Combustion Engineering - Equilibrium

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Combustion Engineering - Equilibrium

  1. 1. Chapter 5: Equilibrium Composition of Flames Name: Niranjwan Chettiar Roll No: 2010213041 Class: ME. Energy Engineering Teacher-In-Charge: Dr. Natarajan
  2. 2. What is Equilibrium? <ul><li>Equilibrium – balancing of forces </li></ul><ul><ul><li>Mechanical Equilibrium – </li></ul></ul><ul><ul><li>static or motionless </li></ul></ul>
  3. 3. What is Equilibrium? (cont’d) <ul><ul><li>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. </li></ul></ul><ul><ul><ul><li>Eg: A metallic object at room </li></ul></ul></ul><ul><ul><ul><li>temperature will feel cool to </li></ul></ul></ul><ul><ul><ul><li>your hand when you first pick </li></ul></ul></ul><ul><ul><ul><li>it up </li></ul></ul></ul>
  4. 4. What is Equilibrium? (cont’d) <ul><ul><li>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. </li></ul></ul><ul><ul><li>Eg: </li></ul></ul><ul><ul><li>H 2 + ½ O 2 H 2 O (exothermic process) </li></ul></ul><ul><ul><li> H 2 O H 2 + ½ O 2 (endothermic process) </li></ul></ul>
  5. 5. What is Equilibrium? (cont’d) <ul><ul><li>We are trying to avoid or postpone </li></ul></ul><ul><ul><ul><li>Thermal Equilibrium - we insulate buildings, perspire in the summer and wear heavier clothing in the winter. </li></ul></ul></ul><ul><ul><ul><li>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 &quot;steady-state&quot; condition which physiologists call homeostasis </li></ul></ul></ul>
  6. 6. Chemical Equilibrium <ul><li>Chemical Reaction – reversible </li></ul><ul><li> - a state of dynamic equilibrium </li></ul><ul><li>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. </li></ul><ul><li>The Law of Mass Acton – states that the rate at which a substance reacts is proportional to its concentrations </li></ul><ul><ul><li>Equilibrium is macroscopically static, but is microscopically dynamic! </li></ul></ul>
  7. 7. Chemical Equilibrium (cont’d)
  8. 8. Chemical Equilibrium (cont’d)
  9. 9. Chemical Equilibrium (cont’d) <ul><ul><li>Let us consider: </li></ul></ul><ul><ul><li>A + B C + D </li></ul></ul><ul><ul><ul><li>According to the Law of Mass Action: </li></ul></ul></ul><ul><ul><ul><li>The rate of forward reaction will be proportional to the concentration </li></ul></ul></ul><ul><ul><ul><li>The concentration of a gas is represented by its partial pressure, p </li></ul></ul></ul><ul><ul><li>Forward Reaction </li></ul></ul><ul><ul><ul><li>R + = K + p A p B </li></ul></ul></ul>
  10. 10. Chemical Equilibrium (cont’d) <ul><ul><li>Reverse Reaction </li></ul></ul><ul><ul><ul><li>R - = K - p C p D </li></ul></ul></ul><ul><ul><li>At Equilibrium, R + = R – </li></ul></ul><ul><ul><li>Dividing the two expressions gives </li></ul></ul><ul><ul><ul><li>Where K is the equilibrium constant </li></ul></ul></ul>
  11. 11. Chemical Equilibrium (cont’d) <ul><ul><li>The number of molecules of each species, affects the form of the expression for the equilibrium constant </li></ul></ul><ul><ul><li>Then equilibrium constant </li></ul></ul>A + B 2C
  12. 12. Chemical Equilibrium (cont’d) <ul><ul><li>Example </li></ul></ul><ul><ul><ul><li>Equilibrium constant </li></ul></ul></ul><ul><ul><ul><li>in terms of moles </li></ul></ul></ul>
  13. 13. Calculation of the Equilibrium Composition <ul><ul><li>Consider the stoichiometric combustion of hydrogen in air: </li></ul></ul><ul><ul><li>H 2 + ½ O 2 + ½ (79/21) N 2 H 2 O + ½ (79/21) N 2 </li></ul></ul><ul><ul><li>H 2 + ½ O 2 + 1.881 N 2 H 2 O + 1.881 N 2 </li></ul></ul><ul><ul><ul><li>Introduce, x as the fraction of the product, this means the flue gas will contain: </li></ul></ul></ul><ul><ul><ul><li>H 2 O: 1 – x moles O 2 : x/2 moles </li></ul></ul></ul><ul><ul><ul><li>H 2 : x moles N 2 : 1.881moles </li></ul></ul></ul>
  14. 14. Calculation of the Equilibrium Composition (cont’d) <ul><ul><ul><li>Total number of mole products </li></ul></ul></ul><ul><ul><ul><li>2.881 + x/2 </li></ul></ul></ul><ul><ul><ul><li>Write expression for partial pressures in terms of system total pressure p T </li></ul></ul></ul>
  15. 15. Calculation of the Equilibrium Composition (cont’d) <ul><ul><ul><li>Equilibrium constant for this reaction: </li></ul></ul></ul><ul><ul><ul><li>Substituting the partial pressures above gives: </li></ul></ul></ul>
  16. 16. Calculation of the Equilibrium Composition (cont’d) <ul><ul><ul><li>After manipulating by squaring both sides and collecting the like terms in x gives </li></ul></ul></ul><ul><ul><ul><li>x 3 (1 – K 2 p T ) + 3.762x 2 – 10.524x + 5.672 = 0 </li></ul></ul></ul>

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