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TZ2            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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TZ1         BASIC NOTIONS

Subsystem flame zone =
opened
SYSTEM
• Insulated- without mass or energy transfer

Subsystem candlewick =   •Closed (without mass transfer)
opened                   • Opened (mass and heat transport through boundary).
Thermal units operating in continuous mode (heat
exchangers, evaporators, driers, tubular reactors, burners)
are opened systems
Thermal units operating in a batch mode (some chemical
Subsystem candle = opened with    reactors) are closed systems
moving boundary

Subsystem stand = closed
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TZ1       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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TZ2   State EquATIONS p-v-T
RT   a
Van der Waals equation isotherms            p~    ~2
v b v

p 2a         RTc
~  v 3  (v  b) 2  0
v ~c       ~
c

2 p     6a    2 RT
~     ~4  ~ c 3  0
v 2     v c (vc  b)
Critical point, solution of these
two equations give a,b
parameters as a function of
critical temperature and critical
Above critical temperature Tc the     pressure
substance exists only as a gas
(liquefaction is not possible even at
infinitely great pressure)
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Pv=RT tutorial Baloon
Example: Calculate load capacity of a baloon filled by hot air. D=20m,
T=600C, Te=200C, p=105 Pa.     M=29 (air)

mg  V (  e   ) g

D                     1 1          1 1       pM air 1     1
 e      M air ( ~  ~ )        (        )
ve v         ve v       R      T20 T60

D3 pM air      1   1   203 105  29 1   1
m                 (  )                 (     )
6      R     T20 T60   6 8314 293 333
m

= 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         P-pressure transducer Kulite XTM 140
(thermally insulated)
p1    v      T v
V                                  p1v1  RT1           ( 2 )  1 2
p1v1  p2 v 2                           p2    v1     T2 v1
p2 v2  RT2      T1   v 2  1
( )
T2   v1
Example: V2/V1=0.5        =cp/cv=1.4
T1=300 K
T2=396 K    temperature increase 96 K!!
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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
 Pressure forces                                               ~105 J/kg
 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]
dq
Thermodynamic definition of entropy s by Clausius     ds  (      ) 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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TZ2   Laws of thermodynamics

Modigliani
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TZ2      Laws of thermodynamics
First law of thermodynamics (conservation of energy)

δq = heat added to              δw = work done by system
expansion work (p.dV) in case of compressible fluids,
system                                surface work (surface tension x increase of surface),

δq = du + δw                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)

δq = heat added to system is Tds
Tds  δq           only in the case of reversible process

Combined first and second law of thermodynamics

Tds = du+pdv
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TZ2
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)
reversible processes are generally defined as

Tds  du  pdv
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).
At a time interval dt the temperature of water Tw decreases by
dTw=-dQwm/(mwcpw) and the temperature Tm increases by
Water           dTm=dQwm/(mmcpm), however the sum of inner energies
Tw=300K         U=ummm+uwmw remains, therefore dU=0. Volume V is constant
and dV=0. Therefore the entropy increase of the whole system
dQwm         dS should be zero (dS=0) TdS  dU  pdV  0
This conclusion is wrong, because the whole process is irreversible
Metal
and dS is actually positive. So that the previous equation could be
Tm=273K
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
dS       wm
    wm
 dQwm   w      m
0
Tm         Tw             TmTw
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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
uinternalenergyof one molecule          kT         1.380 6504×10−23
2
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 calculate specific heat capacity c [J/kg/K]
theoretically. 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 c v 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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TZ2       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)
s          s                 s            s
du  Tds  pdv  T ((          )v dT  ( )T dv)  pdv  T ( ) v dT  (T ( )T  p)dv
T          v   ds
T c          v
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
(        )T  (        )v
v            T
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
p                     ideal gas (pv=RT)
du  cv dT  (T (          ) v  p)dv
T
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TZ2       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
Tds                      T            p
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
v
dh  c p dT  (T ( ) 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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TZ2    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                g
v(      )T     s (      )p
p                T
Because the mixed derivatives equal, the Maxwell equation follows

2g     v         s
(    ) p  ( ) T
Tp    T         p
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TZ2    s(T,v) s(T,p)                     entropy changes

Changes of entropy follow from previous equations for internal energy and enthalpy
changes
p
Tds  du  pdv  cv dT  T (    )v dv
T
v
Tds  dh  vdp  c p dT  T ( ) p dp
T
Special case for IDEAL GAS (pv=RmT, where Rm is individual gas constant)
dT      dv
ds  cv       Rm
T        v
dT      dp
ds  c p     Rm
T       p
Please notice the difference between universal and individual gas constant. And
the difference between molar and specific volume.
~
pv  RT       ~ molar volu me, m3 / mol
v                           R  8.314 J/(mol.K)

pv  RmT        v m 3 / kg    v  v/M M is molecular mass
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TZ2     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 )

T2        v
s 2  s1  cv ln        Rm ln 2
T1        v1
T2        p
s 2  s1  c p ln       Rm ln 2
T1        p1
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TZ2
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

n                           T[0C]   p[Pa]         h[kJ/kg]
 T T           Tc=647 K, T1=373 K,
h  r  c              r=2255 kJ/kg, n=0.38    0               593   2538
 Tc  T1        for water               50         12335      2404

Pressure corresponding to the phase change             100       101384      2255

temperature is calculated from Antoine’s equation      200     1559120       1898
300     8498611       1373
B
ln p  A                 C=-46 K, B=3816.44,
C T           A=23.1964 for water

Entropy change is calculated directly from the enthalpy change
h
s 
T
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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
hLG
dp   h

dT T v
Specific volume changes, e.g. vG-vL
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h finite changes EXAMPLE
There are infinitely many ways how to proceed
Final state       from the state 1 to the state 2. For
T2=500C,          example
p2=2bar
p                                              Red way increases first the pressure to the
L-liquid                   final value and continues by heating.
Water boils at 120C (Antoine’s equation).
S-solid                               1.   Compression hcompres=0
2.   Liquid water h=cpL.120 for cpL=4.21
Initial state
T1=0C,                                       3.   Evaporation hLG=2203 J/kg (from table)
p1=1bar                   G-gas              4.   Steam h=cpG.380, cpG=2.07
Taken together h=3495 J/kg
Green way holds initial pressure when heating
1000C     1200C     T         up to final temperature and isothermal
compression follows.
1.   Liquid water h=cpL.100 for cpL=4.20
Remark: cp depends upon        2.   Evaporation hLG=2255 J/kg (from table)
temperature (use tables)
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                              T2        p
s 2  s1  c p ln       Rm ln 2
p                                                                    T1        p1
L-liquid
S-solid                                            80  273
s2  s1  4.2ln             0.782kJ / kg / K
20  273
Initial state
T1=20C,
p1=1bar               G-gas

T
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TZ2      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)
p
du  cv dT  (T (      ) 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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Check units
It is always useful to check units – all terms in equations must have the same
dimension. Examples
Tds  du  pdv
J                  J                  N    J   m3
K                                       Pa  2  3
kg  K               kg                 m   m    kg

s                              p
( )T                           ( )v
J
v                              T
kg.K   J                           Pa   J
 3                               3
m 3
m K                          K m K
kg

p~  RT
v
J     m3           J
Pa                                  K
m3    mol        mol  K
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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         105                 kg
                       0.457 [ 3 ]
RmT 462  (273  200)          m
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THERMODYNAMIC
DIAGRAMS
Delvaux
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DIAGRAM T-s

isobars
T
s  s0  c p ln
T0
Critical point
isochors
T
s  s0  cv ln
T0

Left curve-
liquid
Right curve-
saturated steam

Implementation of previous equations in the T-s diagram with isobars and
isochoric lines.
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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
T       h
STEAM expansion in a
turbine the enthalpy decrease is
transformed to kinetic energy, entropy is
almost constant (slight increase
corresponds to friction)                        s       s
T       h
Expansion of saturated
steam in a nozzle the same as
turbine (purpose: convert enthalpy to
kinetic energy of jet)
s       s

T       h
Steam compression power
consumption of compressor is given by
enthalpy increase

s       s
Throttling of steam in a valve              T       h
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
T       h
Superheater of steam.
Pressure only slightly decreases
(friction), temperature and enthalpy
increases. Heat delivered to steam is
the enthalpy increase (isobaric process).       s       s
The heat is also hatched area in the Ts
diagram (integral of dq=Tds).

T       h
Boiler (evaporation at the
boiling point temperatrure)
constant temperature, pressure. Density
decreases, enthalpy and entropy
increases. Hatched area is the enthalpy         s       s
of evaporation.

Mixing of condensate and                    T       h
superheated steam purpose of
mixing is to generate a saturated steam
from a superheated steam. Resulting
state is determined by masses of
condensate and steam (lever rule).              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
3
2             T
Mechanical work
W  ( s4  s1 )(T2  T1 )                                   1           4
1
Q  ( s4  s1 )T2                           4


W     T
 1 1                                                   s
Q     T2

3
Clausius Rankine cycle                      2   3
Cycle makes use phase changes.
T
Example POWERPLANTS.
1                    2
1-2 feed pump                                   4                       4
2-3 boiler and heat exchangers                              1
3-4 turbine and generator                                           s
4-5 condenser
3

Ericsson cycle                              2       3
2
John Ericsson designed (200 years                       T
ago) several interesting cycles working
with only gaseous phase. Reversed       1                                   4
4               1
cycle (counterclock orientation) is
applied in air conditioning – see
Brayton cycle shown in diagrams.
s
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Stirling machine
1.   Compression and transport of
Stirling cycle                                                                                cool gas to heater
Gas cycle having thermodynamic
efficiency of Carnot cycle. Casn be
used as engine or heat pump (Stirling
machnines fy.Philips are used in                                                         2.   Expansion of hot gas
cryogenics).
Efficiency can be increased by heat
regenerator (usually a porous insert in
the displacement channel capable to
absorb heat from the flowing gas).
3.   Displacement of gas from hot to
cool section
β-Stirling

4.   Compression (phase 1)
1-2 isothermal compression
2             1

2-3 cooling in
T            regenerator                  4-1 displacement v=const.
and heating in regenerator
-Stirling with
3                4                                regenerator
3-4 isothermal expansion
s
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Thermoacoustical engine
Thermoacoustic analogy of Stirling engine

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 . 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). Analysis is similar
to the analysis of electromagnetic waves in resonant cavity of microwave oven (Helmholtz equation).
Solution of oscillating pressure, temperature and gas velocity is frequently realized by Computer Fluid
Dynamics codes (Fluent).
travelling waves mode:

Tuned resonator
HP2
TZ2
Thermoacoustics
Traveling wave – analogy with Stirling
HP2
TZ2
Thermoacoustics

Principles of thermoacoustics are more than hundred years old.
Lord Rayleigh        | author=Lord Rayleigh | title=The explanation of certain acoustical phenomena
| journal=Nature (London) | year=1878 | volume=18 | pages=319–321]

formulated conclusions as follows: thermoacoustic oscillations are
generated as soon as
 Heat is supplied to the gas at a place of greatest condensation
 Heat is removed at a place of maximum rarefaction (minimum
pressure)
HP2
TZ2
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).
HP2   EXAM

Thermodynamics
HP2      What is important (at least for exam)
Gibbs phase rule

State equations Van der Waals and critical parameters
RT   a
p~    ~2
v b v

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