(TIỂU LUẬN) CHEMICAL REACTION ENGINEERING (CH3347) CLASS CC01—GROUP 2 SEMESTER 212

HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY - VNU OFFICE FOR INTERNATIONAL STUDY PROGRAM FACULTY OF CHEMICAL ENGINEERING      CHEMICAL REACTION ENGINEERING (CH3347) CLASS CC01—GROUP - SEMESTER 212 INSTRUCTOR: Ph.D CHÂU NGỌC ĐỖ UYÊN Student’s Name Student’s ID LÊ PHẠM NHÃ THY 1953016 LÊ NGỌC THẢO 1952123 ĐẬU NGUYỄN ANH THƯ 1952476 VŨ THỊ PHI YẾN 1952542 TRƯƠNG HẠNH THANH TUYỀN 1915806 LÝ THANH THUÝ VY 1852885 Ho Chi Minh city, 2022 SOLUTION 1− ( 0.01 −1 − = (0.04 )  −   = 0.04   {−1  = 0.01  −1  =  = 2000  )  = 100 ⁄  0 = 100 ⁄  =   =  0.04  => = −  0.01 1−   − =   (1 −  ) −     0  0   (1−  ) = 1  (1 −   ) −     0  1−   = 1  [(1 −   ) −   ]  1−0.8 = 1  [(1 −   ) − 0.8  ]  = 1  (1 − 1.25  )     =  ∫ 0  = (− ) 0 =>  = 0.8 = 80%    0 =  2000 100 0  ∫0      (1−1.25  ) = 100 ∫   0.04×100× (1−1.25  )   = 0.506 = 50.6% SOLUTION Half-life time: t1/2= 5.2 days Mean residence time t= 30 days 0.693 k=  1/2 = 0.693 5.2 = 0.1333  −1 For Mixed Flow Reator:  1 = = = 0.2  +  + 0.1333 × 30 0  Fraction of activity = -   0  % of activity = 80% = – 0.2 = 0.8 SOLUTION -r = 0.004 C – 0.01 C A A R V = 2m3 = 2000 liter FA = 100 l/min , CA0 = 100 mmol/l = 0.1 mol/l At equation : X =  −   / 0  +1 Ae   =  =  XAe = 0.8 = 80% Actual conversion   =  − = 2000 100 =  0.04  − 01  CA = CA0 (1-XA) CR = CR0 + CA0 X A Substituting all values, we get XA = 0.072 = 7.2 % 5.9 A specific enzyme acts as catalyst in the fermentation of reactant A At a given enzyme concentration in the aqueous feed stream (25 liter/min) find the volume of plug flow reactor needed for 95% conversion of reactant A (CA0 = mol/liter) The kinetics of the fermentation at this enzyme concentration is given by:  → , 0.1    − = + 0.5     SOLUTION   = 25 ( ) We already have {  0  = 2(  )  = 95% For the volume of plug flow reactor, we have this equation with  = From − = 0.1  1+0.5   =  (1 −  ) = x (1-0.95)=0.1 (mol/l)   →  =−∫   0.1  =−∫ − + 0.5    ≈ 39.457 0.1    →  = 39.457 × 25 ( ) = 986.433( ) 5.10 A gaseous feed of pure A (2 mol/liter, 100 mol/min) decomposes to give a variety of products in a plug flow reactor The kinetics of the conversion is represented by A → 2.5 (products), − = (10 −1 ) Find the expected conversion in a 22-liter reactor Solution A → 2.5 (products) ⇒ Irreversible reaction ( −1 ) (FIRST − = (10 −1 )  ⇒  = 10 ORDER) For this stoichiometry, A → 2.5 (products), the expansion factor is  = 2.5 − = 1.5 We use the equation (5.8) in Handbook Chemical Reaction Engineering by Octave Levenspiel: We use equation (5.21): ⟺ ×  = −(1 +  ) ln(1 −  ) −   0     1  ) × 22() × ⟺10 ( × ( ) = −(1 + 1.5) ln(1 −  ) − 1.5 ×     100 ( )  ⇒  = 0.8997 Example 3.9 The feed contains 20% mole inerts: 0.8A -> 0.4R + 0.4S 0.2 inerts -> 0.2 inerts => εA = (0.8+0.2)−(0.8+0.2) =0 0.8+0.2 pA0 = 0.8P0 = 0.8 x = 0.8 (atm) at t0 = 20 (s)  C = A0  0  0.8 () = 0.082 (.) × (100+273)() = 0.026 ( .   ) Calculate τ:   0 =    τ = tf – t0 =  0   = 208() × 0.026 ( )  100× ( ) 3600 = 194.69 (s) ≈ 195 (s)   tf = τ + t0 = 195 + 20 = 215 (s) According to the data table, using interpolation, at t = 215 (s), pA = 0.125 (atm)  CA = 0.0041 (   ) For mixed flow reactor with εA = 0: -rA = τ=  0−    0 −   = 0.026 − 0041 ( ) => XA =  XA = 84% 195 () ×(− )   = 1.12 x 10-4 (  . ) −4  195() × (1.12×10 )( . ) = =  0.026 ( ) 0.84 ... -> 0 .2 inerts => εA = (0.8+0 .2) −(0.8+0 .2) =0 0.8+0 .2 pA0 = 0.8P0 = 0.8 x = 0.8 (atm) at t0 = 20 (s)  C = A0  0  0.8 () = 0.0 82 (.) × (100 +27 3)() = 0. 026 ( .   ) Calculate... stoichiometry, A → 2. 5 (products), the expansion factor is  = 2. 5 − = 1.5 We use the equation (5.8) in Handbook Chemical Reaction Engineering by Octave Levenspiel: We use equation (5 .21 ): ⟺ ×  =...   = 20 8() × 0. 026 ( )  100× ( ) 3600 = 194.69 (s) ≈ 195 (s)   tf = τ + t0 = 195 + 20 = 21 5 (s) According to the data table, using interpolation, at t = 21 5 (s), pA = 0. 125 (atm) 