..................................APPENDIX A Thermodynamic data A.l Introduction The thermodynamic tables presented here for enthalpy and internal energy differ from those that are usually available, since they incorporate the enthalpy of formation. This means that there is no need for separate tabulations of calorific values, and it will be found that energy balances for combustion calculations are greatly simplified. The enthalpy of formation (t-.Hf) is perhaps more familiar to physical chemists than to engineers. The enthalpy of formation ( t-.Hf) of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference state. The reference state of an element is its most stable state (for example, carbon atoms, but oxygen molecules) at a specified temperature and pressure, usually 298.15 K and a pressure of l bar. In the case of atoms that can exist in different forms, it is necessary to specify their form, for example, carbon is as graphite, not diamond. Combustion calculations are most readily undertaken by using absolute (some­ times known as sensible) internal energies or enthalpy. In steady-flow systems where there is displacement work then enthalpy should be used; this has been illustrated by figure 3.8. When there is no displacement work then internal energy should be used (figure 3.7). Consider now figure 3.8 in more detail. With an adiabatic combustion process from reactants (R) to products (P) the enthalpy is constant, but there is a substantial rise in temperature: (A.l) In the case of the isothermal combustion process (IR ~ IP) the temperature (T) is obviously constant, and the difference in enthalpy corresponds to the isobaric calorific value (-t-.Jfi,): t-.H~ = HR .T- HP,T (A.2) 582 Thermodynamic data A.2 Thermodynamic principles In the following sections, it will be seen how the thermodynamic data for internal energy, enthalpy, entropy and Gibbs function can all be determined from measurements of heat capacity and phase change enthalpies (or internal energies). Furthermore, when the energy change associated with a chemical reaction is measured, the enthalpy of formation can be deduced. This in turn leads to 'absolute' values of internal energy, enthalpy, entropy and Gibbs energy, from which it is possible to derive the equilibrium constant for any reaction. A.2.1 Determination of internal energy, enthalpy, entropy and Gibbs energy Figure 3.8 shows that the enthalpies of both reactants and products are in general non-linear functions of temperature. The Absolute Molar Enthalpy tables presented here use a datum of zero enthalpy for elements when they are in their standard state at a temperature of 25°C. The enthalpy of any substance at 25°C will thus correspond to its enthalpy of formation, t.Hf. The use of these tables will be illustrated, after a description of how they have been developed. Tables are not very convenient for computational use, so instead molar enthalpies and other thermodynamic data are evaluated from analytical functions; a popular choice is a simple polynomial. For species i H;(T) =A;+ B;T + C;T2 + D;T3 + E;T4 + F;T5 (A.3) from which it can be deduced (since dH = Cpd1) that Cp,;(T) = B; + 2C;T + 3D;T2 + 4E;T3 + 5F;T4 (A.4) As U = H- RT U;(T) =A;+ (B;- R0 )T + C;T2 + D,T3 + E;T4 + F;T5 (A.5) and dU = CvdT, so Cv,;(T) = (B;- R0 ) + 2C;T + 3D;T2 + 4E;T3 + 5F;r4 (A.6) and as dH = CpdT = TdS dS = (Cp / T)dT integrating equation (A.4) gives s? = B; ln(T) + 2C;T + 3/2D;T2 + 4/3E;T3 + 5/4F;r4 + G; (A.7) where G; is an integration constant that is used to set the zero datum, for example 0 K by Rogers and Mayhew ( 1988) -the same datum does not have to be used as for enthalpy. A more common choice is to use a polynomial function to describe the specific heat capacity variation, and to divide through by the Molar Gas Constant (R0 ) . Equation (A.4) becomes Cp,;(T)/Ro =a;+ b;T + c;T2 + d;T3 + e;T4 (A.8) where a;= B;/R0 , b; = 2C;/R0 , c; = 3D;/R0 etc. Thus H;(T)/Ro = a;T + b;T2 /2 + c;T3 / 3 + d;T4 /4 + e;T5 / 5 +f; (A.9) 584 Introduction to internal combustion engines and s?(T)/R =a; ln(T) + b;T + c;/2T2 + d;/3T3 + e;/4T4 + g; (A.10) A polynomial fit will only be satisfactory over a small temperature range (300- 1000 K or 1000-5000 K), and such polynomial equations ought never to be used outside their range. As the specific heat capacity variation with temperature has a 'knee' between 900 K and 2000 K, a single polynomial is never likely to be satisfactory. Instead, two polynomials can be used which give identical values of Cp at the transition between the low range and the high range (the transition temperature is usually chosen between 1000 and 2000 K). Examples of such polynomials are presented by Gordon and McBride ( 1971), and were used as the basis for constructing the tables used here. The coefficients for the evaluation of the thermodynamic tables are summarised in table A.1. The tables at the end of this Appendix present the enthalpy (H), internal energy (U), entropy (S) and Gibbs energy (G) for gaseous species (table A.4) and fuels (table A.5). The tables have been extrapolated below 300 K, and these data should be used with caution. The entropy datum of zero at 0 K cannot be illustrated by the evaluation of equation (A.1 0), since there is a singularity in this equation at 0 K and in any case there will be phase changes. Phase changes will lead to isothermal changes in entropy, and each different phase will have a different temperature/ entropy relationship. Instead, use is made of the values of entropy for substances at 25°C in their standard state at a pressure of 1 bar. (WARNING- Many sources use a datum pressure of 1 atm.) The entropy can be evaluated at other pressures from: (A.ll) The internal energy ( U) and Gibbs energy (G) are by definition H = U + pV = U +RoT and G0 = H0 - TS0 (A.l2) with the superscript 0 referring to the datum pressure (p0 ) of 1 bar. The Gibbs energy can be evaluated at other pressures in a similar way to the entropy (equation A.11), and through the use of this equation: (A.l3) The changes inS and G between state 1 (p 1, Td and state 2 (p2 , T2), for a gas or vapour, are given by S2- St = (S2 - S~) +(~-Sf)+ (S?- S1) S2 - S1 = (S~- s?)- Ro ln(p2fpl) (A.l4) and G2 - G 1 = (G2 - G~) + (G ~ - G7) + (c? - G t) G2 - G1 = ( G~ - G?) + RoT2ln(p2jp0 )- RoTiln(ptfp0 ) (A.15) When the entropy and Gibbs energy of a mixture are being evaluated, the properties of the individual constituents are summed, but the pressures (p1 and p2 ) now refer to the partial pressures of each constituent. The use of these tables in combustion calculations is best illustrated by an example. Table A.l Coefficients in equations (A. B), (A.9) and (A. 70), for the evaluation of thermodynamic data from Gordon and McBride (7 971 ), except argon, from Reid et al. (7 987) Nz Oz Hz co C02 HzO NO Ar OH 0 H 1000-5000 K o, 0.28963E1 0.36220E1 0.31002£1 0.29841E1 0.44608E1 0.2?168E1 0.31890E1 2.5016 O.l9106E1 0.254l1E1 2.5 b, 0.15155E-2 0.73618E-3 0.51119£-3 0.14891 E-2 0. 30982E-2 0.29451E-2 0.13382E-2 0 0.95932E-3 -{).27551 E-4 0 q -<l.57235E-6 -<l.19652E-6 0.52644E-7 -0.57900£-6 -0.12393E-5 -<l.80224E-6 -{).52899E-6 0 -{). 1 9442E-6 -<l. 31 028E-8 0 d, 0.99807E-10 0.36202E-10 -<l.34910E-10 0.10365E·9 0.22741E-9 0.10227E-9 0.95919E-JO 0 0.13757E-JO 0.45511£-11 0 e1 -{).652l4E-14 -<l.28946E·14 0.36945E-14 -<l.693S4E·14 -<l.1 5526E-13 -<l.48472E-14 -<l.64848E-14 0 0. 14225E-15 -<l-43681 E-15 0 ~ -<l.90586E3 -<l.12020E4 -<l.87738E3 -0.14l45E5 -{) .48961 E5 -<l.29906E5 0.98283E4 - 745.852 0. 393S4E4 O.l9231E5 0.25472E5 91 0.61615E1 0. 36151 El -0.19629E1 0.63479£1 -<l.98636EO 0.66306E1 0.67458E1 0.43529E1 0.54423E1 0.49203E1 -0.460120EO 300-1000 K 01 0. 36748E1 0. 36256E1 0. 30574E1 0.37101E1 0.24008E1 0.40701£1 2.5016 b, -{).12082£-2 -<l.18782E-l 0.26765E-2 -<l.161.91 E-2 0.87351E-2 -0.11 084E-2 0 Cj 0.23240E-5 0.70555 E-5 -0.58099E-5 0.36924E-5 -0.66071E-5 0.41521E-5 0 d, -<l.6321 SE-9 -<l.67635E.S 0.55210E·8 -<l.20320E·8 0.20022E.S -<l.29637E-8 0 e1 -<l.22577E-1l 0.21 556E-11 -<l.18123E-11 0.23953£:.12 0.63l74E-15 0.80702E-12 0 f1 -<l.l 0612E4 -<l.1 0475E4 -{).98890E3 -{).14356E5 -{).48378E5 -<l.30280E5 - 745.852 91 0.23580El 0.43053E1 -{).22997E1 0.29555E1 0.96951E1 -{).32270EO 0.43529E1 -1 =r ro 3 0 ~ ::J 1:11 3 (')' 0. s1:11 586 Introduction to internal combustion engines EXAMPLE A mixture of carbon monoxide and 10 per cent excess air at 25°( is burnt at constant pressure, and it is assumed that no carbon monoxide is present in the products.