C A0 0.8 mol/dm 3. C T0 1.0 mol/dm 3. r A. v 0 C A0 vc A + r A V V dc A. 4. Stoichiometry (gas phase, P P 0, T T 0 ). From Equation (3-41) we have

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1 Chapter 10 Catalsis and Cataltic Reactors W10-1 Web Example W10 Catalst Deca in a Fluidized Bed Modeled as a CSTR The gas-phase cracking reaction Gas oil (g) Products (g) B C is carried out in a fluidized CSTR reactor. The feed stream contains 80% gas oil () and 20% inert I. The gas oil contains sulfur compounds, which poison the catalst. s a first approximation we will assume that the cracking reaction is first order in the gas oil concentration. The rate of catalst deca is first order in the present actiit, and first order in the reactant concentration. ssuming that the bed can be modeled as a well-mixed CSTR, determine the reactant concentration, actiit, and conersion as a function of time. The olumetric feed rate to the reactor is 5000 m 3 /h. There are 50,000 kg of catalst in the reactor and the bulk densit is 500 kg/m 3. dditional information: C mol/dm 3 C T0 1.0 mol/dm 3 k B k 45 h 1 k d 9 dm 3 /mol h Solution 1. Mole Balance on reactant: dn C 0 C + r W = (WE10-.1) Recalling N C V and r V r W, then for constant olume we hae C 0 C + r V V dc = (WE10-.2) 2. Rate Law: r = kac (WE10-.3) 3. Deca Law: da = k d ac (WE10-.4) 4. Stoichiometr (gas phase, P P 0, T T 0 ). From Equation (3-41) we hae --- = =( 1+ εx) F T F T0 (WE10-.5) For simplicit, gas oil is used to represent the reactie portion of the feed. In actualit, gas oil, distilled from crude, is made up of complex hdrocarbons, which can be cracked, and simple hdrocarbons, which will not crack and are therefore inert in this application.

2 W10-2 Catalsis and Cataltic Reactors Chapter 10 X 1 F C = = F 0 C ε ε C = C 0 C ε = 0 δ = ( ) 0 = 0 0 = C T0 C --- = C T0 (WE10-.6) Soling for ields 1+ = C C T0 (WE10-.7) 5. Combining gies us C 0 ( 1+ 0 ) C C C kac V V dc = T0 (WE10-.8) Diiding both sides of Equation (WE10-.8) b the olume and writing the equation in terms of τ V/, we obtain (WE10-.9) dc dτ C 0 ( ) ( 1+ C C T0 ) + aτk = C τ τ s an approximation we assume the conersion to be F X 0 F C C = = = F 0 C 0 1+ C C T0 C 0 (WE10-.10) Calculation of reactor olume and space time ields V W ,000 = = = 100 m ρ 3 b 500 kg m 3

3 Chapter 10 Catalsis and Cataltic Reactors W a C X t (h) Figure WE10-.1 Variation of C (mol/dm 3 ), a, and X with time in a CSTR. V 100 m t = --- = = 0.02 h 5000 m 3 h C, X, and a time trajectories in a CSTR not at stead state Equations (WE10-.4), (WE10-.9), and (WE10-.10) are soled using Polmath as the ODE soler. The Polmath program is shown in Table WE The solution is shown in Figure WE TBLE WE POLYMTH PROGRM The conersion ariable X does not hae much meaning in flow sstems not at stead state, owing to the accumulation of reactant. Howeer, here the space time is relatiel short ( τ 0.02 h) in comparison with the time of deca t 0.5 h. Consequentl, we can assume a quasi-stead state and consider the conersion as defined b Equation (WE10-.10) alid. Because the catalst decas in less than an hour, a fluidized bed would not be a good choice to carr out this reaction.

4 W10-4 Catalsis and Cataltic Reactors Chapter 10 Web Heat Effects in Moing Beds. We shall consider two cases for modeling the temperature profile in the moing-bed reactor. In one case the temperature of the solid catalst and the temperature of the gas are different and in the other case the are the same. Case 1 (T T s ). The rate of heat transfer between the gas at temperature T and the solid catalst particles at temperature T S is Q P = hã P ( T T S ) (W10-129) where h heat transfer coefficient, kj/m 2 s K ã solid catalst surface area per mass of catalst in the bed, m 2 P /kg cat T S temperature of the solid, K. lso, T a = temperature of heat exchange fluid, K The energ balance on the gas phase is Energ balance dt dw Uã W ( T a T ) + hã P ( T S T ) = Σ F i C Pi (W10-130) If D P is the pipe diameter (m), B is the bulk catalst densit (kg/m 3 ), and is the wall surface area per mass of catalst (m 2 /kg) 4 ã w = D P ρ B The energ balance on the solid catalst is ã w (W10-131) Heat exchange between catalst particle and gas dt S dw hã P ( T S T ) + ( r )( ΔH Rx ) = U S C PS (W10-132) where C PS (J/kg K) is the heat capacit of the solids, U s (kg/s) the catalst loading, and ã P is the external surface area of the catalst pellet per unit mass of catalst bed: where d P is the pellet diameter. 6 ã P = d P ρ b (W10-133) Case 2 (T s T). If the product of the heat transfer coefficient, h, and the surface area, ã P, is er large, we can assume that the solid and gas temperatures are identical. Under these circumstances the energ balance becomes dt Uã W ( T a T ) + ( r )( ΔH Rx ) = dw U S C PS + Σ F i C Pi (W10-134)

5 Chapter 10 Catalsis and Cataltic Reactors W10-5 WP10-21 B The elementar irreersible gas-phase cataltic reaction k 1 B is carried out isothermall in a batch reactor. The catalst deactiation follows a first-order deca law and is independent of the concentrations of both and B. (a) Determine a general expression for catalst actiit as a function of time. (b) Make a qualitatie sketch of catalst actiit as a function of time. Does a(t) eer equal zero for a first-order deca law? (c) Write out the general algorithm and derie an expression for conersion as a function of time, the reactor parameters, and the catalst parameters. Fill in the following algorithm Mole balance Rate law Deca law Stoichiometr Combine Sole 1. Separate 2. Integrate k ns.: X 1 exp 1 W = ( 1 exp( k k d V d t) ) 0 (d) Calculate the conersion and catalst actiit in the reactor after 10 minutes at 300 K. (e) How would ou expect our results in parts (b) and (d) to change if the reaction were run at 400 K? Briefl describe the trends qualitatiel. (f) Calculate the conersion and catalst actiit in the reactor after 10 minutes if the reaction were run at 400 K instead of 300 K. Do our results match the predictions in part (e)? dditional information: C 0 = 1 mol/dm 3 V 0 = 1 dm 3 W = 1 kg k d = 0.1 min 1 at 300 K k 1 = 0.2 dm 3 /(kg cat min) at 300 K E d /R = 2000 K E /R = 500 K