Wednesday, March 15, 2017

Rain when sea temperatures are higher than air temperatures over land?

Some time ago I made and showed graphs indicating there was often rain when sea temperatures were higher than land air temperatures. Yesterday I came across this graph in a physical geography book showing more rain when sea temperatures were higher. I do not know if the author noticed this as well - cannot see where he/she mentions it. See https://books.google.co.za/books?id=iU-pGIOqS1AC&pg=PA38&lpg=PA38&dq=air+and+sea+temperatures+and+rainfall+Canton+Island+Boucher+1975&source=bl&ots=dMMHHLbX5g&sig=rCPunkj6nZuo044iNSUGgIaSdJs&hl=en&sa=X&ved=0ahUKEwiatdO5nNjSAhWsJcAKHS9dC1QQ6AEIGjAA#v=onepage&q=air%20and%20sea%20temperatures%20and%20rainfall%20Canton%20Island%20Boucher%201975&f=false
Something that will give credibility to my idea is the graph below. For my calculations I have used the following conditions: 
1) Temperature of air over the land is 26 deg C. 
2) There is a 2 m/s wind over the sea. 
3) The relative humidity of the land air is 50% and pressure = 100 kPa. 
4) The air blows from the land to the sea and at the sea surface where evaporation occurs, the air cools or warms to the temperature of the sea temperature. 
5) The variable is the sea temperature. 
a) When Tsea=14 deg C the air becomes saturated and condensation occurs (I have used a 100% RH for this on the graph), the evaporation rate from the sea surface is 0 microns of water per hour. 
b) When Tsea=16 deg C, the RH of the air over the sea surface is 92.5% (RH increases when T decreases), the evaporation rate is 55 microns per hour. 
I have put the RH and the evaporation rate in microns on the same vertical axis (not really correct, but convenient). So Tsea runs from 14 deg C to 34 deg C, the RH runs from 100% to 31.6% (see graph), the evaporation rate runs from 0 (at 14 deg C) to 1532 microns of water per hour (if the water were in a basin, water levels would drop from 0 microns per hour when T=14 deg C, to 1532 microns per hour when T=36 degC). As one can see, when sea temperatures are high relative to land temperatures, more evaporation occurs. I used an equation similar to the "Evaporation from a water surface" equation by Engineering toolbox to calculate evaporation rates (they depend on relative humidity and temperature and wind speed).

Tuesday, March 14, 2017

Water source heat pumps for rain.

Water source heat pumps with large moving bodies of water are very efficient – even more so than air and ground source heat pumps.
Idea: 1) Build a huge tank or tanks of air above the sea on steel frames. 2) Use water source heat pumps and solar energy to warm the air in the tanks. 3) Spray water into the tanks to evaporate in the warm air.
http://www.thegreenage.co.uk/tech/water-source-heat-pumps/ describes these heat pumps..
If a big tank of air is situated in the region near a desert and this air is heated using a water source heat pump and solar energy that is reflected by mirrors onto the tank, seawater can be sprayed into the warm air and evaporated by the hot air. The moisture levels and heat levels of the air could be controlled and the air can be manufactured and released when there is a sea breeze (breeze onto land). The warm moist air will rise by convection to form rain.
The heat exchange of the water source heat pump could take place in the warm Arabian Gulf waters, or in the warm Red sea, making the system very efficient.
With regard to the mirrors focusing solar energy onto the tank with air in, a square kilometre of land or sea surface can often provide about 5 000 000 kWh of energy in a day (5 kWh per square metre per day ). This energy could be focused by the mirrors onto the tank described, or onto a number of tanks. Now 1 kWh can evaporate about 1.6 litres (1.6 kg) of water, so this is enough to evaporate about 5 000 000x1.6 = 8 000 000 litres ( 8000 metric tons) of water. This extra 8000 tons of vapour will increase humidity in the general surroundings and moist air could be targeted to various ares to some extent, by controlling temperature (different rate of rising with different temperatures) and calculating sea breeze direction and velocities. 
At a relative humidity of 40% and a temperature of 30 deg C 1 cubic kilometre of air holds 12140 metric tons of water vapour. At the same temperature and an RH of 66% this cubic kilometre holds 20032 metric tons (about 8000 tonnes more).

Monday, March 13, 2017

Artificial heated lagoons for rain

 With colder sea temperatures than land temperatures (eg Western Cape in summer, UAE, etc), cold air from the sea does not hold a lot of moisture and when it blows onto land and heats up, relative humidity decreases. Also this cold air does not easily rise by convection to form rain. One of my ideas is artificial lagoons with solar energy reflected into them to heat the water.
With deep ocean one has a limitless supply of heat from the water (evaporation cools the surface, but there is plenty of heat from the water to heat it up again), but the surface can only be heated to about the sea temperatures below the surface. About 43% of energy from the sun is of visible light frequency. Visible light has the distinct property of being able to penetrate seawater to some depth, whereas the infrared generally gets absorbed quickly. So if you have shallow pools with dark bottoms you suddenly have 43% of the solar energy available in a shallow pool that normally would have heated a huge body of water a little, but there is very little energy that can be used from the water to heat the surface when it cools from evaporation. As I am looking for high temperatures I have to think about shallow pools rather than deep sea. With a shallow pool the energy comes mainly from the solar energy (not the water). If one wants high temperatures (with evaporation and radiation the shallow pool can cool quickly), one needs ways of getting solar energy to the pool or lagoon, etc. So my proposal has been shallow lagoons made by bulldozing out of the sand and with mirrors to reflect solar energy into the lagoon. Say the sun provides 700 W of energy per square metre of water surface. Eventually the temperature of the lagoon water gets so high (at about 26.5 deg C) that all the 700 W per sqare metre is needed for evaporation and so mirrors will be needed for more energy. My graph below shows the 700 W straight line (0.7 kW), the evaporation rate on a 1 sq metre surface (upper curve kg/hour) and the power needed for the evaporation (lower curve kW). Not correct to have kW and kg/hour on the vertical axis, but it works out. The conditions are a 40% relative humidity, a 3 m/s wind over the surface and an atmospheric pressure of 100 kPa.


Friday, March 3, 2017

Growing trees in the deserts

https://www.adn.com/arctic/2017/02/27/scientists-just-measured-a-rapid-growth-in-acidity-in-the-arctic-ocean/
refers to acidification of the Arctic Ocean as well as global warming and ice. It seems to me that a way to take massive amounts of carbon dioxide out of the atmosphere is to grow trees in deserts. First, more rain would have to occur in deserts. One problem with global warming is that the land heats up faster than the oceans. Imagine hotter air flowing from land to cooler ocean. The hot air cools down where air meets water and the relative humidity of this cooling air rises and the air can become saturated immediately above the ocean where evaporation could have taken place. Because no net evaporation can occur into saturated air there is a problem with the land and sea breeze mechanism - no moisture is being picked up by the air. An answer would be to provide hot moist surfaces over the ocean so that the air from land is warmed rather than cooled when flowing over the sea, Warm air can hold more moisture, so evaporation will occur. Now when the air flows back onto land (sea breeze, etc) it will have more moisture than before and rain can result. Clouds absorb thermal infrared and so I presume sea spray will also. So if one puts black objects (that heat up in the sun) in the sea spray, the spray will absorb the thermal infrared radiation from the hot objects and will evaporate. Sea spray will also evaporate by having contact with hot these objects. Black netting hung in the spray should do the trick, because this netting will get hot and radiate thermal infrared. Sunlight itself has almost no thermal infrared in, but does radiate infrared of a higher frequency. My Planck's formula integration tells me that about 0.51% (less than 1%) of solar radiation is thermal infrared (5 to 20 microns in wavelength) radiation. On the other hand a blackbody (something like black netting approximates this) having a temperature of 40 deg C has about 74.2% of its radiated energy in the 5 to 20 micron range. This black netting in the ocean method could be used with deserts that are near cold seas, but have a lot of sunshine. Resulting rain could then get trees growing in deserts to absorb massive amounts of carbon dioxide. Desert soil is often very fertile. Black cement rocks in the sea to create spray from waves and absorb solar energy to heat spray could also be used along with black netting. Another way to make the seawater hotter is to have shallow pools of seawater (preferably with dark bottoms) because shallow pools get hotter in sunny weather.
It seems that the following could occur: 
1) From morning until about midday the air over the sea will become humid because of shallow pools and spray heated on dark objects.
2) At about midday the sea breeze will start (air will blow onto land from the sea). This will bring moist air to the land.
3) At night the air will blow from land to sea, but the air will become more moist than usual over the sea because of the spray.
4) The process of steps 1, 2 and 3 will repeat itself - at about midday moist air will start moving onto the land, etc.
Spray could also be generated by a spray pump using wave motion to drive it.
To get air to rise over the land when it blows from sea to land, use solar air heaters on roofs of the city and elsewhere. Solar air heaters can heat massive volumes of air. The solar air heaters will shade roofs and keep them cool, but will heat air and get it to rise. Another way of getting air to rise over an area is to make that area dark with biochar. The dark land will heat up more relative to lighter coloured land and heat air. When clouds form they will have a cooling effect on the Earth.

Could biochar be used to clear fog?

Could biochar be used to clear fog? After the biochar dust has fallen to the ground the ground should be more fertile.
Fog: Shipping and aviation is sometimes delayed by fog. Fog will reflect some sunlight and absorb some infrared. If you distributed fine biochar particles in the fog these would absorb most of the solar energy (light and infrared) and the fog should heat up and disperse. The Japanese use biochar on snow to help melt snow. The biochar decreases the albedo of the snow and so it absorbs solar energy and melts days sooner.
Calculations: Wikipedia says that in fog there is typically 50 000 kg of water per cubic kilometre of fog (liquid water content of fog is typically 0.05 g per cubic metre). 2257 kJ is needed to vaporise 1 kg of water.
To vaporize this 50 000 kg of water takes about 50 000x2257 kJ=112 850 000 kJ = 112 850 000/3600 kWh = 31347 kWh.
Say the fog is 100 m deep. Then an area of 10 square kilometres has a cubic kilometre of fog. Say the solar energy falling per square metre of ground is 2 kWh per day. Then 2x10x1000000 kWh = 20000000 kWh falls on the 10 square kilometres. This is far more than the 31347 kWh needed to vaporize the fog. Of course not all the solar energy will be absorbed by the fog with biochar dust in and some solar energy must go into heating air as well, but there is a lot of solar energy to spare (31347 kWh needed for vaporization and 20 000 000 kWh available from the sun).