Batch Reactor Cooling Failure: How Much Runaway Potential Is Left
Your site runs a 12 m3 jacketed batch reactor making an acrylic copolymer. Monomer is dosed over three hours at 70 C with jacket cooling water at 12 C, and reaction calorimetry on the full charge gives an adiabatic temperature rise of 118 K. The calorimetry also flags a secondary decomposition with an onset of 145 C and a self-heating rate of 40 K/min at 170 C. The reactor design pressure is 6 barg with a PSV set at 5.5 barg, sized for an external fire only. At 80% of the monomer dose, the cooling water pump trips and header pressure collapses. The reactor is at 78 C and climbing at 1.2 K/min. Operators can inject a shortstop from a 2 m3 tank, but the injection line is rated only to 60 C, and the vessel has a nitrogen blanket at 0.2 barg. The agitator is still running. The shift supervisor must decide within ten minutes.
1. Using the calorimetry data, assess the severity of this cooling failure. Calculate the maximum temperature the mass could reach under two assumptions: (a) the feed is stopped immediately with 25% of the full-charge potential still accumulated, and (b) the feed valve fails open and the rest of the charge reacts. State the implication for the relief system.
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Severity assessment
The relevant comparison is MTSR against the decomposition onset. MTSR is the temperature reached if the accumulated unreacted material reacts with cooling lost; Tonset is where the secondary reaction starts to self-heat.
Case (a): feed stopped, 25% of the total charge potential still accumulated. MTSR = 78 + 0.25 x 118 = 107.5 C. Decomposition onset is 145 C, so there is about 37 K of margin. That is not comfortable, but the scenario is containable if the feed really stops and the jacket goes to maximum cooling.
Case (b): the feed valve fails open and the rest of the charge reacts. Remaining potential = 0.8 x 118 = 94 K, because about 80% of the charge is still unreacted at the trip. MTSR = 78 + 94 = 172 C. That is 27 K above the onset, and the calorimetry shows 40 K/min self-heating at 170 C. Once the mass passes 145 C, the decomposition, not the polymerization, controls the pressure rise.
Implication for the relief system. The PSV is set at 5.5 barg and sized for an external fire. It was not sized for a decomposition gas and vapor load, and a runaway at 172 C would generate far more gas than the fire case. The relief system cannot be counted as the safeguard here; the safeguard has to be stopping the feed and removing heat. On the usual severity classification, MTSR above Tonset means this is a critical scenario, and the process is not inherently safe as designed.
The conclusion is that the answer to the supervisor's question depends entirely on which assumption about the feed valve is true. That is why the feed isolation must be verified, not assumed.
2. Give your immediate actions in order for the next ten minutes, and justify each. Include what you would do about the agitator, the nitrogen blanket, and the shortstop injection.
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First ten minutes
- Stop and physically verify the monomer feed. Close the control valve, close the upstream block valve, and confirm with the flow indication and the dosing pump status. Do not accept the controller output as proof. Every minute of continued dosing adds roughly 1.5 K to the potential MTSR. This is the single highest-leverage action.
- Put the jacket at maximum cooling and start the emergency water supply. Confirm the cooling water valve is fully open mechanically, not just at 100% signal, and check the jacket outlet temperature. If the jacket is already at maximum and the reactor is still climbing at 1.2 K/min, heat removal is already below heat generation and you are in the case (a) or worse scenario.
- Keep the agitator running. Do not stop it to reduce heat input. Agitation keeps the mass homogeneous and lets the jacket remove heat. Losing agitation creates a hot, uninhibited layer at the wall and makes any shortstop ineffective.
- Keep the nitrogen blanket and keep the vessel closed. Do not vent aggressively to reduce pressure; venting can draw air into a vessel containing flammable monomer vapor, and it removes the inert cushion you may need later.
- Prepare the shortstop, but treat the 60 C line rating as real. The line cannot be injected at 78 C without exceeding its rating. Set a trigger: if the temperature passes 85 C, or the self-heating rate rises despite full cooling, evacuate the area and inject. A runaway past 145 C is a far larger hazard than a leaking injection flange, but injecting routinely at 78 C through an under-rated line is not a controlled decision.
- Evacuate non-essential personnel, call the emergency team, and prepare to isolate the reactor from the monomer header.
Do not rely on the PSV. It is a fire-case device and it is not the safeguard for this scenario.
3. What change to the process would you recommend to make this reaction inherently safer, and how would you demonstrate the improvement with a calorimetry number?
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Make accumulation the controlled variable
The hazard is not the reaction heat itself; it is the unreacted monomer inventory that can still react after cooling is lost. The design target is to keep MTSR below Tonset with a documented margin at every point in the batch.
With Tp = 70 C, delta Tad = 118 K and Tonset = 145 C, the allowable accumulated fraction is: f less than (145 - 70)/118 = 0.64 for zero margin. With a 25 K margin (MTSR below 120 C): f less than (120 - 70)/118 = 0.42.
So the dosing rate must be limited so that no more than about 40% of the full-charge potential is ever present as unreacted monomer. In practice:
- Semi-batch dosing with the feed rate on cascade control from reactor heat release or temperature rise, so the feed slows automatically when the reaction slows.
- A smaller initial charge or a pre-reacted seed to shorten the accumulation peak.
- A high-temperature trip that closes the monomer feed at, say, 80 C, independent of the control system.
How to demonstrate the improvement. Run the same calorimetry with the new dosing profile and report the measured accumulation at the worst point, which is normally the end of dosing or the feed trip. Show that the resulting MTSR is at least 25 K below the decomposition onset. The number for the safety case is MTSR = Tp + f_measured x delta Tad. If f_measured is 0.35, MTSR = 70 + 41 = 111 C, which is 34 K below onset, and the scenario moves from critical to acceptable with a margin you can defend.