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=29,1°C. 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.

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).

Molecular weight of the monomer: MW = 100 g/mol

Specific heat of water: CPW = 4,2 kJ/(kg·K)

Specific heat of MMA: CPM = 2 kJ/(kg·K)

 

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To prove that the reaction is thermally stable and will not undergo a thermal runaway, we need to analyze the heat balance and the temperature dependence of the reaction rate.

 

Given information:

- Monomer feed rate (ṁM) = 180 kg/h

- Water feed rate (ṁW) = 270 kg/h

- Inlet temperature (Ti) = 29.1°C

- Reaction temperature (Tr) = 85°C

- Heat of polymerization (Q) = 58 kJ/mol

- Monomer conversion (X) = 80%

- Activation energy (Ea) = 50 kJ/mol

- Molecular weight of the monomer (MW) = 100 g/mol

- Specific heat of water (CPW) = 4.2 kJ/(kg·K)

- Specific heat of MMA (CPM) = 2 kJ/(kg·K)

 

Step 1: Calculate the heat generated by the reaction.

Heat generated = Monomer feed rate × Conversion × Heat of polymerization

Heat generated = (180 kg/h × 0.8 × 58 kJ/mol) / (100 g/mol) = 83.04 kW

 

Step 2: Calculate the heat absorbed by the feed streams.

Heat absorbed = (ṁM × CPM + ṁW × CPW) × (Tr - Ti)

Heat absorbed = (180 kg/h × 2 kJ/(kg·K) + 270 kg/h × 4.2 kJ/(kg·K)) × (85°C - 29.1°C)

Heat absorbed = 83.16 kW

 

Step 3: Analyze the temperature dependence of the reaction rate.

The reaction rate is temperature-dependent and can be described by the Arrhenius equation:

k = A × exp(-Ea/RT)

where:

- k is the reaction rate constant

- A is the pre-exponential factor

- Ea is the activation energy

- R is the universal gas constant

- T is the absolute temperature

 

At the reaction temperature (Tr = 85°C = 358.15 K), the reaction rate constant can be calculated as:

k = A × exp(-Ea/RT)

k = A × exp(-50,000 J/mol / (8.314 J/(mol·K) × 358.15 K))

k = A × exp(-1.69)

 

At a higher temperature (T = Tr + ΔT), the reaction rate constant would be:

k' = A × exp(-Ea/R(Tr + ΔT))

k' = A × exp(-1.69 - Ea/(R × Tr) × ΔT/Tr)

 

The temperature increase (ΔT) would lead to a higher reaction rate, but the heat absorbed by the feed streams would also increase, stabilizing the temperature at 85°C.

 

Since the heat generated by the reaction (83.04 kW) is approximately equal to the heat absorbed by the feed streams (83.16 kW), the reaction is thermally stable and will not undergo a thermal runaway.

 

Therefore, the reaction is thermally stable and will stabilize itself at the reaction temperature of 85°C.