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HEAT PROCESSES
Thermodynamics
processes and cycles
Thermodynamics fundamentals. State variables, Gibbs phase rule, state equations,
internal energy, enthalpy, entropy. First law and the second law of thermodynamics.
Phase changes and phase diagrams. Ts and hs diagrams (example: Ts diagrams for
air). Thermodynamic cycles Carnot, Clausius Rankine, Ericson, Stirling,
thermoacoustics.
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010
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FUNDAMENTALS of
THERMODYNAMICS
Estes
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BASIC NOTIONS
Subsystem flame zone =
opened
SYSTEM
• Insulated- without mass or energy transfer
Subsystem candlewick =
opened
•Closed (without mass transfer)
• Opened (mass and heat transport through boundary).
Thermal units operating in continuous mode (heat
exchangers, evaporators, driers, tubular reactors, burners)
are opened systems
Subsystem candle = opened with
moving boundary
Subsystem stand = closed
Thermal units operating in a batch mode (some chemical
reactors) are closed systems
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StaTE VARIABLES
state of system
is characterized
by
THERMODYNAMIC STATE VARIABLES related with directly
measurable mechanical properties:
 T [K], p [Pa=J/m3], v [m3/kg] (temperature, pressure, specific volume)
Thermodynamické state variables related to energy (could be
derived from T,p,v):
 u [J/kg] internal energy
 s [J/kg/K] specific entropy
 h [J/kg] enthalpy
 g [J/kg] gibbs energy
 e [J/kg] exergy
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Gibbs phase rule
Not all state variables are independent. Number of independent variables
(DOF, Degree Of Freedom) is given by Gibbs rule
NDOF = Ncomponents – Nphases + 2
 1 component, 1 phase (e.g.gaseous oxygen) NDOF=2 . In this case
only two state variables can be selected arbitrarily, e.g. p,v, or p,T or v,T.
 1 component, 2 phases (e.g. equilibrium mixture of water and steam at
the state of evaporation/condensation). In this case only one state variable
can be selected, e.g. pressure (boiling point temperature is determined by p)
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State EquATIONS p-v-T
Van der Waals equation isotherms
RT
a
p~
 ~2
v b v
RTc
p 2a
 ~3  ~
0
2
~
v v c (vc  b)
2 RTc
2 p
6a



0
2
4
3
~
~
~
v
v c (vc  b)
Above critical temperature Tc the
substance exists only as a gas
(liquefaction is not possible even at
infinitely great pressure)
Critical point, solution of these
two equations give a,b
parameters as a function of
critical temperature and critical
pressure
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Pv=RT tutorial Baloon
Example: Calculate load capacity of a baloon filled by hot air. D=20m,
M=29 (air)
T=600C, Te=200C, p=105 Pa.
mg  V ( e   ) g
D
pM air 1
1 1
1 1
1
 e      M air ( ~  ~ ) 
(

)
ve v
ve v
R
T20 T60
m
m
D 3 pM air 1
6
R
1
 203 105  29 1
1
(  )
(

)
T20 T60
6 8314 293 333
= 599 kg
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Pv=RT tutorial Syringe
Record time change of temperature of air compressed in syringe.
x
D
Thermocouple
Adiabatic change
(thermally insulated)
V
p1v1  p2 v2
Example: V2/V1=0.5
P-pressure transducer Kulite XTM 140
p1v1  RT1
p1
v
T v
 ( 2 )  1 2
p2
v1
T2 v1
p2 v2  RT2
T1
v2  1
( )
T2
v1
=cp/cv=1.4
T1=300 K
T2=396 K
temperature increase 96 K!!
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Pv=RT tutorial - expansion
Dynamic inflation test of a viscoelastic pipe (carried out in our laboratory of
cardiovascular mechanics) was based upon recorded oscillation of pressure after a
sudden release of pressure in a container. We did not know a reason for a gradual
pressure increase in the end of experiment. Possible explanation: Temperature of adiabatically expanded air
initially drops and later on slowly increases to the room temperature (and this is accompanied by a gradual increase of pressure).
p1v1  p2 v2
T1=300K,
p1=60kPa
55
45
p1v1  RT1
p2 v2  RT2
p
WT…
35
25
T2=257K,
p2=35kPa
0 0.20.40.60.8 1 1.2
t
Example: p2/p1=35/60=0.58
T3=300K,
p3=41kPa
=cp/cv=1.4
T2
p2 1
( )
T1
p1
T1=300 K
T2=257 K
Slow heating (closed vessel) to the final temperature 300K
tested sample,
viscoelastic tube
(blood vessel)
T3 p3 300


 p3  41kPa
T2 p2 209
In this way only a final value p3 is analysed. It is necessary to know
heat transfer rate (heat transf.coef.) to calculate the time change of p.
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Internal energy u [J/kg]
u-all forms of energy of matter inside the system (J/kg), invariant with respect to
coordinate system (potential energy of height /gh/ and kinetic energy of motion of
the whole system /½w2/ are not included in the internal energy). Internal energy is
determined by structure, composition and momentum of all components, i.e. all
atoms and molecules.
 Nuclear energy (nucleus)
~1017J/kg
 Chemical energy of ionic/covalent bonds in molecule
~107 J/kg
 Intermolecular VdW forces (phase changes)
~106 J/kg
Frequently only the following item (thermal energy) is included into the internal
energy notion (sometimes distinguished as the sensible internal energy)
Thermal energy (kinetic energy of molecules)
~104 J/kg
It follows from energy balances that the change of
internal energy of a closed system at a constant volume
equals amount of heat delivered to the system
du = dq (heat added at isochoric change)
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Enthalpy h [J/kg]
h=u+pv
enthalpy is always greater than the internal energy. The added term pv
(pressure multiplied by specific volume) simplifies energy balancing of
continuous systems. The pv term automatically takes into account
mechanical work (energy) necessary to push/pull the inlet/outlet material
streams to/from the balanced system.
It follows from energy balances that the change of
enthalpy of a closed system at a constant pressure
equals amount of heat delivered to the system
dh = dq (heat added at isobaric change)
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Entropy s [J/kg/K]
Thermodynamic definition of entropy s by Clausius
ds  (
dq
) rev
T
where ds is the specific entropy change of system corresponding to the heat
dq [J/kg] added in a reversible way at temperature T [K].
Boltzmann’s statistical approach: Entropy represents probability of a
macroscopic state (macrostate is temperature, concentration,…). This
probability is proportional to the number of microstates corresponding to a
macrostate (number of possible configurations, e.g. distribution of molecules
to different energy levels, for given temperature).
It follows from energy balances that the change of
entropy of a closed system at a constant temperature
equals amount of heat delivered to the system / T
Tds = dq (heat added at an isothermal and reversible change)
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Laws of thermodynamics
Modigliani
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Laws of thermodynamics
First law of thermodynamics (conservation of energy)
δq = heat added to
system
δw = work done by system
δq = du + δw
expansion work (p.dV) in case of compressible fluids,
surface work (surface tension x increase of surface),
shear stresses x displacement, but also electrical
work (intensity of electric field x current). Later on we
shall use only the p.dV mechanical expansion work.
Second law of thermodynamics (entropy of closed
insulated system increases)
Tds  δq
δq = heat added to system is Tds
only in the case of reversible process
Combined first and second law of thermodynamics
Tds = du+pdv
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THERMODYNAMIC relationships
Be careful at interpretation: First law of thermodynamic was presented in the form
corresponding only to reversible changes (therefore for infinitely slow changes,
without viscous friction, at uniform temperature)
Tds  du  pdv
reversible processes are generally defined as
changes of state (12) which can be recovered
(21) without any change of the system
environment. In an irreversible process it is also
possible to return back (to the initial state) but heat
or work must be added from environment.
This equation enables to calculate the entropy change during a reversible process.
However, entropy is a state variable, and its changes are independent of the way,
how the changes were realized (there are always infinitely many ways how to
proceed from a state 1 to a state 2). So, why not to select the reversible way even in
the case, when the real process is irreversible? It is always simpler and results (for
example calculated entropy changes) hold generally even for irreversible (real)
processes.
In the following it will be demonstrated how to calculate internal energy and enthalpy
changes from the measured changes of temperature, pressure and volume (and
results hold not only for the reversible processes).
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Example: insulated system
Let us consider a perfectly insulated calorimeter (Deware flask, e.g.) containing a
cold metallic block (temperature Tm) and hot water (Tw).
Water
Tw=300K
dQwm
At a time interval dt the temperature of water Tw decreases by
dTw=-dQwm/(mwcpw) and the temperature Tm increases by
dTm=dQwm/(mmcpm), however the sum of inner energies
U=ummm+uwmw remains, therefore dU=0. Volume V is constant
and dV=0. Therefore the entropy increase of the whole system
dS should be zero (dS=0) TdS  dU  pdV  0
This conclusion is wrong, because the whole process is irreversible
and dS is actually positive. So that the previous equation could be
used the whole process must be substituted by an equivalent but
reversible process. For example the water can be used for heating
of a fictive gas at constant temperature Tw (reversibly, entropy of water will be decreased
by dQwm/Tw), followed by an adiabatic expansion cooling down the gas to the temperature
Tm (reversible expansion, not changing the entropy of gas). The gas then isothermally and
reversibly heats the metallic block, thus increasing its entropy by dQwm/Tm. Summing the
entropy changes gives
dQ
dQ
T T
Metal
Tm=273K
dS 
wm
Tm

wm
Tw
 dQwm
w
TmTw
m
0
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Energies and Temperature
The temperature increase increases thermal energy (kinetic energy of molecules).
For constant volume (fixed volume of system) internal energy change is proportional
to the change of thermodynamic temperature (Kelvins)
du = cv dT
where cv is specific heat at constant volume
For constant pressure (e.g. atmospheric pressure) the enthalpy change is also
proportional to the thermodynamic temperature
dh = cp dT
where cp is specific heat at constant pressure.
Specific heat at a constant pressure is always greater than the specific heat at a
constant volume (it is always necessary to supply more heat to increase temperature
at constant pressure, because part of the delivered energy is converted to the
volume increase, therefore to the mechanical work). Only for incompressible
materials it holds cp=cv.
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Energies and Temperature
Internal energy (kinetic energy of translatory motion) of one monoatomic
molecule of an ideal gas is given by
3
where k is Boltzmann constant
uinternal energy of one molecule 
2
kT
1.3806504×10−23
Kinetic energy (1/2mw2) of chaotic thermal motion is therefore independent of
molecular mass (lighter molecules are moving faster)! Knowing molecular mass it
is therefore possible to estimate the specific heat capacity cv [J/kg/K] theoretically
~  3 kT  6.022 10  12.5T [J/mol] , therefore molar heat capacity c  12.5 [J/mol/K]. Specific heat capacity is c  c~ / M
(molar internal energy u
23
where M is molecular mass 2[kg/mol]. For helium (M=0.004) thus calculated specific heat capacity c v=12.5/0.004=3125 J/kg/K. The tabulated cp
value at 300K is cp=5193 J/kg/K. This agrees quite well, because cp-cv=R/M=8.314/0.004=2078 J/kg/K, see the Mayer’s equation).
For more complicated molecules the equipartition principle can be applied,
stating that any mode of motion (translational, vibrational, rotational) has the
same energy kT/2 (monoatomic gas has 3 translational modes in the x,y,z
directions, therefore u=3kT/2).
According to the equipartition theorem the specific heat capacities are constants independent of
temperature. This is not quite true especially at low temperatures, when cv decreases – and in this
case quantum mechanics must be applied. At low temperatures (<100 K) only three translational
degrees of freedom are excited (cv=3/2R), at higher temperatures two additional rotational degrees
increase cv to 5/2R (for diatomic molecules N2, H2, O2) and contribution of vibrational degrees of
freedom is significant at even higher temperatures (>500 K). In water rotational and vibrational
degrees of freedom are excited at very low temperatures.
This nice animated gif (molecule of a peptide with atoms
C,N,O,H) is captured from Wikipedia.org
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u
Energies and Temperature
n
kT  6.022 1023  4.157 nT [J/mol]
2
cv 
4.157 n
4.157 n  R 4.157 n  8.314
, cp 

M
M
M
cvHe 
4.157  (n  3)
J
4.157n  R
J
J
 3118
, c pHe 
 5196
(tabulated c p  5193
)
0.004
kgK
M
kgK
kgK
cvO2 
4.157  (n  5)
J
J
J
 650
, c pO 2  910
(tabulated c p  970
at T  200 K )
0.032
kgK
kgK
kgK
cvN2 
4.157  (n  5)
J
J
J
 742
, c pN 2  1039
(tabulated c p  1050
at T  200 K )
0.028
kgK
kgK
kgK
4.157  (n  6)
J
J
J
 1467
, c pNH 3  1956
(tabulated c p  2420
at T  200 K )
0.017
kgK
kgK
kgK
4.157  (n  8)
J
J
J

 1956
, c pNH 3  2445
(tabulated c p  2420
at T  200 K )
0.017
kgK
kgK
kgK
cvNH3 
cvNH3
cvH 2O 
4.157  (n  6)
J
J
J
 1386
, c pH 2O  1848
(tabulated c p  1960
at T  200 K )
0.018
kgK
kgK
kgK
According to the equipartition theorem the specific heat capacities are constants independent of temperature. This is not quite true
especially at low temperatures, when cv decreases – and in this case quantum mechanics must be applied. At low temperatures
(<100 K) only three translational degrees of freedom are excited (cv=3/2R), at higher temperatures two additional rotational degrees
increase cv to 5/2R (for diatomic molecules N2, H2, O2) and contribution of vibrational degrees of freedom is significant at even
higher temperatures (>500 K). In water rotational and vibrational degrees of freedom are excited at very low temperatures.
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u(T,v) internal energy change
How to evaluate internal energy change? Previous relationship du=cvdT holds only at
a constant volume. However, according to Gibbs rule the du should depend upon a
pair of state variables (for a one phase system). So how to calculate du as soon as
not only the temperature (dT) but also the specific volume (dv) are changing?
Solution is based upon the 1st law of thermodynamic (for reversible changes)
du  Tds  pdv  T ((
s
s
s
s
) v dT  ( )T dv)  pdv  T ( ) v dT  (T ( )T  p)dv
T
v
T c
v
ds
v
where du is expressed in terms dT and dv (this is what we need), coefficient at dT is
known (cv), however entropy appears at the dv term. It is not possible to measure
entropy directly (so how to evaluate ds/dv?), but it is possible to use Maxwell
relationships, stating for example that
s
p
(
v
)T  (
T
)v
Instead of exact derivation I can give you only an idea based upon dimensional analysis: dimension of s/v is
Pa/K and this is just the dimension of p/T!
This term is zero for
So there is the final result
ideal gas (pv=RT)
p
du  cv dT  (T (
T
) v  p)dv
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h(T,p) enthalpy changes
The same approach can be applied for the enthalpy change. So far we can calculate
only the enthalpy change at constant pressure (dh=cpdT). Using definition h=u+pv
and the first law of thermodynamics
c
p
s
s
dh  du  pdv  vdp  Tds  vdp  T ( ) p dT  (T ( ) T  v)dp
T
p
Tds
And the same problem how to express the entropy term ds/dp by something that is
directly measurable. Dimensional analysis: s/p has dimension J/(kg.K.Pa)=m3/(kg.K)
and this is dimension of v/T. Corresponding Maxwells relationship is
(
s
v
) T  ( ) p
p
T
After substuting we arrive to the final expression for enthalpy change
dh  c p dT  (T (
v
) p  v)dp
T
Negative sign in the Maxwell equation is probably confusing, and cannot be derived from dimensional
analysis. Correct derivation is presented in the following slide.
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Maxwell relationships
Basic idea consists in design of a state function with total differential depending
only upon dT and dp . Such a function is Gibbs energy g (previously free enthalpy)
g  h  Ts
g
g
dg  dh  Tds  sdT  vdp  sdT  ( ) T dp  ( ) p dT
p
T
Notice the fact, that using this combination of enthalpy and entropy (h-Ts) the
differentials dh and ds are mutually cancelled. Comparing coefficients at dp and dT
the partial derivatives of Gibbs energy can be expressed as
g
v  ( )T
p
s (
Because the mixed derivatives equal, the Maxwell equation follows
2g
v
s
 ( ) p  ( ) T
Tp
T
p
g
)p
T
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Maxwell relationships
For first law
g  h  Ts (
v
s
) p  ( ) T
T
p
For incompressible elongation
(f force positive when extended)
(
  u  Ts (
u
p
s
) v  ( )T
T
v
T
p
(
) s  ( ) v
v
s
For magnetisation
du  Tds  pdv  0 Hd
H-intensity of magnetic. field
-specific magnetisation
0-magnetic permeability of
vacuum
(
h
(
T
v
)s  ( ) p
p
s
f
S
) L  ( )T
T
L
s

)T   0 (
)H
H
T
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s(T,v) s(T,p)
entropy changes
Changes of entropy follow from previous equations for internal energy and enthalpy
changes
p
)v dv
T
v
Tds  dh  vdp  c p dT  T ( ) p dp
T
Tds  du  pdv  cv dT  T (
Special case for IDEAL GAS (pv=RmT, where Rm is individual gas constant)
dT
dv
 Rm
T
v
dT
dp
ds  c p
 Rm
T
p
ds  cv
Please notice the difference between universal and individual gas constant. And
the difference between molar and specific volume.
pv~  RT
pv  RmT
~
v molar volu me, m3 / mol
v m3 / kg
R  8.314 J/(mol.K)
v  v/M M is molecular mass
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u,h,s finite changes (without phase changes or reactions)
Previous equations describe only differential changes. Finite changes must be
calculated by their integration. This integration can be carried out analytically for
constant values of heat capacities cp, cp and for state equation of ideal gas
u 2  u1  cv (T2  T1 )
h2  h1  c p (T2  T1 )
s 2  s1  cv ln
T2
v
 Rm ln 2
T1
v1
s 2  s1  c p ln
T2
p
 Rm ln 2
T1
p1
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u,h,s finite changes during phase changes
During phase changes (evaporation, condensation, melting,…) both temperature T
and pressure p remain constant. Only specific volume varies and the enthalpy/entropy
changes depend upon only one state variable (for example temperature). These
functions are tabulated (e.g. h-enthalpy of evaporation) as a function of temperature
(see table for evaporation of water), or approximated by correlation
 T T 
h  r  c

 Tc  T1 
n
Tc=647 K, T1=373 K,
r=2255 kJ/kg, n=0.38
for water
Pressure corresponding to the phase change
temperature is calculated from Antoine’s equation
B
ln p  A 
C T
T[0C]
0
593
2538
50
12335
2404
100
101384
2255
200
1559120
1898
300
8498611
1373
C=-46 K, B=3816.44,
A=23.1964 for water
Entropy change is calculated directly from the enthalpy change
h
s 
T
h[kJ/kg]
p[Pa]
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h,s during phase changes (phase diagram p-T)
Melting hSL>0, sSL>0,,
p
L-liquid
Evaporation hLG>0, sLG>0,
S-solid
G-gas
Sublimation hSG>0, sSG>0,
T
Phase transition lines in the p-T diagram are described by the Clausius
Clapeyron equation
dp
h

dT T v
hLG
Specific volume changes, e.g. vG-vL
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h finite changes EXAMPLE
Final state
T2=500C,
p2=2bar
p
L-liquid
S-solid
Initial state
T1=0C,
p1=1bar
G-gas
There are infinitely many ways how to proceed
from the state 1 to the state 2. For
example
Red way increases first the pressure to the
final value and continues by heating.
Water boils at 120C (Antoine’s equation).
1.
Compression hcompres=0
2.
Liquid water h=cpL.120 for cpL=4.21
3.
Evaporation hLG=2203 J/kg (from table)
4.
Steam h=cpG.380, cpG=2.07
Taken together h=3495 J/kg
1000C
1200C
T
Remark: cp depends upon
temperature (use tables)
Green way holds initial pressure when heating
up to final temperature and isothermal
compression follows.
1.
Liquid water h=cpL.100 for cpL=4.20
2.
Evaporation hLG=2255 J/kg (from table)
3.
Steam h=cpG.400, cpG=2.05
4.
Isothermal compression hcompres=0
Sum (1-4) gives the same result h=3495 J/kg
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s finite changes EXAMPLE
Entropy change follows directly from definition
Final state
T2=80C, p2=1bar
p
L-liquid
S-solid
Initial state
T1=20C,
p1=1bar
s 2  s1  c p ln
s2  s1  4.2 ln
G-gas
T
T2
p
 Rm ln 2
T1
p1
80  273
 0.782kJ / kg / K
20  273
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SUMMARY
State equation p,v,T. Ideal gas pV=nRT (n-number of moles, R=8.314 J/mol.K)
First law of thermodynamics (and entropy change)
Tds  du  pdv
Internal energy increment (du=cv.dT for constant volume dv=0)
du  cv dT  (T (
p
) v  p)dv
T
Enthalpy increment (dh=cp.dT for constant pressure dp=0)
v
dh  c p dT  (T ( ) p  v)dp
T
These terms are zero for ideal gas
(pv=RT)
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TZ2
Check units
It is always useful to check units – all terms in equations must have the same
dimension. Examples
Tds  du  pdv
K
J
kg  K
N
J
Pa  2  3
m
m
J
kg
s
( )T
v
J

p
( )v
T
Pa
J
 3
K
m K
J
kg.K
 3
3
m
m K
kg
p~
v  RT
Pa 
J
m3
m3
mol
J
mol  K
K
m3
kg
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Important values
cv=cp ice
= 2 kJ/(kg.K)
cv=cp water = 4.2 kJ/(kg.K)
cp steam
= 2 kJ/(kg.K)
cp air
= 1 kJ/(kg.K)
Δhenthalpyof evaporation water = 2.2 MJ/kg
R = 8.314 kJ/(kmol.K)
Rm water = 8.314/18 = 0.462 kJ/(kg.K)
Example: Density of steam at 200 oC and pressure 1 bar.
p
10 5
kg


 0.457 [ 3 ]
RmT 462  (273  200)
m
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THERMODYNAMIC
DIAGRAMS
Delvaux
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DIAGRAM T-s
isobars
s  s0  c p ln
T
T0
Critical point
isochors
s  s0  cv ln
Left curveliquid
Right curvesaturated steam
Implementation of previous equations in the T-s diagram with isobars and
isochoric lines.
T
T0
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DIAGRAM h-s
Critical point
Left curve
liquid
Right curve
saturated steam
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Thermodynamic processes
Basic processes in thermal apparatuses are
 Isobaric dp=0 (heat exchangers, ducts, continuous reactors)
 Isoentropic ds=0 (adiabatic-thermally insulated apparatus, ideal
flow without friction, enthalpy changes are fully converted to
mechanical energy: compressors, turbines, nozzles)
 Isoenthalpic dh=0 (also adiabatic without heat exchange with
environment, but no mechanical work is done and pressure energy
is dissipated to heat: throttling in reduction valves)
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Thermodynamic processes
STEAM expansion in a
turbine the enthalpy decrease is
transformed to kinetic energy, entropy is
almost constant (slight increase
corresponds to friction)
Expansion of saturated
steam in a nozzle the same as
s
s
h
T
turbine (purpose: convert enthalpy to
kinetic energy of jet)
Steam compression power
h
T
s
s
h
T
consumption of compressor is given by
enthalpy increase
s
Throttling of steam in a valve
s
h
T
or in a porous plug. Enthalpy remains
constant while pressure decreases. See
next lecture Joule Thomson effect.
s
s
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Thermodynamic processes
Superheater of steam.
Pressure only slightly decreases
(friction), temperature and enthalpy
increases. Heat delivered to steam is
the enthalpy increase (isobaric process).
The heat is also hatched area in the Ts
diagram (integral of dq=Tds).
Boiler (evaporation at the
boiling point temperatrure)
s
mixing is to generate a saturated steam
from a superheated steam. Resulting
state is determined by masses of
condensate and steam (lever rule).
s
h
T
constant temperature, pressure. Density
decreases, enthalpy and entropy
increases. Hatched area is the enthalpy
of evaporation.
Mixing of condensate and
superheated steam purpose of
h
T
s
s
h
T
s
s
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Thermodynamic cycles
Periodically repeating processes with working fluid (water, hydrocarbons,
CO2,…) when heat is supplied to the fluid in the first phase of the process
followed by the second phase of heat removal (final state of the working
medium is the same as the initial one, therefore the cycle can be repeated
infinitely many times). Because more heat is supplied in the first phase
than in the second phase, the difference is the mechanical work done by
the working medium in a turbine (e.g.). It follows from the first law of
thermodynamics.
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Thermodynamic cycles
Carnot cycle
2
3
T
2
3
1
4
Mechanical work
W  (s4  s1 )(T2  T1 )
Q  (s4  s1 )T2

1
4
s
W
T
 1 1
Q
T2
3
Clausius Rankine cycle
2
3
T
Cycle makes use phase changes.
Example POWERPLANTS.
1-2 feed pump
2-3 boiler and heat exchangers
3-4 turbine and generator
4-5 condenser
1
2
1
4
4
s
3
Ericsson cycle
John Ericsson designed (200 years
ago) several interesting cycles working
1
with only gaseous phase. Reversed
cycle (counterclock orientation) is
applied in air conditioning – see
Brayton cycle shown in diagrams.
2
3
T
4
2
4
1
s
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TZ2
Thermodynamic cycles
H. Chen et al. / Renewable and Sustainable Energy
Reviews 14 (2010) 3059–3067
ORC – Organic Rankine Cycles
supercritical Rankine cycle (CO2)
Stirling machine
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TZ2
Engine (work produced)
Stirling cycle
Gas cycle having thermodynamic
efficiency of Carnot cycle. Can be
used as engine or heat pump (Stirling
machines fy.Philips are used in
cryogenics).
1.
Efficiency can be increased by heat
regenerator (usually a porous insert in
the displacement channel capable to
absorb heat from the flowing gas).
2.
Compression 1-2 and transport of
cool gas to heater 2-3
Expansion of hot gas
3-4
2
β-Stirling
Refrigeration (work consumed)
1-2 isothermal compression
2
T
1
2-3 cooling in
regenerator
3
4-1 displacement v=const.
and heating in regenerator
4
3-4 isothermal expansion
s
41
3
-Stirling with
regenerator
1
3.
Displacement of gas from hot to
cool section 4-1
4.
Compression (1-2)
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Thermoacoustical engine
Thermoacoustic analogy of Stirling engine
u,p
Standing waves – mutually shifted
pressure and velocity waves (90o)
u
2  u

c
t 2
x 2
2
2
Wave equation for pressure,
velocity. C is speed of sound
c2 
cp p
cv 
Velocity
amplitude
x
pressure
Cold
HE
Stack
(regeneratir)
Hot
HE
Very simple design can be seen on Internet video engines. Cylinder can be a glass test tube with inserted
porous layer (stack). Besides toys there exist applications with rather great power driven by solar energy or
there exist equipments for cryogenics – liquefaction of natural gas.
Mechanical design is simple, unlike theoretical description (N.Rott published series of papers Thermally driven acoustic
oscillations, I.,II.,III.,IV.,V. in Journal of Applied Mathematics and Physics in years 1969,1973, 1975,1976) . Standing temperature and pressure
waves generated inside the cylinder are mutually shifted (phase shift is similar to the Stirling engine, where
compression/expansion phases are shifted with respect to the heating/cooling phase). Solution of oscillating
pressure, temperature and gas velocity is frequently realized by Computer Fluid Dynamics codes (Fluent).
Travelling waves mode – in phase
pressure and velocity waves
Tuned resonator
Travelling waves –
analogy with Stirling
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Thermoacoustics
Phenomena and principles of thermoacoustics are more than hundred years old.
Singing Rijke tube
Sondhauss tube
Taconis oscillations
Rijke P.L. Annalen der Physik 107 (1859), 339
Sondhauss C. Annalen der Physik 79 (1850), 1
Taconis K.W. Physica 15 (1949) 738
Generated sound
waves
Thin tube
Heated wire
screen
Heated bulb
Thin tube
inserted into a
cryogenic
liquid
Liquid helium
Air flowing through
an empty tube
Lord Rayleigh
| author=Lord Rayleigh | title=The explanation of certain acoustical phenomena | journal=Nature (London) |
year=1878 | volume=18 | pages=319–321]
formulated principles as follows:
thermoacoustic oscillations are generated as soon as
Heat is supplied to the gas at a place of greatest condensation (maximum density)
Heat is removed at a place of maximum rarefaction (minimum pressure)
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Thermoacoustics
Scott Backhaus and Greg Swift: New varieties of thermoacoustic engines,
2002
Thermoacoustic machines can operate either as heat pumps (usually refrigerators) when forced oscillations
are driven by an oscillating membrane (usually loadspeaker), or as an engine (prime mover) that turns heat
into mechanical energy (sound).
Spontaneous oscillations (engine mode) exist only if the axial temperature gradient in stack is high enough so
that the Rayleigh criterion will be satisfied (see fig. showing axial gradient Tstack in a stack and Tcrit in a gas
parcel oscillating in the x-direction at a large distance from the wall of stack).
Babaei H.,Siddiqui K.: Design and optimisation of thermoacoustic
devices. Energy Conversion and Management, 49 (2008), 3585-3598
Refrigerator (heat pump if  Tstack   Tcrit )
T
Hot
HE
loadspeaker
Axial temperature
of stack Tstack
stack
Tcrit 
Tp
c p u
-frequency, -therm.expansion [1/K],
p-mean pressure amplitude, u-mean
velocity amplitude, -mean density
Engine (oscillations generated if  Tstack   Tcrit
Heat supplied to gas
parcel at max.density
Hot
HE
Cold
HE
Critical gradient Tcrit
(insulated parcel)
Piston or
piezocrystal
Heat removed
from gas to stack
x
Gas parcel oscillating back and forth in accordance with pressure (motion
left – increasing pressure – compression – gas temperature increases heat is removed from gas to stack and work is consumed by gas parcel)
)
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Thermoacoustics
How to understand the expression for the critical axial temperature gradient?
Tcrit 
 Tpa
 c pua
-frequency, -therm.expansion [1/K],
p-mean pressure amplitude, u-mean
velocity amplitude, -mean density
T describes temperature difference during pressure oscillation of a thermally
insulated gas parcel (adiabatic compression/expansion)
Tds  dh  vdp
Tcrit
T
T

x
x 
ua
T  1

Mean velocity of gas parcel
times time of period
For ideal gas thermal
expansion coefficient =1/T
p
Ts  c p T 

pa
T  
cp
For adiabatic compression Ts=0
T temperature change during adiabatic compression of gas parcel
x displacement of a gas parcel during one oscillation
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Thermoacoustics - modelling
Theoretical background of thermoacoustics developed by Nicolas Rott and published as a serie of articles
(Thermally driven acoustic oscillations) in Journal of Applied Mathematics and Physics is not freely available
in Internet, on contrary to the recent PhD thesis Mathematical Aspects of Thermoacoustics 2009 (Eidhoven)
or the Technical Research Institute of Sweden report written by Freddy Rietdijk Thermoacoustic refrigeration
using a standing-wave device 2010. Some papers are available from Science direct, for example
This paper presents analytical solution of oscillating velocity, pressure and temperature field in a tube
(capillary of a stack). The solution is almost the same as that derived by Nicolas Rott
Continuity equation relating axial (u1) and transversal (v1) velocities and density .  is
prescribed frequency and a is speed of sound, L is length of tube
Momentum balance for axial velocity (pressure is independent of radius and radial velocity
profile is expressed in terms of bessel function J0). v is viscous penetration depth
(therefore velocities are not exactly in phase with oscillation of pressure).
Enthalpy balance of gaseous layer. The first RHS term corresponds to the time derivative of
pressure. Only the radial heat transport characterised by the thermal penetration depth t is
important. Penetration depth shifts temperature oscillations.
Enthalpy balance in the wall of stack must be solved too (stack represents an
regenerator). s is the penetration depth in solid material of wall.
State equation of ideal gas (after expansion, index 1 means the first term of expansion)
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Thermoacoustics
CFD combustion thermoacoustics
Low computational cost CFD analysis of thermoacoustic oscillations
Applied Thermal Engineering, Volume 30, Issues 6-7, May 2010, Pages 544-552
Andrea Toffolo, Massimo Masi, Andrea Lazzaretto
Timing of the pressure fluctuation due to the acoustic mode at 36 Hz and of the heat released by
the fuel injected through the main radial holes (if the heat is released in the gray zones, the
necessary condition stated by Rayleigh criterion is satisfied and thermoacoustic oscillations at this
frequency are likely to grow).
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Magnetic refrigeration
Application of magnetic field upon ferromagnetic material causes orientation of magnetic spin of
molecules therefore decreases the magnetic entropy. Total entropy (sum of magnetisation entropy and
lattice entropy, thermal vibration of molecules in a crystal lattice) remains constant assuming a thermally
insulated (adiabatic) system, therefore the magnetic entropy decrease must be compensated by the
thermal entropy (and temperature) increase.
The first law of thermodynamic can be formulated in terms of internal energy as
du  Tds  pdv  0 Hd
(o magnetic permeability of vacuum, H intensity of magnetic field [T], specific magnetisation)
For constant volume the relationship between entropy, temperature and H is
ds 
c pH
T
dT  0

T
dH
s / H Maxwell
For most materials magnetisation decreases
with temperature therefore d/dT<0 and matter
heats when H increases.
This equation enables to
construct the T-s diagrams and
demonstrate thermodynamic
cycles of refrigeration.
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Magnetic refrigeration
Principle of Active Magnetic Regenerator (AMR) cycle….
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Magnetic refrigeration
Romero Gómez, J., Ferreiro Garcia, R., Carbia Carril, J., Romero Gómez, M., Experimental analysis of a
reciprocating magnetic refrigeration prototype, International Journal of Refrigeration (2013)
The working principle is as follows: during the first semicycle of operation one of
the regenerators (2) is subjected to the magnetic field action generated by the
permanent magnets (6). As a consequence of this field, the material contained
in this part of the regenerator undergoes the MCE. The HTF, consisting of glycol
water, is pumped via the circulation pump (4) through the servo-valve (5a)
activated during the first semi-cycle to absorb the heat generated by the
magnetocaloric material whilst it is cooled. The heat absorbed from the
magnetocaloric material is ceded to the environment in the HHEX(1). Once the
heat is ceded, the HTF is returned through the other regenerator. Upon contact
with the magnetocaloric material, whose temperature has decreased as a result
of being removed in the previous semi-cycle from the presence of the magnetic
field, the HTF is cooled. The cooled HTF passes through the two-position threeway valve (5b) toward the CHEX(3). Once the first semi-cycle is finished, the
valve positions change simultaneously (5a) and (5b) whilst the AMR moves (2)
within the magnetic field, thus inverting the flow direction of the circulating HTF.
(1) HHEX, (2)mobile AMR, (2a) regenerator A, (2b) regenerator B,
(3) CHEX, (4) pump, (5a), (5b) solenoid valve, (6) static NdFeB
magnets, (7) motor, (8) magnet fixed to yoke to compensate F, (9)
magnet fixed to the AMR, (10) AMR displacement guide
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Magnetic refrigeration
Romero Gómez, J., Ferreiro Garcia, R., Carbia Carril, J., Romero
Gómez, M., A review of room temperature linear reciprocating
magnetic refrigerators. Renewable and Sustainable Energy Reviews,
Volume 21, May 2013, Pages 1-12
AMR cycle cannot be described by Ts diagrams.
Rotary AMR
Reciprocating AMR
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Magnetic refrigeration
H.R.E.H. Bouchekara, A. Kedous-Lebouc, J.P. Yonnet: Design of a new magnetic
refrigeration field source running with rotating bar-shaped magnets. International
Journal of Refrigeration 35 (2012) 115-121
A Halbach magnet array is an arrangement of permanent magnets to increase
the magnetic field on one side of the array while canceling it on the other side
This paper is only a theoretical analysis calculated by
MATLAB. Permanent magnets NdFeB (NeodymiumIron-Boron alloy, diameter of 7 rods is 20 mm,
magnetic induction 1.4T). Active blocks are
Gadolinium.
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Laser cooling
Ruan X.L. et al. Entropy and efficiency in laser cooling solids.
Physical Review B, 75 (2007), 214304
A phonon is a quantum of collective excitation in a
periodic, elastic arrangement of atoms or molecules in
condensed matter.
Chu, Steven, Science 253, 861-866, 1991
Lasers to achieve extremely low temperatures has
advanced to the point that temperatures of 10-9 K
have been reached. Idea of Doppler effect
atom Na moving with
velocity about 570 m/s at
300K
collision with
photon from
behind has too low
energy
laser should be tuned below the resonance
frequency (difference f should be the
Doppler frequency given by velocity of atom)
v

f

f
0

c
Doppler
shift
HP2
EXAM
Thermodynamic cycles. Heat pumps and Heat engines
HP2
EXAM
Carnot
T
2
3
1
4
s
Clausius Rankine
3
T
2
1
4
s
3
Ericsson
T
2
4
1
s
HP2
What is important (at least for exam)
regenerator
STIRLING
displacing piston
T
41
3
Thermoacoustic
standing wave
compression/expansion wave
stack
2
Thermoacoustic
travelling wave
1
s
regenerator
N
AMR Active Magnetic
Refrigerator
S