Emissivity of Ball Mill Shell Surface
I would like to calculate or make heat balance for our cement ball mill but i don't know what the emissivity of the mill shell surface is. Shell is made of carbon steel.
My first question: Can you tell me what is the emissivity or how to determinate it? and what about the effect of the coating (paint) on the surface of the shell?.
My second question: Does the radiation calculate from this equation:
R = εσ (Ts^4 - Tsur^4)
σ is the Stefan Boltzmann constant = 5.67 x 10 ^ -8 ( W/m^2 .K^4)
ε is the emissivity of the surface
Ts is the temperature of the surface in K
Tsur is the temperature of the surrounding in K
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Re: Emissivity of Ball Mill Shell Surface
My simple guess to measure the emissivity is the following:
First buy a Infra-Red thermometer with adjustable emissivity. Stop your mill. Measure the surface temperature with a contact thermometer. Then measure the temperature with the IR themometer and adjust the emissivity setting until the reading is the same as the contact temperature. When you obtain a correct fit, you get a measure of the emissivity. If you don't have an adjustable IR thermometer, you can still go on but you will need to use the basic equation, read the technical sheet of your IR thermometer and find the fixed emissivity that it uses. Actually an IR thermometer is a calibrated radiometer.
Typical emissivities for various materials can be found at many places and in many books.
See this link for example: http://www.transmetra.ch/pdf/publikationen/emissivity.pdf
Using the Stefan-Boltzmann law
You should indeed use this law to calculate the emission from a mill. Fortunately The geometry of a mill an its surrounding is very simple. Therefore you do not need to use the full "geometry factors algebra". The surrounding of the mill "views" the whole mill and vice-vers which makes it a very simple problem.
Generally it can become more complicated, like for example inside a clinker kiln where the geometry include the clinker bed makes things a little bit more complicated. And this physics become clearly more difficult when "participating media" must be taken into account like the effect of CO2 and H2O or dust.
Finally, let me note that convection plays generally a greater role at law temperatures. In an example that I calculated recently, where surface temperature was 54°C, the convection accounted for 60% of the heat loss and radiation for 40%. For convection, the rotation speed, the size, and even the bolds on the mill play a role (since the bolts are really coolers).
I assumed that you knew the surface temperature by measurement. Of course, if you are building a computational model of the mill, then the conduction from the inside to the outside must also be taken into account to calculate the external surface temperature. Some mills have an insulation.
And also, don't forget water evaporation in case you have some moisture entering the mill.