Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

This course is approximately at this level CHEMISTRYE182019 CH5 Heat and energies Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010

Systems CH5 • 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]

Energy balance of Closed Systems CH5 du-internal energy increase w=p.dv mechanical work done by system (volume increased by dv) q 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 Systemsat constant VOLUME du-internal energy increase q 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 Systemsat constant PRESSURE du-internal energy increase w=p.dV mechanical work done by system (volume increased by dv, dp=0) q 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 Systemsat constant TEMPERATURE 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). ds-entropy increase dT=0 q heat transferred to system Closed system and constant temperature. Amount of heat is expressed in terms of entropy δq = du+pdv=Tds ds = (du+pdv)/T

Internal energy CH5 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)

Internal energy CH5 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 (kineticenergyof molecules) ~104 J/kg

Internal energy CH5 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 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:

Internal energy/enthalpy CH5 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 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) Mass flowrates at inlet/outlet streams [kg/s] Enthalpy flows at inlet/outlet streams [W] Heating power delivered to system through heat transfer surface [W]

Enthalpy CH5 dh = cp dTh = 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)

Enthalpy changes with T CH5 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 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:

Enthalpy of phase changes CH5 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 enthalpy of evaporation (>0 heat must be supplied) hSL = hL – hS enthalpy of melting (>0 heat must be supplied) hGL = hL – hG enthalpy of condensation (<0 heat must be removed) hLS = hS – hL enthalpy of freezing (<0 heat must be removed)

Enthalpy of phase changes CH5 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 Parameters A,B,C can be found in tables for most substances. Later on Clausius Clapeyron equation will be derived, enabling to evaluate enthalpy of phase changes, for example as Boiling temperature of water at atmospheric pressure 100 kPa is 1000C 0 100 200 300 T [0C]

Enthalpy of phase changes CH5 • 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. 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) .

Enthalpy of chem.reactions CH5 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: Standard temperature T=298K Condensed water Standard pressure p=100kPa The difference between the standard enthalpy changes is the standard molar enthalpy of vaporisation.

Hess law CH5 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) Two consecutive exothermic reactions yield the same result as the summary reaction 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.

Enthalpy of formation CH5 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 . 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 where 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 CH4 Q fg-flue gas . Flue gas production 1kg CH4 needs 4 kg O2. Mass fraction of oxygen in air is 0.233. Heat removed in heat exchanger cp0.001 MJ/kgK

Combustion Baloon CH5 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 D . Heat losses m Fuel consumption

Combustion Baloon CH5 . 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

Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010