Atmospheric humidity for water generation.

Copyright 2015 Robert Clark

*Suggestions for Water Production for California.*

California is still in the midst of a huge drought:

California’s About to Run Out of Water. We Have to Act Now.
Annie Sneed, Science, 03.23.15 7:00 am
http://www.wired.com/2015/03/californias-run-water-act-now/

 The state has now ordered a mandatory 25% drop in water usage.

 There appear to be however relatively simple methods of providing extra water. One is by deriving water from humidity in the air, the other by distilling the water from the ocean. Both would appear to have relatively low cost solutions.

 First the humidity solution. The amount of water in the form of water vapor is substantial, especially at high humidity. A key fact is air can store more water vapor at higher temperatures. But the point is the areas in California with the highest drought level are the areas with routine high temperatures, such as Los Angeles. So lets calculate the amount of water in air in Los Angeles. This page gives a graph of average relative humidity levels in LA:

Average Weather For Los Angeles, California, USA.

The average daily high (blue) and low (brown) relative humidity with percentile bands (inner bands from 25th to 75th percentile, outer bands from 10th to 90th percentile).

https://weatherspark.com/averages/29963/Los-Angeles-California-United-States

 You see that the relative humidity commonly reaches the 80% range and above, especially during the warmer months. The relative humidity is the percentage of the maximum possible water vapor the air can hold based on that temperature, called the saturated vapor density. This page gives a calculator for the saturated vapor density based on temperature.

Relative Humidity Calculation.
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/relhum.html

 For a temperature of 80 degrees F, the saturated vapor density is given as 25.4 gm/m³. And for a relative humidity of 80%, this corresponds to an actual amount of water in the air of 20.3 gm/m³, 0.0203 kg/m³.

 However, in actuality when the temperature is highest such as in the afternoon the humidity is lowest, and when the temperature is lowest such as in the early morning, the humidity is highest. So let's use some more representative temperatures and humidities. This page gives the  Los Angeles weather over a two-week period:

Weather past 2 weeks in Los Angeles, California, U.S.A.
http://www.timeanddate.com/weather/usa/los-angeles/historic

 For the period as of today April 3, 2015, for a time during the day when both the temperature and the humidity are high, say, in the 60's range for both, this results in total amount of water in the air of about 10 gm/m³.

*Residential Solutions.*

I.) Some ceiling fans can move quite large amounts of air at a time resulting in quite large amounts of water vapor inflow at a time, if placed for example in residential windows. For instance some fans on this page at their higher speed settings can move 5,000 cubic feet per minute (CFM) of air while using only 30 watts:

ENERGY STAR Most Efficient 2015 — Ceiling Fans 52 inches and under.
http://www.energystar.gov/index.cfm?c=most_efficient.me_ceiling_fans_under_52_inches

  Lower speeds give better energy efficiency, but at the cost of lowered air flow amounts. We’ll use the high speed numbers since we want large air flow to get high water vapor inflow.

 At this high amount of hot air inflow though we would probably want the fan in a basement or attic since we would also need a window for the hot air, once the water is removed, to exit.

 How much water would need to be produced? An average person uses about 50 gallons per day of water:

City Utilities: Water Tips.
http://www.cityutilities.net/resident/pgms/watertips.htm

 So say a residence for a family of 4 needed 200 gallons of water per day. A gallon is 3.785 liters, so this is 757 liters per day. Water weighs 1 kilo per liter so this is 757 kilos per day. Then how long would it take for a fan blowing in 5,000 cubic feet per minute to bring in this much water(vapor)? At 5,000 cfm this is 5,000/3.28³ = 141.7 meters per minute.

 At the the 80 degree F and 80% humidity level, each of these cubic meters of air would contain 0.0203 kg of water vapor. So this would amount to 2.88 kg per minute of water. This would take 262.8 minutes, about 4.4 hours. However, since the humidity actually goes down as the temperature goes up we would need to add up the total amount of humidity during a full 24-hour day. Looking up the representative amounts on the "Weather past 2 weeks in Los Angeles, California, U.S.A." page, we see the totals will still be above 200 gallons per month.
 This amount of power, only 30 watts used for 4.4 hours, is also quite small in energy costs compared to what the county of Los Angeles charges for water.

 An additional question to be resolved however is how can we convert this water vapor to liquid water? Air conditioners are able to do this, by accident, by chilling the air. Air dehumidifiers also commonly work this way. This causes water to condense out like happens for example in cool morning temperatures with morning dew. On our relative humidity calculator page, at 80 degrees F and 80% relative humidity, the dew point is only 73.4 F.  However, both air conditioners and air humidifiers use quite high power levels. We want to minimize additional power used.

 Some possibilities:

1.)If the water produced this way is an adjunct to the water received from the city, then we can use the cool water coming from the city water supply, typically around 50 degree F, to cool this air and get the water to condense out.

2.)It would be nice though, since the calculation showed this air-produced water alone is sufficient to supply the entire household water needs, to find a way that didn’t use the city water supply.

a.)Another type of air dehumidifier uses desiccants to absorb water vapor. The desiccant material is then heated to release the water as liquid and the same desiccant is used over again. However, this material typically is a silica compound and you would not want remnants of this to be left in the water. This also uses additional power for the heating step. If a desiccant could be found that is a type of mineral you would normally see for example in spring water then this might work. You would though need to find a way to get the water to be released as a liquid. Heating as with air dehumidifiers would work. However it may be at the high temperature of southern California would be sufficient so this would happen naturally.

b.) A similar possibility derives from the fact that rain droplets can frequently condense in the air out of water vapor at temperatures higher than they would normally do by having nucleation sites:

Cloud condensation nuclei.
http://en.wikipedia.org/wiki/Cloud_condensation_nuclei

 Then we could add nucleators into the air stream to get the air to condense. These nucleators though again would have to be a non-toxic if ingested. Ideal would again be some type of mineral commonly found in mineral water.

c.)To get the water to condense we could also expand the air flow. Rapidly expanding the air would cause the temperature to drop thereby chilling the water. A problem here though is the air flow is so large it might require an unreasonable size of expansion needed to get the needed temperature drop.

d.)Another possibility would be by increasing the pressure of the air. Just as increasing the pressure increases the temperature at which water makes the transition from liquid to gas, the boiling point, so also the temperature at which it makes the transitions from gas to liquid, the dew point, also increases. This page gives a calculator for how the pressure changes the dew point:

Dew Point Conversion Calculator.
http://www.howelllabs.com/resources/dew-point-conversion-calculator/

 Enter in 73.4 degrees F in the known dew point field for our 80 degree F and 80% relative humidity scenario. Enter in 0 for the “psig” field, which measures how far this is above standard pressure in psi. Then a psig of only 4 gives a dew point of 80.3 F. That is an increase in pressure of less than 30% results in the water condensing out.

 We might be able to generate this amount of extra pressure by circulating the air around in a circle by centrifugal force. A problem though is the size of the air stream coming from a large ceiling fan size diameter might make the circle size needed impractically large. We could constrict the air coming from the fan into a smaller pipe diameter, but by the Bernoulli principle this would reduce the pressure. It still may be possible though that some combination of restricted pipe diameter and circulation diameter size could provide the needed pressure change at a practical size.

 Another possibility for using centrifugal force to effect the compression is by breaking up the airflow into multiple tubes at the microscale. The compression in this scenario is by the fact the acceleration around a circle is a = v²/r, v being the speed, and r the radius of the circulation. However, for our scenario the speed would only be around 3 to 4 mph, resulting in low acceleration and therefore compression for a large size pipe. However, by breaking the airflow up into multiple microtubes we can make the radius r very small giving high acceleration even for low speed.
 Support for the idea this can provide the needed compression comes from the several papers published on MEMs (micromechanical systems) rotors and turbines at the mini and microscales. Running at the hundreds of thousands of RPM's range these can provide pressure ratio's of from 2 to 3, sufficient for our purposes.

 At a slow airflow speed of 3 to 4 mph though these microtubes might need to be only be at the micron scale diameter to get such high RPM's. We might also constrict the pipe diameter first, which by Bernoulli's principle will increase the speed, before we break the air flow up into multiple microtubes (patent pending).

 For either the large pipe or microtube solutions using centrifugal force we could instead of bending the tubes around in a circle, have rifling inside the tube to cause the air to circulate helical fashion within the tube.

 II.) For the single fan bringing in 5,000 cfm it was surprising this could be done at only 30 watt power level. It would be interesting to find out then how large we could make the air inflow in for a single residential house. The ceiling-fan sized fan we used was about 1.5 square meters. For an average size house of a 100 m² floor plan, this would provide about 66 times as much air flow and therefore water, if it were brought in through the roof. So this one home could provide the water needs for 66 other homes.

 This assumes that power requirements are just for running the fan and you would not need power for instance to do the condensing step. Since the power requirements are about at 1/10th that of the cost of water for Los Angeles, the home owner could sell the water at much reduced rates.

 However, it may be possible to reduce or eliminate the need for power to run the fan by using the wind flowing over the roof. A problem though is by the Venturi effect, the wind blowing horizontally over the roof would have the tendency to draw the air out of the house rather than drawing air in. If a solution could be found to draw the air in rather than out while using the air flow of the wind the costs would then be essentially nothing.

*Water Utility Solution.*

 For the water company using this method to produce extra water it becomes particularly simple. The large ceiling-fan type fans discussed in the residential solution only move air in the range of 3 to 4 miles per hour. But wind speeds commonly are above this speed especially in coastal areas. Then you would not even need to use fans for the water utility solution. You would just collect the air driven by the wind in large tubes for processing. Also, for California communities near the coast, using the ocean to supply the necessary cooling to condense the water vapor becomes especially simple (patent pending).

*Sustainable Water Resources*

  Up to 1 billion people don't have reliable access to clean drinking water:

http://wholeworldwater.org/

 This would provide a simple, low cost means of providing clean water, assuming the condensing step could be done without extra power. If a fan is needed, the low amount of power could be supplied by a wind mill or solar power. However, it may be that the air flow from the wind itself would be sufficient and not even a fan would be needed.
.
*Individual Water Production.*

 For a scenario where one is stranded out in the wild or in a life raft without water, the amount of power needed for the fan is so small that it could probably be generated by hand for a single individual just for enough water to sustain life. And considering the wind speed needed is only in the range of 3 to 4 mph a fan might not be needed at all.

 Also, at least for an 80 F and 80% relativity humidity scenario, the 30% extra pressure needed to allow water to condense could be easily supplied by hand. By the Ideal Gas Law PV = RT, to get a 30% increase in pressure we would need to make less than a 23% decrease in volume, assuming we did the compression slowly so as not to increase the temperature. For instance, a piston in a foot long cylinder would only need to make a 3 inch compression to get the needed pressure.
 The life boat case would also be very simple because the condensing could instead be done by using the cool temperatures of the ocean water.


    Bob Clark