Naked Science Forum
General Science => General Science => Topic started by: neilep on 29/01/2008 23:33:31
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Dear Baloonologists !
This is me on my way to work in Cloud Cuckoo Land !
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No, I'm not the cow balloon ..I'm a passenger !
Anyway,
Why is it best to go ballooning in the morning ? It's not the light...it's something else !.....As far as I know ...ballooning just works best in the A.M rather than the P.M. !!
Do ewe know ?..I don't !!...I want to though !!
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Early morning air is colder, and so denser. Since ballooning depends on the low density of the hot air, or helium, within the balloon being lighter than the air outside the balloon, so the colder and denser the air outside, so the greater the difference, and the better the lift.
It is in essence not so difference from the reason turbo charged engines use intercoolers to cool in air entering the cylinders, so increasing the density of the air (although in that case the object is not buoyancy).
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Just a quickie before i get some beauty sleep - yes i know i don't really need it, but every little helps.
When is the coldest part of the day? Just after dawn, why is that then you may ask. Well as the sun rises, it brings in warmer air as that air arrives it pushes up the cooler air. The temperature continues to fall until there is more warm air arriving than departing, this can take some time to happen.
This will in turn continue to push more cooler air up, thus giving the effect explained by George.
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COOL..SO IT IS BETTER TO RIDE ONE IN THE MORNING...???
I WANT A RIDE...
So what do they do in the afternoons how does it change the adjustments to do it, to get the lift you need then..???
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Early morning air is colder, and so denser. Since ballooning depends on the low density of the hot air, or helium, within the balloon being lighter than the air outside the balloon, so the colder and denser the air outside, so the greater the difference, and the better the lift.
It is in essence not so difference from the reason turbo charged engines use intercoolers to cool in air entering the cylinders, so increasing the density of the air (although in that case the object is not buoyancy).
I don't think that the ambient temperature is the only relevant factor; you have to raise the temperature of the air in the balloon by the same amount to give it the same bouyancy so, afaics, the same amount of energy is needed.
But there is always the temperature gradient to take into account. If the air, high above, is colder under dew - forming conditions than when there is no dew then you will get more bouyancy once you have got further up. The energy you gave your air should take you up faster (/further?) if the air at high altitude is more dense.
I do have a problem with energy conservation laws in mind, tho'. You won't get higher than the limit imposed when you equate GPE gained with Thermal energy put in, however you achieve the lifting.
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WOW !!
I am in awe of the incredible information given to me here.
THANK YOU George, Paul, Sophiecentaur
I consider my question well and truly answered....and I understand it too !! [:o]
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I don't think that the ambient temperature is the only relevant factor; you have to raise the temperature of the air in the balloon by the same amount to give it the same bouyancy so, afaics, the same amount of energy is needed.
273°K + 200°K = 473°K, which is 73% increase in temperature; whereas 313°K + 200°K = 513°K, which is a 63% increase in temperature. Since density is proportionate absolute to temperature, so a 200°K increase in temperature from an ambient temperature of 273°K will give you 13% more relative decrease in ambient density than would be achievable with a 200°K increase at 313°K ambient temperature.
Yes, I realise that 313°K are extreme conditions that one expects in tropical desserts, and the most you are likely to get within the UK will probably be no more than around 303°K on a hot summers day (maybe 305°K), but more likely around 298°K.
The other factor is that the amount of water vapour that warmer air can hold is greater, and this too would reduce the density of the air.
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If you really want to get technical (and I'm sure you do!) then you have to take into account the fact that a hot air balloon has a constant volume. The extra air, after expansion, will have escaped. The actual mass you need to heat up is less on a warm day and so you need less energy to raise its temperature. This needs to be taken into account, too, and I think it works against A-S's calculated difference. I can't be naffed to do it at the mo; anyone else want a go?
I don't understand the statement about water vapour; before it has dropped below dew point, there is the same amount of water in air at any temperature. I think the water is irrelevant - except to add some extra payload as it condenses on the basket and ropes.
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If you really want to get technical (and I'm sure you do!) then you have to take into account the fact that a hot air balloon has a constant volume. The extra air, after expansion, will have escaped. The actual mass you need to heat up is less on a warm day and so you need less energy to raise its temperature. This needs to be taken into account, too, and I think it works against A-S's calculated difference. I can't be naffed to do it at the mo; anyone else want a go?
I don't understand the statement about water vapour; before it has dropped below dew point, there is the same amount of water in air at any temperature. I think the water is irrelevant - except to add some extra payload as it condenses on the basket and ropes.
Sorry, not my understanding of how a hot air balloon is heated.
It may be partly true of a helium balloon - except that high altitude helium balloons start out partly inflated for exactly to allow them room to expand as they rise.
As for a hot air balloon, it is blown up using the hot exhaust gasses of the burners - so it does not actually heat any existing air in the balloon, because before it is heated it only has a minimal amount of air in it in the first place -then the burners are pointed into the balloon, and the hot exhaust flows into the balloon, not having to heat up anything except itself.
In any case, even if what you said was true, that would only apply to the amount of energy required to heat up the balloon (i.e. it would require a little more fuel burn on the ground - not any more balloon volume or weight when in the air).
As for the issue of the amount of water vapour in air (at the same relative humidity)
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fen%2F9%2F91%2FDewpoint.jpg&hash=786630607347fcc9f333600ba6183c62)
http://en.wikipedia.org/wiki/Water_vapor#Air_and_water_vapor_density_interactions_at_equal_temperatures
At the same temperature, a column of dry air will be denser or heavier than a column of air containing any water vapor. Thus, any volume of dry air will sink if placed in a larger volume of moist air. Also, a volume of moist air will rise or be buoyant if placed in a larger region of dry air. As the temperature rises the proportion water vapor in the air increases, its buoyancy will become larger. This increase in buoyancy can have a signicant atmospheric impact, giving rise to powerful, moisture rich, upward air currents when the air temperature and sea temperature reachs 25°C or above. This phenomenon provides a significant motivating force for cyclonic and anticyconic weather systems (tornados and hurricanes).
Since a balloon will be more buoyant in denser air, so a balloon will be more buoyant in air containing less water vapour. You will note from the graph that warmer air is capable of holding more water vapour (although it may reasonably be argued that the fact that it is capable of holding more water vapour does not mandate it does hold more water vapour, but air at 100% humidity at 40°C will contain more water vapour than air at 100% humidity at 10°C).
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You have got me thinking.
The relative humidity is not relevant; the air in the balloon will contain the same mix of air / water at all altitudes (apart from the small amount of very hot air / exhaust you inject from the burner - only a top-up, though).
What you say about the difference between humid and dry air densities must be the clue. I think it's like this: It depends upon a difference in humidities at different altitudes. IF the air at ground level is humid (dewy) then it starts off being less dense than dry air higher up would be if you got a bag of it and brought it down to the ground (if the temperatures were the same).
The balloon would actually experience some upthrust, for free, if it were taken to an altitude because of the inherent density difference you start with even before you heat the air up.
My problem with the energy issue only arises in an ideal case where the atmosphere is a uniform mix and the pressure drops off in a normal fashion. The fact is that there is some energy for free, available, courtesy of the atmospheric situation.
A helium balloon is a different matter - it contains the same mass of Helium all the time. The energy available for lifting its payload comes from the energy actually involved in obtaining the helium (probably an enormously inefficient process, involving huge entropy changes).
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The relative humidity is not relevant; the air in the balloon will contain the same mix of air / water at all altitudes (apart from the small amount of very hot air / exhaust you inject from the burner - only a top-up, though).
I was not talking about the air within the balloon, but the air without it.
Buoyancy is based on the difference in the density of air within and without the balloon, so anything that increases the air without the balloon will increase the buoynacy of the balloon, assuming the same air within the balloon.
What you say about the difference between humid and dry air densities must be the clue. I think it's like this: It depends upon a difference in humidities at different altitudes. IF the air at ground level is humid (dewy) then it starts off being less dense than dry air higher up would be if you got a bag of it and brought it down to the ground (if the temperatures were the same).
Dewy air is not a measure of water vapour content. One gets due when water vapour precipitates out of the air, so reducing the amount of water vapour left in the air.
Nonetheless, you would be right that you would expect the air to be cooler, and dryer, as you climb - but this difference would be even greater at midday, when there is even more water vapour in the low altitude air than it would be at dawn, when the colder air can hold less moisture.
A helium balloon is a different matter - it contains the same mass of Helium all the time. The energy available for lifting its payload comes from the energy actually involved in obtaining the helium (probably an enormously inefficient process, involving huge entropy changes).
Mass is not the issue - density is the issue - which is why helium balloons that are destined for high altitude are only partly inflated, so that they have room to expand their volume, so further reducing their density without any change in mass.
The point is that lift is not from upthrust, but from displacement (the rising balloon displaces air as it rises, and that air is pushed down, and as long as the overall volume of air it displaces carries a greater mass than mass of the balloon, so the air will fall, and the balloon will rise to fill the void).
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Let's clear up the helium balloon thing. A Helium balloon contains the same MASS all the time. A hot air balloon contains the same VOLUME at all times; there's a hole in the bottom. That is the fundamental difference between them. They can't be expected to behave the same way, in detail. If you want to, we can discuss it elsewhere. But I'm not sure why you differentiate between upthrust and displacement - the upthrust is equal to the mass displaced (Archimede's principle). Also it would be lunatic to use a constant volume Helium balloon; its ceiling would only be a few hundred feet and you would waste all that expensive Helium!
Now for some hot air:
Perhaps I was writing too much in shorthand. I am comparing two situations.
1. Where the fraction of water, by mass, is the same at altitude as on the ground (ignoring the saturation, which doesn't seem relevant).
2. Where the fraction of water is higher near the ground.
We have to look at the difference in densities of inside and outside air in order to find the upthrust ~(of course).
At a given height above the ground, the balloon will experience more upthrust in condition 2. (even if the pressures and temperatures were the same) because of the density difference (Imagine hoisting a 'cold' balloon and measuring its weight up there; the upthrust would depend only upon the proportion of water it started off with, relative to the proportion of water in the air around it at altitude). N.B. - volume is always the same inside the balloon. This gives a mechanism which seems to explain what they find.
We now have to find why there should be a higher fraction of water in the air at ground level. As you say, one might expect more mass after the Sun has been boiling off surface water so we have a paradox.
Dew forms when the ground is colder than the air above; this can happen on clear nights when the ground radiates directly into space, rather than into clouds, which radiate back some energy. Ground fog or 'radiation fog' forms under these conditions. Is this relevant? Does the presence or absence of clouds affect the water content in the air below them?
Also, I think they say that dew forms better in still air. This implies that there would be less mixing of air at different altitudes; is that relevant?
One final point- and this is another potential explanation. The dew forms right near the ground. A balloon is a tall structure; 20m high?. Even before it has taken off, it is displacing air which is not, actually on the ground. Assume that the actual water content of air all the way up is constant BUT the air right at ground level contains droplets of (extra) water (could this be due to precipitation from higher up? If it only came from the ground level air then this suggestion doesn't hold water, either.) which actually increases the fractional mass in that air, locally. The burner heats this air up and injects the balloon with air which now has a higher fraction of water than the air surrounding the main part of the balloon. Even on the ground, there would be more upthrust.
I think we need input from a meteorologist. Help please.
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It's going to be easier to directly quote the met.office, than for me to accurately explain dew. Is this what you are after?
Dew
Dew is the condensation of water vapour on a surface whose temperature is at or below the dew-point of the air it is in contact with. Dew appears as innumerable small water droplets less than a millimetre in diameter. The most common natural surface upon which dew forms is vegetation, and in particular, grass.
'Dew point' is defined as the temperature at which the air, when cooled, will become saturated. Let us take as an example a day in which the air temperature reaches 18 °C with a dew point of 8 °C. Late in the afternoon, the air temperature begins to fall, but the dew point will still be around 8 °C. However, the air temperature is measured at 1 metre above the ground and, under a clear sky, the temperature of some objects may be significantly lower, due to loss of heat by radiation. Once the temperature of the object has fallen below the dew point, water vapour begins to condense on to it in the form of dew
Dew forms readily on grass because (a) the temperature falls more rapidly nearer to the grass and (b) the grass leaves produce water vapour, which raises the dew point of the air immediately in contact. Dew does not form as readily on other surfaces, such as soil, brick or stone. This is because these materials absorb heat from the sun which is then slowly emitted during the evening, causing the temperature of air immediately in contact to stay above the dew point for much longer than over grass.
Next morning, as the incoming solar radiation strengthens, the dew evaporates. Metal surfaces, such as car bodies, will dry relatively quickly whereas grass stays damp for considerably longer. In fact, from late autumn to early spring, in some places shaded from the sun, grass may remain damp all day after heavy dew.
The key meteorological factors suitable for the formation of dew are:
1. Night time (to eliminate incoming solar radiation).
2. Clear skies (to allow for maximum energy loss due to long wave radiation).
3. Calm winds (to prevent mixing with warmer air aloft).
4. A moisture source (best with high dew-point, or after daytime rain, to promote condensation).
These conditions are often satisfied under a high pressure area, particularly a high pressure formed in warm tropical maritime air such as an extension of the Azores sub-tropical pressure towards the United Kingdom.
Dew should not be confused with deposits from wet fogs or guttation (the exuding of liquid water from the tips of plants, usually under conditions of a warm, moist soil).
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Let's clear up the helium balloon thing. A Helium balloon contains the same MASS all the time. A hot air balloon contains the same VOLUME at all times; there's a hole in the bottom. That is the fundamental difference between them. They can't be expected to behave the same way, in detail. If you want to, we can discuss it elsewhere. But I'm not sure why you differentiate between upthrust and displacement - the upthrust is equal to the mass displaced (Archimede's principle). Also it would be lunatic to use a constant volume Helium balloon; its ceiling would only be a few hundred feet and you would waste all that expensive Helium!
Depends what you are trying to do.
If you want to send a weather balloon into the stratosphere (where, by the way, hydrogen can be used in place of helium), then I agree that constant volume is nonsense. If you want a manned balloon/airship without pressurised cabins or oxygen masks, then constant volume is often adequate.
Ofcourse there are differences in the limits of what a hot air balloon can do and what a helium/hydrogen balloon can do; but the basic principles of lift are the same.
Whether a hot air balloon is constant volume is an interesting question. Both the helium and the hot air balloon have a maximum volume they can reach, as dictated by the maximum size the envelope they are contained in can stretch. The difference is that there is a clear demarcation between the helium used in a helium balloon and the environmental air; whereas no such clear demarcation exists with a hot air balloon, so the situation with a hot air balloon is quite ambiguous.
Let me give you an example of this ambiguity. If we had within the large open canopy of a hot air balloon, a smaller enclosed canopy that actually contained the hot air, but was allowed to expand into the larger canopy - this would be analogous to the fixed mass situation of a helium balloon. The remaining air within the outer canopy is of no significance, because for the purposes of our analogy we can conider it to be of the same density as the ambient air, and if it happened to be any heavier, it would fall out of the bottom anyway, since there is a hole in the bottom of the outer canopy (but not of the inner canopy).
The fact that there is no actual inner container does not mean that a similar volume of hot air does not exist within the canopy (simply that it is not bounded by a container). So the question is, is there any practical difference between viewing the air within the single large canopy as being a mix of some air that is hot, light, and low density, and a lot of other air that is simply at ambient temperature and density; and viewing it as a single mass of lower temperature air without any separation of hot and cold air. If the two are equivalent, then there is no theoretical difference between constant mass and constant volume (clearly, the structures are different, in that a constant volume hot air balloon would need to manage higher temperatures - but functionally a constant volume hot air balloon can be considered equivalent to a constant mass hot air balloon at a higher temperature).
Now for some hot air:
Perhaps I was writing too much in shorthand. I am comparing two situations.
1. Where the fraction of water, by mass, is the same at altitude as on the ground (ignoring the saturation, which doesn't seem relevant).
Why is saturation irrelevant? Is not condensation that which happens when air is saturated with water at one temperature, and then the temperature falls, thus meaning the amount of water the air can contain reduces, causing precipitation of water out of the air in order to ensure the amount of water vapour remaining in the air is at or below the saturation point for the given temperature?
Possibly you mean saturation is irrelevant insofar as precipitation does not contribute to air density, so what matters is only how much water vapour is left in the air - but the amount of water vapour left will depend on what the saturation level for water vapour is at a given temperature.
2. Where the fraction of water is higher near the ground.
We have to look at the difference in densities of inside and outside air in order to find the upthrust ~(of course).
This is ofcourse the case (at least when looking at very high altitudes) - which is why aeroplanes only need to worry about ice on the wings at low altitude, even though (or even because) temperatures are lower at high altitude (i.e. above the altitude of the clouds, there is not enough water vapour in the air to create ice on the wings).
At a given height above the ground, the balloon will experience more upthrust in condition 2. (even if the pressures and temperatures were the same) because of the density difference (Imagine hoisting a 'cold' balloon and measuring its weight up there; the upthrust would depend only upon the proportion of water it started off with, relative to the proportion of water in the air around it at altitude). N.B. - volume is always the same inside the balloon. This gives a mechanism which seems to explain what they find.
But if you look at the graph above, you will see that at saturation levels, only 1% to 4% of the air is water vapour - so you only have maybe 3% difference to play with. By comparison, as I pointed out, simply in terms of looking at the difference is air density due to temperature alone can add about 13% (in extreme differences - but easily half that) to the lift. I do not say that your mechanism is wrong, only that it is the lesser factor.
We now have to find why there should be a higher fraction of water in the air at ground level. As you say, one might expect more mass after the Sun has been boiling off surface water so we have a paradox.
I don't see a paradox.
Air, at lower altitudes (ignoring the first metre - unless you are looking at micro-balloons) will get colder as you rise, and colder air holds less water vapour.
Dew forms when the ground is colder than the air above; this can happen on clear nights when the ground radiates directly into space, rather than into clouds, which radiate back some energy. Ground fog or 'radiation fog' forms under these conditions. Is this relevant? Does the presence or absence of clouds affect the water content in the air below them?
Also, I think they say that dew forms better in still air. This implies that there would be less mixing of air at different altitudes; is that relevant?
One final point- and this is another potential explanation. The dew forms right near the ground. A balloon is a tall structure; 20m high?. Even before it has taken off, it is displacing air which is not, actually on the ground. Assume that the actual water content of air all the way up is constant BUT the air right at ground level contains droplets of (extra) water (could this be due to precipitation from higher up? If it only came from the ground level air then this suggestion doesn't hold water, either.) which actually increases the fractional mass in that air, locally. The burner heats this air up and injects the balloon with air which now has a higher fraction of water than the air surrounding the main part of the balloon. Even on the ground, there would be more upthrust.
I think we need input from a meteorologist. Help please.
No sane person ever goes ballooning in strong winds (at least, where winds are strong at ground level) - it would be suicidal - you would not be able to manage the balloon on the ground.
Incidentally, the burner injects the balloon with hot exhaust gasses, which is a mixture of hot air, CO2, and water vapour. The CO2 would be the heavier component, so would fall out of the bottom of the balloon first as it cooled, and got replaced by fresh hot gasses.
Another thing to bear in mind is that in the morning, the ground is cold, so it will not create thermals. By later in the day, as the ground heats up, it will create thermals. Thermals are good for gliders, since they can be sucked up by the rising air, but for balloons, the reason the thermals are rising is because of their lower density, so by virtue of that it means the balloon will be more likely to be falling (or at least rising less fast) in the thermal.
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Instead of a divergent conversation, it would be more use if we tried to find the nub of the problem.
Whether a hot air balloon is constant volume is an interesting question.
Once it is full, it displaces exactly the same volume of surrounding air at all altitudes (any extra expansion will just push air out of its bottom). This is unlike a Helium balloon, which expands as it rises and displaces a greater and greater volume of the surrounding air until the envelope goes taught. This happens very high up, though.
Why is saturation irrelevant?
It is irrelevant to the water content once the temperature is above and stays above, dew point. The water has the same proportional effect at all temperatures and pressures once it is in vapor form.
Until condensation occurs, the relative densities of totally dry air and non-dry air will be the same; you are dealing, effectively, with ideal gas mixtures.
SO, for this claimed improvement in lift on a dewy morning to work, there must be, either some odd temperature gradient - resulting in colder air at a given altitude than you normally get - producing more upthrust on your balloon than normal or the air in the balloon must have more water than normal in it, making it less dense than it would be, normally. Either of those two factors could account for the better ballooning conditions.
In considering the addition of CO2 to the mixture in the balloon, increasing its density, you must also add the extra H2O molecules which result from the burning of a hydrocarbon.
My back of an envelope calculations tell me that the 5mols of O2 combine with ,say, one mol of Propane (C3H8) to produce 3mols of Co2 and 4 mols of H2O. The overall result is a less dense mixture than the Oxygen we started off with. The amounts of CO2 and H2O will not change in a hurry like you suggest because the rate of diffusion is extremely slow; CO2 will not 'just fall out of the bottom' for hours or even days.
I suppose my water theory is not likely to be right on the grounds that it would mean you could get extra lift just by spraying water into the flame of the burner and everyone would be doing it. Unless I've just invented a new technique!!! Unlikely, I fear.
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Once it is full, it displaces exactly the same volume of surrounding air at all altitudes (any extra expansion will just push air out of its bottom). This is unlike a Helium balloon, which expands as it rises and displaces a greater and greater volume of the surrounding air until the envelope goes taught. This happens very high up, though.
OK, I am having difficulty explaining what I mean in words, so I thought I'd draw a picture or two.
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So, is the warm air balloon (the image on the bottom right) really so different from the hot air inside the cold air image next to it?
In real life, the difference would anyway not be clear cut, since in the real world, the temperature inside the balloon would not be constant, and the gasses at the top of the balloon would be hotter than those at the bottom of the balloon, even if there is no membrane partition between them.
It is irrelevant to the water content once the temperature is above and stays above, dew point. The water has the same proportional effect at all temperatures and pressures once it is in vapor form.
Until condensation occurs, the relative densities of totally dry air and non-dry air will be the same; you are dealing, effectively, with ideal gas mixtures.
But the whole point is that condensation does occur, so reducing the moisture content of the air. This is what happens in the clouds, and what happens on a dewy morning.
If condensation does not happen at that time, the air could have been exposed to both condensation and evaporative processes before it arrived there, so the amount of moisture would be effected by that.
I do agree that if you have cold, dry, air; and you add heat to that air, it will not increase its moisture content (unless it has liquid water nearby that can evaporate - but, ofcourse, if you have dew lying on the ground, that is liquid water that can evaporate as the sun heat it). What I am saying is that if moist warm air, and to reduce its temperature, some of that moisture will condense out (whether this is large drops of rain/dew, or whether it is imperceptibly small droplets of liquid water, it still condenses out).
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I thought we were trying to explain why hot air balloons behave the way they do. Your two inventions are interesting thought experiments but how do they relate to real balloons? Weight is at a premium in ballooning so why add an extra envelope?
(edited later today) I have re-read this and now I see where you are coming from; however, as far as I can see, this situation (effectively variable volume) only arises after a long time in a hot air balloon.
All my ideas relate to the air IN the balloon; it is the only air that is undergoing a change in conditions as the balloon goes up. If that air starts off at ambient temperature and then gets warmer, how ever can it reach its dew point?
I really don't think that the air that has been introduced into the envelope will be at all still; it has been blow in and will be turbulent, well mixed and the same temperature throughout. As it rises, initially, and expands, warm air will escape out of the bottom - a constant volume of lifting gas. There is no reason for much convection/heat loss to occur once it has stabilised but, eventually, you will end up with cooler air at the bottom - which is why you need to top up with the burner. I don't think this is relevant at takeoff. Stratification may occur when you give a top-up burst to keep your altitude after some while - we would need to seek further advice on this.
So, apart from arguing with me, do you have an answer, other than my two suggestions, for the original question?
btw, the bit about CO2 and H2O giving an overall decrease in density has been verified by colleagues.
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(edited later today) I have re-read this and now I see where you are coming from; however, as far as I can see, this situation (effectively variable volume) only arises after a long time in a hot air balloon.
I don;t think that is the case, but I agree it is a side issue.
All my ideas relate to the air IN the balloon; it is the only air that is undergoing a change in conditions as the balloon goes up. If that air starts off at ambient temperature and then gets warmer, how ever can it reach its dew point?
I don't quite understand this statement.
The air inside the balloon changes no more or less than that outside of the balloon. Certainly, as the balloon rises, it encounters (outside of the balloon) different air; and as the air warms up due to sunshine, the air outside of the balloon changes. By comparison, if the temperature of the air within the balloon is constant, then one would expect fairly little change in the air within the balloon.
I really don't think that the air that has been introduced into the envelope will be at all still; it has been blow in and will be turbulent, well mixed and the same temperature throughout. As it rises, initially, and expands, warm air will escape out of the bottom - a constant volume of lifting gas.
Again, I am not sure I understand what you are saying here.
There is no 'blower' or fan to 'blow' the air into the balloon. The only thing in the basket is a propane burner, no fan. So the only reason the how exhaust gasses enter the balloon is because they rise up from the burner, which they do because they are less dense than the cold air they encounter, which means they have already undergone expansion. Certainly, as fresh hot exhaust gasses are introduced into the canopy, one would expect turbulence, I would not deny that; but they would already be less dense than the gasses they are displacing in the canopy even before the enter the canopy.
There is no reason for much convection/heat loss to occur once it has stabilised but, eventually, you will end up with cooler air at the bottom - which is why you need to top up with the burner. I don't think this is relevant at takeoff. Stratification may occur when you give a top-up burst to keep your altitude after some while - we would need to seek further advice on this.
OK, I see what you are getting at - but I don't actually think that stratification is a prerequisite for my argument, it merely makes the argument clearer.
If you simply have one hot (fast) molecule next to one cold (slow) molecule, even if they are not compartmentalised into different layers, I would suggest my argument still holds true.
But as you point out, it is a side issue to the main question.
btw, the bit about CO2 and H2O giving an overall decrease in density has been verified by colleagues.
I never had any doubt of that - that is the advantage of using a low carbon fuel, such as propane, rather than a higher carbon fuel such as paraffin.
So, apart from arguing with me, do you have an answer, other than my two suggestions, for the original question?
That was what I had started out with:
273°K + 200°K = 473°K, which is 73% increase in temperature; whereas 313°K + 200°K = 513°K, which is a 63% increase in temperature. Since density is proportionate absolute to temperature, so a 200°K increase in temperature from an ambient temperature of 273°K will give you 13% more relative decrease in ambient density than would be achievable with a 200°K increase at 313°K ambient temperature.
Yes, I realise that 313°K are extreme conditions that one expects in tropical desserts, and the most you are likely to get within the UK will probably be no more than around 303°K on a hot summers day (maybe 305°K), but more likely around 298°K.
In other words, if the external temperature is 30°K colder, you can get 13% more lift from the balloon for the same increment in temperature.
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By comparison, if the temperature of the air within the balloon is constant, then one would expect fairly little change in the air within the balloon.
How about the PRESSURE and TEMPERATURE? The gas laws surely apply as the air expands out of the hole in the bottom. The burner can help keep temperature constant but it can't do anything about the pressure.
I never had any doubt of that - that is the advantage of using a low carbon fuel, such as propane, rather than a higher carbon fuel such as paraffin.
You never mentioned the Hydrogen, though - you implied an increase in density.
There is no 'blower' or fan to 'blow' the air into the balloon. The only thing in the basket is a propane burner, no fan.
Do you think the warm air just slips quietly to the top of the balloon? The burner produces a serious jet of flame at high speed, pulling air with it; lots of turbulence, convection and mixing involved.
The air inside the balloon changes no more or less than that outside of the balloon.
The burner keeps the temperature high. The pressure outside is decreasing as the balloon rises - and so does the pressure inside. The density of the air outside is reducing as it rises and the outside temperature drops.
273°K + 200°K = 473°K, which is 73% increase in temperature; whereas 313°K + 200°K = 513°K, which is a 63% increase in temperature. Since density is proportionate absolute to temperature, so a 200°K increase in temperature from an ambient temperature of 273°K will give you 13% more relative decrease in ambient density than would be achievable with a 200°K increase at 313°K ambient temperature.
Actually, density is INVERSELY proportional to absolute temperature.
Remember that expanding air escapes out of the hole in the bottom of the balloon and does not displace surrounding air and provide lift. Your calculations seem to ignore this.
I think the basic statement about ideal conditions refers to how the initial lift on the ground is affected by ambient temperature - all the rest is just added complication.
The maximum temperature that the envelope can stand is about 120 Celsius (figure from Wikkers), due to the limitations of the fabric and desired lifetime. Perhaps this is the underlying reason - very cold air on the ground means you can use a bigger difference for the air injected into the balloon (and, to start with, it is injected - to get the thing inflated) because of the material of the envelope.
Anyway, you can't beat some calculations so here goes:
The relative density of a balloon full of air at 120 Celsius will be
Q(273+0)/(273+120) = Q(273/413 = 0.661Q compared with the ambient air when the ambient temp is zero and,
in comparison,
Q(273+20)/(273+120)=Q293/413 =0.709Q when the ambient air temperature is 20 Celsius.
(Q is a constant and the density of the ambient air is 1)
In both cases the volume displaced is the same so the upthrust will be equal to the difference in densities times the volume, which is the weight of displaced air minus the weight of air inside.
The relative lift at 0 Celsius will be 1-0.661 = 0.339
the relative lift at 20 Celsius will be 1-0.709 = 0.0.291
This seems to represent about 14% greater lift.
That seems to be a very good reason for it to work.
Please check my arithmetic.
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How about the PRESSURE and TEMPERATURE? The gas laws surely apply as the air expands out of the hole in the bottom. The burner can help keep temperature constant but it can't do anything about the pressure.
Pressure, yes - but would the temperature inside the canopy really be effected by altitude?
I never had any doubt of that - that is the advantage of using a low carbon fuel, such as propane, rather than a higher carbon fuel such as paraffin.
You never mentioned the Hydrogen, though - you implied an increase in density.
I mentioned CO2 and water vapour. It may have been unclear that I was talking about water vapour produced by combustion, but that was the intent. In any case, the original comment was only meant as an aside (as, whatever the composition of the gas, that composition would not change significantly at different altitudes or different outside temperatures), so I had not expected to have to be anything but cursory.
There is no 'blower' or fan to 'blow' the air into the balloon. The only thing in the basket is a propane burner, no fan.
Do you think the warm air just slips quietly to the top of the balloon? The burner produces a serious jet of flame at high speed, pulling air with it; lots of turbulence, convection and mixing involved.
Yes, but the jet is not caused by a fan, but by the expansion caused by the heat.
The air inside the balloon changes no more or less than that outside of the balloon.
The burner keeps the temperature high. The pressure outside is decreasing as the balloon rises - and so does the pressure inside. The density of the air outside is reducing as it rises and the outside temperature drops.
No problem with that.
273°K + 200°K = 473°K, which is 73% increase in temperature; whereas 313°K + 200°K = 513°K, which is a 63% increase in temperature. Since density is proportionate absolute to temperature, so a 200°K increase in temperature from an ambient temperature of 273°K will give you 13% more relative decrease in ambient density than would be achievable with a 200°K increase at 313°K ambient temperature.
Actually, density is INVERSELY proportional to absolute temperature.
Sorry, my mistake - but the mistake was in the typing not in the intended content.
On the other hand, my calculations did indicate just such an inverse relationship (it talks about the decrease in density being greater).
Remember that expanding air escapes out of the hole in the bottom of the balloon and does not displace surrounding air and provide lift. Your calculations seem to ignore this.
My calculations refer to density - neither mass nor volume specifically.
I think the basic statement about ideal conditions refers to how the initial lift on the ground is affected by ambient temperature - all the rest is just added complication.
Yes, to some extent I think this is partly true.
There are also issues with people attempting altitude records also desiring cold mornings, but in part that may be because the early morning also gives them time, as well as more lift near the ground.
But there is also the problem higher up, if the heating of the ground later in the day creates thermals.
The maximum temperature that the envelope can stand is about 120 Celsius (figure from Wikkers), due to the limitations of the fabric and desired lifetime. Perhaps this is the underlying reason - very cold air on the ground means you can use a bigger difference for the air injected into the balloon (and, to start with, it is injected - to get the thing inflated) because of the material of the envelope.
Yes, that does sound very plausible (although I am not aware of balloons having thermometers being attached to the top of the canopy, so it does not seem that they must be getting that close to the limit as to have to monitor it closely).
Anyway, you can't beat some calculations so here goes:
The relative density of a balloon full of air at 120 Celsius will be
Q(273+0)/(273+120) = Q(273/413 = 0.661Q compared with the ambient air when the ambient temp is zero and,
in comparison,
Q(273+20)/(273+120)=Q293/413 =0.709Q when the ambient air temperature is 20 Celsius.
(Q is a constant and the density of the ambient air is 1)
In both cases the volume displaced is the same so the upthrust will be equal to the difference in densities times the volume, which is the weight of displaced air minus the weight of air inside.
The relative lift at 0 Celsius will be 1-0.661 = 0.339
the relative lift at 20 Celsius will be 1-0.709 = 0.0.291
This seems to represent about 14% greater lift.
That seems to be a very good reason for it to work.
Please check my arithmetic.
Sorry: 273+120 = 393, not 413; thus (273+0)/(273+120) = 0.695, not 0.661.
Similarly: (273+20)/(273+120) = 293/393 = 0.756, not 0.709.
Thus the difference should be between 0.305 and 0.244, which is a 25% increase.
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Cheers for the correction - I was into fatigue by the time I pressed 'POST'.
Adiabatic expansion of the air in the balloon will cause temperature drop as it rises. It's a gas law thing. PV = RT and all that. The burner offsets this, when required, of course.
I didn't mention a fan; the expansion of the fuel + air on the way in acts like a jet. I imagine that the burner is actually designed with this characteristic accentuated so that it will inflate fast, when setting the thing up on the ground.
Without any explicit temperature measurements they need to leave plenty of headroom in order to avoid damage. that must account for the 120 Celsius design limit when the melting point of nylon is a lot higher. I guess it is important to avoid hot spots!
This 25% increase in lift would apply all the way up; the details just get more complicated and depend on conditions changing on the ascent.
So; problem solved? Near enough for Jazz?
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I didn't mention a fan; the expansion of the fuel + air on the way in acts like a jet. I imagine that the burner is actually designed with this characteristic accentuated so that it will inflate fast, when setting the thing up on the ground.
Not disagreeing with anything here, but the point was if the expansion has already taken place near the burner, then there is little capacity for extra expansion once the exhaust gasses have entered the canopy.
Without any explicit temperature measurements they need to leave plenty of headroom in order to avoid damage. that must account for the 120 Celsius design limit when the melting point of nylon is a lot higher. I guess it is important to avoid hot spots!
It is conceivable that a small patch at the very top is more resistant to heat than the rest; but even allowing for that, if all of this is happening with wide margins applied, then one can imagine there might be some variation in temperature allowed simply because temperature is not that tightly regulated. I am not trying to predict what effect this may have, only that the notion that the temperature is a constant 120°C is probably not totally accurate (although it may work as an approximation - I cannot say).
This 25% increase in lift would apply all the way up; the details just get more complicated and depend on conditions changing on the ascent.
So; problem solved? Near enough for Jazz?
Yes, as a rough approximation, I think it works.
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What about convection? As the skin of the balloon is heated, the layer of air surrounding it is also heated, Conduction. This gives the balloon bouyancy and causes it to rise beacasue the colder air sinks forcing the warmer air to rise.
The density of the air surrounding the skin of the ballon decreases, which causes the air to rise, this then gets replaced by colder air. This gets heated, decreases its density, rises and so on and so on...
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But if the air surrounding the envelope is warmer than it was it will produce less upthrust on the balloon. I doubt that the resulting upwards currents of the surrounding air due to convection would produce enough force to overcome this reduced upthrust.
I think that the less heat loss from the skin the better, in fact.