TechFan said:
Hi:
Very interesting and practical project.
I know nothing about emissivity and/or furnaces; but perhaps this can help you.
Since the main goal of calculating emissivity in this case will be to show your customers how much they can save in fuel with your coating, perhaps you can setup a test calculating the fuel savings the coating introduces and maybe these results, since they are related mainly to the difference in emissivity, may help you calculate the relationship between coated and uncoated furnaces regarding emissivity. In other words, the energy consumption analysis may help you find out some interesting relationships and probably present your point to your customers.
To do that you can built (setup somehow) an electrically driven furnace with the commonly used firebricks for the type of furnaces you are targeting in a small scale. Then you place a temperature control system to make the furnace reach and keep a certain temperature for a certain time you choose. You can then use a data logger and plot temp vs. time and current or watts vs. time from the beginning till the end of the test. That will help you calculate the KW-h or energy consumption for the whole process (plus rate of temp changes, etc). When you have that, then you paint the inside of the furnace with your coating and repeat the experiment. Make sure the ambient temp surrounding the furnace is about the same to keep heat transfer about the same. You can log that temp also with your data logger and cancel somehow its effect later on in your calculations, probably by using the difference between inside and outside. With the data collected in both cases (with and without coating) you will be able to calculate how much energy your coating has saved to the process. Furthermore, if your coating only (or mainly) changes emissivity and nothing else (?) and the firebricks you are using emissivity's is known, then you can attribute the change in energy consumption, to the changes in emissivity and calculate the relationship between them as emissivity and energy are closely related. If emissivity becomes problematic to calculate, you still have the energy analysis which your customers will probably like when you present it to them expressed as percentage of how much your coating can save them.
Of course, these experiments need to be run several times in both cases (without and with your coating) to average, calculate standard deviations, etc.
I don't know if doing this will be non expensive though. It all depends on how much you want to expend on this.
Hope it helps you.
Good luck with your Project.
Hi Techfan,
Thanks for your response.
I had thought about this and no doubt this is one of the more practical ways to do it - link the coated tiles/bricks to energy saved. I had thought about it this way - build a small chamber with run of the mill firebricks (most common), put a container with water in it or put a temperature probe inside, then fire the burner. I can either wait for the water to start boiling or measure the time taken for the probe to reach a particular temperature. Then repeat the experiment with similar but coated bricks & see the difference in T(coated) - T(uncoated) as well as Fuel(uncoated) - Fuel(coated).
This way I don't need to measure the emissivity of the surface & I come directly to the point.
Thing is, however, such an experiment takes time & set up to do it everytime a client asks me to. I can video tape it & document the results and present it to the client but you know how cynical buyers are, they want quantitative data points before they buy an expensive product (and the coating is fairly expensive).
Another thought in my mind was this -
I go to the client's site and paint an area of 2 ft x 2 ft of the furnace inside wall bang in the middle of the wall. So if the furnace sidewall is 10ft x 12 ft, I draw a 2x2 ft square in the center of the wall. Then, I paint it with my coating.
There could be multiple datapoints here. A 2by2 square will not make much of a difference in the fuel consumption which could be noticeable. I could, however, take the readings of the wall from the outside.
So, the area which has been coated is on the inside of the wall, if I measure the temperature of the same area but on the outside of the wall, it's supposed to be lower than it was when the area was uncoated, right? Because more heat is being sent back in than before?
The only problem is that I can't stop conduction from the uncoated area inside the wall to ALL of the outside wall area, which could fudge up the figures possibly.
Is there something I can do with an optical/laser pyrometer here?
One more thing, the coating I am developing is slightly off-white or cream in color. I can't find a black color pigment that will last upto 1800 Celsius. How much am I losing by not making the coating completely black?