MS/PhD Reactor Safety: Throughput Increase and Runaway Risk
A chemical engineering MS candidate reviews a proposal to increase throughput in a 2 m³ jacketed CSTR. The reactor processes an aqueous first-order exothermic reaction A → B. Current feed is 1.0 m³ h⁻¹ with C_A0 = 2.0 kmol m⁻³ at 300 K. At the desired 80% conversion the reactor operates at 350 K. The reaction enthalpy is −80 kJ mol⁻¹, the activation energy is 60 kJ mol⁻¹, and the mixture has ρ = 1000 kg m⁻³ and c_p = 4.0 kJ kg⁻¹ K⁻¹. The jacket uses coolant at 300 K and at maximum flow provides UA = 1.0 kW K⁻¹. The proposal raises C_A0 to 3.0 kmol m⁻³ while keeping feed rate, conversion, and coolant conditions unchanged. The candidate must assess the steady-state and runaway risks, not just the extra heat release.
1. Calculate the heat release and adiabatic temperature rise for the current and proposed feeds at 80% conversion.
Read the worked answer
Current feed: F_A0 = 1.0 × 2.0 = 2.0 kmol h⁻¹. At 80% conversion, 1.6 kmol h⁻¹ reacts. Q_dot = 1.6 × 80,000 = 128,000 kJ h⁻¹ = 35.6 kW. ΔT_ad = Q_dot / (ρ Q c_p) = 128,000 / [1000(1.0)(4.0)] = 32 K.
Proposed feed: F_A0 = 3.0 kmol h⁻¹, reacted = 2.4 kmol h⁻¹. Q_dot = 2.4 × 80,000 = 192,000 kJ h⁻¹ = 53.3 kW. ΔT_ad = 48 K.
The adiabatic rise is 50% larger.
2. At 350 K, what is the maximum heat removal rate? Can the proposed feed be steady at 350 K? If not, predict the direction of temperature change and explain the feedback.
Read the worked answer
At 350 K with coolant at 300 K: Q_rem = UA(350 − 300) = 1.0(50) = 50 kW.
Current generation 35.6 kW < 50 kW, so at 350 K the jacket has excess capacity; coolant flow would be throttled. Proposed generation 53.3 kW > 50 kW, so 350 K cannot be steady at full cooling. The temperature must rise.
As T rises, the Arrhenius rate constant increases, so heat generation rises roughly exponentially, while removal rises only linearly with T. The slope of generation at 350 K is approximately Q_gen E_a/(R T²) = 53.3(60,000)/[8.314(350)²] ≈ 3.1 kW K⁻¹, compared with the removal slope UA = 1.0 kW K⁻¹. Generation grows faster than removal, so the operating point is unstable and runaway is possible.
3. Recommend one operational change that preserves the throughput increase but restores a stable operating margin. Quantify the required UA or coolant temperature, and state one trade-off.
Read the worked answer
To keep the higher throughput, increase heat-transfer capacity rather than simply lowering coolant temperature. At 350 K, removing 53.3 kW with 300 K coolant needs UA ≥ 1.07 kW K⁻¹, but local stability needs UA above the generation slope, about 3.1 kW K⁻¹. So a larger jacket area or higher coolant velocity is required.
Alternatively, use staged feed or feed dilution to keep local C_A and heat release lower. Trade-off: a larger UA means more pump power and capital cost; staged feed adds control complexity and may reduce conversion per pass. Simply raising C_A0 with the existing jacket is not acceptable.