For more on benefits and effects of humidification using evaporative cooling see https://vpdcalc.blogspot.com
For information on evaporative cooling see https://en.wikipedia.org/wiki/Evaporative_cooler
Example: The air temperature is 25 deg C and the wet bulb temperature is 15 deg C. If the efficiency of evaporative cooling is 20% then the air will be cooled (20/100)(25-15) = 2 deg C See https://en.wikipedia.org/wiki/Evaporative_cooler#Performance
You can download my free instrumental tune "Sail Cloth" that could remind people of the coarse sail cloth above the waves idea. Download free at https://clyp.it/l3oanx0t
Wednesday, 1 March 2023
Tuesday, 7 January 2020
Reducing air temperatures by reducing carbon dioxide
One can reduce carbon dioxide levels locally and thereby reduce air temperatures by reducing downwelling sky radiation - see https://www.scientificamerican.com/article/can-carbon-dioxide-domes-affect-health/
If one reduces carbon dioxide levels in the air above Australia then air temperatures will be reduced and fire danger will also be reduced. Cooling air increases the relative humidity and chances of relief rain.
Taking CO2 out of the air: Here is some mathematics for all: Basalt has a density of about 3 tonnes per cubic metre.
A 1 mm thick layer of basalt spread over an area of 1 square km has a volume of (1/1000)x(1000)x(1000) = 1000 cubic metres.
Mass of 1 mm thick basalt layer on 1 square km = volumexdensity = (1000 cubic metres)x(3 tonnes per cubic metre) =3000 tonnes.
1 tonne of basalt can react with about 0.3 tonnes of CO2.
Therefore 3000 tones of basalt can react with about 3000x0.3 = 900 tonnes of CO2.
In a cubic metre of air in a polluted city there could be about 900 tonnes of CO2 in a cubic km.
Conclusion: A 1 mm thick layer of basalt could take out all the CO2 for a km above the basalt layer. Powdered basalt should be used to make the reaction thousands of times faster.
See https://gulfnews.com/world/gulf/oman/oman-rocks-to-help-fight-global-warming-1.1810841 and
https://arctic-news.blogspot.com/2016/07/olivine-weathering-to-capture-co2-and-counter-climate-change.html and
https://arstechnica.com/science/2018/02/spreading-crushed-rock-on-farms-could-improve-soil-and-lower-co₂/ and
https://insideclimatenews.org/news/20022018/global-warming-solutions-carbon-storage-farm-soil-crushed-volcanic-rock-research
If one reduces carbon dioxide levels in the air above Australia then air temperatures will be reduced and fire danger will also be reduced. Cooling air increases the relative humidity and chances of relief rain.
Taking CO2 out of the air: Here is some mathematics for all: Basalt has a density of about 3 tonnes per cubic metre.
A 1 mm thick layer of basalt spread over an area of 1 square km has a volume of (1/1000)x(1000)x(1000) = 1000 cubic metres.
Mass of 1 mm thick basalt layer on 1 square km = volumexdensity = (1000 cubic metres)x(3 tonnes per cubic metre) =3000 tonnes.
1 tonne of basalt can react with about 0.3 tonnes of CO2.
Therefore 3000 tones of basalt can react with about 3000x0.3 = 900 tonnes of CO2.
In a cubic metre of air in a polluted city there could be about 900 tonnes of CO2 in a cubic km.
Conclusion: A 1 mm thick layer of basalt could take out all the CO2 for a km above the basalt layer. Powdered basalt should be used to make the reaction thousands of times faster.
See https://gulfnews.com/world/gulf/oman/oman-rocks-to-help-fight-global-warming-1.1810841 and
https://arctic-news.blogspot.com/2016/07/olivine-weathering-to-capture-co2-and-counter-climate-change.html and
https://arstechnica.com/science/2018/02/spreading-crushed-rock-on-farms-could-improve-soil-and-lower-co₂/ and
https://insideclimatenews.org/news/20022018/global-warming-solutions-carbon-storage-farm-soil-crushed-volcanic-rock-research
Rain enhancement by evaporative cooling
It has been suggested that power barges be used to supply electricity to South Africa - we order them, they set up along the coast and plug into our grid to provide power. Now we could also use pumps and tall pipes on power barges to pump water 200 m up and let it fall as spray back to the sea. This could humidify the air near coastal cities. Some power barges have over 100 MW in power and with only 20 MW of power one could pump 20 000 tonnes of water up to a height of 200 m in a hour. One only needs about 1000 tonnes of water to evaporate in a cubic km to produce significant humidification. When the humidified air blows to a city via sea breezes it will be cleaner and cooler and could bring rain.
BOM (Bureau Of Meteorology, Australia) contacted me to say I had obviously put a lot of thought into my idea of cooling and humidifying air over the sea near coastal cities. They have not said they are considering it as a viable method, but my belief is that if they can use pumps on power barges to increase the relative humidity of a region from ocean surface to 200 m above the ocean, then they will increase chances of rain a lot (there are terrible fires and heat waves in Australia). The cooled layer will eventually heat up via infrared radiation from the sea if the air is calm (seawater has a high emissivity and radiates infrared well). So we can create a moist layer of air 200 m deep with a dryer layer of air above.
If you have the humidified layer at the same temperature as the dry layer above, parcels from the humidified layer will rise (moist air is less dense than drier air). When the air mass (moist layer and drier layer above) blows to land via sea breezes and rises, the moist layer becomes saturated more quickly than the dry layer as both layers rise and cool (cooling on rising is how rain occurs). When the bottom moist layer becomes saturated it cools at about 5 deg C for every km rise (the saturated adiabatic lapse rate). The dry layer above cools at 9.8 deg C for every km rise (the dry adiabatic lapse rate) so that the layer above becomes cooler and more dense than the moist layer below. So parcels from the lower moist layer rise, through the denser and cooler layer above, for considerable distances, creating rain under suitable conditions.
If you have the humidified layer at the same temperature as the dry layer above, parcels from the humidified layer will rise (moist air is less dense than drier air). When the air mass (moist layer and drier layer above) blows to land via sea breezes and rises, the moist layer becomes saturated more quickly than the dry layer as both layers rise and cool (cooling on rising is how rain occurs). When the bottom moist layer becomes saturated it cools at about 5 deg C for every km rise (the saturated adiabatic lapse rate). The dry layer above cools at 9.8 deg C for every km rise (the dry adiabatic lapse rate) so that the layer above becomes cooler and more dense than the moist layer below. So parcels from the lower moist layer rise, through the denser and cooler layer above, for considerable distances, creating rain under suitable conditions.
Thursday, 12 September 2019
Cooling air using vegetation
Spekboom could cool the ground in dry areas and enhance rainfall. Sand has air between the grains and this causes sand to act as an insulator concentrating the solar energy in the top few cm of dry sandy soil. So sand gets very hot and heats the air above it by contact and infrared radiation. While sand could reach a temperatures of 60 deg C or so I think Spekboom would be a lot cooler. Leaves of trees and Spekboom, etc, act like "convection machines" because there is a large leaf area at the top of trees, etc, (where leaves are heated in the sun) in contact with the air. So the tree is air-cooled.
If you cool the ground you cool the air above it and the lifted condensation level (LCL) is reduced making rain more likely when the air is lifted by blowing up mountains, etc. The equation is LCL=125(Tair-Tdew) where LCL is in m, Tair is the air temperature in deg C and Tdew is the dew point temperature in deg C.
If you just heat or cool air the dew point temperature remains the same, so cooling air reduces the LCL (height to which air must be lifted for clouds to form). The dew point temperature depends only on the water vapour pressure in the air. Since the atmospheric pressure remains the same and the mole fraction of water vapour remains the same when air is heated or cooled and vapour pressure in the air is (mole fraction of water vapour)x(atmospheric pressure), the dew point remains the same.
If you cool air the vapour pressure deficit decreases.
If you cool the ground you cool the air above it and the lifted condensation level (LCL) is reduced making rain more likely when the air is lifted by blowing up mountains, etc. The equation is LCL=125(Tair-Tdew) where LCL is in m, Tair is the air temperature in deg C and Tdew is the dew point temperature in deg C.
If you just heat or cool air the dew point temperature remains the same, so cooling air reduces the LCL (height to which air must be lifted for clouds to form). The dew point temperature depends only on the water vapour pressure in the air. Since the atmospheric pressure remains the same and the mole fraction of water vapour remains the same when air is heated or cooled and vapour pressure in the air is (mole fraction of water vapour)x(atmospheric pressure), the dew point remains the same.
If you cool air the vapour pressure deficit decreases.
Saturday, 7 September 2019
Townsville Australia 10 Sept 2019 at 13:00
Weather report for Townsville QLD Australia for 10 Sept 2019 at 13:00:
T=27 deg C, RH=15%
Calculation: For all the following calculations the atmospheric pressure is assumed to be 101.325 kPa and the efficiency of evaporative cooling is 30%:
With evaporative cooling of 30% efficiency the air will be cooled 4.22 deg C to 22.78 deg C.
T=27 deg C, RH=15%
Calculation: For all the following calculations the atmospheric pressure is assumed to be 101.325 kPa and the efficiency of evaporative cooling is 30%:
With evaporative cooling of 30% efficiency the air will be cooled 4.22 deg C to 22.78 deg C.
Volumetric heat capacity of air before cooling is 1.185 kJ/(m^3.degC)
The mass of water required to do the evaporative cooling is 2035 tonnes per cubic km.
RH after evaporative cooling will be 29.38%.
RH after evaporative cooling will be 29.38%.
Danger of fire. The greater the Chandler Burning Index the greater the danger of fire:
Before evaporative cooling: Chandler burning index is 124.9 (extreme danger of fire)
After evaporative cooling: Chandler burning index is 60.6 (moderate risk of fire)
Before evaporative cooling: Chandler burning index is 124.9 (extreme danger of fire)
After evaporative cooling: Chandler burning index is 60.6 (moderate risk of fire)
Dew Point:
Dew point before evaporative cooling is -1.82 deg C
Dew point after evaporative cooling is 4.02 deg C
Using Espy's equation for lifted condensation (height air must rise for clouds to start forming):
Dew point before evaporative cooling is -1.82 deg C
Dew point after evaporative cooling is 4.02 deg C
Using Espy's equation for lifted condensation (height air must rise for clouds to start forming):
Lifted condensation level (LCL) before evaporative cooling is 3603 m
Lifted condensation level after evaporative cooling is 2345 m.
Vapour pressure deficit (VPD). Usually VPD should be between 0.45 kPa and 1.25 kPa: Danger zones are VPD is less than 0.4 kPa and VPD is greater than 1.6 kPa
Before evaporative cooling VPD is 3.03 kPa
After evaporative cooling VPD is 1.96 kPa
Lifted condensation level after evaporative cooling is 2345 m.
Vapour pressure deficit (VPD). Usually VPD should be between 0.45 kPa and 1.25 kPa: Danger zones are VPD is less than 0.4 kPa and VPD is greater than 1.6 kPa
Before evaporative cooling VPD is 3.03 kPa
After evaporative cooling VPD is 1.96 kPa
Subscribe to:
Posts (Atom)