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pset03_soln

# pset03_soln - 10.37 HW 3 Spring 2007 Problem 1 A CSTR of...

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10.37 HW 3 Spring 2007 Problem 1. A CSTR of volume 0.602 liters (constant density liquid phase) is operated in which the following reaction occurs: A + B C + D The feed rate of A is 1.16 liters/hr of a solution at concentration 5.87 mmol/L. The feed rate of B is 1.20 liters/hr at a concentration of 38.9 mmol/L. The outlet concentration of species A is 1.094 mmol/L. Calculate the rate constant for this reaction assuming a mass-action rate law of the form: r = k [ A ][ B ]. First draw a diagram of the problem: q A [A] 0 q B [B] 0 q out [A] [B] The only unknown quantities in the above figure are [B] and q tot . A volume balance on the carrier solvent gives the following relationship (constant fluid density): q A + q B = q out = 1.16 L/h + 1.20 L/h = 2.36 L/h Define the extent of reaction: ξ & = rV [ = ] mol time A material balance on component A yields the following: dn A = F A 0 F A ξ & = [ A ] 0 q A [ A ] q out ξ & . dt Setting the accumulation term to zero (steady state operation) and solving for the reaction rate we find: 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].

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ξ & = [ A ] 0 q A [ A ] q out ξ & = ( 5.87mmol/L )( 1.16L/h ) ( 1.094mmol/L )( 2.36L/h ) = 4.23mmol/h A material balance on component B yields the following: dn B = F B 0 F B ξ & = [ B ] 0 q B [ B ] q out ξ & . dt Setting accumulation to zero, and solving for [B] we find: [ B ] = [ B ] 0 q B ξ & . q out ( 38.9 mmol/L )( 1.20 L/h ) ( 4.23 mmol/h ) [ B ] = = 18.0 mmol/L ( 2.36 L/h ) Using the given rate law and knowing the extent of reaction, we can now calculate the rate constant: ξ & = rV = k [ A ][ B ] V ξ & k = [ A ][ B ] V ( 4.23 mmol/h ) L 1 1 k = = 0.357 = 357 M h ( 1.094 mmol/L )( 18.0 mmol/L )( 0.602 L ) h mmol 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].
Problem 2. Consider the catalyzed reaction: A + B B + C with the second-order rate constant 1.15 x 10 -3 m 3 /mol/ksec. The rate law is r = k [ A ][ B ]. What volume of CSTR would be necessary to give 40% conversion of species A if the feed concentration of A is 96.5 mol/m 3 , the feed concentration of B is 6.63 mol/m 3 , and the flow rate is 0.5 m 3 /ksec? First draw a diagram of the problem: q [A] 0 [B] 0 [A] [B] [C] Define the extent of reaction: ξ & = rV [ = ] mol time Define the conversion in terms of the variables of the problem: mol A reacted r A V ξ & X A = = = mol A fed q [ A ] 0 q [ A ] 0 A material balance on species A gives the following equation: dn A = F A 0 F A ξ & = [ A ] 0 q [ A ] q ξ & dt At steady state: 0 = [ A ] 0 q [ A ] q ξ & ξ & = [ A ] 0 q [ A ] q Thus our conversion definition is equivalent to: 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].

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X A = [ A ] 0 q [ A ] q = 1 [ A ] q [ A ] 0 [ A ] 0 [ A ] = ( 1 X A ) [ A ] 0 A material balance on species [B] gives the following: dn B dt = F B 0 F B + 0 ξ & = [ B ] 0 q [ B ] q
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