Shortening a Semi-Batch Feed Creates a Runaway Risk
A semi-batch reactor with 8 m³ working volume carries out an exothermic reaction A → B in solution. The feed contains 2500 mol/m³ of A, has density 950 kg/m³ and cp 3.8 kJ/kg·K, and the heat of reaction is −180 kJ/mol. The vessel is charged with 2 m³ at the start and fed the remaining 6 m³ over 4 h while held at 65°C by a jacket with 18 m² of area and U = 350 W/m²·K; coolant enters at 20°C. The reaction is fast at 65°C, but the existing procedure relies on feed rate to limit accumulation. Differential scanning calorimetry shows an exothermic decomposition onset at 150°C. Production wants to shorten the feed to 3 h. The high-temperature trip is set at 110°C, the emergency cooling water supply is reliable, and the relief system was last sized for the original 4 h recipe.
1. Calculate the full-charge adiabatic temperature rise and the maximum safe fraction of unreacted A accumulated at 65°C, using a 20 K margin below the decomposition onset.
Read the worked answer
Total A in the full 8 m³ charge:
nA = 2500 × 8 = 20,000 mol.
Total heat release = 20,000 × 180 = 3.6 × 10⁶ kJ = 3.6 GJ.
Reaction mass = 8 × 950 = 7600 kg.
Heat capacity = 7600 × 3.8 = 28,880 kJ/K.
ΔTad = 3.6 × 10⁶/28,880 = 124.7 K.
If all A accumulated unreacted at 65°C and cooling were lost, the maximum temperature reached would be 65 + 124.7 = 189.7°C, which is 39.7 K above the 150°C decomposition onset. That is unacceptable.
To keep the maximum temperature at least 20 K below onset, the allowable rise is 150 − 20 − 65 = 65 K. The maximum accumulated fraction is therefore 65/124.7 = 0.52. More than about half of the total A charge must not be allowed to accumulate unreacted. The 3 h feed increases the risk because the feed rate rises by one-third; the operating procedure must demonstrate accumulation below this limit, not merely rely on a faster feed.
2. Compare the average heat-release rate at 4 h and 3 h feeds with the jacket's heat-removal capacity at 65°C.
Read the worked answer
At the 4 h feed, average heat release is 3.6 GJ/14,400 s = 250 kW. At the 3 h feed, it is 3.6 GJ/10,800 s = 333 kW.
Jacket capacity at 65°C with 20°C coolant:
Qjacket = U A ΔT = 350 × 18 × (65 − 20) = 283.5 kW.
At 4 h, the 250 kW release is below the 283.5 kW capacity, so the existing recipe has a heat-removal margin at 65°C. At 3 h, the 333 kW average release exceeds jacket capacity by about 50 kW even if the reaction keeps up with feed and no accumulation occurs. The reactor would heat up until the jacket ΔT increases: at 75°C, capacity is 350 × 18 × 55 = 346.5 kW. A steady state near 74°C might be possible, but that assumes perfect mixing and immediate reaction. Any accumulation would push the temperature higher and increase the decomposition risk. For 3 h at 65°C, the required coolant temperature would be 65 − 333,000/(350 × 18) = 12.1°C, colder than the available 20°C.
3. Recommend a safe operating limit and the safeguards needed before any 3 h feed is attempted.
Read the worked answer
The jacket at 65°C and 20°C coolant can remove 283.5 kW. If the entire 3.6 GJ charge must be removed at that rate, the minimum feed time is 3.6 × 10⁶/283.5 = 12,700 s, or about 3.5 h. A 3 h feed is therefore not justified on the existing jacket unless cooling is improved (colder coolant, larger area, higher U) or the reaction enthalpy/charge is reduced.
Before changing the recipe: verify accumulation calorimetrically or by heat balance; add a feed interlock that trips on high temperature or low coolant flow; lower the high-temperature trip so it acts before 150°C; provide emergency cooling and a quench; confirm mixing and jacket response; and resize the relief system for the shorter feed and for decomposition scenarios, because the original relief basis was the 4 h recipe. The relief system must handle the worst credible case, including full accumulation followed by loss of cooling.