# Mole Balances

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## 1.4 Continuous-Flow Reactors

Continuous-flow reactors are almost always operated at steady state. We will consider three types: the continuous-stined tank reactor (CSTR), the plug-flow reactor (PFR), and the packed-beâ reactor (PBR). Detailed physical descriptions of these reactors can be found in both the Professional Reference Shelf (PRS) for Chapter 1 and in the Visual Encyclopedia of Equipment, http://encyclopedia.che.engin.umkh.edu/Pages/Reactors/CSTR/CSTR.html, and on the CRE Web site.

### 1.4.1 Continuous-Stirred Tank Reactor (CSTR)

A type of reactor commonly used in industrial processing is the stirred tank operated continuously (Figure 1-7). It is referred to as the continuous-stined tankreactor (CSTR) or vat, or backmix reactor, and is primarily used for

Figure 1-7(a) CSTR/batch reactor. (Photo courtesy of Pfaudler, Inc.)

Figure 1-7(b) CSTR mixing patterns.

liquid-phase reactions. It is normally operated at steady state and is assumed to be perfectly mixed; consequently there is no time dependence or position dependence of the temperature, concentration, or reaction rate inside the CSTR. That is, every variable is the same at every point inside the reactor. Because the temperature and concentration are identical everywhere within the reaction vessel, they are the same at the exit point as they are elsewhere in the tank. Thus, the temperature and concentration in the exit stream are modeled as being the same as those inside the reactor. In systems where mixing is highly nonideal, the well-mixed model is inadequate, and we must resort to other modeling techniques, such as residence time distributions, to obtain meaningful results. This topic of nonideal mixing is discussed on the Web site in PDF Chapters 16, 17, and 18 on nonideal reactors.

When the general mole balance equation

Fj0Fj+Vrj dV=dNjdt(1-4)

is applied to a CSTR operated at steady state (i.e., conditions do not change with time),

dNjdt=0

in which there are no spatial variations in the rate of reaction (i.e., perfect mixing),

VrjdV=Vrj

it takes the familiar form known as the design equation for a CSTR

The CSTR design equation gives the reactor volume V necessary to reduce the entering molar flow rate of species j from Fj0 to the exit molar flow rate Fj, when species; is disappearing at a rate −rj. We note that the CSTR is modeled such that the conditions in the exit stream (e.g., concentration and temperature) are identical to those in the tank. The molar flow rate Fj is just the product of the concentration of species j and the volumetric flow rate υ

Similarly, for the entrance molar flow rate we have Fj0 = Cj0 · υ0. Consequently, we can substitute for Fj0 and Fj into Equation (1-7) to write a balance on species A as

V=v0CA0vCArA(1-9)

The ideal CSTR mole balance equation is an algebraic equation, not a differential equation.

### 1.4.2 Tubular Reactor

In addition to the CSTR and batch reactors, another type of reactor commonly used in industry is the tubular reactor. It consists of a cylindrical pipe and is normally operated at steady state, as is the CSTR. Tubular reactors are used most often for gas-phase reactions. A schematic and a photograph of industrial tubular reactors are shown in Figure 1-8.

Figure 1-8(a) Tubular reactor schematic. Longitudinal tubular reactor. [Excerpted by special permission from Chem. Eng., 63(10), 211 (Oct. 1956). Copyright 1956 by McGraw-Hill, Inc., New York, NY 10020.]

Figure 1-8(b) Tubular reactor photo. Tubular reactor for production of Dimersol G. (Photo courtesy of Editions Techniq Institut Français du Pétrole.)

In the tubular reactor, the reactants are continually consumed as they flow down the length of the reactor. In modeling the tubular reactor, we assume that the concentration varies continuously in the axial direction through the reactor. Consequently, the reaction rate, which is a function of concentration for all but zero-order reactions (cf. Equation 3-2), will also vary axially For the purposes of the material presented here, we consider systems in which the flow field may be modeled by that of a plug-flow profile (e.g., uniform velocity

as in turbulent flow), as shown in Figure 1-9. That is, there is no radial variation in reaction rate, and the reactor is referred to as a plug-flow reactor (PFR). (The laminar-flow reactor is discussed in PDF Chapters 16 through 18 and on the Web site, along with a discussion of nonideal reactors.)

Figure 1-9 Plug-flow tubular reactor.

The general mole balance equation is given by Equation (1-4)

Fj0Fj+VrjdV=dNjdt(1-4)

The equation we will use to design PFRs at steady state can be developed in two ways: (1) directly from Equation (1-4) by differentiating with respect to volume V, and then rearranging the result or (2) from a mole balance on species j in a differential segment of the reactor volume ΔV. Let’s choose the second way to arrive at the differential form of the PFR mole balance. The differential volume, ΔV, shown in Figure 1-10, will be chosen sufficiently small such that there are no spatial variations in reaction rate within this volume. Thus the generation term, ΔGj is

Figure 1-10 Mole balance on species j in volume ΔV.

ΔGj=ΔVrj dV=rj ΔV

Dividing by ΔV and rearranging

[Fj|V+ΔVFj|VΔV]=rj

the term in brackets resembles the definition of a derivative

limΔx0[f(x+Δx)f(x)Δx]=dfdx

Taking the limit as ΔV approaches zero, we obtain the differential form of steady state mole balance on a PFR

We could have made the cylindrical reactor on which we carried out our mole balance an irregularly shaped reactor, such as the one shown in Figure 1-11 for reactant species A. However, we see that by applying Equation (1-10), the result would yield the same equation (i.e., Equation (1-11)). For species A, the mole balance is

Figure 1-11 Pablo Picasso’s reactor.

Consequently, we see that Equation (1-11) applies equally well to our model of tubular reactors of variable and constant cross-sectional area, although it is doubtful that one would find a reactor of the shape shown in 1-11 unless it were designed by Pablo Picasso or one of his followers.

The conclusion drawn from the application of the design equation to Picasso’s reactor is an important one: the degree of completion of a reaction achieved in an ideal plug-flow reactor (PFR) does not depend on its shape, only on its total volume.

Again consider the isomerization A → B, this time in a PFR. As the reac-tants proceed down the reactor, A is consumed by chemical reaction and B is produced. Consequently, the molar flow rate FA decreases, while FB increases as the reactor volume V increases, as shown in Figure 1-12.

Figure 1-12 Profiles of molar flow rates in a PFR

We now ask, “What is the reactor volume V1 necessary to reduce the entering molar flow rate of A from FA0 to FA1?” Rearranging Equation (1-12) in the form

dV=dFArA

and integrating with limits at V = 0, then FA = FA0, and at V = V1, then FA = FA1

V1=FA0FA1dFArA=FA1FA0dFArA(1-13)

V1 is the volume necessary to reduce the entering molar flow rate FA0 to some specified value FA1 and also the volume necessary to produce a molar flow rate of B of FB1.

### 1.4.3 Packed-Bed Reactor (PBR)

The principal difference between reactor design calculations involving homogeneous reactions and those involving fluid-solid heterogeneous reactions is that for the latter, the reaction takes place on the surface of the catalyst (see Figure 10-5). The greater the mass of a given catalyst, the greater the reactive surface area. Consequently, the reaction rate is based on mass of solid catalyst, W, rather than on reactor volume, V. For a fluid-solid heterogeneous system, the rate of reaction of a species A is defined as

rA=mol A reacted / (time×mass of catalyst)

The mass of solid catalyst is used because the amount of catalyst is what is important to the rate of product formation. We note that by multiplying the heterogeneous reaction rate, rA, by the bulk catalyst density, ρb(massvolume) , We can obtain the homogeneous reaction rate, −rA

rA=ρb(rA)

(moldm3s)=(gdm3)(molgs)

The reactor volume that contains the catalyst is of secondary significance. Figure 1-13 shows a schematic of an industrial catalytic reactor with vertical tubes packed with solid catalyst.

Figure 1-13 Longitudinal catalytic packed-bed reactor. [From Cropley, American Institute of Chemical Engineers, 86(2), 34 (1990).

In the three idealized types of reactors just discussed (the perfectly mixed batch reactor [BR], the plug-flow tubular reactor [PFR]), and the perfectly mixed continuous-stirred tank reactor [CSTR]), the design equations (i.e., mole balances) were developed based on reactor volume. The derivation of the design equation for a packed-bed catalytic reactor (PBR) will be carried out in a manner analogous to the development of the tubular design equation. To accomplish this derivation, we simply replace the volume coordinate in Equation (1-10) with the catalyst mass (i.e., weight) coordinate W (Figure 1-14).

Figure 1-14 Packed-bed reactor schematic.

As with the PFR, the PBR is assumed to have no radial gradients in concentration, temperature, or reaction rate. The generalized mole balance on species A over catalyst weight AW results in the equation

InOut+Generation=AccumulationFA|WFA|(w+Δw)+rAΔW=0(1-14)

The dimensions of the generation term in Equation (1-14) are

(rA)ΔWmoles A(time)(mass of catalyst)(mass of catalyst)molesAtime

which are, as expected, the same dimensions of the molar flow rate FA. After dividing by ΔW and taking the limit as ΔW → 0, we arrive at the differential Use the differential form of the mole balance for a packed-bed reactor

When pressure drop through the reactor (see Section 5.5) and catalyst decay (see Section 10.7 in Chapter 10) are neglected, the integral form of the packed-catalyst-bed design equation can be used to calculate the catalyst weight

W is the catalyst weight necessary to reduce the entering molar flow rate of species A, FA0, down to a molar flow rate FA.

For some insight into things to come, consider the following example of how one can use the tubular reactor design in Equation (1-11).

#### Example 1-2 How Large Is the Reactor Volume?

Consider the liquid phase cis – trans isomerization of 2–butene

which we will write symbolically as

AB

The reaction is first order in A (−rA = kCA) and is carried out in a tubular reactor in which the volumetric flow rate, υ, is constant, i.e., υ = υ0.

1. Sketch the concentration profile.

2. Derive an equation relating the reactor volume to the entering and exiting concentrations of A, the rate constant k, and the volumetric flow rate υ0 .

3. Determine the reactor volume, V1, necessary to reduce the exiting concentration to 10% of the entering concentration, i.e., CA = 0.1 CA0, when the volumetric flow rate υ0 is 10 dm3/min (i.e., liters/min) and the specific reaction rate, k, is 0.23 min−1.

Solution

1. Sketch CA as a function of V.

Species A is consumed as we move down the reactor, and as a result, both the molar flow rate of A and the concentration of A will decrease as we move. Because the volumetric flow rate is constant, υ = υ0 , one can use Equation (1-8) to obtain the concentration of A, CA = FA0, and then by comparison with the plot in Figure 1-12, obtain the concentration of A as a function of reactor volume, as shown in Figure E1-2.1.

Figure E1-2.1 Concentration profile.

2. Derive an equation relating V, υ0, k, CA0, and CA.

For a tubular reactor, the mole balance on species A (j = A) was shown to be given by Equation (1-11). Then for species A (j = A)

Mole Balance:

For a first-order reaction, the rate law (discussed in Chapter 3, Eq. (3-5)) is

Rate Law:

rA=kCA(E1-2.1)

Because the volumetric flow rate, υ, is constant (υ = υ0), as it is for most all liquid-phase reactions,

Multiplying both sides of Equation (E1-2.2) by minus one and then substituting Equation (E1-2.1) yields

Combine:

Separating the variables and rearranging gives

v0k(dCACA)=dV

Using the conditions at the entrance of the reactor that when V = 0, then CA = CA0

Carrying out the integration of Equation (E1-2.4) gives

Solve:

We can also rearrange Equation (E1-2.5) to solve for the concentration of A as a function of reactor volume to obtain

CA=CA0exp(kV/v0)

3. Calculate V. We want to find the volume, V1, at which CA=110CA0 for k = 0.23 min−1 and υ0 = 10 dm3/min.

Evaluate:

Substituting CA0, CA, υ0, and k in Equation (E1-2.5), we have

V=10dm3/min0.23min1lnCA00.1CA0=10dm30.23ln10=100dm3(i.e., 100L ; 0.1m3)

Let’s calculate the volume to reduce the entering concentration to CA = 0.01 CA0. Again using Equation (E1-2.5)

V=10dm3/min0.23min1lnCA00.01CA0=10dm30.23ln100=200dm3

Note: We see that a larger reactor (200 dm3) is needed to reduce the exit concentration to a smaller fraction of the entering concentration (e.g., CA = 0.01 CA0).

We see that a reactor volume of 0.1 m3 is necessary to convert 90% of species A entering into product B for the parameters given.

Analysis For this irreversible liquid-phase first order reaction (i.e., −rA = kC_) being carried out in a PFR, the concentration of the reactant decreases exponentially down the length (i.e., volume V) of the reactor. The more species A that are consumed and converted to product B, the larger must be the reactor volume V. The purpose of the example was to give a vision of the types of calculations we will be carrying out as we study chemical reaction engineering (CRE).

#### Example 1-3 Numerical Solutions to Example 1-2 Froblem: How Large is the Reactor Volume?

Now we will turn Example 1-2 into a Living Example Problem (LEP) where we can vary parameters to learn their effect on the volume and/or exit concentrations. We will use Polymath to solve the combined mole balance and rate law for the concentration profile.

We begin by rewriting the mole balance, Equation (E1-2.2), in Polymath notation form

Mole Balances

d(Ca)d(V)=ra/vo(E1-3.1)

d(Cb)d(V)=rb/vo(E1-3.2)

Rate Law

ra=k*Ca(E1-3.3)

rb=ra(E1-3.4)

k = 0.23

v0 = 10

A Polymath tutorial to solve the Ordinary Differential Equations (ODEs) can be found on the Web site (http://www.umich.edu/~elements/5e/tutorials/ODE_Equation_Tutorial.pdf).

The parameter values are k = 0.23 min-1, υο = 10 dm3/s and CA0 = 10 mole/dm3. The initial and final values for the integration wrt the volume V are V = 0 and V = 100 dm3.

The output from the Polymath solution is given in Table E 1-3.1 and the axial concentration profiles from species A and B are shown in Figure E1-3.1.

#### TABLE E1-3.1 POLYMATH PROGRAM AND OUTPUT FOR ISOTHERMAL PFR

Analysis Because Polymath will be used extensively in later chapters to solve non-linear ordinary differential equations (ODEs), we introduce it here so that the reader can start to become familiar with it. Figure E1-3.1 shows how the concentrations of species A and B vary down the length of the PFR.