# 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 Beer-Lambert 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? Condensed-phase 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

> 5-6 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

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