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T2 Laboratories Explosion On December 19 2007, a tremendous explosion shocked the northern Jacksonville, Florida. Tons of sodium, hydrogen, and organics exploded when a reactor ruptured and caught fire at a chemical producer called T2 Laboratories. Four people lost their lives. Thirty-two people within the vicinity suffered injuries. The power of the shock wave was felt 15 miles away. The plant manufactured methylcyclopentadienyl manganese tricarbonyl, a gasoline antiknock additive, for a third party distributor. At the time, a 9000 L batch reactor was producing the first intermediate, sodium methylcyclopentadiene and hydrogen (a by-product) from the reaction of metallic sodium with methylcyclopentadiene. Diethylene glycol dimethyl ether (diglyme) was used as a solvent. Normal operating procedure Under normal operating conditions, first the reactants and the solvents were mixed in the batch reactor. The reactor was then heated to 422 K with only slight reaction occurring during this heating process. On reaching 422 K, the heating was switched off and the temperature continued to rise without further heating as the exothermic reaction was proceeding. When the temperature reached 455 K, the operator initiated cooling using the evaporation of boiling water in the reactor jacket as the heat sink (Ta = 373 K). Figure 1 Schematic of the batch reactor used at T2 Laboratories What happened On the fatal day, a reactor operator contacted one of the plant owners informing him that there was no cooling water flow to the reactor. The temperature in the reactor continued to rise. The pressure also increased as hydrogen was produced at an increased rate to the point where the pressure control valve system on the 1-inch diameter hydrogen venting stream could no longer maintain the operating pressure at 4.4 atm. More complicated, and unknown to the plant employees, a second, stronger, exothermic Cooling jacket vent to atmosphere 1-inch hydrogen vent pipe Pressure control valve 4-inch vent pipe and rupture disk Solid sodium metal loaded to reactor Hot oil in Hot oil out 2-inch city water service line Cooling water control valve and manual bypass Cooling water drain valve and manual bypass reaction kicked in, the decomposition reaction of diglyme either catalyzed or enhanced by the presence of sodium: This rapid decomposition led to the creation of even more hydrogen causing the pressure to rise even faster eventually causing the 4-inch rupture disk to break at 28.4 atm absolute. Even with the relief line open, the rate of production of hydrogen was now much greater than the rate of venting eventually causing the reactor to rupture in a horrific explosion. Simplified model The runaway reaction can be approximately modelled by two reactions: A + B → C + ½ D (gas) (1) S → 3 D (gas) + misc liquids and solids (2) where A = methylcyclopentadiene, B = sodium, C = sodium methylcyclopentadiene, D = hydrogen and S is the solvent diglyme. The rate of reaction, r, for these reactions are given by 1  CCkr BAAA (3) and 2  Ckr SSS (4) where Ci is the concentration of component i and ki is the rate constant. The temperature dependence of the rate constants is given by the Arrhenius law:         RT E kTk a exp)( 0 (5) where k0 is the pre-exponential factor, Ea the activation energy and R the ideal gas constant. The reaction characteristics for the two exothermic reactions are summarised in Table 1. The heat of reactions can be assumed to be constant. Table1 Reaction characteristics for reactions 1 and 2. Reaction 1 Reaction 2 k0 4  1014 L mol-1 h -1 1  1084 h -1 Ea 128,000 J / mol 800,000 J / mol ΔHrxn -45,400 J/mol -3.2  105 J / mol Assuming that the volume of the batch does not change, the mole balances for the reaction liquid yield: A BAA A CCkr dt dC 1  (6) A BAA B CCkr dt dC 1  (7) SSS S Ckr dt dC 2  (8) Assuming that the gas in the head space behaves like an ideal gas, the mole balance for the head space yields (detailed derivation in the lecture):   dt dT V RN V RT FF dt dP H D H  ventD  (9) where VH is the volume of the head space in the reactor, Fvent is the molar flow rate out of the reactor through one or both outlet lines, and FD is the molar flow rate of gas leaving the liquid of volume V0 and entering the head space:   D  35.0 SA VrrF 021 (10) Gas exits the reactor in both the pressure control valve line and the rupture disk line. At low gas production, the pressure control valve maintains set point pressure at the initial pressure by venting all produced gas until the gas production reaches 11,400 mol/h: vent  FF D when FD < 11,400 mol/h (11) As the pressure increases but is still below the rupture disk setting, the pressure control line vents to the atmosphere (1 atm) according to equation 12   Fvent PCV1  1CP v1 when P < 28.2 atm (12) where P is the absolute pressure in the reactor (atm). The downstream pressure is 1 atm and the vent rate, CV1, is 3,360 mol h-1 atm-1 . If the pressure P in the reactor exceeds 28.2 atm, the relief line activated by the rupture disk breaks and vents gas at a rate of CV2 = 53,600 mol h-1 atm-1 and the total vent rate is given by      vent VV 21 1  CCPCCPF Vv 21 when P > 28.2 atm (13) The energy balance gives the change of temperature over time:        iPi a S rxnSArxnA CN HrHrV UA TT dt dT , , , 2 10 21 (14) The sum of the product of the number of moles of each species and their heat capacities is essentially constant at NiCPi = 1.26  107 J/K. Assumptions Assume that the liquid volume, V0, in the reactor remains constant at 4000 L and that the vapour space in the head, VH, above the liquid occupies 5000 L. Any hydrogen produced (D) immediately appears as an input stream FD to the head space of the reactor. The dissolved H2 and the vapour pressures for the liquid components in the reactor can be neglected. Any hydrogen generated obeys the ideal gas law. The reactor vessel will fail when the pressure exceeds 45 atm or 600 K. Initial conditions The initial temperature and initial absolute pressure in the system are 422 K and 4.4 atm. The concentrations in the reactor at the end of the heating period (which you won’t be modelling) are CA0 = 4.3 mol / L, CB0 = 5.1 mol / L and CS0 = 3 mol / L. Tasks: a) Plot the reactor temperature, the head space pressure, and the concentrations of A, B, S, D as function of time if the cooling system fails (UA = 0). When does the reactor explode? Discuss briefly, what you observe for the diglyme concentration. b) Model the reactor under normal operating conditions where the cooling water is turned on (UA = 2.77 106 W/K) once the temperature in the reactor reaches 455 K. Again, plot the reactor temperature, the head space pressure, and the concentrations of A, B, S, D as function of time and discuss your observations briefly. c) What is the maximum time in minutes that the cooling can be lost (UA = 0) so that the reactor will not reach an explosion point? (In other words, how long did the operators have to reinstate the cooling before the reaction reached the point of no return and the reactor would have exploded anyway?) d) Briefly discuss five suggestions about how to change the process and/or plant to avoid this accident.

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function T = ode(t,n)
clear all
clc

%given data
Sigma=1.26*(10)^7;
V0=4000;
Vh=5000;
k0_1=(1/3600)*4*(10)^14;
k0_2=(1/3600)*1*(10)^84;
H_1A=-45400;
H_2S=-3.2*(10)^5;
R=8.314;
Ea_1=128000;
Ea_2=800000;

%initial calculations of parameters
Ca_in=4.300;
Ca(:,1)=Ca_in;
Cb_in=5.100;
Cb(:,1)=Cb_in;
Cs_in=3.000;
Cs(:,1)=Cs_in;
Cd_in=0.5*Ca(:,1)+3*Cs(:,1);
Cd(:,1)=Cd_in;

T_in=422;
T(:,1)=T_in;

kA_in=k0_1*exp(-Ea_1/(R*T_in));
kA(:,1)=kA_in;

kS_in=k0_2*exp(-Ea_2/(R*T_in));
kS(:,1)=kS_in;

r1A_in=-kA_in*Ca_in*Cb_in;
r1A(:,1)=r1A_in;
r2S_in=-kS_in*Cs_in;
r2S(:,1)=r2S_in;

Fd_in = V0*(-0.5*r1A_in-3*r2S_in);
Fd(:,1) = Fd_in;

Fvent_in = Fd_in;
Fvent(:,1) = Fvent_in;

P_in=4.4;
P(:,1)=P_in;

t_in=0;
t(:,1)=t_in;

n=130000;
deltaT=0.1;


for k=1:n,

Ca(:,k+1)=Ca(:,k)-deltaT*kA(:,k)*(Ca(:,k))*(Cb(:,k));
Cb(:,k+1)=Cb(:,k)-deltaT*kA(:,k)*(Ca(:,k))*(Cb(:,k));
Cs(:,k+1)=Cs(:,k)-deltaT*kS(:,k)*(Cs(:,k));
Cd(:,k+1)=0.5*Ca(:,k+1)+3*Cs(:,k+1);

%calculation of Temprerature
T(:,k+1)=T(:,k)+V0*deltaT*(r1A(:,k)*H_1A+r2S(:,k)*H_2S)/Sigma;

kA(:,k+1)=k0_1*exp(-Ea_1/(R*T(:,k+1)));
kS(:,k+1)=k0_2*exp(-Ea_2/(R*T(:,k+1)));
   
r1A(:,k+1)=-kA(:,k+1)*Ca(:,k+1)*Cb(:,k+1);
r2S(:,k+1)=-kS(:,k+1)*Cs(:,k...

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