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Can a black coating increase the efficiency of a heat dissipator?
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I have an aluminium profile that contains too many LEDs and heats up around 70°C, which is about +20°C above my expectation. Would an external black coating increase the efficiency of the heat dissipation by the aluminium?
thermodynamics
asked 4 hours ago
adrienlucca.wordpress.com
284111
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up vote
3
down vote
favorite
I have an aluminium profile that contains too many LEDs and heats up around 70°C, which is about +20°C above my expectation. Would an external black coating increase the efficiency of the heat dissipation by the aluminium?
thermodynamics
asked 4 hours ago
adrienlucca.wordpress.com
284111
No. A black coating would enable the heat sink to adsorb more radiation, but it would not help the heat sink radiate more. You need more surface area.
– MaxW
4 hours ago
4
@MaxW That is incorrect. The more that a blackbody absorbs, the more it radiates.
– HiddenBabel
4 hours ago
@HiddenBabel - It is a heatsink for electronics and the temperature is well above ambient, so there isn't a balance between radiation being absorbed and radiation being radiated away. A black coating will not help the heatsink radiate heat.
– MaxW
3 hours ago
1
@HiddenBabel - Besides, this isn't really about blackbody radiation. A heatsink in loses almost all of its excess heat via convection.
– MaxW
3 hours ago
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I have an aluminium profile that contains too many LEDs and heats up around 70°C, which is about +20°C above my expectation. Would an external black coating increase the efficiency of the heat dissipation by the aluminium?
thermodynamics
asked 4 hours ago
adrienlucca.wordpress.com
284111
I have an aluminium profile that contains too many LEDs and heats up around 70°C, which is about +20°C above my expectation. Would an external black coating increase the efficiency of the heat dissipation by the aluminium?
thermodynamics
thermodynamics
asked 4 hours ago
adrienlucca.wordpress.com
284111
asked 4 hours ago
adrienlucca.wordpress.com
284111
asked 4 hours ago
adrienlucca.wordpress.com
284111
asked 4 hours ago
adrienlucca.wordpress.com
284111
asked 4 hours ago
adrienlucca.wordpress.com
284111
284111
No. A black coating would enable the heat sink to adsorb more radiation, but it would not help the heat sink radiate more. You need more surface area.
– MaxW
4 hours ago
4
@MaxW That is incorrect. The more that a blackbody absorbs, the more it radiates.
– HiddenBabel
4 hours ago
@HiddenBabel - It is a heatsink for electronics and the temperature is well above ambient, so there isn't a balance between radiation being absorbed and radiation being radiated away. A black coating will not help the heatsink radiate heat.
– MaxW
3 hours ago
1
@HiddenBabel - Besides, this isn't really about blackbody radiation. A heatsink in loses almost all of its excess heat via convection.
– MaxW
3 hours ago
add a comment |Â
No. A black coating would enable the heat sink to adsorb more radiation, but it would not help the heat sink radiate more. You need more surface area.
– MaxW
4 hours ago
4
@MaxW That is incorrect. The more that a blackbody absorbs, the more it radiates.
– HiddenBabel
4 hours ago
@HiddenBabel - It is a heatsink for electronics and the temperature is well above ambient, so there isn't a balance between radiation being absorbed and radiation being radiated away. A black coating will not help the heatsink radiate heat.
– MaxW
3 hours ago
1
@HiddenBabel - Besides, this isn't really about blackbody radiation. A heatsink in loses almost all of its excess heat via convection.
– MaxW
3 hours ago
No. A black coating would enable the heat sink to adsorb more radiation, but it would not help the heat sink radiate more. You need more surface area.
– MaxW
4 hours ago
No. A black coating would enable the heat sink to adsorb more radiation, but it would not help the heat sink radiate more. You need more surface area.
– MaxW
4 hours ago
4
4
@MaxW That is incorrect. The more that a blackbody absorbs, the more it radiates.
– HiddenBabel
4 hours ago
@MaxW That is incorrect. The more that a blackbody absorbs, the more it radiates.
– HiddenBabel
4 hours ago
@HiddenBabel - It is a heatsink for electronics and the temperature is well above ambient, so there isn't a balance between radiation being absorbed and radiation being radiated away. A black coating will not help the heatsink radiate heat.
– MaxW
3 hours ago
@HiddenBabel - It is a heatsink for electronics and the temperature is well above ambient, so there isn't a balance between radiation being absorbed and radiation being radiated away. A black coating will not help the heatsink radiate heat.
– MaxW
3 hours ago
1
1
@HiddenBabel - Besides, this isn't really about blackbody radiation. A heatsink in loses almost all of its excess heat via convection.
– MaxW
3 hours ago
@HiddenBabel - Besides, this isn't really about blackbody radiation. A heatsink in loses almost all of its excess heat via convection.
– MaxW
3 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
Possibly. Implicit in your question is the assumption that radiative heat transfer is playing or could play an important role in your configuration (vs. convection). If so, then applying a black coating (and thus increasing the emissivity to essentially 1) could benefit you. However, note that the coating itself may hinder heat transfer across various interfaces.
I would plug the relevant numbers into the various formulas for heat transfer ($hA_mathrmsurface(T-T_infty)$ for convection, $kA_mathrmcross,sectionDelta T/Delta x$ for conduction, $sigmaepsilon A_mathrmsurface,facing,surroundings(T^4-T^4_infty)$ for radiation, as described in any introductory heat transfer textbook and at many locations online) to estimate the relative magnitude of convection vs. radiation before attempting to optimize a heat transfer mechanism that might be unimportant.
As an example, the natural convection coefficient $h$ can be very broadly estimated to around $10,mathrmW,mathrmm^-2,mathrmK^-1$ by order of magnitude (I'm sure exceptions exist in certain geometries). From the numbers you've given, one can estimate that changing the emissivity from 0 to 1 (by anodizing the aluminum, for instance) could potentially boost the outgoing heat flux by a detectable amount, but probably less than 100%. Furthermore, as other posters have noted, you may have other options to increase the outgoing heat flux much more substantially—using fins or a fan, perhaps.
answered 3 hours ago
Chemomechanics
3,1462619
add a comment |Â
up vote
1
down vote
Your heat sink gains heat from its contact with the hot LEDs and looses it to the environment (air) around it. Since it uses convection to lose heat, any coating that has less thermal conductivity than aluminium will act as an insulator and will cause it to get hotter, which will work against you.
As one of the comments point out, you can increase the amount of heat lost to the environment by increasing the surface area of your heat sink.
You can also look for a material with higher thermal conductivity so heat is transported faster inside the material.
answered 4 hours ago
user190081
999
2
You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support.
– Chemomechanics
3 hours ago
@Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air.
– MaxW
2 hours ago
add a comment |Â
up vote
1
down vote
You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $frac1pisigma T^4$, where $sigma=5.670367(13)×10^−8 W⋅m^−2⋅K^−4$. Plugging in $T=343$, we get $250 Wm^-2$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^-2$, for a net of $104 Wm^-2$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.
The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^-2$. (And if it's in an enclosed space, you may be getting less than $10mW$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.
answered 3 hours ago
Acccumulation
1,45719
1
(1) Where does the factor of $1/pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space?
– Chemomechanics
3 hours ago
(1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that.
– Acccumulation
3 hours ago
1
(1) The factor of $1/pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle.
– Chemomechanics
2 hours ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Possibly. Implicit in your question is the assumption that radiative heat transfer is playing or could play an important role in your configuration (vs. convection). If so, then applying a black coating (and thus increasing the emissivity to essentially 1) could benefit you. However, note that the coating itself may hinder heat transfer across various interfaces.
I would plug the relevant numbers into the various formulas for heat transfer ($hA_mathrmsurface(T-T_infty)$ for convection, $kA_mathrmcross,sectionDelta T/Delta x$ for conduction, $sigmaepsilon A_mathrmsurface,facing,surroundings(T^4-T^4_infty)$ for radiation, as described in any introductory heat transfer textbook and at many locations online) to estimate the relative magnitude of convection vs. radiation before attempting to optimize a heat transfer mechanism that might be unimportant.
As an example, the natural convection coefficient $h$ can be very broadly estimated to around $10,mathrmW,mathrmm^-2,mathrmK^-1$ by order of magnitude (I'm sure exceptions exist in certain geometries). From the numbers you've given, one can estimate that changing the emissivity from 0 to 1 (by anodizing the aluminum, for instance) could potentially boost the outgoing heat flux by a detectable amount, but probably less than 100%. Furthermore, as other posters have noted, you may have other options to increase the outgoing heat flux much more substantially—using fins or a fan, perhaps.
answered 3 hours ago
Chemomechanics
3,1462619
add a comment |Â
up vote
5
down vote
Possibly. Implicit in your question is the assumption that radiative heat transfer is playing or could play an important role in your configuration (vs. convection). If so, then applying a black coating (and thus increasing the emissivity to essentially 1) could benefit you. However, note that the coating itself may hinder heat transfer across various interfaces.
I would plug the relevant numbers into the various formulas for heat transfer ($hA_mathrmsurface(T-T_infty)$ for convection, $kA_mathrmcross,sectionDelta T/Delta x$ for conduction, $sigmaepsilon A_mathrmsurface,facing,surroundings(T^4-T^4_infty)$ for radiation, as described in any introductory heat transfer textbook and at many locations online) to estimate the relative magnitude of convection vs. radiation before attempting to optimize a heat transfer mechanism that might be unimportant.
As an example, the natural convection coefficient $h$ can be very broadly estimated to around $10,mathrmW,mathrmm^-2,mathrmK^-1$ by order of magnitude (I'm sure exceptions exist in certain geometries). From the numbers you've given, one can estimate that changing the emissivity from 0 to 1 (by anodizing the aluminum, for instance) could potentially boost the outgoing heat flux by a detectable amount, but probably less than 100%. Furthermore, as other posters have noted, you may have other options to increase the outgoing heat flux much more substantially—using fins or a fan, perhaps.
answered 3 hours ago
Chemomechanics
3,1462619
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Possibly. Implicit in your question is the assumption that radiative heat transfer is playing or could play an important role in your configuration (vs. convection). If so, then applying a black coating (and thus increasing the emissivity to essentially 1) could benefit you. However, note that the coating itself may hinder heat transfer across various interfaces.
I would plug the relevant numbers into the various formulas for heat transfer ($hA_mathrmsurface(T-T_infty)$ for convection, $kA_mathrmcross,sectionDelta T/Delta x$ for conduction, $sigmaepsilon A_mathrmsurface,facing,surroundings(T^4-T^4_infty)$ for radiation, as described in any introductory heat transfer textbook and at many locations online) to estimate the relative magnitude of convection vs. radiation before attempting to optimize a heat transfer mechanism that might be unimportant.
As an example, the natural convection coefficient $h$ can be very broadly estimated to around $10,mathrmW,mathrmm^-2,mathrmK^-1$ by order of magnitude (I'm sure exceptions exist in certain geometries). From the numbers you've given, one can estimate that changing the emissivity from 0 to 1 (by anodizing the aluminum, for instance) could potentially boost the outgoing heat flux by a detectable amount, but probably less than 100%. Furthermore, as other posters have noted, you may have other options to increase the outgoing heat flux much more substantially—using fins or a fan, perhaps.
answered 3 hours ago
Chemomechanics
3,1462619
Possibly. Implicit in your question is the assumption that radiative heat transfer is playing or could play an important role in your configuration (vs. convection). If so, then applying a black coating (and thus increasing the emissivity to essentially 1) could benefit you. However, note that the coating itself may hinder heat transfer across various interfaces.
I would plug the relevant numbers into the various formulas for heat transfer ($hA_mathrmsurface(T-T_infty)$ for convection, $kA_mathrmcross,sectionDelta T/Delta x$ for conduction, $sigmaepsilon A_mathrmsurface,facing,surroundings(T^4-T^4_infty)$ for radiation, as described in any introductory heat transfer textbook and at many locations online) to estimate the relative magnitude of convection vs. radiation before attempting to optimize a heat transfer mechanism that might be unimportant.
As an example, the natural convection coefficient $h$ can be very broadly estimated to around $10,mathrmW,mathrmm^-2,mathrmK^-1$ by order of magnitude (I'm sure exceptions exist in certain geometries). From the numbers you've given, one can estimate that changing the emissivity from 0 to 1 (by anodizing the aluminum, for instance) could potentially boost the outgoing heat flux by a detectable amount, but probably less than 100%. Furthermore, as other posters have noted, you may have other options to increase the outgoing heat flux much more substantially—using fins or a fan, perhaps.
answered 3 hours ago
Chemomechanics
3,1462619
edited 2 hours ago
answered 3 hours ago
Chemomechanics
3,1462619
answered 3 hours ago
Chemomechanics
3,1462619
answered 3 hours ago
Chemomechanics
3,1462619
3,1462619
add a comment |Â
add a comment |Â
up vote
1
down vote
Your heat sink gains heat from its contact with the hot LEDs and looses it to the environment (air) around it. Since it uses convection to lose heat, any coating that has less thermal conductivity than aluminium will act as an insulator and will cause it to get hotter, which will work against you.
As one of the comments point out, you can increase the amount of heat lost to the environment by increasing the surface area of your heat sink.
You can also look for a material with higher thermal conductivity so heat is transported faster inside the material.
answered 4 hours ago
user190081
999
2
You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support.
– Chemomechanics
3 hours ago
@Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air.
– MaxW
2 hours ago
add a comment |Â
up vote
1
down vote
Your heat sink gains heat from its contact with the hot LEDs and looses it to the environment (air) around it. Since it uses convection to lose heat, any coating that has less thermal conductivity than aluminium will act as an insulator and will cause it to get hotter, which will work against you.
As one of the comments point out, you can increase the amount of heat lost to the environment by increasing the surface area of your heat sink.
You can also look for a material with higher thermal conductivity so heat is transported faster inside the material.
answered 4 hours ago
user190081
999
2
You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support.
– Chemomechanics
3 hours ago
@Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air.
– MaxW
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Your heat sink gains heat from its contact with the hot LEDs and looses it to the environment (air) around it. Since it uses convection to lose heat, any coating that has less thermal conductivity than aluminium will act as an insulator and will cause it to get hotter, which will work against you.
As one of the comments point out, you can increase the amount of heat lost to the environment by increasing the surface area of your heat sink.
You can also look for a material with higher thermal conductivity so heat is transported faster inside the material.
answered 4 hours ago
user190081
999
Your heat sink gains heat from its contact with the hot LEDs and looses it to the environment (air) around it. Since it uses convection to lose heat, any coating that has less thermal conductivity than aluminium will act as an insulator and will cause it to get hotter, which will work against you.
As one of the comments point out, you can increase the amount of heat lost to the environment by increasing the surface area of your heat sink.
You can also look for a material with higher thermal conductivity so heat is transported faster inside the material.
answered 4 hours ago
user190081
999
edited 4 hours ago
answered 4 hours ago
user190081
999
answered 4 hours ago
user190081
999
answered 4 hours ago
user190081
999
999
2
You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support.
– Chemomechanics
3 hours ago
@Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air.
– MaxW
2 hours ago
add a comment |Â
2
You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support.
– Chemomechanics
3 hours ago
@Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air.
– MaxW
2 hours ago
2
2
You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support.
– Chemomechanics
3 hours ago
You've immediately assumed that radiation is negligible, but the whole point of the question is about optimizing radiative heat transfer. So I think this assumption needs more detailed support.
– Chemomechanics
3 hours ago
@Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air.
– MaxW
2 hours ago
@Chemomechanics - Read the question again. "Would an external black coating increase the efficiency of the heat dissipation by the aluminium?" There is no restriction to radiative transfer, and in fact most of the heat loss of the heatsink will be by convection of the air.
– MaxW
2 hours ago
add a comment |Â
up vote
1
down vote
You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $frac1pisigma T^4$, where $sigma=5.670367(13)×10^−8 W⋅m^−2⋅K^−4$. Plugging in $T=343$, we get $250 Wm^-2$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^-2$, for a net of $104 Wm^-2$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.
The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^-2$. (And if it's in an enclosed space, you may be getting less than $10mW$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.
answered 3 hours ago
Acccumulation
1,45719
1
(1) Where does the factor of $1/pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space?
– Chemomechanics
3 hours ago
(1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that.
– Acccumulation
3 hours ago
1
(1) The factor of $1/pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle.
– Chemomechanics
2 hours ago
add a comment |Â
up vote
1
down vote
You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $frac1pisigma T^4$, where $sigma=5.670367(13)×10^−8 W⋅m^−2⋅K^−4$. Plugging in $T=343$, we get $250 Wm^-2$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^-2$, for a net of $104 Wm^-2$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.
The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^-2$. (And if it's in an enclosed space, you may be getting less than $10mW$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.
answered 3 hours ago
Acccumulation
1,45719
1
(1) Where does the factor of $1/pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space?
– Chemomechanics
3 hours ago
(1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that.
– Acccumulation
3 hours ago
1
(1) The factor of $1/pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle.
– Chemomechanics
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $frac1pisigma T^4$, where $sigma=5.670367(13)×10^−8 W⋅m^−2⋅K^−4$. Plugging in $T=343$, we get $250 Wm^-2$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^-2$, for a net of $104 Wm^-2$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.
The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^-2$. (And if it's in an enclosed space, you may be getting less than $10mW$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.
answered 3 hours ago
Acccumulation
1,45719
You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $frac1pisigma T^4$, where $sigma=5.670367(13)×10^−8 W⋅m^−2⋅K^−4$. Plugging in $T=343$, we get $250 Wm^-2$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^-2$, for a net of $104 Wm^-2$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.
The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^-2$. (And if it's in an enclosed space, you may be getting less than $10mW$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.
answered 3 hours ago
Acccumulation
1,45719
edited 3 hours ago
answered 3 hours ago
Acccumulation
1,45719
answered 3 hours ago
Acccumulation
1,45719
answered 3 hours ago
Acccumulation
1,45719
1,45719
1
(1) Where does the factor of $1/pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space?
– Chemomechanics
3 hours ago
(1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that.
– Acccumulation
3 hours ago
1
(1) The factor of $1/pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle.
– Chemomechanics
2 hours ago
add a comment |Â
1
(1) Where does the factor of $1/pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space?
– Chemomechanics
3 hours ago
(1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that.
– Acccumulation
3 hours ago
1
(1) The factor of $1/pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle.
– Chemomechanics
2 hours ago
1
1
(1) Where does the factor of $1/pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space?
– Chemomechanics
3 hours ago
(1) Where does the factor of $1/pi$ come from? (2) You seem to say that the surroundings may effectively be at >70°C ("it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile"). How does this work, assuming this is a habitable space?
– Chemomechanics
3 hours ago
(1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that.
– Acccumulation
3 hours ago
(1) en.wikipedia.org/wiki/… (2) The effective black body temperature is the temperature a body would have to reach for the amount of heat it is emitting to equal the heat it is absorbing. If you have a light bulb at 500K, then you would have to reach 500K to be in thermodynamic equilibrium with it. But if you have ways of losing heat other than black body radiation, such as conduction with the air, then you can remain at a temperature below that.
– Acccumulation
3 hours ago
1
1
(1) The factor of $1/pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle.
– Chemomechanics
2 hours ago
(1) The factor of $1/pi$ applies to the radiance, not the heat flux. (2) The filament of a nearby incandescent light bulb occupies a minuscule solid view angle.
– Chemomechanics
2 hours ago
add a comment |Â
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No. A black coating would enable the heat sink to adsorb more radiation, but it would not help the heat sink radiate more. You need more surface area.
– MaxW
4 hours ago
4
@MaxW That is incorrect. The more that a blackbody absorbs, the more it radiates.
– HiddenBabel
4 hours ago
@HiddenBabel - It is a heatsink for electronics and the temperature is well above ambient, so there isn't a balance between radiation being absorbed and radiation being radiated away. A black coating will not help the heatsink radiate heat.
– MaxW
3 hours ago
1
@HiddenBabel - Besides, this isn't really about blackbody radiation. A heatsink in loses almost all of its excess heat via convection.
– MaxW
3 hours ago