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HP2 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 TZ1 HP2 FUNDAMENTALS of THERMODYNAMICS Estes HP2 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 HP2 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 HP2 TZ1 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) HP2 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) TZ2 HP2 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 TZ2 HP2 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!! TZ2 HP2 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) TZ2 HP2 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) TZ2 HP2 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) HP2 TZ2 Laws of thermodynamics Modigliani HP2 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 HP2 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 (12) which can be recovered (21) 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). HP2 TZ2 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 HP2 TZ2 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. HP2 TZ2 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 HP2 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 HP2 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. HP2 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 Tp T p HP2 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 HP2 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 HP2 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 HP2 TZ2 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 HP2 TZ2 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 HP2 TZ2 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 HP2 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) HP2 TZ2 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 HP2 TZ2 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 TZ2 HP2 THERMODYNAMIC DIAGRAMS Delvaux HP2 TZ2 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. HP2 TZ2 DIAGRAM h-s Critical point Left curve liquid Right curve saturated steam HP2 TZ2 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) HP2 TZ2 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 HP2 TZ2 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 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. HP2 TZ2 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 HP2 TZ2 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 HP2 TZ2 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 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