Design a chemical reactor for the production of ethylene oxide through the direct oxidation of ethylene. The reactor should have a production rate of 100 tons per day and maintain a constant temperature of 250°C. Assume a feed composition of 90% ethylene and 10% oxygen. The reactor should also have a conversion efficiency of at least 95%. Compute the necessary reactor volume and the optimal flow rate of ethylene and oxygen to achieve the desired production rate and conversion efficiency.

Question

Design a chemical reactor for the production of ethylene oxide through the direct oxidation of ethylene. The reactor should have a production rate of 100 tons per day and maintain a constant temperature of 250°C. Assume a feed composition of 90% ethylene and 10% oxygen. The reactor should also have a conversion efficiency of at least 95%. Comput

Answer 1

To design a chemical reactor for the production of ethylene oxide, we need to consider the reaction kinetics, stoichiometry, and thermodynamics. The direct oxidation of ethylene to ethylene oxide can be represented by the following reaction:C2H4 + 1/2 O2  C2H4OFirst, we need to calculate the molar flow rates of ethylene and oxygen required to achieve the desired production rate of 100 tons per day.1. Calculate the moles of ethylene oxide produced per day:100 tons/day *  1000 kg/ton  *  1 mol C2H4O / 44.05 g  = 2.27 x 10^6 mol/day2. Calculate the moles of ethylene and oxygen required per day:Ethylene: 2.27 x 10^6 mol/day / 0.95  conversion efficiency  = 2.39 x 10^6 mol/dayOxygen:  2.27 x 10^6 mol/day  / 2 = 1.14 x 10^6 mol/day3. Calculate the molar flow rates of ethylene and oxygen:Assuming continuous operation  24 hours/day , we can calculate the molar flow rates as follows:Ethylene: 2.39 x 10^6 mol/day / 24 hours/day = 99,583 mol/hourOxygen: 1.14 x 10^6 mol/day / 24 hours/day = 47,500 mol/hourNow, we need to determine the reactor volume. For this, we will use the reaction rate constant  k  and the rate equation for the reaction. The rate equation for this reaction can be written as:Rate = k * [C2H4] * [O2]^0.5Assuming a first-order reaction with respect to ethylene and a half-order reaction with respect to oxygen, we can determine the reaction rate constant  k  at the given temperature  250C  using the Arrhenius equation:k = Ae^-Ea/RT Where A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. For this reaction, typical values for A and Ea are:A = 1.0 x 10^6 L^0.5/mol^0.5/hourEa = 80 kJ/molT = 250C + 273.15 = 523.15 KR = 8.314 J/mol/Kk = 1.0 x 10^6 * e^-80,000 /  8.314 * 523.15   = 0.012 L^0.5/mol^0.5/hourNow, we can calculate the reactor volume  V  using the rate equation and the molar flow rates:Rate =  99,583 mol/hour  / V = 0.012 *  99,583 mol/hour  *  47,500 mol/hour ^0.5Solving for V, we get:V = 99,583 /  0.012 * 99,583 * 47,500^0.5  = 3,000 LTherefore, the necessary reactor volume is 3,000 L, and the optimal flow rates of ethylene and oxygen are 99,583 mol/hour and 47,500 mol/hour, respectively, to achieve the desired production rate and conversion efficiency.