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, 20225.6 SOLUTION −𝑟𝐴 = (0.04𝑚𝑖𝑛−1 )𝐶𝐴 − (0.01𝑚𝑖𝑛−1 )𝐶𝑅 { 𝑘1 = 0.04𝑚𝑖𝑛−1 𝑘2 = 0.01𝑚𝑖𝑛−1 𝑉 = 2𝑚3 = 2000𝐿 𝜐0 = 100 𝐿⁄𝑚𝑖𝑛 𝐶𝐴0 = 100 𝑚𝑚𝑜𝑙⁄𝐿 𝐾 = 𝑘1 𝑋𝐴𝑒 0.04 𝑋𝐴𝑒 = => = => 𝑋𝐴𝑒 = 0.8 = 80% 𝑘2 − 𝑋𝐴𝑒 0.01 − 𝑋𝐴𝑒 −𝑟𝐴 = 𝑘1 𝐶𝐴0 (1 − 𝑋𝐴 ) − 𝑘2 𝐶𝐴0 𝑋𝐴 = 𝑘1 𝐶𝐴0 (1 − 𝑋𝐴 ) − 𝑘1 (1− 𝑋𝐴𝑒 ) 𝑋𝐴𝑒 = 𝑘1 𝐶𝐴0 [(1 − 𝑋𝐴 ) − 1− 𝑋𝐴𝑒 = 𝑘1 𝐶𝐴0 [(1 − 𝑋𝐴 ) − 1−0.8 = 𝑘1 𝐶𝐴0 (1 − 1.25𝑋𝐴 ) 𝑋𝐴 𝑉 𝑑𝑋𝐴 𝜏 = = 𝐶𝐴0 ∫ (−𝑟𝐴 ) 𝜐0 𝑋𝐴𝑒 0.8 𝐶𝐴0 𝑋𝐴 𝑋𝐴 ] 𝑋𝐴 ] 𝑉 𝜐0 𝑋 = 𝐶𝐴0 ∫0 𝐴 𝑘 2000 100 𝑑𝑋𝐴 𝐶𝐴0 (1−1.25𝑋𝐴 ) 𝑋 = 100 ∫0 𝐴 𝑑𝑋𝐴 0.04 ×100 × (1−1.25𝑋𝐴 ) 𝑋𝐴 = 0.506 = 50.6% SOLUTION Half-life time: t1/2= 5.2 days Mean residence time t= 30 days 0.693 k= 𝑡1/2 = 0.693 5.2 = 0.1333𝑑 −1 For Mixed Flow Reator: 𝐶𝐴 1 = = = 0.2 𝐶𝐴0 + 𝑘𝑡 + 0.1333 × 30 Fraction of activity = - 𝐶𝐴 𝐶𝐴0 % of activity = 80% 5.8 = – 0.2 = 0.8 SOLUTION -rA = 0.004 CA – 0.01 CR V = 2m3 = 2000 liter FA = 100 l/min , CA0 = 100 mmol/l = 0.1 mol/l At equation : XAe = 𝐾𝑐 = 𝐾𝐶 −𝐶𝑅0 /𝐶𝐴0 𝐾𝐶 +1 𝑘1 𝑘2 =4 XAe = 0.8 = 80% Actual conversion 𝑉 𝐹𝐴0 = 𝑋𝐴 −𝑟𝐴 = 2000 100 = 𝑋𝐴 0.04𝐶𝐴 − 0.01𝐶𝑅 CA = CA0 (1-XA) CR = CR0 + CA0 XA 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: 𝑒𝑛𝑧𝑦𝑚𝑒 𝐴→ 𝑅, −𝑟𝐴 = 0.1𝐶𝐴 𝑚𝑜𝑙 + 0.5𝐶𝐴 𝑙𝑖𝑡𝑒𝑟 𝑚𝑖𝑛 SOLUTION 𝑣0 = 25 ( We already have { 𝐶 = 2( 𝐴0 𝑙 𝑚𝑜𝑙 𝑙 ) ) 𝑋𝐴 = 95% For the volume of plug flow reactor, we have this equation with 𝜀 = From −𝑟𝐴 = 0.1𝐶𝐴 1+0.5𝐶𝐴 𝐶𝐴 = 𝐶𝐴0 (1 − 𝑋𝐴 ) = x (1-0.95)=0.1 (mol/l) → 𝐶𝐴𝑓 0.1 𝑉 𝑑𝐶𝐴 + 0.5𝐶𝐴 = −∫ = −∫ 𝑑𝐶𝐴 ≈ 39.457 𝑣0 0.1𝐶𝐴 𝐶𝐴0 −𝑟𝐴 𝑙 → 𝑉 = 39.457 × 25 ( ) = 986.433 (𝑙𝑖𝑡𝑒𝑟) 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), −𝑟𝐴 = (10 𝑚𝑖𝑛−1 ) 𝐶𝐴 Find the expected conversion in a 22-liter reactor Solution A → 2.5 (products) ⇒ Irreversible reaction −𝑟𝐴 = (10 𝑚𝑖𝑛−1 ) 𝐶𝐴 ⇒ 𝑘 = 10 (𝑚𝑖𝑛−1 ) (FIRST ORDER) For this stoichiometry, A → 2.5 (products), the expansion factor is 𝜀𝐴 = 2.5 − = 1.5 We use the equation (5.8) in Handbook Chemical Reaction Engineering by Octave Levenspiel: We use equation (5.21): ⟺𝑘× 𝑉 = −(1 + 𝜀𝐴 ) ln(1 − 𝑋𝐴 ) − 𝜀𝐴 𝑋𝐴 𝜐0 1 𝑚𝑜𝑙 ⟺ 10 ( ) × 22 (𝐿) × ×2( ) = −(1 + 1.5) ln(1 − 𝑋𝐴 ) − 1.5 × 𝑋𝐴 𝑚𝑜𝑙 𝑚𝑖𝑛 𝐿 ) 100 ( 𝑚𝑖𝑛 ⇒ 𝑋𝐴 = 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) 0.8+0.2 =0 pA0 = 0.8P0 = 0.8 x = 0.8 (atm) at t0 = 20 (s) CA0 = 𝑝𝐴0 𝑅𝑇 = 0.8 (𝑎𝑡𝑚) 𝐿.𝑎𝑡𝑚 ) × (100+273)(𝐾) 0.082 ( 𝑚𝑜𝑙.𝐾 = 0.026 ( 𝑚𝑜𝑙 𝐿 ) Calculate τ: 𝑉 𝐹𝐴0 = 𝜏 𝐶𝐴0 τ = tf – t0 = 𝑉𝐶𝐴0 𝐹𝐴0 = 𝑚𝑜𝑙 ) 𝐿 𝑚𝑜𝑙 ( ) 100 × 3600 𝑠 208 (𝐿) × 0.026 ( = 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 ( 𝑚𝑜𝑙 𝐿 ) For mixed flow reactor with εA = 0: -rA = τ= 𝐶𝐴0 − 𝐶𝐴 𝜏 𝐶𝐴0 𝑋𝐴 −𝑟𝐴 = 𝑚𝑜𝑙 ) 𝐿 0.026 − 0.0041 ( => XA = XA = 84% 195 (𝑠) 𝜏×(−𝑟𝐴 ) 𝐶𝐴0 = = 1.12 x 10-4 ( 𝑚𝑜𝑙 𝐿.𝑠 𝑚𝑜𝑙 ) 𝐿.𝑠 195(𝑠) × (1.12×10−4 )( 𝑚𝑜𝑙 ) 𝐿 0.026 ( ) = 0.84 ... 0 .2 inerts => εA = (0.8+0 .2) −(0.8+0 .2) 0.8+0 .2 =0 pA0 = 0.8P0 = 0.8 x = 0.8 (atm) at t0 = 20 (s) CA0 =