Process Safety · Worked exercises

Worked exercises

Four plant-style problems that combine the lessons. Try them on paper first, then open the solutions. The numbers are invented for teaching.

Educational overview. The principles here are general and simplified, and the figures and rules of thumb are typical values, not limits for your plant. Decisions need process-specific data, the applicable codes and standards, and a qualified assessment.

Exercise 1 · Shortest dosing time for a semi-batch reaction

A + B → P is to run semi-batch at 60 °C in toluene (boiling point 111 °C). In the reaction calorimeter, with 1.2 kg of final reaction mass, B was dosed over 2.5 h and the heat flow stayed roughly constant at 35 W during the addition. c′p = 1.85 kJ/kg·K. In the plant, the final mass is 3,000 kg; the jacket gives U = 250 W/m²·K over A = 5.5 m², with brine at 10 °C. The calorimeter showed a maximum accumulation of 25 %; ARC gives TD24 = 150 °C for the final mass.

  1. What is the specific heat of reaction and ΔTad?
  2. What is the shortest dosing time in the plant, with a 20 % cooling margin?
  3. What are the MTSR and the Stoessel class? What if all of B were charged at once?
Show the solution

1. Qrx = 35 W × 2.5 h × 3600 s/h = 315 kJ for 1.2 kg, so Q′rx = 262.5 kJ/kg. ΔTad = 262.5 / 1.85 = 142 K (medium severity).

2. Total heat = 262.5 × 3000 = 787.5 MJ. Cooling capacity qex = 250 × 5.5 × (60 − 10) = 68.75 kW; with the 20 % margin, 57.3 kW usable. tmin = 787,500 / 57.3 = 13,745 s ≈ 3.8 h → dose over 4 h.

3. MTSR = 60 + 0.25 × 142 = 95 °C. Order: Tp 60 < MTSR 95 < MTT 111 < TD24 150 → 1 Class 1.
All at once: Xacc = 100 %, MTSR = 60 + 142 = 202 °C, above both the boiling point and TD24: Class 5. Also, the whole heat would have to be removed at once, far beyond 69 kW. The same chemistry, run as a batch, is not acceptable.

Check it with the dosing-time calculator and the class calculator.

Exercise 2 · Reading DSC results

An alkylation is to be run at 70 °C in a solvent boiling at 150 °C. c′p = 1.75 J/g·K. DSC (gold-plated high-pressure crucible, 4 K/min) gives:

  1. Interpret the signals. What is the severity of the decomposition?
  2. The chemists propose an all-in batch heated to 70 °C. What is the worst-case temperature?
  3. The DSC onset of the decomposition is 230 °C, so “140 K of margin”. Is that enough to approve the batch process?
Show the solution

1. The first exotherm is the desired reaction (Q′rx = 280 J/g → ΔTad,rx = 160 K). The second, present in both samples, is the decomposition of the reaction mass: ΔTad,dc = 850 / 1.75 ≈ 486 K, high severity (and > 500 J/g is close, so check for explosive properties).

2. All-in batch: Xacc = 1, MTSR = 70 + 160 = 230 °C. That is above the boiling point (150 °C) and right at the decomposition. If it is triggered: 70 + 160 + 486 ≈ 716 °C in theory, with violent gas and vapour release.

3. No. A DSC onset at 4 K/min is not a safe limit; in a 5 m³ near-adiabatic reactor the decomposition can run away from far lower temperatures. And the MTSR (230 °C) already reaches the onset. Get TD24 from isothermal DSC or ARC, and redesign as a semi-batch with controlled dosing (see exercise 1).

Exercise 3 · TMRad and TD24 from ARC data

An ARC test of the final reaction mass shows a self-heating rate corresponding to 0.5 W/kg at 150 °C. From a series of isothermal tests, Ea = 100 kJ/mol. c′p = 1.9 kJ/kg·K. The process gives MTSR = 120 °C, and the solvent boils at 140 °C.

  1. What is TMRad at 150 °C?
  2. Estimate TD24 and TD8.
  3. What is TMRad at the MTSR, and the Stoessel class?
Show the solution

1. TMRad = c′p·R·T²/(q′·Ea) = 1900 × 8.314 × 423.15² / (0.5 × 100,000) ≈ 56,600 s ≈ 15.7 h.

2. Extrapolating q′ with Arrhenius and solving TMRad = 24 h and 8 h gives TD24 ≈ 143 °C and TD8 ≈ 161 °C.

3. At 120 °C, TMRad ≈ 119 h: low probability. Order: MTSR 120 < MTT 140 < TD24 143: 1 Class 1, but with only 3 K between boiling point and TD24. Any change that raises the boiling point (pressure, a different solvent) would make it Class 2, and a hold at reflux would eat into the margin.

Try it in the TMRad calculator: T0 150 °C, q′0 0.5 W/kg, Ea 100, c′p 1.9.

Exercise 4 · Fire-case relief for a solvent vessel

A vertical receiver (D = 1.8 m, 2:1 elliptical bottom head) holds methanol up to 2.2 m in the shell. It is bare (no fire-proof insulation) and the area has no prompt fire-fighting. The relief valve is set at 3.5 barg. Methanol: λ = 1,100 kJ/kg, M = 32, k ≈ 1.2, relieving temperature about 64 °C (use this simplified value), Z = 0.95.

  1. Wetted area and heat input?
  2. Relief rate?
  3. Required relief area and diameter (21 % overpressure, Kd 0.975)?
Show the solution

1. Aw = π·D·h + 1.084·D² = π × 1.8 × 2.2 + 1.084 × 1.8² ≈ 16.0 m². Q = 70,900 × 1 × 16.00.82 ≈ 687 kW.

2. W = 687 kW / 1,100 kJ/kg = 0.625 kg/s ≈ 2,250 kg/h.

3. P1 = 350 × 1.21 + 101.3 ≈ 525 kPa(a). C = 0.03948·√(k·(2/(k+1))(k+1)/(k−1)) ≈ 0.0256. A = W/(C·Kd·P1)·√(T·Z/M) ≈ 543 mm², about 26 mm diameter: select the next standard orifice (API letter “F”, 198 mm², is too small; “H”, 506 mm², is just short; “J”, 830 mm², fits).

Check it with the relief calculators.

← LESSON 9Toxic and flammable releasesNEXT →Process safety quiz