Page 1 Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1.htm[10/5/2012 11:28:54 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry The Lecture Contains: Stoichiometry Stoichiometry Calculation Thermochemistry Page 2 Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1.htm[10/5/2012 11:28:54 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry The Lecture Contains: Stoichiometry Stoichiometry Calculation Thermochemistry Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1a.htm[10/5/2012 11:29:00 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Stoichiometry Stoichiometry: The elemental mass balance in a chemical reaction, describing exactly how much oxidizer has to be supplied for complete combustion of certain amount of fuel. Example: Mass is conserved (Figure 8.1) Lean Mixture: Quantity of oxidizer > Stoichiometric quantity Rich Mixture: Quantity of oxidizer < Stoichiometric quantity Page 3 Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1.htm[10/5/2012 11:28:54 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry The Lecture Contains: Stoichiometry Stoichiometry Calculation Thermochemistry Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1a.htm[10/5/2012 11:29:00 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Stoichiometry Stoichiometry: The elemental mass balance in a chemical reaction, describing exactly how much oxidizer has to be supplied for complete combustion of certain amount of fuel. Example: Mass is conserved (Figure 8.1) Lean Mixture: Quantity of oxidizer > Stoichiometric quantity Rich Mixture: Quantity of oxidizer < Stoichiometric quantity Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1b.htm[10/5/2012 11:29:00 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Stoichiometry Calculation Problem: Gasoline Dry air Products (10.02% ; 5.62% ; 0.88% ; 83.48% ) Determine (i) A/F Ratio; (ii) Equivalence Ratio; (iii) Stoichiometric Air Used Solution Equation: 16.32 ( 3.76 ) 7.37 + 0.65 4.13 61.38 9 ((12 8 ) 18)/(16.32(32 (3.76 28))) = 0.05089 Stoichiometric Equation: 12.5( 3.76 ) 8 9 47 ((12 8 ) 18)/(12.5(32 (3.76 28))) = 0.06643 Equivalence ratio 0.766 Since <1 ; The mixture is lean Stoichiometric Air Used: Stoichiometric air = 100 / = 100/ 0.766 = 130.5% Page 4 Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1.htm[10/5/2012 11:28:54 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry The Lecture Contains: Stoichiometry Stoichiometry Calculation Thermochemistry Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1a.htm[10/5/2012 11:29:00 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Stoichiometry Stoichiometry: The elemental mass balance in a chemical reaction, describing exactly how much oxidizer has to be supplied for complete combustion of certain amount of fuel. Example: Mass is conserved (Figure 8.1) Lean Mixture: Quantity of oxidizer > Stoichiometric quantity Rich Mixture: Quantity of oxidizer < Stoichiometric quantity Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1b.htm[10/5/2012 11:29:00 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Stoichiometry Calculation Problem: Gasoline Dry air Products (10.02% ; 5.62% ; 0.88% ; 83.48% ) Determine (i) A/F Ratio; (ii) Equivalence Ratio; (iii) Stoichiometric Air Used Solution Equation: 16.32 ( 3.76 ) 7.37 + 0.65 4.13 61.38 9 ((12 8 ) 18)/(16.32(32 (3.76 28))) = 0.05089 Stoichiometric Equation: 12.5( 3.76 ) 8 9 47 ((12 8 ) 18)/(12.5(32 (3.76 28))) = 0.06643 Equivalence ratio 0.766 Since <1 ; The mixture is lean Stoichiometric Air Used: Stoichiometric air = 100 / = 100/ 0.766 = 130.5% Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1c.htm[10/5/2012 11:29:01 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Thermochemistry (Figure 8.2) Consider the burner as shown below: (Figure 8.3) Assumption: (i) Negligible change in K.E. & P.E., (ii) No shaft work Page 5 Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1.htm[10/5/2012 11:28:54 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry The Lecture Contains: Stoichiometry Stoichiometry Calculation Thermochemistry Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1a.htm[10/5/2012 11:29:00 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Stoichiometry Stoichiometry: The elemental mass balance in a chemical reaction, describing exactly how much oxidizer has to be supplied for complete combustion of certain amount of fuel. Example: Mass is conserved (Figure 8.1) Lean Mixture: Quantity of oxidizer > Stoichiometric quantity Rich Mixture: Quantity of oxidizer < Stoichiometric quantity Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1b.htm[10/5/2012 11:29:00 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Stoichiometry Calculation Problem: Gasoline Dry air Products (10.02% ; 5.62% ; 0.88% ; 83.48% ) Determine (i) A/F Ratio; (ii) Equivalence Ratio; (iii) Stoichiometric Air Used Solution Equation: 16.32 ( 3.76 ) 7.37 + 0.65 4.13 61.38 9 ((12 8 ) 18)/(16.32(32 (3.76 28))) = 0.05089 Stoichiometric Equation: 12.5( 3.76 ) 8 9 47 ((12 8 ) 18)/(12.5(32 (3.76 28))) = 0.06643 Equivalence ratio 0.766 Since <1 ; The mixture is lean Stoichiometric Air Used: Stoichiometric air = 100 / = 100/ 0.766 = 130.5% Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1c.htm[10/5/2012 11:29:01 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Thermochemistry (Figure 8.2) Consider the burner as shown below: (Figure 8.3) Assumption: (i) Negligible change in K.E. & P.E., (ii) No shaft work Objectives_template file:///D|/Web%20Course/Dr.%20D.P.%20Mishra/Local%20Server/FOC/lecture8/8_1d.htm[10/5/2012 11:29:01 AM] Module 2: Thermodynamics of Combustion Lecture 8: Stoichiometry Thermochemistry Where, H R – Total enthalpy of reactants; H P – Total enthalpy of products n iR – No of moles of i th reactant; n iP – No of moles of i th product h iR – Enthalpy of formation per unit mole of i th reactant h iP – Enthalpy of formation per unit mole of i th product – Standard heat of reaction Heat of reaction depends on temperature (Figure 8.4)Read More

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