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Combustion Engineering - Equilibrium
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Combustion Engineering - Equilibrium

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    Combustion Engineering - Equilibrium Combustion Engineering - Equilibrium Presentation Transcript

    • Chapter 5: Equilibrium Composition of Flames Name: Niranjwan Chettiar Roll No: 2010213041 Class: ME. Energy Engineering Teacher-In-Charge: Dr. Natarajan
    • What is Equilibrium?
      • Equilibrium – balancing of forces
        • Mechanical Equilibrium –
        • static or motionless
    • What is Equilibrium? (cont’d)
        • Thermal Equilibrium - When two objects are brought into contact, heat will flow from the warmer object to the cooler one until their temperatures become identical.
          • Eg: A metallic object at room
          • temperature will feel cool to
          • your hand when you first pick
          • it up
    • What is Equilibrium? (cont’d)
        • Chemical Equilibrium - is in equilibrium when there is no tendency for the quantities of reactants and products to change, as long as the system remains undisturbed.
        • Eg:
        • H 2 + ½ O 2 H 2 O (exothermic process)
        • H 2 O H 2 + ½ O 2 (endothermic process)
    • What is Equilibrium? (cont’d)
        • We are trying to avoid or postpone
          • Thermal Equilibrium - we insulate buildings, perspire in the summer and wear heavier clothing in the winter.
          • Chemical Equilibrium - living organisms are constantly moving toward equilibrium, but are prevented from getting there by input of reactants and removal of products. So we try to maintain a "steady-state" condition which physiologists call homeostasis
      
    • Chemical Equilibrium
      • Chemical Reaction – reversible
      • - a state of dynamic equilibrium
      • Important: The equilibrium state, there is no net change in the quantities of reactants and products. But the forward and reverse reactions continue, and at identical rates.
      • The Law of Mass Acton – states that the rate at which a substance reacts is proportional to its concentrations
        • Equilibrium is macroscopically static, but is microscopically dynamic!
    • Chemical Equilibrium (cont’d)
    • Chemical Equilibrium (cont’d)
    • Chemical Equilibrium (cont’d)
        • Let us consider:
        • A + B C + D
          • According to the Law of Mass Action:
          • The rate of forward reaction will be proportional to the concentration
          • The concentration of a gas is represented by its partial pressure, p
        • Forward Reaction
          • R + = K + p A p B
    • Chemical Equilibrium (cont’d)
        • Reverse Reaction
          • R - = K - p C p D
        • At Equilibrium, R + = R –
        • Dividing the two expressions gives
          • Where K is the equilibrium constant
    • Chemical Equilibrium (cont’d)
        • The number of molecules of each species, affects the form of the expression for the equilibrium constant
        • Then equilibrium constant
      A + B 2C
    • Chemical Equilibrium (cont’d)
        • Example
          • Equilibrium constant
          • in terms of moles
    • Calculation of the Equilibrium Composition
        • Consider the stoichiometric combustion of hydrogen in air:
        • H 2 + ½ O 2 + ½ (79/21) N 2 H 2 O + ½ (79/21) N 2
        • H 2 + ½ O 2 + 1.881 N 2 H 2 O + 1.881 N 2
          • Introduce, x as the fraction of the product, this means the flue gas will contain:
          • H 2 O: 1 – x moles O 2 : x/2 moles
          • H 2 : x moles N 2 : 1.881moles
    • Calculation of the Equilibrium Composition (cont’d)
          • Total number of mole products
          • 2.881 + x/2
          • Write expression for partial pressures in terms of system total pressure p T
    • Calculation of the Equilibrium Composition (cont’d)
          • Equilibrium constant for this reaction:
          • Substituting the partial pressures above gives:
    • Calculation of the Equilibrium Composition (cont’d)
          • After manipulating by squaring both sides and collecting the like terms in x gives
          • x 3 (1 – K 2 p T ) + 3.762x 2 – 10.524x + 5.672 = 0