Example 11-4 Calculating the Adiabatic Equilibrium Temperature and Conversion

For the elementary liquid-phase reaction

A 000 B

make a plot of equilibrium conversion as a function of temperature. The reacting species in this industrially important reaction to Jofostan’s economy have been coded with the letters A and B for proprietary reasons.

(a) Combine the rate law and stoichiometric to write − rA as a function of k, CA0, X, and Xe.

(b) Determine the adiabatic equilibrium temperature and conversion when species A and inert I are fed to the reactor at a temperature of 300 K.

(c) What is the CSTR volume necessary to achieve 90% of the adiabatic equilibrium conversion for υ0 = 5 dm3/min?

Additional information:

HAo(298 K)  = -46,000cal/molHAo(298 K)  = -60,000cal/molCPI = 50 cal/molK,CPA =  25 cal/molK,CPB = 25cal/molK,ΘI = 1k  =  0.000035exp (ER(1298-1T)min-1)     with   E=10,000calmolKe=75,000       at   298  K

Solution

(a) Combine rate law and stoichiometry to write — rA as a function of X

1. Mole Balance:

dXdV=rA/FA0                                                                                                    (E11–4.1)

2. Rate Law:

-rA  =   k(CA-CBKe)                                                                                                 (E114.2)

3. Stoichiometry: Liquid (v = v0)

We follow the algorithm Table 11-2 except we apply it to a liquid phase reaction in this case so that the concentrations are written as

CA=CA0(1X)(E11–4.3)

CB=CA0X(E11–4.4)

rA=kCA0(1XXKe)(E11–4.5)

At equilibrium: — rA = 0; so

Ke=CA0XeCA0(1Xe)=Xe(1Xe)(E11–4.6)

Substituting for Ke in terms of Xe in Equation (E11-4.5) and simplifying

rA=kCA0(1XXe)(E11-4.7)

Solving Equation (E11-4.6) for Xe gives

Xe=Ke(T)1 + Ke(T)(E11-4.8)

(b) Find adiabatic equilibrium temperature and conversion

4. Equilibrium Constant: Calculate ACP , then Ke (T ) as a function of temperature

ΔCP=CPBCPA=2525=0 cal/molK

Jofostan Journal of Thermodynamic Data, Vol. 23, p. 74 (1999).

Images

Following the Algorithm

For ACP = 0 , the equilibrium constant varies with temperature according to the van’t Hoff relation

Ke(T) = Ke(T1) exp[ΔHRxoR(1T1-1T)](E11–4.9)

ΔHRxο=HBοHAο=14,000 cal/molKe(T)=75,000 exp [14,0001.987(12981T)]  Ke=75,000 exp[23.64(T298T)](E11–4.10)

Substituting Equation (E11-4.10) into (E11-4.8), we can calculate the equilibrium conversion as a function of temperature:

5. Equilibrium Conversion from Thermodynamics:

Xe=75,000exp[23.64(T298)/T]1 + 75,000exp[23.64(T298)/T](E11-4.11)

We will use the tutorial in the Chapter 11 LEP “Xe vs. T” on the Web site to generate a figure of the equilibrium conversion as a function of temperature. We do this by “fooling” Polymath by using a dummy independent variable "t" to generate Xe and XEB. Next we change variables in plotting program to make T the independent variable and Xe and XEB the dependent variables to obtain Figure E11-4.1 (http://www.umich.edu/~elements/5e/software/Polymath_fooling_tutorial.pdf).

The calculations are shown in Table E11-4.1.

TABLE E11-4.1 EQUILIBRIUM CONVERSION AS A FUNCTION OF TEMPERATURE

T(K) Ke Xe k (min–1)
298 75,000.00 1.00 0.000035
350 2236.08 1.00 0.000430
400 180.57 0.99 0.002596
450 25.51 0.96 0.010507
500 5.33 0.84 0.032152
550 1.48 0.60 0.080279
620 0.35 0.26 0.225566

Note in this case the equilibrium constant Ke decreases by a factor of 105 and the reaction rate constant k increases by a factor of 104.

6. Energy Balance:

For a reaction carried out adiabatically, the energy balance, Equation (T11-1.A), reduces to

XEB = ΣΘiCPi(T-T0)-ΔHRx = (CPA + ΘICPI)(T-T0)-ΔHRxo(E11–4.12)

XEB=75(T480)14,000=5.36×103(T480)(E11-4.13)

Conversion calculated from energy balance

Data from Table E11-4.1 and the following data are plotted in Figure E11-4.1.

T(K) 480 525 575 620
XEB 0 0.24 0.51 0.75
Graph labeled Adiabatic equilibrium conversion shows a curve labeled X subscript e.

Figure E11-4.1 Finding the adiabatic equilibrium temperature (Te ) and conversion (Xe). Note: Curve uses approximate interpolated points.

Adiabatic equilibrium conversion and temperature

The intersection of XEB(T) and Xc(T) gives Xc = 0.49 and Tc = 572K.

(c) Calculate the CSTR volume to achieve 90% of the adiabatic equilibrium conversion corresponding to an entering temperature of 480 K

V = FA0X-rA = CA0υ0XkCA0(1-XXe) = υ0Xk(1-XXe)(E11–4.14)

k = (3.8×10-5exp[10,0001.987(1298-1T)]) = 3.9×10-5exp[16.89(T-298T)](E11–4.15)

From Figure E11-4.1 we see that for a feed temperature of 480 K, the adiabatic equilibrium temperature is 572 K and the corresponding adiabatic equilibrium conversion is only Xe = 0.49. The exit conversion is

X=0.9  Xe = 0.9(0.49) = 0.44

From the adiabatic energy balance, the temperature corresponding to X = 0.44 is

T = T0 + (-ΔHRxCPA + CPI)X = 480    K + 14,00075ca1molKca1mo1(0.44) = 562  K                                                 (E114.16)at    T = 562  K         Xe = 0.53  and  k  = 0.098  min1

V = (0.44)(5dm3min)0.098min1(10.440.53) = 132.2dm3

Analysis: The purpose of this example is to introduce the concept of the adiabatic equilibrium conversion and temperature. The adiabatic equilibrium conversion, Xe, is one of the first things to determine when carrying out an analysis involving reversible reactions. It is the maximum conversion one can achieve for a given entering temperature, T0, and feed composition. If Xe is too low to be economical, try lowering the feed temperature and/or adding inerts. From Equation (E11-4.6), we observe that changing the flow rate has no effect on the equilibrium conversion. For exothermic reactions, the adiabatic conversion decreases with increasing entering temperature T0, and for endothermic reactions the conversion increases with increasing entering T0. One can easily generate Figure E11-4.1 using Polymath with Equations (E11-4.5) and (E11-4.7).

If adding inerts or lowering the entering temperature is not feasible, then one should consider reactor staging.

11.6 Reactor Staging with Interstage Cooling or Heating

11.6.1 Exothermic Reactions

Conversions higher than those shown in Figure E11-4.1 can be achieved for adiabatic operations by connecting reactors in series with interstage cooling.

Illustration shows a reactor in series.

Figure 11-5 Reactor in series with interstage cooling.

In Figure 11-5, we show the case of an exothermic adiabatic reaction taking place in a PBR reactor train. The exit temperature to Reactor 1 is very high, 800 K, and the equilibrium conversion, Xe, is low as is the reactor conversion X1, which approaches Xe. Next, we pass the exit stream from adiabatic reactor 1 through a heat exchanger to bring the temperature back down to 500 K where Xe is high but X is still low. To increase the conversion, the stream then enters adiabatic reactor 2 where the conversion increases to X2, which is followed by a heat exchanger and the process is repeated.

The conversion-temperature plot for this scheme is shown in Figure 11-6. We see that with three interstage coolers, 88% conversion can be achieved, compared to an equilibrium conversion of 35% for no interstage cooling.

Graph shows a graph of a curve labeled X subscript e.

Figure 11-6 Increasing conversion by interstage cooling for an exothermic reaction.

Note: Lines and curves are approximate.

Interstage cooling X used for exothermic reversible reactions

11.6.2 Endothermic Reactions

Another example of the need for interstage heat transfer in a series of reactors can be found when upgrading the octane number of gasoline. The more compact the hydrocarbon molecule for a given number of carbon atoms is, the higher the octane rating (see Section 10.3.5). Consequently, it is desirable to convert straight-chain hydrocarbons to branched isomers, naphthenes, and aromatics. The reaction sequence is

Figure shows a straight chain at the left. To its right, a right arrow labeled k subscript 1 and a left arrow labeled Cat are shown. To its right, a hexagon labeled Napthenes is shown. To its right, a right arrow labeled k subscript 2 and a left arrow labeled Cat are shown. To its right, a circle within a hexagon labeled Aromatics is shown. To its right, plus 3 H subscript 2 is shown.

Typical values for gasoline composition

Gasoline

C5 10%
C6 10%
C7 20%
C8 25%
C9 20%
C10 10%
C11-C12 5%

The first reaction step ( k1) is slow compared to the second step, and each step is highly endothermic. The allowable temperature range for which this reaction can be carried out is quite narrow: Above 530°C undesirable side reactions occur, and below 430°C the reaction virtually does not take place to any reasonable extent. A typical feed stock might consist of 75% straight chains, 15% naphthas, and 10% aromatics.

One arrangement currently used to carry out these reactions is shown in Figure 11-7. Note that the reactors are not all the same size. Typical sizes are on the order of 10-m to 20-m high and 2 m to 5 m in diameter. A typical feed rate of gasoline is approximately 200 m3/ h at 2 atm. Hydrogen is usually separated from the product stream and recycled.

Schematic shows a heating circuit.

Figure 11-7 Interstage heating for gasoline production in moving-bed reactors.

Summer 2015 $2.89/gal for octane number (ON) ON = 89

Because the reaction is endothermic, the equilibrium conversion increases with increasing temperature. A typical equilibrium curve and temperature conversion trajectory for the reactor sequence are shown in Figure 11-8.

Graph shows a graph of a curve labeled X subscript e.

Figure 11-8 Temperature-conversion trajectory for interstage heating of an endothermic reaction analogous to Figure 11-6.

Interstage heating

Example 11-5 Interstage Cooling for Highly Exothermic Reactions

What conversion could be achieved in Example 11-4 if two interstage coolers that had the capacity to cool the exit stream to 480 K were available? Next, determine the heat duty of each exchanger for a molar feed rate of A of 40 mol/s. Assume that 95% of the equilibrium conversion is achieved in each reactor. The feed temperature to the first reactor is 480 K.

Solution

1. Calculate Exit Temperature

For the reaction in Example 11-4, i.e.,

A ⇄ B

we saw that for an entering temperature of 480 K, the adiabatic equilibrium conversion was 0.49. For 95% of the equilibrium conversion (Xe = 0.49), the conversion exiting the first reactor is 0.47. The exit temperature is found from a rearrangement of Equation (E11-4.12)

T = T0 + (-ΔHRxXCPA + ΘICPI) = 480 + 14,000X25 + 50 = 480 + (186.7)(0.47)                                           (E115.1)T = 568  K

We will now cool the gas stream exiting the reactor at T1 = 568 K back down to T2 = T0 = 480 K in a heat exchanger (Figure E11-5.1).

Graph shows a graph labeled Adiabatic equilibrium conversion with a curve labeled X subscript e.

Figure E11-5.1 Reaching 95% of adiabatic equilibrium conversion, i.e., X = 0.47, and then cooling back down to 480 K.

2. Calculate the Heat Load on the Heat Exchanger

The following calculations may be familiar from your course in Heat Transfer. There is no work done on the reaction gas mixture in the exchanger, and the reaction does not take place in the exchanger. Under these conditions (Fi|in = Fi|out), the energy balance given by Equation (11-10)

Q¨-W¨s + ΣFi0Hi0-ΣFiHi = 0                                                         (1110)

for W¨s = 0 becomes

Q¨ = Σ  FiHi-Σ   Fi0Hi0 = Σ  Fi0(Hi-Hi0)                                                         (E115.2)

 = Σ  FiCPi(T2-T1) = (FACPA + FBCPB + FiCPi)(T2-T1)                                                      (E11-5.3)

But CPA = CPB

Q  ¨ =  ((FA + FB)CPA + FICPI)(T2-T1)                                                                     (E115.4)

Energy balance on the reaction gas mixture in the heat exchanger

Also, for this example, FA0 = FA + FB

Q¨=FA0(CPA + CIΘPI)(T2-T1)=40 mols75 calmolK(480568)K=264kcals                                                                                                                                    (E11-5.5)        

That is, 264 kcal/s must be removed to cool the reacting mixture from 568 K to 480 K for a feed rate of 40 mol/s.

3. Second Reactor

Now let’s return to determine the conversion in the second reactor. Rearranging Equation (E11-4.7) for the second reactor

T2=T20 + ΔX(-ΔHRxoCPA)=480 + 186.7ΔX                                                                    (E11-5.6)

The conditions entering the second reactor are T20 = 480 K and X = 0.47. The energy balance starting from this point is shown in Figure E11-5.2. The corresponding adiabatic equilibrium conversion is 0.72. Ninety-five percent of the equilibrium conversion is 68.1% and the corresponding exit temperature is T = 480 + (0.68 - 0.47)186.7 = 519 K.

Graph shows three reactors in series to the left of a curve labeled Adiabatic equilibrium conversion.

Figure E11-5.2 Three reactors in series with interstage cooling. Note: Curve uses approximate interpolated points.

Bonding with Unit Operations and Heat Transfer

4. Heat Load

The heat-exchange duty to cool the reacting mixture from 519 K back to 480 K can again be calculated from Equation (E11-5.5)

Q  ¨ = FA0(CPA + ΘICPI)(480-519) = (40 mo1s)(75 ca1mol.K)(39)       = 117kcals

5. Subsequent Reactors

For the third and final reactor, we begin at T0 = 480 K and X = 0.68 and follow the line representing the equation for the energy balance along to the point of intersection with the equilibrium conversion, which is Xe = 0.82. Consequently, the final conversion achieved with three reactors and two interstage coolers is X = (0.95)(0.82) = 0.78.

Analysis: For highly exothermic, reversible reactions carried out adiabatically, reactor staging with interstage cooling can be used to obtain high conversions. One observes that the exit conversion and temperature from the first reactor are 47% and 568 K respectively, as shown by the energy balance line. The exit stream at this conversion is then cooled to 480 K where it enters the second reactor. ,n the second reactor, the overall conversion and temperature increase to 68% and 519 K. The slope of X versus T from the energy balance is the same as the first reactor. This example also showed how to calculate the heat load on each exchang e also note that the heat load on the third exchanger will be less than the first exchanger because the exit temperature from the second reactor (519 K) is lower than that of the first reactor (568 K). Consequently, less heat needs to be removed by the third exchanger.

11.7 Optimum Feed Temperature

We now consider an adiabatic reactor of fixed size or catalyst weight, and investigate what happens as the feed temperature is varied. The reaction is reversible and exothermic. At one extreme, using a very high feed temperature, the specific reaction rate will be large and the reaction will proceed rapidly, but the equilibrium conversion will be close to zero. Consequently, very little product will be formed. At the other extreme of low feed temperatures, the equilibrium conversion is high. So the question is, “Why not cool the feed to the lowest possible entering temperature, T0? The answer is that we know at low temperatures the rate constant k is small and it’s possible the reaction would not proceed to any reasonable extent, resulting in virtually no or little conversion. So we have these two extremes, low conversion at high temperatures due to equilibrium limitations and low conversion at low temperatures because of a small rate of reaction. Example 11-6 explores how to find the optimum temperature.

Example 11-6 Finding the Optimum Entering Temperature for Adiabatic Operation

We are going to continue the reaction we have been discussing in Examples 11-4 and 11-5.

A 00000 B

To plot the conversion profiles for different temperatures for the isomerization we will use all the same parameter values and conditions we used in Examples 11-4 and 11-5 and only vary the entering temperature. We start by applying Equations (T11-2.11) through (T11-2.8) to a liquid system so where there is no temperature dependence in the concentration term, i.e., CA = CA0 (1 - X). Starting with the mole balance

dXdV = -rA0FA0                                                                               (E11-6.1)

we combine the mole balance, rate law, and stoichiometry equations to get

dXdV = kCA0(1-X-XKe)                                                           (E11-6.2)

We next consider the energy balance for ACP = 0

T = T0 + X[-ΔHRxo]CPA + ΘCPII = T0 + 14,000X75 = T0 + 187X                                              (E11-6.3)

Xe = Ke1 + Ke                                                                     (E11-6.4)

We will use the data and equations for Ke, k, and XEB as given in Example E11-4.

Changing the Adiabatic Equilibrium Temperature and Conversion

Again using the values Xe as a function of T in Table E11-4.1 and the energy balance

XEB = (CPA + ΘCPII)(T-T0)-ΔHRx = 5.36×10-3(T-T0)                                             (E11-6.5)

Figure E11-6.1 shows a plot Xe and XEB on the same plot for different entering temperatures, i.e., T0

Graph shows a curve labeled Adiabatic equilibrium Conversion.

Figure E11-6.1 Adiabatic equilibrium temperature.

We see that for an entering temperature of T0 = 580 K, the adiabatic equilibrium conversion Xe is 0.24, and the corresponding adiabatic equilibrium temperature is 625 K. However, for an entering temperature of T0 = 380 K, Xe is 0.75 and the corresponding adiabatic equilibrium temperature is 520 K.

Conversion Profiles for X and Xe

To gain insight as to why there is an optimum feed temperature, let’s kok at the conversion profiles for different entering temperatures, T0. First we enter the mole balance, rate law, stoichiometric, and energy balance equations (cf. Equations (E11-6.1)through (E11-6.5)) into Polymath, MATLAB, and Wolfram. The Polymath program and output for T0 = 480 K is shown in Table E11-6.1.

TABLE E11-6.1 POLYMATH PROGRAM AND NUMERICAL OUTPUT

Images

The conversion profiles for X and Xe are shown in Figure E11-6.2 for T0 = 480 K where we see they approach each other up to the end of the 100 dm3 reactor.

Images

Wolfram Sliders

Graph shows two curves that represent Conversion profiles for T subscript 0 equals 480K.

Figure E11-6.2 Conversion profiles.

Because the reactor temperature increases as we move along the reactor, the equilibrium conversion, Xe, also varies and decreases along the length of the reactor, as shown in Figure E11-6.2.

We next plot the corresponding conversion profiles down the length of the reactor for entering temperatures of 380 K, 480 K, and 580 K, as shown in Figure E11-6.3.

For an entering temperature of 580 K, we see that the conversion and temperature increase very rapidly over a short distance (i.e., a small amount of catalyst). This sharp increase is sometimes referred to as the “point” or temperature at which the reaction “ignites.” We also observe that the conversion, which is relatively small at a value of 0.24, remains constant from V = 15 dm3 to the reactor exit. If the inlet temperature were lowered to 480 K, the corresponding equilibrium conversion would increase to 0.49; however, the reaction rate is slower at this lower temperature so that this conversion is not achieved until closer to the end of the reactor. If the entering temperature were lowered further to 380 K, the corresponding equilibrium conversion would be virtually 1.0, but the rate is so slow that a conversion of 0.03 is achieved for the specified reactor volume of 100 dm3. At very low feed temperatures, the specific reaction rate will be so small that virtually all of the reactant will pass through the reactor without reacting—that is, the reaction never “ignites” and little conversion is achieved.

Graph shows four curves that represent Equilibrium conversion for different feed temperatures.

Figure E11-6.3 Equilibrium conversion for different feed temperatures.

It is apparent that with conversions close to zero for both high and low feed temperatures, there must be an optimum feed temperature that maximizes conversion. As the feed temperature is increased from a very low value, the specific reaction rate will increase, as will the conversion. The conversion will continue to increase with increasing feed temperature until the equilibrium conversion is approached in the reaction. Further increases in feed temperature for this exothermic reaction will only decrease the conversion due to the decreasing equilibrium conversion. This optimum inlet temperature is shown in Figure E11-6.4.

If we made similar plots for other entering temperatures we would obtain the conversion at the exit of the reactor shown in Table E11-6.2

TABLE E11-6.2 EXIT CONVERSION AS A FUNTION OF ENTERING TEMPERATURE

T0 380 420 460 500 540 580 620 660
X 0.029 0.12 0.43 0.43 0.33 0.24 0.17 0.11

These exit conversion are plotted as a function of the entering temperature in Figure E11-6.4.

A bell curve represents conversion at exit of reactor.

Figure E11-6.4 Conversion at exit of reactor.

Analysis: We see that at high temperatures, the rate is rapid and equilibrium is reached near the entrance to the reactor and the conversion will be small. At the other extreme of low temperatures, the reaction never “ignites” and the reactants pass through the reactor virtually unreacted. At entering temperature between these two extremes we find there is an optimum entering feed temperature that maximizes conversion. The optimum feed temperature in this example is 474 K, and the corresponding conversion is 0.48.

Closure. Virtually all reactions that are carried out in industry involve heat effects. This chapter provides the basis to design reactors that operate at steady state and that involve heat effects. To model these reactors, we simply add another step to our algorithm; this step is the energy balance. One of the goals of this chapter is to understand each term of the energy balance and how it was derived. We have found that if the reader understands the various steps in the derivation, he/she will be in a much better position to apply the equation correctly. In order not to overwhelm the reader while studying reactions with heat effects, we have broken up the different cases and only consider the case of reactors operated adiabati-cally in this chapter. Chapter 12 will focus on reactors with heat exchange. Chapter 13 will focus on reactors not operated at steady state. An industrial adiabatic reaction concerning the manufacture of sulfuric acid provides a number of practical details is included on the CRE Web site in PRS R12.4 (http://www.umich.edu/~elements/5e/11chap/live.html and http://www.umich.edu/~elements/5e/12chap/pdf/sulfuricacid.pdf).

Building blocks read from bottom to top as follows: Mole Balance, Rate Law, Stoichiometry, Combine, and Energy Balance.

Summary

For the reaction

A + baB000caC + daD

1. The heat of reaction at temperature T, per mole of A, is

ΔHRx(T) = caHC(T) + daHD(T)-baHB(T)-HA(T)                                          (S11-1)

2. The mean heat capacity difference, ACp, per mole of A is

ΔCP = caCPC + daCPD-baCPB-CPA                                                     (S11-2)

where CPi is the mean heat capacity of species i between temperatures TR and T.

3. When there are no phase changes, the heat of reaction at temperature T is related to the heat of reaction at the standard reference temperature TR by

ΔHRx(T) = HRxo(TR) + ΔCP(T-TR)                                                                (S11-3)

4. The steady-state energy balance on a system volume V is

Q˙FA0ΣΘiCPi(TTi0)[ΔHRxo(TR) + ΔCP(TTR)]FA0X = 0(S11-4)

We now couple the first four building blocks Mole Balance, Rate Law, Stoichiometry, and Combine with the fifth block, the Energy Balance, to solve nonisothermal reaction engineering problems as shown in the Closure box for this chapter.

5. For adiabatic operation (Q¨0) of a PFR, PBR, CSTR, or batch reactor (BR), and neglecting Ws¨, we solve Equation (S11-4) for the adiabatic conversion-temperature relationship, which is

X = ΣΘiCPi(TT0)[ΔHRx(TR) + ΔCP(TTR)](S11-5)

Solving Equation (S11-5) for the adiabatic temperature-conversion relationship:

T = X[ΔHRxo(TR)] + ΣΘiCPiT0 + XΔCPTR[ΣΘiCPi + XΔCP] (S11-6)

Using Equation (S11-4), one can solve nonisothermal adiabatic reactor problems to predict the exit conversion, concentrations, and temperature.

Cre Web Site Materials

• Expanded Materials (http://umich.edu/~elements/5e/11chap/expanded.html)

1. Puzzle Problem “What’s Wrong with this Solution?"
(http://umich.edu/~elements/5e/11chap/expanded_ch11_puzzle.pdf)

• Learning Resources (http://umich.edu/~elements/5e/11chap/learn.html)

1. Summary Notes (http://umich.edu/~elements/5e/11chap/summary.html)

2. Self-Tests (http://umich.edu/~elements/5e/11chap/add.html)

A. Exercises (http://umich.edu/~elements/5e/11chap/add.html)

B. i>clicker Questions (http://umich.edu/~elements/5e/11chap/iclicker_ch11_q1.html)

3. PFR/PBR Solution Procedure for a Reversible Gas-Phase Reaction
(http://umich.edu/~elements/5e/11chap/learn-pfrpbr.html)

• Living Example Problems (http://umich.edu/~elements/5e/11chap/live.html)
Table 11-2 Algorithm for Gas Phase Reaction

1. Example 11-3 Adiabatic Isomerization of Normal Butane

2. Example 11-3 Formulated in AspenTech—Download from the CRE Web site (http://www.umich.edu/~elements/5e/software/aspen-example11-3.html)

3. Example 11-6 Finding the Optimum Entering Temperature for Adiabatic Operation

Screenshot shows a 11-3a - Aspen Plus 2006.5 - aspenONE window which displays RPlug Reactors tab.

A step-by-step AspenTech tutorial is given on the CRE Web site.

FAQs (Frequently Asked Questions) (http://umich.edu/~elements/5e/11chap/faq.html)

Professional Reference Shelf (http://www.umich.edu/~elements/5e/11chap/prof.html)

R11.1 Variable Heat Capacities (http://umich.edu/~elements/5e/11chap/pdf/CD-Ch08-VariableHeatCapacities.pdf). Next, we want to arrive at a form of the energy balance for the case where heat capacities are strong functions of temperature over a wide temperature range. Under these conditions, the mean values of the heat capacity may not be adequate for the relationship between conversion and temperature. Combining the heat of reaction with the quadratic form of the heat capacity

CPi = αi + βTi + γT2i

we find that the heat capacity as a function of temperature is

ΔHRx(T) = ΔHRxo(TR) + Δα(TTR) + Δβ2(T2TR2) + Δγ3(T3TR3)

Example 11-4 is reworked for the case of variable heat capacities (http://www.umich.edu/~elements/5e/11chap/prof.html).

Questions and Problems

The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.

A = B = C = D = 
Images

Homework Problems

In each of the questions and problems, rather than just drawing a box around your answer, write a sentence or two describing how you solved the problem, the assumptions you made, the reasonableness of your answer, what you learned, and any other facts that you want to include. See the Preface for additional generic parts (x), (y), (z) to the home problems.

Schematic shows two persons discussing about a sketch using a pencil and a set square.

Creative Problems

Before solving the problems, state or sketch qualitatively the expected results or trends.

Questions

Q11-1A i>clicker. Go to the Web site (http://www.umich.edu/~elements/5e/11chap/iclicker_ch11_q1.html) and view at least five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case: explaining why the question should not be on the next exam. In either case, explain your reasoning.

Q11-2A Suppose the reaction in Table 11-2 were carried out in BR instead of PFR, what steps in Table 11-2 would be different?

Q11-3A Rework Problem P2-9D for the case of adiabatic operation.

Q11-4A Prepare a list of safety considerations for designing and operating chemical reactors. What would be the first four items on your list? (See www.sache.org and www.siri.org/graphics.) The August 1985 issue of Chemical Engineering Progress may be useful.

Q11-5A What if you were asked to give an everyday example that demonstrates the principles discussed in this chapter? (Would sipping a teaspoon of Tabasco or other hot sauce be one?)

Q11-6A Example 11-1: What Additional Information Is Required? How would this example change if a CSTR were used instead of a PFR?

Q11-7A Example 11-2: Heat of Reaction. (1) What would the heat of reaction be if 50% inerts (e.g., helium) were added to the system? (2) What would be the % error if the ACp term were neglected?

Q11-8A Example 11-3: Adiabatic Liquid Phase Isomerization of Normal Butane. Can you explain why the CSTR volume is smaller than the PFR volume? Hint: Equation (E11-3.13) may be helpful to your explanation.

Q11-9A Read over the problems at the end of this chapter. Make up an original problem that uses the concepts presented in this chapter. See Problem P5-1B for guidelines.

Computer Simulations and Experiments

P11-1A Load the following programs from the CRE Web site where appropriate:

(a) Example Table 11.2: Algorithm for Gas Phase Reaction Wolfram

(i) Use the base case for all of the variables. Which two slider variables have the greatest effect on the temperature profile and which two have the least effect?

(ii) At what value of the heat of reaction does the rate near the entrance change from increasing with distance to decreasing with distance down the reactor?

(iii) How does the rate profile change as you vary T0?

(iv) What happens to X and Xe profiles as you vary T0? Can you explain this trend?

(v) Write a conclusion on what you found in experiments (i) through (iv).

(b) Example 11-3: Adiabatic Isomerization of Balance Wolfram

(i) Describe how both X and Xe profiles change for each of the parameter values.

(ii) What is the minimum feed temperature that should be maintained so that maximum possible conversion is achieved at the end of the reactor?

(iii) What is the critical value of KCD at which reaction rate starts to monotonically decrease along the volume of the reactor?

(iv) Vary ΔHRxο and observe the change in the temperature profile. Explain why at a low value of the heat of reaction, −ΔHRxο the temperature becomes virtually constant along the volume of the reactor.

(v) What parameter values affect the reaction rate profile, −rA, the most and in which ways do they change it?

Polymath

(vi) What if the butane reaction were carried out in a 0.8-m3 PFR that can be pressurized to very high pressures?

(vii) What inlet temperature would you recommend?

(viii) Is there an optimum inlet temperature?

(ix) Plot the heat that must be removed along the reactor (Q¨ vs.  V) to maintain isothermal operation.

(c) AspenTech Example 11-3. Download the AspenTech program from the CRE Web site. (1) Repeat using AspenTech. (2) Vary the inlet flow rate and temperature, and describe what you find.

(d) Example 11-4: Calculating the Adiabatic Equilibrium Temperature and Conversion Polymath

(i) Make a plot of the equilibrium conversion as a function of entering temperature, T0.

(ii) What do you observe at high and low T0?

(iii) Make a plot of Xe versus T0 when the feed is equal molar in inerts that have the same heat capacity.

(iv) Compare the plots of Xe versus T0 with and without inerts and describe what you find and write a conclusion.

(e) Example 11-5: Interstage Cooling for Highly Exothermic Reactions. (1) Determine the molar flow rate of cooling water (CPw = 18 cal/mol-K) necessary to remove 220 kcal/s from the first exchanger. The cooling water enters at 270 K and leaves at 400 K. (2) Determine the necessary heat exchanger area A (m2) for an overall heat-transfer coefficient, U, of 100 cal/s-m2-K. You must use the log-mean driving force in calculating Q¨.

Q˙ = UA[Th2Tc2)(Th1Tc1)]ln(Th2Tc2Th1Tc1)(E11-5.7)
Schematic shows countercurrent heat exchanger.

Figure P11-1A (e) Countercurrent heat exchanger.

Images

Hall of Fame

(f) Example 11-6: Optimum Feed Temperature

Wolfram

(i) Explore the optimum feed temperature. Start with the base case, and then change the parameters Ke, k, CA0, AHRx, ΘI, CP1, and CPA one at a time as you vary T0 to find the optimum entering temperature. Write two or more conclusions.

(ii) Set Ke2 at its maximum value, 150,000, and then vary T0. What differences do you observe from the base case?

(iii) Next set ΘI at 3, and again vary the other parameters. Write a sentence or two saying how each variable affects the optimum feed temperature.

(iv) Explain how inert affects the equilibrium conversion and actual conversion. What should be the inert quantity to achieve at least 50% conversion at the end of the reactor?

(v) Analyze how variation in heat of reaction affects the temperature in the reactor. What is the optimum heat of reaction so that maximum possible temperature is attained at the end of reactor?

(vi) Write a conclusion about what you found in experiments (i) through (v).

Problems

P11-2A For elementary reaction

A ⇄ B

the equilibrium conversion is 0.8 at 127°C and 0.5 at 227°C. What is the heat of reaction?

P11-3B The equilibrium conversion is shown below as a function of catalyst weight down a PBR

Graph shows a horizontal axis labeled W and a vertical axis labeled X subscript e that intersect at O. A curve starts from a point on the vertical axis, slides, and extends horizontally.

Please indicate which of the following statements are true and which are false. Explain each case.

(a) The reaction could be first-order endothermic and carried out adiabatically.

(b) The reaction is first-order endothermic and the reactor is heated along its length with Ta being constant.

(c) The reaction is second-order exothermic and cooled along the length of the reactor with Ta being constant.

(d) The reaction is second-order exothermic and carried out adiabatically.

P11-4A The elementary, irreversible, organic liquid-phase reaction

A + B → C

is carried out adiabatically in a flow reactor. An equal molar feed in A and B enters at 27°C, and the volumetric flow rate is 2 dm3/s and CA0 = 0.1 kmol/m3.

Additional information:

HAo(273K)   = -20    kcal/mol,HBο(273 K) = 15 kcal/mol,HCo(273K)   = -41    kcal/mol

CPA = CPB = 15cal/molK                               CPC = 30  cal/molKk = 0.01dm3mols  at 300 K          E =10,000 cal/mol

PFR

(a) Plot and then analyze the conversion and temperature as a function of PFR volume up to where X = 0.85. Describe the trends.

(b) What is the maximum inlet temperature one could have so that the boiling point of the liquid (550 K) would not be exceeded even for complete conversion?

(c) Plot the heat that must be removed along the reactor (Q¨   vs. V) to maintain isothermal operation.

(d) Plot and then analyze the conversion and temperature profiles up to a PFR reactor volume of 10 dm3 for the case when the reaction is reversible with KC = 10 m3/kmol at 450 K. Plot the equilibrium conversion profile. How are the trends different than part (a)? (Ans.: When V = 10 dm3 then X = 0.0051, Xeq = 0.517)

CSTR

(e) What is the CSTR volume necessary to achieve 90% conversion?

BR

(f) The reaction is next carried out in a 25 dm3 batch reactor charged with NA0 = 10 moles. Plot the number of moles of A, NA, the conversion, and the temperature as a function of time.

P11-5A The elementary, irreversible gas-phase reaction

 B + C

is carried out adiabatically in a PFR packed with a catalyst. Pure A enters the reactor at a volumetric flow rate of 20 dm3/ s, at a pressure of 10 atm, and a temperature of 450 K.

Additional information:

CPA = 40J/molK          CPB = 25J/molK            CPC = 15J/molKHAο = 70kJ/mol          HBο = 50kJ/mol            HCο = 40kJ/mol

All heats of formation are referenced to 273 K.

k    =   0.133   exp[ER(1450-1T)]dm3kgcats    with E   =   31.4  kJ/mol

(a) Plot and then analyze the conversion and temperature down the plug-flow reactor until an 80% conversion (if possible) is reached. (The maximum catalyst weight that can be packed into the PFR is 50 kg.) Assume that ΔΡ = 0.0.

(b) Vary the inlet temperature and describe what you find.

(c) Plot the heat that must be removed along the reactor (Q vs. V) to maintain isothermal operation.

(d) Now take the pressure drop into account in the PBR with Pb = 1 kg/dm3. The reactor can be packed with one of two particle sizes. Choose one.

α = 0.019/kg-cat for particle diameter D1      α = 0.0075/kg-cat for particle diameter D2

(e) Plot and then analyze the temperature, conversion, and pressure along the length of the reactor. Vary the parameters α and P0 to learn the ranges of values in which they dramatically affect the conversion.

(f) Apply one or more of the six ideas in Preface Table P-4, page xxvii, to this problem.

P11-6B The irreversible endothermic vapor-phase reaction follows an elementary rate law

CH3COCH3CH2CO+CH4AB+C

and is carried out adiabatically in a 500-dm3 PFR. Species A is fed to the reactor at a rate of 10 mol/min and a pressure of 2 atm. An inert stream is also fed to the reactor at 2 atm, as shown in Figure P11-6B. The entrance temperature of both streams is 1100 K.

Schematic shows a horizontal cylinder. F subscript (A0) equals 10 mol per min and F subscript 10 merge and pass horizontally through the left end of the cylinder labeled C subscript (A01). An arrow is shown at the right end of the cylinder.

Figure P11-6B Adiabatic PFR with inerts.

Additional information:

k = exp(34.34–34,222/T)1/sCPI = 200J/molK(Tin degress Kelvin, K)CPA = 170J/molKCPB = 90J/molKCPC = 180J/molKΔHRx = 80,000J/mol

(a) First derive an expression for CA01 as a function of CA0 and ΘΙ.

(b) Sketch the conversion and temperature profiles for the case when no inerts are present. Using a dashed line, sketch the profiles when a moderate amount of inerts are added. Using a dotted line, sketch the profiles when a large amount of inerts are added. Qualitative sketches are fine. Describe the similarities and differences between the curves.

(c) Sketch or plot and then analyze the exit conversion as a function of 0j. Is there a ratio of the entering molar flow rates of inerts (I) to A (i.e., ΘΙ = FI0/FA0) at which the conversion is at a maximum? Explain why there “is” or “is not” a maximum.

(d) What would change in parts (b) and (c) if reactions were exothermic and reversible with ΔH°Rx = -80 kJ/mol and KC = 2 dm3/mol at 1100 K?

(e) Sketch or plot FB for parts (c) and (d), and describe what you find.

(f) Plot the heat that must be removed along the reactor (Q vs. V) to maintain isothermal operation for pure A fed and an exothermic reaction. Part (f) is “C” level of difficulty, i.e., Pll-6C(f).

P11-7B The gas-phase reversible reaction

AB

is carried out under high pressure in a packed-bed reactor with pressure drop. The feed consists of both inerts I and Species A with the ratio of inerts to the species A being 2 to 1. The entering molar flow rate of A is 5 mol/min at a temperature of 300 K and a concentration of 2 mol/dm3. Work this problem in terms of volume. Hint: V=W/ρB,rA=ρBrA.

Additional information:

FA0 = 5.0mol/minT0 = 300KCA0 = 2mol/dm3T1 = 300KC1 = 2CA0k1 = 0.1min1at 300KCP1 = 18cal/mol/KUa = 150cal/dm3/min/KCPA = 160cal/mol/KTao = 300KE = 10,000cal/molV = 40dm3ΔHRx = 20,000cal/molαρb = 0.02dm3KC = 1,000cal/molCoolantKPs = 160cal/mol/Km˙C = 50mol/minρB = 1.2kg/dm3CPCool = 20cal/mol/K

(a) Adiabatic Operation. Plot X, Xe, p, T, and the rate of disappearance as a function of V up to V = 40 dm3. Explain why the curves look the way they do.

(b) Vary the ratio of inerts to A (0 ¡ < 10) and the entering temperature, and describe what you find. Plot the heat that must be removed along the reactor (Q vs. V) to maintain isothermal operation. Part (c) is “C” level of difficulty.

We will continue this problem in Chapter 12.

P11-8B Algorithm for reaction in a PBR with heat effects

The elementary gas-phase reaction

A + B ⇄ 2C

is carried out in a packed-bed reactor. The entering molar flow rates are FA0 = 5 mol/s, FB0 = 2FA0, and FI = 2FA0 with CT0 = 0.3 mol/dm3. The entering temperature is 330 K.

Additional information:

CPA = CPB = CPC = 20cal/mol/K,CPI = 40  cal/mol/K,  E = 25kcalmol,

ΔHRx = 20kcalmol@298K

α = 0.0002kg1,k = 0.004dm6kgmols@310K,

KC = 1000@303K

Note: This problem is continued in P12-1B with some of the entering conditions (e.g., FB0) modified.

(a) Write the mole balance, the rate law, KC as a function of T, k as a function of T, and CA, CB, CC as a function of X, p, and T.

(b) Write the rate law as a function of X, p, and T.

(c) Show the equilibrium conversion is

Xe = 3KC4(3KC4)22KC(FC41)2(KC41)

and then plot Xe vs. T.

(d) What are X© Cp , ACp, T0, entering temperature T1 (rate law), and T2 (equilibrium constant)?

(e) Write the energy balance for adiabatic operation.

(f) Case 1 Adiabatic Operation. Plot and then analyze Xe, X, p, and T versus W when the reaction is carried out adiabatically. Describe why the profiles look the way they do. Identify those terms that will be affected by inerts. Sketch what you think the profiles Xe, X, p, and T will look like before you run the Polymath program to plot the profiles. (Ans.: At W = 800 kg then X = 0.3583)

(g) Plot the heat that must be removed along the reactor ( Q vs. V) to maintain isothermal operation. Part (g) is “C” level of difficulty, i.e., P11-8C(g).

P11-9A The reaction

A + B ⇄ C + D

is carried out adiabatically in a series of staged packed-bed reactors with interstage cooling (see Figure 11-5). The lowest temperature to which the reactant stream may be cooled is 27°C. The feed is equal molar in A and B, and the catalyst weight in each reactor is sufficient to achieve 99.9% of the equilibrium conversion. The feed enters at 27°C and the reaction is carried out adiabatically. If four reactors and three coolers are available, what conversion may be achieved?

Additional information:

ΔHRx = 30,000cal/mol ACPA = CPB = CPC = CPD = 25cal/g molKKe(50C) = 500,000        FA0 = 10mol A/min

First prepare a plot of equilibrium conversion as a function of temperature. (Partial ans.: T = 360 K, Xe = 0.984; T = 520 K, Xe = 0.09; T = 540 K, Xe = 0.057)

P11-10A Figure P11-10A shows the temperature-conversion trajectory for a train of reactors with interstage heating. Now consider replacing the interstage heating with injection of the feed stream in three equal portions, as shown in Figure P11-10A:

Three squares with an X sign across it and curved edges are arranged horizontally and connected with arrow labeled 450 degrees Celsius. A downward arrow from 520 degrees Celsius is labeled X subscript e equals 0.3.

Figure P11-10A

Sketch the temperature-conversion trajectories for (a) an endothermic reaction with entering temperatures as shown, and (b) an exothermic reaction with the temperatures to and from the first reactor reversed, i.e., T0 = 450°C.

Supplementary Reading

1. An excellent development of the energy balance is presented in

ARIS, R., Elementary Chemical Reactor Analysis. Upper Saddle River, NJ: Prentice Hall, 1969, Chaps. 3 and 6.

A number of example problems dealing with nonisothermal reactors may or may not be found in

BURGESS, THORNTON W., The Adventures of Old Man Coyote. New York: Dover Publications, Inc., 1916.

BUTT, JOHN B., Reaction Kinetics and Reactor Design, Revised and Expanded, 2nd ed. New York: Marcel Dekker, Inc., 1999.

WALAS, S. M., Chemical Reaction Engineering Handbook of Solved Problems. Amsterdam: Gordon and Breach, 1995. See the following solved problems: 4.10.1, 4.10.08, 4.10.09, 4.10.13, 4.11.02, 4.11.09, 4.11.03, 4.10.11.

For a thorough discussion on the heat of reaction and equilibrium constant, one might also consult

DENBIGH, K. G., Principles of Chemical Equilibrium, 4th ed. Cambridge: Cambridge University Press, 1981.

2. The heats of formation, H¡ (T), Gibbs free energies, G¡ (TR), and the heat capacities of various compounds can be found in

GREEN, D. W. and R. H. PERRY, eds., Chemical Engineers’ Handbook, 8th ed. New York: McGraw-Hill, 2008.

REID, R. C., J. M. PRAUSNITZ, and T. K. SHERWOOD, The Properties of Gases and Liquids, 3rd ed. New York: McGraw-Hill, 1977.

WEAST, R. C., ed., CRC Handbook of Chemistry and Physics, 94th ed. Boca Raton, FL: CRC Press, 2013.

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