How can we get clothes out of a vacuum chamber without condensing water on them?

[Johnny Reb]

Note that the vacuum doesn't eliminate the problem of needing to apply heat to evaporate/boil the water. Perhaps you could blast the clothes with microwaves while running the vacuum pump?

Could you explain why the vacuum wouldn't eliminate the need to apply heat to evaporate or boil water? From my limited understanding of water, boiling point is just when the gas molecules reach the temperature required to overcome atmospheric pressure and escape the liquid.

 

[sophiecentaur]

Could you explain why the vacuum wouldn't eliminate the need to apply heat

There's a simple argument: Why would standard atmospheric pressure be the only pressure where latent heat is needed for evaporation? Work is still done to break the bonds which hold the surface molecules in place so energy still needs to be put into the process (or the temperature will drop and the vapour pressure will reduce)

 

[Jeff Rosenbury]

Curved non contact black body radiant heating panels .

Moderately powerful suck down which reduces chamber pressure to about 90% atmospheric but not to full vacuum .

Constant small inflow of atmospheric air .

Tumbling action .

Energy recovery system .

Why non contact?

 

[Vanadium 50]

Could you explain why the vacuum wouldn't eliminate the need to apply heat to evaporate or boil water?

See Post #14.

 

[Jeff Rosenbury]

I just had to bump that post!

The condensing design with a heat exchanger system has been available for some while and, refrigeration units being what they are, you can expect it to work for just as long as one with heating elements in it. It is on our shopping list. But my wife is a real enthusiast for the outdoor washing line. It has the great advantage that UV from the Sun acts as a pretty good bleach and keeps 'whites' good. However, towels can feel a bit rough when dried outside and not tumbled.

You forgot that clothes last three times as long as when tumble dried. Clothes lines are great where you can get away with them. Zoning laws sometimes interfere, and industrial cleaners can forget clotheslines.

 

[Jeff Rosenbury]

There's a simple argument: Why would standard atmospheric pressure be the only pressure where latent heat is needed for evaporation? Work is still done to break the bonds which hold the surface molecules in place so energy still needs to be put into the process (or the temperature will drop and the vapour pressure will reduce)

This is true, but with lower pressure, the heat can be supplied by lower temperature sources. (Added savings because clothes aren't damaged by overheating as sometimes happens in traditional dryers.)

Most commercial water distillation is done in stages of lower pressure with the re-condensation of one stage providing the boiling heat for the next.

The same ideas could be applied to clothes, saving energy and time.

 

[sophiecentaur]

The same ideas could be applied to clothes, saving energy and time.

That's true if the latent heat is supplied via a heat pump but not if an electric heater is used.
I was re-thinking the original proposal and it strikes me that faster spin speeds are very good value for water extraction. High spin speeds are very stressful on bearings and there will be design issues with water seals and bearing sizes for a conventional front loader washing machine. I wonder if the fast spinning could be done in the dryer unit instead. I realize that the traditional 'bent tin' construction of all the tumble dryers I have seen, would not be adequate but the lifetime of a beefed up spin / tumbler could be very long.

 

[jbriggs444]

That's true if the latent heat is supplied via a heat pump but not if an electric heater is used.
I was re-thinking the original proposal and it strikes me that faster spin speeds are very good value for water extraction. High spin speeds are very stressful on bearings and there will be design issues with water seals and bearing sizes for a conventional front loader washing machine. I wonder if the fast spinning could be done in the dryer unit instead.

Interesting concept. If you have a nice high spin speed you may not even need a fan motor. The whole dryer assembly begins to resemble a squirrel cage fan with hot air being forced through the clothing by centrifugal force.

 

[PAllen]

Note that the vacuum doesn't eliminate the problem of needing to apply heat to evaporate/boil the water. Perhaps you could blast the clothes with microwaves while running the vacuum pump?

Right, and it is also not instant. The speed is proportional to the rate energy is supplied. While the clothes start at a temperature above boiling for vacuum, they will cool and the process will slow down unless energy is added. In vacuum, you won't have conduction or convection from the outside, only radiant transfer which is really slow.

 

[CrunchBerries]

Probably already mentioned, but what about a very fast spin along with the vacuum? The tumbler could be riddled with holes similar to a washing machine, but the centrifugal force would propel the vapor outwards. The vacuum creating mechanism would be hooked up to the outside of the outer tube and collect the vapors at a high speed. At the exit of the vacuum would be liquid water which would be nearing the amount of water extracted from fabrics.. It would be a high powered and loud system, but for a very short period of time.. like maybe 30 seconds at most.

 

[sophiecentaur]

Probably already mentioned, but what about a very fast spin along with the vacuum? The tumbler could be riddled with holes similar to a washing machine, but the centrifugal force would propel the vapor outwards. The vacuum creating mechanism would be hooked up to the outside of the outer tube and collect the vapors at a high speed. At the exit of the vacuum would be liquid water which would be nearing the amount of water extracted from fabrics.. It would be a high powered and loud system, but for a very short period of time.. like maybe 30 seconds at most.

Two problems there. Firstly, the best evaporation would be with the clothes being separate and tumbling. Secondly, high speed spinning can introduce creases. So the two operations, spinning and tumbling, would need to be done consecutively, I think.
We could perhaps take a lesson from the Dyson Airblade, which uses a very fast air blast and ' blows the water off the hands'. I can't think you to apply the principle to clothes - but possibly a drum with a very fine mesh would allow air to be forced through the fabric as it is spinning. (You would need a gentle cycle for some weak fabrics).
It gets more and more complicated, though. Any of the systems we are discussing involve small loads so we'd also need a way of automatically feeding a few items at a time into the system without getting some of them torn up by the violent ride.

 

[jim hardy]

but the centrifugal force would propel the vapor outwards.

Think about that one.

Vapor ? Water ?

Suspended droplets will be slung outward
but water in its vapor phase being less dense than air (18 vs 29 g/mol) will be displaced toward center.

You see this when your kids have a helium balloon in the car - watch and you'll see it lean into a curve not out as you'd expect.

 

[sophiecentaur]

Oh yes, of course. What a smartypants!
So the rotation would keep the clothes against the sides of the drum and a small amount of air, drawn in from the outer case would keep them from sticking to the sides. Sounds like a good idea (with a bit of regulation involved). You could start the vapour removal once the liquid stopped being thrown out.

 

[sophiecentaur]

You see this when your kids have a helium balloon in the car - watch and you'll see it lean into a curve not out as you'd expect.

You can also get warm and cold air sloshing about in that way as you go round a bend, when the heater has just started to operate on a very cold morning (turn off the jet to the feet). Steering into an 'offside' bend will make your head warmer (as the driver).

 

[jbriggs444]

T
Suspended droplets will be slung outward
but water in its vapor phase being less dense than air (18 vs 29 g/mol) will be displaced toward center.

Not when it is dissolved in air. Otherwise we'd be suffocating in argon and CO2 or burning our lungs out in oxygen.

 

[mfb]

The mass difference is just too small, so air is mixed well. All the air can get pushed outwards, including water vapor. No need to consider molar masses.

 

[jim hardy]

The mass difference is just too small, so air is mixed well. All the air can get pushed outwards, including water vapor. No need to consider molar masses.

C'mon now guys.

why is MW of air 29, weighted average of N2(28) and O2(32) ? Throw some 18 in the mix and ...

That's why summertime cumulus clouds build vertically- moist air rises because it's lighter. When they reach dewpoint they become visible and get a real boost from latent heat of vaporization.
I grew up just east of the Everglades where you see it every summer afternoon .

 

[sophiecentaur]

C'mon now guys.

etc.

But we have been busy evaporating the water and pumping out the air at the same time (or have I missed the plot at this stage?) That means there will be a much lower partial pressure of the air gases. Allowing, also for the turbulence, there won't be much displacement going on. Nonetheless, it still sounds like a good idea to pump from the inside of the drum to keep the clothes aerated more

 

[mfb]

C'mon now guys.

why is MW of air 29, weighted average of N2(28) and O2(32) ? Throw some 18 in the mix and ...

That's why summertime cumulus clouds build vertically- moist air rises because it's lighter. When they reach dewpoint they become visible and get a real boost from latent heat of vaporization.
I grew up just east of the Everglades where you see it every summer afternoon .

Clothes dryers are a bit smaller than the typical scale of cloud formation. Let's give it 2000 rpm and an inner diameter of 40 cm. That corresponds to a rim speed of about 40 m/s, or 0.09 meV difference between water and nitrogen/oxygen. Compare this to the 40 meV thermal energy, and we get a ratio difference of something like 2 parts in 1000 in equilibrium.

Also, cloud formation directly starts with different regions of air with different humidity. We don't have to wait for it to unmix, which takes much longer than convection.

 

[Jeff Rosenbury]

Clothes dryers are a bit smaller than the typical scale of cloud formation. Let's give it 2000 rpm and an inner diameter of 40 cm. That corresponds to a rim speed of about 40 m/s, or 0.09 meV difference between water and nitrogen/oxygen. Compare this to the 40 meV thermal energy, and we get a ratio difference of something like 2 parts in 1000 in equilibrium.

Also, cloud formation directly starts with different regions of air with different humidity. We don't have to wait for it to unmix, which takes much longer than convection.

The point is that the plan goes against nature. That means putting in more energy to offset the problem. How much more? Run the numbers. (The numbers depend on your system, so I'll let you do it.)

That doesn't make it unworkable, particularly when the energy can be stolen elsewhere. A low pressure system lowers the boiling point so perhaps the waste heat from the motor is enough to offset the drive to the center or the heat of vaporization (OK, that second one is dreaming).

All things being equal, spinning the air will drive water vapor to the center of the spin, not the outside. But things are never equal, so possibly. Try to view the problem as a thermodynamics problem as well as a physical one.

I do like the idea of a hard spin to extract water, but it's been done. Commercial dryers already use the technique.

 

[mfb]

I calculated if water vapor and air would un-mix in a dryer: they do not. This is not necessary, we can just pump both together.
There is a much stronger but still quite weak effect of both going towards the outside. Or just install a fan.

 

[jim hardy]

But we have been busy evaporating the water and pumping out the air at the same time (or have I missed the plot at this stage?) That means there will be a much lower partial pressure of the air gases. Allowing, also for the turbulence, there won't be much displacement going on. Nonetheless, it still sounds like a good idea to pump from the inside of the drum to keep the clothes aerated more

Yes !
Sure I was picking nits .
But that's how we keep our thinking straight - consider the details THEN decide which ones to discard as insignificant.

I couldn't let that blanket statement about water vapor being hurled to the outside go unchallenged.
Maybe it's because i was imprinted early by watching Grandma's hand-crank cream separator, who knows, but that's where i learned to envision how In a centrifuge the lighter stuff moves toward the center.

old jim

 

[OrangeDog]

All this fancy talk. Let your clothes dry in the sun and theyll be good to go and sterilized(due to UV radiation) in about 20 minutes. All you need is a clothes pin, string, and to live in the Cayman Islands.

 

[jbriggs444]

All this fancy talk. Let your clothes dry in the sun and theyll be good to go and sterilized(due to UV radiation) in about 20 minutes. All you need is a clothes pin, string, and to live in the Cayman Islands.

You can, of course, dry clothes in low temperatures as well. Frozen clothes take longer but they do dry on the line.

 

[russ_watters]

Could you explain why the vacuum wouldn't eliminate the need to apply heat to evaporate or boil water? From my limited understanding of water, boiling point is just when the gas molecules reach the temperature required to overcome atmospheric pressure and escape the liquid.

If you heat water up to 100C/212F in a closed container at atmospheric pressure, it will just sit there at the boiling point and not boil. Additional heat is needed to convert it from a liquid to a gas: as said, it is chemical bonds that make a liquid a liquid and energy is required to break those bonds. And as it turns out, the energy required to boil water is much, much larger than the energy required to heat it. Of course, in this situation, the temperature is dropping...

Let's say you have 1kg of water in your clothes at room temperature and you reduce the pressure to 0.01 atmospheres. Here's a steam table that gives the properties:
http://www.efunda.com/materials/water/steamtable_sat.cfm
Boiling point: 7C
Latent heat of vaporization: 2484 kJ. As it boils, the energy to reduce the temperature from 18C to 7C is 75.4-29.4=46 kJ. The other 2438 kJ needs to be provided externally.

If you use a conventional 1200W microwave to provide the necessary heat (assuming that is output power) it will take a whopping 34 minutes to boil-off the rest of the water! It's no wonder dryers use large electric heating coils or gas furnaces to provide the heat.

 

[Jeff Rosenbury]

You can, of course, dry clothes in low temperatures as well. Frozen clothes take longer but they do dry on the line.

There's a big problem here: I don't get my own Caribbean island.

 

[Jeff Rosenbury]

If you heat water up to 100C/212F in a closed container at atmospheric pressure, it will just sit there at the boiling point and not boil. Additional heat is needed to convert it from a liquid to a gas: as said, it is chemical bonds that make a liquid a liquid and energy is required to break those bonds. And as it turns out, the energy required to boil water is much, much larger than the energy required to heat it. Of course, in this situation, the temperature is dropping...

Let's say you have 1kg of water in your clothes at room temperature and you reduce the pressure to 0.01 atmospheres. Here's a steam table that gives the properties:
http://www.efunda.com/materials/water/steamtable_sat.cfm
Boiling point: 7C
Latent heat of vaporization: 2484 kJ. As it boils, the energy to reduce the temperature from 18C to 7C is 75.4-29.4=46 kJ. The other 2438 kJ needs to be provided externally.

If you use a conventional 1200W microwave to provide the necessary heat (assuming that is output power) it will take a whopping 34 minutes to boil-off the rest of the water! It's no wonder dryers use large electric heating coils or gas furnaces to provide the heat.

Fortunately when you recondense the water you get the energy back (excepting minor losses, or likely gains in this case). So boiling at 7ºC, adding some pressure, then recondensing on the other side of the boiling chamber a few degrees higher uses the heat given off by condensate to boil the next batch of water.

Of course that doesn't completely solve the time issues, but a little targetted microwave heating might be enough to keep ice from forming an insulating layer, keeping things moving.

Target the microwaves at a frequency where they are absorbed by ice rather than liquid water. This should reduce ice formation and keep the drying speed faster.

 

[russ_watters]

Fortunately when you recondense the water you get the energy back (excepting minor losses, or likely gains in this case). So boiling at 7ºC, adding some pressure, then recondensing on the other side of the boiling chamber a few degrees higher uses the heat given off by condensate to boil the next batch of water.

That can be done efficiently (it is, in fact, a technique for desalinating water) but it would be very difficult to do quickly because of the low delta-T's involved.

 

[Jeff Rosenbury]

That can be done efficiently (it is, in fact, a technique for desalinating water) but it would be very difficult to do quickly because of the low delta-T's involved.

Always bringing reality to the discussion. It's not enough to rain with my heater/condenser, you have to rain on my parade as well. Good job.

I can think of ways around the problem, but they go beyond the practical. At best, I think would be a low(er) pressure intercooler to pre-dry the warm air going into the dryer. This might give a significant decrease in drying time, but isn't really revolutionary. I can't imagine it hasn't been done on big, commercial dryers.

 

[Johnny Reb]

A dryer has two possible outputs (apart from dry clothes) - water vapor or liquid water. If it's water vapor, somewhere you need to provide the latent heat of vaporization. If it's liquid water, you don't, but you are also going to a much lower entropy state, and that means you need a heat engine, especially if you want to do it quickly. These constrain what can be done by pumping only on the surrounding air.

There's a simple argument: Why would standard atmospheric pressure be the only pressure where latent heat is needed for evaporation? Work is still done to break the bonds which hold the surface molecules in place so energy still needs to be put into the process (or the temperature will drop and the vapour pressure will reduce)

If you heat water up to 100C/212F in a closed container at atmospheric pressure, it will just sit there at the boiling point and not boil. Additional heat is needed to convert it from a liquid to a gas: as said, it is chemical bonds that make a liquid a liquid and energy is required to break those bonds. And as it turns out, the energy required to boil water is much, much larger than the energy required to heat it. Of course, in this situation, the temperature is dropping...

I apologize for getting off topic but this was a subject I thought I understood and am now realizing my knowledge might be more flawed than previously thought.'
first, I realize now I should have used the term "local ambient pressure" instead of "atmospheric pressure.
Now all of your responses to my question seem to be pointing to the need for additional heat in order to break the bonds. This would be the latent heat of evaporation, right? Now my question is, if you were to lower the pressure to a point where the latent heat of evaporation was lower than room temperature, shouldn't the heat from the room temperature be enough to boil the water without adding any additional heat? As I am writing this I realize that the ambient heat would be adding heat to the system, but I feel that that was implied in the Original post.
I thought this was along the same principles as the cause of cavitation in a pump, where the pump doesn't have enough supply and the low pressure on the back side of the pump causes the water to vaporize and create air bubbles that can really bugger up your pump, and that doesn't receive any additional heat from anything more than ambient heat.

 

[russ_watters]

Now all of your responses to my question seem to be pointing to the need for additional heat in order to break the bonds. This would be the latent heat of evaporation, right? Now my question is, if you were to lower the pressure to a point where the latent heat of evaporation was lower than room temperature, shouldn't the heat from the room temperature be enough to boil the water without adding any additional heat?

You are mixing together temperature and energy (heat), but yes, if the boiling point is lower than ambient temperature, heat will flow from the room into the dryer and you can boil-off the water without providing heat of your own.

The difficulty is in making that happen quickly, since you've sucked-out most of the air, which is the traditional medium for heat transfer in a dryer...

I feel that that was implied in the Original post.

Yes, the mechanism appears correctly conceived in the OP. The problem is with the "instantly" part.

 

[mfb]

This would be the latent heat of evaporation, right?

Right.

Now my question is, if you were to lower the pressure to a point where the latent heat of evaporation was lower than room temperature, shouldn't the heat from the room temperature be enough to boil the water without adding any additional heat?

The latent heat is an amount of energy (per mass), not a temperature.

For every temperature there is a partial pressure of water where evaporation and condensation are in equilibrium. If the actual partial pressure is below this equilibrium, you get net evaporation. If this equilibrium partial pressure is equal to the total pressure, you get boiling.

You can get boiling at room temperature by reducing pressure - but then the energy is extracted from the clothes, and they get colder, which reduces evaporation. At some point the water freezes, and sublimation from a very cold surface will take a really long time.

 

[OmCheeto]

Curved non contact black body radiant heating panels .
Moderately powerful suck down which reduces chamber pressure to about 90% atmospheric but not to full vacuum .
Constant small inflow of atmospheric air .
Tumbling action .
Energy recovery system .

This is pretty close to what I've come up with, after much head scratching.
Though I'm not sure about letting in atmospheric air.

If you radiantly heat the drum of clothes to above freezing, a cold plate will be able to condense the water vapor.

Since I have no idea how to do the maths on evaporation rates, I of course cheated, and found a site where a similar question was asked:

https://van.physics.illinois.edu/qa/listing.php?id=1440
Q: Can you calculate the rate at which water will flash to steam given the temperature and psia? For example, assuming approximately .8 PSIA[5516 pascals] and 100 degrees F[311 K], at what rate would the water evaporate? Would lowering the pressure or increasing the temperature change the evaporation rate significantly? Thanks.
...

The person answering said it was possible, but quite complicated. I kind of agree with that now, as I'm still not sure what I'm doing. There are way too may "this and that" pressures, and variables and things.

Anyways, he gave an equation: (mass loss rate)/(unit area) = (vapor pressure - ambient partial pressure)*sqrt( (molecular weight)/(2*pi*R*T) )
and he came up with an answer of 1.1 kg/(m² sec)

Which, from my interpolated wet blob of filthy couch blankets and things:

liquid water = 4.6 kg
surface area = 6 m²
Temp_initial = 311K (100°F or 38°C)
Pressure_final = 5516 pascals (0.8 psia)
​

I calculated the drying time would be about 1.12 seconds.
Eureka! The instant dryer!

But as everyone has already mentioned, maintaining the temperature at 311K is quite problematic.
And just for kicks, I determined that my dirty couch stuff would be at about -230K at t_final.[see note 1]

Back to the drawing board.

note 1: It would probably most likely be lower than that, as this is one nightmarish multidimensional fill and drain problem. As more water is evaporated, it leaves less and less water to be cooled. I'm guessing it might be around -1000 Kelvin in reality/unreality. But that's why I think I like it, and will continue working on it.

ps. I've completely forgotten how to do matrix maths, and curve fitting, based on converting the data from the following: http://www.engineeringtoolbox.com/relative-humidity-air-d_687.html
into an equation form. I ended up with: pascals = e^{(-0.0002578K^2 + 0.2137K - 32.73)}
(based on the chart data from -18°C to 52°C)
I think it might be correct.
But I have no idea what it means. I'll try and fix that.

 

[epenguin]

How about prepping the clothes by rinsing in ethanol? That would displace a good portion of the water and evaporate more readily. Crystal formation is the enemy of quick drying. Ethanol would resist crystalization under the temperature loss that would tend to accompany evaporation.

Don't use acetone. ⇒ #3

 

[Johnny Reb]

Thanks MFB and Russ_waters for helping with that. makes a lot more sense now.