5. Isothermal Reactor Design: Conversion

Why, a four-year-old child could understand this. Someone get me a four-year-old child.

—Groucho Marx

Schematic shows the knots made by tying ropes together.

Tying everything together

5.1 Design Structure for Isothermal Reactors

One of the primary goals of this chapter is to solve chemical reaction engineering (CRE) problems by using logic rather than memorizing which equation applies where. It is the author’s experience that following this structure, shown in Figure 5-1, will lead to a greater understanding of isothermal reactor design. We begin by applying our general mole balance equation (level ①) to a specific reactor to arrive at the design equation for that reactor (level ②). If the feed conditions are specified (e.g., NA0 or FA0), all that is required to evaluate the design equation is the rate of reaction as a function of conversion at the same conditions as those at which the reactor is to be operated (e.g., temperature and pressure). When –rA = f(X) is known or given, one can go directly from level ③ to the last level, level ⑨, to determine either the batch time or reactor volume necessary to achieve the specified conversion.

Isothermal-reaction design algorithm for conversion is depicted.

Figure 5-1 Isothermal-reaction design algorithm for conversion.

Logic vs. Memorization

Use the algorithm rather than memorizing equations.

When the rate of reaction is not given explicitly as a function of conversion, we must proceed to level ④, where the rate law must be determined by either finding it in books or journals or by determining it experimentally in the laboratory. Techniques for obtaining and analyzing rate data to determine the reaction order and rate constant are presented in Chapter 7. After the rate law has been established, one has only to use stoichiometry (level ⑤) together with the conditions of the system (e.g., constant volume, temperature) to express concentration as a function of conversion.

For liquid-phase reactions and for gas-phase reactions with no pressure drop (P = P0), one can combine the information in levels ④ and ⑤ to express the rate of reaction as a function of conversion and arrive at level ➅. It is now possible to determine either the time or reactor volume necessary to achieve the desired conversion by substituting the relationship linking conversion and rate of reaction into the appropriate design equation (level ⑨).

For gas-phase reactions in packed beds where there is a pressure drop, we need to proceed to level ⑦ to evaluate the pressure ratio p (that is, (p = P / P0)) in the concentration term using the Ergun equation (Section 5.5). In level ⑧, we combine the equations for pressure drop in level ⑦ with the information in levels ④ and ⑤ to proceed to level ⑨, where the equations are then evaluated in the appropriate manner (i.e., analytically using a table of integrals, or numerically using an ODE solver). Although this structure emphasizes the determination of a reaction time or reactor volume for a spec-ified conversion, it can also readily be used for other types of reactor calculations, such as determining the conversion for a specified volume. Different manipulations can be performed in level ⑨ to answer the different types of questions mentioned here.

The structure shown in Figure 5-1 allows one to develop a few basic concepts and then to arrange the parameters (equations) associated with each concept in a variety of ways. Without such a structure, one is faced with the possibility of choosing or perhaps memorizing the correct equation from a multitude of equations that can arise for a variety of different combinations of reactions, reactors, and sets of conditions. The challenge is to put everything together in an orderly and logical fashion so that we can arrive at the correct equation for a given situation.

The Algorithm

1. Mole balance

2. Rate law

3. Stoichiometry

4. Combine

5. Evaluate

Fortunately, by using the algorithm to formulate CRE problems shown in Figure 5-2, which happens to be analogous to the algorithm for ordering dinner from a fixed-price menu in a fine French restaurant, we can eliminate virtually all memorization. In both of these algorithms, we must make choices in each category. For example, in ordering from a French menu, we begin by choosing one dish from the appetizers listed. Analogous to the French menu, in Step 1 of the CRE algorithm shown in Figure 5-2, we begin by choosing the appropriate mole balance for one of the three types of reactors. After making our reactor choice (e.g., PFR), we go to the small oval ready to make our second choice in Step 2, Rate Laws. In Step 2 we choose the rate law (entrée), and in Step 3 we specify whether the reaction is gas or liquid phase (cheese or dessert). Finally, in Step 4 we combine Steps 1, 2, and 3 and either obtain an analytical solution or solve the equations using an ODE solver. The complete French menu is on the CRE Web site (www.umich.edu/~elements/5e/index.html), Chapter 5 Summary Notes.

Algorithm for isothermal reactors is shown.

Figure 5-2 Algorithm for isothermal reactors.

We now will apply this algorithm to a specific situation. Suppose that we have, as shown in Figure 5-2, mole balances for three reactors, three rate laws, and the equations for concentrations for both liquid and gas phases. In Figure 5-2, we see how the algorithm is used to formulate the equation to calculate the PFR reactor volume for a first-order gas-phase reaction. The pathway to arrive at this equation is shown by the ovals connected to the dark lines through the algorithm. The dashed lines and the boxes represent other pathways for solutions to other situations. The algorithm for the pathway shown in Figure 5-2 is

1. Mole balances, choose species A reacting in a PFR

2. Rate laws, choose the irreversible first-order reaction

3. Stoichiometry, choose the gas-phase concentration

4. Combine steps 1, 2, and 3 to arrive at Equation A

5. Evaluate. The combine step can be evaluated either

a. Analytically (Appendix A.1)

b. Graphically (Chapter 2)

c. Numerically (Appendix A.4)

d. Via software (e.g., Polymath, Wolfram, MATLAB)

Substitute parameter values in steps 1–4 only if they are zero.

We can solve the equations in the combine step either

1. Analytically (Appendix A.1)

2. Graphically (Chapter 2)

3. Numerically (Appendix A.4)

4. Using software (e.g., Polymath).

In Figure 5-2, we chose to integrate Equation A for constant temperature and pressure to find the volume necessary to achieve a specified conversion (or calculate the conversion that can be achieved in a specified reactor volume). Unless the parameter values are zero, we typically do not substitute numerical values for parameters in the combine step until the very end.

For the case of isothermal operation with no pressure drop, we were able to obtain an analytical solution, given by Equation B in Figure 5-2, which gives the reactor volume necessary to achieve a conversion X for a first-order gas-phase reaction carried out isothermally in a PFR. However, in the majority of situations, analytical solutions to the ordinary differential equations appearing in the combine step are not possible. Consequently, we include Polymath, or some other ODE solver such as MATLAB, in our menu in that it makes obtaining solutions to the differential equations much more palatable.

5.2 Batch Reactors (BRs)

One of the jobs in which chemical engineers are involved is the scale-up of laboratory experiments to pilot-plant operation or to full-scale production. In the past, a pilot plant would be designed based on laboratory data. In this section, we show how to analyze a laboratory-scale batch reactor in which a liquid-phase reaction of known order is being carried out.

In modeling a batch reactor, we assume there is no inflow or outflow of material and that the reactor is well mixed. For most liquid-phase reactions, the density change with reaction is usually small and can be neglected (i.e., V = V0). In addition, for gas-phase reactions in which the batch reactor volume remains constant, we also have V = V0.

5.2.1 Batch Reaction Times

The time necessary to achieve a specific conversion depends upon how fast the reaction takes place, which in turn is dependent on the rate constant and the reactant concentration. To get a feel of how long it takes to carry a batch reaction, we shall calculate the batch reaction times for different values of the reaction rate constant, k, for a first- and for a second-order reaction. First, let’s solve for the time to achieve a conversion X for the second-order reaction

2AB + C

Schematic shows shoe prints labeled, Following the Algorithm.

Following the Algorithm

The Algorithm

1. The mole balance on a constant-volume, V = V0, batch reactor is

Mole balance

NA0dXdt=rAV0(2-6)

Dividing by NA0 and recognizing CA0 = NA0/V0 we obtain

dXdt=-rACA0(5-1)

2. The rate law is

-rA=k2CA2(5-2)

Rate Law

3. From stoichiometry for a constant-volume batch reactor, we obtain

CA=CA0(1-X)(4-12)

Stoichiometry

4. Combining the mole balance, rate law, and stoichiometry we obtain

-rA=k2CA02(1-X)2

Combine

Next, we substitute for in Equation (5-1)

dXdt=k2CA0(1-X)2(5-3)

5. To evaluate, we separate the variables and integrate

dX(1-X)2=k2CA0dt

Initially, if t = 0, then X = 0. If the reaction is carried out isothermally, k will be constant; we can integrate this equation (see Appendix A.1 for a table of integrals used in CRE applications) to obtain

Evaluate

0tdt=1k2CA00XdX(1-X)2

Second-order, isothermal, constant-volume batch reactor

tR=1k2CA0(X1X)(5-4)

First-order isothermal constant-volume batch reactor

This time is the reaction time t (i.e., tR) needed to achieve a conversion X for a second-order reaction in a batch reactor. In a similar fashion, we can apply the CRE algorithm to a first-order reaction to obtain the reaction time, tR, needed to achieve a conversion X

tR=1k1ln(11X)(5-5)

It is important to have a grasp of the order of magnitudes of batch reaction times, tR, to achieve a given conversion, say 90%, for different values of the product of specific reaction rate, k, and initial concentration, CA0. Table 5-1 shows the algorithm to find the batch reaction times, tR, for both first- and second-order reactions carried out isothermally. We can obtain these estimates of tR by considering the first- and second-order irreversible reactions of the form

TABLE 5-1 ALGORITHM TO ESTIMATE REACTION TIMES

Mole Balance dXdtR=rANA0V
Rate Law First-Order Second-Order
rA = k1CA rA=k2CA2
Stoichiometry (V = V0) CA=NAV0=CA0(1X)
Combine dXdtR=k1(1X) dXdtR=k2CA0(1X)2
Evaluate (Integrate) tR=1k1ln11X tR=Xk2CA0(1X)

For first-order reactions, the reaction time to reach 90% conversion (i.e., X = 0.9) in a constant-volume batch reactor scales as

tR=1k1ln11X=1k1ln11–0.9=2.3k1

If k1 = 10−4 s1,

tR=2.3104s1=23,000s=6.4h

The time necessary to achieve 90% conversion in a batch reactor for an irreversible first-order reaction in which the specific reaction rate, k1, is (10−4 s−1) is 6.4 h.

For second-order reactions, we have

tR=1k2CA0X1X=0.9k2CA0(1–0.9)=9k2CA0

If k2CA0 = 10−3 s−1,

tR=910-3s-1=9000s=2.5h

We note that if 99% conversion had been required for this value of kCA0, the reaction time, tR, would jump to 27.5 h.

Table 5-2 gives the order of magnitude of time to achieve 90% conversion for first- and second-order irreversible batch reactions. Flow reactors would be used for reactions with characteristic reaction times, tR, of minutes or less.

Schematic shows an hourglass and a digital clock below it that reads 0:00:00.

Estimating reaction times

TABLE 5-2 BATCH REACTION TIMES

First-Order k1 (s−1)

Second-Order k2CA0 (s−1)

Reaction Time tR

10−4

10−3

Hours

10−2

10−1

Minutes

1

10

Seconds

1000

10,000

Milliseconds

The times in Table 5-2 are the reaction time to achieve 90% conversion (i.e., to reduce the concentration from CA0 to 0.1 CA0). The total cycle time in any batch operation is considerably longer than the reaction time, tR, as one must account for the time necessary to fill (tf) and heat (te) the reactor together with the time necessary to clean the reactor between batches, tc. In some cases, the reaction time calculated from Equations (5-4) and (5-5) may be only a small fraction of the total cycle time, tt.

tt = tf + te + tc + tR

Typical cycle times for a batch polymerization process are shown in Table 5-3. Batch polymerization reaction times may vary between 5 and 60 hours. Clearly, decreasing the reaction time with a 24-hour reaction, as is the case in Example 13-2, is a promising endeavor. As the reaction time is reduced (e.g., 2.5 h for a second-order reaction with k2CA0 = 10–3 s–1), it becomes important to use large lines and pumps to achieve rapid transfers and to utilize efficient sequencing to minimize the cycle time.

TABLE 5-3 TYPICAL CYCLE TIME FOR A BATCH POLYMERIZATION PROCESS

Activity

Time (h)

1. Charge feed to the reactor and agitate, tf

0.5–2.0

2. Heat to reaction temperature, te

0.5–2.0

3. Carry out reaction, tR

(varies)

4. Empty and clean reactor, tc

1.5–3.0

______

    Total time excluding reaction

2.5–7.0

Batch operation times

Usually, one has to optimize the reaction time with the processing times listed in Table 5-3 to produce the maximum number of batches (i.e., pounds or kilograms of product) in a day.

In the next four examples, we will describe the various reactors needed to produce 200 million pounds per year of ethylene glycol from a feedstock of ethane. We begin by finding the rate constant, k, for the hydrolysis of ethylene oxide to form ethylene glycol.

Example 5-1 Determining k from Batch Data

It is desired to design a CSTR to produce 200 million pounds of ethylene glycol per year by hydrolyzing ethylene oxide. However, before the design can be carried out, it is necessary to perform and analyze a batch-reactor experiment to determine the specific reaction-rate constant, k. Because the reaction will be carried out isothermally, the specific reaction rate will need to be determined only at the reaction temperature of the CSTR. At temperatures above 80ºC, there is a significant by-product formation, while at temperatures below 40ºC, the reaction does not proceed at a significant rate; consequently, a temperature of 55ºC has been chosen. Because water is present in excess, its concentration (55.5 mol/dm3) may be considered constant during the course of the reaction. The reaction is first-order in ethylene oxide.

The equation for the liquid-phase reaction is shown.

Check 10 types of homework problems on the CRE Web site for more solved examples using this algorithm. (http://www.umich.edu/~elements/5e/probsolv/tentypes/index.htm)

In the laboratory experiment, 500 mL of a 2 M solution (2 kmol/m3) of ethylene oxide (A) in water was mixed with 500 mL of water (B) containing 0.9 wt % sulfuric acid, which is a catalyst. The temperature was maintained at 55ºC. The concentration of ethylene glycol (C) was recorded as a function of time (Table E5-1.1).

(a) Derive an equation for the concentration of ethylene glycol as a function of time.

(b) Rearrange the equation derived in (a) to obtain a linear plot of a function concentration versus time.

(c) Using the data in Table E5-1.1, determine the specific reaction rate at 55ºC.

Illustration shows an equipment that is rotated in the clockwise direction with a mixture labeled A, B, C within it. Two sets of blades are shown within the equipment.

TABLE E5-1.1 CONCENTRATION-TIME DATA

Time (min)

Concentration of Ethylene Glycol (C) (kmol/m3)*

0.0

0.000

0.5

0.145

1.0

0.270

1.5

0.376

2.0

0.467

3.0

0.610

4.0

0.715

6.0

0.848

10.0

0.957

* 1 kmol/m3 = 1 mol/dm3 = 1 mol/L.

Solution

Part (a)

1. The mole balance on ethylene oxide (A) given in Equation (1-5) for a constant volume, V0, well-mixed batch reactor can be written as

1V0dNAdt=rA(E5-1.1)

Schematic shows shoe prints labeled, Following the Algorithm.

Following the Algorithm

Taking V0 inside the differential and recalling that the concentration is

CA=NAV0

then the differential mole balance becomes

d(NA/V0)dt=dCAdt=rA(E5-1.2)

2. The rate law for the ethylene oxide hydrolysis is

Images
Images
Algorithm for isothermal reactors is shown.

-rA=kCA(E5-1.3)

Because water is present in such excess, the concentration of water at any time t is virtually the same as the initial concentration, and the rate law is independent of the concentration of H2O (CBCB0 = 55.5 moles per dm3).

3. Stoichiometry. Liquid phase, no volume change, V = V0 (Table E5-1.2):

TABLE E5-1.2 STOICHIOMETRIC TABLE

Species Symbol Initial Change Remaining Concentration
CH2CH2O A NA0 NA0X NA=NA0(1X) CA=CA0(1X)
H2O B ΘBNA0 NA0X NB=NA0(ΘBX) CB=CA0(ΘBX)
CBCA0ΘB=CB0
(CH2OH)2 C 0 NA0X NC=NA0X CC=CA0X=CA0CA
NT0¯ NT=NT0NA0X¯

Recall that ΘB is the ratio of the initial number of moles of B to A (i.e., ΘB=NB0NA0=CB0CA0).

For species B, i.e., water,

CB = CA0BX)

We quickly see that water is in excess, as the molarity of water is 55 moles per liter. The initial concentration of A after mixing the two volumes together is 1 molar. Therefore,

ΘB=CB0CA0=55mol/dm31mol/dm3=55

The maximum value of X is 1, and ΘB≫1, therefore CB is virtually constant

CBCA0 Θ = CB0

For species C, i.e., ethylene glycol, the concentration is

CC=NCV0=NA0XV0=NA0NAV0=CA0CA(E5-1.4)

4. Combining the rate law and the mole balance, we have

dCAdt=kCA(E5-1.5)

5. Evaluate. For isothermal operation, k is constant, so we can integrate this equation (E5-1.5)

CA0CAdCACA=0tkdt=k0tdt

Schematic shows mole balance, rate law, and stoichiometry within three triangles.

Combining mole balance, rate law, and stoichiometry

using the initial condition that

when t = 0, then CA = CA0 = 1 mol/dm3 = 1 kmol/m3.

Integrating yields

lnCA0CA=kt(E5-1.6)

The concentration of ethylene oxide (A) at any time t is

CA=CA0ekt(E5-1.7)

The concentration of ethylene glycol (C) at any time t can be obtained from the reaction stoichiometry, i.e., Equation (E5-1.4)

CC=CA0CA=CA0(1ekt)(E5-1.8)

Part (b)

We are now going to rearrange Equation (E5-1.8) for the concentration of ethylene glycol in such a manner that we can easily use the data in Table E5-1.1 to determine the rate constant k

lnCA0CCCA0=kt(E5-1.9)

Two graphs are shown.

Part (c)

We see that a plot of ln[(CA0CC)/CA0] as a function of t will be a straight line with a slope −k. Using Table E5-1.1, we can construct Table E5-1.3 and use Excel to plot ln(CA0CC)/CA0 as a function of t.

TABLE E5-1.3 PROCESSED DATA

t (min)

CC (kmol/m3)

CA0CCCA0

ln(CA0CCCA0)

00.0

0.000

1.000

–0.0000

00.5

0.145

0.855

–0.1570

01.0

0.270

0.730

–0.3150

01.5

0.376

0.624

–0.4720

02.0

0.467

0.533

–0.6290

03.0

0.610

0.390

–0.9420

04.0

0.715

0.285

–1.2550

06.0

0.848

0.152

–1.8840

10.0

0.957

0.043

–3.1470

From the slope of a plot of ln[(CA0 = CC)/CA0] versus t, we can find k, as shown in the Excel plot in Figure E5-1.1.

Graph shows points connected by a line segment.

Figure E5-1.1 Excel plot of data.

Slope = –k = –0.311 min–1

k = 0.311 min–1

The rate law becomes

rA=0.311min1CA(ES-1.10)

Evaluating the specific reaction rate from batch-reactor concentration– time data

The rate law can now be used in the design of an industrial CSTR. For those who prefer to find k using semilog graph paper, this type of analysis can be found at www.physics.uoguelph.ca/tutorials/GLP. Tutorials are also given in the Summary Notes for Chapter 3 (last example) and Chapter 7 (Excel).

Analysis: In this example, we used our CRE algorithm

(mole balance → rate law → stoichiometry → combine)

to calculate the concentration of species C, CC, as a function of time, t. We then used experimental batch data of CC versus t to verify the reaction as a first-order reaction and to determine the specific reaction-rate constant, k.

Images

5.3 Continuous-Stirred Tank Reactors (CSTRs)

Continuous-stirred tank reactors (CSTRs), such as the one shown here schematically, are typically used for liquid-phase reactions.

A design of CSTR reactor is shown.

Figure 5-4 CSTR.
(http://encyclopedia.che.engin.umich.edu/Pages/Reactors/CSTR/CSTR.html)

In Chapter 2, we derived the following design equation for a CSTR

Mole balance

V=FA0X(-rA)exit(2-13)

which gives the volume V necessary to achieve a conversion X. As we saw in Chapter 2, the space time, τ, is a characteristic time of a reactor. To obtain the space time, τ, as a function of conversion, we first substitute for FA0 = υ0CA0 in Equation (2-13)

V=v0CA0X(-rA)exit(5-6)

and then divide by υ0 to obtain the space time, τ, to achieve a conversion X in a CSTR

τ=Vυ0=CA0X(-rA)exit(5-7)

This equation applies to a single CSTR or to the first reactor of CSTRs connected in series.

5.3.1 A Single CSTR

5.3.1.1 First-Order Reaction in a CSTR

Let’s consider a first-order irreversible reaction for which the rate law is

rA = kCA

Rate law

For liquid-phase reactions, there is no volume change during the course of the reaction, so we can use Equation (4-12) to relate concentration and conversion

CA=CA0(1-X)(4-12)

Stoichiometry

We can combine the mole balance Equation (5-7), the rate law, and the concentration equation (4-12) to obtain

Combine

τ=1k(X1-X)

Rearranging

CSTR relationship between space time and conversion for a first-order liquid-phase reaction

Images

A plot of conversion as a function of τk using Equation (5-8) is shown in Figure 5-5.

Graph shows a curve representing the first-order reaction in a CSTR reactor.

Figure 5-5 First-order reaction in a CSTR.

We can increase τk by either increasing the temperature to increase k or increasing the space time τ by increasing the volume V or decreasing the volu-metric flow rate υ0. For example, when we increase the reactor volume, V, by a factor of 2 (or decrease the volumetric flow rate υ0 by a factor of 2) as we go from τk = 4 to τk = 8, the conversion only increases from 0.8 to 0.89.

We could also combine Equations (4-12) and (5-8) to find the exit reactor concentration of A, CA,

CA=CA01+τk(5-9)

5.3.1.2 A Second-Order Reaction in a CSTR

For an irreversible second-order liquid-phase reaction e.g.,

-rA=kCA2

being carried out in a CSTR, the combination of the second-order rate law and the design equation (i.e., combined mole balance Eq. (2-13) and rate law) yields

V=FA0X-rA=FA0XkCA2

Using our stoichiometric table for constant density υ = υ0, CA = CA0(1 – X), and FA0X = υ0CA0X, then

V=υ0CA0XkCA02(1-X)2

Dividing both sides of the combined rate law and design equation for a second-order reaction by υ0 gives

τ=Vυ0=XkCA0(1-X)2(5-10)

We solve Equation (5-11) for the conversion X

Images

Conversion for a second-order liquid-phase reaction in a CSTR

The minus sign must be chosen in the quadratic equation because X cannot be greater than 1. Conversion is plotted as a function of the Damköhler parameter for a second-order reaction, Da2 = τkCA0, in Figure 5-6. Observe from this figure that at high conversions (say 67%), a 10-fold increase in the reactor volume (or increase in the specific reaction rate by raising the temperature) will only increase the conversion up to 88%. This observation is a consequence of the fact that the CSTR operates under the condition of the lowest reactant concentration (i.e., the exit concentration), and consequently the smallest value of the rate of reaction.

Graph shows a curve representing Conversion as a function of the Damköhler number (D a subscript 2 equals tau kC subscript (A0)) for a second-order reaction in a CSTR.

Figure 5-6 Conversion as a function of the Damköhler number (Da2 = τkCA0) for a second-order reaction in a CSTR.

Da=rA0VFA0

5.3.1.3 The Damköhler Number

For a first-order reaction, the product is often referred to as the reaction Damköhler number, Da1, which is a dimensionless number that can give us a quick estimate of the degree of conversion that can be achieved in continuous-flow reactors. The Damköhler number is the ratio of the rate of reaction of A to the rate of convective transport of A evaluated at the entrance to the reactor.

Da=rA0VFA0=Rate of reaction at entranceEntering flow rate of A="A reaction rate""A convection rate"

The Damköhler number for a first-order irreversible reaction is

Da1=-rA0VFA0=k1CA0Vv0CA0=τk1

The Damköhler number for a second-order irreversible reaction is

Da2=-rA0VFA0=k2CA02Vv0CA0=τk2CA0

It is important to know what values of the Damköhler number, Da, give high and low conversion in continuous-flow reactors. For irreversible reactions, a value of Da = 0.1 or less will usually give less than 10% conversion, and a value of Da = 10.0 or greater will usually give greater than 90% conversion; that is, the rule of thumb is

0.1 ≤ Da ≤ 10

if Da<0.1,thenX<0.1if Da>10,thenX>0.9

Equation (5-8) for a first-order liquid-phase reaction in a CSTR can also be written in terms of the Damköhler number

X=Da11+Da1(5-12)
Schematic shows a thumbs up symbol.

5.3.2 CSTRs in Series

A first-order reaction with no change in the volumetric flow rate (υ = υ0) is to be carried out in two CSTRs placed in series (Figure 5-7).

Figure shows two CSTR reactors in series labeled negative r subscript A1: V subscript 1 and negative r subscript A2: V subscript 2.

Figure 5-7 Two CSTRs in series.

The effluent concentration of reactant A from the first CSTR can be found using Equation (5-9)

CA1=CA01+τ1k1

with τ1 = V10. From a mole balance on reactor 2

V2=FA1-FA2-rA2=υ0(CA1-CA2)k2CA2

Solving for CA2, the concentration exiting the second reactor, we obtain

First-order reaction

CA2=CA11+τ2k2=CA0(1+τ2k2)(1+τ1k1)(5-13)

If both reactors are of equal size (τ1 = τ2 = τ) and operate at the same temperature (k1 = k2 = k), then

CA2=CA0(1+τk)2

If, instead of two CSTRs in series, we had n equal-sized CSTRs connected in series (τ1 = τ2 = ⋯ = τn = τi = (Vi0) operating at the same temperature (k1 = k2 = ⋯ = kn = k), the concentration leaving the last reactor would be

CAn=CA0(1+τk)n=CA0(1+Da1)n(5-14)

Substituting for CAn in terms of conversion

CA0(1-X)=CA0(1+Da1)n

and rearranging, the conversion for these identical n tank reactors in series will be

CSTRs in series

X=11(1+Da1)n11(1+τk)n(5-15)

Conversion as a function of the number of tanks in series

Recall from Chapter 2, Figure 2-6, that as n becomes large, the conversion approaches that of a PFR.

A plot of the conversion as a function of the number of CSTRs (i.e., tanks) in series for a first-order reaction is shown in Figure 5-8 for various values of the Damköhler number τk. Observe from Figure 5-8 that when the product of the space time and the specific reaction rate is relatively large, say, Da1 ≥ 1, approximately 90% conversion is achieved in two or three reactors; thus, the cost of adding subsequent reactors might not be justified. When the product is τk small, Da1 ~ 0.1, the conversion continues to increase significantly with each reactor added.

Economics

Graph shows a curve representing the Conversion as a function of the number of CSTRs (that is, tanks) in series for different Damköhler numbers for a first-order reaction.

Figure 5-8 Conversion as a function of the number of CSTRs (i.e., tanks) in series for different Damköhler numbers for a first-order reaction.

The rate of disappearance of A in the nth reactor is

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Example 5-2 Producing 200 Million Pounds per Year in a CSTR

Close to 16 billion pounds of ethylene glycol (EG) were produced in 2013. It previously ranked as the twenty-sixth most produced chemical in the nation on a total pound basis. About one-half of the ethylene glycol is used for antifreeze, while the other half is used in the manufacture of polyesters. In the polyester category, 88% was used for fibers and 12% for the manufacture of bottles and films. The 2014 selling price for ethylene glycol was $0.55 per pound.

It is desired to produce 200 million pounds per year of EG. The reactor is to be operated isothermally. A 16.1 mol/dm3 solution of ethylene oxide (EO) in water is mixed (see Figure E5-2.1) with an equal volumetric solution of water containing 0.9 wt % of the catalyst H2SO4 and fed to a CSTR. The specific reaction-rate constant is 0.311 min–1, as determined in Example 5-1. Practical guidelines for reactor scale-up are given by Mukesh.1

(a) If 80% conversion is to be achieved, determine the necessary CSTR volume.

(b) If two 800-gal reactors were arranged in parallel with the feed equally divided, what would be the corresponding conversion?

(c) If two 800-gal reactors were arranged in series, what would be the corresponding conversion?

1 D. Mukesh, Chemical Engineering, 46 (January 2002), www.CHE.com.

Uses and economics

Schematic shows anti-freeze liquid in a container, a thermometer, and a car partly covered by snow.

Scale-up of batch reactor data

Solution

Assumption: Ethylene glycol (EG) is the only reaction product formed.

The equation for the liquid-phase reaction is shown.
Schematic shows a mixture in an equipment with a rotating stirrer rotates in the counterclockwise direction.

Figure E5-2.1 Single CSTR.

The specified ethylene glycol (EG) production rate in mol/s is

FC=2×108lbmyr×1yr365days×1day24h×1h3600s×454glbm×1mol62g=46.4molsec

From the reaction stoichiometry

FC = FA0X

we find the required molar flow rate of ethylene oxide for 80% conversion to be

FA0=FCX=46.4mo1/s0.8=58.0mol/s

(a) We now calculate the single CSTR volume to achieve 80% conversion using the CRE algorithm.

1. CSTR Mole Balance:

V=FA0X-rA(E5-2.1)

2. Rate Law:

rA=kCA(E5-2.2)

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Following the Algorithm

3. Stoichiometry: Liquid phase (υ = υ0):

CA=FAυ0=FA0(1-X)υ0=CA0(1-X)(E5-2.3)

4. Combining:

V=FA0XkCA0(1-X)=υ0Xk(1-X)(E5-2.4)

5. Evaluate:

The entering volumetric flow rate of stream A, with CA01 = 16.1 mol/dm3 before mixing, is

υA0=FA0CA01=58mol/s16.1mol/dm3=3.6dm3s

From the problem statement υB0 = υA0

FB0=υB0CB01=3.62dm3s×[1,000gdm3×1mol18g]=201mols

The total entering volumetric flow rate of liquid is

υ0=υA0+υB0=3.62dm3s+3.62dm3s=7.2dm3s

Substituting in Equation (E5-2.4), recalling that k = 0.311 min−1, yields

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A tank 5 ft in diameter and approximately 10 ft tall is necessary to achieve 80% conversion.

(b) CSTRs in parallel. What would you guess would happen if two 800-gal CSTRs arranged in parallel as shown in Figure E5-2.2 had a volumetric flow rate υ0 = 3.62 dm3/s fed to each reactor? Would the conversion increase, decrease, or remain the same? To find out, let’s begin by rearranging Equation (E5-2.4)

Figure shows a CSTR vessel reactor with a rotating stirrer immersed inside the mixture. The height, width, and volume of the equipment are respectively 10 feet, 5 feet, and 1500 gallons. An inlet at the top left and an outlet at the bottom right is shown.

1500 gallon tank

Vυ0k=τk=X1-X

to obtain

X=τk1+τk(E5-2.5)

where

τ=Vυ0/2=800gal×3.785dm3gal13.62dm3/s=836.5s

The Damköhler number for a first-order reaction is

Da1 = τk = 836.5s×0.0052s−1 = 4.35

Substituting into Equation (E5-2.5) gives us

X=Da11+Da1=4.351+4.35=0.81

Figure shows two CSTRs are connected in parallel.

Figure E5-2.2 CSTRs in parallel.

The conversion exiting both of the CSTRs in parallel is 81%.

Problem P5-2(b) asks you to generalize the result for n equal-size reactors Vi in parallel with equal feed rates (FA0/n) and show that the conversion would also be the same if everything were fed to one big reactor of volume V = nVi.

(c) CSTRs in series. Recalling what we found in Chapter 2 when we sequenced reactors, what do you guess will happen to the conversion in Part (a) if the reactors are placed in series? Will it increase or decrease? If the 800-gal reactors are arranged in series, the conversion in the first reactor [cf. Equation (E5-2.5)] is

X1=τ1k1+τ1k(E5-2.6)

where

τ=V1υ0=(800gal×3.785dm3gal)×17.24dm3/s=418.2s

The Damköhler number is

Da1=τ1k=418.2s×0.0052s=2.167X1=2.1671+2.167=2.1673.167=0.684

To calculate the conversion exiting the second reactor, we recall that V1 = V2 = V and υ01 = υ02 = υ0; then

τ1 = τ2 = τ

First CSTR

Conversion in the series arrangement is greater than in parallel for CSTRs. From our discussion of reactor staging in Chapter 2, we could have predicted that the series arrangement would have given the higher conversion.

Figure shows two CSTRs are connected in series.

Figure E5-2.3 CSTRs in series.

A mole balance on the second reactor is

The flow rate F subscript A1 indicated as “In” minus flow rate F subscript A2 indicated as “Out” plus the rate of formation of A2 (r subscript A2) times volume V indicated as “Generation” equals 0.
InOut+Generation=0FA1FA2+rA2V=0

Basing the conversion on the total number of moles reacted up to a point per mole of A fed to the first reactor

FA1 = FA0 (1 – X1) and FA2 = FA0 (1 – X2)

Rearranging

V=FA1-FA2-rA2=FA0X2-X1-rA2

-rA2=kCA2=kFA2υ0=kFA0(1-X2)υ0=kCA0(1-X2)

Second CSTR

Combining the mole balance on the second reactor [cf. Equation (2-24)] with the rate law, we obtain

V=FA0(X2-X1)-rA2=CA0υ0-(X2-X1)kCA0(1-X2)=υ0k(X2-X11-X2)(E5-2.7)

Solving for the conversion exiting the second reactor yields

X2=X1+Da11+Da1=X1+τk1+τk=0.684+2.1671+2.167=0.90

The same result could have been obtained from Equation (5-15)

X2=11(1+τk)n=11(1+2.167)2=0.90

Over two hundred million pounds of EG per year can be produced using two 800-gal (3.0-m3) reactors in series.

Analysis: The CRE algorithm was applied to a first-order irreversible liquid-phase reaction carried out isothermally in a single CSTR, two CSTRs in series, and also two CSTRs in parallel. The equations were solved algebraically for each case. When the entering molar flow rate was equally divided between the two CSTRs in parallel, the overall conversion was the same as that for a single CSTR. For two CSTRs in series, the overall conversion was greater than that of a single CSTR. This result will always be the case for isothermal reactions with power-law rate laws with reaction orders greater than zero.

Safety considerations

We can find information online about the safety of ethylene glycol and other chemicals from Table 5-4. One source is the Vermont Safety Information Resources, Inc., Web site (Vermont SIRI, www.siri.org). For example, we can learn from the Control Measures that we should use neoprene gloves when handling the material, and that we should avoid breathing the vapors. If we click on “Dow Chemical USA” and scroll the Reactivity Data, we would find that ethylene glycol will ignite in air at 413ºC.

Safety information MSDS

TABLE 5-4 ACCESSING SAFETY INFORMATION

Screenshot displays the information in 2 rows and 4 columns. Row 1 reads, 5001-19 Macron Fine Chemicals, Ethylene Glycol, 107-21-1, and a button labeled "View SDS." Row 2 reads, 5387-03 J.T.Baker, Ethylene Glycol, 107-21-1, and a button labeled "View SDS."

5.4 Tubular Reactors

Gas-phase reactions are carried out primarily in tubular reactors where the flow is generally turbulent. By assuming that there is no dispersion and there are no radial gradients in either temperature, velocity, concentration, or reaction rate, we can model the flow in the reactor as plug flow.2

The differential form of the PFR design equation such as

FA0dXdV=-rA(2-15)

must be used when there is a pressure drop in the reactor or heat exchange between the PFR and the surroundings.

2 Laminar flow reactors (LFTs) and dispersion effects are discussed in Web PDF Chapter 17. As a general rule, the conversion calculated for a PFR will not be significantly different than that for an LFR. For example, from Table E17-1.1 in the Web Chapter 17 PDF (available here: http://www.umich.edu/~elements/5e/17chap/Fogler_Web_Ch17.pdf), we see on page 17-9 that when τk = 0.1 then XPFR = 0.095 and XLFR = 0.09, when τk = 2.0 then XPFR = 0.865 and XLF = 0.78 and when τk = 4, then XPFR = 0.98 and XLF = 0.94.

Use this differential form of the PFR/PBR mole balances when there is ΔP or heat effects.

Illustration shows plug-flow tubular reactor.

Figure 1-9 (Revisited) tubular reactor. (http://encyclopedia.che.engin.umich.edu/Pages/Reactors/PBR/PBR.html)

In the absence of pressure drop or heat exchange, the integral form of the plug-flow design equation can be used,

V=FA00XdXrA(2-16)

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Following the Algorithm

As an example, consider the elementary reaction

2 A → Products

for which the rate law is

rA=kCA2

Rate Law

We shall first consider the reaction to take place as a liquid-phase reaction and then as a gas-phase reaction.

5.4.1 Liquid-Phase Reactions in a PFR ∴ υ = υ0

The combined PFR mole balance and rate law is

dXdV=kCA2FA0

If the reaction is carried out in the liquid phase, the concentration of A is

Stoichiometry (liquid phase)

CA = CA0(1 − X)

and for isothermal operation, we can bring k outside the integral

V=FA0kCA020XdX(1-X)2=υ0kCA0(X1-X)

Combine

This equation gives the reactor volume to achieve a conversion X. Dividing by υ0 (τ = V0) and solving for conversion, we find

X=τkCA01+τkCA0=Da21+Da2

Evaluate

where Da2 is the Damköhler number for a second-order reaction, i.e. τkCA0.

5.4.2 Gas-Phase Reactions in a PFR υ = υ0 (1 + εX) (T/T0)(P0/P)

For constant-temperature (T = T0) and constant-pressure (P = P0) gas-phase reactions, the concentration is expressed as a function of conversion

Stoichiometry (gas phase)

CA=FAυ=FAυ0(1+ϵX)=FA0(1-X)υ0(1+ϵX)=CA0(1-X)(1+ϵX)

and then combining the PFR mole balance, rate law, and stoichiometry

Combine

V=FA00X(1+εX)2kCA02(1-X)2dX

The entering concentration CA0 can be taken outside the integral sign since it is not a function of conversion. Because the reaction is carried out isothermally, the specific reaction rate constant, k, can also be taken outside the integral sign.

For an isothermal reaction, k is constant. V

V=FA0kCA020X(1+εX)2(1-X)2dX

From the integral equations in Appendix A.1, we find that

Evaluate Reactor volume for a second-order gas-phase reaction

V=υ0kCA0=[2ε(1+ε)ln(1X)+ε2X+(1+ε)2X1X](5-17)

5.4.3 Effect of ε on Conversion

We now look at the effect of the change in the total number of moles in the gas phase on the relationship between conversion and volume. For constant temperature and pressure, Equation (4-23) becomes

υ = υ0 (1 + εX)

T = T0
P = P0 (∴p = 1)

Let’s now consider three types of reactions, one in which ε = 0 (i.e., δ = 0), one in which ε < 0 (δ < 0), and one in which ε > 0 (δ > 0). When there is no change in the number of moles with reaction, (e.g., A → B) δ = 0 and ε = 0, then the fluid moves through the reactor at a constant volumetric flow rate (υ = υ0) as the conversion increases.

When there is a decrease in the number of moles (δ < 0, ε < 0) in the gas phase, the volumetric gas flow rate decreases and the conversion increases. For example, when pure A enters for the reaction 2A → B, and taking A as the basis of calculation, then A → B/2 and we have ϵ=yA0δ=1(121)=0.5

υ = υ0(1 − 0.5X)

Consequently, the gas molecules will spend more time in the reactor than they would if the flow rate were constant, υ = υ0. As a result, this longer residence time would result in a higher conversion than if the flow rate, υ, were constant at υ0.

On the other hand, if there is an increase in the total number of moles (δ > 0, ε > 0) in the gas phase, then the volumetric flow rate, υ, will increase as the conversion increases. For example, for the reaction A → 2B, then ε= yA0δ=1(2–1)=1

υ = υ0 (1+ X)

and the molecules will spend less time in the reactor than they would if the volumetric flow rate were constant. As a result of this smaller residence time in the reactor, the conversion will be less than what would result if the volumetric flow rate, υ, were constant at υ0.

Graph shows the change in gas-phase volumetric flow rate down the length of the reactor for the three cases.

Figure 5-9 Change in gas-phase volumetric flow rate down the length of the reactor.

The importance of changes in volumetric flow rate (i.e., ε ≠ 0) with reaction

Figure 5-9 shows the volumetric flow rate profiles for the three cases just discussed. We note that, at the end of the reactor, virtually complete conversion has been achieved.

Example 5-3 Producing 300 Million Pounds per Year of Ethylene in a Plug-Flow Reactor: Design of a Full-Scale Tubular Reactor

Ethylene ranks first in the United States in total pounds of organic chemicals produced each year, and it is the number-one organic chemical produced each year. Over 60 billion pounds were produced in 2016, and it sold for $0.28 per pound. Sixty-five percent of the ethylene produced is used in the manufacture of fabricated plastics, 20% for ethylene oxide, 16% for ethylene dichloride and ethylene glycol, 5% for fibers, and 5% for solvents.

Determine the plug-flow reactor volume necessary to produce 300 million pounds of ethylene a year by cracking a feed stream of pure ethane. The reaction is irreversible and follows an elementary rate law. We want to achieve 80% conversion of ethane, operating the reactor isothermally at 1100 K and at a pressure of 6 atm. The specific reaction rate at 1,000 K is 0.072 s–1 and the activation energy is 82,000 cal/mol.

Solution

C2H6 → C2H4 + H2

Let A = C2H6, B = C2H4, and C = H2. In symbols,

A → B + C

The economics

The uses

Because we want the reader to be familiar with both metric units and English units, we will work some of the examples using English units. Trust me, a number of companies still use concentrations in lb-mol/ft3. To help you relate English and metric units, the corresponding metric units will be given in parenthesis next to the English units. The only step in the algorithm that is different is the evaluation step.

The molar flow rate of ethylene exiting the reactor is

FB=300×1061bmyear×1year365days×1day24h×1h3600s×lb-mol28lbm=0.340lb-mols(FB=154.4mols)

Next, calculate the molar feed rate of ethane, FA0, to produce 0.34 lb mol/s of ethylene when 80% conversion is achieved,

FB=FA0XFA0=0.34lb mol/s0.8=0.425lb-mols(FA0=193mol/s)

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Following the Algorithm

1. Plug-Flow Mole Balance:

FA0dXdV=-rA(2-15)

Rearranging and integrating for the case of no pressure drop and isothermal operation yields

V=FA00xdXrA (E5-3.1)

Mole balance

2. Rate Law:3

rA=kCA with k = 0.072s1at 1000 k(E5-3.2)

The activation energy is 82 kcal/mol.

3 Ind. Eng. Chem. Process Des. Dev., 14, 218 (1975); Ind. Eng. Chem., 59(5), 70 (1967).

Rate Law

3. Stoichiometry: For isothermal operation and negligible pressure drop, the concentration of ethane is calculated as follows:

Gas phase, constant T and P:

υ=υFTFT0=υ0(1+ϵX)

CA=FAυ=FA0(1X)υ0(1+εX)=CA0(1X1+εX)(E5-3.3)

Stoichiometry

CC=CA0X(1+εX)(E5-3.4)

4. Combine Equations (E5-3.1) through (E5-3.3) to obtain

V=FA00XdXkCA0(1-X)/(1+εX)=FA00X(1+εX)dXkCA0(1-X)   =FA0CA00X(1+εX)dXk(1-X)(E5-3.5)

Combining the design equation, rate law, and stoichiometry

5. Evaluate:
Since the reaction is carried out isothermally, we can take k outside the integral sign and use Appendix A.1 to carry out our integration.

V=FA0kCA0=0X(1+εX)dX1X=FA0kCA0[(1+ε)ln11XεX](E5-3.6)

6. Parameter evaluation:

CA0=yA0CT0=yA0P0RT0=(1.0)(6atm(0.73Ft3atm/1b-mo1Ro)×(1980oR))      =0.004151b-molFt3(0.066mol/dm3)ε=yA0δ=(1)(1+1-1)=1

Oops! The rate constant k is given as 0.072 s–1 but that is at 1000 K, and we need to calculate k at reaction conditions, which is 1100 K.

Analytical solution

k(T2)=k(T1)exp[ER(1T11T2)]=k(T1)exp[ER(T2T1T1T2)](E5-3.7)=0.072sexp[82,000cal/mol(1100-1000)K1.987cal/(molK)(1000K)(1100K)]=3.07s1

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Following the Algorithm

Substituting FA0, k, CA0, and into Equation (E5-3.6) yields

V=0.425lb mol/s(3.07/s)(0.00415lb-mol/ft3)[(1+1)ln11X(1)X]=33.36ft3[2ln(11X)X](E5-3.8)

For X = 0.8,

V=33.36ft3[2ln(11–0.8)0.8]=80.7ft3=(2280dm3=2.28m3)

It was decided to use a bank of 2-inch schedule 80 pipes in parallel that are 40 feet in length. For schedule 80 pipe, the cross-sectional area, AC, is 0.0205 ft2. The number of pipes, n, necessary is

n=Reactor volumeVolume of 1 pipe=80.7ft3(0.0205ft2)(40ft)=98.4(E5-3.9)

To determine the concentrations and conversion profiles down the length of the reactor, z, we divide the volume Equation (E5-3.8) by the cross-sectional area, AC,

z=VAC(E5-3.10)

Equation (E5-3.9) was used along with the volume of one pipe, 0.81 ft3, the cross-sectional area of one pipe AC = 0.0205 ft2, and Equations (E5-3.8) and (E5-3.3) to obtain Figure E5-3.1. Using a bank of 100 pipes will give us the reactor volume necessary to make 300 million pounds per year of ethylene from ethane. The concentration and conversion profiles down any one of the pipes are shown in Figure E5-3.1.

Graph shows the conversion and concentration profiles.

Figure E5-3.1 Conversion and concentration profiles.

Analysis: The CRE algorithm was applied to a gas-phase reaction that had a change in the total number of moles during the reaction. A bank of 100 PFRs in parallel, each with a volume of 0.81 ft3, will give the same conversion as 1 PFR with a volume of 81 ft3. The conversion and concentration profiles are shown in Figure E5-3.1. You will note that the profiles change more rapidly near the entrance to the reactor where the reactant concentrations are high and change more slowly near the exit where most of the reactants have been consumed, resulting in a smaller rate of reaction.

The number of PFRs in parallel 100 pipes in parallel

Figure shows two PFRs tube at the top and two PFRs tube at the bottom are connected in parallel.

100 pipes in parallel

We will also solve this problem using Polymath and Wolfram in order to turn it into a Living Example Problem (LEP). To view .cdf files, all you need is Wolfram CDF Player, which can be downloaded for free at http://www.wolfram.com/cdf-player/. We will use the m, kg, s units and the values shown above in parenthesis (e.g., CA0 = 0.066 mol/dm3) in the LEP. The reader should use the “sliders” in Wolfram for this LEP on the CRE Web site (http://www.umich.edu/~elements/5e/05chap/live.html) to vary the parameters and determine their effect on the conversion X.

5.5 Pressure Drop in Reactors

Pressure drop is ignored for liquid-phase kinetics calculations.

In liquid-phase reactions, the concentration of reactants is insignificantly affected by even relatively large changes in the total pressure. Consequently, we can totally ignore the effect of pressure drop on the rate of reaction when sizing liquid-phase chemical reactors. However, in gas-phase reactions, the concentration of the reacting species is proportional to the total pressure; therefore, proper accounting for the effects of pressure drop on the reaction system can, in many instances, be a key factor in the success or failure of the reactor operation. This fact is especially true in microreactors packed with solid catalyst. Here, the channels are so small (see Problem 5-21B) that pressure drop can limit the throughput and conversion for gas-phase reactions.

5.5.1 Pressure Drop and the Rate Law

For gas-phase reactions, pressure drop may be very important.

We now focus our attention on accounting for the pressure drop in the rate law. For an ideal gas, we recall Equation (4-25) to write the concentration of reacting species i as

Ci=CA0(Θi+viX1+εX)PP0T0T(5-18)

where Θi=Fi0FA0,ϵ=yA0δ, and νi is the stoichiometric coefficient (e.g., νA = –1, νB = –b/a). We now must determine the ratio pressure (p = P/P0) as a function of the PFR reactor volume, V, or the PBR catalyst weight, W, to account for pressure drop. We then can combine the concentration, rate law, and design equation. However, whenever accounting for the effects of pressure drop, the differential form of the mole balance (design equation) must be used.

If, for example, the second-order reaction

2A → B + C

is being carried out in a packed-bed reactor, the differential form of the mole balance equation in terms of catalyst weight is

When PP0, one must use the differential forms of the PFR/PBR design equations.

FA0dXdW=rA(Gram molesGram catalystmin)(2-17)

The rate law is

rA=kCA2(5-19)

From stoichiometry for gas-phase reactions (Table 3-5)

CA=CA0(1X)1+ϵXpT0T

and the rate law can be written as

-rA=k[CA0(1-X)1+εXpT0T]2(5-20)

Note from Equation (5-20) that the larger the pressure drop (i.e., the smaller P) from frictional losses, the smaller the reaction rate!

Combining Equation (5-20) with the mole balance (2-17) and assuming isothermal operation (T = T0) gives

FA0dXdW=k[CA0(1-X)1+εX]2p2

Dividing by FA0 (i.e., υ0CA0) yields

dXdW=kCA0υ0(1-X1+εX)2p2

For isothermal operation (T = T0), the right-hand side is a function of only conversion and pressure

dXdW=F1(X,p)(5-21)

Another equation is needed (e.g., p = f (W)).

We now need to relate the pressure drop to the catalyst weight in order to determine the conversion as a function of catalyst weight (i.e., catalyst mass).

5.5.2 Flow Through a Packed Bed

The majority of gas-phase reactions are catalyzed by passing the reactant through a packed bed of catalyst particles.

Figure shows a packed bed of catalyst particles in a packed porous bed. A leftward arrow pointing to the left end and a rightward arrow from the right end is shown.

The equation used most often to calculate pressure drop in a packed porous bed is the Ergun equation:4,5

4 R. B. Bird, W. E. Stewart, and E. N. Lightfoot, Transport Phenomena, 2nd ed. (New York: Wiley, 2002), p. 191.

5 A slightly different set of constants for the Ergun Equation (e.g., 1.8G instead of 1.75G) can be found in Ind. Eng. Chem. Fundamentals, 18 (1979), p. 199.

dPdZ=GρgcDP(1ΦΦ3)[150(1Φ)μTerm 1DP+1.75GTerm 2](5-22)

Ergun equation

Term 1 is dominant for laminar flow, and Term 2 is dominant for turbulent flow, where

P

= pressure, lbf/ft2 = or (kPa)

ϕ

= porosity=Volume of voidtotal bed volume=void fraction

1 − ϕ

= Volume of solidtotal bed volume

gc

= 32.174 lbm · ft/s2 lbf (conversion factor)

= 4.17 × 108 lbm · ft/h2 · lbf

(Recall that for the metric system gc = 1.0)

DP

= diameter of particle in the bed, ft or (m)

μ

= viscosity of gas passing through the bed, lbm/ft·h or (kg/m · s)

z

= length down the packed bed of pipe, ft or (m)

u

= superficial velocity = volumetric flow rate ÷ cross-sectional area of pipe, ft/h or (m/s)

ρ

= gas density, lbmft3 or (kg/m3)

G

= ρu = superficial mass velocity, (lbm/ft2 · h) or (kg/m2 · s)

In calculating the pressure drop using the Ergun equation, the only parameter that varies with pressure on the right-hand side of Equation (5-22) is the gas density, ρ. We are now going to calculate the pressure drop through a packed-bed reactor.

Because the PBR is operated at steady state, the mass flow rate at any point down the reactor, m˙(kg/s), is equal to the entering mass flow rate, m˙(i.e., equation of continuity)

m˙0=m˙ρ0υ0=ρυ

Recalling Equation (4-16), we have

υ=υ0P0P(TT0)FTFT0(4-16)

ρ=ρ0υ0υ=ρ0PP0(T0T)FT0FT(5-23)

Combining Equations (5-22) and (5-23) gives

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Simplifying yields

dPdZ=β0P0P(TT0)FTFT0(5-24)

where β0 is a constant that depends only on the properties of the packed bed (ϕ, DP) and the fluid properties at the entrance conditions (i.e., μ, G, ρ0, T0, P0). Typical units of β0 are (atm/ft) or (Pa/m).

β0=G(1Φ)ρ0gcDPΦ3[150(1Φ)μDP+1.75G](e.g.,kPam,atmft)(5-25)

For tubular packed-bed reactors, we are more interested in catalyst weight rather than the distance z down the reactor. The catalyst weight up to a distance of z down the reactor is

W=(1Φ)Acz×ρc[Weighted of catalyst]=[Weighted of solids]×[Density of solid catalyst](5-26)

where Ac is the cross-sectional area. The bulk density of the catalyst, (mass of catalyst per volume of reactor bed), is just the product of the density of the solid catalyst particles, ρi, and the fraction of solids, (1 − ϕ):

Bulk density

ρb=ρc(1-ϕ)

Using the relationship between z and W [Equation (5-26)], we can change our variables to express the Ergun equation in terms of catalyst weight:

dPdW=-β0Ac(1-ϕ)ρcP0P(TT0)FTFT0

Use this form for multiple reactions and membrane reactors.

Further simplification yields

dPdW=-α2TT0P0P/P0(FTFT0)(5-27)

Again, let p = (P/P0), then

dpdW=α2pTT0FTFT0 (5-28)

Used for multiple reactions

where

α=2β0Acρc(1Φ)P0(5-29)

Differential form of Ergun equation for the pressure drop in packed beds

and where typical units of = might be (kg−1) or (lb−1m).

We will use Equation (5-28) when multiple reactions are occurring or when there is pressure drop in a membrane reactor. However, for single reactions in packed-bed reactors, it is more convenient to express the Ergun equation in terms of the conversion X. Recalling Equation (4-20) for FT,

FTFT0=1+εX(4-20)

where, as before

ε=yA0δ=FA0FT0δ(4-22)

Substituting for the ratio (FT/FT0), Equation (5-28) can now be written as

dpdW=α2p(1+εX)TT0 (5-30)

We note that when is negative, the pressure drop =P will be less (i.e., higher pressure) than that for ε = 0. When is positive, the pressure drop ΔP will be greater than when ε = 0.

For isothermal operation, Equation (5-30) is only a function of conversion and pressure

Use for single reactions

dpdw=F2(X,p)(5-31)

Recalling Equation (5-21), for the combined mole balance, rate law, and stoichiometry

For relaxation, see YouTube video on pressure drop, “Chemical Engineering Gone Wrong,” accessible through the CRE Web site home page.

dXdW=F1(X,p)(5-21)

Two coupled equations to be solved numerically

we see that we have two coupled first-order differential Equations, (5-31) and (5-21), that must be solved simultaneously. A variety of software packages (e.g., Poly-math) and numerical integration schemes are available for this purpose.

Analytical Solution. If ε = 0, or if we can neglect (εX) with respect to 1.0 (i.e., 1 ≫ εX), we can obtain an analytical solution to Equation (5-30) for iso-thermal operation (i.e., T = T0). For isothermal operation with = 0, Equation (5-30) becomes

dpdW=α2p(5-32)

Rearranging gives

2pdpdW=α

Isothermal with ε = 0

Taking p inside the derivative, we have

dp2dW=α

Integrating with p = 1 (P = P0) at W = 0 yields

p2 = (1 − αW)

Taking the square root of both sides gives

p=PP0=(1αW)1/2(5-33)

Pressure ratio only for ε = 0 and isothermal

Icon shows a caution symbol.

Caution

Be sure not to use this equation if ε ≠ 0 or if the reaction is not carried out isothermally. The pressure drop parameter α is

α=2β0Acρc(1Φ)P0(kg1orlbm1)(5-29)

Equation (5-33) can be used to substitute for the pressure in the rate law, in which case the mole balance can then be written solely as a function of conversion and catalyst weight. The resulting equation can readily be solved either analytically or numerically.

If we wish to express the pressure in terms of reactor length z, we can use Equation (5-26) to substitute for W in Equation (5-33). Then

p=PP0=(1-2β0zP0)1/2(5-34)

from which one can plot the pressure along the length of the reactor.

5.5.3 Pressure Drop in Pipes

Normally, the pressure drop for gases flowing through pipes without packing can be neglected. For flow in pipes, the pressure drop along the length of the pipe can be approximated by

p=(1-αpV)1/2(5-35)

where

αp=4fFG2Acp0P0D(5-36)

and where fF usually is the Fanning friction factor, D is the pipe diameter, and the other parameters are the same as previously defined.

For the flow conditions given in Example 5-4 in a 1000-ft length of 1½-inch schedule-40 pipe (αp =p = 0.05 m–3), the pressure drop is usually less than 5%. However, for high volumetric flow rates through microreactors, the pressure drop might be significant. See sample calculation on the Web site at http://www.umich.edu/~elements/5e/05chap/pdf/alpha-P-calculation-Example-5-4.pdf.

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