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Department of Mechanical Engineering ME 322 – Mechanical Engineering Thermodynamics Lecture 37 Heat of Reaction 1 st Law Analysis of Combustion Systems.

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Presentation on theme: "Department of Mechanical Engineering ME 322 – Mechanical Engineering Thermodynamics Lecture 37 Heat of Reaction 1 st Law Analysis of Combustion Systems."— Presentation transcript:

1 Department of Mechanical Engineering ME 322 – Mechanical Engineering Thermodynamics Lecture 37 Heat of Reaction 1 st Law Analysis of Combustion Systems

2 Combustion System Analysis 2 Consider the complete combustion of octane in 150% theoretical air, Combustion Chamber C 8 H 18 PTA = 150% Products (p) In the previous lecture, we found the balanced reaction, The First Law applied to the system identified above is,

3 Combustion System Analysis 3 Combustion Chamber C 8 H 18 PTA = 150% Products (p) reactants (air and fuel) combustion products Potential issue: There are no  h values (except for O 2 and N 2 ). This has the potential to cause a datum state problem.

4 Resolving the Datum State Problem 4 In combustion calculations, the enthalpy of all stable* elements is defined as zero at the standard reference state (SRS), *‘Stable’ means chemically stable at the SRS. For example, diatomic oxygen (O 2 ) is stable at the SRS. Monatomic oxygen (O) is not stable at the SRS.

5 Enthalpy of Formation 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 SRS) The enthalpy of a compound at the standard reference state Example – Methane C 2H 2 CH 4 25°C 1 atm 25°C 1 atm 5

6

7 Enthalpy of Formation Values Using EES* … *Unit setting = molar 7

8 Enthalpy of Formation Values Conclusion: EES uses the SRS as the datum state for enthalpy for the ideal gases! Therefore, enthalpy of formation values can be calculated from EES using the ideal gas substances (except AIR) 8 Results...

9 Enthalpy Values in Combustion 9 What do we know so far? 1.The enthalpy of a stable element at the SRS is 0 2.The SRS is 25°C, 0.1 MPa 3.The enthalpy of a compound at the SRS is the enthalpy of formation (Table 15.1 or from EES)

10 Enthalpy Values at Other States 10 The enthalpy of a component at any temperature in a combustion process can be evaluated by, Accounts for the enthalpy difference relative to the SRS How is the enthalpy difference in brackets determined??

11 1.If the heat capacity of the component can be assumed constant, 2.If the constant heat capacity assumption is not accurate enough, then use the ideal gas tables (Table 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 SRS. Does such a thing exist? Enthalpy Values at Other States 11 Three possibilities...

12 Enthalpy Values at Other States 12 Exploring Option 3 from the previous slide... If a thermodynamically consistent set of tables exists, then Therefore the enthalpy of the component could simply be looked up in a table at the given temperature, If something like this were available... combustion calculations would be EESy!

13 Enthalpy Values at Other States 13 ALL of the ideal gas enthalpy reference states (except for the ideal gas ‘AIR’) in EES are referenced to the SRS! This is from the EES Help Menu for the ideal gas CO2... All other ideal gases in EES (except AIR) say the same thing! Significance: Combustion calculations just became EESy!

14 Heat of Reaction 14 Combustion Chamber Fuel Air Products (P) Consider an aergonic combustion process as shown below The First Law applied to this system results in, Reactants (R) Dividing by the molar flow rate of the fuel,

15 Heat of Reaction 15 The molar flow rate ratios are the molar coefficients from the balanced combustion reaction! Therefore, This is known as the molar heat of reaction.

16 Heating Values of Fuels 16 Given: Gaseous octane (C 8 H 18 ) is burned completely in 100% theoretical air. The reactants and the products are at the SRS. Find: The 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

17 The system boundary is drawn around the combustion chamber. Applying the First Law results in, Heating Values 17 Dividing both sides of this equation by the molar flow rate of the fuel, Number of moles of reactant per mole of fuel Number of moles of product species per mole of fuel How are these found?

18 The molar flow rate ratios on the previous slide are the molar coefficients from the balanced combustion reaction (for one mole of fuel)! Therefore, Heating Values 18 Observations... 1.The 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! Notice: PTA = 100% means stoichiometric combustion

19 For the complete combustion of normal octane in 100% theoretical air, we previously found, Heating Values 19 Applying the First Law to the system,

20 Heating Values 20 Now we have an interesting problem. Is the water liquid or gas (or both)?

21 Let’s consider both extremes (1) the H2O is all vapor and (2) the H2O is all liquid. Heating Values 21 All vapor water... All liquid water...

22 Heating Values 22 All vapor water... All liquid water... Lower Heating Value (LHV) Higher Heating Value (HHV) Observations... 1.The reactants and products are at the SRS 2.The reaction occurs with PTA = 100% (  i = i ) 3.The difference between the HHV and the LHV is the enthalpy of vaporization of water!

23 Heating Values 23 1 mol fuel Stoichiometric air Products (vapor H 2 O) Products (liquid H 2 O) T SRS, P SRS T SRS P SRS

24 Heating Values 24 The heating values represent the maximum possible heat transfer that can occur per mole of fuel. The reactants and products are at the SRS The HHV represents fully condensed water vapor The LHV represents all water vapor These values provide a basis for the combustion efficiency,

25 Example 25 Back to our problem... Is there liquid water in the products at the SRS? If so, how much? Will water condense? Since T SRS < T dp, water will condense

26 Example 26 How much water will condense? At T SRS = 25°C, the mole fraction of water vapor in the products is, The mole fraction of the water vapor at 25°C can be found,


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