Exercise 3 Text

Pagina 1

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 co nversion 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 + H 2O ↔ 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 T OUT. Determine the exit temperature T OUT 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 (Nm 3/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 Totale 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 (Nm 3/h) n in (mol/h) n out (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: 04.61100120264 73411 2 OH The specific on the amount of…