A reasonably general energy balance for a flow reactor can be written in English as
Enthalpy of input streams enthalpy of output streams þ heat generated by reaction heat transferred out
¼ accumulation of energy and in mathematics as
QininHin QoutoutHout VHRRR ^^ UUAextð ^TT TextÞ ¼
dðV ^ ^HHÞ
This is an integral balance written for the whole system. The various terms deserve discussion. The enthalpies are relative to some reference temperature, Tref: Standard tabulations of thermodynamic data (see Chapter 7) make it convenient to choose Tref¼ 298 K, but choices of Tref¼ 0 K or Tref ¼ 0C are also common. The enthalpy terms will normally be replaced by temperature using
H¼ ZT Tref
CPdT ð5:15Þ
For many purposes, the heat capacity will be approximately constant over the range of temperatures in the system. Then
H¼ CPðT TrefÞ ð5:16Þ
where CP is the average value for the entire reactant mixture, including any inerts. It may be a function of composition as well as temperature. An additional term—e.g, a heat of vaporization—must be added to Equations (5.15) and (5.16) if any of the components undergo a phase change. Also, the equations must be modified if there is a large pressure change during the course of the reaction. See Section 7.2.1.
By thermodynamic convention,HR< 0 for exothermic reactions, so that a negative sign is attached to the heat-generation term. When there are multiple reactions, the heat-generation term refers to the net effect of all reactions. Thus, the HRR term is an implicit summation over all M reactions that may be occurring: HRR ¼ X Reactions ðHRÞIRI¼ XM I¼1 ðHRÞIRI ð5:17Þ The reaction rates in Equation (5.17) are positive and apply to ‘‘the reaction.’’ That is, they are the rates of production of (possibly hypothetical) components having stoichiometric coefficients ofþ1. Similarly, the heats of reaction are per mole of the same component. Some care is needed in using literature values. See Section 7.2.1.
Chapter 7 provides a review of chemical thermodynamics useful for estimating specific heats, heats of reaction, and reaction equilibria. The examples here in Chapter 5 assume constant physical properties. This allows simpler illustrations of principles and techniques. Example 7.16 gives a detailed treatment of a rever- sible, gas-phase reaction where there is a change in the number of moles upon reaction and where the equilibrium composition, heat capacities, and reaction rates all vary with temperature. Such rigorous treatments are complicated but should be used for final design calculations. It is better engineering practice to include phenomena than to argue on qualitative grounds that the phenomena
are unimportant. Similarly, high numerical precision should be used in the calculations, even though the accuracy of the data may be quite limited. The object is to eliminate sources of error, either physical or numerical, that can be eliminated with reasonable effort. A sensitivity analysis can then be confined to the remaining sources of error that are difficult to eliminate. As a practical matter, few reactor design calculations will have absolute accuracies better than two decimal places. Relative accuracies between similar calculations can be much better and can provide justification for citing values to four or more decimal places, but citing values to full computational precision is a sign of naivete´.
The heat transfer term envisions convection to an external surface, and U is an overall heat transfer coefficient. The heat transfer area could be the reactor jacket, coils inside the reactor, cooled baffles, or an external heat exchanger. Other forms of heat transfer or heat generation can be added to this term; e.g, mechanical power input from an agitator or radiative heat transfer. The reactor is adiabatic when U¼ 0.
The accumulation term is zero for steady-state processes. The accumulation term is needed for batch reactors and to solve steady-state problems by the method of false transients.
In practice, the integral formulation of Equation (5.14) is directly useful only when the reactor is a stirred tank with good internal mixing. When there are temperature gradients inside the reactor, as there will be in the axial direction in a nonisothermal PFR, the integral balance remains true but is not especially useful. Instead, a differential energy balance is needed. The situation is exactly analogous to the integral and differential component balances used for the ideal reactors discussed in Chapter 1.
5.2.1 Nonisothermal Batch Reactors
The ideal batch reactor is internally uniform in both composition and tempera- ture. The flow and mixing patterns that are assumed to eliminate concentration gradients will eliminate temperature gradients as well. Homogeneity on a scale approaching molecular dimensions requires diffusion. Both heat and mass diffuse, but thermal diffusivities tend to be orders-of-magnitude higher than molecular diffusivities. Thus, if one is willing to assume compositional uniformity, it is reasonable to assume thermal uniformity as well.
For a perfectly mixed batch reactor, the energy balance is dðVHÞ
dt ¼ VHRR UAextðT TextÞ ð5:18Þ For constant volume and physical properties,
dT dt ¼ HRR CP UAextðT TextÞ VCP ð5:19Þ
Suppose that there is only one reaction and that component A is the limiting reactant. Then the quantity
Tadiabatic¼H Rain CP
ð5:20Þ gives the adiabatic temperature change for the reaction. This is the temperature that the batch would reach if the physical properties really were constant, if there were no change in the reaction mechanism, and if there were no heat transfer with the environment. Despite all these usually incorrect assumptions, Tadiabatic provides a rough measure of the difficulty in thermal management of a reaction. If Tadiabatic¼ 10 K, the reaction is a pussycat. If Tadiabatic¼ 1000 K, it is a tiger. When there are multiple reactions,HRR is a sum accord- ing to Equation (5.17), and the adiabatic temperature change is most easily found by setting U¼ 0 and solving Equation (5.19) simultaneously with the component balance equations. The long-time solution givesTadiabatic.
The N component balances are unchanged from those in Chapter 2, although the reaction rates are now understood to be functions of temperature. In matrix form,
dðaVÞ
dt ¼ m
R
V ð5:21ÞThe design equations for a nonisothermal batch reactor include Nþ1 ODEs, one for each component and one for energy. These ODEs are coupled by the temperature and compositional dependence of
R
. They may also be weakly coupled through the temperature and compositional dependence of physical properties such as density and heat capacity, but the strong coupling is through the reaction rate.Example 5.5: Ingredients are quickly charged to a jacketed batch reactor at an initial temperature of 25C. The jacket temperature is 80C. A pseudo-first- order reaction occurs. Determine the reaction temperature and the fraction unreacted as a function of time. The following data are available:
V¼ 1 m3 Aext¼ 4:68 m2 U¼ 1100 J=ðm2EsEKÞ ¼ 820 kg=m3 Cp¼ 3400 J=ðkgEKÞ k ¼ 3:7 108expð6000=TÞ HR¼ 108,000 J=mol ain¼ 1900:0 mol=m3
Physical properties may be assumed to be constant. Solution: The component balance for A is
da dt¼ ka
and the energy balance is dT dt ¼ HRR CP UAextðT TextÞ VCP ¼ Tadiabatic ka ain UAextðT TextÞ VCP where Tadiabatic¼ 73.6 K for the subject reaction. The initial conditions are a¼ 1900 and T ¼ 298 at t ¼ 0: The Arrhenius temperature dependence prevents an analytical solution. All the dimensioned quantities are in consistent units so they can be substituted directly into the ODEs. A numerical solution gives the results shown in Figure 5.2.
The curves in Figure 5.2 are typical of exothermic reactions in batch or tub- ular reactors. The temperature overshoots the wall temperature. This phenom- enon is called an exotherm. The exotherm is moderate in Example 5.2 but becomes larger and perhaps uncontrollable upon scaleup. Ways of managing an exotherm during scaleup are discussed in Section 5.3.
Advice on Debugging and Verifying Computer Programs. The computer programs needed so far have been relatively simple. Most of the problems can
120 100 80 60 40 20 0 0 0.5 1 Time, h 1.5 2 Temperature, °C 0 0 0.5 1 Time, h (a) (b) 1.5 2 Fraction unreacted 0.2 0.4 0.6 0.8 1
FIGURE 5.2 (a) Temperature and (b) fraction unreacted in a nonisothermal batch reactor with jacket cooling.
be solved using canned packages for ODEs, although learning how to use the solvers may take more work than writing the code from scratch. Even if you use canned packages, there are many opportunities for error. You have to spe- cify the functional forms for the equations, supply the data, and supply any ancillary functions such as equations of state and physical property relation- ships. Few programs work correctly the first time. You will need to debug them and confirm that the output is plausible. A key to doing this for physically motivated problems like those in reactor design is simplification. You may wish to write the code all at once, but do not try to debug it all at once. For the non- isothermal problems encountered in this chapter, start by running an isothermal and isobaric case. Set T and P to constant values and see if the reactant concen- trations are calculated correctly. If the reaction network is complex, you may need to simplify it, say by dropping some side reactions, until you find a case that you know is giving the right results. When the calculated solution for an isothermal and isobaric reaction makes sense, put an ODE for temperature or pressure back into the program and see what happens. You may wish to test the adiabatic case by setting U¼ 0 and to retest the isothermal case by setting Uto some large value. Complications like variable physical properties and vari- able reactor cross sections are best postponed until you have a solid base case that works. If something goes wrong when you add a complication, revert to a simpler case to help pinpoint the source of the problem.
Debugging by simplifying before complicating is even more important for the optimization problems in Chapter 6 and the nonideal reactor design problems in Chapters 8 and 9. When the reactor design problem is embedded as a subroutine inside an optimization routine, be sure that the subroutine will work for any parameter values that the optimization routine is likely to give it. Having trouble with axial dispersion? Throw out the axial dispersion terms for heat and mass and confirm that you get the right results for a nonisothermal (or even isother- mal) PFR. Having trouble with the velocity profile in a laminar flow reactor? Get the reactor program to work with a parabolic or even a flat profile. Separately test the subroutine for calculating the axial velocity profile by sending it a known viscosity profile. Put it back into the main program only after it works on its own. Additional complications like radial velocity components are added still later.
Long programs will take hours and even days to write and test. A systematic approach to debugging and verification will reduce this time to a minimum. It will also give you confidence that the numbers are right when they finally are produced.
5.2.2 Nonisothermal Piston Flow
Steady-state temperatures along the length of a piston flow reactor are governed by an ordinary differential equation. Consider the differential reactor element shown in Figure 5.3. The energy balance is the same as Equation (5.14) except
that differential quantities are used. TheQH terms cancel and z factors out to give: dðQHÞ dz ¼ Q dH dz ¼ Acuu dH dz ¼ HRR Ac UA 0 extðT TextÞ ð5:22Þ Unlike a molar flow rate—e.g, aQ—the mass flow rate, Q, is constant and can be brought outside the differential. Note that Q¼ uuAc and that A0ext is the external surface area per unit length of tube. Equation (5.22) can be written as dH dz ¼ HRRc uu UA0ext uuAc ðT T extÞ ð5:23Þ
This equation is coupled to the component balances in Equation (3.9) and with an equation for the pressure; e.g., one of Equations (3.14), (3.15), (3.17). There are Nþ2 equations and some auxiliary algebraic equations to be solved simulta- neously. Numerical solution techniques are similar to those used in Section 3.1 for variable-density PFRs. The dependent variables are the component fluxes , the enthalpy H, and the pressure P. A necessary auxiliary equation is the thermodynamic relationship that gives enthalpy as a function of temperature, pressure, and composition. Equation (5.16) with Tref¼ 0 is the simplest example of this relationship and is usually adequate for preliminary calculations.
With a constant, circular cross section, A0ext¼ 2R (although the concept of piston flow is not restricted to circular tubes). If CPis constant,
dT dz ¼ HRR uuCP 2U uuCPRðT T extÞ ð5:24Þ
This is the form of the energy balance that is usually used for preliminary calculations. Equation (5.24) does not require that uu be constant. If it is con- stant, we can set dz¼ uudt and 2/R ¼ Aext/Acto make Equation (5.24) identical to Equation (5.19). A constant-velocity, constant-properties PFR behaves
qlost= UA¢ext(T _ Text) Dz qgenerated= _DHR4 AcDz
qout= rQH + Dz
qin= rQH
z z + Dz
d( rQH) dz
identically to a constant-volume, constant-properties batch reactor. The curves in Figure 5.2 could apply to a piston flow reactor as well as to the batch reactor analyzed in Example 5.5. However, Equation (5.23) is the appropriate version of the energy balance when the reactor cross section or physical properties are variable.
The solution of Equations (5.23) or (5.24) is more straightforward when tem- perature and the component concentrations can be used directly as the depen- dent variables rather than enthalpy and the component fluxes. In any case, however, the initial values, Tin, Pin, ain, bin,. . . must be known at z ¼ 0. Reaction rates and physical properties can then be calculated at z¼ 0 so that the right-hand side of Equations (5.23) or (5.24) can be evaluated. This gives T, and thus Tðz þ zÞ, directly in the case of Equation (5.24) and implicitly via the enthalpy in the case of Equation (5.23). The component equations are evaluated similarly to give aðz þ zÞ, bðz þ zÞ, . . . either directly or via the concentration fluxes as described in Section 3.1. The pressure equation is evaluated to give Pðz þ zÞ: The various auxiliary equations are used as neces- sary to determine quantities such as uu and Ac at the new axial location. Thus, T, a, b,. . . and other necessary variables are determined at the next axial position along the tubular reactor. The axial position variable z can then be incremented and the entire procedure repeated to give temperatures and compo- sitions at yet the next point. Thus, we march down the tube.
Example 5.6: Hydrocarbon cracking reactions are endothermic, and many different techniques are used to supply heat to the system. The maximum inlet temperature is limited by problems of materials of construction or by undesir- able side reactions such as coking. Consider an adiabatic reactor with inlet temperature Tin. Then T(z)< Tinand the temperature will gradually decline as the reaction proceeds. This decrease, with the consequent reduction in reac- tion rate, can be minimized by using a high proportion of inerts in the feed stream.
Consider a cracking reaction with rate
R ¼ 10 14expð24,000=TÞa, g=ðm3EsÞ
where a is in g/m3. Suppose the reaction is conducted in an adiabatic tubular reactor having a mean residence time of 1 s. The crackable component and its products have a heat capacity of 0.4 cal/(gEK), and the inerts have a heat capacity of 0.5 cal/(gEK); the entering concentration of crackable component is 132 g/m3 and the concentration of inerts is 270 g/m3; Tin¼ 525C. Calculate the exit concentration of A given HR¼ 203 cal/g. Physical properties may be assumed to be constant. Repeat the calculation in the absence of inerts.
Solution: Aside from the temperature calculations, this example illustrates the systematic use of mass rather than molar concentrations for reactor
calculations. This is common practice for mixtures having ill-defined molecular weights. The energy balance for the adiabatic reactor gives
dT dt ¼ HRR CP ¼ T adiabatic ka ain
Note that and CP are properties of the reaction mixture. Thus, ¼ 132þ 270¼ 402 g/m3 and CP¼ [0.4(132) þ 0.5(270)]/402 ¼ 0.467 cal/(gEK). This gives Tadiabatic¼ 142:7 K. If the inerts are removed, 132 g/m3, CP¼ 0.4 cal/(gEK), and Tadiabatic¼ 507:5 K:
Figure 5.4 displays the solution. The results are aout¼ 57.9 g/m3 and Tout¼ 464.3C for the case with inerts and aout¼ 107.8 g/m3 and Tout¼ 431.9C for the case without inerts. It is apparent that inerts can have a remarkably beneficial effect on the course of a reaction.
In the general case of a piston flow reactor, one must solve a fairly small set of simultaneous, ordinary differential equations. The minimum set (of one) arises for a single, isothermal reaction. In principle, one extra equation must be added for each additional reaction. In practice, numerical solutions are some- what easier to implement if a separate equation is written for each reactive component. This ensures that the stoichiometry is correct and keeps the physics and chemistry of the problem rather more transparent than when the reaction coordinate method is used to obtain the smallest possible set of differential
150 100 50 0 0 0.2 0.4 0.6 0.8 1 Residence time, s Concentration, g/m 3 Without inerts With inerts
equations. Computational speed is rarely important in solving design problems of this type. The work involved in understanding and assembling and data, writing any necessary code, debugging the code, and verifying the results takes much more time than the computation.
5.2.3 Nonisothermal CSTRs
Setting ^TT ¼ Tout, ^HH¼ Hout, and so on, specializes the integral energy balance of Equation (5.14) to a perfectly mixed, continuous-flow stirred tank:
dðVoutHoutÞ
dt ¼ QininHin QoutoutHout VHRR UAextðTout TextÞ ð5:25Þ where HRR denotes the implied summation of Equation (5.17). The corre- sponding component balance for component A is
dðVaÞ
dt ¼ Qinain Qoutaoutþ VRA ð5:26Þ and also has an implied summation
RA¼ A,IRIþ A,IIRIIþ ð5:27Þ The simplest, nontrivial version of these equations is obtained when all physical properties and process parameters (e.g., Qin, ain, and Tin) are constant. The energy balance for this simplest but still reasonably general case is
ttdTout
dt ¼ Tin Tout
HRR tt CP
UAextðTout TextÞtt
VCp ð5:28Þ
The time derivative is zero at steady state, but it is included so that the method of false transients can be used. The computational procedure in Section 4.3.2 applies directly when the energy balance is given by Equation (5.28). The same basic procedure can be used for Equation (5.25). The enthalpy rather than the temperature is marched ahead as the dependent variable, and then Toutis calculated from Houtafter each time step.
The examples that follow assume constant physical properties and use Equation (5.28). Their purpose is to explore nonisothermal reaction phenomena rather than to present detailed design calculations.
Example 5.7: A CSTR is commonly used for the bulk polymerization of styrene. Assume a mean residence time of 2 h, cold monomer feed (300 K), adiabatic operation (UAext¼ 0), and a pseudo-first-order reaction with rate constant
where T is in kelvins. Assume constant density and heat capacity. The adiabatic temperature rise for complete conversion of the feed is about 400 K for undiluted styrene.
Solution: The component balance for component A (styrene) for a first- order reaction in a constant-volume, constant-density CSTR is
ttdaout
dt ¼ ain aout kttaout The temperature balance for the adiabatic case is
ttdTout dt ¼ Tin Tout HRR tt CP ¼ Tin Toutþ Tadiabatic ktta ain
Substituting the given values,
daout
d ¼ ain aout 2 1010expð10,000=ToutÞaout ð5:29Þ and
dTout
d ¼ Tin Toutþ 8 10
12expð10,000=T
outÞaout=ain ð5:30Þ where ¼ t=tt and Tin¼ 300 K. The problem statement did not specify ain. It happens to be about 8700 mol/m3for styrene; but, since the reaction is first order, the problem can be worked by setting ain¼ 1 so that aout becomes equal to the fraction unreacted. The initial conditions associated with Equations (5.29) and (5.30) are aout¼ a0 and Tout¼ T0 at ¼ 0. Solutions for a0¼ 1 (pure styrene) and various values for T0are shown in Figure 5.5.