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I'm building a sublimation heat press. I have a 1000x250x8 mm sheet of cast aluminum. I will be using a thermostatically controlled silicon heat mat to regulate the heat up to a maximum 392 degrees Fahrenheit (200°C). I want to be able to reach my maximum heat in under 3 minutes. Can someone please tell me how much wattage the silicon mat needs to be?

Transistor
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tonycarr
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  • The last time I was looking around for this stuff, Tempco (https://www.tempco.com/) was a convenient source: https://www.tempco.com/Tempco/Resources/16-Engineering-Data/HeatRequirementCalcs.pdf Also: https://www.tempco.com/Tempco/Resources/Engineering-Data/Wattage-and-Watt-Density-Information/Silicone-Rubber-Heater-Wattage-and-Watt-Density-Information.htm – Chris Knudsen Jan 08 '24 at 15:26
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    Heating that volume of aluminum from 70F to 392F takes 780kJ. If you want to do it in 3 minutes, that means a minimum of 4334W. This is a lower bound and assumes no losses of any kind. – vir Jan 08 '24 at 15:28
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    Silicon or silicone? – winny Jan 08 '24 at 15:30
  • @winny pretty sure silicone is intended. – Arsenal Jan 08 '24 at 15:36
  • Welcome to EE.SE! In the future, please try to think of a more descriptive question title. "I'm building a sublimation heat press" is not a question, and does not provide a clue what the question is really about. – marcelm Jan 08 '24 at 16:09
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    I spent time measuring the emissivity of aluminum (FAB work) and found values from about 0.15 for a non-oxidized specular surface (hard to get) to about 0.4-ish for a specular sheet that's oxidized, which this one *will* be. That's in open-air, so to speak. (Inside a integrating sphere it is obviously then just 1. A non-specular sheet will have higher emissivity than 0.15 to 0.4 range.) At the peak temperature of 200 C, Stefan-Boltzmann says about 525 W to 1300 W net radiation/absorption (including external environment temp acting on sheet, as well.) That's only the radiation part. – periblepsis Jan 08 '24 at 16:46
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    Aside from radiation losses, a press will have some mounting points for the platen. You can use (relatively) thermally insulating materials such as ceramic, G10 or titanium but there will still likely be substantial losses. Though at only 3 minutes to 200°C, I suspect a majority of the energy will go into heating the alumin[i]um. – Spehro Pefhany Jan 08 '24 at 17:00
  • Everyone is assuming, quite logically, that you're coming up from room temp. Can you idle at an intermediate temperatures? If not, perhaps because it needs to be handled, you'll also have to consider how long it take to cool down after a heating cycle. – Chris H Jan 09 '24 at 09:03
  • Also, can you reduce the thermal mass of the plate? Obviously just making it half as thick it would probably be too flexible, but half as thick with a supporting frame backing it, (slightly) thermally decoupled from the mat and plate, would do you a lot of good – Chris H Jan 09 '24 at 09:05
  • The answer to this question is found the FAQ of this company: https://www.tempco.com/Tempco/Resources/Engineering-Data/Wattage-and-Watt-Density-Information/Silicone-Rubber-Heater-Wattage-and-Watt-Density-Information.htm – D Duck Jan 09 '24 at 19:50
  • Insulating the back and sides of the aluminum plate can reduce the radiated heat a lot and potentially reduce your heating requirements. Obviously, the insulation is over the heating element also and not between the aluminum and the heater. – user4574 Jan 10 '24 at 01:46

3 Answers3

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Switching to Celsius (so that it's relevant to those of us not in the colonies), 392°F = 200°C (which makes it look as you converted the other way!).

Mass of aluminium (2700 kg/m3) = L × W × T × ρ = 1 × 0.25 × 0.008 × 2700 = 5.4 kg.

Area of aluminium = L × W = 0.25 m2.

$$ P = \frac {m \times SHC \times \Delta T} t $$

where SHC of aluminium is 902 J/kg/K, ΔT = 200°C - 20°C = 180°C and t = 3 minutes = 180 s. $$ P = \frac {5.4 \times 902 \times 180} {180} = 4.87 \ \text {kW} $$

Power density of the silicone mat (assuming full coverage):

$$ PD = \frac P A = \frac {4.87} {0.25} = 19.5 \ \mathrm {kW/m^2} = 1.95 \ \mathrm {W/cm^2} $$

jonathanjo
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Transistor
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Recommended watt density for silicone heaters is in the 2.5 to 5 watts per square inch range.

If we use vir's number of approximately 5kW (which are based on heat capacity of the metal ignoring losses) and your dimensions of 390 square inches area you'll see that represents almost 13W/in^2, so it's unlikely a conventional silicone heater can reliably do the job in the time required. Something like tubular heaters (as in an oven) would be able to do it.

Spehro Pefhany
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You're quite likely going to be out of spec for your heater mat, not just on power but on its own maximum temperature.

Recalling that heat flow is proportional to ΔT, and knowing that the element within the mat isn't in brilliant thermal contact to the plate (silicone isn't a very good thermal conductor and the interface won't be perfect), for the top end of your rapid heating requirement, the mat will have to be considerably over 200°C. Most mats aren't rated beyond 200°C continuous, and those that are seem to have a lower power output. I've recently designed a sample oven pushing the limits of my components (200°C maximum target, wire-wound resistors rated to 200°C) with good thermal contact between my resistors and my sample mount. That wasn't production but a one-off experiment, so pushing the heaters to 210°C (measured with a thermal camera) was acceptable. But even that slight deviation from specifications requires slow heating to the maximum.

You can buy small metal-bodied heater elements with a considerably higher maximum temperature rating and better thermal contact to the plate. You may think these would make for uneven heating but spreading plenty of small ones out, with such a thick heat spreader plate, it won't be any more of an issue than with a heater mat (that won't get the edges as hot as the middle because of heat losses). They will also allow you to work around the mounting points better than a mat would.

Chris H
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