\(v_{boom\,lignite}=3.4354*\cfrac{801}{6}=458.63\,ms^{-1}\)
and
\(T_{p}=\cfrac{458.63^2*12.0107*10^{-3}}{2*8.3144}=151.93\,K\)
Lignite will spontaneously ignite at \(151.93-273.15=-121.22\,^oC\)! What happened? Apparently, lignite low density value is due to bulk (air space) and is not structural. The quoted density value is bulk density.
In the case of Anthracite, another form of coal, with a density value of \(1506\,kgm^{-3}\),
\(v_{boom\,anthracite}=3.4354*\cfrac{1506}{6}=862.29\,ms^{-1}\)
and
\(T_{862}=\cfrac{862.29^2*12.0107*10^{-3}}{2*8.3144}=537.04\,K\)
It will ignite at \(263.90\,^oC\)
And bituminous coal of a density \(1346\,kgm^{-3}\),
\(v_{\small{boom\,bituminous}}=3.4354*\cfrac{1346}{6}=770.67\,ms^{-1}\)
and
\(T_{770}=\cfrac{770.67^2*12.0107*10^{-3}}{2*8.3144}=428.98\,K\)
It will ignite at \(155.83\,^oC\).
And another quoted value of density for coal at \(1400\,kgm^{-3}\)
\(v_{\small{boom\,brown}}=3.4354*\cfrac{1400}{6}=801.59\,ms^{-1}\)
and
\(T_{770}=\cfrac{801.59^2*12.0107*10^{-3}}{2*8.3144}=464.10\,K\)
It ignites at \(190.95\,^oC\)
Crystalline (structural) density should be used and not bulk density in the calculations for \(v_{boom}\) and \(T_{v\,boom}\).
Do we need to differentiate \(v_{rms}\) and \(v_p\) and so, \(T_{v\,boom}\) and \(T_{boom\,p}\)?
Apparently you need to have most of the particles at \(v_{boom}\) (\(v_{p}=v_{boom}\)) to ignite the matter, and afterwards the matter settles to a temperature where the particles has a \(v_{rms}\) at \(v_{boom}\). ie. (\(v_{rms}=v_{boom}\)). In the former case, we are concerned with velocities at which most particles collide and in the latter, we are concerned with the mean of velocities squared of the particles as indicative of its energy content; the measured temperature according to Kinetic Theory for Ideal gases.
Speculative... still go put out the coal fire.
and
\(T_{770}=\cfrac{801.59^2*12.0107*10^{-3}}{2*8.3144}=464.10\,K\)
It ignites at \(190.95\,^oC\)
Crystalline (structural) density should be used and not bulk density in the calculations for \(v_{boom}\) and \(T_{v\,boom}\).
Do we need to differentiate \(v_{rms}\) and \(v_p\) and so, \(T_{v\,boom}\) and \(T_{boom\,p}\)?
Apparently you need to have most of the particles at \(v_{boom}\) (\(v_{p}=v_{boom}\)) to ignite the matter, and afterwards the matter settles to a temperature where the particles has a \(v_{rms}\) at \(v_{boom}\). ie. (\(v_{rms}=v_{boom}\)). In the former case, we are concerned with velocities at which most particles collide and in the latter, we are concerned with the mean of velocities squared of the particles as indicative of its energy content; the measured temperature according to Kinetic Theory for Ideal gases.
Speculative... still go put out the coal fire.
