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[Luc1]

Hi everybody,

I am an Italian chemical engineering student and I am currently dealing with many problems regarding process safety - like LEL, UEL, LOC, AIT, etc, etc,....

My question are:

1)In one exam they ask me to find LEL and UEL of a organic fuel at different tempratures: 25°C and 300°C,and i am in posses of the following formula for the LEL at 25°C and LEL at other temperature:

-LEL_25°C=0.55*C_{stoichiometric}
C_{stoichiometric}=[(moles of Fuel)/(moles of Fuel+moles of Air)]_{stoichiometric}

-LEL_T=LEL(25°C)-[0.75/|ΔH_combustion|]*(T-25°C) where T is the unknown temperature

for UEL in the first equation instead of 0.55 is 3.5, and in the second instead of minus after LEL(25) there is + .Using these equations i get negative LEL at 300°C, so i suspect that these are not the real equation i need.Are they correct or im badly wrong?If i am wrong can you tell me the correct equations for finding LEL and UEL?


2)In another exam i got a acrylic acid plant conversion from propylene, i got 3 types of feedstock at the reactor: the propylene,water/steam and Air.The text says that the amount of oxygen in the Air feedstock is equal to the 60% of MOC.

This is the MOC formula that i know:

- MOC=LEL*(moles of O₂/moles of Fuel)_{stoichiometric} -which LEL is used is unspecifed,and besides I do not know if "stoichiometric" it is related to the quantity of fuel in the feedstock or the quantity indicated the chemical formula of partial oxidation of propylene.

I found some data of LEL and UEL at 20°C on EngineeringToolbox.com and i used this values in the MOC forumula,assuming that moles of fuel are the one i have in the feedstock, but I got the result that the moles of oxygen are lower than those required by the reaction.So even this formula is wrong?Or am I doing something wrong?


3)I do not have any formula to determine the temperature of auto ignition, there is one?



Extra question: in my tests, I often deal with plants of steam reforming, in the priamry reformer the two main reactions occur, the reaction of methane reforming and CO shift reaction.I know the feedstock fed to the reactor, and assuming that the reactor works at the equilibrium temperature (about 850 degrees celsius) and knowing the two equilibrium constants at 850°C i can calculate the output of the reactor using a system of two equations of chemical equilibrium.

Here is the system, i used wolfram|alpha site to solve it: The system

I posses a TI-89 titanium calculator, when i try to solve this system of equation using the "solve(eqn1 and eqn2,{x , y})" syntax my calculator dies in the infinte "busy" status ,and even after three hours of calculations it continue to show "busy" status, and no results is shown.Wolfram took just 4 secs to solve it,but i can took my pc in my classroom druing the exam .So is there any soltuion to avoid the calculator blocks in this way?


Tanks for answers,and excuse my poor english.Bye