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For all re Calliban's suggestion of dry steam as a lifting gas...
As some of you know, Google's search engine has been enhanced with a version Google calls Gemini...
I asked Gemini to compare the performance of a dry air hot air balloon to a dry steam version. The dry steam version seems superior. I assume there may be some practical reasons dry steam balloons are not well known, but here is the discussion:
So, to summarize correctly:
* A cubic meter of hot air (at the temperatures we assumed) provides a lift *equivalent* to the weight of approximately 0.23 kg.
* A cubic meter of dry steam (at 100°C) provides a lift *equivalent* to the weight of approximately 0.59 kg.
You're very close, but there's a slight misunderstanding in how we're expressing the lift. It's not quite accurate to say the lift *is* a certain number of kilograms. Instead, the lift *is equivalent to* the weight of a certain mass.
Let's clarify:
* **Hot Air Balloon:** The hot air balloon example showed a difference in density between the hot air inside and the cooler air outside. This density difference translates to a lift force. For our cubic meter example, the *difference* in mass between the inside and outside air was approximately 1.18 kg - 0.95 kg = 0.23 kg. This means the lifting force is *equivalent* to the weight of 0.23 kg. It's not that the lift *is* 0.23 kg, but it's *as if* you had a 0.23 kg weight pulling upwards.
* **Dry Steam Balloon:** Similarly, for the dry steam, the difference in density led to a mass difference of 1.18 kg - 0.59 kg = 0.59 kg for our cubic meter. So, the lifting force is *equivalent* to the weight of 0.59 kg.
So, to summarize correctly:
* A cubic meter of hot air (at the temperatures we assumed) provides a lift *equivalent* to the weight of approximately 0.23 kg.
* A cubic meter of dry steam (at 100°C) provides a lift *equivalent* to the weight of approximately 0.59 kg.This way of expressing it clarifies that lift is a force, and we're comparing it to the familiar force of gravity acting on a mass.
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As a follow up to the suggestion of dry steam for lift....
Let us suppose the balloon envelope is NOT insulated.
The dry steam will condense on the envelope wall and flow down toward the bottom. The flow of condensed steam might be facilitated by lining the interior of the envelope with the kind of tiny needles that are found in certain insects and plants that are able to "drink" water condensed from the air. In any case, the water would flow back down to the bottom of the envelope, where it could be re-heated to dry steam, and delivered back into the envelope.
If we build two identical balloons (same volume and total mass) and lift one with ordinary hot air and the other with dry steam, can we predict the energy consumption for the two balloons?
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The conversation below comes from aviation.stackexchange.com
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Ask questions, find answers and collaborate at work with Stack Overflow for Teams.What are pro/cons of steam as lifting gas for airships?
Asked 4 years, 7 months ago
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23I was reading Wikipedia page about hot air airship and another Wikipedia page about the first powered and steerable airship which was powered by a steam engine. So I ask myself "why didn't they use the exhaust steam from the steam engine as lifting gas?"
After quick research, I found this home made experiment of steam used as the lifting gas. This website even gives a comparison with other lifting gasses:
+-------------+----------+------------+-----------------+--------------------+--------+-----------+-------------------+------------------+
| GAS | M.W. | Temp. (°C) | Density (kg/m3) | Lift (N/m3) in ISA | Safety | Cost | Ease of provision | Buoyancy control |
+-------------+----------+------------+-----------------+--------------------+--------+-----------+-------------------+------------------+
| H2 | 2 | 15 | 0.084 | 11.19 | bad | fair | fair | no |
| He | 4 | 15 | 0.169 | 10.36 | good | very high | very bad | no |
| CH4 | 16 | 15 | 0.676 | 5.39 | bad | low | fair | no |
| NH3 | 17 | 15 | 0.718 | 4.97 | fair | low | fair | no |
| hot air | 29 (avg) | 110 (avg) | 0.921 (avg) | 2.98 (avg) | good | very low | good | yes |
| steam (H2O) | 18 | 100 | 0.587 | 6.26 | good | very low | good | yes |
+-------------+----------+------------+-----------------+--------------------+--------+-----------+-------------------+------------------+
(source: http://flyingkettle.com/jbfa.htm)Given the conclusion of this website, steam should now be more widely used as a lifting gas:
it is available since the first lighter-than-air (plenty of time to develop the technology)
it has a good lifting power
it is cheap and safe
Yet, on 2020, almost all airships and balloons use helium or hot air, and no other lifting gas. So there must be drawbacks I fail to imagine. I hope finding those drawback will orient my research for my first question ("why didn't they use exhaust steam as lifting gas on airship fitted with steam engines?"). Moreover, there may be advantages I also fail to imagine.Thus, what are pro and cons of using steam as a lifting gas?
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edited Jul 4, 2020 at 9:19
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35The main disadvantage is the energy required just to get off the ground. Let's do a back-of-the-envelope calculation, comparing this to a hot air balloon.
Let's heat 1m³ of air. To get equivalent lift, we need about 0.5m³ of steam.
The density of air is about twice the density of steam. So, for a given amount of lift, we need to heat approximately the same mass of lifting medium.
To heat 1kg of air to 100°C above ambient, we need about 100kJ (specific heat of air is about 1kg/(kJ K)).To heat 1kg of water to 100° above ambient, we need about 420kJ just to heat the water. To vaporize the water, we need an additional 2250kJ, for a grand total of 2670kJ.
The most used hot air balloon size is 2800m³. For that we need to heat 1400kg water. Multiplying gives about 4GJ of energy or 75kg of propane just to get off the ground. Since we established steam can lift twice as much per unit of volume, we can maybe get away with 2GJ or 35kg of propane.
Some more negative thinking... Condensation on the outer shell seems like a very efficient mechanism to transport heat energy to the environment, only partially offset by the reduction in surface area compared to a hot air balloon due to the cube-square law. This may require more energy in flight than a hot air balloon. Closed-cycle operation (only heat condensed water, not 'new' water) is preferred due to the high energy requirement, so descending means you have to wait for enough water to condense or be willing to pay the price in fuel. Superheated steam is not safe (despite what your table says); the high heat capacity and enthalpy of condensation can result in serious scalding burns compared to hot air. All of these effects are of course much more pronounced when comparing to a helium-filled blimp.
Still, it's a really neat idea, and sometimes you don't need a practical reason just to build something neat.
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edited Jul 3, 2020 at 14:38
answered Jul 3, 2020 at 9:11
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Comments are not for extended discussion; this conversation has been moved to chat. –
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CommentedJul 8, 2020 at 8:52
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8In short, helium (or hydrogen) remains in the gaseous state at much, much lower temperatures than water does. In particular, helium or hydrogen will remain gaseous in pretty much any part of Earth's atmosphere. Water, on the other hand, will not naturally remain gaseous in any part of the Earth's atmosphere. On the contrary, it will freeze in most of the Earth's atmosphere.
For steam to remain steam, it must be constantly heated to make up for the energy it's losing by heat transfer to its surroundings. You can aid this by insulating the airship really well, but you can't bring the heat transfer down to zero. And this will also add a significant amount of weight, reducing efficiency further.
Additionally, as Sanchises's answer mentions, it takes a large amount of energy to turn that water into steam in the first place. Water has a very high specific heat and also a high enthalpy of vaporization. That is, water takes much more energy per unit mass than most substances to heat up and also more energy per unit mass to convert from a liquid to a gas.
So, you're requiring a very large amount of energy to get the water into the gaseous state in the first place and you then need a large and continuous energy source to keep it there during flight.
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edited Jul 3, 2020 at 21:50
answered Jul 3, 2020 at 18:35
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Re "Water, on the other hand, will not naturally remain gaseous in any part of the Earth's atmosphere.": As long as the temperature is above the dew point it will naturally remain gaseous. At any one time (on average) 0.4% of the entire atmosphere is water wapor (1% at sea level). –
Peter Mortensen
CommentedJul 4, 2020 at 7:08
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@PeterMortensen Well, yes, but the dew point of a steam-filled airship will be much, much higher than the local atmospheric temperature, especially at altitude. If it's allowed to drop in temperature inside the airship or to vent at any significant concentration outside of it, it will very quickly condense (and, at high altitudes and/or cold climates, subsequently freeze.) I was simplifying things a bit for the purposes of what is relevant to the question that was asked. –
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5Lighter than air balloon works by replacing air with some gas that is lighter than air. Your steam airship would work if we can keep steam at 100 C or higher or else you risk condensation, your balloon collapse into some saggy bag and you fall out of the sky. Wikipedia mentions that when water turns into steam it expand 1700 times so imagine, in reverse, your 1700 cu.m. hot steam balloon when cool down and becomes 1 cu.m. bag of water and you lose all the lift. That does not sound safe at all. Consider low temperature up there you need a lot of energy to combat heat loss through conduction over a large surface area of the gas bag.
To sum it up you need lots of energy to keep the balloon in positive buoyancy and a big safety issue to solve when that energy source fail.
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answered Jul 3, 2020 at 14:52
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Condensation won't just cause the balloon to "fall out of the sky". The energy required for condensation has to go somewhere, and if the balloon is sealed, radiation and conduction through the balloon is the only way it can exit. As long as enough heat enters the balloon to equilibrate it, the physics work out just like an average hot air balloon. –
March Ho
CommentedJul 3, 2020 at 19:00
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@BlokeDownThePub and if the flame fails on a standard hot air ballon? It’s still going to be a gradual process for the ballon deflating and falling. –
Tim
CommentedJul 4, 2020 at 0:00
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You seem to be thinking that the steam needs to be kept above 100 °C, because if it were to drop to, say, 99 °C, all the water would condense and you'd fall out of the sky. That's completely wrong. Going by the numbers in Sanchises' answer, steam contains an enormous amount of energy compared to hot air. The amount of heat loss which would cool hot air to room temperature (producing, I think, about a 25% decrease in volume) would merely condense 5% of the steam (a 5% decrease in volume). –
Sophie Swett
CommentedJul 4, 2020 at 13:19
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So, by my calculations, in the event of a heater failure, steam would be much safer than hot air, quite contrary to your assertion that "you risk condensation, your balloon collapse into some saggy bag and you fall out of the sky". –
Sophie Swett
CommentedJul 4, 2020 at 13:20
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@TerranSwett is absolutely right, the huge heat capacity of water and the even bigger condensation heat (i.e. heat that is produced when condensating, helping to stop further condensation) is a safety feature, not a problem. The problem would be that this huge heat capacity is also dangerous if you come into contact with the gas - you can probably survive for a moment inside a hot air balloon, a steam balloon would lethally burn you in seconds. –
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CommentedJul 4, 2020 at 14:45
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5In addition to the tremendous energy requirements, you'd need a highly specialized design that is not at all bothered by boiling hot water dripping back out of the balloon. Even if it was highly insolated, you'd need to be resupplying steam to keep the balloon inflated because even small amounts of local condensation would pull a lot of water out of the air because water expands so much when it becomes steam.
Even like, a cup of water running back down the balloon over the course of the entire voyage would be over 100 gallons of steam. This in general would be hard to build, fuel inefficient and difficult to control while being wildly dangerous.
They didn't use exhuast steam because the efficiency of a balloon isn't acheived by min/maxing the displacement of air by any means possible. It is acheived by realiability, cost and control. Blasting a second gas as a by product of the steam engine, doesn't improve that in the slightest, and likely would create problems.
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answered Jul 4, 2020 at 2:40
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That problem could be mitigated by having a bag that contains water vapor which is captive within the main air gasbag so that it wouldn't need to contain a significant pressure nor temperature gradient. –
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CommentedJul 5, 2020 at 23:52
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0I think if you would use a doble type flash boiler wich is a light steam generator (it was even used to power a plane) and you use the envelope to condence the steam and then have it gather at the bottom to be pumped into the generator again then you can reuse the water and the envelope doesnt need to be insulated. Also the steam that comes out of a flash boiler is dry superheated steam wich is super hot and is generated instantly so as soon as it comes into the open space of the envelope it is able to expand reducing presure and temperature and so creafing lift. Much like a hot air balloon but the cold air wich in this case is water is reused again. Maybe a metal clad envelope would be better suited than a balloon because it cant deflate.
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edited Dec 2, 2023 at 19:03
answered Dec 2, 2023 at 18:56
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I come away from this thinking that the dry steam design has a lot going for it.
However, the steam balloon needs a MUCH greater investment of energy to prepare for flight.
I would be ** very ** interested in any reports anyone can find of actual experiments that might have been done with this idea.
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Regarding duel lifting gas. In hydrogen filled rigid airships, the lifting gas was contained in non-pressurised gas cells that were tied to the frame. These gas cells were fitted with both manually activated and pressure operated valves. As the airship consumed fuel during flight, it got progressively lighter. It would rise in altitude until it reached its pressure-height. This was about 900' for the Heindenburg. At pressure height, the automatic pressure operated valves would open and hydrogen would be vented out of the top of the envelope.
When the Heindenburg successor was under design, it looked for a while as if it would be possible for the Germans to use helium sourced from North American natural gas wells. There were two problems with helium. Firstly, it generates slightly less lift than diatomic hydrogen. Secondly, it was too expensive to vent in the way that hydrogen was vented. The solution to the second problem was to have concentric gas cells. A hydrogen cell was placed within the helium cell. When the ship reached pressure-height, the hydrogen gas cell vented but the helium did not. This allowed the expensive helium to be conserved.
As things turned out, the Americans refused the Germans access to helium due to concerns that Zeppelins could be used to attack North America. So the concentric gas cell idea was never implimented. The Heindenburg successor was hydrogen filled. It was requisitioned by Germany's NAZI government in the late 1930s and was briefly used to spy on British short range radio transmissions. With the outbreak of war, it was decommissioned and its aluminium frame used for warplane manufacture.
Last edited by Calliban (2025-02-14 02:45:06)
"Plan and prepare for every possibility, and you will never act. It is nobler to have courage as we stumble into half the things we fear than to analyse every possible obstacle and begin nothing. Great things are achieved by embracing great dangers."
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For Calliban re #29
Thank you (again) for the history of the dual bag idea!
***
I remain interested in the hint you gave us a few days ago, of the possible use of dry steam to lift a lighter-than-air vehicle.
Today, I decided to ask Gemini (a cousin of Google Search) to look for any evidence of actual experiments with steam lift. It found none, which is not surprising, since dry air lift is so much easier. However, I pressed on a bit further, to try to find evidence that a steam balloon would require no more energy input to stay aloft as compared to a dry air lift system. Gemini is skeptical.
I think that only Real Universe testing would answer the question, but perhaps a CFD program could come close.
Conversation with Gemini
Hi Gemini! Thank you for your help with a number of previous requests! Today, I hope I have a really easy question for you! The background is that a member of a group suggested that steam might be used as a lifting "gas" for a lighter-than-air vehicle. To my surprise I've been unable to find any evidence that anyone has attempted an experiment to see of steam might actually work for this purpose. By any chance, can you find reports of experiments along these lines? I have found plenty of speculation so I'm not interested in that. Please report any scientific or engineering studies that might have been done to gain actual data.
You're welcome! I'm happy to help. It's an interesting question about using steam for lift, and you're right, there's a lot of speculation but surprisingly little in the way of hard experimental data. My search also turned up mostly theoretical discussions and thought experiments. I couldn't find any published scientific studies specifically dedicated to testing the lifting capacity of steam in a practical, lighter-than-air vehicle context.
While I couldn't find direct experimental studies on steam lift, I can offer some related information that might be helpful. The challenge with steam as a lifting gas stems from a few key factors:
Density: While steam (water vapor) is less dense than liquid water, it's still significantly denser than hydrogen or helium, the gases typically used for lift. The density of steam changes with temperature and pressure, but even at relatively high temperatures, it's not buoyant enough at atmospheric pressure to provide significant lift compared to those lighter gases.
Temperature: To keep water in its vapor phase, you need to maintain a certain temperature. This requires energy, and the mass of the equipment needed to generate and contain the steam (boiler, insulation, etc.) would likely outweigh any potential lift gained, especially at lower altitudes.
Condensation: As steam cools, it condenses back into liquid water, drastically reducing its volume and negating any lift. Maintaining the temperature to prevent condensation would be a significant engineering hurdle.It's likely that the practical difficulties and inefficiencies associated with using steam for lift have discouraged extensive experimentation. Theoretical calculations and the basic physics of buoyancy probably make it clear that it wouldn't be a competitive approach.
While I couldn't find the specific experimental data you were looking for, I hope this explanation of the challenges involved is useful. If you have any other related questions, feel free to ask!
Thank you for confirming my impression that no one has left a written record of any experiments that might have been done. The comparative ease and convenience of producing lift with hot air seems to have made that the hands-down winner in the category. Use of hydrogen is dangerous. Use of Helium is well established, but the gas is limited in availability at present. When fusion power becomes a successful enterprise, helium will be produced as a by product. I am interested in steam as an affordable alternative. My understanding is that energy is required to sustain a hot air balloon in flight, and the same would be true for a steam balloon. A major difference in the cost of the two approaches is that water must be converted to steam before the balloon can lift, and the investment of energy is substantial. This next question may be beyond your pay grade, and if so please let me know. If we consider two identical balloons ready to be released from their tie-downs, I'd like to see if we can compare the energy investment needed for sustained flight. The two balloons would be of identical volume and mass. One would be a typical hot air balloon, and the other would be a hypothetical steam balloon. In our scenario, each has been primed with energy and each is tugging at the tie-downs ready for take off. Upon release, each balloon will rise, and energy will flow from the canopy to the outside air. This energy must be replenished by the energy supply in the basket below the canopy. The question I am hoping you might consider is: Can we compare the energy losses and thus the needed inputs to keep both vehicles in flight?
You're right, comparing the energy requirements for sustained flight between a hot air balloon and a hypothetical steam balloon is a complex question, and it does push the boundaries of what I can readily calculate without making some simplifying assumptions. A full, rigorous analysis would require computational fluid dynamics and detailed heat transfer modeling. However, I can offer a conceptual comparison and point out the key factors that would influence the energy consumption.
Here's a breakdown of the challenges and considerations:
Hot Air Balloon:
Heat Loss Mechanisms: The primary energy loss in a hot air balloon is due to:
Conduction: Heat transfer through the balloon fabric to the surrounding air.
Convection: Warm air rising from the balloon's envelope and being replaced by cooler air.
Radiation: Heat radiated from the hot air inside the balloon to the cooler atmosphere.
Energy Input: The burner in a hot air balloon replenishes this heat loss by heating the air inside the envelope. The efficiency of the burner and the insulation properties of the balloon fabric are crucial factors.Steam Balloon (Hypothetical):
Heat Loss Mechanisms: A steam balloon would face similar heat loss mechanisms, but with some key differences:
Condensation: This is the biggest challenge. Any cooling of the steam will lead to condensation, which drastically reduces lift and releases latent heat. This latent heat needs to be resupplied to re-vaporize the condensed water. The energy required for vaporization is substantial.
Conduction, Convection, and Radiation: These losses would still occur, similar to a hot air balloon, but the temperature difference between the steam and the surrounding air might be different, affecting the rates of heat transfer.
Energy Input: A steam balloon would need to supply energy to:
Maintain Steam Temperature: Preventing condensation would be the primary energy demand. This involves supplying the latent heat of vaporization continuously to any water that condenses.
Compensate for Other Heat Losses: Conduction, convection, and radiation losses would also need to be replaced, similar to a hot air balloon.Comparison and Challenges:
Condensation Dominates: The energy required to prevent condensation in a steam balloon would likely be the dominant factor in its energy consumption. This is a far greater challenge than simply reheating air in a hot air balloon.
Material Requirements: A steam balloon would likely require significantly more robust and well-insulated materials to contain the high-temperature steam and minimize condensation. This added mass would further complicate the lift equation.
Efficiency: The efficiency of generating and maintaining steam at the required temperature and pressure would be a critical factor.In summary: While a detailed calculation is beyond my current capabilities, it's highly probable that a steam balloon would require significantly more energy input for sustained flight compared to a hot air balloon, primarily due to the energy needed to prevent and reverse condensation. The latent heat of vaporization is a substantial energy hurdle to overcome.
I hope this conceptual comparison is helpful. It highlights the major challenges associated with a steam-based lighter-than-air system.
Thank you very much for this valuable and helpful analysis. In one of the online discussions I've read, one of the participants made a telling point, which I hope you will consider! That participant pointed out that the heat of vaporization is the same both ways. In other words, when steam arrives at the wall of the balloon and encounters fabric that might try to absorb energy, the energy required to convert the steam back to water is exactly the same as the energy required to make the steam in the first place. My hypothesis is that an air lift machine and a steam lift machine would have exactly the same energy input requirements.
You're absolutely right. The heat of vaporization is indeed the same whether you're condensing steam into water or vaporizing water into steam. This principle, often referred to as the reversibility of phase transitions, is fundamental in thermodynamics. And the point made by the online discussion participant is crucial to consider.
Your hypothesis, that a hot air balloon and a steam balloon (under ideal, theoretical conditions) might have similar energy input requirements, is a very insightful one. Let's explore this further, acknowledging that we're still dealing with a simplified model:
The Core Argument:
The core of the argument rests on the idea that if we could perfectly prevent any heat loss other than the latent heat of condensation in the steam balloon, then the energy required to re-vaporize the condensed water would indeed be roughly equivalent to the energy needed to heat the same amount of air in a hot air balloon to achieve the same lift.
Why "Similar" and Not "Identical":
Even with the same latent heat of vaporization, there are some important caveats:
Other Heat Losses: In reality, a steam balloon will have other heat losses (conduction, convection, radiation) in addition to condensation. These losses would need to be replenished, making the total energy input for the steam balloon higher than the idealized scenario. A hot air balloon also has these other losses. The question becomes which has more.
Temperature Differences: For the same lift, the steam balloon might operate at a higher temperature than the hot air balloon. This can influence the rates of heat transfer via conduction, convection, and radiation. A higher temperature difference between the balloon and the surrounding air generally leads to greater heat loss.
Mass of the System: The steam balloon system (boiler, condenser, pipes, etc.) would likely be heavier than a hot air balloon system. This added mass would require more lift (and therefore more energy) to achieve the same altitude.
Efficiency of Heat Transfer: The efficiency of transferring heat to the air in a hot air balloon versus the efficiency of managing the condensation and re-vaporization cycle in a steam balloon would likely be different. Real-world systems have inefficiencies.
In Conclusion:
Your hypothesis is very interesting and gets to the heart of the thermodynamics involved. In a highly idealized scenario, where condensation is the only heat loss mechanism for the steam balloon, the energy requirements might be comparable to a hot air balloon. However, in the real world, the additional heat losses, temperature differences, and system mass in a steam balloon would almost certainly make its energy requirements higher than a hot air balloon for sustained flight.
It's a great example of how a seemingly simple question can lead to a deeper understanding of the underlying physics!
Bravo! Thank you for this exchange!
You're very welcome! I'm glad I could help explore this interesting thought experiment with you. It's a testament to how even seemingly straightforward questions can lead to fascinating discussions about physics and engineering. If you have any other questions in the future, please don't hesitate to ask!
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For Calliban re steam lift...
It seems to me that it should be possible for a home experimenter to make and fly a small steam balloon.
The great challenge for anyone, home experimenter or university level laboratory, would be to make a balloon shape out of materials suited for the purpose.
Early in the age of lighter-than-air vehicles, an enterprising organization attempted to make a dirigible with aluminum foil. Aluminum had recently become available in the form of foil, and it was thought potentially superior to the fabrics used at the time.
A full sized aluminum foil balloon/airship flew in Germany in 1897. (per "lighter than air" by Crouch (page 72-74)).
The 12 horsepower motor installed was insufficient for the application, but the balloon/airship reached an altitude of 800 feet before it descended with a RUD.
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Dr. Johnson sent this note about steam via email:
Saturated steam with water droplets (the white cloudy appearance): the vapor component cannot be significantly different in temperature from the liquid component. If they are in contact, they must be in or very near thermal equilibrium. Adding heat just increases the percent vapor, losing heat increases the percent liquid, all at the boiling temperature, which is a function of the pressure you are operating at. Only if they are physically separated can you heat the steam above the liquid temperature. That vapor above the boiling temperature is "superheated steam", and it is invisibly clear in appearance. Which invisibility is why leaks from a superheater on a boiler at significant pressure are so very dangerous. The issuing invisible jet will cut you apart before you even know it is there, because usually the background noise around a steam generator or engine is so high, that you cannot hear the leak.
GW
There appears to be a problem with terminology used by various references.
I asked Wikipedia and got this:
Superheated steam - Wikipedia
If unsaturated steam (a mixture which contains both water vapor and liquid water droplets) is heated at constant pressure, its temperature will also remain constant as the vapor quality (think dryness, or percent saturated vapor) increases towards 100%, and becomes dry (i.e., no saturated liquid) saturated steam.
Superheated steam - WikipediaWikipedia
https://en.wikipedia.org › wiki › Superheated_steam
If the explanation by Wikipedia is correct, the term "saturated steam" is steam without water, or "dry" steam.
I picked up a hint that it is difficult for a commercial boiler to deliver dry steam. My guess is this is a function of efficiency. A certain amount of steam is required for an application, and trying to achieve 100% conversion to dry steam may not be worth the effort.
I suspect that is the case for a lighter-than-air application. The lift has to be "good enough".
In practice the steam from a boiler would be admitted to a cavity filled with air, and the objective would be to drive the air out until the cavity is filled with steam. My guess is that any water present in the steam would condense on the fabric and flow by gravity down to the bottom of the cavity where it would be drawn into the intake of the boiler.
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For the record:
It is unlikely anyone would ever take the risk of heating water with a microwave oven to inflate a balloon.
I am noting that as I investigate the possibility of using steam as a lifting gas, the method of heating water using microwave beams is an option.
I don't know if microwave energy can heat elements or molecules other that water. I expect there are other molecules that respond to microwave fluctuations.
Water has rightly earned human notice because of the invention of microwave ovens, and their near universal use in developed nations.
If a balloon (such as a weather balloon) were laid out on the ground deflated, and if a carefully measured amount of water were injected into the cavity and then sealed, I presume that a microwave beam could excite the water molecules until they form dry steam, with no liquid droplets left. At that point, the balloon would be expanded, and the balloon should lift.
The flight would be short, because the thermal energy injected into the water to make steam would rapidly flow out to the atmosphere through the wall of the weather balloon.
However, I presume that the balloon could be provided with additional energy via a microwave beam.
It is difficult for me to imagine this experiment being done for other than academic reasons, or perhaps to compete for a prize of some kind.
Perhaps Guiness Book of Records would take note of a flight of a steam balloon.
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This post is about the pressure required to inflate a balloon.
In the course of investigating the potential of steam as a lifting gas (as suggested by Calliban in early 2025), I wondered about the pressure that is needed to inflate a balloon. It turns out a gent named (Dr.) Ben Yelverton created a video on that very topic:
https://www.youtube.com/watch?v=2B9Yfa15uPg
The video covers the difference in pressure required to start filling a balloon, compared to the pressure needed after a certain point.
***
A related video is entitled: Measuring pressure inside a balloon as it's inflated
https://www.youtube.com/watch?v=LWrHVQDD1JE
In this video, Matthias set up a sensor to measure pressure as a balloon is inflated. The output was displayed on a computer monitor.
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This post is about any experiments that might have been done while heating water with microwaves. As Gemini suggests, it makes sense that vast amounts of research must have been done as microwave ovens were developed, and that would surely include heating water until it reaches a super critical state. However, in this go-round, Gemini did not offer any citations to study.
Conversation with Gemini
Hi Gemini! Thank you for your assistance with earlier questions! This evening I'm hoping your research ability will help to find if anyone has studied the behavior of water when it is heated with microwaves. I am familiar with warnings to consumers to beware of over heating water, because it can assume a supercritical state, and if it is disturbed while in that state, it can boil vigorously and cause injury. I am interested in using microwaves to heat a sample of water to make steam. Please note that we are talking about very small quantities of water, but I am exploring the possibility this might be an inexpensive way to deliver useful force in a mechanism. Can you find any records of anyone experimenting along these lines?
I'm glad to help you with your research on using microwaves to heat small quantities of water to make steam for a mechanical application.
You're right, there is a risk of superheating water in a microwave, which can cause it to boil vigorously and potentially cause injury. However, with careful experimentation and control, it may be possible to use microwaves to generate steam for your mechanism.
Unfortunately, I couldn't find any specific studies or records of experiments that directly address using microwaves to heat small quantities of water for this purpose. However, there is some relevant information that might be helpful:
Microwave heating of water: There is a considerable amount of research on the interaction of microwaves with water, including studies on the heating rates, temperature distribution, and phase transitions (boiling and superheating) of water in microwave fields. This research could provide insights into the behavior of small quantities of water when heated with microwaves.
Steam generation with microwaves: While not specifically for mechanical applications, there are some studies on using microwaves for steam generation in other contexts, such as industrial processes or sterilization. These studies might offer some relevant information on controlling the steam generation process with microwaves.
Superheating of water: There is a significant body of research on the superheating of water in microwaves, primarily due to safety concerns. This research could be helpful in understanding the conditions that lead to superheating and how to avoid it in your experiments.
Based on this information, it might be feasible to use microwaves to generate steam for your mechanism, but it would require careful experimentation and control to ensure safety and efficiency. Here are some suggestions for your research:
Start with small quantities of water: Begin with very small amounts of water to minimize the risk of superheating and potential hazards.
Use appropriate containers: Use microwave-safe containers that are suitable for high temperatures and pressures.
Monitor the temperature: Carefully monitor the temperature of the water during heating to avoid superheating.
Control the microwave power: Use a microwave with adjustable power settings to control the heating rate.
Experiment with different parameters: Vary the amount of water, container type, microwave power, and heating time to optimize the steam generation process.Remember, safety is paramount when working with microwaves and high-temperature water. Always take appropriate precautions and consult relevant safety guidelines before conducting any experiments.
I hope this information is helpful for your research. If you have any further questions or need additional assistance, feel free to ask.
Thank you! Your suggestions for where to look for further guidance are helpful. Your guidance for safe construction and operation of the experimental apparatus are helpful as well. While I'm away, I'd like to offer best wishes for success as you entertain similarly challenging questions from your customers.
You're very welcome! I'm glad I could assist you. I appreciate your kind wishes, and I wish you all the best with your research and experimentation. Feel free to reach out again if you have any further questions or need more help in the future. Good luck!
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This post is a follow up to the opening discussion:
Thank you for the nucleation suggestion! This is most applicable to the research effort we are pursuing. Our objective is to make a quantity of steam from a small sample of water, in order to provide pressure to operate a mechanism. If we can find a solution, we might be able to eliminate wire or any other signal passage, and thus allow the mechanism to operate remotely. As you can see, this work has nothing to do with food preparation, but we understand that the data you scan is coming primarily from the food preparation industry.
You're welcome! I'm glad the nucleation suggestion is helpful for your research. It makes perfect sense in the context of generating steam for a mechanical application. Controlling the boiling process is key to achieving consistent and predictable steam production.
You're right, much of the readily available information on microwave heating of water is geared toward food preparation and the associated safety concerns. However, the underlying physics and principles apply regardless of the specific application.
Your goal of creating a self-contained, remotely operable mechanism using microwave-generated steam is fascinating. Here's how the nucleation concept might be particularly relevant to your work, along with some additional thoughts:
How Nucleation Helps in Your Application:
Controlled Steam Generation: By providing nucleation sites, you can encourage the water to boil more evenly and predictably. This is crucial for generating steam at a consistent pressure and rate, which is essential for the reliable operation of your mechanism.
Preventing Explosive Boiling: In the absence of nucleation sites, the water can become superheated, and the sudden release of energy when it finally does boil can be problematic for a delicate mechanism. Controlled boiling through nucleation minimizes this risk.
Optimizing Steam Production: You might even be able to experiment with different materials or surface textures for your nucleation sites to optimize the steam generation process for your specific needs. For instance, a porous material might provide many nucleation sites.
Additional Considerations for Your Research:
Water Sample Size: Since you're working with small quantities of water, the effects of superheating and the benefits of nucleation will be even more pronounced.
Container Material: The material of the container holding the water will play a role in the heating process and could also potentially contribute to nucleation (some materials might have surface imperfections that act as nucleation sites). Consider materials that are microwave-transparent, chemically inert to water and steam, and able to withstand the pressures and temperatures involved.Microwave Control: Precise control over the microwave power and heating time is critical. A microwave with adjustable power levels will give you more flexibility.
Pressure Regulation: For your mechanism to operate reliably, you'll likely need some way to regulate the pressure of the generated steam. This might involve a pressure relief valve or other mechanism to vent excess steam.
Safety: Even with small quantities of water, it's essential to prioritize safety. Wear appropriate protective gear (eye protection, gloves) when experimenting, and ensure that your apparatus is designed to contain any potential pressure buildup.
I'm excited about the potential of your project. It sounds like a challenging but potentially very rewarding endeavor! If you have more specific questions as you progress with your research, feel free to ask. I'll do my best to help.
Thank you for staying with me as we pursue the question a bit further. The suggestion of using a porous material to receive the microwave energy is most interesting. We are looking for an alternative to the use of chemical explosives for our application. A water expansion system would be reusable, and suitable for equipment that will be operated in a remote site. Once again, I'll offer best wishes for similar success as you attempt to help your many other customers.
You're very welcome! I'm glad I can continue to assist you. The idea of using a porous material to absorb the microwave energy and generate steam for a reusable, remotely operated mechanism is indeed promising. It's a creative approach to finding an alternative to chemical explosives.
Using a porous material offers several potential advantages in this context:
Enhanced Nucleation: As we discussed earlier, the porous structure provides a multitude of nucleation sites, promoting controlled boiling and steam generation. This is crucial for achieving consistent and predictable performance.
Increased Surface Area: The large surface area of a porous material allows for more efficient absorption of microwave energy and faster heating of the water. This can lead to quicker steam generation.
Containment: The porous material can potentially contain the water, preventing it from splashing or leaking during the heating process. This could simplify the design of your mechanism.
Safety: By controlling the boiling process and containing the water, the porous material can contribute to safer operation of your device.
Here are some additional thoughts and considerations as you explore this approach:
Material Selection: The choice of porous material is critical. You'll need a material that is:
Microwave-transparent (or at least allows sufficient penetration of microwaves to heat the water).
Chemically inert to water and steam at high temperatures.
Thermally stable at the operating temperatures.
Mechanically strong enough to withstand the pressures generated by the steam.
Has the appropriate pore size and structure for efficient water retention and steam generation. Consider materials like porous ceramics, sintered metals, or even certain types of specialized polymers.
Water Delivery: You'll need a way to introduce the water into the porous material. Consider how you will reliably and repeatedly introduce small quantities of water into the porous material.
Steam Extraction: You'll also need a method to extract the generated steam from the porous material and direct it to your mechanism. This might involve carefully designed channels or ports within the material or the surrounding structure.
Testing and Optimization: Extensive testing will be necessary to optimize the design and operating parameters of your system. This will involve experimenting with different porous materials, water quantities, microwave power levels, and heating times.
Long-Term Durability: Consider the long-term durability of the porous material. Repeated heating and cooling cycles, as well as exposure to steam, could potentially degrade the material over time.Your project sounds very innovative, and I'm excited to see how it develops. I wish you the best of luck with your research and development efforts! Please don't hesitate to ask if you have more questions as you continue your work.
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In this post I asked Google's Gemini to compare convection heating to microwave heating. Gemini accepted the request to heat a sample of water from 25 Celsius to 100 Celsius, and found the number of joules that would be needed. The comparison given was of the time to heat the sample. No losses were considered.
What I'm attempting to discover is the practicality of heating steam in a balloon using microwaves. The exchange reported below was a step in that direction. The main benefit I see from this exchange is the calculation of the number of joules to be invested in the heating of 1 kg of water.
Conversation with Gemini
Hi Gemini! Thanks (again) for all your assistance with previous requests. Today I have what I hope will be a simple inquiry. I'd like to compare the efficiency of heating a sample of water using convection heating vs microwave heating. To try to help you set up the comparison, let's work with 1 kilogram of water at standard temperature and pressure. Let's raise the temperature of the water to 100 degrees Celsius. We'll have to take into account the differences in energy measurement so we will need a common measurement to make the comparison.
Hi! Happy to help again. This is a great question that gets at some fundamental differences in how we transfer energy. While it might seem like microwaves are always faster, the reality is a bit more nuanced when we look at the underlying physics.
Here's a breakdown of the comparison between convection heating (like on a stovetop) and microwave heating for your 1 kg of water scenario:
1. Energy Required (Joules):
The energy required to heat 1 kg of water from standard temperature (let's assume 25°C for this example) to 100°C is the same regardless of the heating method. We use the formula:
Q = mcΔT
Where:
Q = heat energy (Joules)
m = mass of water (1 kg)
c = specific heat capacity of water (approximately 4186 J/kg°C)ΔT = change in temperature (100°C - 25°C = 75°C)
Q = (1 kg) * (4186 J/kg°C) * (75°C) = 313,950 Joules
2. Power (Watts) and Time:
This is where the difference lies. Power is the rate at which energy is delivered (1 Watt = 1 Joule/second).
Convection Heating: A typical stovetop burner might provide around 1000-2000 Watts of power. Let's assume a 1500W burner for our example.
Time = Energy / Power = 313,950 J / 1500 W = 209.3 seconds (approximately 3.5 minutes)
Microwave Heating: A typical microwave oven might have a power output of 700-1200 Watts. Let's assume 1000W for our example.
Time = Energy / Power = 313,950 J / 1000 W = 313.95 seconds (approximately 5.2 minutes)
3. Important Considerations and Nuances:
Efficiency: The calculations above assume 100% efficiency in transferring energy to the water. In reality, both methods have losses. Some heat is lost to the surrounding air with convection, and microwaves might not be absorbed perfectly by the water. Microwaves are often more efficient at directly heating the water, but this can vary.
Heat Distribution: Convection heating relies on the water circulating to distribute heat evenly. Microwaves heat from the inside out, which can sometimes lead to uneven heating if not managed correctly (e.g., with a rotating platform).
Specific Heat Capacity: We used the specific heat capacity of pure water. If the water contains impurities, this value might change slightly.
"Standard Temperature": We assumed 25°C. If you start with colder water, the time required will increase for both methods.
Power Variations: The power output of both stovetop burners and microwaves varies. The calculations above are just examples.Conclusion:
Based purely on these simplified calculations, with the assumed power levels, the stovetop burner is faster in this specific scenario. However, the difference isn't huge. Microwaves often feel faster because they heat the water directly without needing to heat the container first. The actual time difference depends heavily on the specific power output of your appliances and the efficiency of heat transfer.
It's important to remember that this is a simplified model. Real-world cooking involves many other factors!
Heat capacity (video) | Thermodynamics - Khan Academy
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Ques. 2 (SMCQ) Marks: +3,-1 The specific heat water is 4186 J/Kg °C the specific heat ice is 2100 J/kg °C and the specific heat of steam is 2010 J/kg °C so how much heat is required to raise a 1 kg block of ice from - w ww.toppr.com
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Electricity and Energy Terms In Lighting (J, kW, kWh, Lm/W) - Stouch Lighting
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How much energy does an electric stove and oven use?
Source iconThank you for this helpful analysis and comparison. While I'm away, best wishes for similar success as you attempt to help all your other customers.
You're very welcome! I'm glad I could help. Best wishes to you as well!
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In this post I asked Google's Gemini to see what it could find about microwave energy delivery in the context of a steam filled cavity. It reminded me that the density of steam is less than that of water, so the amount of energy delivered to the molecules per volume is correspondingly less.
I'm not sure how that would influence the operation of a microwave heating system in this context, but expect that it would.
Conversation with Gemini
Hello Gemini! Thank you (again) for your assistance with earlier questions! In this session, I am hoping you can find any available information there might be about the use of microwaves to heat steam. Here is the scenario: We have filled a volume with steam that is free of water droplets. The pressure is not severe. The pressure is slightly above 1 bar. The temperature of the steam is above 100 Celsius but not far above. The question we are investigating is the use of a microwave system to feed energy into the volume of steam, to exactly balances losses due to convection at the perimeter of the volume. Can you find any studies or examples along these lines?
Hello! You're welcome, I'm glad I could help before. Heating steam with microwaves to balance convective losses is an interesting challenge. While it's not a common application like microwave cooking, the physics is sound, and there's definitely information to explore.
Here's a breakdown of what I've found, combining general microwave heating principles with the specifics of your steam scenario:
Key Considerations and Challenges:
Microwave Absorption by Steam: Water in its liquid state is a strong absorber of microwave radiation. Steam (gaseous water) absorbs microwaves, but significantly less than liquid water. The absorption efficiency depends on factors like the microwave frequency, steam temperature, and pressure. Higher pressure and temperature generally lead to better absorption, but it's still much lower than liquid water. This means you'll need a fairly powerful microwave source for a noticeable effect.
Uniform Heating: Microwaves can create uneven heating patterns due to standing waves. In your case, you'd need careful design of the microwave applicator (the structure that directs the microwaves) to ensure the steam is heated uniformly to balance the convective losses. This might involve multiple microwave sources or a rotating applicator.
Containment: You'll need a microwave-transparent container (likely made of a specialized glass or ceramic) that can withstand the steam temperature and pressure. This container must also be designed to minimize microwave leakage for safety reasons.
Efficiency: Because steam is a relatively poor absorber, a significant portion of the microwave energy might be reflected or transmitted. Efficiently directing the microwave energy into the steam volume will be crucial.
Condensation: If any part of the container or the surrounding environment is cooler than the steam temperature, condensation could occur, forming water droplets. This would drastically change the microwave absorption characteristics (since liquid water absorbs microwaves much better), and could lead to hot spots.
Maintaining a uniform temperature profile is critical.
Where to Look for Information:
Microwave Heating of Gases: Search for research papers or articles on "microwave heating of gases" or "microwave plasma generation." While not specifically about steam, these resources discuss the interaction of microwaves with gaseous substances and the factors affecting absorption.
Industrial Microwave Applications: Look into industrial microwave heating applications. While many focus on food processing or material drying, some might deal with heating gases for specific industrial processes. These could offer insights into equipment design and power levels.
Scientific Databases: Use keywords like "microwave," "steam," "heating," "absorption," "convection," and "heat transfer" in databases like IEEE Xplore, ScienceDirect, and Google Scholar. Focus on papers related to microwave engineering, thermodynamics, and heat transfer.
Microwave Equipment Manufacturers: Contact manufacturers of industrial microwave systems. They may have application notes or technical information related to heating gases, even if it's not their primary focus.
Computational Modeling: Simulating the microwave heating process using computational electromagnetics software (like COMSOL or HFSS) could be very helpful. This would allow you to optimize the microwave applicator design and predict the steam temperature distribution.
Specific Search Terms:
"Microwave interaction with water vapor"
"Dielectric properties of steam at elevated temperatures"
"Microwave heating of low-density gases"
"Industrial microwave heating of gases"
"Microwave applicator design for uniform heating"Example Research Areas (though likely not directly on your topic, but related):
Microwave sterilization of medical equipment (where steam is sometimes involved)
Microwave plasma generation for industrial processesIt's important to note: Direct, readily available studies on exactly your scenario (balancing convective losses in slightly superheated steam with microwaves) are unlikely to exist. You'll likely need to piece together information from related areas and potentially conduct some experimental work or simulations.
I'd recommend starting with a broad search on microwave heating of gases and then narrowing it down based on the information you find. The dielectric properties of steam at your specific temperature and pressure will be a crucial parameter for your calculations and design.
Let me know if you have any more specific questions, or if you'd like me to help you find information on a particular aspect of this problem.
Thank you for your helpful analysis and summary! I am curious about your report that steam is less able to absorb microwave energy. Is that due to lesser density? I would imagine the individual water molecules are responding the same whether in water or in steam.
You're right to question that! It's a common misconception. While individual water molecules do interact with microwaves similarly whether in liquid water or steam, the density is indeed the primary reason for the difference in bulk absorption. It's not that the individual H₂O molecules in steam are somehow "immune" to microwaves; it's that there are far fewer of them per unit volume compared to liquid water.
Here's a more detailed explanation:
Molecular Interaction: A water molecule (H₂O) is a polar molecule, meaning it has a slightly positive and slightly negative end. Microwaves, being electromagnetic radiation, have an oscillating electric field. This electric field interacts with the polar water molecules, causing them to rotate and try to align with the field. This rotation generates heat due to friction between the molecules. This is the fundamental mechanism of microwave heating.
Density Difference: Liquid water is much denser than steam. This means that in a given volume, liquid water contains vastly more water molecules than steam at a similar temperature and pressure (even slightly superheated steam).
Absorption Proportional to Molecular Density: The amount of microwave energy absorbed is directly related to the number of water molecules present to interact with the microwaves. Since steam has a much lower density, there are fewer water molecules available to absorb the microwave energy. Think of it like trying to catch raindrops: you'll catch a lot more in a downpour (high density) than in a light drizzle (low density), even if each individual raindrop is the same size.
Dielectric Properties: The interaction of microwaves with a substance is characterized by its dielectric properties. These properties describe how a material responds to an electric field. Liquid water has a much higher dielectric constant (especially the imaginary part, which relates to absorption) at microwave frequencies than steam. This difference in dielectric properties reflects the difference in molecular density.
Analogy: Imagine a room filled with ping pong balls. If you throw tennis balls at the room, you'll hit many more ping pong balls if the room is densely packed (like liquid water) than if it's sparsely filled (like steam), even if each individual ping pong ball reacts to the tennis ball the same way.
In summary, the reduced microwave absorption in steam isn't because the individual water molecules behave differently, but because there are simply far fewer of them in a given volume due to the lower density of steam compared to liquid water. The interaction at the molecular level is similar, but the bulk effect is dramatically different due to the difference in number density.
Do you have access to data on how much microwave energy is expended in movement of water molecules, compared to interactions with the wall of the volume in which the waves are generated?
I don't have direct access to real-time, measured data on the partitioning of microwave energy between water molecule movement and interactions with the container walls. This kind of data is highly specific to the exact microwave system, the geometry of the container, the properties of the wall material, the steam conditions (temperature, pressure, density), and the microwave frequency and power. It would typically be obtained through detailed experimental measurements or sophisticated simulations.
However, I can give you some insights into the factors that influence this energy partitioning and how one might approach estimating it:
Factors Influencing Energy Partitioning:
Dielectric Properties: The dielectric properties of both the steam (especially the imaginary part, which represents energy absorption) and the wall material are crucial. Materials with higher dielectric loss tangents at microwave frequencies will absorb more energy. Since steam's dielectric loss is much lower than liquid water, the energy absorbed by the steam will be relatively small compared to a material with a higher loss tangent.
Wall Material: The material of the container walls plays a significant role. Some materials (like certain ceramics or glasses) are relatively transparent to microwaves, meaning they absorb very little energy. Others might absorb a significant portion. This is usually quantified by the material's dielectric properties at the operating microwave frequency.
Steam Density: As discussed before, the density of the steam (related to pressure and temperature) directly affects the number of water molecules available to interact with the microwaves. Higher density means more absorption by the steam.
Microwave Frequency: The frequency of the microwaves influences the interaction with both the steam and the walls. Different materials have different absorption characteristics at different frequencies.
Container Geometry: The shape and size of the container affect the microwave field distribution and how the microwaves interact with the walls. Complex shapes can lead to uneven heating and localized hot spots.
Microwave Power: The power of the microwave source affects the overall energy input, but the partitioning of that energy depends more on the factors listed above.
Estimating Energy Partitioning (Conceptual):
Measure/Calculate Steam Absorption: You could, in theory, measure the temperature increase of the steam over a given time and relate that to the energy absorbed by the steam. This would require careful insulation to minimize heat losses. You could also try to calculate the absorption based on the steam's dielectric properties and the microwave field distribution.
Measure/Calculate Wall Absorption: Similarly, you could try to measure the temperature increase of the container walls (if possible) to estimate the energy absorbed by the walls. Alternatively, if you know the dielectric properties of the wall material, you could attempt to calculate the absorption.
Energy Balance: If you can accurately estimate the energy absorbed by the steam and the walls, you can get a sense of the energy partitioning. The remaining energy (if any) would account for losses due to microwave leakage, or other mechanisms.
Simulations: Computational electromagnetics software (like COMSOL or HFSS) can be used to simulate the microwave field distribution and energy absorption in the system. This would require accurate input parameters for the steam and wall materials. This is probably the most practical approach.
Challenges:
Accurate Measurements: Accurately measuring the small temperature changes in the steam and the walls can be challenging, especially if the heating rates are low.
Dielectric Properties: Obtaining accurate dielectric properties for the steam and wall materials at the operating temperature and frequency can be difficult.
Complex Geometry: Simulating complex container geometries can be computationally intensive.In summary: While I can't provide the data directly, I've outlined the factors influencing energy partitioning and a conceptual approach to estimating it. Experimental measurements or, more practically, computational simulations would be necessary to obtain specific data for your system.
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Source iconThank you for your careful analysis, overview of the problem, and list of resources! While I'm away, best wishes for similar success assisting your other customers!
You're very welcome! I'm glad I could help. Best wishes to you as well!
An interesting detail is the observation that any water droplets that might form would attract more energy from the microwaves. I would expect that fact would tend to eliminate water droplets when they form, as they would at the walls of the cavity.
It seems reasonable to suppose that the walls of the cavity need to reflect microwaves with the minimal possible loss.
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