CyberTruck SS Heat

ajdelange

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Dids has explained everything well in his latest post.
I assume you wrote that without looking at the curves in my last post. It is obvious that at the shiny end of the scale the gradient has a much larger reflectcitivity component than emissivity. You may find the explanation eloquent and perhaps it is but it is dead wrong. At 40 °C surface temperature it only takes a 7% change in reflectivity to cover a 77% drop in emissivity. Please look at the data.

A lot of us (older folks) have noticed shiny chrome bumpers get very hot to the touch in direct sun.
Then they aren't that shiny or the air is very hot.


"Pickling" the SS could be a way to achieve the best of all worlds. A very shiny CT will be quite annoying for other drivers to look at depending on the sun angle, etc.
Clearly they aren't going to polish it to a mirror fininish. Probably have an albedo of about 0.8 when it leaves the factory. Probably have a temperature of 60 - 70°C (no wind) or 45 - 55 with moderate breeze.
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ajdelange

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Again, wrong.
To say that you have to ignore what the model clearly shows. What's your justification for that? If you have one let's hear it and have an intelligent discussion. "Again, wrong" without any explanation as to why you think it's wrong isn't what we hope to do here.
 

John K

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regarding colorant issue

White pigment/colorant objects reflect the visible wavelength. Energy transmitted to the surface minus the visual wavelength equals less energy striking/absorbed into object.

Black pigment/ color is absorbs the visual wavelength. The absorbed visual wavelength is converted into heat In the object.

If you used a up brightener coating, part of the us spectrum wavelength is absorbed and reflected back in the visual wavelength spectrum. Thus, the object appears brighter in the visual spectrum.

The heat retained and transmitted is based on the object‘s characteristics.
 

John K

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When energy is transmitted to a mirrored surface, the energy reflects. If the energy overexposes the ability to reflect, heat transfers into the object.
 


ajdelange

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White pigment/colorant objects reflect the visible wavelength...spectrum wavelength is absorbed and reflected back in the visual wavelength spectrum. Thus, the object appears brighter in the visual spectrum...
Not quite sure what you are trying to say here but the facts are that black paint on a stainless substrate will absorb more vis/NIR than white paint and thus pass more energy into the substrate causing it to get warmer. Because the black paint also has high emissivity the whole shooting match looks very much like a Planckian body. With white paint less energy is absorbed and transferred to the substrate. It doesn't get as hot. But as the white paint has high emissivity too the system still looks like a Planckian body - just one at a lower temperature.

Just as a side comment I used to put SS NEMA boxes in the Middle East. We always painted them white.
 
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John K

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Not quite sure what you are trying to say here but the facts are that black paint on a stainless substrate will absorb more vis/NIR than white paint and thus pass more energy into the substrate causing it to get warmer. Because the black paint also has high emissivity the whole shooting match looks very much like a Planckian body. With white paint less energy is absorbed and transferred to the substrate. It doesn't get as hot. But as the white paint has high emissivity the system still looks like a Planckian body - just one at a lower temperature.

Just as a side comment I used to put SS NEMA boxes in the Middle East. We always painted them white.
Black paint contains pigment absorbs visible spectrum wavelengths. The absorbs ion process reducing the wavelength into the IR. The object‘ characteristics come into play whether that or energy is retained or conducted.

white reflects visible spectrum, does not enter the object. Unless energy level exceeds the pigments color to reflect.

This is the reason why black absorbs heat. The objects thermal charateristics take over from there.

Every color we see is the visual spectrum being affected by pigments absorbing part of the visible spectrum. The reflective energy interacts with our red,green and blue cones in our eyes. Based on how much our cones react, we interpret color.

The absorbed energy in the pigment is still there and converts to thermal energy.
 

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ajdelange

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When energy is transmitted to a mirrored surface, the energy reflects. If the energy overexposes the ability to reflect, heat transfers into the object.
Again not sure what you are trying to say here. How could "the energy overexpose the ability to reflect? Gettiing it so hot it burns or chars? What in fact is likely to happen is that a highly polished piece of stainless might have emissivity 0.07 and albedo 0.95 suggesting a surface temperature od 45 °C. Over time it would get weathered, and pick up a patina so that it's albedo drops to say 80% but its emissivity increases to say 0.4 in which case its temperature might go up to 65 °C.
 


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This thread has officially decended into mansplaining silliness. Lol

None of you are going to affect the CT design by being internet scientists ? or by proving the other wrong.
 

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We are discussing practical consumer products, none of these will have perfect reflection or be an absolute black body.

In a reflective surface

energy Transmitted > (reflected energy - substrate emissivity)

The material will build heat.

In a black surface

(energy transmitted + absorbed wavelengths) > substrate emissivity

note: formulas overly simplified

What I was describing at the very beginning is the reason for the absorption when the wavelength is slowed into wider IR wavelengths. The substrate emissary characteristics takes over from that point.
 

ajdelange

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This is a wrong assumption. The emissivity of any paint is based on the paint‘s characteristics and not the black pigment.

https://www.design1st.com/Design-Resource-Library/engineering_data/ThermalEmissivityValues.pdf
Again I'm not quite following you. Yes, the emissivity of any two black paints, in fact between any two paints, will be different but they are all about the same. Let's look at the data you linked. The black pigments listed there have an average emissivity of 0.845 with standard deviation of 0.10. The white pigments, 0.89 with standard deviation 0.03. Statistically speaking they are all about the same. What's interesting about this data is that paint in general, irrespective of color, is usually listed as having emissivity of 0.95 or 0.96. Combining all the black and white paints listed on your site we find a mean emissivity of 0.872, median of 0 .88 and standard deviation of 0.068. What this means is that the assumption is not wrong. To illustrate that this is the practical case take a non contact thermometer and go about your house measuring all the objects in the room, especially painted ones. Your thermometer assumes emissivity of 0.95 (unless you set in a different number). Even though some paints are more emissive than others you will get temperatures that are quite close to one another. If you try to measure a piece of stainelss steel, however, you will read a temperature substantially colder than the ambient (unless you put a piece of tape on it as I suggested in an earlier post).
 

ajdelange

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In a reflective surface

energy Transmitted > (reflected energy - substrate emissivity)
That "equation" says, in words, that the energy transmitted is greater than the reflected energy minus the substrate emissivity. I don't think that is really what you are trying to say. The facts are that a body which is being impinged with radiation will come to equilibrium at such temperature that the energy taken away from it (by re-radiation, conduction, convection) is equal to the energy it absorbs.
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