caloric equation of state
• First law – existence of an extensive state variable U called internal energy
• the internal energy U is therefore a property of the system
• For a certain state of the system there is a unique value of U....relationship
to other state variables (p,T,...) that determine the state of the system?
 Caloric equation of state that relates the internal energy to temperature and
other mechanical (p,v,...) (chemical /electromagnetical) state variables
Simple single-phase system in thermodynamic equilibrium: U=u*m, u=u(T,v)
dv
v
u
dT
T
u
du
T
v
*
* 

















Incremental change in u associated
with incremental temperature rise T
when v is kept constant
Incremental change in u associated with
incremental specific volume rise v when T is
kept constant
„specific isochoric heat capacity“ cv

44. 44
Thermodynamic
March 2008
Caloric properties of ideal gases
For ideal gases the internal energy is just a function of temperature so that
with cv=cv(T) (constant only for monatomic gases). For moderate temperature
differences
where cv is to be taken as a suitable average value. For the enthalpy of an ideal
gas
and hence where even for variable cp and cv.
Note: specific heat capacities are properties of a system
Note: cpcv, for solids and liquids cp cv=c
dT
c
du v *

 
1
2
1
2 * T
T
c
u
u v 


T
R
u
v
p
u
h *
*
: 



R
c
T
h
c v
p
p 










: R
c
c v
p 

„specific isobaric heat capacity“

45. 45
Thermodynamic
March 2008
Historical question: How much heat Q has to be added to a given amount of
Substance (single phase!!) to increase its temperature by a certain amount T?
Answer by means of first law:
Case differentiation:
• isochoric case V=const., i.e. dV=0: dU=deQ deQ=m cv dT
• isobaric case p=const, i.e. dp=0: dU=dH-p dV-V dp=deQ-p dV
dH=deQ deQ=m cP dT
Heat capacities – historical origin
Q
T=?
Q
d
dV
p
Q
d
W
d
dU e
e
e 





46. 46
Thermodynamic
March 2008
Quasistatic, adiabatic compression of ideal gas:
For constant cv and cp („calorically ideal gas“):
Expressed in terms of T, p:
Expressed in terms of p,v:
with the
isentropic coefficient (ratio of specific heats) k=cp/cv
Important applications of 1st law
dv
p
du 

dv
v
T
R
dT
cv
*


v
dv
R
T
dT
cv 

1
2
1
2
ln
)
(
ln
v
v
c
c
T
T
c v
p
v 


const
v
T
v
T 


 1
1
1
1
2
2 *
*
k
k
const
p
T
p
T 





k
k
k
k 1
1
1
1
2
2 *
*
const
v
p
v
p 

k
k
2
2
2
2 *
*
isentropic p v k =const
isothermal p v =const
p
v

47. 47
Thermodynamic
March 2008
Important applications of 1st law
Steady, adiabatic compression process of thermally and calorically ideal gas
(kinetic and potential energy neglected)
For a reversible compression:
Note: Losses increase the discharge temperature, i.e. increase
the work required for an adiabatic compression from ps to pd!
   
1
2
2
1 *
*
*
0 T
T
c
m
P
h
h
m
P p 




 

k
k 1
*










s
d
s
d
p
p
T
T
P
M
p1, T1, h1
p2, T2, h2
























1
*
*
*
1
k
k
s
d
s
p
p
p
T
c
m
P 

48. 48
Thermodynamic
March 2008
Important applications of 1st law
Steady, adiabatic throttling of gas:
Ideal gas: T2=T1
Real gases: Whether gas heats up or cools down for certain inflow conditions
determined by
Joule-Thomson coefficient
Note: The pressure always decreases from 12! (2nd law)
p1, T1, h1
p2<p1, T2, h2
  1
2
2
1
*
0 h
h
h
h
m 


 
h
p
T











49. 49
Thermodynamic
March 2008
Important applications of 1st law
For an ideal gas, the gas
temperature does not
change during throttling.
Question: Why does the
gas temperature increase
during idling motion?
p
ps
ps-pSV
ps+pSV
Ts
T increases
v
p
V

50. 50
Thermodynamic
March 2008
The polytropic process
• for an isothermal change of state of an ideal gas
• for an isobaric change of state
• for an isentropic change of state of a calorically ideal gas
• for an isochoric change of state
During all processes, one state variable stays constant  „simple“ changes of
state
„polytropic“ process:
Attempt to describe real-world processes in approximate manner (heat
transfer, losses,...)
const
v
p 
*
const
v
p 
0
*
const
v
p 
k
*
const
v
p 

*
exponent
ropic
.....polyt
, n
const
v
p n

*

51. 51
Thermodynamic
March 2008
The second law of TD
• All real-world processes are irreversible with reversible processes being just
limiting cases of differing applicability to analysis of such real world
processes.
• Temperature, density, and pressure differences tend to equalize over time
 there is a definite arrow of time!
• According to the first law, the energy of an isolated system stays constant.
• Interchangeability of different energy forms (heat  work)?
• Maximum achievable efficiencies of heat engines („available anergy“)?
Answers and explanations provided by „Second law of TD“

52. 52
Thermodynamic
March 2008
The second law of TD
• Each system possesses an extensive state variable called entropy that
accounts for effects of irreversibility in a thermodynamic system.
• The entropy balance equation reads with
Note:
• In an isolated system, the entropy cannot decrease
• In a closed system, the entropy may either increase or decrease depending on whether
the entropy production rate is larger or smaller than the entropy transferred away
from the system by means of heat removal from the system (effect of temperature
level!)
• The 2nd law introduces a specific distinction between heat and work!
• The 2nd law imposes restrictions on the conversion of heat to work (exergy, anergy)
• For a reversible, adiabatic change of state, the entropy stays constant
 isentropic process
S
d
S
d
dS i
e 

S
d
T
Q
d
S
d m
e
e
e
)
(

 processes
reversible
le
irreversib
for






 0
S
di

53. 53
Thermodynamic
March 2008
p-V diagram of real compressors
n polytropic exponent
P compression ratio p2/p1
s0 clearance volume referred to swept
volume
L loss factor accounting for losses
associated with leakage, heating up
of gas during suction,….
Adiabatic mass flow Mad=rs*VSW*l*rpm/60
swept volume VSW
Clearance
volume
s
0
TDC
s0
BDC
p2
p1
pressure
V1
A
B C
D
volume
VSW hv
V2
1
l s n
  
1 1
0
( )
P - L
Mad adiabatic mass flow (kg/s)
rpm speed (min-1)
VH swept volume
l filling efficiency= volumetric
efficiency hV minus losses L

54. 54
Thermodynamic
March 2008
p-V diagram of real compressors
Effect of compression ratio p2/p1 and clearance volume s0
Dependence of volumetric efficiency on
compression ratio p2/p1:
4, 6, 8
Clearance volume s0 constant 10%VH
Dependence of volumetric efficiency
on clearance volume s0:
a: 6%,
b: 10%,
c: 15 % of VH
Compression ratio p2/p1 constant
p TDC
V
BDC
Vh
V
a
b
b
c

55. 55
Thermodynamic
March 2008
p-V diagram of real compressors
Discharge valve leakage
p
v
p
v
Suction valve leakage
capacity is not increased!
Effect of valve leakage

56. 56
Thermodynamic
March 2008
Isothermal compression
Adiabatic compression
2nd stage
Adiabatic compression 1st stage
Adiabatic compression
Energy savings due to 2-stage
compression with (lossless) intercooling
Stage 1
Stage 2
p
V
Multistage compression

57. 57
Thermodynamic
March 2008
Compression Losses
P P P P P
mech mech th fr mech V SV V DV
   
, , , ,
In real compression processes there are frictional losses and valve losses
A
B C
D
Suction valve losses
Discharge valve losses
so that a measure for the efficiency is given by the
isentropic efficiency
that refers the work for an ideal isentropic compression to the actual one.
P
his
mech, th
mech
P

,

58. 58
Thermodynamic
March 2008
Discharge state:
Partial pressure pg,2 increased in proportion to pressure ratio;
because of T2>T1, ps,g(T2) has also increased….
component g is gaseous.
State after cooler:
Saturation pressure ps,g(T1) the same as in suction state but partial
pressure has increased pg,2 > pg,1 ….
possibility that certain amount of g condenses (so that pg,2 =ps,g(T1) )
Suction state
Partial pressure pg,1 of a condensable gas component g is smaller than
corresponding saturation pressure ps,g(T1) ….
component g is gaseous
Condensation
p T
1 1
,
p T
2 2
,
p T
2 1
,