Thursday, January 19, 2017

Removing fog at airports

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Two methods that could be used to heat foggy air at airports. THE MOST IMPORTANT ASPECT OF THESE METHODS IS: To heat air, air must be brought into contact with warm surfaces. When this happens (as in a solar air heater) air can be heated by 10 deg C very quickly. The air should have intimate contact with large surface areas. This can happen when air is moved through gauze, for example.
Method 1: Solar thermal collector with a large surface area that air will come in contact with in passing through the system. The evacuated tube collector supplies significantly more energy under cloudiness than the flat plate collector.  So use evacuated tube collectors to heat water in an apparatus that has a large surface area, for air passing through it to come into contact with. The water will retain heat and air can be heated throughout the night. The wind could blow foggy air through this hot water system and onto the airport. These collectors do not loose much by radiation (as black surfaces will). You can heat a huge volume of air with warm water. Water has a volumetric heat capacity of about 4200 kJ per cubic metre per 1 deg C rise in temperature and air has a volumetric heat capacity of about 1.2 kJ per cubic metre per 1 deg C rise in temperature.

Method 2: What about placing a gauze fence round foggy airports (but not in places that will affect air traffic)? Hopefully fog particles would collect on the gauze as foggy air moves through it. 
Another variation:  It appears that London uses about 35 GWh of electricity every hour (power is 35 GW). 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). 
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 it would take 35000 kWh to evaporate the fog and this is done in an hour. 
Then this would use (35000000 Wh)/(35000000000 Wh) = 1/1000 of the energy used by the city of London in an hour.
 If one had gauze shaped heaters between gauze fences around Heathrow, one could clear fog like this (air has to come into contact with large surface areas to heat up efficiently and gauze would provide this as air blows through it).
If you are going to have to heat the air as well as vaporise the liquid water content, that will require some more energy. You need roughly 1.2 kJ of energy to heat one cubic metre of air by 1 deg C (1.2 includes water vapour).
The question remains: If the air is saturated at, say, 2 deg C and there is 0.05 g of water content per cubic metre, how much would a cubic metre of air need to be heated before it could absorb the 0.05 g?
Well if the temperature is 2 deg C then the air will hold 5.56 g of water vapour per cubic metre and if the temperature is 2.13 deg C, then the air will hold 5.61 g of water vapour (5.56+0.05=5.61). So it would have to be heated only 0.13 deg C. 
For one cubic kilometre of air with 0.05 g of water per cubic metre the following holds:The energy just for vaporization is approximately 31300 kWh and if you include heating the air 0.13 deg C, the energy required will be about 75000 kWh. 
The graph below shows the number of Megawatt-hours needed to vaporize one cubic kilometre of fog in saturated air for different liquid water contents in the fog. The energy needed is of the form 
(V)(LWC)(2257) + (V)(Tinc)(1.2), where V=volume of air, Tinc=temperature increase of the air needed so that it can take up the extra vapour generated by vaporisation and LWC is the liquid water content. Since, for the small range of temperature increases needed, Tinc is approximately proportional to LWC, the form becomes
V(2257xLWC + 1.2xTinc) = constant volumex(2257xLWC + 1.2xconstantxLWC) = 
constantx(LWC)(2257 + constant) = LWCxconstant. Therefore Energy needed=LWC times constant and so the graph is linear.

Wednesday, January 18, 2017

Note on calculations

The calculations are made easier if you have a table of saturation pressures for water vapour and perform linear interpolation to get in between values. The partial vapour pressure is given by PsatxRH, where RH is the relative humidity and Psat is the saturation vapour pressure for water vapour. If you heat a parcel of air, the surrounding air remains at more or less the same pressure (unless there are other factors). On heating the parcel the mole ratio of all components in the parcel remains the same, the total pressure P remains the same, so the pressure of all components remains the same (including the partial pressure of the water vapour). Another way to look at is to say that the partial pressure of any component is its mole ratio multiplied by the total pressure.

Cooling Earth Using Clouds says, "For third straight time, Earth sets hottest year record"
One way to get rid of heat is to enhance formation of low level clouds in warm climate zones. To make clouds you need at least these two aspects: 
1) Air with moisture in and the more the better. 
2) A means to make this air rise.
If you just make air moist, but it is cold air relative to the surrounding air, it will not rise and form clouds. To heat air, air must come into contact with warm surfaces. So here is an idea: Put black netting in the sea where there is spray from waves. The black netting will absorb solar energy and when air passes through the netting the hot netting will heat the air and with the right construction you will have hot moist air rising to form low level clouds.

Thursday, January 5, 2017

Air Cleaner

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The blog owner T E Miller (Swayseeker), known as Eddie will not accept liability or responsibility for any problems arising from the use of this blog and its calculations. I try to provide good calculations and analysis, but cannot guarantee that there are no mistakes. Here is a site that tells you how to build your own solar air heater:
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Have been doing the following calculations: Los Angeles (latitude 34.05 deg N) has a maximum (always facing the sun) 11.2 kWh of solar energy per square metre on a good day on 1 July. On a horizontal surface it has 8.8 kWh of solar energy per sq metre per day (assuming a sunny day). It takes about 1.2 kJ of solar energy to heat 1 cubic metre of air 1 deg C (volumetric heat capacity of air is about 1.2 kJ per cubic metre per 1 deg C temperature rise - depends on pressure, etc). I have done this for the first day of each month and give a graph of how much air could be heated 5 deg C by one sq metre of horizontal surface by solar energy in one day. These are theoretical values and solar air heaters are certainly not 100% efficient, but the volume is enormous. I have been promoting this idea in Africa, China, US, India, via the Internet, etc, and am hoping it will have a good effect on the world.
If people generally knew the following facts about air the world might have been been different: 

Air is very little affected by radiation (sun shining through it, radiation from fires, heaters and so on). But air is heated by coming into contact with hot surfaces (casing of heaters, hot tar and so on) and the hot ground heats air and causes upward movement of this less dense air on a grand scale. The ground only has fairly superficial contact with air. On the other hand a solar air heater (a sort of greenhouse with a solar absorber to heat up in the sun and large hot surfaces to make contact with the air) is a different matter - it will heat air efficiently and a solar air heater on each rooftop could get warm air rising and out of polluted cities and also cause more rain to fall when vapour condenses in the cooler regions. 
People make their own solar air heaters and I believe India and China, with their pollution, could get polluted air moving out of their cities with them. It takes about 1.2 kilojoules of solar energy to heat 1 cubic metre of air by 1 degree C and every second 0.8 kilojoules of solar energy can easily fall on every square metre of some locations at noon.
The graph shows the number of cubic metres of air in a day that can theoretically be heated 5 deg C, using solar energy falling on a square metre of horizontal surface in Los Angeles. The x-axis shows 1 July, 1 Aug, etc.

Wikipedia says that rain dust (alkaline rainfall deposits, caused by particles from Saharan dust, etc) could help combat acid rain. 
An idea of mine: If one had huge solar air heaters and put Saharan dust in them, the hot air could carry the alkaline dust into sulfur dioxide-polluted air and neutralize acidity.
Wikipedia also says," Acid rain does not directly affect human health. The acid in the rainwater is too dilute to have direct adverse effects. However, the particulates responsible for acid rain (sulfur dioxide and nitrogen oxides) do have an adverse effect. Increased amounts of fine particulate matter in the air do contribute to heart and lung problems including asthma and bronchitis." 

Therefore the rain itself is good, because sulfur oxides, etc, are washed out, improving health prospects (reducing asthma, etc). Rain also washes out ozone, so if one could neutralize rain and get more rain, that would generally be good.
Graph below: The number of cubic metres of air in a day that can theoretically be heated 5 deg C, using solar energy falling on a square metre of horizontal surface in Los Angeles. The x-axis shows 1 July, 1 Aug, etc.

Wednesday, January 4, 2017


The first two graphs are for Cape Town, South Africa (lat 33.9 deg S). These two graphs show 
1) the solar energy falling per day on one square metre, maximum (always facing the sun - upper graph) and for a horizontal surface (lower graph). Solar energy in kWh 

2) the theoretical volume of air per day (cubic metres) that could be heated 5 deg C by a solar air heater of dimensions 1m by 1m always facing the sun (upper graph) and on a horizontal surface (lower graph).

The graph below is for Delhi, India. Lat 28.4 deg N. The graph shows the theoretical volume of air per day (cubic metres) that could be heated 5 deg C by a solar air heater of dimensions 1m by 1m on a horizontal surface. The x-axis shows 1 July, 1 Aug, etc. So on 1 July, in a day, the 1m by 1m solar heater on a horizontal surface in Delhi could heat 5241 cubic metres of air by 5 deg C.
Looking at temperatures and relative humidities for Delhi (India) and taking a low rainfall month of November, with an average RH of 55% and daily average temperature of 20.8 deg C, the graph uses figures as follows: The surrounding air temperature is 20.8 deg C and air is heated to the temperature shown on the T-axis (parcel of heated hotter air is at temperature T deg C), using solar air heaters. The line with the steeper slope shows the height to which the parcel will rise, using a dry adiabatic lapse rate of 9.8 deg per 1000 m rise and an environmental lapse rate of 6.5 deg C every 1000 m (fairly standard sort of figures). The line with less steep slope shows how high the heated air parcel must rise before clouds start to form (uses Espy's equation). When the parcel is heated to 28 deg C it will rise further than it needs to before clouds start to form. Before about 28 deg C it will not rise far enough for clouds to form. Actual lapse rates for Delhi would have to be taken into consideration for accurate conclusions. You can also work this out yourselves.I will tell you the near ground dew point for the parcel - it is 11.43 deg C. Espy's equation says, for clouds to form, the air parcel must rise 125(T-Tdew) where T is the near ground level temperature of the parcel and Tdew is the near ground level dew point of the parcel. As for how high it can rise, after it has risen 1 km, starting at T=27 deg C, say, the temperature of the parcel is 27-9.8 deg and the surrounding air is at 20.8-6.5 deg, etc. When the parcel and surrounding air are at the same temperature the parcel will stop rising (this is modified a bit because water vapour is less dense than air)