13. Unsteady-State Nonisothermal Reactor Design

Chemical engineers are not gentle people, they like high temperatures and high pressures.

—Prof. Steven LeBlanc

Overview. Up to now we have focused on the steady-state operation of nonisothermal reactors. In this chapter the unsteady-state energy balance will be developed and then applied to CSTRs, as well as well-mixed batch and semibatch reactors. We will use the difference between the rate of “heat generated,” Qg, and the rate of “heat removed,” Qr, to arrive at the user-friendly form of the energy balance

dTdt=Qg.Qr.ΣNiCPi

Section 13.1 shows how to arrange the general energy balance (Equation (11-9)) in a more simplified form for unsteady-state operation.

Section 13.2 discusses the application of the energy balance to the operation of batch reactors and discusses reactor safety and the reasons for the explosion of an industrial batch reactor.

Section 13.3 discusses the startup of a CSTR and how to avoid exceeding the practical stability limit.

Section 13.4 shows how to apply the energy balance to a semibatch reactor with a variable ambient temperature.

Section 13.5 closes the chapter with a case study of the T2 Laboratory explosion involving multiple reactions in a batch reactor.

Safety is another focus of this chapter, and the example problems and homework problems were chosen to emphasize the hazards of runaway reactors.

13.1 The Unsteady-State Energy Balance

We start by deriving the user-friendly form of the energy balance that can be easily used in making reactor calculations. As I have stated before, the reason I derive the equations rather than just giving the final result is (1) the reader can see what assumptions are used at what points along the way in the derivation to the user-friendly energy balance, and (2) it is my experience that if the reader goes line-by-line through the derivation, he/she will be less likely to insert the incorrect numbers into symbols in the equation when making a calculation.

We begin by recalling the unsteady-state form of the energy balance developed in Chapter 11.

Q.-W.s+i=1mFiHi|in-i=1mFiHi|out=(dÊsysdt)(11-9)

We shall first concentrate on evaluating the change in the total energy of the system wrt time, (dEsys/dt). The total energy of the system is the sum of the products of specific energies, (dEˆsys/dt). of the various species in the system volume and the number of moles, Ni (mol i), of that species

Eˆsys=Σmi=1NiEi=NAEA+NBEB+NCEC+NDED+NIEI(13-1)

In evaluating Eˆsys, as before we neglect changes in the potential and kinetic energies and substitute for the internal energy Ui in terms of the enthalpy Hi

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We note the last term on the right-hand side of Equation (13-2) is just the total pressure times the total volume, i.e., PV, and this term is virtually always smaller than the other terms in Equation (13-2), and thus will be neglected† For brevity, we shall write all summations as

Σ=Σmi=1

unless otherwise stated.

When no spatial variations are present in the system volume, and time variations in the product of the total pressure and volume (PV) are neglected, the energy balance, after substitution of Equation (13-2) into (11-9), gives

Q.-W.S+ΣFi0Hi0|in-ΣFiHi|out=[ΣNidHidt+ΣHidNidt]sys(13-3)

† Marat Orazov while a student at University of California, Berkeley, pointed out that the last term in Equation (13-2) need not be neglected for the special case of an ideal gas with a constant total number of moles. The denominator on the r.h.s. of Equation (13-9) in this case is just Ni(CPi-R).

Recalling Equation (11-19)

Hi=Hio(TR)+TRTCPidT(11-19)

and differentiating with respect to time, we obtain

dHidt=CPidTdt(13-4)

Then, substituting Equation (13-4) into (13-3) gives

Q.-W.S+ΣFi0Hi0-ΣFiHi=ΣNiCPidTdt+ΣHidNidt(13-5)

The mole balance on species is

dNidt=νirAV+Fi0Fi(136)

Using Equation (13-6) to substitute for dNi/dt, Equation (13-5) becomes

Q.-W.S+ΣFi0Hi0-ΣFiHi

=ΣNiCPidTdt+ΣνiHi(-rAV)+ΣFi0Hi-ΣFiHi(13-7)

Rearranging, and recalling vHii=ΔHRx, we have

This form of the energy balance should be used when there is a phase change.

dTdt=Q.-W.S-ΣFi0(HiHi0)+(ΔHRx)(rAV)ΣNiCPi(138)

Substituting for Hi and Hi0 for the case of no phase change gives us

dTdt=Q.-W.S-ΣFi0CPi(T-Ti0)-[ΔHRx(T)](-rAV)ΣNiCPi(139)

Energy balance on a transient CSTR or semibatch reactor

Equation (13-9) applies to a semibatch reactor, as well as to the unsteady-state operation of a CSTR, and is also shown in Table 11-1 as Equation (T11-1.I).

For liquid-phase reactions where ΔCP is usually small and can be neglected, the following approximation is often made

ΣNiCPiΣNi0CPi=NA0ΣΘiCPiCPs=NA0CPs

PCs: Heat capacity of solution

where CPc is the heat capacity of the solution. The units of the batch term (NAo CPs) are (cal/K) or (J/K) or (Btu/°R), and for the flow term

ΣFi0CPi=FA0CPs

the units are (J/s · K) or (cal /s · K) or (Btu/h · °R) .1 With this approximation and assuming that every species enters the reactor at the same temperature T0, we have

dTdt=Q.-W.S-FA0CPs(T-T0)+--ΔHRx(T)](-rAV)NA0CPs(1310)

13.2 Energy Balance on Batch Reactors (BRs)

A batch reactor is usually well mixed, so that we may neglect spatial variations in the temperature and species concentrations.

Schematic shows a CSTR at the left and a heat exchanger at the right.

The energy balance on batch reactors (BR) is found by setting the inlet and outlet flows to zero, i.e., Fi0 = Fi = 0, in Equation (13-9)

dTdt=Q.W.s+(ΔHRx)(rAV)ΣNiCpi(1311)

Next we use the heat-exchanger energy balance to obtain Q., Equation (12-19), and realize that the heat “added” to the batch reactor, Q., is just minus the heat “removed” from the batch reactor, Q.rbQ.=-Q.rb Neglecting shaft work (we could not do this in problem p12-6B) we get

dTdt=Q.gbQ.rbWsNeglectΣNiCpi

dTdt=QgbQrbΣNiCpi(1312)

1 If the heat capacities were given in terms of mass (i.e., Cp = J/g. K), then both FA0 and NA0 would have to be converted to mass

for batch mA0Cpsm=NA0CPS(g)(J/gK)=(mol)J(molK)=JK

and for flow m.A0CPsm=FA0CPS(g/s)(J/gK)=(mols)(J/mo1K)=JKs

However, we note that the units of the product of mass flow rate and mass heat capacities would still be the same as the product of molar flow and molar heat capacities (e.g., cal/s · K), respectively.

where the terms for heat generated, Q.gb, and heat removed, Q.rb, for a batch system are

Q.gb=(-ΔHRx)(-rAV)(13-13)

Q.rb=m.CPC(T-Ta1)[1-exp(-UAmCPC)](1314)

with an added bonus we obtained in Equation (12-17)

Ta2=T-(T-Ta1)[exp(-UAmCPC)](1315)

Reminder: The sign convention

Heat Added Q. = +10J/s

Heat Removed Q. = -10J/S

Work Done by System Ws = +10J/S

Work Done on System Ws = -10J/S

Equation (13-12) is the preferred form of the energy balance when the number of moles, Ni, is used in the mole balance, rather than the conversion, X. The number of moles of species i at any X is

Ni=NA0(Θi+viX)

Consequently, in terms of conversion, the user-friendly form of the energy balance becomes

dTdt=Qgb˙Qrb˙NA0(ΘiCPi+ΔCPX)(1316)

Batch reactor energy balance

Equation (13-12) must be coupled with the mole balance

NA0dXdt=-rAV(26)

Batch reactor mole balance

13.2.1 Adiabatic Operation of a Batch Reactor

Batch reactors operated adiabatically are often used to determine the reaction orders, activation energies, and specific reaction rates of exothermic reactions by monitoring the temperature-time trajectories for different initial conditions. In the steps that follow, we will derive the temperature-conversion relationship for adiabatic operation.

For adiabatic operation (Q. = 0) of a batch reactor (Fi0 = 0) and when the work done by the stirrer can be neglected (W.S0), Equation (13-10) can be written as

dTdt=(-ΔHRx)(-rAV)ΣNiCPi(13-17)

It is shown in the Chapter 13 Summary Notes on the CRE Web site (http://www.umich.edu/~elements/5e/13chap/RelationshipBetweenXandT.pdf) that if we combine Equation (13-17) with Equation (2-6), we can do a lot of rearranging and integrating to arrive at the following user-friendly Equation for an adia-batic batch reactor.

X=ΣΘiCPi(T-T0)-ΔHRx(T)(1318)

T=T0+[-ΔHRx(T0)]XΘiCPi+XΔCP(1319)

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Temperature-conversion relationship for any reactor operated adiabatically

We note that for adiabatic conditions, the relationship between temperature and conversion is the same for BRs, CSTRs, PBRs, and PFRs. Once we have T as a function of X for a batch reactor, we can construct a table similar to Table E11-3.1 and use techniques analogous to those discussed in Section 11.3.2 to evaluate the following design Equation that determines the time necessary to achieve a specified conversion.

t=NA00XdX-rAV(2-9)

However, if you do not have that much time on your hands to form a table and use Chapter 2 integration techniques, then use Polymath, Wolfram, or MATLAB to solve the coupled differential forms of mole balance Equation (2-6) and the energy balance Equation (13-19) simultaneously.

Time on your hands.

Photograph shows a clock on a hand.

NA0dXdt=-rAV(26)

Example 13-1 Batch Reactor

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

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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 Jofos-tan. You charge (i.e., fill) 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 T2 Laboratory plant explosion in Florida, see page 696.

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

E=1R,000cal /mol,ΔHRX=-20,202cal/mol,CPA=35cal/mol/K,CPB=18cal/mol/CpC=46cal/mol/K,cal/ andCPM=19.5cal/mol/K.

Prof. Dr. Sven Köttlov objected to this Internet purchase and said that it should be recorded that he is skeptical of these parameter values. 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

A: Ethylene oxide: NA0 = (13.7 mol/dm3) (4 dm3) = 54.8 mol

B: Water:              NB0 = (55.5 mol/dm3) (10 dm3) = 555 mol

M: Methanol:        NM = (24.7 mol/dm3) (4 dm3) = 98.8 mol

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) Adiabatic Operation: Plot conversion and temperature X and Τ 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) Heat Exchange: Plot the temperature and conversion as a function of time when a heat exchanger is added. The product of the overall heat-transfer coefficient and exchange surface area is UA = 10 cal/s/K with TA1 = 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 (2009).

As before, there is approximately a 10°C (17°F) rise in temperature immediately after mixing.

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

dXdt=-rAVNA0(E13-1.1)

2. Rate Law:

-rA=kCA(E13-1.2)

3. Stoichiometry:

NA=NA0(1-X)(2-4)

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Recall that for a liquid-phase batch reaction V = V0

CA=NAV=NAV0=NA0(1-X)V0=CA0(1-X)(E13.1.3)

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

dXdt=k(1-X)(E13-1.4)

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

k=(4.71×109)exp[(ER)1T]s-1(E13-1.5)

With ER=9059K

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

k=(2.73×10-4)exp[9059K(1297-1T)]s-1(E13-1.6)

5. Energy Balance.

Part (a) Adiabatic Operation.

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

T=T0+[-ΔHRx(T0)]XΘiCpi+ΔCpX(E13-1.7)

6. Parameters:

Cps=ΣΘiCpi=CpA+CpBΘB+CpMΘM(E13-1.8)

CPS=35+18(55554.8)+19.5(98.854.8)=35+182.3+35.2=252.5calmolK

The lumped heat capacity of the solution, CPs is

CPS=252.2ca1mo1K

In the Equation for the heat of reaction (i.e., Eq. (E13-1.9)), we are going to neglect the second term on the right-hand side, i.e.,

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

ΔHRx(T)=ΔHRxo=-20cal/mol(E13-1.10)

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 that are initially at 3°C

T0=276+10=286K

T=T0+X(-ΔHRx)CPS=286K+X(-1(-20,202))calmol252.5cal/molK

T = 286 + 80X(E13-1.11)

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


dXdt=k(1-X)(E13-1.4)k=(2.73×10-4)exp[9059K(1297-1T)]s-1(E13-1.6)T=286+80X(E13-1.11)

where T is in K and t is in seconds.

A table similar to that used in Example 11-3 can now be constructed or one can make better use of his or her time using Polymath.


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 a f 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


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The Polymath equations do not include the runaway reaction that “ignites” at 350 K. A switching technique, SW, can be used to prevent the temperature from numerically going to infinity and could be used here as described in detail in Example 13-6.

We note for the initial temperature of 286 K that the reaction starts off relatively slowly then at approximately 1200 sec (20 minutes) “ignites” and the temperature rises rapidly to 370 K, at which point we reach complete conversion.

Graph depicts temperature–time trajectory.

Figure E13-1.1 Temperature-time trajectory.

Graph depicts conversion–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, m C, 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

dXdt=-rAVNA0(E13-1.1)

Neglecting the energy balance is

dTdt=-Q.gbQ.rbNA0CPs(E13-1.2)

Energy Balance

with

Q.rb=mcCPC{(T-Ta1)[1-exp(-UAmcCPC)]}(E13-1.13)

Q.gb=(-rAV)(-ΔHRxo)(E13-1.14)

-rAV=NA0k(1-X)(E13-1.15)

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

Parameters

CPS=252.5cal/mol/KCPC=4.16cal/g/Kmc=10g/s

NA0=54.8mo1,UA=10cal/K/s

Tal=290K,T286K

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

One notes after comparing Figure E13-1.1 and Figure E13-1.1 that the temperature profiles are not as steep as the adiabatic case. As expected the temperature of this exothermic reaction first increases as (Qgb>Qrb), goes through a maximum and then as the reactants have been consumed the temperature decreases because the heat “removed” is greater that the heat “generated,” i.e., (Qrb>Qgb). As a result the specific reaction-rate constant goes through a maximum as the fluid becomes heated at first and then later cooled. Figure E13-1.6 shows the trajectories of the “heat generated,” Qgb and “heat removed,” Qrb A plot of Qgb and Qrb as a function of time can often be useful in understanding the reaction dynamics. Whenever (Qgb>Qrb)

TABLE E13-1.3 POLYMATH PROGRAM

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Polymath Solution

Graph depicts polymath temperature–time trajectory.

Figure E13-1.3 Polymath temperature—time trajectory.

Graph depicts polymath conversion–time trajectory.

Figure E13-1.4 Polymath conversion—time trajectory.

Wolfram sliders are shown.

Figure E13-1.5 Wolfram sliders.

Graph depicts Wolfram heat-generated as heat removed trajectories.

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

dTdt=QgbQrbNA0ΣΘiCPi(E13-1.16)

the temperature will increase when (Qgb>Qrb) and the temperature will decrease when Qrb>Qgb..

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, Tai, T0, mc, etc., given in the problem statement. However, if you change these values as suggested in LEP Problem P13-1B (a), you will find situations where the conversion remains very low and cases where the temperature curve is extremely steep.

13.2.2 Case History of a Batch Reactor with Interrupted Isothermal Operation Causing a Runaway Reaction

In Chapters 5 and 6 we discussed the design of reactors operating isothermally. This operation may be achieved by efficient control of a heat exchanger. The following example shows what can happen when the heat exchanger suddenly fails.

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

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 a deep sleep. The explosion occurred in a batch reactor that was used to produce nitroaniline from ammonia and o-nitrochlorobenzene (ONCB):

Reaction involves o-nitrochlorobenzene and ammonia.

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.

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.

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.

A decision was made to triple production.

Living Example Problem

Photograph shows aftermath of an explosion at Monsanto plant with water sprayed to extinguish fire in the destructed remains of the plant.

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

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 H20. Normally, the reactor is charged with 3.17 kmol of ONCB, 103.6 kmol of H20, 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, after which 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: -rONCB =KCONCBCNH3

with k=0.00017m3kmo1min at 1 8 8°C (461K) 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:

V = 3.26 m3

ΔHRx=-5.9×105kcal/kmol

CPONCB=CPA=40cal/molK

CPH2=cPW=18CN1/mo1KCPNH3=CPB=8.38cal/molK

Assume that ΔCP0

UA=35.85kca1minCwith Ta=298K

Solution

A+2BC+D

Mole Balance:

dXdt=-rAyNA0=-rACA0(E13.2.1)

Rate Law:

-rA=kCACB(E13-2.2)

Stoichiometry (liquid phase):

CA=CA0(1-X)(E13-2.3)

CB=CA0(ΘB-2X)(E13-2.4)

with

ΘB=NB0NA0

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Combine:

-rA=kCA02(1-X)(ΘB-2X)(E13-2.5)

Substituting our parameter values into Equation (3-21)

k=k(T0)exp[ER(1T0-1T1)](3-21)

We obtain

k=0.00017exp[112731.987(1461-1T)]m3kmo1min

Energy Balance:

dTdt=UA(Ta-T)+(rAV)(ΔHRx)NiCPi(E13-2.6)

For ΔCP = 0,

ΣNiCPi=NCP=NA0WPA+NB0CPB+NWCPW

Parameter evaluation for day of explosion:

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

NCp = 2504 kcal/K

Qg=(rAV)(ΔHRx)

Qr=UA(T-Ta)

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

Living Example Problem

Again, let Qg be the heat generated, i.e., Qg = (rAV)(ΔHRx), and let Qr be the heat removed, i.e., Qr = UA(T- Ta).

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Then

dTdt=QgQrNCP(E13-2.8)

A. Isothermal Operation Up to 45 Minutes

The reaction takes place isothermally at 175°C (448 K) up to the time the heat exchanger fails and cooling stops at 45 min. we will calculate the conversion, X, the temperature, T, as well as Qr and Qg at the point that the heat exchanger fails. Next we calculate T, X, Qr, and Qg again 10 minutes later at the point when cooling is restored, and then again compare Qg and Qr.

Combining Equations (E13-2.1) with (E13-2.5) and canceling yields

dXdt=kCA0(1-X)(ΘB-2X)(E13-2.9)

ΘB=339.04=3.64

Integrating Equation (E13-2.9) gives us

t=[VkNA0](1ΘB-2)1n[ΘB-2XΘB(1-X)](E13-2.10)

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

Substituting for k and the other parameter values

45min=[5.119.m30.0001167m3/kmo1.min(9.044kmo1)]×(11.64)ln[3.64-2X3.64(1-X)]

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

we can now calculate the rate of generation, Qg, at this temperature and conversion and compare it with the maximum rate of heat removal, Qr, that is available for a constant coolant temperature of Ta = 298 K. The rate of generation, Qg, is

Qg=rAVΔHRx=kNA0(1-X)NA0[(NB0/NA0)-2X]V(-ΔHRx)V2(E13-2.11)

Comparing Qr and Qg at the Time Cooling Stops

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

Qg=(0.0001167)(9.0448)2(10.033)5.119[33(9.0448)2(0.033)]5.9×105

Qg=3830kcal/min

The corresponding maximum cooling rate is

Qr=UA(T298)=35.85(448298)(E13-2.12)

Qr=5378kcal/min

Everything is OK.

Therefore

Qr>Qg(E132.13)

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

Between t = 45 and t = 55 min, Qr = 0. The Polymath program was modified to account for the time of adiabatic operation by using an "if statement" for Qr in the program, i.e.,

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

Qr = if (t > 45 and t < 55) then (0) else (UA(T - 298))

A similar "if statement" is used for isothermal operation between t = 0 and t = 45 minutes, i.e., (dT/dt) = 0.

Comparing Qr and Qg After 10 minutes of Adiabatic Operation

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 Qg at 55 min and see that it has increased to

Qg= 6591 cal/min

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

Qr = 6093 kcal/min

we can also use the LEP Polymath Program to find these values for X, T, Qr, and Qg. In addition we note that by changing the run time from t = 45 minutes to t = 55 min-utes and by including an “if statement,” we can include the complete calculation of trajectory temperature and conversion versus time. The “if” statement will include the time of isothermal operation up to 45 min,

dTdt=if (t<45) than (0)rlse((QgQr)/NCp

The point of no return

Living Example Problem

and adiabatic operation between 45 and 55 and heat exchange thereafter.

Qr = if (45 < t < 55) then (0) else (UA (T - 298))

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

Qg>Qr(E132.14)

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, Qr = UA(T - 298), at 55 min. Table E13-2.4 gives the Polymath program to determine the temperature-time trajectory. (Note that one can change Naq and Nbq 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, Qr will still be greater than Qg 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 Qg and Qr at 55 min and substituting into Equation (E13-2.8), we find that

dTdt=(6591kca1/min-(6093kca1/min)2504kca1/C=0.2oC/min

TABLE E13-2.1 POLYMATH PROGRAM


Images

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Figure E13-2.2 Temperature—time trajectory.

The explosion occurred shortly after midnight.

A cartoon shows a person running away from smoke out of a cylindrical can with two other persons looking at him.

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 causes the temperature to increase exponentially. The reaction is running away! One observes in Figure E13-2.2 that 119 min after the batch reaction 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 NCp 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 X 109 J, which is enough energy to lift the entire 2500-ton building 300 m (the length of three football fields) straight up. In problem P13-1B (b) you are asked to use the Wolfram sliders to explore this example, i.e., E13-2, and find such things as the value of amount of reactant fed below which no explosion would have o ;urred.

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, then 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, mvap, 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

Images
A reactor is shown with spring relief valve at the top of the reactor, pressure gauge in the middle, and rupture disk at the bottom of the reactor.

Qr=m.vapΔHvap+UA(T-Ta)=830kgmin×540kca1kg+35.83kca1minK(538-298)K=4.48×105kca1min+8604kca1min=456,604kca1min

This value of Qr is much greater than the “heat generated” Qg

(Qg = 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.

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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) described in the Professional Reference Shelf (http://www.umich.edu/~elements/5e/13chap/pdf/CD_ARSST_ProfRef.pdf).

13.3 Batch and Semibatch Reactors with a Heat Exchanger

In our past discussions of reactors with heat exchangers, we assumed that the ambient temperature Ta was spatially uniform throughout the exchangers. This assumption is true if the system is a tubular reactor with the external pipe surface exposed to the atmosphere, or if the system is a CSTR or BR where the coolant flow rate through the exchanger is so rapid that the coolant temperatures entering and leaving the exchanger are virtually the same.

We now consider the case where the coolant temperature varies along the length of the exchanger while the temperature in the reactor is spatially uniform. The coolant enters the exchanger at a mass flow rate, rhc, at a temperature, Ta1, and leaves at a temperature, Ta2 (see Figure 13-1). Figure 13-1 could represent a CSTR as shown, or a batch reaction (BR) if the inlet and outlet flow rate were set to zero, i.e., FA = FA0 = 0, or a semibatch if the outlet streams are set equal to zero, i.e., FA = FB = 0.

A tank reactor with heat exchanger is shown.

Figure 13-1 Tank reactor with heat exchanger.

As a first approximation, we assume a quasi-steady state for the coolant flow and neglect the accumulation term (i.e., dTa I dt = 0 ). As a result, Equation (12-19) will give the rate of heat transfer from the exchanger to the reactor

Q.=m.CCpc(Ta1-T)[1-exp(-UA/m.CCpc)](12-19)

The energy balance on a semibatch reactor is

dTdt=QWsΣFi0CPi(TTi0)[ΔHRx(T)](rAV)ΣNiCPi(139)

Using a slight rearrangement of the heat exchanger energy balance equation, i.e., Equation (12-9) and neglecting shaft work, W.s

dTdt=Qgs˙Qrs˙ΣNiCPi(1320)

where the “heat generated,” Q.gs, for a the semibatch reactor is the same as that for a BR,

Qgs˙=(rAV)(ΔHRx)(1321)

however, the “heat removed” for a semibatch reactor, Q.gs, is

Qrs˙=ΣFi0CPi(TTi0)+m˙CPC[TTa1][1exp(UAm˙CPC)](1322)

Note: The subscript “s” in Qrs and Qgs means these equations only apply to a semibatch reactor.

For large coolant flow rates, Equation (13-22) reduces to

dTdt=(rAV)(ΔHRx)Qgs[ΣFi0CPi(TTi0)+UA(TTa)]QrsΣNiCPi(1323)

One added complexity for the semibatch reactor is that the heat transfer area, A, could vary with time. For example, as the tank in Figure 13-1 fills up, the reacting fluid that comes in contact with the heat exchange area could change significantly as the fluid level rises. This change in area will depend on the reactor geometry, how the exchanger is placed in the reactor, and the initial and final fluid volumes in the reactor. In the semibatch reactor examples that follow, we will assume the surface area, A, remains virtually constant, although this assumption will be analyzed in problem P13-1 (d) (vii).

13.3.1 Startup of a CSTR

Startup of a CSTR

In reactor startup, it is often very important how temperature and concentrations approach their steady-state values. For example, a significant overshoot in temperature may cause a reactant or product to degrade, or the overshoot may be unacceptable for safe operation, such as initiating a secondary reaction that causes runaway. If either case were to occur, we would say that the system exceeded its practical stability limit. The practical limit is specific to the specific reaction and conditions under which the reaction is carried out, and is usually determined by the reaction safety engineer. Although we can solve the unsteady temperature-time and concentration-time equations numerically to see if such a limit is exceeded, it is often more insightful to study the approach to steady state by using the temperature-concentration phase plane. To illustrate these concepts, we shall confine our analysis to a liquid-phase reaction carried out in a CSTR.

Images

A qualitative discussion of how a CSTR approaches steady state is given in PRS R13.5 on the CRE Web site (http://www.umich.edu/~elements/5e/13chap/prof-steadystate.html). This analysis, summarized in Figure PRS13.5 in the Sum-mary is developed to show the four different regions into which the phase plane is divided and how they allow one to sketch the approach to the steady state.

Example 13-3 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, Cwi = 55.3 kmol/m3, at T = 297 K and 0.1 wt % H2SO4 in the 1.89 m3 reactor. The feed stream consists of 36.3 kmol/h of propylene oxide (A), 453.6 kmol/h of water (B) containing 0.1 wt % H2SO4, and 45.4 kmol/h of methanol (M).

The water coolant flows through the heat exchanger at a rate of 2.27 kg/s (453.6 kmol/h). The molar densities of pure propylene oxide (A), water (B), and methanol (M) are pA0 = 14.8 kmol/m3, pB0 = 55.3 kmol/ m3, and pM0 = 24.7 kmol/ m3, respectively.

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

Additional information:

UA=7262kca1hKwithTa1=288.7Kwith  CPW=18kcal/kmol.kCPA=35kcal/kmol.k, CPB=18 kcal/kmol.k CPC=46kcal/kmol.k, CPM=19.5 kcal/kmol.k 

Again, the temperature of the mixed reactant streams entering the CSTR is T0 = 297K.

A CSTR with heat exchanger is shown.

Solution

A + BC

Mole Balances:

Images

A:dCAdt=rA+(CA0-CA)v0VCAi=0(E13-3.1)B:dCBdt=rB+(CB0-CB)v0VCBi=55.3kmo1m3(E13-3.2)C:dCCdt=rA+-CCv0VCCi=0(E13-3.3)M:dCMdt=rA+v0(CM0-CM)VCMi=0(E13-3.4)

Rate Law:

-rA=kCA(E13-3.5)

Stoichiometry:

-rA=-rB=rC(E13.3.6)

Energy Balance:

dTdt=Q.gs-Q.rsNCP(E13-3.7)

where

Qgs=(rAV)(ΔHRx)(E13.3.8)

using

Fi0CPi=FA0ΘiCPi(E13-3.9)

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)

Qrs=FA0ΣΘiCPiQrs1(T-T0)+mCCPC(T-Ta1)[Qrs21-exp(-UAmcCPC)](E13-3.10)

Q.rs=Q.rs1+Q.rs2

Ta2=T-(T-Ta1)exp(-UAm.cCPW)(12-17)

Evaluation of Parameters:

NCP=NiCPi=CPANA+CPBNB+CPCNC+CPMNM(E13-3.11)=35(CAV)+18(CBV)+46(CCV)+19.5(CMV)

ΘiCPj=CPA+FB0FA0CPB+FM0FA0CPM=35+18FB0FA0+19.5FM0FA0(E13-3.12)

υ0=FA0pA0+FB0pB0+FM0pM0=(FA.0148+FB055.3+FM.02471m3h(E13-3.13)

Neglecting ACP 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

ΔHRx=-20,013kca1kmo1A

The Polymath program is shown in Table E13-3.1.

Figures E13-3.1 and E13-3.2 show the concentration of propylene oxide and reactor temperature as a function of time, respectively, for an initial temperature of Ti = 297 K and only water in the tank (i.e., CAi = 0). One observes that both the temperature and concentration oscillate around their steady-state values (T = 331 K, CA = 0.687 kmol/m3) as steady state is approached.

Graph depicts propylene oxide concentration as a function of time for C subscript Ai equals 0 and T subscript i equals 297 K.

Figure E13-3.1 Propylene oxide concentration as a function of time for CAi = 0 and Ti = 297 K.

Graph depicts temperature–time trajectory for CSTR startup for C subscript Ai equaling 0 and T subscript i equaling 297 K.

Figure E13-3.2 Temperature—time trajectory for CSTR startup for CAi = 0 and Ti = 297 K.

Graph depicts concentration–temperature phase-plane trajectory.

Figure E13-3.3 Concentration—temperature phase-plane trajectory using Figures E13-3.1 and E13-3.2.

Graph depicts concentration–temperature phase-plane for three different initial conditions.

Figure E13-3.4 Concentration—temperature phase-plane for three different initial conditions.

Figure E13-3.3 combines Figures E13-3.1 and E13-3.2 into a phase plot of CA versus T. The final steady state operating concentration of A is 0.658 kmol/m3 at a temperature of 331.5 K. The arrows on the phase-plane plots show the trajectories with increasing time. The maximum temperature reached during startup is 337.5 K, which is below the practical stability limit of 355 K.

Unacceptable startup

Next, consider Figure E13-3.4, which shows three different trajectories for three different sets of initial conditions:

(1) Ti = 297 K      CAi = 0 (same as Figure E13-3.3)

(2) Ti = 339 K      CAi = 0

(3) Ti = 344 K      CAi = 2.26 kmol/m3

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Oops! The practical stability limit was exceeded

After three hours, the reaction is operating at steady state and all three trajectories converge on the final steady-state temperature of 331 K and the corresponding concentrations

CA=0.658kmol/m3CC=2.25kmol/m3CB=34kmol/m3CM=3.63kmol/m3T=331.5K

One can use the Wolfram sliders in the LEP for Example 13-4 to observe how the parameters change the trajectories as the temperature and concentration approach steady state.

For this reaction system, the plant safety office believes that an upper temperature limit of 355 K 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 = 344 K and an initial concentration of 2.26 kmol/m3, the practical stability limit of 355 K would be exceeded as the reactor approached its steady-state temperature of 331 K. See the concentration-temperature trajectory in Figure E13-3.4.

TABLE E13-3.1 POLYMATH PROGRAM FOR CSTR STARTUP


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Figures E13-3.1 through E13-3.4 show the concentration and temperature time trajectories for the start up of a CSTR for different initial conditions.

Living Example Problem

The reader is encouraged to use Wolfram to explore this problem to learn the answers to questions such as: (1) what parameter values lead to the greatest and smallest number of oscillations before steady state is reached (see Problem P13-1B (c)).

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Analysis: One of the purposes of this example was to demonstrate the use of phase plots, e.g., T versus CA, 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.

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