Re: Calculate emissivity of flame in gases mixtures
 From: "Zigoteau" <zigoteau@xxxxxxxxx>
 Date: 14 May 2005 08:35:31 0700
Hi, Rohas,
I owe you an apology. Because you gave an extremely simplified form of
the problem, I thought you were a student asking for answers to a
homework problem.
> > But then you see that if you are dealing with a gas, the
> > emissivity is a function of the optical path length
> > through the gas. You have even written down the equation.
> The emissivity i am looking for is about the heat that transfered
> through radiation. Is a factor and one of the characteristics,
> properties of a gas.
One thing you haven't mentioned is wavelength (or frequency). Hydrogen
and methane flames are definitely blue, but have temperatures at which
a black body would glow yellow, so that the emissivity is definitely
not uniform over the visible range. Most of the energy transfer will
occur at the absorption lines of hydrogen and ethane. There are
electronic transitions in the visible and ultraviolet, and particularly
methane has vibrational transitions in the infrared. If you are
interested in the total radiation from your hot gas, I do not think
that it is justified to say "hydrogen's emissivity is 0.085 and
methane's 0.35" as you did in your initial post.
> Although i have the equation it still is useless to me, though in a
> mixture i have to calculate, find a lot more data
> than i need (like kappa factor or l)
But l is the path length in the gas, i.e. the size of your combustion
chamber. Kappa is not a constant but a function of wavelength. When you
have two gases, the BeerLambert law says that the overall kappa is the
sum of the kappas for the components, and the kappa for each component
is proportional to the partial pressure.
Are you familiar with the concept of oscillator strength for an
obsorption band? Condensedphase absorption bands tend to be broad and
weak, while gas phase absorption bands are intense and narrow. The
oscillator strength, which is proportional to the intensity times the
width, is comparable in both. Radiation within the absorption band
reaches equilibrium within a very short distance (emissivity~1), while
other wavelengths go through the flame essentially unchanged
(emissivity~0). Hence the total amount of radiation emitted depends
essentially on the width of the absorption band.
I did a web search on those two authors Felske and Tien you mentioned,
and it came up with lots of papers on radiation from flames. All the
ones I found cost money to download. Apparently soot  nanoparticles of
carbon  is a major factor in radiation, because it is black and
absorbs over a much broader spectral range than gases do. The
combustion of methane may possibly produce some soot, depending on the
flame temperature and the amount of oxygen.
> The problem is that i don't know where to search since these
> experiments took place  and still take place  in the last
> 56 years, especially with the mix of hydrogen and ethane
> and the use of it as a fuel. It's important to
> estimate the emissivity of this mixture since we need to calculate
the
> heat transfer in the combustion chamber.
As well as a web search, do a literature search. As well as combustion
research, your question is important for atomic absorption
spectroscopy. Your Felske and Tien did their work fify years ago.
Cheers,
Zigoteau.
.
 References:
 Calculate emissivity of flame in gases mixtures
 From: Rohas
 Re: Calculate emissivity of flame in gases mixtures
 From: Zigoteau
 Re: Calculate emissivity of flame in gases mixtures
 From: Rohas
 Re: Calculate emissivity of flame in gases mixtures
 From: Zigoteau
 Re: Calculate emissivity of flame in gases mixtures
 From: Rohas
 Calculate emissivity of flame in gases mixtures
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