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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.
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.
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).
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).
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).
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.
Whether a hot air balloon is constant volume is an interesting question.
Why is saturation irrelevant?
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.
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.
(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.
btw, the bit about CO2 and H2O giving an overall decrease in density has been verified by colleagues.
So, apart from arguing with me, do you have an answer, other than my two suggestions, for the original question?
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.
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 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.
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.
The air inside the balloon changes no more or less than that outside of the balloon.
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.
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.
QuoteI 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.
QuoteThere 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.
QuoteThe 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.
Quote273°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 = 0.339the relative lift at 20 Celsius will be 1-0.709 = 0.0.291This seems to represent about 14% greater lift.That seems to be a very good reason for it to work.Please check my arithmetic.
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?