English term
stoechiometric completion
Could you explain what it actually means.
Thanks!
The sentence which was used this word is as follows
"However, their formulation of the combustion process was restricted to the simple and common assumption that the entrained air mixes with the fuel and burns to stoechiometric completion instantaneously."
Oct 29, 2015 09:08: Neil Ashby changed "Level" from "Non-PRO" to "PRO"
PRO (3): Edith Kelly, Shera Lyn Parpia, Neil Ashby
When entering new questions, KudoZ askers are given an opportunity* to classify the difficulty of their questions as 'easy' or 'pro'. If you feel a question marked 'easy' should actually be marked 'pro', and if you have earned more than 20 KudoZ points, you can click the "Vote PRO" button to recommend that change.
How to tell the difference between "easy" and "pro" questions:
An easy question is one that any bilingual person would be able to answer correctly. (Or in the case of monolingual questions, an easy question is one that any native speaker of the language would be able to answer correctly.)
A pro question is anything else... in other words, any question that requires knowledge or skills that are specialized (even slightly).
Another way to think of the difficulty levels is this: an easy question is one that deals with everyday conversation. A pro question is anything else.
When deciding between easy and pro, err on the side of pro. Most questions will be pro.
* Note: non-member askers are not given the option of entering 'pro' questions; the only way for their questions to be classified as 'pro' is for a ProZ.com member or members to re-classify it.
Responses
Poland
Do it properly, or don't do it at all.
Native in: Polish
Works in: English to Polish, Polish to English, English
complete combustion
stoichiometric
There are plenty of definitions of this on the internet, e.g.
Stoichiometric or Theoretical Combustion is the ideal combustion process where fuel is burned completely.
A complete combustion is a process burning all the carbon (C) to (CO2), all the hydrogen (H) to (H2O) and all the sulphur (S) to (SO2).
With unburned components in the exhaust gas such as C, H2, CO, the combustion process is uncompleted and not stoichiometric .
http://www.engineeringtoolbox.com/stoichiometric-combustion-...
Spain
DrNeil TechTrans, Precision & Clarity
Native in: English
Works in: Spanish to English, English
the entrained air and the fuel are both 100% oxidised (burnt). All matter entering is 100% burnt
https://en.wikipedia.org/wiki/Stoichiometry
Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products leading to the insight that ...
Stoichiometry and Balancing Reactions - Chemwiki
chemwiki.ucdavis.edu › Analytical Chemistry › Chemical Reactions
Oct 14, 2015 - Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means element and metron means measure, so stoichiometry literally translated means the measure of elements.
Stoichiometric Combustion - Engineering ToolBox
www.engineeringtoolbox.com/stoichiometric-combustion-d_399....
Stoichiometric or Theoretical Combustion is the ideal combustion process where fuel is burned completely. A complete combustion is a process burning all the carbon (C) to (CO2), all the hydrogen (H) to (H2O) and all the sulphur (S) to (SO2).
--------------------------------------------------
Note added at 4 hrs (2015-10-29 09:16:53 GMT)
--------------------------------------------------
They have taken the assumption that all matter entering the "tunnel fire experiment" has been burnt to the fullest extent, i.e., that every single atom has been fully oxidised. If that were the case then the reaction (fire) would have proceeded to "stoichiometric completion" (every single C, N, S, etc. atom in the original starting material would have been completely oxidised to CO2, NO2, SO2, etc...not just partially oxidised to CO or N2O, etc.). Of course in reality this is not the case and so any calculations based on this assumption will be errant (or "restricted").
Something went wrong...