Solution

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CRE – 03.02.15 - Exam A (Solution) 096116 Chemical Reaction Engineering 3 February 2015 Exam A Family name _________________________________________________ First name _________________________________________________ ID number _________________________________________________ Signature _________________________________________________ 1. Adiabatic CSTR with reversible, exothermic reaction (40%) The elementary, exothermic, reversible gas phase reaction 𝐴𝐴→ 2𝐵𝐵 𝑟𝑟= 𝑘𝑘� 𝐶𝐶𝐴𝐴− 𝐶𝐶𝐵𝐵 2 𝐾𝐾𝑒𝑒𝑒𝑒 � is to be carried out adiabatic ally. Pure A is fed at a rate of 𝐹𝐹𝐴𝐴 0 = 2 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚𝑚𝑚𝑚𝑚, a temperature of 𝑇𝑇0 = 100°𝐶𝐶, and a concentration of 𝐶𝐶𝐴𝐴 0 = 0.01 𝑚𝑚𝑚𝑚𝑚𝑚/𝑚𝑚. The equilibrium constant is a weak decreasing function of temperature, but, for simplicity, we assume it constant: 𝐾𝐾𝑒𝑒𝑒𝑒= 5.5 ∙ 10−3 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚 a) What is the adiabatic equilibrium temperature and conversion? b) If you know that the kinetic constant (measured in min-1) is function of temperature (measured in K): 𝑘𝑘= 24 𝑒𝑒−2100 𝑇𝑇 what is the CSTR volume necessary to achieve 80% of the adiabatic equilibrium conversion? Thermodynamic data: Reaction heat at 100°C Specific heats ∆𝐻𝐻� 𝑅𝑅= −3500 cal/mol 𝐶𝐶𝑃𝑃𝐴𝐴� = 5 cal/mol/°C 𝐶𝐶𝑃𝑃𝐵𝐵� = 2.5 cal/mol/°C 1 CRE – 03.02.15 - Exam A (Solution) Solution Part a Basically, we have to find the adiabatic conditions in terms of conversion and temperature. This means that we have to couple the mass balance with the energy balance. No kinetic information is needed for this part. Let’s start with the mass balance (the objective is to write the conversion as a function of temperature). Let’s write the concentration of species A and B as a function of conversion, under the assumption of constant pressure. We have to take into account that the number of boles of products is different than the…