5 ms·

You also need power to take the heat away that is created when the water in the air turns into a liquid. Best case you need to cool for about 0.7kwh per litre o

You also need power to take the heat away that is created when the water in the air turns into a liquid. Best case you need to cool for about 0.7kwh per litre of water you make. That means for your 10 litre example you need best case 7kwh. That is a decent sized solar array and if you have any inefficiencies in your heat pump setup it gets much worse. Also you assume per day - solar only works during the day so you need storage as well so your system can work during the night.

kragen 5y ago
Yes, that's why I said moving the air takes "a totally insignificant amount of power compared to the actual refrigeration involved." You're talking about the actual refrigeration. The terminology you need to google for "the heat that is created when the water in the air turns into a liquid" is "enthalpy of vaporization of water"; around room temperature this is about 2.4 MJ/kg, so 10 liters/day is 24 MJ/day, which is 6.8 kWh/day, or, in SI units, 280 watts. However, remember that a heat engine operating at the Carnot limit is reversible. The coefficient of performance of a typical heat pump at these temperatures, the kind you might buy off the shelf at a big-box hardware store, is about 2, so you only need about 12 MJ/day (3.4 kWh/day, 140 watts). At a typical desert capacity factor of 25% this means you need a 560-watt solar array, about US$120 and three square meters. (California's utility-scale PV average capacity factor was 29% last I looked.) Very cloudy and polar places can have PV capacity factors as low as 10%, but they also have easier sources of drinking water. Like a rain barrel. It's easier to store water or to "store coldness" than to store electricity, so you don't need electrical storage, you just need a heat pump sized for your peak throughput instead of your average throughput. Heat pumps are pretty expensive, so you might think this is a big problem, but the cheapest air conditioners I can find for sale around here are about 2000 watts, not 500 watts. I've been noodling on desiccant-powered heat pumps for this and other uses, which may be able to reduce the cost of such systems, gather a larger fraction of solar energy than the 21% of high-efficiency PV panels, and provide built-in energy storage.
kragen 5y ago
It's worth noting that the Nature paper abstract https://www.nature.com/articles/s41586-021-03900-w.pdf https://www.nature.com/articles/s41586-021-03900-w.pdf claims they're spending as little as 0.4kWh/ℓ, which is less than the 0.68kWh/ℓ of the enthalpy of vaporization, though more than the 0.34kWh/ℓ of the enthalpy of vaporization divided by the CoP of a commonplace vapor-compression heat pump. (They point out that "solar-driven cooler–condenser devices suffer from a steep loss in electric energy conversion", by which I think they mean photovoltaic panels are only about 21% efficient, so you lose 79% of your sunlight before it can start driving your compressor.) They're using a sorbent-based cycle rather than a vapor-compression cycle. I should have read this before commenting. They cite a paper by Kim et al. that calculates the thermodynamic limits on the specific yield of atmospheric water harvesting as 5–50 ℓ/kWh (0.02–0.2 kWh/ℓ), which I suppose depends on the air temperature and humidity.

kragen 5y ago

Yes, that's why I said moving the air takes "a totally insignificant amount of power compared to the actual refrigeration involved." You're talking about the actual refrigeration. The terminology you need to google for "the heat that is created when the water in the air turns into a liquid" is "enthalpy of vaporization of water"; around room temperature this is about 2.4 MJ/kg, so 10 liters/day is 24 MJ/day, which is 6.8 kWh/day, or, in SI units, 280 watts. However, remember that a heat engine operating at the Carnot limit is reversible. The coefficient of performance of a typical heat pump at these temperatures, the kind you might buy off the shelf at a big-box hardware store, is about 2, so you only need about 12 MJ/day (3.4 kWh/day, 140 watts). At a typical desert capacity factor of 25% this means you need a 560-watt solar array, about US$120 and three square meters. (California's utility-scale PV average capacity factor was 29% last I looked.) Very cloudy and polar places can have PV capacity factors as low as 10%, but they also have easier sources of drinking water. Like a rain barrel. It's easier to store water or to "store coldness" than to store electricity, so you don't need electrical storage, you just need a heat pump sized for your peak throughput instead of your average throughput. Heat pumps are pretty expensive, so you might think this is a big problem, but the cheapest air conditioners I can find for sale around here are about 2000 watts, not 500 watts. I've been noodling on desiccant-powered heat pumps for this and other uses, which may be able to reduce the cost of such systems, gather a larger fraction of solar energy than the 21% of high-efficiency PV panels, and provide built-in energy storage.

kragen 5y ago

It's worth noting that the Nature paper abstract https://www.nature.com/articles/s41586-021-03900-w.pdf https://www.nature.com/articles/s41586-021-03900-w.pdf claims they're spending as little as 0.4kWh/ℓ, which is less than the 0.68kWh/ℓ of the enthalpy of vaporization, though more than the 0.34kWh/ℓ of the enthalpy of vaporization divided by the CoP of a commonplace vapor-compression heat pump. (They point out that "solar-driven cooler–condenser devices suffer from a steep loss in electric energy conversion", by which I think they mean photovoltaic panels are only about 21% efficient, so you lose 79% of your sunlight before it can start driving your compressor.) They're using a sorbent-based cycle rather than a vapor-compression cycle. I should have read this before commenting. They cite a paper by Kim et al. that calculates the thermodynamic limits on the specific yield of atmospheric water harvesting as 5–50 ℓ/kWh (0.02–0.2 kWh/ℓ), which I suppose depends on the air temperature and humidity.