Lecture 37. Heat of Reaction. 1 st Law Analysis of Combustion Systems
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1 Department of Mechanical Engineering ME 322 Mechanical Engineering hermodynamics Heat of eaction Lecture 37 1 st Law Analysis of Combustion Systems
2 Combustion System Analysis Consider the complete combustion of octane in 150% theoretical air, C 8 H 18 A = 150% Combustion Chamber roducts (p) In the previous lecture, we found the balanced reaction, C8H (O N 2) 8CO 2 + 9H2O O N 2 he First Law applied to the system identified above is, n h n h n h a a f f p p p 2
3 Combustion System Analysis C 8 H 18 A = 150% Combustion Chamber roducts (p) C8H (O N 2) 8CO 2 + 9H2O O N 2 n h n h n h a a f f p p p reactants (air and fuel) combustion products otential issue: here are no Dh values (except for O 2 and N 2 ). his has the potential to cause a datum state problem. 3
4 esolving the Datum State roblem In combustion calculations, the enthalpy of all stable* elements is defined as zero at the standard reference state (SS), 25C K at 0.1 Ma 1 atm 77F at 14.5 psia 1 atm * Stable means chemically stable at the SS. For example, diatomic oxygen (O 2 ) is stable at the SS. Monatomic oxygen (O) is not stable at the SS. 4
5 Enthalpy of Formation he enthalpy of a compound at the standard reference state 0 hf Heat released in an exothermic reaction (or absorbed in an endothermic reaction) when a compound is formed from its elements. (Elements and compound at the SS) Example Methane 25 C 1 atm C 2H 2 CH 4 25 C 1 atm n h n h n h CH CH C C H H n n n n n C 2 hch h 4 C hh 2 CH CH CH n CH 4 h CH h 0 f,ch 4 4 H 5
6
7 Enthalpy of Formation Values Using EES* *Unit setting = molar 7
8 Enthalpy of Formation Values esults... Conclusion: EES uses the SS as the datum state for enthalpy for the ideal gases! herefore, enthalpy of formation values can be calculated from EES using the ideal gas substances (except AI) 8
9 Enthalpy Values in Combustion What do we know so far? 1. he enthalpy of a stable element at the SS is 0 2. he SS is 25 C, 0.1 Ma 3. he enthalpy of a compound at the SS is the enthalpy of formation (able 15.1 or from EES) 9
10 Enthalpy Values at Other States he enthalpy of a component at any temperature in a combustion process can be evaluated by, Accounts for the enthalpy difference relative to the SS 0, h h Dh i f i i 0 h h h h i f, i i i SS How is the enthalpy difference in brackets determined?? 10
11 Enthalpy Values at Other States hree possibilities... 0 h h h h i f, i i i SS 1. If the heat capacity of the component can be assumed constant, 0 h h c i f, i p, i SS 2. If the constant heat capacity assumption is not accurate enough, then use the ideal gas tables (able C.16c). In this case the datum state for the table does not have to match the enthalpy of formation. 3. Use a set of property tables for all components that has all enthalpy values referenced to the SS. Does such a thing exist? 11
12 Enthalpy Values at Other States Exploring Option 3 from the previous slide... 0 h h h h i f, i i i SS If a thermodynamically consistent set of tables exists, then h h 0 f, i i SS herefore the enthalpy of the component could simply be looked up in a table at the given temperature, h i i h If something like this were available... combustion calculations would be EESy! 12
13 Enthalpy Values at Other States ALL of the ideal gas enthalpy reference states (except for the ideal gas AI ) in EES are referenced to the SS! his is from the EES Help Menu for the ideal gas CO2... All other ideal gases in EES (except AI) say the same thing! Significance: Combustion calculations just became EESy! 13
14 Heat of eaction Consider an aergonic combustion process as shown below eactants () Fuel Air Combustion Chamber roducts () he First Law applied to this system results in, n h n h i i i i Dividing by the molar flow rate of the fuel, n n i i q hi hi n fuel n fuel n fuel 14
15 Heat of eaction n n i i q hi hi n fuel n fuel n fuel he molar flow rate ratios are the molar coefficients from the balanced combustion reaction! herefore, q h h n fuel i i i i his is known as the molar heat of reaction. 15
16 Heating Values of Fuels Given: Gaseous octane (C 8 H 18 ) is burned completely in 100% theoretical air. he reactants and the products are at the SS. CH 8 18 SS A 100% Find: he heat released during this combustion process per mole of fuel for the following cases, (a) the water in the products is all vapor (b) the water in the products is all liquid Combustion Chamber roducts SS 16
17 Heating Values CH 8 18 he system boundary is drawn A 100% around the combustion chamber. Applying the First Law results in, n h n h i i i i SS Combustion Chamber roducts SS Dividing both sides of this equation by the molar flow rate of the fuel, n i n i hi h n fuel n fuel n fuel i Number of moles of reactant per mole of fuel Number of moles of product species per mole of fuel How are these found? 17
18 Heating Values CH 8 18 A 100% he molar flow rate ratios on the previous slide are the molar coefficients from the balanced combustion reaction (for one mole of fuel)! herefore, n n n n n SS Combustion Chamber i i hi h i q ihi ihi fuel fuel fuel n fuel Observations... roducts Notice: A = 100% means stoichiometric combustion 1. he importance of being able to balance the combustion reaction is evident! 2. As long as the combustion process is aergonic, the First Law will be as written above, independent of the conditions in and out of the combustion chamber! SS 18
19 Heating Values CH 8 18 For the complete combustion of A 100% normal octane in 100% theoretical air, we previously found, Applying the First Law to the system, q h h n fuel fuel i i i i SS Combustion Chamber q h h h h h h h n fuel C H (O N ) 8CO + 9H O + 47N CO 2 H O 4 O 5 N C H 0 O 3.67 N q h h h n 1 f,co 2 f,h O f,c H roducts SS 19
20 Heating Values CH 8 18 SS A 100% Combustion Chamber roducts SS C H (O N ) 8CO + 9H O + 47N q h h h n fuel 1 f,co 2 f,h O f,c H Now we have an interesting problem. Is the water liquid or gas (or both)? 20
21 Heating Values CH 8 18 A 100% Let s consider both extremes (1) the H2O is all vapor and (2) the H2O is all liquid. SS Combustion Chamber roducts SS All vapor water... q h h h n fuel 1 f,co 2 f,h O f,c H MJ MJ MJ MJ q , n fuel kmol kmol kmol kmol All liquid water... C H (O N ) 8CO + 9H O + 47N MJ MJ MJ MJ q , n fuel kmol kmol kmol kmol 21
22 Heating Values All vapor water... q n fuel All liquid water... q n fuel MJ 5, kmol MJ 5, kmol CH 8 18 SS A 100% Combustion Chamber Lower Heating Value (LHV) Higher Heating Value (HHV) roducts SS Observations he reactants and products are at the SS 2. he reaction occurs with A = 100% ( i = i ) 3. he difference between the HHV and the LHV is the enthalpy of vaporization of water! 22
23 Heating Values q SS, SS 1 mol fuel SS SS roducts (vapor H 2 O) LHV Stoichiometric air roducts (liquid H 2 O) H 2 Oh fg, H 2 O HHV HHV LHV H 2 Ohfg, 2 H O 23
24 Heating Values he heating values represent the maximum possible heat transfer that can occur per mole of fuel. he reactants and products are at the SS he HHV represents fully condensed water vapor he LHV represents all water vapor hese values provide a basis for the combustion efficiency, comb Actual heat transferred per mole of fuel HHV or LHV 24
25 Example CH 8 18 SS Combustion Chamber Back to our problem... Is there A 100% liquid water in the products at the SS? If so, how much? C H (O N ) 8CO + 9H O + 47N Will water condense? roducts SS y w w y Ma Ma ka w 52.6C dp Since SS < dp, water will condense 25
26 Example CH 8 18 SS A 100% Combustion Chamber How much water will condense? At SS = 25 C, the mole fraction of water vapor in the products is, C H (O N ) 8CO + 9H O + 47N roducts SS y v nv 8 n 47 v he mole fraction of the water vapor at 25 C can be found, y w w bar bar 26
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