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 TZ2            HEAT PROCESSES

                                       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,
Rudolf Žitný, Ústav procesní a
zpracovatelské techniky ČVUT FS 2010



           Subsystem flame zone =
                                     • 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
  state of system
  is characterized

             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
          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)
 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       ~

                                                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)
         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

                                                         = 599 kg
         Pv=RT tutorial Syringe
   Record time change of temperature of air compressed in syringe.


                          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!!
         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)
         Enthalpy h [J/kg]

   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)
         Entropy s [J/kg/K]
   Thermodynamic definition of entropy s by Clausius     ds  (      ) rev
   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)
 TZ2   Laws of thermodynamics

 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
         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)
 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).
         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
                         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
                                dS       wm
                                                   wm
                                                          dQwm   w      m
                                        Tm         Tw             TmTw
         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.
         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
   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
 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

 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
 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
                                dh  c p dT  (T ( ) p  v)dp
 Negative sign in the Maxwell equation is probably confusing, and cannot be derived from dimensional
 analysis. Correct derivation is presented in the following slide.
 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
 TZ2    s(T,v) s(T,p)                     entropy changes

  Changes of entropy follow from previous equations for internal energy and enthalpy
               Tds  du  pdv  cv dT  T (    )v dv
              Tds  dh  vdp  c p dT  T ( ) p dp
  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
 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
         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
       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
           s 
          h,s during phase changes (phase diagram p-T)
                                                             Melting hSL>0, sSL>0,,

                                                              Evaporation hLG>0, sLG>0,

                                  G-gas              Sublimation hSG>0, sSG>0,

       Phase transition lines in the p-T diagram are described by the Clausius
       Clapeyron equation
                dp   h
                dT T v
                                          Specific volume changes, e.g. vG-vL
       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
        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
       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
                 S-solid                                            80  273
                                                  s2  s1  4.2ln             0.782kJ / kg / K
                                                                    20  273
       Initial state
         p1=1bar               G-gas

  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 (      ) v  p)dv
  Enthalpy increment (dh=cp.dT for constant pressure dp=0)

               dh  c p dT  (T ( ) p  v)dp
                                  T                         These terms are zero for ideal gas
        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
                                       v                              T
                                   kg.K   J                           Pa   J
                                         3                               3
                                    m 3
                                         m K                          K m K

                                                p~  RT
                                           J     m3           J
                                    Pa                                  K
                                           m3    mol        mol  K
          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

        DIAGRAM T-s

                                                                          s  s0  c p ln
                      Critical point
                                                                          s  s0  cv ln

                 Left curve-
                                                             Right curve-
                                                             saturated steam

       Implementation of previous equations in the T-s diagram with isobars and
       isochoric lines.
       DIAGRAM h-s

              Critical point

         Left curve
                               Right curve
                               saturated steam
       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)
          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
          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
       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
          Thermodynamic cycles
       Carnot cycle                                                    2           3
                                                 2             T
       Mechanical work
           W  ( s4  s1 )(T2  T1 )                                   1           4
           Q  ( s4  s1 )T2                           4

                W     T
                   1 1                                                   s
                Q     T2

       Clausius Rankine cycle                      2   3
       Cycle makes use phase changes.
       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

       Ericsson cycle                              2       3
       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.
           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
       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

                                                                                                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
           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).
                                               More advanced arrangement operates at
                                               travelling waves mode:

                                                                                                  Tuned resonator
       Traveling wave – analogy with Stirling

        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
  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      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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