Lecture 8 - Stoichiometry Notes | EduRev

: Lecture 8 - Stoichiometry Notes | EduRev

 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)
  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
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