A chimney is a bistable fluidic structure, sort of like a siphon: once the chimney fills with hot air, the higher pressure from outside the chimney forces the newly produced hot air up the chimney. Hot air is a lot less dense than cold air. Air at 275°C is half the density of air at 0°C, because it's twice the temperature.
Drying things out is a major problem in day-to-day life: laundry or dishes after you wash them, fruit or other food you want to preserve, parts of your building you don't want to grow Stachybotrys mold and poison you, and so on. Dryness, cold, and poison are the three major ways to inhibit life on media that could otherwise support it (e.g. mold, bacteria, insect predation), and of these, dryness is by far the preferred option in many cases.
Room-temperature evaporation is usually adequate to dry things out enough that they can't support life. Usually, the main determinant of the speed of room-temperature evaporation is air circulation. With no air circulation, relative humidity rapidly reaches 100%, and evaporation stops. Even a fairly weak draft is often sufficient to reach adequate evaporation speeds.
But evaporation, like heat, decreases the density of air. My calculations below suggest that you could construct an evaporation-driven chimney that would provide adequate air circulation even without a source of above-ambient temperature, probably using laminar rather than turbulent airflow.
In a sense, this is an evaporative downdraft cooltower, meant to make things dry rather than cold and wet. The only difference is that you saturate it with heat instead of humidity.
The vapor pressure of water increases exponentially with temperature, which means that any such chimney device will have a critical temperature below which it will basically not work at all, and it's much harder to build one that will work at 0° (0.6 kPa vapor pressure) than 16° (1.8 kPa), 25° (3.2 kPa), or 35° (5.6 kPa) — you have an entire order of magnitude of density difference here.
It's probably a good idea to shoot for 16°, since 0° sounds too hard, and higher ambient temperatures in places where people live are often a result of humidity-induced greenhouse effect, so in practice the dryer may not work that much better.
Of course, evaporation also cools the air and the thing from which the water evaporates, but you can passively warm the cooled air back to room temperature — it's just a matter of providing sufficiently high thermal coupling. At sufficiently high airflow, this heat transfer can be the limiting factor in drying things out. For this exercise, I will simply assume that heat transfer is adequate.
With these assumptions, the device seems quite practical. It seems that you should be able to maintain 25 cm/s airflow with a 5.4-centimeter evaporation chimney, numbers which don't really depend on the chimney's height, according to the calculations below. However, it turns out you can get even stronger airflow by operating it as an evaporative downdraft cooltower!
Water molecules weigh 18 daltons, while the other major air molecules weigh 30 and 32 daltons. So pure water vapor weighs about 41% less than dry air.
So at 16°, air at 100% relative humidity and one sea-level atmosphere (100 kPa) will consist of roughly 1.8% water, which means its density is decreased by 41% × 1.8% = 0.74%. If your incoming air already has 50% relative humidity, which is common, you're down to 0.37%; if the air density is 1.23 kg/m³, that's about 4.5 mg/ℓ, 4.5 μg/cc, or 4.5 g/m³.
Okay, how much pressure can you get out of that? At Earth surface gravity, it works out to about 45 mPa/m, which is a pretty tiny pressure!
How much laminar airflow can you get out of that? Well, it depends on your air duct size. The Darcy-Weisbach equation says your pressure loss to friction is f[D] ρLV²/(2D), where f[D] is the Darcy friction factor (64/Re in the laminar case), ρ is the density of the fluid, L is the duct length, D is the diameter, and V is the velocity.
Conveniently, or not, the chimney length here drops out: our pressure difference is L × 45 mPa/m, proportional to the chimney length, and the Darcy–Weisbach pressure drop is also proportional to the chimney length. So we have 45 mPa/m = f[D] ρV²/(2D); if we divide both sides by the density of air, we get 0.037 m/s/s = f[D] V²/(2D).
Will it be laminar? Laminar flow dominates at Reynolds numbers below about 2000 or 3000. The Reynolds number Re in a pipe is VD/ν, where V is velocity, D is diameter, and ν is kinematic viscosity, measured in e.g. m²/s. The kinematic viscosity of air is about 1.5 × 10⁻⁵ m²/s, so the laminar-turbulent transition takes place (assuming Re = 2000) at around VD = .03 m²/s. That is, if D is 3 cm, it takes place at 1 m/s; if D is 30 cm, it takes place at 10 cm/s; if D is 1 m, it takes place at 3 cm/s. So you can get it to be turbulent by using a big enough duct, even without changing the velocity.
If we figure on a Reynolds number in the middle of the laminar range, giving us a comfortable safety factor, f[D] is about 0.06. Solving for V², we get
V² = 2D × 0.037 m/s/s / 0.06
or simplified
V² = D × 1.2 m/s/s
so our actual air velocity is
V = √(D × 1.2 m)/s
assuming that Reynolds number.
So, uh, that assumption gives us a duct size, no? I was assuming VD = .015 m²/s, so .015 m² = D√(D × 1.2 m). Square both sides (adding meaningless complex solutions) and you get .000225 m⁴ = D³ × 1.44 m, so D = (.000225 m³ / 1.44)^⅓, or about 5.4 cm.
Okay, so if you have a 5.4 cm wide evaporation chimney, you get √(.054 m × 1.2 m)/s = 0.25 m/s airflow from the tiny pressure difference from evaporation, which is pretty respectable.
You can make the duct a bit bigger (up to twice the Reynolds number, which is growing as D×√D, so you can make it up to 59% wider: up to 8.6 cm, which would have 2½× the cross-sectional area and 1.26× the flow velocity, for a total of three times the volumetric flow) before you hit a slowdown from turbulent airflow, at which point you have to make it a lot bigger to compensate. Eventually, the total turbulent airflow will exceed the possible laminar airflow, but it's probably better to plan to stay well into either the laminar or the turbulent regime on the Moody diagram, avoiding the much lower airflow just above the transition. If you want to stay in the laminar regime but need more total airflow volume, you can have a number of small chimneys, perhaps a grid of partitions inside a larger space.
Such partitions would also help prevent the moist air coming up the chimney from mixing with dry air descending from above, which will reduce the pressure difference, but not the Darcy–Weisbach loss, thus reducing the flow rate. In the limit of an infinitely large chimney, you will have a plume of water vapor rising slowly from your moist object and gradually diffusing into the air, while nearby air remains stationary or sinks.
0.25 m/s × π × (.054 m/2)² = 0.00056 m³/s total airflow in one of these chimneys. That's about 0.69 g/s of total air per second, of which 0.37% is the water we're extracting: 2.5 mg/s per chimney, or about 9.2 grams of water per hour per chimney. So to evaporate a liter of water in four hours — suitable for drying laundry or fruit — you might need 27 such chimneys, perhaps a 6×5 grid inside a larger 30×30 cm duct, for a total of .015 m³/s or 32 cfm, somewhat less than a standard bathroom ventilation fan.
As I said, chimneys are bistable: once they have a draft, they keep sucking (much like standing armies), but until they're full of the lower-density fluid, they don't suck. So how do you start the chimney?
The usual approaches are using a fan and using an inverted funnel.
The fan works in the obvious way, but if you have the fan, you don't really need the chimney. The chimney is quieter and uses less energy, though.
An inverted funnel leading up into the chimney can capture individual chimneyless updrafts of moist air from below and fill the chimney with them. This will only work if the moist air updrafts are sufficient.
It barely matters at all what materials an evaporative chimney is made from. It could be sheet metal, paper (as long as it doesn't get rained on, or if it's greased to protect it), silk cloth, brick, or polyethylene sheeting. Soft materials will need some kind of external support.
If you're venting it outside, you might consider it desirable to curve the chimney at the top, or put a roof over it, so that rain doesn't fall into it.
In many cases it might be beneficial to use a transparent material.
Earlier I said I assumed heat flow was adequate. Is that reasonable?
Heat flow is really critical for two reasons: first, the 0.5%/° change in air density with temperature could easily overwhelm the 0.4% change in air density with humidity, making your "chimney" work backwards; and second, if the thing you're trying to dry out is cold, water on it will evaporate really slowly, which will give it more time to rot before it dries out.
Evaporating 2.5 mg/s of water is about 10.4 mW of heat dissipation, so you suck up about 10.4 mW of heat per chimney. The earlier-suggested 27-chimney assembly thus sucks up about 280 milliwatts.
A glance at a psychrometric chart shows that the temperature drop of the air at 15° and 50% humidity is about 5°. Air is about 1.01 J/g/K, so that's about 5 J/g, and at 0.69 g/s, that's 3.4 watts. This differs from my calculation above by a factor of 12, so one of them must be completely wrong. Maybe both.
You can deliver the heat from the environment by convection, conduction, and radiation.
Room-temperature radiation is more than adequate to replace a few hundred milliwatts (it's on the order of tens of watts per square meter — Stefan–Boltzmann 57 nW/m²/K⁴ gives you 390 W/m² at 15°C, and perhaps more relevantly the derivative is about 5 W/m²/K), but only if the objects you're drying is mostly exposed to radiation from other objects that aren't drying. That means they have to be spread out flat, and you don't want to put them under glass.
Convection will replace the missing heat, but only at the cost of the five- or ten-degree temperature drop in the air, which will keep the chimney from sucking air up.
Conduction is likely to be the best approach in many cases. It can take several forms: metal plates maintained at room temperature by a flow of water or air on the non-wet side, say by running water through pipes or by funneling wind past them; thick metal fins connected to a large temperature reservoir; or even flat heat pipes.
But wait! If you have to struggle to overcome a 5° temperature drop, which produces a countervailing effect that's four times as strong, why not use the temperature drop? Instead of a moist-warm-air chimney above your drying objects, you have a cold-air drainpipe below. Your drying objects don't dry as fast, since they're colder.
As a bonus, you get free cold air, which can be used by itself to cool things to retard decay, or to improve the efficiency and delta-T of a refrigerative cooler.
Yeah, if you have the chance, some sunlight on whatever you're trying to dry will help a lot. Then you can get not only more-than-adequate heat flow, on the order of 1000 watts per square meter instead of tens, but also enough heat to make the chimney run purely on heat.