Part 2.: introduction to thermochemistry SB&VW-14.1-14.6 Until now, we have specified the heat given to the devices analyzed, and not concerned ourselves with how this heat might be produced. In this section, we examine the issue of how obtain the heat needed for work production. For the most part, this is from converting chemical energy into heat, so the discussion will be on reacting mixtures of gas which are involved in chemical combustion processes The topic addressed is"thermochemistry, which is the combining of thermodynamics with chemistry to predict such items as how much heat is released from a chemical reaction. This is the Q or"q that we have used in the cycle analysis. The principal components of the approach are use of a chemical balance plus the steady flow energy equation(SFEE)which equates the sum of shaft work(from) and heat transfer(to)a control volume to the difference in control volume inlet and exit enthalpy fluxes 2.C. Fuels There are a wide variety of fuels used for aerospace power and propulsion. a primary one is jet fuel(octane, essentially kerosene) which has the chemical formula C&H,o. Other fuels we consider are hydrogen(H2 )and methane(CH4) The chemical process in which a fuel, for example methane, is burned consists of (on a very basic level--there are many intermediate reactions that need to be accounted for when computations of the combustion process are carried out) CH4+202→CO2+2H2O (Reactants)(Products) The reactions we describe are carried out in air, which can be approximated as 21%O2 and 79% N,. This composition is referred to as" theoretical air. There are other components of air (for example Argon, which is roughly 1%), but the results given using the theoretical air approximation are more than adequate for our purposes. With this definition, for each mole of O2 3.76(79/21)moles of n2 are involved: CH4+202+2(3.76N2→CO2+2H2O+752N2 Even if the nitrogen is not part of the combustion process, it leaves the combustion chamber at the same temperature as the other products, and this change in state(change in enthalpy) needs to be accounted for in the steady flow energy equation. At the high temperatures achieved in internal combustion engines(aircraft and automobile)reaction does occur between the nitrogen and oxygen, which gives rise to oxides of nitrogen, although we will not consider these reactions The condition at which the mixture of fuel and air is such that both completely participate in the reaction is called stoichiometric. In gas turbines, excess air is often used so that the 2C-1
Part 2.C: Introduction to Thermochemistry [SB&VW-14.1-14.6] Until now, we have specified the heat given to the devices analyzed, and not concerned ourselves with how this heat might be produced. In this section, we examine the issue of how we obtain the heat needed for work production. For the most part, this is from converting chemical energy into heat, so the discussion will be on reacting mixtures of gas which are involved in chemical combustion processes. The topic addressed is “thermochemistry”, which is the combining of thermodynamics with chemistry to predict such items as how much heat is released from a chemical reaction. This is the “Q” or “q” that we have used in the cycle analysis. The principal components of the approach are use of a chemical balance plus the steady flow energy equation (SFEE) which equates the sum of shaft work (from) and heat transfer (to) a control volume to the difference in control volume inlet and exit enthalpy fluxes. 2.C.1 Fuels There are a wide variety of fuels used for aerospace power and propulsion. A primary one is jet fuel (octane, essentially kerosene) which has the chemical formula C H18 . Other fuels we 8 consider are hydrogen (H2 )and methane ( CH4 ). The chemical process in which a fuel, for example methane, is burned consists of (on a very basic level—there are many intermediate reactions that need to be accounted for when computations of the combustion process are carried out): CH4 + 2O2 → CO2 + 2H O . 2 (Reactants) (Products) The reactions we describe are carried out in air, which can be approximated as 21% O2 and 79% N2 . This composition is referred to as “theoretical air”. There are other components of air (for example Argon, which is roughly 1%), but the results given using the theoretical air approximation are more than adequate for our purposes. With this definition, for each mole of O2 , 3.76 (79/21) moles of N2 are involved: CH4 + 2O2 + 2 3 76 )N → CO2 + 2H O + 7.52N2 ( . 2 2 Even if the nitrogen is not part of the combustion process, it leaves the combustion chamber at the same temperature as the other products, and this change in state (change in enthalpy) needs to be accounted for in the steady flow energy equation. At the high temperatures achieved in internal combustion engines (aircraft and automobile) reaction does occur between the nitrogen and oxygen, which gives rise to oxides of nitrogen, although we will not consider these reactions. The condition at which the mixture of fuel and air is such that both completely participate in the reaction is called stoichiometric. In gas turbines, excess air is often used so that the 2C-1
temperatures of the gas exiting the combustor is kept to within desired limits(see Figures A-8, A 9, A-ll in Part 1 for data on these limits. Muddy poin Why is there 3.76 N,?(MP 2C. 1) What is the most effective way to solve for the number of moles in the reactions?(MP 2. 2 Fuel-Air Ratio The reaction for aeroengine fuel at stoichiometric conditions is CH1s+12502+12.5(376)N2→8CO2+9H2O+470N On a molar basis, the ratio of fuel to air is [1/(12. 5+47.0)]=1/59.5=0.0167 To find the ratio on a mass flow basis, which is the way in which the aeroengine industry discusses it, we need to "weight the molar proportions by the molecular weight of the components. The fuel molecular weight is 1 14, the oxygen molecular weight is 32 and the nitrogen molecular weight is(approximately) 28. The fuel/air ratio on a mass flow basis is thus Fuel-air ratio =0.0664 12.5×32+12.5×3.76×28 If we used the actual constituents of air we would get 0.0667. a value about 0.5% different Muddy points Do we always assume 100% complete combustion? How good an approximation is this? (MP 2C.3 2. C3 Enthalpy of formation The systems we have worked with until now have been of fixed chemical composition. Because of this, we could use thermodynamic properties relative to an arbitrary base, since all comparisons could be made with respect to the chosen base. For example, the specific energy u (0.01C)=0.0 for steam. If there are no changes in composition, and only changes in properties of given substances, this is adequate. If there are changes in composition, however, we need to have a reference state so there is consistency for different substances The convention used is that the reference state is a temperature of 25 C (298 K)and a pressure of 0. 1 MPa. (These are roughly room conditions. At these reference conditions, the enthalpy of the elements(oxygen, hydrogen, nitrogen, carbon, etc. )is taken as zero The results of a combustion process can be diagrammed as below. The reactants enter at standard conditions; the combustion(reaction) takes place in the volume indicated. Downstream of the reaction zone there is an appropriate amount of heat transfer with the surroundings so that the products leave at the standard conditions. For the reaction of carbon and oxygen to produce O2, the heat that has to be extracted is Ocv =-393, 522 k/kmole; this is heat that comes out of the control volume 2C-2
temperatures of the gas exiting the combustor is kept to within desired limits (see Figures A-8, A- 9, A-11 in Part 1 for data on these limits.) Muddy points Why is there 3.76 N2? (MP 2C.1) What is the most effective way to solve for the number of moles in the reactions? (MP 2C.2) 2.C.2 Fuel-Air Ratio The reaction for aeroengine fuel at stoichiometric conditions is: CH +12.5 O2 +12 . . 5 (3 76 ) N → 8 CO2 + 9 H O + 47.0 N 8 18 2 2 2 On a molar basis, the ratio of fuel to air is [1/(12.5+47.0)] = 1/59.5 = 0.0167. To find the ratio on a mass flow basis, which is the way in which the aeroengine industry discusses it, we need to “weight” the molar proportions by the molecular weight of the components. The fuel molecular weight is 114, the oxygen molecular weight is 32 and the nitrogen molecular weight is (approximately) 28. The fuel/air ratio on a mass flow basis is thus 1 ×114 Fuel-air ratio = = 0 0664 . 12 5 . . . × 32 +12 5 × 3 76 × 28 If we used the actual constituents of air we would get 0.0667, a value about 0.5% different. Muddy points Do we always assume 100% complete combustion? How good an approximation is this? (MP 2C.3) 2.C.3 Enthalpy of formation The systems we have worked with until now have been of fixed chemical composition. Because of this, we could use thermodynamic properties relative to an arbitrary base, since all comparisons could be made with respect to the chosen base. For example, the specific energy uf ( . o 0 01 C) = 0 0 . for steam. If there are no changes in composition, and only changes in properties of given substances, this is adequate. If there are changes in composition, however, we need to have a reference state so there is consistency for different substances. The convention used is that the reference state is a temperature of 25o C (298 K) and a pressure of 0.1 MPa. (These are roughly room conditions.) At these reference conditions, the enthalpy of the elements (oxygen, hydrogen, nitrogen, carbon, etc.) is taken as zero. The results of a combustion process can be diagrammed as below. The reactants enter at standard conditions; the combustion (reaction) takes place in the volume indicated. Downstream of the reaction zone there is an appropriate amount of heat transfer with the surroundings so that the products leave at the standard conditions. For the reaction of carbon and oxygen to produce CO2, the heat that has to be extracted is QCV = −393 522 kJ/kmole , ; this is heat that comes out of the control volume. 2C-2
I mole C I mole CO2 250C.0.1 MPa 25°C,0.MPa Volume I mole CO? ecv=-393,522 KJ, heat is out of control volume Figure C-1: Constant pressure combustion There is no shaft work done in the control volume and the first law for the control volume(SFEE) mass flow of enthalpy in rate of heat addition=mass flow of enthalpy out We can write this statement in the form >, h,+Ocv=>mthe (C.3.1) In Eq(C3. 1)the subscripts"R"and"P"on the summations refer to the reactants(r)and products (P)respectively. The subscripts on the mass flow rates and enthalpies refer to all of the components at inlet and at exit The relation in terms of mass flows can be written in molar form which is often more convenient for reacting flow problems, by using the molecular weight, Mi, to define the molar mass flow rate, ni, and molar enthalpy, hi, for any individual i(or e )component as ni; =m; /M;; mass flow rate in terms of kmoles/sec h i,=M hi; enthalpy per kmole The sfee is. in these terms n:h:+ >nh (C.3.2) The statements that have been made do not necessarily need to be viewed in the context of flow have one unit of C and one unit of O2 at the initial conditions and we carry out a constant pressure reaction at ambient pressure, Pamb. If so, 2C-3
C + 02 C02 1 kmole C 1 kmole C02 25o C, 0.1 MPa 25o C, 0.1 MPa Volume 1 kmole C02 Q = -393,522 KJ, heat is out of control volume cv Figure C-1: Constant pressure combustion There is no shaft work done in the control volume and the first law for the control volume (SFEE) reduces to: mass flow of enthalpy in + rate of heat addition = mass flow of enthalpy out. We can write this statement in the form ∑mh + Q ˙ ˙ CV = ∑m he ˙ (C.3.1) ii e R P In Eq. (C.3.1) the subscripts “R” and “P” on the summations refer to the reactants (R) and products (P) respectively. The subscripts on the mass flow rates and enthalpies refer to all of the components at inlet and at exit. The relation in terms of mass flows can be written in molar form, which is often more convenient for reacting flow problems, by using the molecular weight, Mi , to define the molar mass flow rate, n˙i , and molar enthalpy, hi , for any individual ith (or eth) component as n˙i = m˙i / M ; mass flow rate in terms of kmoles/sec i hi = M hi ; enthalpy per kmole i The SFEE is, in these terms, ∑nh + Q ˙ ˙ CV = ∑n he ˙ . (C.3.2) ii e R P The statements that have been made do not necessarily need to be viewed in the context of flow processes. Suppose we have one unit of C and one unit of O2 at the initial conditions and we carry out a constant pressure reaction at ambient pressure, Pamb . If so, 2C-3
Q-w P since P=Pf=Pamb. Combining terms, U final+Pinal V(Initial + Initial Initial )=ecv initial In terms of the numbers of moles and the specific enthalpy this is ∑n1h1+Qcv=∑nlh (C.3.3) The enthalpy of CO2, at 25C and o 1 MPa, with reference to a base where the enthalpy of the elements is zero, is called the enthalpy of formation and denoted by h Values of the heat of formation for a number of substances are given in Table A. 9 in SB& vw. The enthalpies of the reactants and products for the formation of CO2 are ho =hc For one kmole: Qcv-Enhe=Hp=(n9)co--393, 522 kJ/kmole The enthalpy of Co2 in any other state(T, P)is given by h7-(19)a1+( 298K,0.1Mp→T,P These descriptions can be applied to any compound. For elements or compounds that exist more than one state at the reference conditions( for example, carbon exists as diamond and as graphite), we also need to specify the state Note that there is a minus sign for the heat of formation. The heat transfer is out of the control volume and is thus negative by our convention. This means that elements>hco,=-393,522 kJ Muddy points Is the enthalpy of formation equal to the heat transfer out of the combustion during the formation reaction?(MP 2C. 4) Are the enthalpies of H, and (monoatomic hydrogen) both zero at 298K?(MP 2C.5) 2C-4
U final −Uinitial = Q − W = QCV − Pamb (Vfinal − Vinitial ), since P = Pf = Pamb . Combining terms, i U final + PfinalVfinal − (Uinitial + PinitialVinitial ) = QCV , or, = QCV H . final − Hinitial In terms of the numbers of moles and the specific enthalpy this is ∑nh + QCV = ∑n he (C.3.3) ii e R P The enthalpy of CO2, at 25o C and 0.1 MPa, with reference to a base where the enthalpy of the o elements is zero, is called the enthalpy of formation and denoted by hf . Values of the heat of formation for a number of substances are given in Table A.9 in SB&VW. The enthalpies of the reactants and products for the formation of CO2 are: = hC = 0 h hO2 o f For one kmole: Q , CV = ∑n he = HP = ( ) = −393 522 kJ/kmole. e P CO2 The enthalpy of CO2 in any other state (T,P)is given by hTP = ( ) + (∆h ) . , h0 f 298K,. 0 1 MPa 298K,. , 0 1 Mpa→T P These descriptions can be applied to any compound. For elements or compounds that exist in more than one state at the reference conditions (for example, carbon exists as diamond and as graphite), we also need to specify the state. Note that there is a minus sign for the heat of formation. The heat transfer is out of the control volume and is thus negative by our convention. This means that h , elements > h = −393 522 kJ . CO2 Muddy points Is the enthalpy of formation equal to the heat transfer out of the combustion during the formation reaction? (MP 2C.4) Are the enthalpies of H2 and H (monoatomic hydrogen) both zero at 298K? (MP 2C.5) 2C-4
2. C 4 First Law Analysis of Reacting Systems The form of the first law for the control volume is(there is no shaft work) ∑nh1+Qcv=∑nh This is given in terms of the moles of the different constituents, and it reduces to the more familiar form for a single fluid(say air)with no reactions occurring. We need to specify one parameter as the basis of the solution 1 kmole of fuel. 1 kmole of air. 1 kmole total etc. We use 1 kmole of fuel as the basic unit and examine the burning of hydrogen. H2+202→2H20 The reactants and the products are both taken to be at o 1MPa and 25C, so the inlet and exit P and Tare specified. The control volume is the combustion chamber. There is no shaft work done and the sfee is in the form of Equation(C 1. 2). The enthalpy of the entering gas is zero for both the hydrogen and the oxygen(elements have enthalpies defined as zero at the reference state) If the exit products are in the gaseous state, the exit enthalpy is therefore related to the enthalpy of formation of the product by H2oneH2O-n H2o(g) 2x( -241, 827)kJ=-483, 654 k; gaseous state at exit If the water is in a liquid state at the exit of the process 2x(-285,783)kJ=-571,676 There is more heat given up if the products emerge as liquid. The difference between the two values is the enthalpy needed to turn the liquid into gas at 25C: h= 2442 kJ/kmole A more complex example is provided by the burning of methane(natural gas )in oxygen, CH4+2O2→CO2+2H2O) The components in this reaction equation are three ideal gases(methane, oxygen, and cO2) and liquid water. We again specify that the inlet and exit states are at the reference conditions so that ∑n再-(列2m=-4873J n-()2+212)m =-393,522+2(-285,838)=-965,198kJ Qcv=-965,198kJ-(-74,873)=-890,325kJ 2C-5
2.C.4 First Law Analysis of Reacting Systems The form of the first law for the control volume is (there is no shaft work): ∑n h + Q ˙ ˙ ˙ CV = ∑nh . ii e e R P This is given in terms of the moles of the different constituents, and it reduces to the more familiar form for a single fluid (say air) with no reactions occurring. We need to specify one parameter as the basis of the solution; 1 kmole of fuel, 1 kmole of air, 1 kmole total, etc. We use 1 kmole of fuel as the basic unit and examine the burning of hydrogen. H2 + 20 → 2HO 2 2 The reactants and the products are both taken to be at 0.1MPa and 25o C, so the inlet and exit P and T are specified. The control volume is the combustion chamber. There is no shaft work done and the SFEE is in the form of Equation (C.1.2). The enthalpy of the entering gas is zero for both the hydrogen and the oxygen (elements have enthalpies defined as zero at the reference state). If the exit products are in the gaseous state, the exit enthalpy is therefore related to the enthalpy of formation of the product by: o hf n˙eH O he HO = n˙e ( ) 2 HO HOg ( ) 2 2 2 = 2 x (- 241,827)kJ = - 483,654 kJ; gaseous state at exit. If the water is in a liquid state at the exit of the process: o hf n˙eH O he HO = n˙e ( ) 2 HO HOl ( ) 2 2 2 = 2 x (- 285,783) kJ = - 571, 676. There is more heat given up if the products emerge as liquid. The difference between the two values is the enthalpy needed to turn the liquid into gas at 25o C: hfg = 2442 kJ/kmole. A more complex example is provided by the burning of methane (natural gas) in oxygen, producing . CH4 + 2O2 → CO2 + 2H O ( )l 2 The components in this reaction equation are three ideal gases (methane, oxygen, and CO2) and liquid water. We again specify that the inlet and exit states are at the reference conditions so that: ∑nh hf o , ii = ( ) = −74 873 kJ R CH4 ∑nh h ee = ( ) f + 2( ) o CO2 hf o HOl P 2 ( ) Q =-393,522 + 2(-285,838) = -965,198 kJ CV = −965 198 , kJ – (-74,873) = -890,325 kJ. 2C-5