capt 10 reatores

Prévia do material em texto

10. Catalysis and Catalytic Reactors
Learning Resources
Example CD10-1: Analysis of a Heterogeneous Reaction
[Class Problem, Winter 1997]
	 
	 
	 
	 
	 
	 
	 
	 
	Experimental data for the gas-phase catalytic reaction
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	are shown below. The limiting step in the reaction is known to be irreversible, so that the overall reaction is irreversible. The reaction was carried out in a differential reactor to which A, B, and C were all fed.
	 
	 
	 
	 
	 
	(a) Sketch as a function of PA , as a function of P B , and as a function of P C. 
(b) From your observations in part (a), which species would appear in the numerator of the rate expression? Which species would appear in the denominator of the rate expression? To what power is the denominator raised?
(c) From your conclusions from part (b), suggest a rate law consistent with the experimental data.
	 
	 
	(d) Evaluate the rate law parameters.
(e) From your rate expression, which species can you conclude are adsorbed on the surface?
(f) From your conclusions in part (e), suggest a mechanism and rate-limiting step for this reaction.
(g) For an entering partial pressure of A of 2 atm in a PBR, what is the ratio of sites of A to C sites at 80% conversion of A? 
(h) At what conversion are the number of A and C sites equal?
	 
	 
	 
	 
	 
	
Solution
(a) Compare runs 1, 3, 6, and 7. P B and P C are fixed. As P A increases the rate first increases then levels off. Therefore, P A must be in both the numerator and denominator.
	 
	 
	 
	 
	 
	Compare runs 4 and 5. PA and PB are fixed. Because the reaction is irreversible, PC must be in the denominator. 
	 
	 
	 
	 
	 
	Compare runs 2 and 7. PA and PC are fixed.  increases by a factor of 10 when PB increases by a factor of 10. PB is only in the numerator. 
	 
	 
	 
	 
	 
	
(b) A and B are in the numerator, C and A are in the denominator. 
(c) 
(d) For fixed P C and P B ,
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	 
	For fixed P A and P B
	 
	 
	 
	 
	 
	
 
	 
	 
	 
	 
	 
	
We note that the non-linear regression techniques described in Chapter 5 could have easily been used.
Find the Mechanism and Rate Limiting Step
 (f) A and C are on the surface
	
10. Catalysis and Catalytic Reactors
Learning Resources
Example CD10-2: Least Squares Analysis to Determine the Rate Law Parameters k, K T, and K B
(Example 6-2 in 2nd ed.)
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	Use the data in Table 10-6 together with Equation (10-81) to evaluate the rate law parameters. Then calculate the catalyst weight necessary for these rate law parameters.
	 
	 
	 
	 
	 
	 
	 
	
Solution
The calculations are shown in Table CD10-2.1. A plot ofversus P T at constant P B will be linear with slope 1/k (see Figure CD10-2.1).
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	
Figure CD10-2.1
	 
	 
	 
	 
	 
	 
	 
	Rearranging Equation (10-81) for PB = 0, we have
	 
	 
	 
	 
	 
	 
	 
	
	(CD10-2.1)
	 
	 
	 
	 
	 
	 
	From Figure CD10-2.1,
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	Substituting these values into Equation (CD10-2.1), we find that
	 
	 
	 
	 
	 
	 
	 
	
	(CD10-2.2)
	 
	 
	
	 
	 
	 
	The constant K B can be evaluated from the slope of the plot ofversus P B for constant P T . From Figure CD10-2.2,
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	
Figure CD10-2.2
	 
	 
	 
	 
	 
	 
	 
	In addition to the graphical determination, we can use linear regression to determine the rate law parameters. We will use Equations (5-31) through (5-33) along with Table CD10-2.2 to determine k, K T , and
K B. in Equations (10-81) and (5-22).
Recall Equation (10-81),
	 
	 
	 
	 
	 
	 
	 
	
	(10-81)
	 
	 
	 
	 
	 
	 
	and Equation (5-30),
	 
	 
	 
	 
	 
	 
	 
	
	(5-30)
	 
	 
	 
	 
	 
	 
	where Y=, a 0 = 1/kK T, a 1 = K B /kK T , a 2 = 1/k, X 1 =P B , and X 2 = P T . Next, recall the linear regression equations and apply them for the 16 runs of this example.
	 
	 
	 
	 
	 
	 
	 
	
	(CD10-2.3)
(CD10-2.4)
(CD10-2.5)
	 
	 
	If we were to use a software package such as POLYMATH, we would simply enter Y, X 1 X 2 , and for each run and the parameters a 0, a 1 , and a 2 would be displayed in a few seconds. Alternatively, we can form Table CD10-2.2 and carry out the numerical operations ourselves to determine the rate law parameters. We can either form the table using a calculator or by using a spreadsheet such as Excel or Lotus 123.
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	Equation (CD10-2.3) becomes
	 
	 
	 
	 
	 
	 
	 
	6.62 X 109 = 16a0 + 12a1 + 61.5a2
	(CD10-2.6)
	 
	 
	 
	 
	 
	 
	Equation (CD10-2.4) becomes
	 
	 
	 
	 
	 
	 
	 
	5.67 X 109 = 12a0 = 44a = 12a21
	(CD10-2.7)
	 
	 
	Equation (CD10-2.5) becomes
	 
	 
	 
	 
	 
	 
	 
	6.0 X 1010 = 61.5a0 + 12a1 + 761.25a2
	(CD10-2.8)
	 
	 
	 
	 
	 
	 
	Solving these three equations simultaneously, we obtain a0 = 7.12 x 10 7 , a 1 = 9.0 x 10 7 , and a 2 = 7.16 x 10 7 . The corresponding rate law parameters are
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	Using POLYMATH, we obtain
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	with, of course, the parameter values being the same as those obtained from Table CD10-2.2 and Equation (CD10-2.5).
After substituting the numerical values of k, KB , and K T into Equation (10-80), the rate law at 600°C for hydrodemethylation of toluene,
is given by the equation
	 
	 
	 
	 
	 
	 
	 
	
	(CD10-2.9)
	 
	 
	 
	 
	 
	 
	whereP i is in atm.
	 
	 
	 
	 
	 
	 
	 
	After we have the adsorption constant K T and K B , we could calculate the ratio of sites. For example, the ratio of toluene sites to benzene sites at 40% conversion is
	
	 
	 
	We see that at 40% conversion there are approximately 20% more sites occupied by toluene than by benzene.
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	Calculate the catalyst weight necessary for these parameters:
	 
	 
	 
	If we were to neglect pressure drop, we can solve for the catalyst weight necessary to achieve 65% conversion with the aid of Simpson's five-point formula.
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	Substituting for k, K T , K B ,P B , PH 2, and PT in Equation (E10-3.2) yields
	 
	 
	 
	 
	 
	 
	 
	
	(CD10-2.10)
	 
	 
	 
	 
	 
	 
	We can now proceed to solve for the catalyst weight necessary to obtain this conversion by using Equation (CD10-2.10).
	 
	 
	 
	 
	 
	 
	 
	Hand calculations: The calculations for X versus (1/) are displayed in Table CD10-2.3.
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	Using numerical integration (Appendix A.5), after dividing the area under the curve into two parts: X = 0 to X = 0.52 with h 1 = 0.13 and X = 0.52 to X = 0.65 with h 2 = 0.13, we have
	 
	 
	 
	 
	 
	Numerical 
evaluation
	 
	
	
	 
	 
	 
	 
	 
	 
	Since the reciprocal of the rate of reaction increases sharply as the conversion approaches unity, the numerical integration probably should have included a greater number of integrals than the six used in the calculation.
	 
Example CD10-3: Decay in a Straight-Through Reactor
	 
	 
	 
	 
	 
	 
	 
	 
	 
	 
	The gas-phase cracking of a Salina light gas-oil reaction
	 
	 
	 
	 
	 
	A typical cost
of the catalyst is
$1 million
	 
	gas-oil (g)  products (g) + coke (s)
A  B
	 
	 
	 
	is to be carried out in a straight-through transport reactor containing a catalyst that decays as a result ofcoking. The reaction is carried out at 750°F.
The entering concentration of A is 0.2 kmol/m 3 . The catalyst particles are assumed to move with the mean gas velocity (U g =U s = 7.5 m/s). The rate law is
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	with K A = 3 m 3 /kmol and K B = 0.01 m 3 /kmol. The maximum value of the term K B C B is small (0.002), so it can be neglected with respect to the other terms (e.g., 1). Using this fact and the bulk density,b , the rate law becomes
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	with k = 8 s -1 . At 750°F, the catalyst activity for light gas-oil over a synthetic catalyst for short contact times (i.e., less than 100 s) is
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	with A ´= 7.6 s -1/2 . As a first approximation, neglect volume change with reaction, pressure drop, and temperature variations.
Determine the conversion as a function of distance down the reactor. The reactor length is 6 m.
	 
	 
	 
	 
	 
	 
	 
	
Solution
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	For a catalyst particle traveling with a velocity of U, the time that the catalyst particle has been in the reactor when it reaches a height is just
	 
	 
	 
	 
	 
	 
	 
	
	(CDE10-4.4)
	 
	 
	 
	 
	 
	 
	and
	 
	 
	 
	 
	 
	 
	 
	
	(CDE10-4.5)
	 
	 
	 
	 
	 
	 
	where
	 
	 
	 
	 
	 
	 
	 
	Stoichiometry. Gas-phase reaction with= 0, T = T 0 , and P = P 0.
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	Combining yields
	 
	 
	 
	 
	 
	 
	 
	
	(CDE10-4.6)
	 
	 
	 
	 
	 
	 
	POLYMATH Solution. In using POLYMATH to obtain a solution, it is usually easiest to write the mole balance, rate law, and stoichiometry separately rather than combining them into a single equation. Therefore, starting with the mole balance
	
	 
	 
	 
	 
	 
	 
	
	(CDE10-4.7)
	 
	 
	 
	 
	 
	 
	we substitute for -r A and a using Equations (CDE10-4.2) and (CDE10-4.5). The POLYMATH program and solution are shown in Table CDE10-4.1 and Figure CDE10-4.1 respectively.
	 
	 
	 
	
Table CDE10-4.1
POLYMATH Program Coking in a Straight-Through Transport Reactor
	 
	 
	
	
	 
	 
	 
	 
	 
	 
	 
	 
Figure CDE10-4.1
Conversion and Activity Profiles
	
10. Catalysis and Catalytic Reactors
Learning Resources
Example CD10-4: Catalyst Poisoning in a Constant-Volume Batch Reactor
Develop a decay Equation for catalyst poisoning solely as a function of catalyst activity.
Solution
	 
	 
	For the case of a constant-volume batch reactor, a mole balance of the poison in the gas phase combined with the rate law for poisoning [Equation (10-109)] gives
	 
	 
	 
	 
	 
	 
	 
	
	
(CD10-4.1)
	 
	 
	 
	 
	 
	 
	where k A =kC T0 is the initial total concentration of unpoisoned site, C P•S is the concentration of the poisoned sites, and ƒ is the fraction of the poisoned sites. Taking the ratio of Equation (10-111) to Equation (CD10-4.1)
	 
	 
	 
	 
	 
	 
	 
	
	 
	 
	 
	 
	 
	 
	 
	and then integrating with limits a = 1 and C P =C P0
	 
	 
	 
	 
	 
	 
	 
	
	(CD10-4.2)
	 
	 
	 
	 
	 
	 
	or
	 
	 
	 
	 
	 
	 
	 
	
	(CD10-4.3)
	 
	 
	 
	 
	Constant-
volume batch
	 
	The decay can now be written in the form
	 
	 
	 
	 
	 
	
	
	
	(CD10-4.4)
	
	
	
	
	
	
	which is of the form:
	
	
	
	
	
	
	
	
	(CD10-4.5)
	
	
	
	
	
	
	where