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Internal Combustion Engines – ME422 Yeditepe Üniversitesi THERMODYNAMICS of COMBUSTION Prof.Dr. Cem Soruşbay Internal Combustion Engines Thermodynamics of Combustion ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ ¾ Introduction Properties of mixtures Combustion stoichiometry Combustion in IC-engines Chemical energy Heat of reaction Heat of formation Chemical equilibrium Adiabatic flame temperature Dissociation reactions 1 Introduction Thermodynamics deals with equilibrium states and how chemical composition can be calculated for a system of known atomic/molecular composition if two independent thermodynamic properties are known. Systems undergoing chemical reactions are not in general, in chemical equilibrium - reactions are rate-controlled Chemical composition at a given instant in time is controlled by the thermodynamic properties, chemical reaction rates and fluid dynamics of the system. Introduction First Law of Thermodynamics for a thermodynamic system, the time rate of change of energy of the system is equal to the rate at which work is done on the system + the rate at which heat is transferred to the system. Neglecting PE and KE within system, energy consists of internal energy due to, - thermal energy due to translation, rotation and vibration of molecules - so called sensible energy, - chemical energy due to chemical bonds between atoms in the molecules, For system of mass m, work done W and heat transfer rate to the system q, first law can be written as, d (mu ) & dt =W + q 2 Introduction mechanical work - power is, dV dx W& = − pA = − p dt dt dV d (mu ) = −p +q dt dt first law becomes, thermodynamic properties can be evaluated by assuming that the whole system is uniform or by dividing the system into subsystems that are assumed uniform, j number of cells with volumes Vj J mu = ∑ ρ j V j u j j =1 Introduction For a closed system energy balance can be obtained integrating first law equation with-respect-to time, m(u2 − u1 ) = −W12 + Q12 where t2 t2 dV W12 = ∫ p dt dt t1 for constant pressure, integrating, Q12 = ∫ q dt t1 d (mu + pV ) =q dt d (mh) =q dt m(h2 − h1 ) = Q12 3 Introduction If the chemical composition is constant, chemical energy does not change - for ideal gases, T2 u2 − u1 = ∫ cv dT T1 T2 h2 − h1 = ∫ c p dT T1 c p = cv + R Properties of Mixtures For mixtures of gases, system mass is obtained from the sum of masses of separate species, m = ∑ mi i density of system is the sum of species densities, ρ = ∑ ρi i mass fraction is mass of species i, divided by total mass, yi = and by definition mi ρ i = m ρ ∑y i =1 i 4 Properties of Mixtures Similarly mole fraction is mole of species i, divided by total number of moles, or ratio of molar concentration of species i to total molar concentration, xi = ∑x and i N i ni = N n =1 i M =∑ Molecular weight is, i mi =∑ xi M i N i relation between mole fraction and mass fraction, xi = N i mi / M i M yi = = N m/ M Mi Properties of Mixtures mixture internal energy and enthalpy per unit mass, u = ∑ yi u i i h = ∑ yi hi i internal energy and enthalpy per mole of mixture, uˆ = ∑ xi uˆi hˆ = ∑ xi hˆi i i pressure of mixture – for an ideal gas, is equal to sum of partial p of component gases if each existed alone in the mixture volume at mixture temperature, ∑ p =∑x i i i p= p i 5 Properties of Mixtures volume of mixture – for an ideal gas, is equal to sum of partial volumes which the component gases would occupy if each existed alone at the pressure and temperature of the mixture, ∑V = ∑ x V = V i i i i Combustion Stoichiometry When molecules undergo chemical reaction, the reactant atoms are rearranged to form new combinations. For example, hydrogen and oxygen react to form water : 1 H 2 + O2 → H 2O 2 two atoms of hydrogen and one atom of oxygen form one molecule of water - number of atoms of H and O must be the same on both sides Such reaction equation represents initial and final states and does not indicate actual path of reaction, which may involve many intermediate steps and intermediate species. 6 Combustion Stoichiometry Relative masses of molecules are obtained by multiplying number of moles of each species by the molecular weights (kg/kg-mol) For hydrogen-oxygen reaction, 1 H 2 + O2 → H 2O 2 ⎛ ⎞ ⎛1 2 kg ⎞⎛ 32 kg ⎞ ⎟⎟ + ⎜ mol O 2 ⎟⎜⎜ ⎟⎟ kg mol H 2 kg mol O − ⎝ ⎠ 2 ⎠ 2 ⎠ ⎝ ⎝ (1 mol H 2 )⎜⎜ ⎞ ⎛ 18 kg ⎟⎟ = (1 mol H 2 O )⎜⎜ ⎝ kg − mol H 2 O ⎠ mass of reactants equals mass of products, although mols of reactants do not equal moles of products - fixed p and T, ideal gas, 1 volume H 2 + 1 / 2 volume O 2 = 1 volume H 2 O Combustion Stoichiometry Stoichiometric calculations are done by performing atom balance for each of the elements in mixture. The theoretical amount of air required to burn a fuel completely to products with no dissociation is defined as stoichiometric air. In most combustion calculations dry air is assumed as a mixture of 79% (vol) N2 and 21% (vol) O2 or 3.764 moles of N2 per mole O2 Molecular weight of pure air is 28.96, as it also contains small amounts of argon, carbon dioxide, hydrogen etc. 7 Dry Air N2 O2 Ar CO2 % (vol) 78.09 20.95 0.93 0.03 % (mass) 75.02 23.15 1.28 0.05 Ne , neon He , helium Kr , kripton Xe , xenon H2 , hydrogen IC Engine Combustion In IC engines, complete combustion of the fuel with air, under ideal conditions - stoichiometric mixture with no dissociation, gives, m⎞ ⎛ Cn H m + ⎜ n + ⎟ (O2 + 3.76 N 2 ) → 4⎠ ⎝ m m⎞ ⎛ n CO2 + H 2O + 3.76 ⎜ n + ⎟ N2 2 4⎠ ⎝ Under real engine conditions, CO2 , CO , H2 , O2, H , O , NO are also produced as a result of the dissociation reactions (parçalanma reaksiyonları). Sulphur in the fuel produces SO2 etc. 8 Fuel / Air Ratio Mols of stoichiometric air per mol of fuel, nair − stoich m⎞ ⎛ = 4.76 ⎜ n + ⎟ n fuel 4⎠ ⎝ Stoichiometric Fuel / Air ratio by weight, f stoich = m fuel mair The percent excess air is, % excess air = = = M fuel n fuel M air nair − stoich = M fuel 29.0 (n + m / 4) 4.76 (mair actual air used, mair-stoich stoichiometric air) 100 (mair − mair − stoich ) mair − stoich ( 100 (nair − nair − stoich ) 100 nO2 − nO2 − stoich = nair − stoich nO2 − stoich ) Equivalence Ratio The fuel/air equivalence ratio is defined as the actual fuel / air mass ratio, divided by stoichiometric fuel / air mass ratio, φ= f f stoich % excess air is defined as, air excess ratio is defined as, (hava fazlalık katsayısı) % excess air = λ= 100 (1 − φ ) φ 1 φ 9 Equivalence Ratio Stoichiometric mixture, φ = λ = 1.0 Lean mixture, (fakir karışım) φ <1 λ >1 φ >1 λ <1 Rich mixture, (zengin karışım) Stoichiometric Combustion Fuel composition, for 1 kg of fuel, c h o n s w a kg kg kg kg kg kg kg of of of of of of of carbon hydrogen oxygen nitrogen sulphur water ash, etc c + h + o + n + s + w + a = 1 kg 10 Stoichiometric Combustion For complete combustion of the fuel the following reactions take place carbon : C + O2 → CO2 1 mol C + 1 mol O2 12 kg C + 32 kg O2 1 kg C + (32/12) kg O2 1 mol CO2 44 kg CO2 (44/12) kg CO2 hydrogen : H2 + 1 O2 → H 2O 2 2 kg H2 + 16 kg O2 1 kg H2 + 8 kg O2 18 kg H2O 9 kg H2O Stoichiometric Combustion sulphur : S + O2 → SO2 32 kg C + 32 kg O2 1 kg C + 1 kg O2 64 kg CO2 1 kg CO2 oxygen required will be, for c kg carbon for h kg hydrogen for s kg sulphur 8c/3 kg 8h kg s kg minimum amount of O2 needed for complete combustion of 1 kg of fuel mO − min = 8c + 8h + s − o 3 [kg-O2 / kg-fuel] 11 Stoichiometric Combustion mO −min = 8cσ 3 [kg-O2 / kg-fuel] where 3 ⎛ 8h + s − o ⎞ ⎟ 8⎝ c ⎠ σ = 1+ ⎜ The amount of oxygen in air is 23.3 % by mass, mair −min = mo − min 0.233 [kg-air / kg-fuel] [kg-air / kg-fuel] mair −min = 11.44cσ Stoichiometric Combustion For IC engine fuels mair −min = 14 − 15 [kg-air / kg-fuel] the ratio of air included in the combustion process, to the theoretically mimimum air required gives the air excess ratio, λ= λ mair mair − min equivalence ratio φ= 1 λ 12 Chemical Energy To describe the chemical energy released when fuel reacts with air to form products, chemical species in the reactants and products and their states are specified. Heat of vaporization of liquid fuels and heat of pyrolysis of solid fuels is small compared with the chemical energy relased by combustion. But effect of water condensation can be important. For lean HC-air mixtures with low T, products may be assumed to be complete (usually CO2, H2O, O2 and N2). But with high product T and rich mixtures it is generally necessaary to include other species and assume chemical equilibrium to determine species mole fractions. If products are not in chemical equilibrium, then chemical kinetic analysis (or measurements) is required to determine end state. Heat of Reaction Consider F and A mixture of mass m, constant-volume combustion with heat transfer, Qv ,from state 1 to state 2 m[(u2 − u1 )sensible + (u2 − u1 )chemical ] = Qv for n species, u at state 1 can be obtained by, n (u1 ) sensible = (u (T1 )) sensible = ∑ yi (ui (T1 )) sensible i =1 n T1 T1 i =1 To To = ∑ yi ∫ (cvi ) r dT = ∫ (cv ) r dT To is ref temperature, (cv)r is specific heat of reactant mixture 13 Heat of Reaction Similarly, T2 (u2 ) sensible = ∫ (cv ) p dT To If heat heat transfer is just large enough to bring products’ temperature back to the reactants temperature, and if this T is taken as the ref T, To for sensible energy, then (u2 − u1 ) sensible = 0 and Qv is the chemical energy released by the reaction (sabit hacimdeki reaksiyon ısısı) The quantity [(1+f)/f](-Qv/m) is the lower heating value (LHV) of the fuel for constant-volume combustion. If the water in the products is condensed, that quantity becomes the higher heating value (HHV) of the fuel for constant-V combustion Heat of Reaction If reaction takes place at constant pressure, m[(h2 − h1 )sensible + (h2 − h1 )chemical ] = Q p again if T1 = T2 = To , then Qp is the chemical energy released. For constant-pressure case, if moles of gaseous products Np are larger than moles of gaseous reactants Nr , then some of the chemical is expended to push aside the ambient pressure. Thus, Q p − Qv = ∆( pV ) = ( N p − N r ) Rˆ To = ∆NRˆ To Here “overbar” shows quantity per mole 14 Heat of Reaction m⎞ ⎛ Cn H m + ⎜ n + ⎟ (O2 + 3.76 N 2 ) → 4⎠ ⎝ m m⎞ ⎛ n CO2 + H 2O + 3.76 ⎜ n + ⎟ N2 2 4⎠ ⎝ For the reaction, assume fuel is in gas phase and gaseous water, ∆N = m −1 4 if m > 4, ∆N>0 , since both Qv and Qp are negative, |Qv| > |Qp| for most cases of interest, the difference is negligable. Heat of Reaction Heat of reaction can be obtained for reactions at T other than To To T2 T1 To Q p = m ∫ (c p ) r dT + Q p (To ) + m ∫ (c p ) p dT Q p − Q p (To ) = m(hs 2 − hs1 ) hs is the sensible enthalpy Qp(To) is negative for an exothermic reaction 15 Heat of Reaction Enthalphy of reactants and products Heat of Formation Heat of reaction of fuels combusting in air or oxygen with starting and ending points at 25 OC and 1 atm gives fuel heating value. These values are tabulated for common fuels. This cannot be applied to all reactions - needs enormous amounts of tabulated data. Instead selected reactions and their heat of reactions can be added to obtain any given reaction and its heat of reaction. Heat of formation of a particular species is defined as the heat of reaction per mole of product formed isothermally from elements in their standard states. Standard state is chosen as the most stable form of the element at 1 atm and 25 OC Heat of Formation of elements, ∆ho in their standard state is assigned a value of zero. These are given in JANAF tables. 16 Heat of Formation Absolute enthalpy of a substance can be calculated from, sensible enthalpy relative to reference T plus the heat of formation at ref T, T hˆ = ∫ cˆ p dT + ∆hˆ o To where ∆ĥ Here o is the heat of formation overbar, “ ^ ” shows quantity per mole Adiabatic Flame Temperature at constant-pressure combustion ( H R )TR = ( H P )TP H R (TR ) − H P (TP ) + ∆H o = 0 at constant-volume combustion U R (TR ) − U P (TP ) + ∆U o = 0 17 Adiabatic Flame Temperature Products Chemical Equilibrium Chemical equilibrium is achieved for constant T and p systems when rate change of concentrations goes to zero for all species - very fast reaction rates or very small change in concentrations. For a system of J species in chemical equilibrium, p and T do not change, which may be specified by stating that Gibbs free energy of the system (G = H – TS) does not change : (dG )T , p = 0 J G = ∑ N j gˆ j j =1 gˆ j = hˆ j − Tsˆ j 18 Chemical Equilibrium Considering the following reaction, aA + bB = cC + dD for chemical equilibrium, the equilibrium constants, Kp = pCc pDd xCc xDd c + d − a −b = p p Aa pBb x Aa xBb Dissociation Reactions For example, CO + H 2O ⇔ H 2 + CO2 k f = A f exp − ( E f / RT ) kb = Ab exp − ( Eb / RT ) here kf and kb are the forward and backward reaction rate constants, A is the pre-exponential factor, E is the activation energy. 19 Dissociation Reactions λ =1 Combustion with dissociation, m⎞ ⎛ Cn H m + ⎜ n + ⎟ (O2 + 3.76 N 2 ) → 4⎠ ⎝ β1 CO2 + β2CO + β3 H 2O + β4 H 2 + β5O2 + β6 N 2 K p1 = CO + H 2O ⇔ H 2 + CO2 1 CO + O2 ⇔ CO2 2 K p2 2 = β1 β 4 β 2 β3 β12 β22 β5 NR pR Dissociation Reactions Carbon balance, n = β1 + β 2 Hydrogen balance, m / 2 = β3 + β4 Oxygen balance, 2(n + m / 4) = 2 β1 + 2 β 2 + β 3 + 2 β 5 Equation of state, pV = NRT main reaction equation + 6 equations to find the values of 6 unknowns βi i=1,2, … 6 20 Dissociation Effects on Flame Temperature Flame temperature reduces as a result of dissociation reactions. H With dissociation R P Tb T Temperature 21