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The electric heating wire of the methanol synthesis heater is broken

2009-03-18View Original

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Customers have reported that in methanol synthesis, among two production units with towers of the same size, one unit is able to shut down its heater while the other cannot. My analysis suggests that this might be due to a low percentage of CO and CO2 in the gas entering the tower, an excessively high space velocity, poor catalyst activity, or low catalyst hotspot temperatures, which result in a low temperature of the gas entering the tower and thus prevent the heater from being turned off. According to the electric heater manufacturers, 2 of the heating filaments broke. So why did only 2 of the filaments break and not all of them? :P
Reply #22009-03-18
CO is exothermic in the reaction, while CO2 is endothermic in the reaction. Poor catalyst activity is possible; this kind of situation might occur when the catalyst reaches the end of its life cycle, right? The low hotspot temperature results in a low temperature of the gas entering the tower, preventing the heater from being turned on; under normal operating conditions, the electric heater is not used.
Reply #32009-03-18
1. The synthesis towers of the two manufacturers’ identical units: one operates without an electric furnace, while the other requires it to be kept on continuously; The main reasons for this are (1) the CO/co2 levels at the tower inlet may differ, and (2) the inlet gas flow rate and the circulation rate may vary ; (3) The catalytic activity varies, and so do the production operations and maintenance practices, which results in different synthesis rates. (4) There are differences in the heat exchange efficiency of the heat exchangers inside (or outside) the tower; there are many reasons for this, and a case-by-case analysis is required. (5) Leaks can also occur in the synthetic internals such as catalyst frame covers and quench packing, and these differences lead to variations in whether the synthesis tower is self-heating or not. 2. The electric furnace wire is burned out, which is indicated by the presence of voltage but no current. In the case of two-phase open circuit, there is voltage in all three phases, but current flows only in one phase. This post was last edited by FEIJING on 2009-3-18 17:03]

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