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Comment by Mirioron | Hacker News Reader

It's a cool idea and all, but how do the calculations look for actually moving enough air through the "net"?

One cubic meter of air at 100% humidity and 25C only holds 22g of water. For the 8 liters per day value it doesn't seem like an unreachable amount: 363 cubic meters of air need to pass through the filter. You probably need to more than double that number due to relative humidity not being at 100% and we probably won't get all the water in the air, but it seems reachable.

On the other hand, 22500 liters per day seems quite crazy. I'm interested to see how they're going to tackle this.

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At 100% humidity and 20 C air holds 18 g H2O per cubic metre so to deliver 22 500 kg of water per day they must entirely dessicate 1.25e6 m3 of air. Air is about 4 kg per m3 so this means moving 5 000 tonne of air through the system.

If the energy cost of moving the air is equivalent to lifting it is, say, two metre then the energy required to do this is 5e6 * 9.8 * 2 joule, about 2e9 J or 555 kWh so it would need something like 250 m2 of solar panels at 20 5 efficiency for 10 hours.

At higher temperatures 100 % is more water but that only changes the yield by at most a factor of about three.

Of course the humidity in the areas where this would be used is much lower so the amount of air to be processed is correspondingly higher.

I'm not saying it can't work, or guaranteeing that I haven't slipped a decimal place somewhere but these figures suggest to me that it would be very expensive and might be outcompeted by piping in water from desalination plants.

1. If there is wind, obviously you need less energy to move the air through the device.

2. Second, your estimate based on gravity is off. Apparently, the fan power to move air is given by [1]

Power = (pressure change of air) x (volume/time of air)

Volume/air is 1.25e6/10hours = 1.25e5. Assuming that the pressure change through the device is minimal, we can look at the bottom most line (100 Pa) on the first figure, to estimate ideal power to be ~5KW of power. Even with loses, having 10-20 KW power systems is equivalent to the energy consumption of 2-5 large houses.

[1] https://www.engineeringtoolbox.com/fans-efficiency-power-con...

Not to mention the energy you need to remove 14 MWh of energy from the water in order to condense 22,500 kg of it.

This is usually the more important part and the part most projects hide. 14MWh in 24h is about .5MW = 500,000W. For comparison a Pentium has to dissipate about 100W, so it's like 5000 Pentiums.

The Pentium has a big dissipator that is directly under a fan, that has another fan(s) nearby to get more fresh air. And the Pentium runs at ~50°C that is like 30°C(15°F) over the room temperature. I doubt this absorbed gets so hot, so the temperature difference is probably only a tenth of a degree, so the dissipator must be much bigger. (Dissipation is not linear in fluids, so a 1/30 of difference of temperature usually means more than a x30 bigger dissipator.)

”Air is about 4 kg per m³”

https://en.wikipedia.org/wiki/Density_of_air gives around 1.3 kg/m³ for dry air. If you add the 18 grams of water you give, you only get around 1.32 kg/m³.

⇒ I think 4 kg/m³ is too high.

Humid air is actually less dense than dry air, as any aircraft pilot should be able to tell you. That's because the water molecule H2O has a molecular weight of 18, while O2 has a molecular weight of 32 and N2 has a molecular weight of 28.

Thanks. I misremembered that one.

So it reduces the energy cost by a factor of three. Still needs a heap of solar panels to get about 200 kWh per day. I have a friend in southern France who has eight panels and it has taken him fifteen weeks this summer (105 days) to get 1000 kWh this summer, say 10 kWh per day so you need 160 panels. to get 200 kWh per day.

My purpose was to point out that the size of the installation is likely to be quite a bit bigger than a cursory glance at the article would suggest.

555 kWh is about 55 USD at $0.10/kWh, so for 22.5 m3 that's about $2.5 per m3 (or 1000 l) of water, cheaper than EU residential prices (3 EUR per 1000l in France) and also than than what I can find about US residential prices (NYC: $10/100 cuft or $3.5/1000l) Cheaper is good, because this is just marginal electricity cost, total cost per 1000l is going to be higher, it will include maintenance and equipment depreciation.

The assumption above is 100% humidity. Which you won't get even close to in a desert near Amman, where it's around 50%, and only because of the ocean proximity.

If you look at the humidity map, it's rather clear that most warm areas away from the ocean/sea are screwed - it's around 18% humidity.

https://www.windy.com/-Show-add-more-layers/overlays?rh,31.9...

They aren't using grid power, they are using solar so you need to consider the capital cost over the lifetime of the system.

And remember that my figures assume that you extract all of the water from 100% humidity. But you don't have that and anyway a condenser would work perfectly well if you did.

A household air dehumidifier can pull 8 litres per day out of a 500 sqft home - seems possible.

https://www.directenergy.com/learning-center/energy-efficien...

That’s not in a desert though.

the link is about humidifiers, not dehumidifiers

https://blog.constellation.com/2018/09/21/energy-efficient-d... gives a ballpark number:

"In order to be approved by ENERGY STAR®️, energy-efficient dehumidifiers must have an energy factor of at least 2.00, or an energy factor greater than 2.80 for dehumidifiers that remove more than 75 pints of moisture a day."

So 2-3 litres water extracted per kWh electricity consumed (And I'm guessing this is probably in best case damp conditions).

Is that with all the windows closed?

It's a cool idea and all

I see what you did there...

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