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April 9

Natural Gas Blue Flame

I have an off-the-wall or off-the-stove question. The natural gas, which is primarily methane, burns with a blue flame. I have two possible simplistic explanations for why the flame is blue and would like to know whether either of them is right or whether there is a third explanation, or what. The first is that the column in which methane is combining with oxygen, being very hot, is radiating black-body radiation, and the flame (column of chemical activity) is hot enough that it is past white to blue. (The hottest stars are blue, for instance.) The second is that there is a spectral emission process that has one or more bright lines in the visible blue range. So can someone tell me whether either of these simplistic explanations is right or provide another simplistic explanation? Robert McClenon (talk) 02:44, 9 April 2018 (UTC)[reply]

Turning your Bunsen burner's air supply down produces a yellow (and cooler) flame so that at least shows it can't be an inherent property of natural gas fires. Sagittarian Milky Way (talk) 03:00, 9 April 2018 (UTC)[reply]

Does Flame#Flame color answer your questions? --Guy Macon (talk) 03:54, 9 April 2018 (UTC)[reply]

The substances mainly responsible for the blue flame are dicarbon and methylidyne. There is also a bit of ultraviolet from hydroxyl radical and Cyano radical. The yellow flame is glowing soot. Graeme Bartlett (talk) 08:18, 9 April 2018 (UTC)[reply]

Just to elaborate on the blackbody radiation angle: transparent bodies (such as most gases at atmospheric pressure) do not radiate anything (that's a consequence of the more general Kirchhoff's law of thermal radiation). In no usual ~~Earthly~~ combustion context will you ever encounter blackbody radiation that looks blue, because the temperatures needed would be too high. Flames can go up to the adiabatic flame temperature which is below 3000K for all common fuels, yet the Sun (whose spectrum has a peak that more or less defines the visible spectrum, because evolution) has a photosphere temperature of about or above 5000K. Any shade of blue in a flame must be the result of some form of chemiluminescence. As noted by SMW below though, electrical arcs (lightning, spark plug etc.) can create extremely high (though localized) temperatures with blackbody radiation in the blue (and ultraviolet). Tigraan^{Click here to contact me} 12:21, 9 April 2018 (UTC)[reply]

What about lightning? Sagittarian Milky Way (talk) 19:19, 9 April 2018 (UTC)[reply]

You're right. Post amended. Tigraan^{Click here to contact me} 15:52, 10 April 2018 (UTC)[reply]

Here are a couple videos on the makeup and properties of flames that I think are fairly good: [1] [2] --47.146.63.87 (talk) 08:42, 11 April 2018 (UTC)[reply]

Feynman Lectures. Exercises. Exercise 20-10 PNG

...

The equatorial and polar radii of the earth are 6378.388 km and 6356.912 km. Its specific gravity ρ at various depths D below the surface are shown in the Table below (* denotes a discontinuity).
D(km) ρ
0......2.60
30*....3.0
.......3.3
100....3.4
200....3.5
400....3.6
1000...4.7
2000...5.2
2900*..5.7
.......9.4
3500...10.2
5000*..11.5
.......16.8
6000...17.1

Using these values, estimat:
a) the moment of inertia of the earth,
b) its rotational angular momentum,
c) its rotational kinetic energy, and
d) the time required for the rotational axis to precess about the pole of the ecliptic due to the torques of the moon and the sun.

(The tilt of the earth's axis is 23 1/2° .)

— R. B. Leighton , Feynman Lectures on Physics. Exercises

Can't understand what does the discontinuity mean.
The Solutions gives a formula
$I=\sum I_{i}={\tfrac {8}{15}}\pi \rho _{i}(r_{i}^{5}-r_{i-1}^{5})$ ,
$r_{0}=0$
And they got
$I=\sum _{i=1}^{11}I_{i}=8.1\cdot 10^{37}kg\cdot m^{2}$ .

Solution author finds the shell moment of inertia as difference of two spheres
$I={\tfrac {2}{5}}M_{1}r_{1}^{2}-{\tfrac {2}{5}}M_{2}r_{2}^{2}={\tfrac {8}{15}}\pi \rho (r_{1}^{5}-r_{2}^{5})$

I 've got $7.4\cdot 10^{37}kg\cdot m^{2}$ , using mean value of specific gravity, e.g. for D= 30 km ρ= 0.5(3 + 3.3). Username160611000000 (talk) 18:53, 9 April 2018 (UTC)[reply]

You took the wrong average. the density of the shell from 0 to 30 km is probably 2.8, etc Greglocock (talk) 19:26, 9 April 2018 (UTC)[reply]

I get 8.061E37 kg m^2. Greglocock (talk) 21:17, 9 April 2018 (UTC)[reply]

I've rearranged ρ-array and got 7.6 PNG.

Ah. think I understood PNG . But then it is not clear what is the specific gravity on depth interval 6000 - 6400 km.Username160611000000 (talk) 04:56, 10 April 2018 (UTC)[reply]

Discontinuity means that there is a sudden change in density, e.g. rock on iron core. So 0-30 km is assumed to steadily increase from 2.6 to 3.0, then 30-100 steadily increases from 3.3 to 3.4, etc. As for the rest ... ugh. Have fun. ;) (note that they gave two radii, so this presumably isn't a strictly spherical approximation or shell theorem exercise?) Wnt (talk) 00:32, 10 April 2018 (UTC)[reply]

For more info on discontinuity, see (for example): "Moho" (Mohorovičić) discontinuity. —2606:A000:4C0C:E200:0:0:0:3 (talk) 03:57, 10 April 2018 (UTC)[reply]

Ok. How to solve d) question? From ex. 20-8 PNG1 PNG2 and answers PNG3, we know formula for torque. We should find r and θ. For half of sphere the CM is on distance X_CM = 3R/8 = 0.375R. For disk X_CM = 4R/3π = 0.4244R. Besides, even for tilted sphere torque = 0 always, and even for tilted disk torque is changed from τ_max to 0 within 1/4 year.

So we can approximate r as 0.375R and find θ from proportion :
23.5° ~ 6378.388 km
(23.5°-θ) ~ 6356.912 km
θ=0.08°

From Eq. 20.15
$\mathbf {\tau } =\mathbf {\Omega } \times \mathbf {L_{0}} .$
$\tau =\Omega L_{0}\sin(\Omega ;L_{0}).$
$\Omega ={\tfrac {\tau }{L_{0}\sin(\Omega ;L_{0})}}.$
$T=2\pi /\Omega ={\tfrac {2\pi L_{0}\sin(\Omega ;L_{0})}{\tau }}={\tfrac {2\pi L_{0}\sin(23.5^{\circ })}{{\tfrac {2GM_{sun}mr^{2}sin(2\theta )}{R_{sun-earth}^{3}}}+{\tfrac {2GM_{moon}mr^{2}sin(2\theta )}{R_{moon-earth}^{3}}}}}$
$={\tfrac {\pi L_{0}\sin(23.5^{\circ })}{GM_{\text{earth}}r^{2}\sin(2\theta )({\tfrac {M_{\text{sun}}}{R_{\text{sun-earth}}^{3}}}+{\tfrac {M_{\text{moon}}}{R_{\text{moon-earth}}^{3}}})}}$
$T=20000years.$

But answers give 25725 years. Why?Username160611000000 (talk) 05:41, 10 April 2018 (UTC)[reply]

The precision of these types of astrophysical calculation should not be overstated. After all, you began with two significant figures - why should you end with 5?

Here is an amusing list of astrophysics questions from the Ohio State University short-course: Order-of-Magnitude Astrophysics - to help get you into the mindset of a space scientist. It's really pretty silly to presume that we can get better accuracy than nearest-order-of-magnitude when our input data is so coarse, and our pure mathematical equations multiplicatively amplify errors and imprecisions.

Nimur (talk) 17:53, 10 April 2018 (UTC)[reply]