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CH5 CHEMISTRY E182019 Heat and energies This course is approximately at this level Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010 CH5 Systems SYSTEM Insulated – without mass, energy, and heat transfer through the system boundary (no exchange with environment) Closed – without mass transport (impermeable boundary for mass transfer), but heat and energy exchange with surrounding is possible (e.g.batch chemical reactor, Papine’s pot) Opened – both mass and energy exchange between system and surroundings is possible. State of matter inside a system is characterised by state variables T,p,v Internal energy u [J/kg] Enthalpy h [J/kg] Entropy s [J/kg.K] CH5 Energy balance of Closed Systems q du -internal energy increase w =p.dv mechanical work done by system (volume increased by dv) heat transferred to system First law of thermodynamics δq = du + δw Only du-is a state variable, while q (heat) and w (work) depend upon previous history and are NOT state variables. CH5 Energy balance of Closed Systems at constant VOLUME q du -internal energy increase heat transferred to system Closed system and constant volume (no mechanical work). Internal energy increase is related to temperature increase δq = du du = cv dT cv – specific heat capacity at constant volume [J/kg.K] CH5 Energy balance of Closed Systems at constant PRESSURE q du -internal energy increase w =p.dV mechanical work done by system (volume increased by dv, dp=0) heat transferred to system Closed system and constant pressure. Amount of heat is expressed in terms of enthalpy h=u+pv δq = du+pdv=dh-vdp=dh dh = cp dT cp – specific heat capacity at constant pressure [J/kg.K] CH5 Energy balance of Closed Systems at constant TEMPERATURE q heat transferred to system ds -entropy increase dT=0 w =p.dV mechanical work done by system (at constant temperature dT=0). However w can be also surface work (surface tension x increase of surface), shear stresses acting at surface x displacement, or even electrical work (intensity of electric field x electrical current). Later on we shall consider only the p.dV term representing the mechanical expansion work of compressible substances (e.g. gases). Closed system and constant temperature. Amount of heat is expressed in terms of entropy δq = du+pdv=Tds ds = (du+pdv)/T CH5 Internal energy du = cv dT ...heat dq transferred to system at constant volume (more correctly, as soon as no other form of work is done). Internal energy changes with changing temperature (small, kinetic energy) Internal energy changes during phase changes (medium, VdW forces) Internal energy changes during chemical reactions (high, bonding forces) CH5 Internal energy 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) Chemical energy of ionic/covalent bonds in molecule Intermolecular VdW forces (phase changes) Pressure forces Thermal energy (kinetic energy of molecules) ~1017J/kg ~107 J/kg ~106 J/kg ~105 J/kg ~104 J/kg CH5 Internal energy Internal energy is a state variable and according to Gibb’s phase rule it depends not only upon temperature T, but also upon specific volume or pressure (even in the case of no phase or chemical changes). The previous statement du=cv.dT holds only at the particular case, when volume is constant (or dv=0). More general formula p du cv dT (T ( ) v p)dv T describes variation of internal energy with temperature and specific volume. The second term is zero not only if the volume is constant, but also in the case that the substance can be characterized by state equation of ideal gas: T( p R )v p T p p p 0 T v CH5 Internal energy/enthalpy Internal energy u is suitable for design (and energy balancing) of closed batch systems, e.g. reactors operating periodically. Volume of system is constant no mechanical work is done and amount of transferred heat is equivalent to internal energy change. Enthalpy h=u+pv is more suitable for energy balancing of continuous (opened) systems operating at steady state. The term pv automatically takes into account mechanical work necessary for delivering/removal of matter to inlet/outlet streams. Therefore energy balance reduces to 2 h2 m 1h1 Q m H 1 m 1h1 1, m 2 Mass flowrates at m inlet/outlet streams [kg/s] Q [W ] H 2 m 2 h2 H 1 , H 2 Enthalpy flows at inlet/outlet streams [W] Q [W ] Heating power delivered to system through heat transfer surface [W] Enthalpy is preferred for energy balancing of processes carried out at constant pressure, e.g. chemical reactions, phase changes,…when volume is changing but pressure remains constant (typical situation in technologies) CH5 Enthalpy dh = cp dT h = u + pv ...heat dq transferred to system at constant pressure Enthalpy changes with changing temperature (small, kinetic energy) Enthalpy changes during phase changes (medium, VdW forces) Enthalpy changes during chemical reactions (high, bonding forces) CH5 Enthalpy changes with T Enthalpy is a state variable and according to Gibb’s phase rule it depends not only upon temperature T, but also upon specific volume or pressure (even in the case of no phase or chemical changes). The previous statement dh=cp.dT holds only at the particular case, when pressure is constant (or dp=0). More general formula dh c p dT (T ( v ) p v)dp T describes variation of enthalpy with temperature and pressure. The second term is zero not only if pressure is constant, but also in the case that the substance can be characterized by state equation of ideal gas: T( v R ) p v T v v v 0 T p CH5 Enthalpy of phase changes Evaporation of liquids and melting of solids are examples of processes when heat must be added. The heat is necessary to break the VdW forces and make molecules „free“. Reverse processes are condensation of steam and solidification of liquids – heat must be removed from the system to complete the phase change. In both cases phase changes proceed at constant temperature and constant pressure. Therefore heat supplied/removed is the enthalpy change of phase changes. hLG = hG – hL hSL = hL – hS enthalpy of evaporation (>0 heat must be supplied) enthalpy of melting (>0 heat must be supplied) hGL = hL – hG hLS = hS – hL enthalpy of condensation (<0 heat must be removed) enthalpy of freezing (<0 heat must be removed) CH5 Enthalpy of phase changes Phase changes proceed at equlibrium between two phases and according to Gibb’s rule there is only one degree of freedom (2-phases, 1-component). Therefore enthalpy change depends only upon one state variable, e.g. upon temperature T (temperature of boiling, …) and corresponding pressure is determined by some thermodynamic equations. Antoine’s equation describes relationship between T,p of saturated vapours B ln p A C T Parameters A,B,C can be found in tables for most substances. ln p A Later on Clausius Clapeyron equation will be derived, enabling to evaluate enthalpy of phase changes, for example as B C T RT 2 d ln p hLG M dT Boiling temperature of water at atmospheric pressure 100 kPa is 1000C T 1004 p 0 100 200 300 T [ 0C] CH5 Enthalpy of phase changes Maybe that you are little bit confused by previous slide (Antoine’s and Clausius Clapeyron equations). Details will be discussed later when analysing equilibrium states using different techniques. So far it is sufficient to know, that Temperature of phase changes increases with pressure Enthalpy of evaporation decreases with pressure. The higher is pressure, the higher is boiling point temperature and the lower is amount of heat necessary for evaporation of 1 kg of substance. As soon as temperature is increased up to the critical temperature Tc, the enthalpy of evaporation drops to zero. Example: Enthalpy of evaporation of water is hLG = 2500 kJ/kg at 00C hLG = 2400 kJ/kg at 500C hLG = 2250 kJ/kg at 1000C hLG = 1900 kJ/kg at 2000C Enthalpy of chem.reactions CH5 During chemical reactions some covalent bonds are released and some are created. Knowing Lewis structure of reactants and products it is possible to estimate energy released during chemical reaction. -C -N -O -H =C =N =O C N Example: CH4+2O22H2O+CO2 C 348 292 351 413 611 615 741 837 891 CH4 (4single bonds 413)=1652 kJ.mol-1 N 292 161 200 391 615 418 481 891 946 2O2 2(1double bond 498)=996 kJ.(2mol)-1 O 351 200 139 463 741 481 498 H 413 391 463 436 2H2O 2(2single bond 463)=1852 kJ.(2mol)-1 CO2 (2 double bonds 741)=1482 kJ.mol-1 This example demonstrates that breaking the bonds of the reactants (one mole of CH4 and two moles of O2) requires 2648=1652+996 kJ, while when the bonds of the product are formed 3334=1852+1482 kJ of energy are released. The net profit is 3334-2648=686 kJ, which is approximately 15% less than the molar enthalpy change predicted before (802.8 kJ.mol-1). Because most chemical reactions proceed at a constant pressure the heat exchanged in the course of reactions is equal to the enthalpy change hR. If enthalpy of products is less than the enthalpy of reactants heat is released (hR<0) and reaction is called EXOTHERMIC. The reaction when the heat must be supplied is ENDOTHERMIC (hR>0) . CH5 Enthalpy of chem.reactions The enthalpy change depends on pressure, temperature and also on the physical state of the reactants and products (s-solid, l-liquid, g-gas). Data corresponding to the standard state (p=100 kPa at constant temperature T=298 K), are published, e.g., by the National Institute of Standards and Technology. The enthalpy change for different or varying temperature must be recomputed by an integration of the specific (or molar) heat capacities of the reactants and products. Examples of chemical reactions at standard state: 1 H2 ( g) O2 ( g) H2 O( g ) 2 1 H2 ( g) O2 ( g) H2 O(l ) 2 ~ ( h 0 ) 298 24183 . kJ . mole 1 ~ ( h 0 ) 298 28584 . kJ . mole 1 Standard temperature T=298K Condensed water Standard pressure p=100kPa The difference between the standard enthalpy changes is the standard molar enthalpy of vaporisation. CH5 Hess law The enthalpy change of the reaction is a function of state and is independent of the path, that is, of any intermediate reaction that may have occurred. Example: Combustion of carbon (symbols s,g,l means solid, gas, liquid) 1 C( s) O2 ( g) CO( g) 21 CO( g) O2 ( g) CO2 ( g) 2 ~ ( h 0 ) 298 111 kJ . mole 1 ~ ( h 0 ) 298 283 kJ . mole 1 Two consecutive exothermic reactions yield the same result as the summary reaction C( s) O2 ( g ) CO2 ( g ) ~ ( h 0 ) 298 394 kJ . mole 1 Consequence: While specific reaction mechanism is important for evaluation of chemical reaction rate, it can be substituted by any other sequence of intermediate reactions (may be fictive) when calculating enthalpy cheanges. CH5 Enthalpy of formation It can be assumed that the analysed chemical reaction proceeds in two steps, the first is the decomposition of all reactants to free elements, and the second step is the addition of the free elements into the products. Both the steps can be considered as chemical reactions accompanied by the enthalpy changes, corresponding to the formation of compounds from elements in their standard state. This is the standard enthalpy of formation ~0 ( h f ,compound ) 298 The standard molar enthalpy of formation of free elements, e.g. C, or molecules O2, N2, H2, etc., is equal to zero (molecule O2 and not the radical O is assumed to be in the most stable state). The standard enthalpy change of the calculated reaction is ~ ~ ( Hr0 )298 P ( h f0, P )298 R ( h f0, R )298 P R where R 0, P 0 are stoichiometric coefficients of reactants (R) and products (P). Combustion of Methane CH5 Example: Gas burner operating at atmospheric pressure consumes mass flowrate of CH4 0.582 kg/s. Calculate mass flowrate of flue gas. Assuming temperature of fuel and air 298K and temperature of flue gases 500 K calculate heating power. Air Q fg-flue gas CH4 (h ) 0 R 298 CH 2O CO 2H O 4 2 2 2 CH 4 1, O 2 2, CO 2 1, H 2O 2 (h 0f ,CH 4 )298 75, (h f0,CO2 ) 298 393, (h f0, H 2O ) 298 242 i (h ) 0 f ,i 298 (hR0 )298 802 MJ hR 50 M CH4 0.016 kg 75 1 393 2 242 802 kJ / mol CH 4 i Flue gas production 1kg CH m fg mCH4 mO2 mN2 mCH4 mO2 (1 needs 4 kg O2. Mass fraction of oxygen in air is 0.233. N N 0.767 ) mCH (5 4 ) 0.582(5 4 ) 10.57kg / s O O 0.233 4 2 2 4 2 2 Heat removed in heat exchanger Q hR mCH4 m fg c p (Tfg T0 ) 50 0.582 0.01057 (500 298) 27MW cp0.001 MJ/kgK CH5 Combustion Baloon Example: Calculate fuel consumption (propane burner) of a baloon filled by hot air. D=20m, T=600C, Te=200C, heat transfer coefficient estimated to =1 W/m2.K. Reaction C H 5O 3CO 4H O 3 8 2 2 2 D C 3 H 8 1, O 2 5, CO 2 3, H 2O 4 ~ ~ ~ (h f0,C3 H 8 ) 298 103, (h f0,CO2 ) 298 393, (h f0, H 2O ) 298 242 ~ ~ (hR0 ) 298 i (h f0,i ) 298 103 3 393 4 242 2044 kJ / mol C3 H 8 i ~ (hR0 ) 298 2044 MJ hR 46.5 M C3 H 8 0.044 kg m Heat losses Q D 2 (T Te ) 400 40 50kW Fuel consumption mf Q 50 0.001kg / s hR 46500 CH5 Combustion Baloon The burner unit gasifies liquid propane, mixes it with air, ignites the mixture, and directs the flame and exhaust into the mouth of the envelope. Burners vary in power output; each will generally produce 2 to 3 MW of heat