# Peak Wavelength of Emission Spectrum of a Flame vs. Actual Flame Temperature

I am looking at a lab test of two variants of mixed hydrocarbon gas (57.8% $H_2$, 16.9% $N_2$, 12.4% $CH_4$, 1.11% water vapor, and 11.8% "unknown" gases) burning in oxygen which shows a discontinous emission spectra. The authors are claiming that since their graph shows a peak wavelength at 471 nm, the gas therefore has a peak flame temperature at 6132 K / 5859 °C / 10578 °F according to Wien's Displacement Law, which is absurdly higher than the adiabatic flame temperature of any known hydrocarbon gas.

Now, I know that the spikes shown correspond to the $C_2$ and $CH$ Swan Band. Can someone provide a detailed explanation as to why this is not an accurate measurement of flame temperature?

I should also mention the graph only ranges from 200 nm to 1100 nm, and that they fitted the emission spectrum to a blackbody curve. Could the peak wavelength actually be further into the infrared spectrum? Is fitting the emission spectrum to a blackbody curve even appropriate for a non-blackbody emitter such as a flame?

• Your skepticism seems very well founded. Wikipedia has "an acetylene/oxygen flame burns at about 3,773 K" en.wikipedia.org/wiki/Oxy-fuel_welding_and_cutting // Fitting to a peak in the spectra seems totally weird // Could you provide a reference to the paper? – MaxW Jan 7 '16 at 16:18
• magnegas.com/docs/MG-Flame-report.pdf – Koatsen Jan 7 '16 at 16:25
• My suspicion is the authors chose an arbitrary peak and fitted it to that point. There is no rationale or equation they used in justifying this action. – Koatsen Jan 7 '16 at 16:30
• I'm not an expert on measuring flame temperatures, but the method used seems totally wrong. You don't curve fit the whole spectra but rather take background measures (i.e not on a peak) on the high energy side and extrapolate to zero intensity. At least that is what I did for x-ray fluorescence and it worked well. – MaxW Jan 7 '16 at 18:37
• Another possible issue is perhaps they are confusing "flame temperature" with "color temperature" due to emission lines which I think is more likely. Just because a blue flame is blue doesn't mean it burns over 6000 K. – Koatsen Jan 7 '16 at 20:26

There seems to be a number of problems here.

1. spectrometers need to be calibrated in intensity on a wavelength-by-wavelength basis. Each part of the system has an efficiency which can vary with wavelength, but the grating response tends to be one of the strongest. Here is a random example of a grating efficiency curve from a Thorlabs catalog :

You can see right away that a grating efficiency curve can impose it's shape on the measured spectrum, generating a shape superficially similar to a blackbody, but totally instrumental. Without calibration for efficiency, you just have ADC counts vs wavelength raw data!

1. Fitting of a Plank shape should be done to the spectrum without ignoring the peaks. In other words, a good fitting program will fit the shape you are looking for (Plank) PLUS fit the peaks at the same time. And don't forget to calibrate intensity first (see 1.)!

1. " Is fitting the emission spectrum to a blackbody curve even appropriate for a non-blackbody emitter such as a flame?" Probably not unless you really know what you are doing. Flames are beautiful and complicated. There's thermal emission from gas in addition to peaks, but stronger is the blackbody-like emission from hot soot particles. Since there will be a distribution in temperatures within the 3D volume of a flame, you have a spectrum recorded from a significant fraction of the flame volume will include a variety of Plank spectra corresponding to different temperatures.

So the answer is: you have to do fitting, remove peaks by fitting them simultaneously, and more importantly, you must calibrate the spectrometer intensity, because grating efficiency curves look strangely Plank-like, althouth they fall to zero much faster above and below their useful wavelength range. Also have to deal with considering the sample volume of the flame and whether or not it's true black-body emission or not.

Those "peak" values, are closer to peak efficiency wavelengths of the spectrometer, and not really related to Wein's displacement law at all!

I calculated a small table of values using the formula for Wien's law: $$\lambda_{max,nm} = \dfrac{2898000}{T}$$ where T is the temperature in Kelvin and the wavelength is in nanometers.

I can't reconcile my table, the graph, nor the calculated temperature. For the top graph it looks like the cutoff is something like 350nm which would be rough 8250 Kelvin which is much greater than 6132 Kelvin which the authors cited, and much hotter than an acetylene/oxygen flame ($\approx$3773K).

  T(Kelvin)   W(max) in nanometers
2000        1449
2250        1288
2500        1159
2750        1054
3000         966
3250         892
3500         828
3750         773
4000         725
4250         682
4500         644
4750         610
5000         580
5250         552
5500         527
5750         504
6000         483
6250         464
6500         446
6750         429
7000         414
7250         400
7500         386
7750         374
8000         362
8250         351
8500         341
8750         331
9000         322
9250         313
9500         305
9750         297

• This is Ref. 1 used in the report: tempe.mi.cnr.it/zizak/tutorial/cairol06-flame-emission.pdf Fig. 7 shows "Emission spectrum of a rich pre-mixed butane-air flame." By the logic of fitting to a blackbody curve, then the peak flame temperature of butane-air becomes 9500 K, which is definitely not the actual flame temperature. I think my assertion over the confusion between "color temperature" and "flame temperature" is probably correct. – Koatsen Jan 7 '16 at 22:37
• The lack of a control gas in the testing such as butane or acetylene make it difficult to accept their results. They referenced a butane spectrum off Wikipedia and it is nearly identical to the second graph I posted. – Koatsen Jan 7 '16 at 23:09
• Revised comment: I'd add that 350nm or so is probably getting near to the UV cutoff for air. The oxygen in the air will absorb short wave UV, so UV radiation shorter than 350 would be attenuated quickly by air. – MaxW Jan 7 '16 at 23:12
• The authors stated that their spectrometer wasn't sensitive enough to detect emissions in the UV range. Mind you, the Ocean Optics HR2000 spectrometer is the same model I used when I took introductory physics and chemistry to analyze discharge tube spectra. You would think a dedicated optics lab would have access to better equipment. I also found different experimental setups for measuring flame temperatures but none of them involved using a single spectrometer. – Koatsen Jan 7 '16 at 23:20
• Is the background luminosity (not on a spectral "line" of some sort) of a flame really black body radiation?!? – MaxW Jan 8 '16 at 0:24