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-2temperatures 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