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To generate heat electrically inside a balloon for lift would a 20 watt LED light or a 20 watt heating element make more heat to lift a small closed hot air balloon droid and solar cell?

Having a black balloon would convert the light into heat and a 20 watt LED does get hot and heat-sync would add to the total heat output. I don't know if the balloon and parts being black would matter with a heating element or material would be best to use unlike traditional open hot air balloon material?

enter image description here In a transparent balloon all the parts of it can be inside of it with the exception of the propellers.

https://robotics.stackexchange.com/questions/15323/could-a-hot-air-balloon-be-powered-from-the-ground-like-a-drone

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    $\begingroup$ The reason that LEDs are so popular for light is because they're so efficient at turning electricity into light. Very little of the electricity is turned into heat. Thus, if you want heat and light, use an old-style incandescent bulb. $\endgroup$ – RonJohn Mar 12 '18 at 23:03
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    $\begingroup$ @RonJohn "Very little of the electricity is turned into heat." is not true. At the very highest end, you'd only see some 45-50% energy conversion efficiency from an LED, which still means 50% lost as heat. More common household LEDs are down to only some 20-30% (70-80% heat), near fluorescent lighting at around 15-25%. The reason LEDs are claimed as high efficiency is what they're compared to: halogen is down to 10-15%, and traditional incandescent to under 5%. $\endgroup$ – Bob Mar 13 '18 at 0:45
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    $\begingroup$ @RonJohn If you want heat, you're probably better off using a resistive heater that doesn't generate light at all. If you want infrared radiation (vs conduction/convection), you'd probably get a special low-temperature incandescent bulb (well, a radiative heater) that doesn't put out much visible light. $\endgroup$ – Bob Mar 13 '18 at 1:21
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    $\begingroup$ If you put a black box around the LED, it should not make a difference. $\endgroup$ – fishinear Mar 13 '18 at 21:00
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    $\begingroup$ ©Muze I meant that if you put a black box around the 20W LED, it would not be different from a 20W electric heater. $\endgroup$ – fishinear Mar 16 '18 at 11:52
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Incandescent lights turn about 1-2% of the electric power into light, the rest is lost as heat and infrared radiation. They're pretty good as infrared heaters, though, but you don't want an infrared heater as infrared goes through air and would heat the balloon itself, not the air inside.

LEDs aren't very efficient, modern ones only turn about 25% of the electric power into visible light, and the remaining 75% of electric power becomes heat.

So LEDs are a lot more efficient than incandescents, but saying "very little of the electricity is turned into heat" is wrong, since 75% isn't "very little".

If you want to heat the air inside your balloon with electricity, the best would be a lightweight heating element, as a LED would let a significant fraction of the power escape outside the balloon as light.

This would only work if you have a very lightweight electrical power source. If you want heat, current technology isn't competitive with a butane/propane torch. Compare energy densities:

Current LiIon batteries contain 150-200 Wh/kg.

Propane contains 12000 Wh/kg thermal energy.

Batteries are great because electrical energy is a lot more useful than the heat you get when burning propane, but this doesn't apply to your case.

If your balloon is tethered, then battery weight doesn't matter, but the weight of the cable is an issue.

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    $\begingroup$ +1 for not falling into the "little heat from LEDs" trap. There's a reason cooling them is such a big issue (and so many die from overheating). Also a good point re: energy densities, but the question linked from the OP assumes a tether (which can be an electrical cable). $\endgroup$ – Bob Mar 13 '18 at 0:48
  • $\begingroup$ I've done a very quick calculation and it looks like Li-ion batteries do contain enough energy to lift their own weight by heating air (neglecting losses which scale with the surface area while everything else scales with volume, so a big system is more likely to work). Start at room temp (300K). 200Wh/kg (7.2E5J/kg) is enough to heat 1m³ of air (1.225kg) by 587K.That expands it by a factor of 2.95, giving 1.95×1.225kg of lift -- more than the mass of the battery $\endgroup$ – Chris H Mar 13 '18 at 12:08
  • $\begingroup$ @Chris H You neglected to mention how much energy is in a lithium ion battery, and how much it masses. $\endgroup$ – Justin Thyme Mar 13 '18 at 13:40
  • $\begingroup$ @JustinThyme I was running short of characters. I just used a 1kg battery at 200Wh, and neglected the mass of the resistor/controls etc. -- it was only a ballpark estimate (but now I want to try it!). This has also been discussed at electronics.se without conclusion $\endgroup$ – Chris H Mar 13 '18 at 14:07
  • $\begingroup$ @Chris H I presume I do not need to warn you, but discretion says I must. Me careful of LiIon batteries. Short one out, and you really don't need to question if it will lift its own weight through an exothermic reaction. $\endgroup$ – Justin Thyme Mar 13 '18 at 15:37
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You would want to use a heating element. If you use a light (whether visible light from a LED or infrared from a quartz heating element), some of the energy is turned into light, rather than heat. If you have a transparent balloon, it escapes. If you have a black balloon, it gets turned into heat when it reaches the edge. But there's a catch. That heat is being placed right on the edge of the balloon, which is the part that is easiest to cool from the outside. To minimize heat flow out of your balloon, you want the balloon envelope to be the coolest part, not the warmest.

Thus, the solution is to have a heater element on the inside of the balloon, with a small fan blowing air across it. That way you have one hot point in the middle of the balloon (at the heater), and the heat has to slowly move to the edges (via convection) until it reaches the envelope and can exchange the heat with the outside world.

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    $\begingroup$ wont need fan if heater at bottom, hot air will rise up and cold descends to bottom. $\endgroup$ – Sampo Sarrala - codidact.org Mar 13 '18 at 19:28
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    $\begingroup$ @SampoSarrala Potentially. My engineering instincts suggest to me that a compact heater like you would want on a hot air balloon would have trouble outputting enough heat without parts melting off unless it had some forced air, but it is possible that when you finally run the numbers, convection will be sufficient to move the hot air without a fan. $\endgroup$ – Cort Ammon Mar 13 '18 at 22:41
  • $\begingroup$ "some of the energy is turned into light, rather than heat." Light is still Electromagnetic Radiation, so it would still cause a (slight) increase in temperature, right? Like, the fact that it's on the visible spectrum doesn't mean it can't generate small amounts of heat. $\endgroup$ – jmite Mar 13 '18 at 23:08
  • $\begingroup$ @jmite Hmm, I see I need to be more precise in that wording for clarity. I should change that wording to "some of the energy is turned into light (radiation), rather than non-radiative heat." Plain ol' visible light does indeed cause a slight increase in temperature when it is absorbed, but that's what the rest of the paragraph talks to -- you don't want the increase in temperature to occur near the edges of the volume. $\endgroup$ – Cort Ammon Mar 13 '18 at 23:11
  • $\begingroup$ not sure either but i bet that good radiator is more efficient with less added weight than fan which will also need additional power. I'm not engineer either and might be wrong but heat should still provide at least some air circulation. $\endgroup$ – Sampo Sarrala - codidact.org Mar 14 '18 at 1:02
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One can well discuss various LED efficiencies here, but this is rather missing the fundamental point in which this question is a bit absurd.

Heat is the dumbest form of energy.

Whatever you do when handling energy / converting it between different forms, it will always tend to slip through your fingers by just dissipating into heat, which is generally pretty hard to make use of. Conversely, electric energy is an pretty “sophisticated” form that can easily be used for all kind of things.

You may wonder why I say that – do not most power plants generate electricity from heat? Well, not actually, that's impossible according to the second law of thermodynamics. What coal power plants really do is, converting heat flow along a temperature differential into electricity. That is, they let heat flow into some cooling dump (usually a river's water), and through turbines are able to pull out some of this energy into a more useful form (rotation movement and ultimately electricity). But this process is inherently wasteful: even a well-designed modern coal plant only has a conversion efficiency of about 33%.

The other way around is always trivial: if you let current flow through pretty much any circuit, the energy will basically always end up as mostly heat. Sometimes, a bit is also converted to other energy, notably, an antenna can send out electromagnetic energy (radio waves). Mind, these are eventually also converted to heat, but before that happens they can travel a long way. And for LEDs it's basically the same thing: they convert a not insignificant amount of energy to light instead of straight to heat, but all that accomplishes is that you give the energy a bit of an extra chance to escape from the ballon unused. So, this really is a bad idea: LEDs are, by design, “unusually bad at converting electricity into heat”.

But the bad idea actually starts much earlier, by proposing electrical heating at all. As I said, electricity is precious. If you went through all the trouble of bringing energy in heavy, inefficient batteries instead of cheap, energy-packed gas cans, the last thing you want to do is just burn it to heat. There is however one trick you could pull that might theoretically make it worthwhile: use a heat pump. That's basically the inverse of a power plant: instead of converting electricity into heat, it forces a heat flow from the ambient air into the balloon. That way, you can on paper get a “conversion efficiency” of more than 100%.

Practically speaking, that's unfortunately not going to offset the drawbacks: heat pumps are themselves not ideally efficient, and they're heavy, adding to the weight of the batteries.

Be sensible, heat your balloon with gas or with solar heat, not with electricity.

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  • $\begingroup$ I wonder if you could use something like thermoelectric cooling, except instead of radiating the heat out into the air, you use it to heat your balloon. $\endgroup$ – AndyD273 Mar 13 '18 at 15:05
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    $\begingroup$ @AndyD273 the Peltier effect is substantially less efficient yet than mechanically-driven heat pumps. It's conceivable that thermocouples might get better in the future though, with advances in material science. $\endgroup$ – leftaroundabout Mar 13 '18 at 15:29
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    $\begingroup$ Hmm... I wonder what the trade off looks like with the weight of a refrigerant/compressor system for the heat pump vs a lightweight heating element. Might see big airships relying on more efficient systems than small ones, mimicking oceangoing ships today. $\endgroup$ – Joe Bloggs Mar 15 '18 at 15:33
  • $\begingroup$ @JoeBloggs It would mostly depend on the required range. For long distances, you'd eventually anyways need batteries weighing more than even a sophisticated heat-pump system, so the efficiency gain of such a system would definitely pay off. For short distances, you might be ok with inefficient heating elements, though those would use up the smaller batteries even more quickly. $\endgroup$ – leftaroundabout Mar 15 '18 at 15:51
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Here's another idea, but it's line-of-sight: Use a black balloon, and heat with an IR laser beam from a power source on the ground (over short distances you could actually use a follow spot). You could deliver 1kW without much trouble. Note that you'd need to expand the beam to almost as big as the balloon to avoid melting it. This also provide altitude control by varying the heating power.

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  • $\begingroup$ Isn't that exactly what's being depicted in the photo the OP used to illustrate their question? $\endgroup$ – Tom Auger Apr 22 '20 at 19:12
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    $\begingroup$ @TomAuger check the edit history on the question - it looks like the OP added the photo and the mention of a laser after I answered. Even so my answer still adds the idea of a follow-spot at close range, and specifically an IR laser $\endgroup$ – Chris H Apr 22 '20 at 19:26
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    $\begingroup$ Thanks @chrisH right you are there. $\endgroup$ – Tom Auger Apr 22 '20 at 20:04
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(a) Power consumption (e.g., your 20 watts) describes everything the circuit is doing, including generating heat. All electric circuits generate some heat because (at least at room temperature), all circuit paths have resistance, and where there is resistance, there's heat.

(b) LEDs are not filament lights. They're diodes (Light Emitting Diode = LED). Diodes have very, very low resistivity. Consequently, very little of the energy used is lost to heat. This is what makes them so efficient for lighting a room. Most of your power is used to generate light, not heat.

(c) Your old filament lights ("incandescent" lights) were much less efficient. For example, a 75 watt incandescent light would generate about 1,100 lumens. An LED light generating 1,100 lumens needs only about 10 watts. It's not quite this simple in reality, but you can generally say that the incandescent light is generating 65 watts of heat to light the room (it's actually only losing about 50 watts, if I remember correctly, due to illumination inefficiencies in the filament itself — but we don't care about that).

Frankly, what you really want is a quartz heating element. You want to waste as few watts as possible generating visible light. You want infrared light — check out infrared heaters.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Monica Cellio Mar 13 '18 at 21:29
  • $\begingroup$ @MonicaCellio I can't chat please reinstate comments so I can interact. $\endgroup$ – Muze Mar 14 '18 at 3:32
  • $\begingroup$ @Muze you have 191 rep on this site (plus whatever else you have on other sites). It only takes 20 rep to chat. The discussion here was getting long and was almost entirely between two people, one of whom asked me to move it to chat after misclicking on the automatic prompt to do so. $\endgroup$ – Monica Cellio Mar 14 '18 at 4:14
  • $\begingroup$ @MonicaCellio yes but some how I'm blocked from all chat and I dont know why really. $\endgroup$ – Muze Mar 14 '18 at 12:45
  • $\begingroup$ @Muze your chat profile says that you are suspended on your parent site, which also affects chat. Chat isn't per-site, unfortunately, so that affects everything. I don't know the reasons for your suspension and can't get involved in that; you'll have to ask the mods on the site that suspended you if they can/will do anything about the chat effects. Meanwhile, sorry, we're not going to continue a chat in comments; that's not what comments are for. $\endgroup$ – Monica Cellio Mar 14 '18 at 14:09
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Here are some numbers to play with.

10W of energy produces 34.121416 BTU/hr if it is all converted to heat.

It takes 0.24 BTU of heat to change the temperature of one pound of air by one degree F.

So 10 W of electricity will heat 14 pounds of air by 10 degrees F. if my calculations are correct.

1 cubic foot of air at standard temperature and pressure assuming average composition weighs approximately 0.0807 lbs.

So we are talking theoretically about heating 173 cubic feet of air (a balloon 5.5 ft by 5.5 ft by 5.5 ft) by 10 degrees F. assuming no heat loss at STP.

I am not sure if raising the temperature by 10 degrees F will lift very much, very high.

EDIT

Just to lift an adult man's weight, you'd need a balloon about 4m (13ft) in radius with the air inside heated to a temperature of about 120°C (250°F). That explains why hot-air balloons are generally so large.

From Hot-air balloons

Given an average human mass of 80 kg, a 1 kg LiIon battery would need perhaps 1/80th the volume, but you still have to provide a temperature differential of about 100 degrees C.

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  • $\begingroup$ I doubt that all of the air inside a regular sized hot air balloon is 120°C (250°F)... so increasing the volume of heated air should reduce the temperature requirement. $\endgroup$ – Phil M Mar 13 '18 at 23:25
  • $\begingroup$ @Phil M I have no doubt that the heated air would be stratified. I assumed the 250 figure was averaged. $\endgroup$ – Justin Thyme Mar 14 '18 at 3:08
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The quick answer is that a 20W LED light and a 20W heating element both produce 20W of heat, and inside a black (opaque) balloon, they would both produce the same amount of heat to lift the balloon.

Now, it is obvious this doesn't tell the whole story; intuition seems to suggest a heating element would be more fit for producing heat than a device intended to produce light, not to mention the differences in the weight between the devices.

A lot of people have mentioned how LED lighting isn't as efficient as many people believe.

While it is inaccurate to say LED lighting turns most of its input power into light, it's not like the claim has no basis in reality whatsoever, and I feel like this makes it a bit of a red herring in the reasoning behind the answers.

LED's usually do turn most of its consumed power into light (if more than half counts as most). It's the power circuitry that drives the LED that accounts for most of the power lost as heat.

I have more experience working with laser diodes than LED's, but most often over 60% of the power used in the diode is turned into radiation, while the process of modulating the current flowing through the LED is hugely less efficient.

In the simplest designs, a 20W light might have a power stage that dissipates 18W as heat while supplying 2W to the actual LED, which might produce 1.5W as visible light. Conceivably, you could argue both that the efficiency is 75% (1.5/2) and that it is 7.5% (1.5/20).

(The issues that arise when dealing with thermal dissipation are more attributable to the relatively small size of LED's rather than the amount heat being produced. I realize laser diodes are significantly different than LED's, but I'm assuming most LED's are more efficient, but I may be wrong here)

So in many cases the comparison between a 20W LED light and a 20W (resistive) heating element basically boils down to one between an expensive, unnecessarily convoluted 20W heating element and a 20W heating element, and as people have discussed already, convection would be the more important issue. I would imagine the optimal ohmic heating element in this situation would be a motor of some size with a fan attached to its spindle in a way that it cools the motor.

But, and I think this may be closer to the intent of the question, if you are asking, "assuming you could turn 20W of electricity into either 20W of light or 20W of heat, which would be more effective in producing lift in a hot air balloon?", the answer is, it depends.

You've already taken the color of the balloon into account by saying it's black. This would contain the light produced inside the balloon, but as you've stated, a black balloon would absorb the light and heat up.

To produce lift, we don't want the balloon to heat up. We want the air inside the balloon to heat up.

So let's make the balloon shiny, i.e. 100% reflectivity (this could be approached by using metallized mylar). Then, you keep all the light inside the balloon, and it doesn't waste heat by heating up.

Now that we have all this light energy trapped in the balloon, we just need to turn it into heat in the air. Just kidding, no we don't. The light, not being able to escape the balloon would eventually be absorbed by the molecules in the air itself. It's as it's a law or something, light will eventually turn into heat at some point, in some thing. This heats up the entire volume of the air simultaneously. No convection required. Light clearly wins in terms of speed in this case.

Depending on the level of material sciences, creating a balloon with near perfect reflectivity that is both light and strong enough to support whatever load it carries may be impossible.

Let's go back to black. Probably nylon, or some other fabric. Then, the balloon would heat up, absorbing light better than air, and while it would heat the air inside the balloon, it would also heat the air outside (i.e. lose heat).

But really, this is unavoidable no matter what you do. Even if you used a heating element to heat up the air inside the balloon, once the air is hot, it would heat up the balloon, which would lose heat to the outside.

The main difference would be in how long it takes initially to get the balloon off the ground. It would take some ugly calculus to model (if you heat the air with light via the balloon, you have more surface area = faster heat conduction, but you lose half to the outside and the temperature is lower, and fluid dynamics I don't even want to start thinking about), but put simply, using light would mean a longer wait before the balloon lifted off than using heat, but once the air was at the target temperature, there would be no difference in efficiency.

These thought experiments seemingly pointing to the conclusion that light and heat would be equally as efficient can probably be traced (in a very roundabout way) to one of these facts: light is heat, or eventually becomes heat; where there is heat, there is light; a large part of heat (or all of it, depending on the scale you think at) transmission happens through light; and most fundamentally, heat is such a vague word that it is used in countless contexts and meanings, yet it is difficult to define clearly. Also c.f. black body radiation.

None of this would work with a 20W anything, by the way.

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LEDs produce very little heat to begin with. Considering wattage is just power expenditure, you'd be better off focusing all those 20 watts on heat, not light.

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So many answers and they are all good. I am taking a compilation of these answers and posting my own answer.

No one here said it could not be done. Finding a 20 watt heater element is slim, LED are not efficient for heat, but a UV LED produces more heat than visible light LEDs. UV LEDs allow it to be visible at night with the right material.

Using a UV LED and UV absorbing black material enclosure to surround the UV LED light will make it visible with luminescent parts, light weight, reliable, cheap and abundant as a heat supply. I still am not sure if a solar cells can transfer enough light into heat from inside the balloons in place of batteries. Limits of Solar cells is 18% ish of light absorbed by solar cells is converted in electricity that would be used to produce heat from the sun. Depending on the surface area of all the parts(all in black) and the conversion of sunlight would contribute to the total heat produced. More surface area of the parts with the least amount of volume and weight of the parts is best for capturing sunlight and UV light. Like a black heat sink designed capture light and convert it into heat.

To conclude it is the inefficiency of the UV LED light and efficiency of the absorption of the parts center of the balloon from the Sun externally and UV light internally together heating the air within the balloon for lift by electricity and Sun light.

At night a black balloon would be best with a black box surrounding the UV LED. I may experiment in heating a helium/air mixture in a sealed balloon if air alone does not create enough buoyancy to create lift.

enter image description here Plus 1 everyone.....Thanks

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