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 LÊ NGỌC THẢO VŨ THỊ PHI YẾN Ho Chi Minh city, 20225.6 download by : skknchat@gmail.co − { =2 =100 −1 = (0.04 = 0.04 = 2000 ) − (0.01 −1 SOLUTION −1 ) −1 = 0.01 ⁄ = 100 ⁄ = − =1 = = = )− 0[(1− [(1 (1 − )− 1− 1− 8 ] ] − 1.25 ) = download by : skknchat@gmail.co  = 2000  100  = 0.506 = 50.6% download by : skknchat@gmail.co SOLUTION Half-life time: t1/2= 5.2 days Mean residence time t= 30 days k= 0.693 = 0.693 1/2 = 0.1333 −1 5.2 For Mixed Flow Reator: = Fraction of activity = - = – 0.2 = 0.8  % of activity = 80% 5.8 download by : skknchat@gmail.co SOLUTION -rA = 0.004 CA – 0.01 CR V = 2m = 2000 liter FA = 100 l/min , CA0 = 100 mmol/l = 0.1 mol/l At equation : − 0/ XAe = = =4  XAe = 0.8 = 80% Actual conversion CA = CA0 (1-XA) CR = CR0 + CA0 XA Substituting all values, we get XA = 0.072 = 7.2 % download by : skknchat@gmail.co 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: → SOLUTION 0= = 95% 25 (min) For the volume of plug flow reactor, we have this equation with =0 From − = = 0.1 1+0.5 0(1 ) = x (1-0.95)=0.1 (mol/l) − → → = 39.457 × 25 (min) = 986.433 ( ) download by : skknchat@gmail.co 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 − = (10 −1 ) ⇒ −1 = 10 ( ) (FIRST ORDER) For this stoichiometry, A → 2.5 (products), the expansion factor is = We use the equation (5.8) in Handbook Chemical Reaction Engineering by Octave Levenspiel: We use equation (5.21): ⟺ × = −(1 + ) ln(1 − )− download by : skknchat@gmail.co ⟺10( ⇒ = 0.8997 download by : skknchat@gmail.co )×22( )× 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) = 0.8+0.2 pA0 = 0.8P0 = 0.8 x = 0.8 (atm) at t0 = 20 (s)  CA0= Calculate τ: = 0  τ = tf – t0 = = 208 ( ) × 0.026 ( ) = 194.69 (s) ≈ 195 (s) 100 ×36001 ( )  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: download by : skknchat@gmail.co -rA = τ = => XA = ×(− − 00.026 ( ) )= 195( ) × (1.12×10−4)( ) = 0.84  XA = 84% download by : skknchat@gmail.co ... 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) = 0.8+0 .2 pA0 = 0.8P0 = 0.8 x = 0.8 (atm) at t0 = 20 (s)  CA0= Calculate τ: = 0  τ = tf – t0 = = 20 8... Handbook Chemical Reaction Engineering by Octave Levenspiel: We use equation (5 .21 ): ⟺ × = −(1 + ) ln(1 − )− download by : skknchat@gmail.co ⟺10( ⇒ = 0.8997 download by : skknchat@gmail.co )? ?22 ( )×... SOLUTION Half-life time: t1 /2= 5 .2 days Mean residence time t= 30 days k= 0.693 = 0.693 1 /2 = 0.1333 −1 5 .2 For Mixed Flow Reator: = Fraction of activity = - = – 0 .2 = 0.8  % of activity = 80%