# Variations of colour in fire

1. Aug 19, 2004

### devoy

Is heat always the factor that determines the colour of a flame?
Is there a way of approximating the variations in colour that would occur within flames from a wood fire?

2. Aug 19, 2004

### chroot

Staff Emeritus
The color of a flame is mostly determined by the dominant spectral emission lines of the material being heated up by the flame. If you hold a bit of copper in a flame, for example, you'll see green light, because the dominant spectral emission line from copper is green.

- Warren

3. Aug 20, 2004

### devoy

Hi,
I realise I haven't made myself very clear...
Within say a flame from a wood fire there is usally a variation of colour from red to yellow to perhaps white,similarly within a natural gas flame orange to blue etc etc
Is this caused by...
1) Variations in the intensity of heat?
2) density of photons ?
3) density of flammable material at iginition point?
4) something else or all of the above?

Is there a way to model these colour variations?

Thanks

4. Aug 20, 2004

### BobG

Wood isn't a constant composition.

Look at the tree rings. When water was scarce, the tree grew very slowly, making the wood very dense and having a different composition of elements than during the wet periods (the dark 'rings' are the dense slow growth periods, with the fat light regions the fast growth periods). The knots where tree limbs formed also have a slightly different composition.

Bottom line, within one wooden log, different parts of the log have different percentages of each element and the variation in composition means not all parts of the log will burn at the same temperature (i.e. you have a certain percentage of element A, element B which are burning at too low a temperature for element C to combust - at a different point on the log, which has a different percentage of element A, element B, etc, you might be burning at a temperature high enough for element C to combust, as well).

In other words, you have quite a bit of variation in the material you're burning even in one log. The predominate color is going to be determined by how much of each element is being burned at any given point at any given time.

To test your theories about intensity of heat, etc. you need to burn something that does have a constant composition throughout. In open air, I don't think you're going to be able to vary the combustion temperature by much. The intensity and the density might change the brightness, the intensity of the flame, but I don't think it will change the color.

5. Aug 20, 2004

### HallsofIvy

Staff Emeritus
No, you were clear to begin with, and chroot's answer was correct. "Wood" contains a large number of elements, including many minerals (not to mention the minerals that just happen to be lieing on top of the wood). The color (yellow, red, green, blue) of the flames depends on the material burning at that time (ordinary salt, which is most common, gives a yellow flame- green and blue, while possible, are much less common).

Now, if you were talking about heating, not burning, a specific metal, then the color (yellow, red, white) would depend upon the heat.

6. Aug 20, 2004

### LURCH

I think he was talking about the different colours of a wood fire at a given point in time. I.E., yellow at the base, redder near the edges. Or of a flame from a gass stove, whichtends to be almost lear at the bottom, then blue near the middle, turning red at the edges. I would venture to say that, because these colours tend to follow the rainbow's spectrum, with the lowest energy wavelengths being in the same location as the collest temps, yes the temperature determines the colour.

7. Aug 20, 2004

### devoy

Yep, this is what I mean.

8. Aug 20, 2004

### Clausius2

Sure there is a way to simulate such colours!!!.

The variety of colours is caused by variations of Temperature field. As you know, the radiation of a particle is a function of its temperature, as Planck principle states. The only thing to reproduce the colours, is to solve the temperature field. I'm currently trying to simulate numerically a diffusion flame (it's a mixing of Fuel and a coaxial stream of Air).
The problem is written with the Navier-Stokes equations for Reactive flows. So I have to practise in latex:

$$\frac{\partial \rho u}{\partial x} +\frac{\partial \rho v}{\partial y}=0$$

$$\frac{\partial \rho u^2+E_{u}P}{\partial x} +\partial \frac{ \rho uv-\rho \frac{\partial u}{\partial y}}{\partial y}=0$$

$$\frac{\partial \rho uT}{\partial x} +\partial \frac{ \rho vT-\frac{\rho \partial T}{Pr \partial y}}{\partial y}=Da T' Y_{O} Y_{F} exp(-\frac{\beta}{T})$$

$$\frac{\partial \rho uY_{O}}{\partial x} +\partial \frac{\rho vY_{O}-\frac{\rho \partial Y_{O}}{PrLe \partial y}}{\partial y}=\frac{DaS}{1+S} Y_{O} Y_{F} exp(-\frac{\beta}{T})$$

$$\frac{\partial \rho uY_{F}}{\partial x} +\partial \frac{\rho vY_{F}-\frac{\rho \partial Y_{F}}{PrLe \partial y}}{\partial y}=\frac{Da}{1+S} Y_{O} Y_{F} exp(-\frac{\beta}{T})$$

This set of five equations with the equation of state (ideal gas), solves the fields of Temperature, velocity (u,v), pressure(P) and density(rho), mass fraction of oxidyzer (Yo) and Fuel (Yf), for a steady diffusion flame at low Mach and Reynolds numbers.
You may be able to recognize between this equations the famous numbers of Combustion theory:
Da=Damkholer number
Eu=Euler number
Pr=Prandtl number
Le=Lewis number

So here is the response to your question. If you integrate this, please call me.

9. Aug 20, 2004

### LURCH

...And then I found http://www.coolquiz.com/trivia/explain/docs/flame.asp [Broken] .

Last edited by a moderator: May 1, 2017
10. Aug 20, 2004

### kurious

Think of a bunsen burner burning methane gas.
Keep air out of the base and flame is yellow.
Let air into base and flame is blue.
The blue flame is hotter because more oxygen
mixes with methane gas.
No oxygen - very low temperature!

11. Aug 20, 2004

### vsage

I think the key words here are "emission spectroscopy".