koti Posted January 17, 2018 Share Posted January 17, 2018 I'm loosely considering different materials for flashlight building and I would like to understand more about the physics of heat conductivity and convection. Most mainstream flashlights are built from aluminium, the more expensive ones for collectors are copper, titanium and rarely some exotic alloys. Copper being the best conductor of heat (price wise) is often used as a shell of high end flashlights which produce high amounts of heat. My question is this...comparing Lead and Aluminium (for ease of discussion, I'm not actually considering building a flashlight out of lead). Lead is roughly 5 times denser than Aluminium and is also roughly 6 times less thermally conductive than Aluminum. These are not the exact numbers as you can see by the numbers in the links I provided but lets keep them like that for ease of the experiment. Say we have a plate of Lead weighing 1kg and a plate of Aluminum weighing also 1kg where both plates have the same thickness of say 1cm - lets assume they are square shaped. Aluminum plate will have roughly 5 times the surface area of the Lead plate (right?) If I solder the same LED emitter onto each plate and provide same, steady current to them: 1. Which plate will heat up quicker to its final, maxed out temperature (same?) 2. Which plate will keep the LED emitter cooler for longest? 3. Which plate will transfer more heat to the ambient air around due to convection and at which point in time for each plate? Link to comment Share on other sites More sharing options...
koti Posted January 24, 2018 Author Share Posted January 24, 2018 1 week and not a single reply...could someone get me on the right track so I have the units correct when digging into this? I should be able to have the thermal behaviour of the emitters from the datasheets. Should I take into account something else besides what I included in the OP? Link to comment Share on other sites More sharing options...
Bender Posted January 25, 2018 Share Posted January 25, 2018 I don't have time to work out the details, but here are some hints: - Heat transfer is proportional to surface area. - heat transfer is proportional to temperature difference - materials with a good conductivity will have a more uniform temperature - at equilibrium, energy input equals energy output - for the transient behaviour, you need to take the specific heat into account. - convection depends on the orientation of the plate and on anything that can obstruct or guide air flow - transient heat transfer problems are very complex without finite element software. A decent approximation of the equilibrium situation can probably be reached by asuming uniform temperature. Link to comment Share on other sites More sharing options...
studiot Posted January 25, 2018 Share Posted January 25, 2018 (edited) Hi Koti, this is an engineering problem. Are you familiar with the electrical/electronic engineering term 'Thermal Resistance' ? The answer is not as simple as a Physicist might offer since there are several bottlenecks or restrictions in the heat path. This is rather like a series of valves or taps in the flowline of a fluid. This is why manufacturers publish values of thermal resistance, which are added up. For your heat source the manufacturer will publish the controlling figure for the controlling bottleneck . This is the junction to case resistance of the LED source. You then have the case to mounting plate resistance Then the mounting plate to heatsink resistance Then the heatsink to ambient resistance which will depend upon the colour and orientation of the heatsink plate. Good values for all of these are available, though I doubt that there would be any for a lead heatsink. Formerly these were empirically measured, but modern finite element techniques provide very close estimates. Sorry I no longer have access to such software. Here are some references, let me know if you want more help. https://www.sparkfun.com/tutorials/314 https://www.electronics-cooling.com/1995/06/how-to-select-a-heat-sink/ https://www.electronics-cooling.com/1995/06/how-to-select-a-heat-sink/ Edited January 25, 2018 by studiot Link to comment Share on other sites More sharing options...
MigL Posted January 27, 2018 Share Posted January 27, 2018 First rule of heat sinks... Highest thermal conductivity and largest possible surface area. And who builds their own flashlights ? Link to comment Share on other sites More sharing options...
studiot Posted January 27, 2018 Share Posted January 27, 2018 (edited) A heat sink the size of a football pitch composed of the highest possible thermal conductivity would be of no help if you can't get the heat into it. This is the same as saying that connecting a 6 foot diameter pipe to the end of my hosepipe will not get me any more water through if the faucet is turned right down. Edited January 27, 2018 by studiot Link to comment Share on other sites More sharing options...
Sensei Posted January 27, 2018 Share Posted January 27, 2018 (edited) On 25.01.2018 at 12:50 AM, koti said: Should I take into account something else besides what I included in the OP? What will be source of heat? Just light from LEDs.. ? How many LEDs will be installed.. ? What will be their total power? How about thinking also about the most reflective material for surface (even tiny layer of it (electroplating?), if it's expensive).. ? Instead of bothering about heat, reflect photons, as much as you can (highly polished surface?) You have here table of reflectance coefficient per wavelength for couple metals Au, Ag and Al. https://en.wikipedia.org/wiki/Reflectance Search for and check materials reflectance table Edited January 27, 2018 by Sensei Link to comment Share on other sites More sharing options...
studiot Posted January 27, 2018 Share Posted January 27, 2018 Sensei, koti is building a flashlight (torch). Even LED light sources are less than 50% efficient at conversion. The rest of the energy is dissipated as heat. Link to comment Share on other sites More sharing options...
John Cuthber Posted January 27, 2018 Share Posted January 27, 2018 On 1/17/2018 at 8:35 PM, koti said: 1. Which plate will heat up quicker to its final, maxed out temperature (same?) As far as I can tell, nobody has yet mentioned the property you need to work that out. https://en.wikipedia.org/wiki/Thermal_diffusivity Link to comment Share on other sites More sharing options...
studiot Posted January 27, 2018 Share Posted January 27, 2018 3 minutes ago, John Cuthber said: As far as I can tell, nobody has yet mentioned the property you need to work that out. https://en.wikipedia.org/wiki/Thermal_diffusivity Actually that is contained in the electrical engineer's version of themal resistance I mentioned, as is the emissivity of the surface and other (heat) transport parameters. Link to comment Share on other sites More sharing options...
John Cuthber Posted January 27, 2018 Share Posted January 27, 2018 Resistance (Thermal or otherwise) is a property of a thing. Resistivity is a property of a material. The discussion started off with a collection of different metals etc. You can alter the resistance by changing the design of the torch- for example, by putting fins on it. But that's a different story. Link to comment Share on other sites More sharing options...
studiot Posted January 27, 2018 Share Posted January 27, 2018 47 minutes ago, John Cuthber said: Resistance (Thermal or otherwise) is a property of a thing. Resistivity is a property of a material. The discussion started off with a collection of different metals etc. You can alter the resistance by changing the design of the torch- for example, by putting fins on it. But that's a different story. Yes that is true and the OP gave us a specific size and shape for the thing so it has a particular resistance. To investigate the effect of varying that (plate) size, standard tables of thermal resistance are available for different sized plates, based on actual measurements. Link to comment Share on other sites More sharing options...
koti Posted February 11, 2018 Author Share Posted February 11, 2018 (edited) First off I'd like to apologize, I opened a thread then complained that nobody answers and when quite a few people answered I sat quiet for 2 weeks. I'm treating this as a thought experiment, I'm not going to buy 1kg plates of Lead and Aluminium, it's enough that I have to buy Copper all the time for parts which need good heat transfer and times are thin. On 25.01.2018 at 9:37 PM, studiot said: Hi Koti, this is an engineering problem. Are you familiar with the electrical/electronic engineering term 'Thermal Resistance' ? The answer is not as simple as a Physicist might offer since there are several bottlenecks or restrictions in the heat path. This is rather like a series of valves or taps in the flowline of a fluid. This is why manufacturers publish values of thermal resistance, which are added up. For your heat source the manufacturer will publish the controlling figure for the controlling bottleneck . This is the junction to case resistance of the LED source. You then have the case to mounting plate resistance Then the mounting plate to heatsink resistance Then the heatsink to ambient resistance which will depend upon the colour and orientation of the heatsink plate. Good values for all of these are available, though I doubt that there would be any for a lead heatsink. Formerly these were empirically measured, but modern finite element techniques provide very close estimates. Sorry I no longer have access to such software. Here are some references, let me know if you want more help. https://www.sparkfun.com/tutorials/314 https://www.electronics-cooling.com/1995/06/how-to-select-a-heat-sink/ https://www.electronics-cooling.com/1995/06/how-to-select-a-heat-sink/ I am familiar with thermal resistance ofcourse, here's a datasheet of one of the emitters which I often used in my builds last year:http://www.cree.com/led-components/media/documents/ds-XPL.pdf As you can see, this particular emitter has a thermal resistance of 2.2 °C/W @ 3000 mA. The thing is that I overdrive these emitters heavily, and by heavily I mean 200%-250% which is nuts heat-wise but above a certain level of amperage the lumen surplus starts to grow very slowly so I push these emitters to their limit to get the max output. I mainly build small, pocketable flashlights (single and multi-emitter) so these things get hot and they get hot fast. I do this only in my own builds - the torches I build for people are less overkill. Thanks for the links, I will find use for them. On 27.01.2018 at 9:33 PM, MigL said: First rule of heat sinks... Highest thermal conductivity and largest possible surface area. And who builds their own flashlights ? This explains my condition in short: I buy lights to mod them to sell them to buy more lights to mod - I'm a junkie. On 27.01.2018 at 10:10 PM, Sensei said: What will be source of heat? Just light from LEDs.. ? How many LEDs will be installed.. ? What will be their total power? How about thinking also about the most reflective material for surface (even tiny layer of it (electroplating?), if it's expensive).. ? Instead of bothering about heat, reflect photons, as much as you can (highly polished surface?) You have here table of reflectance coefficient per wavelength for couple metals Au, Ag and Al. https://en.wikipedia.org/wiki/Reflectance Search for and check materials reflectance table The source are mainly CREE and Nichia LED emitters, varying from model they are stock between 3,5v-12v and 3A-20A amperage. These tiny ~10m^2 emitters give out ridiculous amounts of heat when driven high. I build single, triple and quad emitter builds from these. The higher amp emitters are mainly single emitter builds. Electroplating (especially Ag) is the holy grail of high-end torch building, I haven't had a chance to do my own project yet (cost). In small form factor pocketable torches the reflection quality factor is not significant, the emitter output is what does the job. In larger reflectors it becomes a very significant factor ofcourse. I use TiR optics as well and aspheric lenses instead of reflectors for more concentrated beam (more throw instead a wall of light) On 27.01.2018 at 10:31 PM, John Cuthber said: As far as I can tell, nobody has yet mentioned the property you need to work that out. https://en.wikipedia.org/wiki/Thermal_diffusivity After reading through the above it struck me that in order to get any decent heatsinking increase (time-wise) on one of my very small, high amp torches I will need to increase the material volume significantly - make the whole thing a lot thicker. This way I'll at least keep the emitter cooler for a few seconds more before it heats up the torch body. Fins are a thing for larger torches. I will try to work out the calculations on my own, thanks. Edit: A CNC maschine and a few hundred kg’s of copper for trial and error is what I need I guess Edited February 11, 2018 by koti Link to comment Share on other sites More sharing options...
John Cuthber Posted February 11, 2018 Share Posted February 11, 2018 9 hours ago, koti said: Edit: A CNC maschine and a few hundred kg’s of copper for trial and error is what I need I guess Silver is a slightly better conductor of heat (and has a much higher thermal diffusivity)... Link to comment Share on other sites More sharing options...
Bender Posted February 12, 2018 Share Posted February 12, 2018 For a mobile application such as flashlights, the much lighter aluminium seems the better choice. It has the additional benefit of being much cheaper. For the same heat resistance, its mass is about half that of copper. Link to comment Share on other sites More sharing options...
koti Posted February 13, 2018 Author Share Posted February 13, 2018 5 hours ago, Bender said: For a mobile application such as flashlights, the much lighter aluminium seems the better choice. It has the additional benefit of being much cheaper. For the same heat resistance, its mass is about half that of copper. Yes, majority of flashlights are made from HAIII aluminium. The more expensive ones are copper, titanium or ceramic. Link to comment Share on other sites More sharing options...
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