TomBrooklyn
Specifier/Regulator
Would air blowing over a hot lamp (lightbulb) reduce any heat effect due to radiation?
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Does Forced Convection Reduce Radiated Heat?
- Thread starter TomBrooklyn
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Tom
I'm not quite sure how to answer your question. A light bulb produces a certain amount of heat energy (dependent on the type) over a certain period of time. This energy is then transferred to the surrounding air through a combination of radiative, conductive, and convective heat transfer. All three will occur. By using forced convection versus natural convection, you could increase the amount of cooling that occurs and might be able to reduce the temperature of the bulb. I guess this could be taken as reducing the "heat effect due to radiation" as the formula for radiative heat transfer is (1.714E-9)(A)(e)(T**4) where T is the surface temperature in degrees Rankin, A is the area of the light bulb, e is the emissivity of the glass, and 1.714E-9 is the Stefan-Boltzmann constant.
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I'm not quite sure how to answer your question. A light bulb produces a certain amount of heat energy (dependent on the type) over a certain period of time. This energy is then transferred to the surrounding air through a combination of radiative, conductive, and convective heat transfer. All three will occur. By using forced convection versus natural convection, you could increase the amount of cooling that occurs and might be able to reduce the temperature of the bulb. I guess this could be taken as reducing the "heat effect due to radiation" as the formula for radiative heat transfer is (1.714E-9)(A)(e)(T**4) where T is the surface temperature in degrees Rankin, A is the area of the light bulb, e is the emissivity of the glass, and 1.714E-9 is the Stefan-Boltzmann constant.
Is this what you were looking for?? Patricia Lougheed
Please see FAQ731-376 for tips on how to make the best use of the Eng-Tips Forums.
Because blowing air (forced convection) carries heat away, the surface temperature of the bulb will be less. The emitted heat is proportional to the surface temperature to the fourth power. Therefore, a lower surface temperature means less radiated heat to the fourth power! This is obvious by intuition but the math is good to know as well.
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- #5
TomBrooklyn
Specifier/Regulator
Thanks for the replies.
I think I need to refine the question a little. Does the forced convection cooling *significantly* affect the amount of radiated heat falling on a nearby surface? I guess I'm unclear about whether the heat is being radiated by temperature of the lamp, the light of the lamp, or both. And if both, to what degree the warming of a nearby object would be due to the light, and to what degree the temperature of the lamp.
I Hope that makes enough sense. Thanks.
I think I need to refine the question a little. Does the forced convection cooling *significantly* affect the amount of radiated heat falling on a nearby surface? I guess I'm unclear about whether the heat is being radiated by temperature of the lamp, the light of the lamp, or both. And if both, to what degree the warming of a nearby object would be due to the light, and to what degree the temperature of the lamp.
I Hope that makes enough sense. Thanks.
TB-
An incandescent light bulb emits energy primarily in the ultraviolet (UV), infrared (IR), and visible portions of the electromagnetic spectrum. The portion of the total energy in the UV region amounts to practically nothing. The portion of the total energy in the visible portion is typically less than 20%. The portion of the total energy in the IR portion is typically >80%. As for flourescent bulbs, the proportions are tilted further towards the visible and UV portions of the spectrum.
The heat emitted is dependant on the surface temperature which depends on:
A. the temperatures, locations, geometries, and finishes of nearby objects,
B. the surface area of the bulb,
C. the power consumed by the bulb,
D. the emittance of the bulb's surface finish, ideally determined over the visible and IR bands, and
E. altitude (density).
With no air forced over the bulb, then you have natural convection cooling. In that case, the surface temperature will also depend on the bulb's orientation with respect to gravity and the available airspace for air plume development.
With forced air over the bulb, then you have forced convection cooling. In this case, the surface temperature will also depend on the airflow distribution around the bulb's (velocity and temperature).
As you can see, there are many variables and they all play together to influence the surface temperature.
To answer your question more directly, blowing air over the bulb may dramatically lower the bulb's surface temperature. By lowering the surface temperature of the bulb, you may have a dramatic effect on nearby objects. Naturally, the more air you blow over the bulb, the closer the surface temperature comes to the air temperature.
Lastly, by using high emittance/low absorptance finishes on nearby objects and by providing them with good heat sinks, you can keep those object temperatures to a minimum.
The only way to really know what kind of temperatures to expect is to do some handcalculations (the equations are non-linear and will require iterative solution methods using a spreadsheet program, math program, or the like). If the problem is critical, get an experienced thermal analyst to look at the design with a CFD (computational fluid dynamics) code. He will need to be empowered to get the information he needs to do a reasonable simulation so he/she will need funding to determine (by research and test) emittance and relectance (absorptance=1=reflectance) values of candidate surface finishes for the nearby objects.
It sounds like you may need the help of a thermal guy. Good luck!
An incandescent light bulb emits energy primarily in the ultraviolet (UV), infrared (IR), and visible portions of the electromagnetic spectrum. The portion of the total energy in the UV region amounts to practically nothing. The portion of the total energy in the visible portion is typically less than 20%. The portion of the total energy in the IR portion is typically >80%. As for flourescent bulbs, the proportions are tilted further towards the visible and UV portions of the spectrum.
The heat emitted is dependant on the surface temperature which depends on:
A. the temperatures, locations, geometries, and finishes of nearby objects,
B. the surface area of the bulb,
C. the power consumed by the bulb,
D. the emittance of the bulb's surface finish, ideally determined over the visible and IR bands, and
E. altitude (density).
With no air forced over the bulb, then you have natural convection cooling. In that case, the surface temperature will also depend on the bulb's orientation with respect to gravity and the available airspace for air plume development.
With forced air over the bulb, then you have forced convection cooling. In this case, the surface temperature will also depend on the airflow distribution around the bulb's (velocity and temperature).
As you can see, there are many variables and they all play together to influence the surface temperature.
To answer your question more directly, blowing air over the bulb may dramatically lower the bulb's surface temperature. By lowering the surface temperature of the bulb, you may have a dramatic effect on nearby objects. Naturally, the more air you blow over the bulb, the closer the surface temperature comes to the air temperature.
Lastly, by using high emittance/low absorptance finishes on nearby objects and by providing them with good heat sinks, you can keep those object temperatures to a minimum.
The only way to really know what kind of temperatures to expect is to do some handcalculations (the equations are non-linear and will require iterative solution methods using a spreadsheet program, math program, or the like). If the problem is critical, get an experienced thermal analyst to look at the design with a CFD (computational fluid dynamics) code. He will need to be empowered to get the information he needs to do a reasonable simulation so he/she will need funding to determine (by research and test) emittance and relectance (absorptance=1=reflectance) values of candidate surface finishes for the nearby objects.
It sounds like you may need the help of a thermal guy. Good luck!
There are two sources of radiation from a light bulb:
1) The extremely hot filament, passing uneffectedly through the bulb.
2) Re-radiated from the bulb because it is not completely transparent to the filament's radiation. This is obviously more significant for a "frosted" bulb.
Cooling the bulb will, to a first approximation, effect only the second part of the radiation.
You could also crack the bulb from thermal stresses induced by the cooling air.
poetix99 is right about the radiation originating from two places. The filament will emit as a black body source. Some of that radiation will be absorbed by the glass and some will transmit through. The amount absorbed by the glass will then radiate away to the ambient air. I don't know how much is absorbed and reradiated but the end result is the same; practically all of the energy emits as infrared heat.
Make your life simple. Assume that a 100W bulb will provide 100W of heat (IR) to the surrounding surfaces. For radiation calculations, assume the bulb is a point in space (having no size). That assumption makes view factor calculations easier and makes analysis easier in general.
Cracking the bulb is unlikely, however, unless the cooling air is very cold and the bulb is heated and then suddenly exposed to the cold air. I suspect your application will require an environmentally-stable bulb for other reasons anyway. Use of this kind of bulb will make thermal cracking a remote possibility unless, as I said, the cooling air is VERY cold.
Make your life simple. Assume that a 100W bulb will provide 100W of heat (IR) to the surrounding surfaces. For radiation calculations, assume the bulb is a point in space (having no size). That assumption makes view factor calculations easier and makes analysis easier in general.
Cracking the bulb is unlikely, however, unless the cooling air is very cold and the bulb is heated and then suddenly exposed to the cold air. I suspect your application will require an environmentally-stable bulb for other reasons anyway. Use of this kind of bulb will make thermal cracking a remote possibility unless, as I said, the cooling air is VERY cold.
TB-
The answer to your question is YES. It stands to reason (and calculation) that moving air reduces the temperature of the bulb, directly at the outer glass portion and indirectly at the filament.
Without running any numbers, my educated guess is that you can reduce the temperature of the bulb by one half using forced convection.
The answer to your question is YES. It stands to reason (and calculation) that moving air reduces the temperature of the bulb, directly at the outer glass portion and indirectly at the filament.
Without running any numbers, my educated guess is that you can reduce the temperature of the bulb by one half using forced convection.
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- #11
TomBrooklyn
Specifier/Regulator
What I'm looking to do is reduce the heat but not the light.
You WILL be reducing the heat because the moving air cools the surface of the bulb and other surfaces it moves over. My assumption all along is that the air is permanently moved away from the bulb. If you are talking about circulating the air in a closed volume, then that's a different matter.
If the tranmissivity of the bulb is high (as, for example, with CLEAR glass), the filament will directly radiate that transmitted amount of the energy directly onto the "nearby object", unchanged by the amount of air that is blown around the room.
Yes, the "direct radiation" is also (slightly) absorbed by the intervening air, which is then heated (slightly).
Yes, there will be convective heat transfer from the hot bulb to the surrounding air. This air will become quite hot in an enclosed space. (Does anybody else remember the "Easy Bake Ovens" like the one my sister had? Just one small light bulb was the heat source.)
Forced convective cooling of the bulb surface will lower the bulb temperature, but unless the situation involves an enclosed space with bulb and "nearby object", my hunch is that the convective heating is nil compared to the radiative heating (which depends greatly, as noted by others, upon the characteristics of the surface of the object). All of this depends to some degree upon geometry (as noted by others).
Yes, the "direct radiation" is also (slightly) absorbed by the intervening air, which is then heated (slightly).
Yes, there will be convective heat transfer from the hot bulb to the surrounding air. This air will become quite hot in an enclosed space. (Does anybody else remember the "Easy Bake Ovens" like the one my sister had? Just one small light bulb was the heat source.)
Forced convective cooling of the bulb surface will lower the bulb temperature, but unless the situation involves an enclosed space with bulb and "nearby object", my hunch is that the convective heating is nil compared to the radiative heating (which depends greatly, as noted by others, upon the characteristics of the surface of the object). All of this depends to some degree upon geometry (as noted by others).
TomBrooklyn,
Sounds like you need to look at efficiency rather than heat output by itself.
In buildings, the lamp heat is actually factored into the HVAC for the building, by providing re-heat for humidity control.
All of your light fixture suppliers have or should have access to the data you want.
The air over the fixture does not alter the total heat released into your enclosure for a given wattage input.
Sounds like you need to look at efficiency rather than heat output by itself.
In buildings, the lamp heat is actually factored into the HVAC for the building, by providing re-heat for humidity control.
All of your light fixture suppliers have or should have access to the data you want.
The air over the fixture does not alter the total heat released into your enclosure for a given wattage input.
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- #15
TomBrooklyn
Specifier/Regulator
The light I want to use is a High Pressure Sodium lamp, which is about as efficient as lights get. I don't want to use flourescent as they do not put out enough light in a point source.
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