Six Types of Critical Thinking Questions 1. Write a Critical Thinking Question for each type of CTQ

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Six Types of Critical Thinking Questions
1. Write a Critical Thinking Question for each type of CTQ for the Monsanto Incident, Example 13-2.
2. Write another question for each CTQ for the case history concerning ethylene oxide in the insulated storage tank.

Example 13-2

Adapted from the problem by Ronald Willey, Seminar on a Nitroaniline Reactor Rupture. Prepared for SAChE, Center for Chemical Process Safety, American Institute of Chemical Engineers, New York (1994). Also see Process Safety Progress, vol. 20, no. 2 (2001), pp. 123–129. The values of ΔHRx and UA were estimated from the plant data of the temperature–time trajectory in the article by G. C. Vincent, Loss Prevention, 5, 46–52. A serious accident occurred at the Monsanto plant in Sauget, Illinois, on August 8, 1969, at 12:18 A.M. (see Figure E13-2.1). (Sauget (pop. 200) is the home of the 1988 Mon-Clar League Softball Champions.) The blast was heard as far as 10 mi away in Belleville, Illinois, where people were awakened from a deep sleep. The explosion occurred in a batch reactor that was used to produce nitroaniline from ammonia and o-nitrochlorobenzene (ONCB):NO + 2NH3 NO -NH + NH4Cl

This reaction is normally carried out isothermally at 175°C and about 500 psi over a 24-hour period. The ambient temperature of the cooling water in the heat exchanger is 25°C. By adjusting the coolant flow rate, the reactor temperature could be maintained at 175°C. At the maximum coolant rate, the ambient temperature is 25°C throughout the heat exchanger. Also review the T2 Laboratories Safety Modules
(http://umich.edu/~safeche/assets/pdf/courses/Problems/344ReactionEngrModule(1)PS-T2.pdf).

Let me tell you something about the operation of this reactor. Over the years, the heat exchanger would fail from time to time, but the technicians would be “Johnny on the Spot” and run out and get it up and running within 10 minutes or so, and there was never any problem. It is believed that one day someone in management looked at the reactor and said, “It looks as if your reactor is only a third full and you still have room to add more reactants and to make more product and more money. How about filling it up to the top so we could triple production?” They did, and started the reactor up at 9:45 P.M. Around midnight the reactor exploded and the aftermath is shown in Figure E13-2.1. On the day of the accident, two changes in normal operation occurred.
1. The reactor was charged with 9.044 kmol of ONCB, 33.0 kmol of NH3, and 103.7 kmol of H2O. Normally, the reactor is charged with 3.17 kmol of ONCB, 103.6 kmol of H2O, and 43 kmol of NH3. A decision was made to triple production.

Figure E13-2.1image

2. The reaction is normally carried out isothermally at 175°C over a 24-hour period. However, approximately 45 minutes after the reaction was started, cooling to the reactor failed, but only for 10 minutes, after which cooling was again up and running at the 55-minute mark. Cooling may have been halted for 10 minutes or so on previous occasions when the normal charge of 3.17 kmol of ONCB was used and no ill effects occurred. The reactor had a rupture disk designed to burst when the pressure exceeded approximately 700 psi. If the disk would have ruptured, the pressure in the reactor would have dropped, causing the water to vaporize, and the reaction would have been cooled (quenched) by the latent heat of vaporization.
a. Plot and analyze the temperature–time trajectory up to a period of 120 minutes after the reactants were mixed and brought up to 175°C (448K).
b. Show that all of the following three conditions had to have been present for the explosion to occur: (1) increased ONCB charge, (2) cooling stopped for 10 minutes at a time early in the reaction, and (3) relief-system failure.
Additional information:Rate law: -FONCB = KCONCBCNH3 with k=0.00017m3kmo1 min at 188C (461 K) and E= 11273 cal/mol The reactionThe reaction volume for the previous 3.17 kmol ONCB charge was Assume that V = 3.9 m +0.42 m = 4.32 mAHRX=-5.9x105kcal/kmol CPONCBCPAX-X140x CPH20x=cPWx=18 Assume that ACp = 0 UA=35.85 kca1min Cx with Tax=298x

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