Sep 28 2015

Pagina 1

CRE – 28.09.15 - Exam E (Solution) 1 096116 Chemical Reaction Engineering 28 September 2015 Exam E Family name _________________________________________________ First name _________________________________________________ ID number _________________________________________________ Signature _________________________________________________ 1. Second order reaction in a CSTR (25%) The following second-order liquid-phase reaction is taking place in a CSTR: 𝐴𝐴+ 𝐵𝐵→ 2𝐶𝐶+ 𝐷𝐷 A and B are fed to the reactor at rates of 4 mol/min and 2 mol/min respectively at a temperature of 300 K. The total volumetric flow rate is 10 l/min. Specific heats (in J/mol/K) of A, B, C and D are 125, 100, 130 and 135 respectively. The reactor is jacketed by water at a temperature of 40°C. The overall heat transfer coefficient has been estimated at 200 J/m2/s/K, while the heat transfer area is 0.5 m2. Mixing is ensured through an agitator, which contributes a work of 10 kW to the reactor. The heats of formation of A, B, C and D (at 298 K) are -45 kJ/mol, -30 kJ/mol, -50 kJ/mol and -60 kJ/mol respectively. The rate constant at 300 K is 0.1 l/mol/min and the activation energy is 30,000 J/mol . Find the steady -state temperature in the re actor for 90% consumption of the limiting reactant. Find the volume of the reactor to achieve this conversion. Solution Let us write the energy balance for a CSTR in non -isothermal conditions, according to what we studied for single reactions: 𝑄𝑄̇ − 𝑊𝑊̇ 𝐹𝐹𝐵𝐵 0 − 𝑋𝑋∆𝐻𝐻𝑟𝑟𝑟𝑟= � 𝜃𝜃𝑖𝑖𝐶𝐶̃𝑃𝑃𝑖𝑖(𝑇𝑇− 𝑇𝑇𝑖𝑖0) 𝑈𝑈𝐴𝐴(𝑇𝑇𝑒𝑒𝑟𝑟𝑒𝑒− 𝑇𝑇) − 𝑊𝑊̇ 𝐹𝐹𝐵𝐵 0 − 𝑋𝑋�∆𝐻𝐻𝑟𝑟𝑟𝑟(𝑇𝑇𝑅𝑅) + ∆𝐶𝐶̃𝑃𝑃(𝑇𝑇− 𝑇𝑇𝑅𝑅)� = � 𝜃𝜃𝑖𝑖𝐶𝐶̃𝑃𝑃𝑖𝑖(𝑇𝑇− 𝑇𝑇𝑖𝑖0) We used B as the limiting species because of the stoichiometry and the initial number of moles. Thus, we can easily calculate the reactor temperature: CRE – 28.09.15 - Exam E…