Exercise 3 Text Outlet Temperature of a WGS reactor

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1 Department of Energy Politecnico di Milano Via Lambruschini 4 – 20156 MILANO Exercises of “Fundamentals of Chemical Processes” Prof. Gianpiero Groppi Exercise 3 Outlet temperature of a WGS reactor (Stage I) for the conversion of CO, applied for the abatement of CO to a fixed value. The gas stream exiting the second stage of a Reformer supplied with air is sent to a two-stage reactor for the conversion of CO, where the following reaction occurs (Water Gas Shift, WGS): CO + H2O ↔ CO2 + H2 With reference to the first stage of the conversion of CO, it can be assumed that the outlet gases reach the thermodynamic equilibrium composition at the exit temperature TOUT. Determine the exit temperature TOUT that allows for the amount of CO in the outlet gas stream to be equal to 0.39% (volume %) with respect to the dry current. DATA: Stream 1: T = 360°C P = 30 atm Stream 2: TOUT = ? P = 30 atm CO/Σdry = 0.39% Hypothesis: Ideal gas, WGS is at the chemical equilibrium 2 Species Flow Rate (Nm3/h) % Vol. Dry Basis H2 56.36 N2 22.24 CO 12.56 CO2 8.27 Ar 0.27 CH4 0.3 H2O 73411 dry 120264 100.00 Total 193675 3 Solution The approach with the extent of reaction is chosen to write the material balances. The basis for the material balances can be set to 100 mol/h of incoming dry gas. The extent of the reaction λ is defined and the composition of the outcoming species can be written, in terms of molar flow rates. Species Flow Rate (Nm3/h) nin (mol/h) nout (mol/h) H2 56.36 56.36 +  N2 22.24 22.24 CO 12.56 12.56 -  CO2 8.27 8.27 +  Ar 0.27 0.27 CH4 0.3 0.3 H2O 73411 61.04 61.04 -  dry 120264 100.00 100.00 +  Total 193675 161.04 161.04 Given the 100 mol/h inlet dry gas basis, the amount of H2O is calculated as follows: 2 73411100 61.04120264H On    mol/h The specific on the amount…