Thermal emissivity of an object
- Thread starter Rubycon
- Start date
Jaepheth
Platinum Member
- Apr 29, 2006
- 2,572
- 25
- 91
I would think most paints would act more as an insulator than helping with heat emission.
perhaps a metal electroplating would be more useful.
I would also guess that the difference in surface area between something with a mirror finish and that with a matte finish would be negligible.
perhaps a metal electroplating would be more useful.
I would also guess that the difference in surface area between something with a mirror finish and that with a matte finish would be negligible.
It depends on the composition of the paint. I had an offer to do my masters work on new stealth paints for the USAF, but chose another project so I only know very little about it. Essentially what I gathered is that they change the chemical formulation to alter the radioactive tendencies of the paint. I believe the paints they use are non-glossy (flat?), so I would speculate that non-glossy paints generally have lower emissivity, though this may not be true.
Back in the day we used to anodize aluminum motorcycle cylinders using a DC power supply and a sulphuric acid bath and then dye them with clothes dye. It looked neat anyway.
I am with the biker through experience with air cooled engines also.
I would agree with the tendency for application of a thin layer of black paint to an engine cooling surface for improved radiative properties. However, this site (Link)
suggests no support to the notion that a black color has on thermal emissivity.
In my opinion, I would think that a thin carbon film (in the form of soot) would have better thermal emission than a bare metal surface however. The key is the thin film and a thick layer of carbon (or paint) would seem to have insulative properties.
In aircooled engine building books I have read of arguments that mirror finishes hinder radiation. The argument presents the polished tea kettle and how it holds heat ("Keep your VW alive" John Muir if remembered correctly).
The principal of the black body radiator needs to be considered for a material though (paint coatings and composition, surface characteristics--mirrored or rough) and surfaces seem different from the interiors of the material. This is where analysis seems tricky to me because black body approximations behave similarly only at high temperatures (Astronomy: i.e. Wiens displacement law).
---------------------------------------------------------------------------------------------
Cyclo wizard may know. For the Stealth technologies and the use of thin film interference, however, I am unsure of thermal (Infrared) applications in addition to radar (radio). Very interesting indeed! "Never say never"
Destructive interference requirements of long wavelength radio absorbsion and reemission make thin films a "coating." In my Physics text, a recent work problem gave me a stealth coating thickness around a half centimeter for a "antireflective polymer" with refractive index (n) of 1.5
I wonder where the progress is in stealth. I have heard that the current stealth coatings have problems with being hygroscopic, and it is why the f-117's are housed at low humidity environments at Holloman.
On the note of visibility, applications of antireflective coatings, and thin-film interference I am uncertain of the question for Glossy painted vs. flat.
Although excitement exists over laser treatments of metallic surfaces which point to the nature of surface characteristics and the meaning of glossy and flat when investigation enters the nanoscale.
Dark colors emit more radiation. However, radiation is a tiny fraction of the cooling capacity of an object in the majority of cases we see on a day-to-day basis. You need very high temperatures (visible light and above) for it to be a non-negligible factor.
Painting something dark will normally decrease cooling capability due to the extra thermal barrier provided by the paint. Anodization is usually thin enough to not have much of an effect (usually just done for looks).
Painting something dark will normally decrease cooling capability due to the extra thermal barrier provided by the paint. Anodization is usually thin enough to not have much of an effect (usually just done for looks).
Exactly right. The group that develops this stuff (link) specializes in coatings/thin films. I really don't know much about it other than what I already said. I don't know anything about the relationship between formulation and radiative properties, but someone like f95toli (the guy knows everything
) can probably give us more info.
silverpig
Lifer
- Jul 29, 2001
- 27,703
- 12
- 81
We did a simple 2nd year experiment where we had 3 aluminum rods which we heated in a bath to 100 C. They had a small hole bored halfway through them lengthwise and each had a thermometer inserted. We then pulled all three out and let them cool in the air. One was polished aluminum, one had a rougher finish, and one had some kind of translucent yellow coating on it.
The one with the coating cooled quickest. The polished one cooled slowest.
The one with the coating cooled quickest. The polished one cooled slowest.
- Aug 10, 2005
- 17,768
- 485
- 126
Not painting anything - This thread at CPF got a discussion going because we're getting some new heads for our laser cannons and they are polished instead of HA3+ matte. :shocked:
The temps are supposed to be cooler but this newer head has different TEC setups and the PID is completely different from the older ones.
The temperature differences in the CPF thread have to be an anomaly. It's unbelievable.
The temps are supposed to be cooler but this newer head has different TEC setups and the PID is completely different from the older ones.
The temperature differences in the CPF thread have to be an anomaly. It's unbelievable.
FWIW. Teledyne Continental Aircraft engines have cylinders that are treated with Alodyne which gives a yellow stain. Lycoming Textron Aircraft engines have painted cylinders.
After reading the first post in that thread and thinking a little bit, I think I understand what is going on. The radiative flux, as given by the Stefan-Boltzmann law, is F = e*s*(T1^4-T2^4), where F is the flux (energy per unit surface area per unit time), e is the emissivity, s is the Stefan-Boltzmann constant (usually sigma), T1 is the hot object, and T2 is the temperature of the sink. Since this is in terms of a flux, a rough surface will give a higher net energy transfer rate (that is, the flux integrated over the entire surface area). Thus, if two samples of one material are rough and polished, respectively, they will have the same emissivity but the former will have a higher specific surface area and, therefore, a higher energy transfer rate. If, however, a suitable coating is applied, the emissivity of the coating is much higher than that of the substrate and the net transfer rate is higher still. The coating will also increase the surface area, of course, because it increases the dimension of the part that it is coating, further increasing the transfer rate.
I'll run through some numbers and see if I can figure out how he can get those results. I was also surprised to see such a drastic difference at relatively low temperatures where I wouldn't intuitively think that radiation would be the controlling heat transfer mechanism.
I believe we can use heatsinks as a case study here. When talking heatsinks there is little thing called "Film Resistance". Any coatings, even anodizing, can have a detrimental effect on this since it is comprised of oxides & pigments which have a different thermal efficiency value than the base material.
We can think of a heatsink as a "thermal network". Put simply, the heat should flow through this "network" from the heatsource to the surrounding environment. Resistance at any key point will result in a reduction in thermal efficiency of the overall design.
We can think of a heatsink as a "thermal network". Put simply, the heat should flow through this "network" from the heatsource to the surrounding environment. Resistance at any key point will result in a reduction in thermal efficiency of the overall design.
edcarman
Member
- May 23, 2005
- 172
- 0
- 71
Radiation may well be the predominant mechanism in this case. You can see this by using a convection approximation for radiation (using Cyclowizard's notation, all temps in K):
F = hrad*(T1-T2), where hrad = e*s*(T1^2 + T2^2)*(T1 + T2)
NOTE: multiplying this out will give the same formula as Cyclowizard's
Taking surface temperature of 80°C (353K) and environment of 20°C (293K) gives:
hrad = 7.7*e W/m^2.K
Comparing this with a natural convection coefficient that would be around 1-5 W/m^2.K, shows that, for more emissive bodies, radiation will play a significant role.
F = hrad*(T1-T2), where hrad = e*s*(T1^2 + T2^2)*(T1 + T2)
NOTE: multiplying this out will give the same formula as Cyclowizard's
Taking surface temperature of 80°C (353K) and environment of 20°C (293K) gives:
hrad = 7.7*e W/m^2.K
Comparing this with a natural convection coefficient that would be around 1-5 W/m^2.K, shows that, for more emissive bodies, radiation will play a significant role.
edcarman
Member
- May 23, 2005
- 172
- 0
- 71
As far as the resitance calculation goes, you would weigh the benefits of decreasing the radiation resitance against increasing the conduction resistance.
Rcond = t/kA (t = thickness, k = conductivity, A = heat flow area)
Rrad = 1/(hrad.A)
So it would basically depend on your film thickness vs. emissivity. Many heatsinks will have an oxide film forming anyway which can increase the emissivity.
'Film resistance' is related to convection (whether natural or forced) at the surface of the fin. It doesn't really have anything to do with radiation. It depends on the geometry of the fin assembly and the flow/material properties of the heat transfer fluid (generally air), not on the fin properties. However, you're right that any coating will necessarily decrease this rate of heat transfer by adding an additional conductive resistance. However, if the thermal conductivity of the coating is sufficiently high, this additional resistance will be negligible, so it would be worth it if the higher emissivity can cover this deficit.
I believe the below excerpt from a wiki article is instructive:
"Kirchhoff's law in thermodynamics, also called e.g. Kirchhoff's law of thermal radiation, is a general statement equating emission and absorption in heated objects, proposed by Gustav Kirchhoff in 1859 (and proved in 1861), following from general considerations of thermodynamic equilibrium. (See Kirchhoff's laws for other laws named after Kirchhoff.)
An object at some non-zero temperature radiates electromagnetic energy. If it is a perfect black body, absorbing all light that strikes it, it radiates energy according to the black-body radiation formula. More generally, it is a "grey body" that radiates with some emissivity multiplied by the black-body formula. Kirchhoff's law states that:
At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.
Here, the absorptivity (or absorbance) is the fraction of incident light (power) that is absorbed by the body/surface. In the most general form of the theorem, this power must be integrated over all wavelengths and angles. In some cases, however, emissivity and absorption may be defined to depend on wavelength and angle, as described below.
Kirchhoff's Law has a corollary: the emissivity cannot exceed one (because the absorptivity cannot, by conservation of energy), so it is not possible to thermally radiate more energy than a black body, at equilibrium. This has two caveats, however. First, if the surface is diffractive, so that incident energy at one angle is partially reflected to another angle, then the emissivity at one angle can exceed unity, but not the emissivity of power integrated over all angles. Second, if the object is nonlinear (e.g. fluorescent), so that incident power at one wavelength is re-emitted at another wavelength, then the emissivity at some wavelengths can exceed unity, but not the emissivity of power integrated over all wavelengths. In negative luminescence the angle and wavelength integrated absorption exceeds the material's emission, however, such systems are powered by an external source and are therefore not in thermal equilibrium.
This theorem is sometimes informally stated as a poor reflector is a good emitter, and a good reflector is a poor emitter. It is why, for example, lightweight emergency thermal blankets are based on reflective metallic coatings: they lose little heat by radiation."
So....Matte finish = generally poor reflector but good emitter, and so matte finishes likely have emissivity higher than that of a flat finish.
"Kirchhoff's law in thermodynamics, also called e.g. Kirchhoff's law of thermal radiation, is a general statement equating emission and absorption in heated objects, proposed by Gustav Kirchhoff in 1859 (and proved in 1861), following from general considerations of thermodynamic equilibrium. (See Kirchhoff's laws for other laws named after Kirchhoff.)
An object at some non-zero temperature radiates electromagnetic energy. If it is a perfect black body, absorbing all light that strikes it, it radiates energy according to the black-body radiation formula. More generally, it is a "grey body" that radiates with some emissivity multiplied by the black-body formula. Kirchhoff's law states that:
At thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.
Here, the absorptivity (or absorbance) is the fraction of incident light (power) that is absorbed by the body/surface. In the most general form of the theorem, this power must be integrated over all wavelengths and angles. In some cases, however, emissivity and absorption may be defined to depend on wavelength and angle, as described below.
Kirchhoff's Law has a corollary: the emissivity cannot exceed one (because the absorptivity cannot, by conservation of energy), so it is not possible to thermally radiate more energy than a black body, at equilibrium. This has two caveats, however. First, if the surface is diffractive, so that incident energy at one angle is partially reflected to another angle, then the emissivity at one angle can exceed unity, but not the emissivity of power integrated over all angles. Second, if the object is nonlinear (e.g. fluorescent), so that incident power at one wavelength is re-emitted at another wavelength, then the emissivity at some wavelengths can exceed unity, but not the emissivity of power integrated over all wavelengths. In negative luminescence the angle and wavelength integrated absorption exceeds the material's emission, however, such systems are powered by an external source and are therefore not in thermal equilibrium.
This theorem is sometimes informally stated as a poor reflector is a good emitter, and a good reflector is a poor emitter. It is why, for example, lightweight emergency thermal blankets are based on reflective metallic coatings: they lose little heat by radiation."
So....Matte finish = generally poor reflector but good emitter, and so matte finishes likely have emissivity higher than that of a flat finish.
Having worked on low observable technology for many years (Including the B-2 and the F-117a before I decided to go to the patent Office, I can say that the paint of the F-117 is engineered to minimize RF return, that is, reflectance of incoming radiation. Most ground based IR seeking missiles are initially targeted by radar and switch to IR seeking once they are in relatively close range.
Right, which is why the B2's shape and engine exhausts are shaped so strangely. Paint can only do so much to limit the IR signal given off by jet engines.
Originally posted by: CycloWizard
Right, which is why the B2's shape and engine exhausts are shaped so strangely. Paint can only do so much to limit the IR signal given off by jet engines.
Correct, the exhaust from the engines of the B-2 is emitted from the top rear portion of the plan so as to minimize IR contrast with the sky, which is important for a huge, slow target. As for the shape of the B-2, the design is completely devoid of any "corner" reflectors, which are the bane of any low RF observable system.
One thing to note is that making something hard to see with an IR camera is a much more difficult problem than making something hard to see with radar. With radar, the only concern is minimizing RF return, as the detectors are looking for a signal bounced back from an object. With IR, the detectors are looking at contrast, and thus are able to detect objects which are different (either hotter or colder) than their background. For example, if a ground based missile silo is equipped with an IR detector, it will be trying to spot a plane against a veryc old background, namely space. Thus, for a plane suchg as the B-2 to be low IR observable to ground based IR, the bottom of the plane has to appear to have the same thermal properties as the background. Thats why the B-2 has its engine exhausted vents from the top/rear of the plane. In addition, the engines actually actively mix colder air in with the hot exhaust so as to reduce the IR "plume" trailing behind the plane (this is also done in modern nuclear aircraft carriers, as it is quite possible with modern IR detectors to follow an aircraft carrier by the trail of slightly warmer water it leaves behind it)
TRENDING THREADS
-
Discussion Zen 5 Speculation (EPYC Turin and Strix Point/Granite Ridge - Ryzen 9000)- Started by DisEnchantment
- Replies: 24K
-
Discussion Intel Meteor, Arrow, Lunar & Panther Lakes Discussion Threads
- Started by Tigerick
- Replies: 19K
-
-
-
Facebook
Twitter