Wednesday, January 31, 2018

If coastal cities need more rain...

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If coastal cities need more rain they could try heating water in greenhouses to add water vapour to the air. Here is a graph (below) using the average of six evaporation equations. You can get 20 or more times the evaporation by increasing water temperatures. 
Studies on evaporation from windy sea conditions show more rain with more evaporation of spray. 
In many cities you can evaporate over 100 000 litres every day with a 100m by 100m greenhouse and add it to the atmosphere, humidifying it and increasing chances of rain. 1kWh can evaporate about 1.5 litres and many cities can easily get about 8 kWh of solar energy on every horizontal square metre in a day.
When the air is hotter than the sea the water cools the air immediately above it and the relative humidity (RH) of this air increases and water can even condense out of the air so that "negative evaporation" occurs. If the seawater is hotter than the air the evaporation increases substantially, because the water heats the air above it, the RH of the air decreases, and the hotter air can take up moisture faster. This principle of high evaporation rate with water being hotter than air can be used to increase the humidity of the air and increase chances of rainfall. Therefore, heating seawater in greenhouses and so on can be very effective. Evaporation equations do not all give the same rate of evaporation and I use an average from six evaporation equations. Example: Wind speed=10 km/h, Air temperature above water is 32 deg C, temperature of water is 20 deg C, pressure = 1 atm. The average of the six equations is -1 mm per day so water condenses out and we could get a mist immediately above the water. Now we keep everything the same except we heat the water in a greenhouse to 37 deg C. The rate of evaporation (given by the average result of the 6 equations) is now 28 mm per day and we can increase the humidity of the air substantially, increasing chances of rain. Calculations show, if the area of greenhouses were one square kilometre, one could evaporate over 10 million litres a day in sunny areas.

Air tends to move back and forth with sea breezes and land breezes, so the humidity could accumulate every day from the greenhouse evaporation.

Here is another graph (below) where air temperature is increased by only 1 deg C. So, as the air heats more relative to the sea, so the evaporation from the sea decreases. Decreased evaporation means decreased chances of rain.

Tuesday, January 9, 2018

Rain by humidification above the ocean.

Above shark nets, about 20 m above the ocean, have thick pipes with thousands of holes in that water streams out. Water can be pumped into the pipes using wind turbines. If the wind is blowing at 10 km per hour and the pipes are 1 km long, then the volume of air humidified in an hour is 1 km x 10 km x 0.02 km = 0.2 cubic kilometres.
In a day this is 4.8 cubic km of air. Nights are warmer with more humid air and I am going to use this as an example: RH=65% and Tair=20 deg C before humidification. RH=80% and Tair=22 deg C after humidification (the air will be blowing back and forth with land and sea breezes). Before humidification the water vapour content of the air is 11.2 grams/cubic metre and after humidification it is 15.5 grams/cubic metre. This is an increase of 38%. This relies on the fact that more humid air will keep in heat from the ocean when air is colder than the ocean. See

The heat from the ocean will help humidify. Air is usually colder than the ocean at night. During the day any mist will absorb solar energy and heat up and evaporation will occur. 

Thursday, January 4, 2018

Easy wet bulb temperature determination

People have been searching on the Internet for an easy way to calculate wet bulb temperature (and so have I). Experts give various long calculations and I wanted an accurate value easily calculated. I had a lot of trouble searching on various forums and eventually found a site that uses an equation that can be solved numerically that gives me an accurate answer, but the formula did not work unless I changed the P units from hectopascals to atmospheres. So instead of 1000 hPa I use 1000/1013.25. I solve numerically using a computer program I wrote to get Tw (wet bulb temperature). The formula is at
and I notice that P is included in one equation and left out in the next equation. However if you use atm it should work well with P included in both. The formula gives RH, but you can find Tw numerically. Say you are trying to find the wet bulb temperature for Td=45 deg C and RH=67%. Then let RHS=formula given and start with Tw=45 (represents RH=100%) and decrease Tw iteratively until RHS<=67. Find Tw at that point.

You can use a wet bulb calculator to check how accurate the above formula is (pretty good).
Your inputs into the program will be Td, RH and P. (P in atm). (Td is dry bulb T and Tw is wet bulb T.)

Code in Pascal:
until (RHS<=rh);

{Now print Tw}