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Unformatted text preview: Problem Set 4 Problem 1. (RC(O)OCH 2 ) 3 (RCH 2 ) 3 + 3 CO 2 r = k[(RC(O)OCH 2 ) 3 ] A B + 3 C r = k[A] Given: T = 150 C k(T ) = k = 5*10-3 (min-1 ) Ea = 85 kJ/mol F A,0 = 2.5 mol/min y A,0 = 1 T = 227 C X = 0.9 P = 10 atm First, find k at the reaction temperature using Eq 3-21 from Fogler: E 1 1 85 kJ / mol 1 1 R T 0 T 3 1 8.314 10 3 kJ /( mol K ) 423.15 K 500.15 K 1 k ( T ) = k ( T ) e = 5 10 e = 0.206255min min Next, make a stoichiometric table for the flow system (see Table 3-4 in Fogler). This table applies to both a PFR and CSTR reactor. Species A B C Total Feed Rate to Reactor (mol/min) F A0 0 0 F A0 Change within Reactor (mol/min) -F A0 X F A0 X 3F A0 X Effluent Rate from Reactor (mol/min) F A = F A0 (1 X) F B = F A0 X F C = 3F A0 X F T = F A0 (1 + 3 X) Since this is a gas-phase reaction, with a change in the total number of moles, the volumetric flow rate ( ) will not be constant. Simplify Eq 3-41 in Fogler for the steady state (constant P and T) ideal gas case to: = o F T = 0 F A 0 ( 1 + 3 X ) = 0 ( 1 + 3 X ) = F A RT ( 1 + 3 X ) F T 0 F A 0 P a) CSTR The design equation for CSTR volume in terms of conversion is (Eq 2-13 in Fogler): F A 0 X F A 0 X F A 0 X 0 ( 1 + 3 X ) F A 0 X F A RT ( 1 + 3 X ) X V CSTR , a = ( r A ) exit = k [ ] A = k F A = kF A 0 ( 1 X ) = Pk ( 1 X ) Plugging in numbers: mol .082 L atm ( 500.15 K )( 1 + 3 0.9 0.9 min mol K 3 2.5 ( ) ) V CSTR , a = = 1655.36 L 1.7 10 L 1 ( 10 atm ) 0.206255 min ( 1 0.9 ) Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. b) PFR Neglect pressure drop, so equation is the same as above. The design equation for a PFR in terms of conversion is (Eq 2-16 in Fogler): X X X X X V = F dX = F dX = F dX = F F A RT ( 1 + 3 X ) dX...
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- Fall '04