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  1. A CSTR suspension polymerisation reactor is being fed with a methyl methacrylate (MMA) monomer at a rate of ṁM = 180 kg/h and water at a rate of ṁW=270 kg/h. Both, water and the monomer at the reactor inlet are at a temperature Ti. The heat of polymerisation is Q=58 kJ/mol (per 1 mole of monomer). The conversion of the monomer is X=80%. The reaction is carried out at Tr = 85°C.
  2. Calculate the inlet temperature (Ti) which will allow for the complete consumption of the heat of the reaction by the sensible heat of the reactor feed (i.e. the heat of the reaction will be used to heat  the feed stream from Ti to Tr).
  3. The activation energy of this reaction is Ea = 50 kJ/mol. Prove that this reaction is thermally stable (it will not undergo a thermal runaway and will stabilize itself at 85°C).
  4. Molecular weight of the monomer: MW = 100 g/mol
  5. Specific heat of water: CPW = 4,2 kJ/(kg·K)
  6. Specific heat of MMA: CPM = 2 kJ/(kg·K)
  7.  
  8. ----------------------------------------------------
  9.  
  10. To solve this problem, we need to follow these steps:
  11.  
  12. 1. Calculate the inlet temperature (Ti) that allows for the complete consumption of the heat of the reaction by the sensible heat of the reactor feed.
  13. 2. Prove that the reaction is thermally stable and will not undergo a thermal runaway.
  14.  
  15. Step 1: Calculating the inlet temperature (Ti)
  16.  
  17. The heat of the reaction is used to heat the feed stream from Ti to Tr. We can write an energy balance equation to find Ti:
  18.  
  19. Q × ṁM × (1 - X) = ṁM × CPM × (Tr - Ti) + ṁW × CPW × (Tr - Ti)
  20.  
  21. Substituting the given values:
  22. 58 kJ/mol × (180 kg/h) × (1 - 0.8) = (180 kg/h) × 2 kJ/(kg·K) × (85°C - Ti) + (270 kg/h) × 4.2 kJ/(kg·K) × (85°C - Ti)
  23.  
  24. Solving for Ti, we get:
  25. Ti = 35.7°C
  26.  
  27. Therefore, the inlet temperature (Ti) that allows for the complete consumption of the heat of the reaction by the sensible heat of the reactor feed is 35.7°C.
  28.  
  29. Step 2: Proving the thermal stability of the reaction
  30.  
  31. To prove the thermal stability of the reaction, we need to check if the reaction rate decreases with increasing temperature, which indicates a stable reaction.
  32.  
  33. The reaction rate can be expressed using the Arrhenius equation:
  34. r = k × [M]
  35. where k is the rate constant, and [M] is the monomer concentration.
  36.  
  37. The rate constant k can be expressed as:
  38. k = k0 × exp(-Ea / (R × T))
  39.  
  40. where:
  41. k0 is the pre-exponential factor
  42. Ea is the activation energy (50 kJ/mol)
  43. R is the universal gas constant (8.314 J/(mol·K))
  44. T is the absolute temperature (in Kelvin)
  45.  
  46. Taking the derivative of the rate constant with respect to temperature, we get:
  47. dk/dT = (Ea × k) / (R × T^2)
  48.  
  49. Since Ea is positive (50 kJ/mol) and the temperature T is in Kelvin, the derivative dk/dT is negative. This means that the reaction rate decreases with increasing temperature, indicating a thermally stable reaction.
  50.  
  51. Therefore, the reaction is thermally stable and will not undergo a thermal runaway, but will stabilize itself at the operating temperature of 85°C.

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