How to do a thermal calculation? - Page 1
Forum section: electronics / general

author: Analog Kid
author_rank: Super Contributor
timestamp: January 23, 2025, 05:05:20 pm
content: Hopefully not too-too off-topic here. Was thinking of posting this in Thermal Imaging, but it's not an imaging topic.OK: I know almost nothing about how to do any kind of thermal calculations. What I'm trying to figure out is how much heat I need to heat a small enclosed space.The space is a cold frame I made for starting seedlings. It encloses a volume of ~3.9 ft.3. Poorly insulated, with thin polyethylene sheeting on all sides.I got a small cup warmer which I stuck inside this space. It says it produces 18 watts, so let's take that at more or less face value.My question is what temperature I can expect to get from this warmer. There's an aluminum measuring cup about half full of water sitting on top of it. So far the space is getting warm, but not very: only about 2°F above ambient (which is ~63°).I was hoping for more like 75° to encourage my seeds to sprout. Don't think this warmer is going to get there.Is there any simple (hopefully) formula I can use to figure 1) what temperature increase I can expect from this li'l cup warmer, but more importantly 2) how much energy input (in watts) I'd need to achieve my target temperature? (Assuming some heat loss)Hopefully this discussion won't get too far into the weeds with all kinds of extraneous factors, quantum effects, complicated formulae, etc., but if it does, it does.

author: Ian.M
author_rank: Super Contributor
timestamp: January 23, 2025, 05:19:55 pm
content: First assumption: heat flow is proportional to temperature difference. This is *NOT* always true, especially when significant natural convection occurs. Ref:https://en.wikipedia.org/wiki/Newton%27s_law_of_coolingYou have measured 2° F temperature rise with 18W of heat input. You need 12° F temperature rise to reach your target temperature. Therefore, ignoring the uncertainties in your measurement and heat input, you need (at least) 108W of heater power.However the coffee warmer may use a PTC element or a thermostat to limit the max. temperature it can reach so unless the container of liquid on it is large enough for it to stay well below that temperature, you may not have 18W of average actual heat input. You need to measure the power it draws, preferably with an energy meter over a minimum of an hour period to find the actual heat input, then put cold water in the cup and see if the power draw temporarily increases.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 23, 2025, 05:24:04 pm
content: Quote from: Ian.M on January 23, 2025, 05:19:55 pmSo it's linear? If so, thanks; that was easy.QuoteRight-o. Now to cobble up something using a multimeter ...

author: Ian.M
author_rank: Super Contributor
timestamp: January 23, 2025, 05:38:37 pm
content: No. The ASSUMPION is its linear, but as even Newton knew natural convection was nonlinear when he published his results, that's only a first approximation that needs to be confirmed in practice. Also note that thin polyethylene film is near transparent to IR so if your heater surface temperature is high and there is line of sight to outside your cold frame, you may be loosing a significant amount of heat as IR radiation.If you can rig a bigger heater e.g. a 60W or 100W incandescent bulb in a large inverted tin can to trap light and shortwave IR, with holes punched in what was the bottom for air flow*you will get a greater temperature rise, that's easier to measure accurately and will get a second point on your graph to check linearity. Halve the input power with a diode#(assuming an AC supply to the bulb) and get a third point and you should have a fairly good idea if Newton's Law of Cooling is applicable.* An ordinary incandescent bulb is less than 5% efficient (in terms of visible light output vs electrical power) so if you trap most of the light without overheating its envelope or base, makes a cheap >99% efficient heater!# You will get a little more than half power on half-wave rectified AC as the filament will run cooler so will have a bit lower resistance.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 23, 2025, 06:07:05 pm
content: That (light bulb in a can) will be my backup scheme.Other possibility is a small heating mat.

author: IanB
author_rank: Super Contributor
timestamp: January 23, 2025, 06:41:31 pm
content: There are two sides to a heat equation: there is insulation, and there is heat source. If the insulation side of the equation is poor, then the heat requirement will be disproportionate.So consider whether poly sheeting is adequate? Maybe have two layers with an air gap between them? Possibly consider glass instead. Make sure there are no gaps where drafts can get through. Just as in a house, drafts will let heat out.Only when you have a nice, solid, well insulated enclosure, then consider heat sources.With a well designed enclosure, solar heating can give a temperature rise of 10-20 degrees. Larger enclosures are probably better than smaller ones in this regard.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 23, 2025, 06:44:49 pm
content: I have two layers of very thin poly, so some air between layers. No plans for further insulation or glass; this is a non-optimized makeshift project. I think an incandescent bulb would supply more than enough heat. (They use them in chicken coops in winter.)

author: Siwastaja
author_rank: Super Contributor
timestamp: January 23, 2025, 07:17:57 pm
content: Linear heat loss as predicted by thermal conduction value ("lambda") of the material * temperature difference is good enough first-order approximation, it gets you in the right ballpark.Air-gapped two sheets is much better than no insulation of any kind. If possible, partitioning the gap even to just a few separate segments by making horizontal "air stoppers", like just a piece of tape holding the two sheets together, improves insulation while still keeping the solution cheap and makeshifty enough. But of course it would be nearly impossible to know thermal conduction value for such material, so you need to test. Try 0.2-0.3 W/(mK) for starters, such that you use the total thickness including the air gap in the calculation.

author: Ice-Tea
author_rank: Super Contributor
timestamp: January 23, 2025, 07:36:23 pm
content: How to do a thermal calculation? Start by understanding it's just electricity but with other units. Same laws and formulae apply.* if you make abstraction of the earlier mentioned non-linearity ;-)

author: IanB
author_rank: Super Contributor
timestamp: January 23, 2025, 07:55:02 pm
content: Quote from: Siwastaja on January 23, 2025, 07:17:57 pmThis actually gives very poor predictions in common situations. The result can be off by an order of magnitude.With thin layers of conductive material like window panes or poly sheets, the insulation value is dominated by the so-called boundary layer of air at the surface. Engineering heat transfer calculations will often use an overall heat transfer coefficient "U" for conduction between the inside space and the outside space. Values of U can be looked up in various places*, with units given either in W/m²/K or Btu/hr/ft²/°F. The equation is Q = UAΔT, with Q = heat transfer rate, A = area, ΔT = temperature difference. A typical U value for a single glazed window in a house is ~6 W/m²/K or ~1 Btu/hr/ft²/°F. If you are doing calculations with electric heating, it is obviously more convenient to use metric units.Example calculation: U = 5 W/m²/K, A = 4 m², inside temperature = 20°C, outside temperature = 5°C. Heat loss = 5 x 4 x (20 − 5) = 300 W.*For U values in Btu/hr/ft²/°F, see for example:https://www.greenhousecatalog.com/greenhouse-insulation

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 23, 2025, 08:06:46 pm
content: it takes a lot of heat to significantly warm a polyethylene enclosure. This is empirical knowledge. Now on to the theory.There are three kinds of heat transfer in any heat flow problem. Radiation, conduction and convection. Radiation and conduction are straightforward mathematically. Convection not so much. There are two types of convection - forced and thermal (think fans and hot air rising).But for first order approximations you can make simplifying assumptions.Assume that convection keeps the air inside your cloche isothermal. Similarly assume that convection keeps the air outside your enclosure isothermal. In this situation these are relatively good approximations. This approximation is only good if your enclosure is leak tight. If the breeze blows through this type of convection will dominate the heat loss. Tight means something on the order of a couple of air exchanges per day or so.So now you have two terms. Conduction and radiation.Conduction between two isothermal regions is proportional to the area of the conducting region divided by its thickness and multiplied by the conductivity of the material. This condition applies for the conduction through the polyethylene walls and roof of the enclosure. The structure can normally be neglected. When you are looking up conductivity make sure the units are appropriate. You will find several variants out there.Unfortunately there will be significant heat loss to the ground. That isn't all bad, you want your seeds to get warm. But the thermal equation is a bit more complex here, unless you do something more exotic like putting a Styrofoam insulating layer under your seed bed. In which case you can neglect the losses to the ground if the foam is thick enough. I would estimate something in the 2-4 inch (5-10 cm) class.Radiation is also going to be significant, and a bit complex. First, it is proportional to the difference of the fourth power of the absolute temperature of the radiation source and the fourth power of the destination. The polyethylene is effectively transparent to thermal radiation so the temperature will be the sky which can vary from around 273 K on clear winter nights up to over 300 K on cloudy nights. There are radiation losses to the ground also. But it is typically warmer than a clear winter night. There are two radiation sources, your heater and the ground. Both are significant. There is a simple fix for this. Your seeds don't need light until they germinate. If you cover you cloche with aluminized mylar the heat is reflected back in and reduces these losses to near zero. One of the most common ways to get this stuff is in the space blankets sold by camping suppliers. The radiation control is why these things are effective.All of this can get you to the steady state temperature, but doesn't tell you how long it will take to get there. For that you need a time domain model that has heat capacity of the important items. In this case the soil mass will be dominant.

author: IanB
author_rank: Super Contributor
timestamp: January 23, 2025, 08:18:46 pm
content: IMHO it is always best to do this stuff by experimentation.Put a remote sensing thermometer in the enclosure, monitor the temperature over day and night. If it doesn't get warm enough, insulate it better and/or add more heating. Adjust until desired result is obtained.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 23, 2025, 08:30:25 pm
content: Quote from: IanB on January 23, 2025, 08:18:46 pmBingo!You win the prize.(The thermometer that's in there came from the late, great All Electronics. Shoulda ordered more of those.)

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 24, 2025, 04:09:41 am
content: Quote from: IanB on January 23, 2025, 08:18:46 pmTests are the right way to go for answers. At least answers that can be trusted. Theory and analysis helps in figuring what tests to run, and what to do when you don't like the results. And sometimes can save time and money when the concept just isn't practical.Two examples to illustrate. Smudge pots were once commonly used in citrus and other orchards to ward off frost damage. Smudge parts are oil burners that were placed every few meters in the orchard.But the heat of burning oil wasn't why they worked. It was the thick smoke that limited radiation losses from the ground. So generating more heat with a smokeless burner that utilizes all of the stored energy in the fuel doesn't work. Another strategy used in some orchards was placing large fans blowing horizontally across the tops of the trees. Only works when there are local cold zones where colder air settles, and works be bringing slightly warmer air from nearby.The other good source is to look at what the people that have been solving the problems for years are doing. Greenhouses, commercial growers, etc. Can give you some idea of what you will need. If you haven't seen your answer in operation there is a good chance that has been tried and rejected.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 25, 2025, 10:54:53 pm
content: I've got a rig being tested. Bought a big can of V-8 vegetable juice at the market ($3.49, and I get to drink the juice), punched holes in the top, made some cutouts at the bottom, screwed a lamp socket to a piece of wood and put a cord on it. (Bulb is burning base-down.)I'm starting with a 25-watt flame-tip bulb. If that doesn't give enough heat the next step is a 40-watt bulb.Later I'll spray the inside of the can flat black.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 25, 2025, 11:19:24 pm
content: Since you want to go the experimental route, do some more measurements to inform the next steps of your journey. In addition to the air temperature in your enclosure, measure the bulb temperature (tough measurement but at least you can estimate by how long you can touch it steady state, open air) the temperature of the V-8 can, and the air temperature. You might also want to grab some measurements to understand the variables you may need to control, or at least understand. Wind speed and air temperature difference with known area openings in your enclosure.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 25, 2025, 11:21:13 pm
content: I'll do what I can and report back here.

author: IanB
author_rank: Super Contributor
timestamp: January 26, 2025, 02:06:19 am
content: Quote from: Analog Kid on January 25, 2025, 10:54:53 pmSpray the outside too (more so the outside than the inside). You want the heat to radiate inside the enclosure and not be trapped inside the can. Spraying the inside will make little difference, because the heat is already trapped inside the can no matter what you do to the inside surfaces.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 26, 2025, 02:10:06 am
content: Righto.Although wouldn't painting the inside black help absorb some of the light and convert it into heat?Probly not much, I suppose, but something?

author: IanB
author_rank: Super Contributor
timestamp: January 26, 2025, 03:49:45 am
content: Quote from: Analog Kid on January 26, 2025, 02:10:06 amThe light has nowhere to go except through any holes you might have made in the walls of the can. So even with a mirror finish inside all the light will be absorbed eventually, it just depends how many times it bounces off the walls before it all gets absorbed.If you put an incandescent lamp inside an enclosed space, the enclosed space will get hot, which will shorten the life of the bulb. Painting the outside of the space black will help it stay cooler, which will improve the bulb life.

author: Haenk
author_rank: Super Contributor
timestamp: January 26, 2025, 08:01:47 am
content: Quote from: IanB on January 23, 2025, 08:18:46 pmI had courses on thermodynamics, and "doing it right" is extremely complex. Even simplyfied physical models are a PITA. Therefor I too suggest the experimental approach. Or just use something ready-made: I purchased a PCB preheater for under 30 EUR - this comes with a temperature sensor, several heating elements and an electronica temperature control. Pretty nifty for that little money. Instead of the heating plate maybe install some older PC coolers, maybe a fan as well. No calculations needed, let the regulation due the work.

author: Sorama
author_rank: Frequent Contributor
timestamp: January 26, 2025, 08:30:48 am
content: Quote from: Ian.M on January 23, 2025, 05:38:37 pmSome correction…For long waves (as the inside temperature is low), the polyethylene acts as a mirror, like glass.So line of sight is irrelevant and polyethylene not near transparant for IR.It’s thermal conductivity is around 0,4W/m.°C

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 27, 2025, 12:06:30 am
content: Some empirical data on polyethylene. I don't have a high end thermal camera. Just a cheap handheld bolometer based job. But like all the others it operates in the 8-12 micrometer band which corresponds both to a range of wavelengths where the atmosphere is relatively transparent and the peak wavelengths for radiation from ~300K blackbody.Two pictures. (edit, I can't make the image order come out right. Read the file names). First is the view of my woodstove with nothing between the camera and the scene. The second with a polyethylene bag used to hold zip ties. First thing I grabbed in my shop. The bag was measured with a micrometer to be made of 10 mil (0.25 mm) material, so the image is taken through two 10 mil layers with a small air gap between.My conclusion? There is substantial transmission of heat by radiation through this type of covering. The change in apparent temperature could be just a minor shift in aim, but the change in the image scale indicates there are some losses going through the polyethylene, but it in no way acts like a mirror, reflecting most of the incoming light. Based on the index of refraction of the material and near normal incidence I would expect reflection losses on the order of 10%.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 27, 2025, 12:28:12 am
content: Well, thanks for all that, but my intuitive, seat-of-the-pants conclusion is that polyethylene is a very poor insulator.In this application, it really matters not: the cold frame is indoors, in front of a window.So far the 25W bulb gives about a 6-7° F increase over ambient temperature; not quite enough.I just spray-painted the outside of my can flat black, waiting for it to dry.I'll try that with the 25W bulb, then with a 40W bulb to see if I can get it up to about 75°, at which point I'll be happy.(I'm just trying to germinate some seeds indoors during winter.)

author: IanB
author_rank: Super Contributor
timestamp: January 27, 2025, 01:07:02 am
content: The situation is complicated. Firstly, the IR camera is looking at much higher temperatures in a stove than are found in a cold frame, so the wavelengths being measured are much shorter. That could make a difference? But also, there must be many factors at play in a greenhouse, including blocking the wind and preventing convective heat removal.For instance, Morgan Gold of Gold Shaw Farm on YouTube has a giant polytunnel greenhouse, and it can evidently maintain an internal temperature of 30°F with an outside temperature of −3°F, and with no additional heating than some livestock living inside it. Given the surface area of the walls, they must be doing a fairly decent job of keeping the heat in. Also, I suspect size matters and that a bigger inside volume is more efficient than a smaller volume.Quote from: Analog Kid on January 27, 2025, 12:28:12 amIf indoors, in front of a window, why not just cover the front facing wall with the poly film, and the other walls with polystyrene insulating sheets? If they are not facing the window, they do not need to let light through.QuotePainting the can black won't change the heat equation, since all heat from the bulb must escape no matter what. It has nowhere else to go. The idea of painting walls black is to help the bulb run cooler, since enclosed bulbs tend to burn out rather quickly by overheating.Also, if you are trying to germinate some seeds, wouldn't it be better to heat the soil from underneath with a heating mat, like one of these?https://www.amazon.com/Seedling-Starting-Waterproof-Heating-Germination/dp/B0B82J9JD6Here's the Gold Shaw Farm illustration. Start at the beginning for the outside temperature, and jump to 12 minutes in for the inside temperature.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 27, 2025, 01:35:26 am
content: Quote from: IanB on January 27, 2025, 01:07:02 amThat's the alternative, but I'm trying for a $cheaper$ solution first to see if it'll even work.Now the whole house smells like spray paint, as it bakes with the bulb on inside it ...

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 27, 2025, 03:17:34 am
content: Quote from: IanB on January 27, 2025, 01:07:02 amAgree on the complexity. I was merely supporting your assertion that thin poly films have little absorbtion and contradicting the post asserting that they are near perfect mirrors. There is a lot of published data on poly transmission, often missing key points like exact chemistry and sample thickness. But all assert high transmission in the 8-12 band. The pictures communicate better to someone who rejects theory in favor of measurements. In my pictures the distance from camera to heat source is short enough to limit atmospheric transmission. This is definitely not true for an outdoor enclosure radiating to the sky. It is not clear to me what the spectral sensitivity of these cheap bolometer cameras are. In my professional background bandpass filters were often employed to avoid the potentially varying background from wavelengths that wouldn't make it through the atmosphere anyway. But some ads for these cameras mention sensitivity from 5-12 micrometers. Who knows what any specific camera does.All of that matters much more in an outdoor greenhouse. As does wind proofing. It is always nice when the whole problem is described so that only the pertinent answers are generated.For another cheap solution to OPs problem, go to a thrift store and find a used electric blanket. Covered with a polyethylene sheet and placed under the seedbed it should work perfectly for this application.

author: Sorama
author_rank: Frequent Contributor
timestamp: January 27, 2025, 07:53:01 am
content: It seems no one really knows how to use an ir camera here.@catalinawow,Did you determine the emission of that bag?Did you set the camera according to this emission?Did you measure and set the ambient temperature on the camera?Die you measure and set T reflected on the camera?No?Then how the hell can you say what you said?Thermography is more than just holding a camera on front of an object…

author: Sorama
author_rank: Frequent Contributor
timestamp: January 27, 2025, 08:11:31 am
content: Also;Atmospheric ‘transmission’ is never an issue in the 8-12 micrometer spectrum and certainly has no impact on the distance , not even with dozens of meters in between.Just take a decent Ir camera and set the distance between you and the object way out of order and you will notice it won’t change the temperature.So what you mention about the outdoor box is simply incorrect.The only difference when measuring outdoors is that the T reflected plays a major role in correctly measuring the temperature by using thermography.

author: Sorama
author_rank: Frequent Contributor
timestamp: January 27, 2025, 08:16:46 am
content: Quote from: Analog Kid on January 27, 2025, 01:35:26 amPainting it dark is useless.The absorption/reflection of an object does not change if there is no sunlight involved.Only in the shorter IR waves -which are present in sunlight- the color is relevant.If these seeds are put outside, you want to prevent very low temperatures of the sky at night.So you need to put something reflective (like alu paper) on the inside of outside of that box. that will reflect the low temperatures radiation at night.But it will also prevent heating during daytime, something you want to happen probably.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 27, 2025, 09:48:02 am
content: Quote from: Sorama on January 27, 2025, 07:53:01 amYou perhaps will notice the extremely limited claims I made about what this test showed. Substantial transmission. No claims beyond saying that it wasn't a mirror. Which I will agree is technically incorrect since I am quite sure there is some reflection.Perhaps you also will notice that I said that I don't own a proper instrument. None of the things you mention, which are absolutely correct for precision quantitative thermography are possible with this viewer. Nevertheless, just as you can use a two and a half digit voltmeter to measure voltage even though the results are not as definitive as a much higher precision instrument, I will stand by the limited claims I made about the result. And would be very interested in seeing results from a "properly" performed measurement that contradict that limited conclusion.

author: Sorama
author_rank: Frequent Contributor
timestamp: January 27, 2025, 10:27:12 am
content: Quote from: CatalinaWOW on January 27, 2025, 09:48:02 amthere is a huge difference between measuring a voltage with a 2,5 digit dmm as the accuracy might very well be be the same as a 6 digit dmm, and doing a quantitative IR measurement.Like I said, the T reflected has a huge impact on the result, just like a proper measured en set emission coefficient will.You will perfectly be able to measure correctly the voltage with this 2,5 digit because you do not need to measure and set parameters that will influence this voltage measurement, while you won't be able to do quantitive IR measurements at all without proper certification and knowledge about how to do quantitative thermography as there are several environmental parameters to be set.Not to mention that most (low cost) IR camera's don't even have the possibility to set these parameters.

author: Someone
author_rank: Super Contributor
timestamp: January 27, 2025, 12:58:49 pm
content: Quote from: Sorama on January 27, 2025, 10:27:12 amWell you started with a pretty bold claim:Quote from: Sorama on January 26, 2025, 08:30:48 am"like a mirror" would imply to most english speakers that you mean extremely high reflectivity and low loss.

author: Sorama
author_rank: Frequent Contributor
timestamp: January 27, 2025, 06:02:23 pm
content: Quote from: Someone on January 27, 2025, 12:58:49 pmTnx for the remark.What I wanted to explain is that a window/glass behaves transparent for sunlight ( so less reflective) and when the inside walls and floor are re-emitting that same heat by means of IR but this time at a way lower temperature (hence more long wave IR) , this glass will reflect most of the IR ( coming from the walls and floor, or its surroundings in general) and thus behaves as a mirror.They call it the ‘orangery’ effect ( hope this is the right translation).It’s not a perfect mirror, but it does reflect more than it absorbs or emits.People often think that glass is transparant for IR, which is not the case when this IR is coming from walls , floor, furniture.But is is way more transparant for IR waves of sunlight!Poly ethylene turns out to have a pretty low emmission, thus a high reflectance.

author: IanB
author_rank: Super Contributor
timestamp: January 27, 2025, 06:20:27 pm
content: Quote from: Sorama on January 27, 2025, 06:02:23 pm(In English, it's called the greenhouse effect.)

author: Sorama
author_rank: Frequent Contributor
timestamp: January 27, 2025, 06:26:44 pm
content: Quote from: IanB on January 27, 2025, 06:20:27 pmok, tnx.We learned something today.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 27, 2025, 09:46:08 pm
content: With a 40-watt bulb, my li'l cold frame is now ~9-10° F above ambient. Good enough for now.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: January 27, 2025, 11:23:09 pm
content: I typed two lengthy entries which both got lost in the electron shuffle, so will just post a short note here.Low emissivity does not imply high reflectivity. Transparent objects have low emissivity. Google and you will find that polyethylene is pretty transparent in the LWIR. My pictures, crude though the instrumentation is, implies that because the image of the stove appears when viewed through the bag. The high temperature of the stove is not particularly important, I have imaged scenes outdoors (ambient temperature approximately 0 C) through the bag also but can't be pfaffed to post them. They do show reduced contrast, the transparency is not perfect and the crumpled bag surfaces don't help either.This does imply that polyethylene would not give much of a greenhouse effect. I haven't really looked into why the greenhouses that Tim Fox references work, but suspect there is something special is going on. A thin film coating on the polyethylene that is reflective in the LWIR for example. Which might also explain the results Sonoma seems to have found. Depends on what samples he is observing.

author: Analog Kid
author_rank: Super Contributor
timestamp: January 28, 2025, 12:11:23 am
content: Possibly useful data point for this discussion:According to a cursor online search, polyethylene sheeting (assuming ~6 mil here) has an R value of 0.85.Not a terribly good insulator.

author: IanB
author_rank: Super Contributor
timestamp: January 28, 2025, 01:05:43 am
content: Quote from: Analog Kid on January 28, 2025, 12:11:23 amIt pains me as an engineer to see resistance values quoted everywhere without a unit of measure.In the USA, the above quoted R value would be 0.85 ft²·°F·h/Btu. But it's not your fault. Everyone is doing it.As to whether or not that is a good insulator, we should compare it to other things. That value is about the same as 4 inch thick brick wall, 3/4 inch thick plywood sheeting, better than a sheet of 1/2 inch drywall, and about the same as a pane of window glass. I would say that is not bad for something that is only 6 mil thick?

author: Sorama
author_rank: Frequent Contributor
timestamp: January 28, 2025, 06:14:45 am
content: Quote from: IanB on January 28, 2025, 01:05:43 amIt is indeed.But let’s not forget that the most important factor in terms of thermal performance is airtightness.I’m talking about magnitudes.Thermal conductivity is a molecular process, while convection is a physical displacement of large amounts of energy.This convection/loss of hot air can only be limited by airtightness.