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.

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