I am trying to obtain ferrous sulfate heptahydrate crystal from a liquor which is rich in Fe2+ and Fe3+ sulfate and sulfuric acid. However I can only achieve crystal (more likely powder) form of FeSO4.7H2O (37%) mixed with 58% Rhomboclase (H5O2)+Fe+++(SO4)2•2(H2O) with minor impurities. How can...
Oh and don't forget mass and energy balances. Probably you will get it next semester. The real chemical engineering classes start at the 4th semester generally.
None :D. We are expected to solve problems related to production processes of chemicals. Yes of course they are related to chemistry but as an engineer you will be expected to design a factory for example. And this will be about your mathematical skills not your chemistry. I suggest you to...
Not really, if you like mathematics and physics, then chemical engineering is a good option for you. It is not chemistry or you won't become a chemist. Chemical engineering is the art of applying math and physics on chemistry and chemical processes. You don't need to be super interested in...
It's like traveling by plane. Better than traveling by car or bus but once it crashes survival rates are quite low. However, the possibility of the accidents is actually lower than that of cars. Planes are also faster and cleaner in terms of air pollution of cities.
I see but our professors want us to find an article that explains the reaction kinetics and conversion. We can't start doing mass balance because of the lack of information about the conversion (we do not need the kinetics right now). But thanks for the the source. It seems like a useful article.
And how can I find a detailed process information of KAAP Kellogg modern process with Ruthenium catalyst? It's quite similar to the Kellogg process with the iron catalysts one. However I need to know what changes with the modern one.
I need to design an ammonia production process but I can't find the kinetic info from the articles online. The catalyst will be iron. Can somebody explain how to find kinetics data and conversion rate? I found some articles but there was no conversion info on them.
Even though a fuel cell is more efficient at lower temperatures as shown in Table 3.1, the voltage losses are much less in higher temperature fuel cells. Therefore, it is more advantageous to run a fuel cell at a higher temperature yet lower efficiency to produce higher operating voltages...
I know theoretically fuel cell efficiency decreases as temperature increases. But in practice it is vice versa. What is the reason of that? I couldn't find enough sources. Any kind of help is much appreciated.
Don't know exactly. I didn't touch it some other students did. But it was at 24 degrees which was the room temperature before it was immersed into bath. Water in bath was at 80 btw.