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

[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.

 

[OmCheeto]

Temperature is a huge problem. Let's take @OmCheeto's dirty towel (I found a towel the same mass, and could reproduce the numbers):
We have 130 grams of water. It will start to freeze as soon as we lower the temperature by 20 K. That is sufficient to evaporate ~1/25 of the total mass, or about 10 grams. We get roughly 20-30 grams more to evaporate before all the water is frozen. Doesn't work.

We can start by heating the towel, but not enough to get rid of the whole water by evaporation without freezing some. Multiple vacuum cycles with hot air in between could work, but that doesn't sound efficient.
A pressure below atmospheric pressure, but sufficient to still deliver heat all the time?

I'm not even sure if evaporation at room temperature and vacuum can beat evaporation at dryer temperatures and atmospheric pressure.

I have completed my calculations, and came up with a similar conclusion.
The amount of energy to dry clothes in a vacuum is roughly the same as at atmospheric pressure.
For 4.6 kg of water, I had to add 2.9 kwh of energy to maintain the temperature.

Fun project!

I think I'll look into dryer heat exchangers now.
This guy seems to have only one serious design flaw:

http://www.builditsolar.com/Experimental/DryerHX/DryerHXTest1.htm

Dryer into HX, 136°F
Dryer out of HX, 88°F
Room into HX, 79°F
Room out of HX, 95°F​

But even with the flaw, I think it's pretty impressive; "The heat exchanger, as it is, saves about 32% on the total energy used"​

 

[Buckleymanor]

I have completed my calculations, and came up with a similar conclusion.
The amount of energy to dry clothes in a vacuum is roughly the same as at atmospheric pressure.
For 4.6 kg of water, I had to add 2.9 kwh of energy to maintain the temperature.

Would you not dry the clothes in a vacuum shortly after the hot cycle used in the wash.
Eliminating some of the 2.9 kwh to maintain the temperature.
Most wash cycles have a hot portion to remove stains etc, this will be wasted as heat at the end of the cycle anyway or do your figures take this into account.

 

[sophiecentaur]

Would you not dry the clothes in a vacuum shortly after the hot cycle used in the wash.

They would be all soapy! There are several rinses with cold water after the hot wash. You would need to store the drained hot water and pump out the heat later.

 

[eltodesukane]

I've had the thought for years that it should be possible to dry damp clothes instantly (or almost instantly): You place them into an airtight container, then pump all the air out. In the vacuum, the boiling point of water drops to room temperature, so the water would all boil away, leaving perfectly dry clothes.
The only problem is that now the vacuum chamber is full of water vapor. If you open the door to the chamber to get your clothes out, the vapor would condense again, getting your clothes wet all over.

-- If you pump the air out, would that not also pump the water vapor out?
-- Also, I often dream of a microwave with a cooling switch, so that it could boil or freeze food.

 

[OmCheeto]

Would you not dry the clothes in a vacuum shortly after the hot cycle used in the wash.
Eliminating some of the 2.9 kwh to maintain the temperature.
Most wash cycles have a hot portion to remove stains etc, this will be wasted as heat at the end of the cycle anyway or do your figures take this into account.

I see at least one problem with this idea; "You see, back in 1988, the Consumer Product Safety Commission recommended that water heater manufacturers preset the maximum temperature at 120°F [ = 48.9 °C = 322 K] to prevent burns. And the manufacturers voluntarily followed those recommendations — supposedly". [ref: NPR]

But this did give me another idea. Just use the dryer to initially heat the clothes, just as we do now.
In a dryer experiment I did on Oct 5, 2008, I came up with the following information:

The dryer initially heated up to 145°F in all 3 heat settings. From there, the temperature was maintained at a lower range:

High 122-134 °F __ 50.0-56.7 °C __ 323-330 K
Med 115-125 °F __ 46.1-51.7 °C __ 319-325 K
Low 100-110 °F __ 37.8-43.3 °C __ 311-316 K​

From this, I'm guessing that some scientist determined that clothing has been shown to have an effective maximum, non-disintegrating temperature of 330 K.
Replacing 311 K with 330 K shows no appreciable change in energy requirement.

So I've decided that I will next look at Russ's idea, of using a microwave.
Somewhat of a minor inconvenience, as I assume I will have to separate my clothes: whites, darks, included metal.
A quick google search indicates someone made one 15 years ago:

Tech Notes; Using Microwaves to Dry Clothes
By CLIFFORD J. LEVY
Published: September 15, 1991​

And although wiki claims; "Japanese manufacturers^{[citation needed]} have developed highly efficient clothes dryers that use microwave radiation...", I can find no evidence of a production model.

Though this recent article; http://www.2450mhz.com/PDF/TechRef/Mw%20Clothes%20Drying.pdf
lists a possible solution; "...Noting how consumers have accepted and adapted to the microwave oven, the most likely scenario will be an evolution in consumer laundry habits and the birth of an entirely new industry of clothing and laundry products developed expressly for the microwave clothes dryer."

ie, microwavable safe undies!

But current technical problems aside, no one mentioned drawing a vacuum on the system, and the goal was "instant drying", so I'll continue with my research.
...

Still not instant. I come out with 35 minutes to dry clothing with 4.6 kg of water, @ 5000 watts.

double check on maths:
4.6 kg water * 2,264,800 J/kg latent heat of vaporization = 10,418,080 Joules to maintain constant temperature
= 10,418,080 watt seconds
10,418,080 watt seconds / 5,000 watt microwave = 2,084 seconds

=34.7 minutes



ps. I spent several hours on the beach yesterday with an 80 year old retired chemist, trying to get him to explain "partial pressure" to me, but all I got was an hour long "Oh my god, it was so cool when programmable calculators came out. It made figuring that out so much easier. .........." speech.

I learned nothing.
But it was a nice day.

 

[gjonesy]

You may put a chilled wall inside of chamber, at -50C. At 10 Pa, this value of temperature is below of the triple point of water,
so you can stop the vacuum pump and the water vapors will be collected by the chilled wall.

If that's the winning Idea I'll accept the royalty check...lol

 

[Scott C.]

I've had the thought for years that it should be possible to dry damp clothes instantly (or almost instantly): You place them into an airtight container, then pump all the air out. In the vacuum, the boiling point of water drops to room temperature, so the water would all boil away, leaving perfectly dry clothes.

The only problem is that now the vacuum chamber is full of water vapor. If you open the door to the chamber to get your clothes out, the vapor would condense again, getting your clothes wet all over.

It's a kooky idea, but I'm just wondering if anybody has a bright idea for getting the clothes out of the vacuum chamber without water condensing on them? If the idea makes a million dollars, I'll split the royalties with you.

The real problem is getting "perma-press" to work and the "spring fresh" scent of dryer type fabric softeners (both of which which require heat).

 

[A N Madhavan]

I've had the thought for years that it should be possible to dry damp clothes instantly (or almost instantly): You place them into an airtight container, then pump all the air out. In the vacuum, the boiling point of water drops to room temperature, so the water would all boil away, leaving perfectly dry clothes.

The only problem is that now the vacuum chamber is full of water vapor. If you open the door to the chamber to get your clothes out, the vapor would condense again, getting your clothes wet all over.

It's a kooky idea, but I'm just wondering if anybody has a bright idea for getting the clothes out of the vacuum chamber without water condensing on them? If the idea makes a million dollars, I'll split the royalties with you.

When you create vacuum in chamber, water vapour too comes-out along with air so purpose would be served.

 

[jim hardy]

When you create vacuum in chamber, water vapour too comes-out along with air so purpose would be served.

and

Still not instant. I come out with 35 minutes to dry clothing with 4.6 kg of water, @ 5000 watts.

double check on maths:
4.6 kg water * 2,264,800 J/kg latent heat of vaporization = 10,418,080 Joules to maintain constant temperature
= 10,418,080 watt seconds
10,418,080 watt seconds / 5,000 watt microwave = 2,084 seconds

=34.7 minutes

Has anyone figured the mechanical work to pump 4.7 kilograms of this air-water vapor mix from , just say 0.1bar to 1bar ?

 

[gjonesy]

Could you install a condenser coil in the vacuum chamber, perhaps separated by a thick perforated metal baffle with the condenser at the opposite end. Keeping the coolant circulating through the camber totally isolated from the vacuum process, perhaps pulling the moisture away from the clothing?

Did a little rethink on this, what if on the cloths side of the vacuum chamber you could pump in hot air say175F , keeping an open vent until the ambient temperature of the pressure vessel were 175F then pulled the vacuum from the condenser side quickly?

 

[Jeff Rosenbury]

My understanding of partial pressures is that the amount of water vapor is nearly independent of the amount of other gasses. So raising the temperature raises the amount of water vapor as well as providing the heat of vaporization. Reducing pressure only forces boiling when it drops below the partial pressure. The other gasses play one role, they provide extra heat for vaporization. So pumping them out slows the process of heat transfer.

To dry quickly, add extra air, of a form that carries lots of heat (molar heat capacity?). Increase the gas flow so water vapor is cycled out of the drying chamber quickly.

To save energy, use a condenser/heat exchanger. Have a large, low pressure (equal to the working pressure) gas storage area.

A nearly ideal situation would be to hang the clothes in a large, dry area and blow warm air over them. Perhaps use solar energy to add heat, and wind power to provide gas exchange.

I wonder if I can patent a clothesline?

But to speed the process, increase pressure and/or use a gas with higher heat capacity. As the OP pointed out, sucking the vapor out is fast. As others have pointed out, heat transfer is what slows the process.

I don't see a reason to reduce pressure. I see good arguments for increasing pressure and airflow.

In short, more hot dry airflow dries faster.

 

[OmCheeto]

...
Has anyone figured the mechanical work to pump 4.7 kilograms of this air-water vapor mix from , just say 0.1bar to 1bar ?

I could have sworn I was done with this...
The mechanical work was actually one of the very first things I calculated.
But given that I've learned a little bit since last Thursday, I'm pretty certain the answer is wrong.

I used the wiki value for water vapor density:

wiki; "The density (mass/volume) of water vapor is 0.804 g/L, which is significantly less than that of dry air at 1.27 g/L at STP".​

I determined that the volume of my spherical dry was 70 liters, and the volume of 4.6 kg of water vapor was 5800 liters, at STP.
The energy to pump that out is 592,000 joules. (=0.16 kwh = $0.016 electricity)
What's that phrase? "Not even wrong"?

But STP is 0°C at 1 bar, and we were trying to stay away from frozen water.
Anyone know how to calculate the density of water vapor at 0.1 bar @ 100°F? ( = 10,000 pascals @ 311 K)

hmmmm... Maybe Russ's link...

Yippie!
0.069067 kg/m³ with a saturation temperature of 46.1°C ( = 319 K = 115°F )

From that I get 1.73 kwh, as an answer.
But it's going to take my $500, 23 liter/min pump, more than 2 days to evacuate the system. (49.4 hours)

hmmm... Not very instantaneous.

 

[anorlunda]

I think I'll look into dryer heat exchangers now.

I looked into that many years ago. I had a house in a cold climate and with electric baseboard resistance heating. I thought that I could just vent the dryer into the house, achieving 100% efficiency for the drying energy because (during winter months) it would subtract from baseboard electric power demand. Ditto for any other interior energy wasting devices; in winter they caused zero waste.

I was also running an electric humidifier in winter because electric heat made internal humidity uncomfortably low. Eliminating the humidifier made the efficiency for internal dryer venting more than 100%.

But the down side was dust. A dust filter is only partially effective. Worse, if the filter came loose or had a leak, the accumulated dust would create a serious fire hazard. In the end, I reluctantly rejected the internal venting idea, and looked into heat exchangers instead. I had to reject the heat exchanger too because it would force me to move the dryer away from the wall, and there wasn't enough room for that.

 

[jim hardy]

From that I get 1.73 kwh, as an answer.
But it's going to take my $500, 23 liter/min pump, more than 2 days to evacuate the system. (49.4 hours)

You sir, are just amazing !

Bravo -I like the thought of a counterflow heat exchanger to recover whatever latent heat of vaporization could be recovered to warm incoming outside air. As simple as incoming and outgoing tube ducts being large concentric tubes.
But i suppose if one lived nearer Arctic Circle there'd be an upside to venting that warm moist air right into the living space.

old jim

 

[Jeff Rosenbury]

But the down side was dust. A dust filter is only partially effective. Worse, if the filter came loose or had a leak, the accumulated dust would create a serious fire hazard. In the end, I reluctantly rejected the internal venting idea, and looked into heat exchangers instead. I had to reject the heat exchanger too because it would force me to move the dryer away from the wall, and there wasn't enough room for that.

I've seen this done with nylon hose (panty hose) as a filter. It still produces lots of dust, but perhaps not dangerous levels. It also adds lots of humidity, but so does hanging clothes on a drying rack (my current solution).

 

[OmCheeto]

I looked into that many years ago. I had a house in a cold climate and with electric baseboard resistance heating. I thought that I could just vent the dryer into the house, achieving 100% efficiency for the drying energy because (during winter months) it would subtract from baseboard electric power demand. Ditto for any other interior energy wasting devices; in winter they caused zero waste.

I was also running an electric humidifier in winter because electric heat made internal humidity uncomfortably low. Eliminating the humidifier made the efficiency for internal dryer venting more than 100%.

But the down side was dust. A dust filter is only partially effective. Worse, if the filter came loose or had a leak, the accumulated dust would create a serious fire hazard. In the end, I reluctantly rejected the internal venting idea, and looked into heat exchangers instead. I had to reject the heat exchanger too because it would force me to move the dryer away from the wall, and there wasn't enough room for that.

Well, I live in Oregon, so what you've described is exactly what I do.
Though I ignore the dust and fire hazard(?). I have a 10ft dryer hose so it's pretty far away from the dryer inlet.
In the winter season only, of course.

In the summer I hang my clothes on my covered back porch for a day. It takes about 15 minutes to get rid of the residual moisture, and "crunchiness".
I'd put them in the sun, but there are so many birds around here, I'd just have to rewash them when they were done drying.

You sir, are just amazing !

Bravo -

I like the thought of a counterflow heat exchanger to recover whatever latent heat of vaporization could be recovered to warm incoming outside air. As simple as incoming and outgoing tube ducts being large concentric tubes.
But i suppose if one lived nearer Arctic Circle there'd be an upside to venting that warm moist air right into the living space.

old jim

Amazing? I wouldn't call "only 2 days to dry your clothes" amazing.
It strikes me as very peculiar, that pumping out all that vapor, has no effect, other than to waste energy.
No matter what I do, I always end up having to add 2265 kJ, for every kg of evaporated water.

Anyways, yes, I think a counterflow heat exchanger would work admirably. And people who live in warm humid environments should probably look into condensing dryers. But then again. I lived in Florida for a year, circa 1979, and 3 minutes after putting on dry clothes, they were soaking wet.

 

[jim hardy]

It strikes me as very peculiar, that pumping out all that vapor, has no effect, other than to waste energy.
No matter what I do, I always end up having to add 2265 kJ, for every kg of evaporated water.

My get rich schemes usually end similarly.

https://en.wikipedia.org/wiki/Critical_point_(thermodynamics)
At the critical point, only one phase exists. The heat of vaporization is zero.

but it's hard to conceive of a supercritical clothes dryer , ~3200psi & 700F .

 

[OmCheeto]

...
but it's hard to conceive of a supercritical clothes dryer , ~3200psi & 700F .

Good grief!
hmmmm... (google google google)
Well, there's a solution.
Send stevendaryl to Venus. Not quite supercritical, but the pressure and temperature are way above the liquid line: 872°F & 1350 psi
No water, no dryer.

ps. I had another idea. Heat up about 50 kg of those little desiccant bags. This place says you can heat them to 347°F.
Engineering toolbox says;

Silica gel - SiO2
blah blah blah... and adsorbs water up to 40% of its own mass. The bulk density of silica gel is 480 - 720 kg/m3. The specific heat is 1.13 kJ/kgK.​

They absorb best when at 70°F, so 347-70°F = 448-294 K = 154 K
154 K * 50 kg * 1.13 kJ/kgK = 8,700,000 joules. (heat energy provided by silica gel desiccant packs)

What did I come up with earlier?

4.6 kg water * 2,264,800 J/kg latent heat of vaporization = 10,418,080 Joules to maintain constant temperature​

Pretty close. Just reduce dryer load by 20%.

Benefits:
No additional heat needed.
No vacuum needed. (A vacuum might make it worse. I have no idea how desiccants work.)

Drying time: unknown. Might be several hours.

 

[mfb]

The silica will release heat when it absorbs water vapor. Probably more than vaporizing the water in the clothes needs.
The downside: you have to dry the silica afterwards.

 

[Nidum]

An efficient fast drying system would be one that needs little or no evaporation of the water in order to function . A system where water is more nearly displaced / transported rather than evaporated .

A spin dryer works this way . Common designs though seem to be effective in removing the bulk of the water but not in achieving total dryness .

Possibly arranging for a reasonably high velocity positive pressure air flow through the clothes in one direction would work better . Cold air initially changing to warm air for final stages of drying .

Essentially the same drying action as on a clothes line but achieved in a smaller space and much faster .