Example 13–1 Adiabatic Batch Reactor

It is still winter, and although you were hoping for a transfer to the plant in the tropical southern coast of Jofostan, unfortunately you are still the engineer of the CSTR from Example 12-3, in charge of the production of propylene glycol.

You are considering the installation of a new, attractive-looking, glass-lined 1,000-dm³ CSTR, and you decide to make a quick check of the reaction kinetics and maximum adiabatic temperature. You have a stylish, nicely decorated and instrumented 40-dm³ (~10 gal) stirred batch reactor you ordered from a company in Jofostan. You charge this reactor with 4 dm³ (~1 gal) of ethylene oxide, 4 dm³ (~1 gal) of methanol, and 10 dm³ (~2.5 gal) of water containing 0.1 wt % H₂SO₄. For safety reasons, the reactor is located on a boating pier on the banks of Lake Wobegon (you don’t want the entire plant to be destroyed if the reactor explodes). At this time of year in northern Minnesota, the initial temperature of all materials is 276 K (3˚C). We have to be careful here! If the reactor temperature increases above 350 K (77˚C), a secondary, more exothermic reaction will take over, causing runaway and subsequent explosion, similar to what happened in the T-2 Laboratory plant explosion in Florida, see page 660.

Although you requested obtaining the data for this reaction from the Jofostan national laboratory, the purchasing department decided to save money and buy it off the Internet. The values it purchased are

The concentration of pure ethylene oxide and methanol are 13.7 mol/dm³ and 24.7 mol/dm³ respectively. Consequently, the initial number of moles added to the reactor are

The sulfuric acid catalyst takes up negligible space, so the total volume is 18 dm³. Use the data and the reaction-rate law given in Example 12-3. We are going to carry out two scenarios: (1) to learn how fast the temperature rises and how long it takes to reach 350 K for adiabatic operation, and (2) how long would it take to reach 345 K if we added a heat exchanger.

(a) Plot conversion and temperature X and T as a function of time for adiabatic operation. How many minutes should it take the mixture inside the reactor to reach a conversion of 51.5%? What is the corresponding adiabatic temperature?

(b) Plot the temperature and conversion when the heat exchange is added. The product of the overall heat-transfer coefficient and area is UA = 10 cal/s/K with T_a1 = 290 K and the coolant rate is 10 g/s, and it has a heat capacity of 4.16 cal/g/K.

Solution

The initial temperature is 3˚C, and if the reactor temperature increases to above 77˚C, a dangerous exothermic side reaction can occur, as was reported in Jofostan Journal of Chemical Safety, Vol. 19, p. 201 (1988).

As before, there is approximately a 10˚C (17˚F) rise in temperature immediately after mixing. The activation energy is 32,400 cal/gmol • K.

1. Mole Balance, Chapter 2, rearranging Equation (2-6) we have

2. Rate Law:

3. Stoichiometry:

Recall that for liquid batch V = V₀

4. Combining Equations (E13-1.1), (E13-1.2), and (2-6), we have

Changing the data in Example 12-3 from English units to mks units, we have

Choosing 297 K as a reference temperature for k₁, and putting Equation (E13-1.5) in the form of Equation (3-21), we get

5. Energy Balance.

Part (a) Adiabatic Operation.

Using Equation (13-19), the relationship between X and T for an adiabatic reaction is given by

6. Parameters:

The lumped heat capacity of the solution is

In the expression for the heat of reaction, we are going to neglect the second term on the right-hand side, i.e.,

We can do this because ΔC_P is –7 cal/mol/K and for a 50-K temperature difference, ΔC_P (T–T_R) = 350 cal/mol, which is negligible with respect to the heat of reaction of –20,202 or cal/mol

In calculating the inlet temperature after mixing, T₀, we must include the temperature rise 10˚C (17˚F) resulting from the heat of mixing the two solutions initially at 3˚C

Adiabatic Energy Balance

A summary of the heat and mole balance equations is given in Table E13-1.1.

TABLE E13-1.1 SUMMARY FOR FIRST-ORDER ADIABATIC BATCH REACTION

The software package Polymath will be used to combine Equations (E13-1.4), (E13-1.6), and (E13-1.11) to determine conversion and temperature as a function of time. Table E13-1.2 shows the program, and Figures E13-1.1 and E13-1.2 show the solution results.

TABLE E13-1.2 POLYMATH PROGRAM

Adiabatic Batch Operation

Figure E13-1.1 Temperature–time trajectory.

Figure E13-1.2 Conversion–time trajectory.

Part (b) Heat Exchanger in a Batch Reactor.

We now consider the case where a heat exchanger is added to the batch reactor. The coolant enters at 290 K and the flow rate, , through the exchanger is 10 g/s. The mole balance, Equation (E13-1.1), and the physical properties, Equations (E13-1.2) through (E13-1.6) and (E13-1.8), remain the same.

Mole Balance

Neglecting ΔC_P the energy balance is

Energy Balance

with

The Polymath Program is shown in Table E13-1.3 with parameters

and the other terms remain the same as in Part (a).

TABLE E13-1.3 POLYMATH PROGRAM

Figure E13-1.3 Temperature–time trajectory.

Figure E13-1.4 Conversion–time trajectory.

Figure E13-1.5 Specific reaction-rate trajectory.

Figure E13-1.6 Heat-generated as heat removed trajectories.

One notes from Figure E13-1.3(a) and (b) that the temperature profiles are not as steep as the adiabatic case. The specific reactor-rate constant, Figure E13-1.5, goes through a maximum as the fluid becomes heated at first then later cools. Figure E13-1.6 shows the trajectories of the “heat generated,” Q_g, and “heat removed,” Q_r. Whenever Q_g > Q_r

the temperature will increase when Q_g > Q_r and the temperature will decrease when Q_r > Q_g.

Analysis:

(a) Adiabatic. The initial temperature is rather low, so the reaction is slow at first. However, as the exothermic reaction proceeds, it heats up and becomes virtually autocatalytic as it goes from a small conversion at 1600 seconds to complete conversion just a few seconds later. The conversion reaches X = 1.0 and the temperature reaches its maximum value, where it remains.

(b) Heat Exchange. Because the temperature is maintained at a lower value in the heat exchange case than in the adiabatic case, there will be less conversion for the parameter values Ua, T_ai, T₀, , etc., given in the problem statement. However, if you change these values as suggested in Problem P13-1_A (a), you will find situations where the conversion remains very low and cases where the temperature curve is extremely steep.

Example 13–2 Safety in Chemical Plants with Exothermic Runaway Reactions²

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 ΔH_Rx 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 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 miles away in Belleville, Illinois, where people were awakened from their sleep. The explosion occurred in a batch reactor that was used to produce nitroaniline from ammonia and o-nitrochlorobenzene (ONCB):

This reaction is normally carried out isothermally at 175°C and about 500 psi. The ambient temperature of the cooling water in the heat exchanger is 25°C. By adjusting the coolant 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.

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.

Figure E13-2.1 Aftermath of the explosion. (St. Louis Globe/Democrat photo by Roy Cook. Courtesy of St. Louis Mercantile Library.)

A decision was made to triple production.

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.

2. The reaction is normally carried out isothermally at 175°C over a 24-h period. However, approximately 45 min after the reaction was started, cooling to the reactor failed, but only for 10 min, and cooling was again up and running at the 55-minute mark. Cooling may have been halted for 10 min 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 min 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 min at a time early in the reaction, and (3) relief-system failure.

Additional information:

Rate law: –r_ONCB = kC_ONCBC_NH3

with and E = 11,273 cal/mol

The reaction volume for the new charge of 9.0448 kmol of ONCB

V = 3.265 m³ ONCB/NH₃ + 1.854 m³ H₂O = 5.119 m³

Case History

The reaction volume for the previous charge of 3.17 kmol of ONCB:

Assume that ΔC_P ≈ 0

Solution

Mole Balance:

Rate Law:

Stoichiometry (liquid phase):

with

Combine:

Substituting our parameter values into Equation (3-21)

We obtain

Energy Balance:

For ΔC_P = 0,

∑ N_i C_Pi = NC_P = N_A0 C_PA + N_B0C_PB + N_WC_PW

Parameter evaluation for day of explosion:

NC_P = (9.0448)(40) + (103.7)(18) + (33)(8.38)

NC_P = 2504 kcal/K

Again, let Q_g be the heat generated, i.e., Q_g = (r_AV)(ΔH_Rx), and let Q_r be the heat removed, i.e., Q_r = UA (T – T_a).

Q_g = (r_AV)(ΔH_Rx)

Q_r = UA (T – T_a)

A. Isothermal Operation Up to 45 Minutes

We will first carry out the reaction isothermally at 175°C (448 K) up to the time the cooling was turned off at 45 min. Combining and canceling yields

Integrating Equation (E13-2.9) gives us

At 175°C = 448 K, k = 0.0001167m³/kmol · min.

Substituting for k and the other parameter values

The calculation and results can also be obtained from the Polymath output

Solving for X, we find that at t = 45 min, then X = 0.033.

We will calculate the rate of generation, Q_g, at this temperature and conversion and compare it with the maximum rate of heat removal, Q_r, that is available for a constant coolant temperature of T_a = 298 K. The rate of generation, Q_g, is

At this time (i.e., t = 45 min, X = 0.033, T = 175°C), we calculate k, then Q_r and Q_g . At 175°C, k = 0.0001167 m³/min · kmol.

The corresponding maximum cooling rate is

Therefore

Everything is OK.

The reaction can be controlled. There would have been no explosion had the cooling not failed.

B. Adiabatic Operation for 10 Minutes

Unexpectedly, the cooling failed from 45 to 55 min after the reaction was started. We will now use the conditions at the end of the period of isothermal operation as our initial conditions for the adiabatic operation period between 45 and 55 min

t = 45 min   X = 0.033   T = 448 K

Interruptions in the cooling system have happened before with no ill effects.

Between t = 45 and t = 55 min, Q_r = 0. The Polymath program was modified to account for the time of adiabatic operation by using an “if statement” for Q_r in the program, i.e., Q_r = if (t > 45 and t < 55) then (0) else (UA (T – 298)). A similar “if statement” is used for isothermal operation, i.e., (dT/dt) = 0.

For the 45- to 55-min period without cooling, the temperature rose from 448 K to 468 K, and the conversion increased from 0.033 to 0.0424. Using this temperature and conversion in Equation (E13-2.11), we calculate the rate of generation Q_g at 55 min and see that it has increased to

Q_g = 6591 kcal/min

The maximum rate of cooling at this reactor temperature is found from Equation (E13-2.12) to be

Q_r = 6093 kcal/min

Here, we see that at the end of the 10-minute down time, the heat exchange system is now operating again, but now

The point of no return

and the temperature will continue to increase. We have a Runaway Reaction!! The point of no return has been passed and the temperature will continue to increase, as will the rate of reaction until the explosion occurs.

C. Batch Operation with Heat Exchange

Return of the cooling occurs at 55 min after startup. The values at the end of the period of adiabatic operation (T = 468 K, X = 0.0423) become the initial conditions for the period of restored operation with the heat exchange. The cooling is turned on at its maximum capacity, Q_r = UA (T – 298), at 55 min. Table E13-2.1 gives the Polymath program to determine the temperature–time trajectory. (Note that one can change N_A0 and N_B0 to 3.17 and 43 kmol in the program and show that, if the cooling is shut off for 10 min, at the end of that 10 min, Q_r will still be greater than Q_g and no explosion will occur.)

The complete temperature–time trajectory is shown in Figure E13-2.2. One notes the long plateau after the cooling is turned back on. Using the values of Q_g and Q_r at 55 min and substituting into Equation (E13-2.8), we find that

TABLE E13-2.1 POLYMATH PROGRAM

Figure E13-2.2 Temperature–time trajectory.

The explosion occurred shortly after midnight.

Consequently, even though dT/dt is positive, the temperature increases very slowly at first, 0.2°C/min. By 11:45, the temperature has reached 240°C and both the “heat generated” and the temperature begin to increase more rapidly. This rapid increase is a result of the Arrhenius temperature dependence, which caused the temperature to increase exponentially. The reaction is running away! One observes in Figure E13-2.2 that 119 min after the batch was started, the temperature increases sharply and the reactor explodes at approximately midnight. If the mass and heat capacity of the stirrer and reaction vessel had been included, the NC_p term would have increased by about 5% and extended the time until the explosion occurred by 15 or so minutes, which would predict the actual time the explosion occurred, at 12:18 A.M.

When the temperature reached 300°C, a secondary reaction, the decomposition of nitroaniline to noncondensable gases such as CO, N₂, and NO₂, occurred, releasing even more energy. The total energy released was estimated to be 6.8 × 10⁹ J, which is enough energy to lift the entire 2500-ton building 300 m (the length of three football fields) straight up.

D. Safety Disk Rupture Failure

We note that the pressure-safety-relief disk should have ruptured when the temperature reached 265°C (ca. 700 psi) but did not and the temperature continued to rise. If it had ruptured and all the water had vaporized, 10⁶ kcal would have been drawn from the reacting solution, thereby lowering its temperature and quenching the runaway reaction.

If the disk had ruptured at 265°C (700 psi), we know from fluid mechanics that the maximum mass flow rate, , out of the 2-in. orifice to the atmosphere (1 atm) would have been 830 kg/min at the time of rupture. The corresponding heat removed would have been

This value of Q_r is much greater than the “heat generated” Q_g (Q_g = 27,460 kcal/min), so that the reaction could have been easily quenched.

Analysis: Runaway reactions are the most deadly in the chemical industry. Elaborate safety measures are usually installed to prevent them from occurring. However, as we show in this example, the back-up plan failed. If any one of the following three things had not occurred, the explosion would not have happened.

1. Tripled production

2. Heat-exchanger failure for 10 minutes early in the operation

3. Failure of the relieving device (rupture disk)

In other words, all the above had to happen to cause the explosion. If the relief valve had operated properly, it would not have prevented reaction runaway but it could have prevented the explosion. In addition to using rupture disks as relieving devices, one can also use pressure relief valves. In many cases, sufficient care is not taken to obtain data for the reaction at hand and to use it to properly size the relief device. This data can be obtained using a specially designed batch reactor called the Advanced Reactor Safety Screening Tool (ARSST), as shown in Chapter 13, PRS R13.1.

Example 13–3 Heat Effects in a Semibatch Reactor

The second-order saponification of ethyl acetate is to be carried out in a semibatch reactor shown schematically below in Figure E13-3.1.

Aqueous sodium hydroxide is to be fed at a concentration of 1 kmol/m³, a temperature of 300 K, and a volumetric rate of 0.004 m³/s to an initial volume of 0.2 m³ of water and ethyl acetate. The concentration of water in the feed, C_W0, is 55 kmol/m³. The initial concentrations of ethyl acetate and water in the reactor are 5 kmol/m³ and 30.7 kmol/m³, respectively. The reaction is exothermic and it is necessary to add a heat exchanger to keep its temperature below 315 K. A heat exchanger with UA = 3000 J/s · K is available for use. The coolant enters at a mass flow rate of 100 kg/s and a temperature of 285 K.

Are the heat exchanger and coolant flow rate adequate to keep the reactor temperature below 315 K? Plot the reactor temperature, T, and the concentrations, C_A, C_B, and C_C as a function of time.

Additional information:³

³ Value for k from J. M. Smith, Chemical Engineering Kinetics, 3rd ed. (New York: McGraw-Hill, 1981), p. 205. Note that ΔH_Rx and K_C were calculated from values given in Perry’s Chemical Engineers’ Handbook, 6th ed. (New York: McGraw-Hill, 1984), pp. 3–147.

Figure E13-3.1 Semibatch reactor with heat exchange.

Solution

1. Mole Balances: (See Chapter 6.)

Initially,

N_Wi = V_iC_W0 = (0.2 m³) (30.7 kmol/m³) = 6.14 kmol

2. Rate Law:

3. Stoichiometry:

4. Energy Balance:

The user-friendly energy balance is

Only species B (NaOH) and water flow into the reactor, so Equation (13-22) becomes

Q_rs1 is the heat removed by mass flow, and Q_rs2 is the heat removed through heat exchange.

In Jofostan we have the added bonus that we can ask Prof. Dr. Sven Köttlov to calculate the temperature of the heat exchange fluid at its exit

5. Evaluation of Parameters:

C_PA = 170,700 J/kmol/K, C_P = 75,246 J/kmol/K

We note the heat capacities for B, C, and water are essentially the same at C_P

NC_P = C_PAN_A + C_P (N_B + N_C + N_D + N_W)

The Polymath program is given in Table E13-3.1. The solution results are shown in Figures E13-3.2 and E13-3.3.

TABLE E13-3.1 POLYMATH PROGRAM AND OUTPUT FOR SEMIBATCH REACTOR

Figure E13-3.2 Temperature–time trajectories in a semibatch reactor.

Figure E13-3.3 Concentration–time trajectories in a semibatch reactor.

Analysis: From Figure E13-3.3 we see that the concentration of species B is virtually zero, owing to the fact that it is consumed virtually as fast as it enters the reactor up to a time of 252s. By the time we reach 252s, all species A has been consumed, and the reaction rate is virtually zero and no more of species C or D is produced and no more B is consumed. Because species B continues to enter the reactor at a volumetric flow rate υ₀, after 252 seconds, the fluid volume continues to increase and the concentrations of C and D are diluted. The figure shows that before 252 s, Q_g > Q_r, and the reactor temperature and the coolant temperature increase. However, after 252 s, the reaction rate, and hence Q_g, are virtually zero so that Q_r > Q_g and the temperature decreases. Because of the impractical short reaction time (252 seconds), a semibatch reactor would never be used for this reaction at this temperature; instead, we would most likely use a CSTR or PFR. See Problem P13-1_B (c) to reflect on this example.

Wrong Reactor Choice!

Example 13–4 Startup of a CSTR

Again, we consider the production of propylene glycol (C) in a CSTR with a heat exchanger in Example 12-3. Initially there is only water, C_wi = 3.45 lb-mol/ft3, at T_i = 75°F and 0.1 wt % H₂SO₄ in the 500-gallon reactor. The feed stream consists of 80 lb-mol/h of propylene oxide (A), 1000 lb-mol/h of water (B) containing 0.1 wt % H₂SO₄, and 100 lb-mol/h of methanol (M).

The water coolant flows through the heat exchanger at a rate of 5 lb_m/s (1000 lb-mol/h). The molar densities of pure propylene oxide (A), water (B), and methanol (M) are ρ_A0 = 0.923 lb-mol/ft³, ρ_B0 = 3.45 lb-mol/ft³, and ρ_M0 = 1.54 lb-mol/ft³, respectively.

Plot the temperature and concentration of propylene oxide as a function of time, and a concentration vs. temperature graph for different entering temperatures and initial concentrations of A in the reactor.

Additional information:

Again, the temperature of the mixed reactant streams entering the CSTR is T₀ = 75°F.

Solution

Mole Balances:

Energy Balance:

where

using

the “heat removed” term from the unsteady startup of a CSTR is similar to that for the “heat removed” term from a semi-batch reactor. Recalling Equations (12-12) and (12-19)

Evaluation of Parameters:

Neglecting ΔC_P because it changes the heat of reaction insignificantly over the temperature range of the reaction, the heat of reaction is assumed constant at its reference temperature

The Polymath program is shown in Table E13-4.1.

TABLE E13-4.1 POLYMATH PROGRAM FOR CSTR STARTUP

Figures E13-4.1 and E13-4.2 show the concentration of propylene oxide and reactor temperature as a function of time, respectively, for an initial temperature of T_i = 75°F and only water in the tank (i.e., C_Ai = 0). One observes that both the temperature and concentration oscillate around their steady-state values (T = 138°F, C_A = 0.039 lb-mol/ft³) as steady state is approached.

Unacceptable startup

Figure E13-4.3 combines Figures E13-4.1 and E13-4.2 into a phase plot of C_A versus T. The final operating concentration of A is 0.0379 lb-mol/ft³ at a temperature of 138°F. The arrows on the phase-plane plots show the trajectories with increasing time. The maximum temperature reached during startup is 152°F, which is below the practical stability limit of 180°F.

Figure E13-4.1 Propylene oxide concentration as a function of time for C_Ai = 0 and T_i = 75°F.

Figure E13-4.2 Temperature–time trajectory for CSTR startup for C_Ai = 0 and T_i = 75°F.

Figure E13-4.3 Concentration–temperature phase-plane trajectory using Figures E13-4.1 and E13-4.2.

Figure E13-4.4 Concentration–temperature phase-plane for three different initial conditions.

Next, consider Figure E13-4.4, which shows three different trajectories for three different sets of initial conditions:

After three hours, the reaction is operating at steady state and all three trajectories converge on the final steady-state temperature of 138°F and the corresponding concentrations

Oops! The practical stability limit was exceeded.

For this reaction system, the plant safety office believes that an upper temperature limit of 180°F should not be exceeded in the tank. This temperature is the practical stability limit. The practical stability limit represents a temperature above which it is undesirable to operate because of unwanted side reactions, safety considerations, secondary runaway reactions, or damage to equipment. Consequently, we see that if we started at an initial temperature of T_i = 160°F and an initial concentration of 0.143 mol/dm³, the practical stability limit of 180°F would be exceeded as the reactor approached its steady-state temperature of 138°F. See the concentration–temperature trajectory in Figure E13-4.4.

Figures E13-4.1 through E13-4.4 show the concentration and temperature time trajectories for the start up of a CSTR for different initial conditions.

Analysis: One of the purposes of this example was to demonstrate the use of phase plots, e.g., T versus C_A, in analyzing CSTR startup. Phase plots allow us to see how the steady state is approached for different sets of initial conditions and if the practical stability limit is exceeded causing a secondary, more exothermic reaction to set in.

Example 13–5 Multiple Reactions in a Semibatch Reactor

The series reactions

are catalyzed by H₂SO₄. All reactions are first order in the reactant concentration. However, Reaction (1) is exothermic and Reaction (2) is endothermic. The reaction is to be carried out in a semibatch reactor that has a heat exchanger inside with UA = 35,000 cal/h·K and a constant exchanger temperature, T_a, of 298 K. Pure A enters at a concentration of 4 mol/dm³, a volumetric flow rate of 240 dm³/h, and a temperature of 305 K. Initially, there is a total of 100 dm³ of liquid in the reactor, which contains 1.0 mol/dm³ of A and 1.0 mol/dm³ of the catalyst H₂SO₄. The reaction rate is independent of the catalyst concentration. The initial temperature inside the reactor is 290 K.

Plot and analyze the species concentrations and reactor temperature as a function of time.

Additional information:

Solution

1. Mole Balances:

2. Rates:

(a) Rate Laws:

(b) Relative Rates:

(c) Net Rates:

3. Stoichiometry (liquid phase): Use C_A, C_B, C_C

4. Energy Balance:

Semibatch reactor:

A slight rearrangement of (13-23) is

Expanding

Substituting the parameter values into Equation (E13-5.14)

Equations (E13-5.1) through (E13-5.3) and (E13-5.8) through (E13-5.12) can be solved simultaneously with Equation (E13-5.14) using an ODE solver. The Polymath program is shown in Table E13-5.1. The concentration–time and temperature– time trajectories are shown in Figures E13-5.1 and E13-5.2.

TABLE E13-5.1 POLYMATH PROGRAM FOR STARTUP OF A CSTR

Figure E13-5.1 Concentration–time.

Figure E13-5.2 Temperature (K)–time (h).

Analysis: At the start of the reaction, both C_A and T in the reactor increase because C_A0 and T₀ are greater than C_Ai and T_i. This increase continues until the rate of consumption of the reacting species is greater than the feed rate to the reactor. We note that at about 0.2 h, the reactor temperature exceeds the feed temperature (i.e., 305 K) as a result of the heat generated by the exothermic reaction (Reaction 1). The temperature continues to rise until about 0.6 h, at which point reactant A is virtually all consumed. After this point the temperature begins to drop for two reasons: (1) the reactor is cooled by the heat exchanger and (2) heat is absorbed by the endothermic reaction rate (Reaction 2). The question is, does the maximum temperature (435 K) exceed a temperature that is too high, resulting in a high vapor pressure that then results in evaporation losses or causes a highly exothermic secondary reaction to set in?

Example 13–6 T2 Laboratories Explosion⁴

⁴ This example was coauthored by Professors Ronald J. Willey, Northeastern University, Michael B. Cutlip, University of Connecticut, and H. Scott Fogler, University of Michigan, and published in Process Safety Progress, 30, 1 (2011).

Figure E13-6.1 Aerial photograph of T2 taken December 20, 2007. (Courtesy of Chemical Safety Board.)

T2 Laboratories manufactured a fuel additive, methylcyclopentadienyl manganese tricarbonyl (MCMT), in a 2,450-gallon, high-pressure batch reactor utilizing a three-step batch process.

Step 1a. The liquid-phase metalation reaction between methylcyclopentadiene (MCP) and sodium in a solvent of diethylene glycol dimethyl ether (diglyme) to produce sodium methylcyclopentadiene and hydrogen gas is

Hydrogen immediately comes out of the solution and is vented at the top in the gas head space.

Step 1b. At the end of Step 1a, MnCl₂ is added. The substitution reaction between sodium methylcyclopentadiene and manganese chloride produced manganese dimethylcyclopentadiene and sodium chloride is

Step 1c. At the end of Step 1b, CO is added. The carbonylation reaction between manganese dimethylcyclopentadiene and carbon monoxide produces the final product, methylcyclopentadienyl manganese tricarbonyl (MCMT):

We will only consider Step 1a as this step is the one in which the explosion occurred.

Procedure

First, solid sodium is mixed in the batch reactor with methylcyclopentadiene dimer and a solvent diethylene glycol dimethyl ether (diglyme). The batch reactor is then heated to about 422 K (300˚F) with only slight reaction occurring during this heating process. On reaching 422 K, the heating is turned off, as the exothermic reaction is now proceeding, and the temperature continues to increase without further heating. When the temperature reaches 455.4 K (360˚F), the operator initiates cooling using the evaporation of boiling water in the reactor jacket as the heat sink (T_a = 373.15 K) (212˚F).

What Happened

On December 19, 2007, when the reactor reached a temperature of 455.4 K (360˚F), the process operator could not initiate the flow of cooling water to the cooling jacket shown in Figure E13-6.2. Thus, the expected cooling of the reactor was not available and the temperature in the reactor continued to rise. The pressure also increased as hydrogen continued to be produced at an increased rate, to the point that the reactor’s pressure control valve system on the 1-inch diameter hydrogen venting stream could no longer maintain the operating pressure at 50 psig (4.4 atm). As the temperature continued to increase further, a previously unknown exothermic reaction of the diglyme solvent that was catalyzed by sodium accelerated rapidly.

Figure E13-6.2 Reactor.

This reaction produced even more hydrogen, causing the pressure to rise even faster, eventually causing the rupture disk which was set at 28.2 atm absolute (400 psig), to break, in the 4-inch diameter relief line of H₂. Even with the relief line open, the rate of production of H₂ was now far greater than the rate of venting, causing the pressure to continue to increase to the point that it ruptured the reactor vessel initiating a horrific explosion. The T2 plant was completely leveled and four personnel lives were lost. Surrounding businesses were heavily damaged and additional injuries were sustained.

Before continuing with this example, it might be helpful to view the 9-minute Chemical Safety Board (CSB) video, which you can access directly from the CRE Web site (under YouTube videos), or you can read the supporting reports (http://www.csb.gov/videos/runaway-explosion-at-t2-laboratories/). You can also search the Web for “T2 explosion video.”

Simplified Model

Summarizing the important reactions for Step 1

Let A = methycylcopentadiene, B = sodium, S = solvent (diglyme), and D = H₂. This runaway reaction can be approximately modeled with two reactions. These reactions are

In Reaction (1), A and B react to form products. Reaction (2) represents the decomposition of the liquid-phase solvent S catalyzed by the presence of B, but this reaction only begins to proceed once a temperature of approximately 470 K is reached.

The rate laws, along with the specific reaction-rate constants at the initial temperature of 422 K, are

The heats of reaction are constant.

ΔH_Rx1A = –45,400 J/mol

ΔH_Rx2S = –320,000 J/mol

The sum of products of the moles of each species and their corresponding heat capacities in the denominator of Equation (13-12) is essentially constant at

∑ N_jC_Pj = 1.26 × 10⁷ J/K

Assumptions

Assume that the liquid volume, V₀, in the reactor remains constant at 4,000 dm³ and that the vapor space, V_H, above the reactor occupies 5,000 dm³. Any gas, H₂ (i.e., D), that is formed by Reactions (1) and (2) immediately appears as an input stream F_D to the head-space volume. The dissolved H₂ and the vapor pressures for the liquid components in the reactor can be neglected. The initial absolute pressure within the reactor is 4.4 atm (50 psig). During normal operation, the H₂ generated obeys the ideal gas law. The pressure control system on the H₂ vent stream maintains the pressure, P, at 4.40 atm up to a flow of 11,400 mol/hr. The reactor vessel will fail when the pressure exceeds 45 atm or the temperature exceeds 600 K.

Additional information:

UA = 2.77 x 10⁶ J hr^–1 K^–1. The concentrations in the reactor at the end of the reactor heating at 422 K are C_A0 = 4.3 mol/dm³, C_B0 = 5.1 mol/dm³, C_I0 = 0.088 mol/dm³, and C_S0 = 3 mol/dm³. The sensible heat of the two gas venting streams may be neglected.

Problem Statement

Plot and analyze the reactor temperature and head-space pressure as a function of time along with the reactant concentrations for the scenario where the reactor cooling fails to work (UA = 0). In Problem P13-1(f) you will be asked to redo the problem when the cooling water comes as expected whenever the reactor temperature exceeds 455 K.

Solution

(1) Reactor Mole Balances:

Reactor (Assume Constant Volume Batch)

Liquid

(2) Head-Space Mole Balance:

Let N_D = moles of gas D in the reactor vapor space V_v. The flows in and out of the head space are shown in the figure below.

A balance on species D (H₂) in the head-space volume V_H yields

where F_vent is the molar flow rate of gas out of the head space through one or both outlet lines, and F_D (i.e., F_H2) is the molar rate of gas leaving the liquid and entering the head space and is equal to the hydrogen generated in the liquid (see relative rates and net rates in the algorithm where V₀ is the liquid reactor volume).

The assumptions of a perfect gas in the head-space volume and modest changes in T allow Equation (E13-6.4) to be written in terms of total gas pressure in the reactor head space

Substituting for N_D in Equation (E13-6.3) and rearranging

(3) How the Venting Works:

Gas exits the reactor through the pressure control valve line. At low gas production, the pressure control valve maintains set-point pressure at the initial pressure by venting all produced gas until the rate of gas production reaches 11,400 mol/hr.

We now need to know a little more about the venting system for H₂ and the condition for flow out the vent. As the pressure increases, but is still below the rupture disk setting, the pressure control line vents to the atmosphere (1 atm) according to the equation

where P is the absolute pressure in the reactor (atm), 1 atm is the downstream pressure, and the pressure correlation constant C_v is 3,360 mol/hr • atm. If the pressure P within the reactor exceeds 28.2 atm (400 psig), the relief line activated by the rupture disk breaks and vents gas in the reactor at the rate given F_vent = (P–1)C_v2 where C_v2 = 53,600 mol/atm • hr.

After the rupture disk blowout at P = 28.2 atm, both the pressure control line and the rupture disk lines vent the reactor according to the equation

Equations (E13-6.7) through (E13-6.10) can then be used to describe the F_vent flow rate with time for the appropriate logic for the values of F_D and P.

(4) Rates:

Laws:

Relative Rates:

Net Rates:

(5) Stoichiometry:

Neglect reactor-liquid volume change form loss of product gases.

(6) Energy Balance:

Applying Equation (E13-2.3) to a batch system (F_i0 = 0)

Substituting for the rate laws and ∑N_jC_Pj

(7) Numerical Solutions: “Tricks of the Trade”

A rapid change of temperature and pressure is expected as Reaction (2) starts to run away. This typically results in a stiff system of ordinary differential equations, which can become numerically unstable and generate incorrect results. This instability can be prevented by using a software switch (SW) that will set all derivates to zero when the reactor reaches the explosion temperature or pressure. This switch can have the form of Equation (E13-6.26) in Polymath and can be multiplied by the right-hand side of all the differential equations in this problem. This operation will halt (or freeze) the dynamics when the temperature T becomes higher than 600 K or the pressure exceeds 45 atm.

We now will solve the essential equations from (E13-6.1) through (E13-6.26) for the scenario where there is no cooling and thus UA = 0. Also the switch SW1 must be implemented in all the differential equations as discussed above.

TABLE E13-6.1 POLYMATH PROGRAM

Note: The second reaction took off about 3.6 hours after starting the reactor.

Figure E13-6.3(a) Temperature (K) versus time (h) trajectory.

Figure E13-6.3(b) Pressure (atm) versus time (h) trajectory.

Figure E13-6.3(c) Concentration (mol/dm³) versus time (h) trajectory.

We note from Figures E13-6.3(a)–(b) that the explosion occurred at approximately 3.6 h after startup and the concentration of diglyme begins to drop sharply before that point. We also note that numerical instabilities occur at about the point of the arrows in these figures because of the rapid increase in temperature.

Analysis: Runaway would not have occurred if (1) the cooling system had not failed, causing the reactor temperature to rise and initiate a second a reaction, and (2) the solvent dygline had not decomposed at the higher temperature to produce hydrogen gas (D). The rate of production of H₂ gas was greater than the removal of H₂ from the head space, causing the pressure to build up to the point that it ruptured the reactor vessel.