Advanced Reaction Engineering
——Water Gas Shift Reaction in a Membrane Bed Reaction
Dr. Idem Dr. Liang Dr. Pan
Group 5 Name: Su Liusong Number: S1411W0694 Group Members: Cui ding, Li Na
2014.11.10
1. Introduction
The water-gas shift(WGS)reaction is important in the chemical and petrochemical industry for the production of hydrogen using synthesis gas as a feedstock. As the reaction equation can be wrote that:
CO?H2OcatalystsCO2?H2 ; ?H0??41kJ/mol
There are many kinds of catalysts to catalyze the above-mentioned reaction. Reaction kinetics is one of the most important parameters for water-gas shift reaction over many kinds of catalysts such as Fe-base and Rh-base catalysts because it indicates that how fast CO reacts with water, and it is one of the key parameters for simulating and design the reactor. The kinetics of the water-gas shift over three newly developed chromium-free
catalysts
Fe2O3/ZrO2,
Rh/ZrO2
and
Rh/Fe2O3/ZrO2 were investigated in a membrane packed bed plug flow reactor in the temperature range of 623-773K at a total pressure of 21 bar. Different kinetic models based on redox and associative WGS mechanisms depending on the type of catalyst were evaluated to represent the kinetic data. In order to explore the advantage of using this highly active catalyst in a process, development of a kinetic model is essential in the industry.
Figure 1 is a model of plug flow reactor, we can conclude that the composition of the fluid varies from point to point along a flow path, as a consequence, the material balance for a reaction component must be made for a differential element of volume dV. Thus, as for reactant A, we can know that:
Input = output + disappearance by reaction + accumulation (1) We can see for volume dV that:
Input of A is marked to be FA, moles/time; Output of A is marked to be FA+dFA,moles/time; Disappearance of A by reaction is (-rA)dV,
and in the plug flow reactor, the accumulation of the material is zero, so we can obtain:
FA?FA?dFA???rA?dV (2) -rA is the disappearant rate of the reactant, From the equation (2), we can obtain:
?dFA???rA?dV
And we know that:
dFA?d??FA0?1?XA?????FA0dXA So we can conclude that:
FA0dXA???rA?dV (3)