Web Solved Problems Web Example SP-8.1 Hydrodealkylation of Mesitylene in a PFR CH 3 H 2. m-xylene can also undergo hydrodealkylation to form toluene:

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1 Chapter 8 Multiple Reactions W8-1 Web Solved Problems Web Example SP-8.1 Hydrodealkylation of Mesitylene in a PFR The production of m-xylene by the hydrodealkylation of mesitylene over a Houdry Detrol catalyst 1 involves the following reactions: H 2 CH 4 (WESP-8.1.1) m-xylene can also undergo hydrodealkylation to form toluene: H 2 CH 4 (WESP-8.1.2) A significant economic incentive The second reaction is undesirable, because m-xylene sells for a higher price than toluene ($1.32/lb m vs. $0.30/lb m ). 2 Thus we see that there is a significant incentive to maximize the production of m-xylene. The hydrodealkylation of mesitylene is to be carried out isothermally at 1500 R and 35 atm in a packed-bed reactor in which the feed is 66.7 mol% hydrogen and 33.3 mol% mesitylene. The volumetric feed rate is 476 ft 3 /h and the reactor volume (i.e., V = W ρ ) is 238 ft 3 b. The rate laws for reactions 1 and 2 are, respectively, 0.5 r 1M = k 1 C M C H 0.5 r 2T = k 2 C X C H (WESP-8.1.3) (WESP-8.1.4) where the subscripts are: M mesitylene, X m-xylene, T toluene, Me methane, and H hydrogen (H 2 ). At 1500 R, the specific reaction rates are Reaction 1: k 1 Reaction 2: k 2 ( ft 3 lb mol) 0.5 h ( ft 3 lb mol) 0.5 h The bulk density of the catalyst has been included in the specific reaction rate (i.e., k 1 = k 1 ρ b ). Plot the concentrations of hydrogen, mesitylene, and xylene as a function of PFR space time. Calculate the space time where the production of xylene is a maximum (i.e., ). τ opt 1 Ind. Eng. Chem. Process Des. Dev., 4, 92 (1965); 5, 146 (1966). 2 November 2004 prices, from Chemical Market Reporter (Schnell Publishing Co.), 265, 23 (May 17, 2004). Also see and

2 W8-2 Multiple Reactions Chapter 8 Solution Reaction 1: M+ H X+ Me (WESP-8.1.1) Mole balance on each and every species 1. Mole balances: Reaction 2: X+ H T+ Me (WESP-8.1.2) Hydrogen: H r (WESP-8.1.5) H Mesitylene: M r M (WESP-8.1.6) Xylene: X r X (WESP-8.1.7) Toluene: T r T (WESP-8.1.8) Me Methane: r (WESP-8.1.9) Me 2. Rate laws and net rates: Given Reaction 1: Reaction 2: = k 1 C H r 1M r 2T = k 2 C H C M C X (WESP-8.1.3) (WESP-8.1.4) Relative rates: (1) r 1H = r 1M = r 1Me = r 1X (WESP ) (2) r 2T = r 2Me = r 2H = r 2X (WESP ) Net rates: r M = r 1M = k 1 C 1/2 H C M r H = r 1H + r 2H = r 1H r 2T = k 1 C 1/2 H C M k 2 C 1/2 r T = r 2T = k 2 C 1/2 H C X 3. Stoichiometry The volumetric flow rate is H C X r X = r 1X + r 2X = r 1H r 2T = k 1 C 1/2 H C M k 2 C 1/2 H C X r Me = r 1Me + r 2Me = r 1H + r 2T = k 1 C 1/2 H C M + k 2 C 1/2 H C X (WESP ) (WESP ) (WESP ) (WESP ) (WESP ) F v v T P 0 = T 0 P T 0

3 Chapter 8 Multiple Reactions W8-3 Because there is no pressure drop P = P 0 (i.e., y = 1), the reaction is carried out isothermally, T = T 0, and there is no change in the total number of moles; consequently, v = v 0 Flow rates: F H F M F X v 0 C H v 0 C M v 0 C X F Me v 0 C Me = F H0 F H = v 0 ( C H0 C H ) F M0 F M F X = v 0 ( C M0 C M C X ) (WESP ) (WESP ) (WESP ) (WESP ) (WESP ) 4. Combining and substituting in terms of the space-time yields V τ = --- v 0 If we know C M, C H, and C X, then C Me and C T can be calculated from the reaction stoichiometry. Consequently, we need only to solve the following three equations: The emergence of user-friendly ODE solvers favors this approach over fractional conversion. dc H k dτ 1 C H C M k 2 C X C H dc M k dτ 1 C M C H dc X k dτ 1 C M C H k 2 C X C H 5. Parameter evaluation: At T 0 = 1,500º R and P 0 = 35 atm, the total concentration is (WESP ) (WESP ) (WESP ) P C 0 35 atm T0 = = = lb mol/ft 3 RT 0 atm ft ( 1, 500 R) lb mol R C H0 = y H0 C T0 = ( 0.667) ( lb mol/ft 3 ) = lb mol/ft 3 1 C M0 = -- C 2 H0 = lb mol/ft 3 C X0 = 0 V 238 ft t = --- = = 0.5 h 476 ft 3 h v 0

4 W8-4 Multiple Reactions Chapter 8 0 C T0 v lb mol ft 3 ft 3 = = h 0 = lb mol/h We now solve these three equations, (WESP ) to (WESP ), simultaneously using Polymath. The program and output in graphical form are shown in Table WESP and Figure WESP-8.1.1, respectively. However, I hasten to point out that these equations can be solved analytically and the solution was given in the first edition of this text. Figure WESP Concentration profiles in a PFR. TABLE WESP POLYMATH PROGRAM Web Example SP-8.2 Hydrodealkylation of Mesitylene in a CSTR For the multiple reactions and conditions described in Web Example SP-8.1, calculate the conversion of hydrogen and mesitylene along with the exiting concentrations of mesitylene, hydrogen, and xylene in a CSTR.

5 Chapter 8 Multiple Reactions W8-5 Solution As in Web Example SP-8.1, we assume v = v 0 ; for example, 1. CSTR Mole Balances: F A = vc A = v 0 C A, etc. Hydrogen: v 0 C H0 v 0 C H = r H V (WESP-8.2.1) Mesitylene: v 0 C M0 v 0 C M = r M V (WESP-8.2.2) Xylene: v 0 C X = r X V (WESP-8.2.3) Toluene: v 0 C T = r T V (WESP-8.2.4) Methane: v 0 C Me = r Me V (WESP-8.2.5) 2. Net Rates The rate laws and net rates of reaction for these reactions were given by Equations (WESP ) through (WESP ) in Web Example SP Stoichiometry: Same as in Web Example SP-8.1. Combining Equations (WESP ) through (WESP ) with Equations (WESP ) through (WESP-8.2.3) and after dividing by v 0, (τ = V/v 0 ), yields C H0 C M0 ( k 1 C H C M + k 2 C H )τ C H ( k 1 C H )τ C M C M C X C X ( k 1 C H C M k 2 C H C X )τ (WESP-8.2.6) (WESP-8.2.7) (WESP-8.2.8) Next, we put these equations in a form such that they can be readily solved using Polymath. f( C H ) 0 = C H C H0 +( k 1 C H C M + k 2 C H C X )τ f( C M ) 0 = C M C M0 + k 1 C H C M τ f( C X ) 0 = ( k 1 C H C M k 2 C H C X )τ C X (WESP-8.2.9) (WESP ) (WESP ) The Polymath program and solution to equations (WESP-8.2.9), (WESP ), and (WESP-8.2.1) are shown in Table WESP The problem was solved for different values of τ and the results are plotted in Figure WESP For a space time of τ 0.5, the exiting concentrations are C H , , and all in lb mol/ft 3. The overall conversion is C M C X Overall conversion Hydrogen: Mesitylene: F X H0 F H C H H0 C H = = = = F H0 C H0 F X M0 F M C M M0 C M = = = = F M0 C M0

6 W8-6 Multiple Reactions Chapter 8 TABLE WESP POLYMATH PROGRAM AND SOLUTION To vary τ CSTR, one can vary either v 0 for a fixed V or vary V for a fixed v 0. Figure WESP Concentrations as a function of space time. We resolve Equations (WESP-8.2.6) through (WESP ) for different values of τ to arrive at Figure WESP The moles of hydrogen consumed in reaction 1 are equal to the moles of mesitylene consumed. Therefore, the conversion of hydrogen in reaction 1 is C X M0 C M H = = = X H = 0.36 C H0 The conversion of hydrogen in reaction 2 is just the overall conversion minus the conversion in reaction 1; that is, X 2H = X H X 1H = = 0.22 The yield of xylene from mesitylene based on molar flow rates exiting the CSTR for τ 0.5 is

7 Chapter 8 Multiple Reactions W8-7 Overall selectivity, S, and yield, Ỹ. Ỹ M/X F X F M0 F M C X C M0 C M = = = Ỹ M/X 0.41 mol xylene produced = mole mesitylene reacted The overall selectivity of xylene relative to toluene is S X/T F X C X C T C X = = = = C M0 C M C X S X/T 0.7 mol xylene produced = mole toluene produced Recall that for a CSTR the overall selectivity and yield are identical with the instantaneous selectivity and yield. Web Example SP-8.3 Calculating Concentrations as Functions of Position for NH 3 Oxidation in a PFR The following gas-phase reactions take place simultaneously on a metal oxide supported catalyst: NH 3 + 5O 2 4NO+ 6H 2 O 2NH O 2 N 2 + 3H 2 O 2NO+ O 2 2NO 2 4NH 3 + 6NO 5N 2 + 6H 2 O Writing these equations in terms of symbols yields 2 Reaction 1: 4A 5B 4C 6D = k 1A C A C B (WESP-8.3.1) Reaction 2: 2A 1.5B E 3D r 2A = k 2A C A C B (WESP-8.3.2) Reaction 3: 2C B 2F r 3B = k 3B C 2 C C B (WESP-8.3.3) 23 Reaction 4: 4A 6C 5E 6D = k 4C C C C A (WESP-8.3.4) r 1A r 4C with 3 k 1A = 5.0 ( m 3 kmol) 2 min k 2A = 2.0 m 3 kmol min k 3B = 10.0 ( m 3 kmol) 2 min k 4C = 5.0 ( m 3 kmol) 23 min 3 Reaction orders and rate constants were estimated from periscosity measurements for a bulk catalyst density of 1.2 kg/dm 3.

8 W8-8 Multiple Reactions Chapter 8 Note: We have converted the specific reaction rates to a per unit volume basis by multiplying the k on a per mass of catalyst basis by the bulk density of the packed bed (i.e., k = k ρ B ). Determine the concentrations as a function of position (i.e., volume) in a PFR. Additional information: Feed rate 10 dm 3 /min; volume of reactor 10 dm 3 ; and C A0 = C B0 = 1.0 mol dm3, C T0 = 2.0 mol dm 3 Solution Mole balances: Species A: A r A (WESP-8.3.5) Species B: B r B (WESP-8.3.6) Solutions to these equations are most easily obtained with an ODE solver Species C: C r C (WESP-8.3.7) Species D: D r D (WESP-8.3.8) Species E: E r E (WESP-8.3.9) Species F: F r F (WESP ) Total: F A + F B + F C + F D + F E + F F (WESP ) Rate laws: See above for r 1A, r 2A, r 3B, and r 4C. Stoichiometry: A. Relative rates r Reaction 1: 1A r B r C r = = = D (WESP ) r Reaction 2: 2A r B r E r = = = D (WESP ) r Reaction 3: 3C r B r = = F (WESP ) r Reaction 4: 4A r C r E r = = = D (WESP )

9 Chapter 8 Multiple Reactions W8-9 B. Concentrations: For isothermal operation and no pressure drop, the concentrations are given in terms of the molar flow rates by C j F j = C T0 Next substitute for the concentration of each species in the rate laws. Writing the rate law for species A in reaction 1 in terms of the rate of formation, r 1A, and molar flow rates, and, we obtain F A F B 2 r 1A k 1 C A C B k 1A C F A T C F 2 B = = T Thus Similarly for the other reactions, r 1A r 2A k 1A C 3 F F 2 = A B T = k 2A C 2 F F A B T (WESP ) (WESP ) r 3B r 4C k 3B C 3 F 2 = F C B T F k 4C C C F = A T (WESP ) (WESP ) Next, we determine the net rate of reaction for each species by using the appropriate stoichiometric coefficients and then summing the rates of the individual reactions. Net rates of formation: 2 Species A: r A r 1A + r 2A +-- r (WESP ) 3 4C Species B: r B 1.25 r 1A r 2A + r 3B (WESP ) Species C: r C r 1A + 2 r 3B + r 4C (WESP ) Species D: r D 1.5 r1a 1.5 r 2A r 4C (WESP ) 5 Species E: r E r (WESP ) 2 6 4C Species F: (WESP ) r F Combining: Rather than combining the concentrations, rate laws, and mole balances to write everything in terms of the molar flow rate as we did in the past, it is more convenient here to write our computer solution (either Polymath or our own program) using equations for,, and so on. Consequently, we shall write r 1A F A r 2A 2r 3B

10 W8-10 Multiple Reactions Chapter 8 Equations (WESP ) through (WESP ) and (WESP-8.3.5) through (WESP ) as individual lines and let the computer combine them to obtain a solution. The corresponding Polymath program written for this problem is shown in Table WESP and a plot of the output is shown in Figure WESP One notes that there is a maximum in the concentration of NO (i.e., C) at approximately 1.5 dm 3. However, there is one fly in the ointment here: It may not be possible to determine the rate laws for each of the reactions. In this case, it may be necessary to work with the minimum number of reactions and hope that a rate law can be found for each reaction. That is, you need to find the number of linearly independent reactions in your reaction set. In Web Example SP-8.3, there are four reactions given [(WESP-8.3.5) through (WESP-8.3.8)]. However, only three of these reactions are independent, as the fourth can be formed from a linear combination of the other three. Techniques for determining the number of independent reactions are given by Aris. 4 4 R. Aris, Elementary Chemical Reactor Analysis (Upper Saddle River, N.J.: Prentice Hall, 1969).

11 Chapter 8 Multiple Reactions W8-11 TABLE WESP POLYMATH PROGRAM Figure WESP Molar flow rates profiles.

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