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Thermodynamics proces ses a nd cycles

HP2. TZ 2. HEAT PROCESSES. Thermodynamics proces ses a nd cycles.

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Thermodynamics proces ses a nd cycles

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  1. HP2 TZ2 HEAT PROCESSES Thermodynamicsprocesses 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

  2. FUNDAMENTALS of THERMODYNAMICS HP2 TZ1 Estes

  3. BASIC NOTIONS HP2 TZ1 Subsystem flame zone = opened • SYSTEM • Insulated- without mass or energy transfer • 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 • Thermal units operating in a batch mode (some chemical reactors) are closed systems Subsystem candlewick = opened Subsystem candle = opened with moving boundary Subsystem stand = closed

  4. StaTE VARIABLES HP2 TZ1 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

  5. Gibbs phase rule HP2 TZ1 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)

  6. State EquATIONS p-v-T HP2 TZ2 Van der Waals equation isotherms Critical point, solution of these two equations give a,b parameters as a function of critical temperature and critical pressure AbovecriticaltemperatureTcthe substance existsonly as a gas (liquefactionis not possibleevenatinfinitelygreatpressure)

  7. HP2 TZ2 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) D m = 599 kg

  8. HP2 TZ2 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 • Thermal energy (kineticenergyof molecules) ~104 J/kg Only the following item (thermal energy) is often included into the internal energy concept (sometimes distinguished as the sensible internal energy) It follows from energy balances that the change ofinternal energy of a closed system at a constant volume equals amount of heat delivered to the system du = dq (heat added at isochoric change)

  9. HP2 TZ2 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/outletmaterial 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)

  10. HP2 TZ2 Entropy s[J/kg/K] Thermodynamic definition of entropy s by Clausius 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: Entropyrepresents 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)

  11. HP2 TZ2 Entropy s[J/kg/K] statistical (Boltzmann) taken from Wikipedia

  12. Laws of thermodynamics HP2 TZ2 Modigliani

  13. Laws of thermodynamics HP2 TZ2 First law of thermodynamics (conservation of energy) δw = work done by system δq = heat added to system 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. δq = du + δw Second law of thermodynamics (entropy of closed insulated system increases) δq = heat added to system is Tds only in the case of reversible process Tds δq Combined first and second law of thermodynamics Tds = du+pdv

  14. Entropy and Helmoltz energy HP2 TZ2 Gough-Joule effect – stretching rubber band Derived Maxwell equation using Helmholtz energy A f L use your lips to feel the temperature increase during stretching entropy decreases with stretching tension f increases with temperature at fixed length (fibre contracts with increasing temperature!) f+df df>0 ds<0 L+dL f s=0(only 1 microstate -possible arrangement) Lmax

  15. Energies and Temperature HP2 TZ2 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.

  16. u(T,v) internal energy change HP2 TZ2 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) 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 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! So there is the final result cv ds This term is zero for ideal gas (pv=RT)

  17. h(T,p) enthalpy changes HP2 TZ2 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 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 After substuting we arrive to the final expression for enthalpy change Negative sign in the Maxwell equation is probably confusing, and cannot be derived from dimensional analysis. Correct derivation is presented in the following slide. cp Tds

  18. HP2 TZ2 s(T,v) s(T,p) entropy changes Changes of entropy follow from previous equations for internal energy and enthalpy changes Special case for IDEAL GAS (pv=RmT, where Rm is individual gas constant) Please notice the difference between universal and individual gas constant. And the difference between molar and specific volume.

  19. HP2 TZ2 u,h,sfinite 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, cpand for state equation of ideal gas

  20. u,h,sfinite changes during phase changes HP2 TZ2 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 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 C=-46 K, B=3816.44, A=23.1964 for water Entropy change is calculated directly from the enthalpy change

  21. h,sduring phase changes (phase diagram p-T) HP2 TZ2 p L-liquid S-solid G-gas T MeltinghSL>0, sSL>0,, Evaporation hLG>0, sLG>0, SublimationhSG>0, sSG>0, Phase transition lines in the p-T diagram are described by the Clausius Clapeyron equation hLG Specific volume changes, e.g. vG-vL

  22. HP2 TZ2 SUMMARY State equation p,v,T. IdealgaspV=nRT(n-number of moles, R=8.314 J/mol.K) First law of thermodynamics (and entropy change) Internal energy increment (du=cv.dT for constant volume dv=0) Enthalpy increment (dh=cp.dT for constant pressure dp=0) These terms are zero for ideal gas (pv=RT)

  23. HP2 TZ2 Check units It is always useful to check units – all terms in equations must have the same dimension. Examples

  24. HP2 TZ2 Important values cv=cpice= 2 kJ/(kg.K) cv=cpwater = 4.2 kJ/(kg.K) cpsteam = 2 kJ/(kg.K) cpair = 1 kJ/(kg.K) Δhenthalpyof evaporation water = 2.2 MJ/kg R = 8.314 kJ/(kmol.K) Rmwater = 8.314/18 = 0.462 kJ/(kg.K) Example: Density of steam at 200 oC and pressure 1 bar.

  25. HP2 TZ2 THERMODYNAMICDIAGRAMS Delvaux

  26. DIAGRAM T-s HP2 TZ2 isobars Critical point isochors Left curve-liquid Right curve-saturated steam Implementation of previous equations in the T-s diagram with isobars and isochoric lines.

  27. HP2 TZ2 DIAGRAM h-s Critical point Left curve liquid Right curve saturated steam

  28. HP2 TZ2 Thermodynamicprocesses • Basic processes in thermalapparatuses are • Isobaricdp=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)

  29. HP2 TZ2 h h h T T T s s s s s s Thermodynamicprocesses STEAM expansion ina turbinethe enthalpy decrease is transformed to kinetic energy, entropy is almost constant (slight increase corresponds to friction) h T Expansion of saturated steam in anozzlethe same as turbine (purpose: convert enthalpy to kinetic energy of jet) s s Steam compressionpower consumption of compressor is given by enthalpy increase Throttling of steamin a valve or in a porous plug. Enthalpy remains constant while pressure decreases. See next lecture Joule Thomson effect.

  30. HP2 TZ2 h h h T T T s s s s s s Thermodynamic processes Superheaterof 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)constant temperature, pressure. Density decreases, enthalpy and entropy increases. Hatched area is the enthalpy of evaporation. Mixing of condensate and 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).

  31. HP2 TZ2 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.

  32. HP2 TZ2 3 2 3 T 2 1 4 1 4 s 3 2 1 4 3 2 T 4 1 s Thermodynamic cycles Carnot cycle Mechanical work 3 Clausius Rankine cycle 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 T 2 4 1 s Ericsson cycle John Ericsson designed (200 years ago) several interesting cycles working with only gaseous phase. Reversed cycle (counterclock orientation) is applied in air conditioning – see Brayton cycle shown in diagrams. 3 2 1 4

  33. HP2 TZ2 2 1 1-2 isothermal compression T 2-3 cooling in regenerator 4-1 displacement v=const. and heating in regenerator 4 3 s 3-4 isothermal expansion Stirling machine Stirling ENGINE Stirling cycle Gas cycle having thermodynamic efficiency of Carnot cycle. Casn be used as engine or heat pump (Stirling machnines fy.Philips are used in cryogenics). Efficiency can be increased by heat regenerator (usually a porous insert in the displacement channel capable to absorb heat from the flowing gas). • Compression and transport of cool gas to heater • Expansion of hot gas • Displacement of gas from hot to cool section • Compression (phase 1) β-Stirling Stirling HEAT PUMP -Stirlingwith regenerator

  34. HP2 TZ2 u,p x Velocity amplitude Stack (regeneratir) Cold HE Hot HE pressure Thermoacoustical engine Thermoacoustic analogy of Stirling engine Very simple design can be seen on Internetvideoengines. 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. Standing waves – mutually shifted pressure and velocity waves (90o) Wave equation for pressure, velocity. C is speed of sound

  35. Thermoacoustics HP2 TZ2 Generated sound waves Heated wire screen Air flowing through an empty tube Phenomena and principles of thermoacoustics are more than hundred years old. Taconis oscillations Taconis K.W. Physica 15 (1949) 738 Sondhauss tube Sondhauss C. Annalen der Physik 79 (1850), 1 Singing Rijke tube Rijke P.L. Annalen der Physik 107 (1859), 339 Thin tube Thin tube inserted into a cryogenic liquid Heated bulb Liquid helium • 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)

  36. Thermoacoustics HP2 TZ2 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). -frequency, -therm.expansion [1/K], p-mean pressure amplitude, u-mean velocity amplitude, -mean density Babaei H.,Siddiqui K.: Design and optimisation of thermoacoustic devices. Energy Conversion and Management, 49 (2008), 3585-3598 Engine(oscillations generated if ) Refrigerator(heat pump if ) T Heat supplied to gas parcel at max.density stack Hot HE Cold HE Hot HE loadspeaker Critical gradient Tcrit (insulated parcel) Piston or piezocrystal Axial temperature of stackTstack 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)

  37. Magnetic refrigeration HP2 TZ2 Application of magnetic field upon ferromagnetic material causes orientation of magnetic spin of molecules therefore decreases the magnetic entropy. Total entropy (sum of the magnetisation entropy and the lattice entropy, thermal vibration of molecules in a crystal lattice) remains constant assuming a thermally insulated (adiabatic) systém.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 (o magnetic permeability of vacuum, H intensity of magnetic field [T], specific magnetisation) For constant volume the relationship between entropy, temperature and H is This equation enables to construct the T-s diagrams and demonstrate thermodynamic cycles of refrigeration.

  38. Laser cooling HP2 TZ2 Lasers illuminating crystals achieve extremely low cryogenic temperatures of 10-9 K. 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 laser should be tuned below the resonance frequency (difference f should be the Doppler frequency given by velocity of atom) collision with photon from behind has too low energy

  39. EXAM HP2 Thermodynamics

  40. What is important (at least for exam) HP2 Gibbs phase rule State equations Van der Waals and critical parameters NDOF = Ncomponents – Nphases + 2 First law of thermodynamics Tds = du+pdv

  41. 3 2 T 1 4 s 3 T 2 4 1 s 3 2 T 4 1 s What is important (at least for exam) HP2 Carnot Clausius Rankine Ericsson

  42. 4 3 1 1 2 N S What is important (at least for exam) HP2 regenerator T STIRLING displacing piston Thermoacoustic standing wave compression/expansion wave stack s Thermoacoustic travelling wave regenerator AMR Active Magnetic Refrigerator

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