Solution to Global Warming (Part 12

3.9.8 HEAT FROM EXPLOSIVES

The biggest volume being produced by the explosive industry is fired during the seasons of New Year. About 20% is exploded before and after New Year’s Day. The next in rank is the annual celebrations of the Christmas season, Town Fiesta, Independence Days, Corporate Organization Anniversaries, Golden Anniversaries and other celebrations. The third is in Mining Industries and fourth in Illegal Fishing. The fifth perhaps is in the military and terrorist activities.

Gunpowder was first discovered by the Chinese in the 11th century and records shows that the Chinese used gunpowder as war rockets against the Mongols as far back as 1279. It then spread to the Arabs and eventually reached Europe in the early part of the 13th century, more specifically in 1241,.
OCCASIONS Tons/Yr
o New Year 6,000,000
o Annual Celebrations 500,000
o Mining Industries 200,000
o Illegal Fishing /Hunting 60,000
o Military Operations/Exercise 40,000
o Demolitions, Etc. 20,000
Total 6,820,000

The early explosive of potassium per chlorate and later replaced by potassium nitrate. The gunpowder that is used in military weapon is made up of Potassium nitrate “KNO3”, Sulfur and charcoal. In Pyrotechnics or Fireworks, the major ingredients consist of a large number of flammable substances such as starch, gums. sugar, shellac, and various petroleum derivatives are frequently used in the mixture in place of charcoal and sulfur. Different colors are given to the fire by incorporating compounds of various metals in powdered form. Other explosive materials used in Fireworks include Nitro-glycerine “C2H5(NO3)2” trinitrotoluene “C6H2K(NO2)3” or TNT. Chemical smokes that produce heavy clouds include Hexachloroethane “ C2H2Cl6 “ Chloro-sulfunic acid “H2ClSO3 “ and titanium tetrachloride “T.Cl4.”.

TYPES OF FIREWORKS:
FIRECRACKERS: Fired primarily to produce sound .

ROMAN CANDLES: Cylindrical containers that are placed on ground and emits balls or stars of fires at intervals.

CATHERINE WHEELS:
Group of fireworks mounted around the periphery of a wheel that, when discharged, force the wheel to rotate by rocket effect

SUN BLAST
Fire works installed in a wheel in which the fireworks are discharged outward from the center and hence does not produce any rotation.

PASTILLES:Spirally coiled tube that rotates when lighted.

SKY ROCKETS
A rocket propelled projectile that explode high in the air to produce several types of display.


Fireworks
A fireworks display at the opening of the New York City Vietnam Veteran’s Memorial uses pyrotechnic materials. Modern fireworks are composed of perchlorate, organic substances like starch or sugar, petroleum products, and trace amounts of metals to give color.
Wesley Bocxe/Photo Researchers, Inc.
Microsoft ® Encarta ® 2007. © 1993-2006 Microsoft Corporation. All rights reserve

Fireworks has become a part of modern culture in all parts of the world that provides the thrills, excitements, exhilarations and enjoyable moments of ecstasy and happiness but its adverse effect to the environment is tremendous that facilitates the accelerated momentum of Global Warming. Mankind is like a moth and fireworks are like candles that out of enthusiasm and excitement the moth dives into the flame to burn itself and eventually die in the same manner that mankind creates Global Warming out of fireworks that will eventually destroy itself.

If we can only realize that about six million tons of fireworks being exploded every New Year produce 3.92 billion kilograms of Nitrogen Oxide which adverse effect is more than 300 times that of carbon dioxide in its contribution to Global Warming, we may not enjoy those fireworks by looking at them as the process of destroying the earth.

TRINITROGLYCERINE REACTION:
362 114 176 90 240
2C2H5(NO3)2” + 4.5 O2 = 4 CO2 + 5 H2O + 4 NO2

PRODUCTS OF COMBUSTION:

176
CO2 = ----------- = 0.486188
362
240
NO2 = ---------- = 0.662983
362
90
H2O = ----------- = 0.286188
362

Every New Year 3.98 billion kilogram of Nitrogen Oxide that is deposited into the atmosphere has an adverse effect that is equivalent to 1.2 billion tons of Carbon dioxide which is about 60% of 2.0 ppm annual increase of CO2 we are experiencing. If the world population ever realized that 20 % of Global Warming that man creates is derived from fireworks, every person should start thinking about the earth before igniting ant firework.

Aside from carbon dioxide and nitrogen dioxide that is produced the explosion emits heat that is added into the atmosphere in the equivalent amount of heat if and when 168 thousand cubic meters of gasoline is burned.

HEAT EMISSION:

24 14,000 BTU 2.2 lb 252 Cal
C = (-------) (-----------------) (----------) (---------------) = 514,581.2155 Cal/Kg
362 lb Kg BTU

10 60,000 BTU 2.2 lb 252 Cal
H = (-------) (--------------) (-----------) (------------) = 918,895.0276 Cal/Kg
362 lb Kg BTU
Total = 1,433,376.2431 Cal/Kg

HEAT PRODUCED DURING CHRISMAS SEASON:
HEAT = 1,433,376.2431 Cal/Kg (6,000,000) (1,000)
= 433,476,243,124,309,000.4 Cal


EQUIVALENT HEAT OF GASOLINE WHEN BURNED

433,476,243,124,309.4 Cal
Liters = -------------------------
8,517,042

= 168,306,818.6 Liters
= 0.00016306818 Km3

Other than Christmas Season and New Year, the amount of explosives fired per day all over the world is very minimal and estimated at 0.03742% of what is fired during Christmas Season and New Year is only equal to the heat if and when 62,975.85729 liters of gasoline is burned. .

The greatest and most destructive explosive of all times were the two Hydrogen Bombs that instantly killed more than 150 innocent civilians in Nagasaki and Hiroshima, in Japan in 1945.

3.9.9 HEAT FROM FRICTIONS & INDUSTRIAL MACHINERIES

Industrial Machineries that consume large amount of energy and emit large amount of heat may be summarized as follows:
ACTIVITIES TONS M3GASOLINE EQIVALENT
PER DAY PER DAY
1. Cement Industries 600,000 45,000
2. Food Industries 744,000 17,440
3. Steel Industries 120,000 11,250
4. Grain Drying & Milling 250,000 3,126
5. Paper Industries 130,200 3,052
6. Wood Industries 250,000 2,930
7. Home Appliances Mfr --------- 2,400
8. Heavy Equipment Mfr --------- 1,800
9. Transport. Mfg & Repairs --------- 1,200
10. Ceramics Industries 9,000 844
11. Sugar Industries 31,000 725
12. Construction Industries -------- 420
13. Plastic Industries 17,360 407
14. Chemical Industries -------- 400
15. Rubber Industries 15,370 360
16. Irrigation & Water Pumping -------- 297
17. Glass Industries 3,074 230
18. Textile Industries 6,329 148
19. Leather/Tanning Industries 2,040 36
20. Others --------- 22,505
Total 112,526 M3/day

The biggest consumer of energy among different industries is the manufacture of cement and the largest consumer of cement in the construction of residential houses. An average size of a residential house at 36 M2 consumes 200 bags of 40 Kg for a total of 8 tons.

About 60% 0f the world populations build concrete residential houses in urbanized area within the period of 40 years. At the average of 5 members in every family, there are 744 million concrete houses built in 40 years or 50,925 a day, which consumes 407,400 tons of cement per day. Assuming that about 50% more is used in the construction of roads and bridges, factories and commercial buildings, the daily consumption of cement worldwide may be placed at 600,000 tons per day.

To manufacture a ton of cement would require the equivalent amount of heat in 75 liters of gasoline making a total world energy consumption for cement manufacture equivalent to the heat of 45,000 M3 of gasoline /day.

SOURCE METRO MANILA NEW YORK
3.10.1 ORGANIC COMPONENTS
1. Pulp & Paper 36.00 % 41.00%
2. Wooden Slabs 12.00% 10.80%
3. Raw Food Wastes 9.80% 9.50%
4. Leaves, Grasses, stems 8.00% 7.25%
5. Plastic Bottles & Caps 6.50% 6.40%
6. Plastic Films, Bags, Sheets 4.80% 4.60%
7. Rubber Materials, Tires 4.25% 4.40%
8. Cooked Food Wastes 2.45% 2.50%
9. Clothing, Fibers 1.75% 2.50%
TOTAL 85.55% 88.95%

3.10.2 INORGANIC COMPONENTS
1. Tin Cans 6.50% 4.90%
2. Concrete Slabs 2.45% 1.85%
3. Iron Bars, Plates 1.75% 1.80%
4. G.I Sheets 1.60% 1.75%
5. Glass 0.85% 0.60%
6. Ceramics, Bricks 0.05% 0.10%
7. Other Inorganic 1.25% 0.05%
TOTAL Inorganic 14.45% 11.05%

TOTAL ORGANIC 85.55% 88.95%
TOTAL INORGANIC 14.45% 11.05%
GRAND TOTAL 100.00% 100.00%

Total Production of Rice worldwide in 1994 is 44 million metric tons which can support the staple food requirement of 336 million people of 6.1 % of world population.

3.11.0 HEAT LOSSES

3.11.1 EARTH AS BLACK BODY RADIATION INTO OUTER SPACE

“JOSEPH STEFAN AND LUDWIG BOLTZMANN (1879 -1884”) WERE THE SCIENTISTS WHO FORMULATED THE EQUATION ON HEAT IRRADIATION:

ACCORDING TO STEFAN-BOLTZMANN LAW, THE AMOUNT RADIATION FROM A BLACK BODY TO ANOTHER BLACK BODY IS EQUAL TO THE FOURTH POWER OF THE DIFFRENCE IN TEMPERATURE TIMES A CONSTANT.

Radiation = a (T4 – t04)
WHERE a = 5.67032 x 10-8 W/m2K4

The radiation per square meter per second from a black body (which is a perfect radiator) at thermodynamic temperature T to surrounding at thermodynamic temperature t0, = å T4 – to4

Research was made but the author failed to obtain the prevailing temperature in outer space so that radiation for a black body would be determined. Instead, the amount of radiation of the earth as a black body was obtained from the Microsoft Corporation. It is recorded at 240 Watts or 4,960,765.55 Cal./M2/day. Backward computation revealed that the prevailing temperature in outer space is -16.33 0C.

T1 = - 15.0 0C = 258.0 0K
t0 = - 154.35 0C = 118.65 0K
Radiation = å T4 – to4 = 5.67032 x 10-8 W/m2K4 (258.0 0K)4 – (241.8 0K)4

T14 = (2580K)4 = 4,430,766,095.00
t04 = (118.650K)4 = 198,200,737.
= 4,232,565,357.86
T14 = (293.00)4 = 7,370,050,801.00
to = (236.67)4 = 3.137,485443.14 4,232,565,357.86
Td = 236.67 – 273.00
= -16.330C
The Outer Space temperature is therefore at -16.330C
= 5.67032 x 10-8 (4,232,565,357.86
= 240.00000000 W/4.18 Joules /Cal/M2
= 57.41626794 Calories/ M2/Sec
= 4,960,765.55. Calories/M2/day
240 Watts = 240/4.18 = 57.41626794 Cal/sec
= 4,960,765.55 Cal/M2/Day
IN DEPTH ANALYSIS OF HEAT EMITTED BY EARTH AS BLACK BODY INTO OUTER SPACE.

Many scientific sources specified that the amount of heat emitted by earth into outer space is 240 watts or 4,960,765 55 Cal/M2/day but it must be understood that this figure applies only to heat emission near the equator. In-depth analysis and common sense would tell us that the emission at the North Pole and South Pole if there is any would be much lesser than what is emitted near the equator. This observation is supported by the application of Stefan-Boltzmann Law and mathematical equation.

In view of the fact that the 240 watts emission has the reference temperature of -16.33 0C for the outer space, it must be understood that a large portion of the ice cap in the North Pole and South Pole which temperature is way below -16.33 0C does not emit any amount of heat. Since that the amount of heat emission is relative to the fourth power of the prevailing temperatures in a pair reference point, the existing and prevailing temperature in each of the particular spot on earth has its proportionate amount of emission into outer space.


THE EARTH AS BLACK BODY = 240 WATTS/M2
= 4,960,765.55 Cal./M2/day

SOLAR RADIATION = 1,267 WATTS/M2.
= 402.489547 BTU/FT2/HR
= 1,091,196.17 CAL./M2/HR

TOTAL HEAT EMITTED INTO OUTER SPACE:

Calories = 4,960,765.55 Cal./M2/day (400,167,499.248166 KM2 (106) (0.3183098861)
= 631,888,778,442,487,659,789.804181 Cal

Km3 OF GASOLINE HEAT EQUIVALENT:

631,888,778,442,487,659,789.804181 Cal
Km3 = (--------------------------------------------) = 74.19110748 Km3
8,517,042 X 1012

74.19110748 Km3 of Gasoline/day
PERCENT EMISSION = ----------------------------------------- (100)
360.333842 Km3 of Gasoline/day
= 20.589547% OF TOTAL HEAT FROM THE SUN

3.11.2 HEAT LOSSES ATTRIBUTED TO OZONE LAYER AS HEAT PROTECTIVE DIAPHRAGM

Scientists claimed that the Ultra-violet component of solar radiation amounts to approximately nine percent but only one percent reached the ground because the ozone found at the Ozone Layer located at 40 Km to 50 Km altitude blocks the ozone. The intensity of solar radiation as it enters at the upper layer of the Atmosphere is measured at 1,267 Watts/M2 but only 1165.64 Watts reach the ground surface.

Solar Intensity at ground surface = 1,267 Watts/M2 (0.92)
= 1,165.64 Watts/M2.

1,165.64 Watts Joules Calories 3,600 Sec
Calories = (------------------) (---------) (-----------) (------------)
M2 Sec. 4.18 J Hr

= 1,003,900.478469 Cal./Hr

1,003,900.478469 Cal/M2 BTU
BTU = (------------------------------) (-------------)
Hr 252 Cal
= 3,983.732057 BTU/M2/Hr

3,983.732057 BTU/M2 778 Ft-Lb
Foot Pounds = (--------------------------) (----------------)
Hr BTU
= 3099343.540670 Ft-Lb/M2/Hr

3,099.343.540670 Ft-Lb/M2 Sec Hr
Horse Power = (--------------------------------) (---------------) (---------------)
Hr 550 Ft-Lb 3,600 Sec
= 1.565325 HP/M2

3.11.3 HEAT LOSSES ATTRIBUTED TO WATER EVAPORATION
On earth surfaces that is being covered by water such as lakes, rivers and Ocean, the average rate of evaporation near the equator is 4.0 millimeters per day or 4 liters/M2/day. The rate of evaporation which is dependent upon atmospheric temperature and humidity decreases from the equator down to zero at the North Pole and the South Pole. The average all over the earth is 2.3874 millimeters per day with about 2.0293 mm during daytime and 0.3581 mm during nighttime.

SURFACE PERCENT AREA (Km2) RATE EVAPORATION
Ocean 53.57% 214,369,729.35 Km2 4.0000 mm 857.478917 Km3
Island Exc. Des. 23.93% 95,760,082.57 Km2 1.0000 mm 95.760083 Km3
Iceberg 15.00% 60,025,124.89 Km2 0.0010 mm 0.060251 Km3
Dessert 7.50% 30,012,562.44 Km2 0.0020 mm 0.060251 Km3
Total 100.00% 400,167,499.25 Km2 2.3874 mm 955.358997 Km3

The above figure for total evaporation could only be true if the intensity of solar radiation is the same as those areas near the equator. In view of the fact that as we move from the equator towards either the North Pole or the South Pole, the intensity of light gradually reduces and eventually come to zero and even below zero degrees where evaporation is also zero. To rectify this figure, it needs to be multiplied by a factor that should take cares the variation of solar radiation intensity. This factor is equal to the reciprocal of “Pi” or 1/3.14592654 or 0.3183098861.

Daily Evaporation = 955.358997 Km3 (0.3183098861)
= 304.100214 Km3

Unlike evaporation by boiling water that requires 1,040 Calories per gram at the equator, the process of evaporation by humidification of air requires but approximately 650 Calories per Gram. By computation, the amount of heat that is lost to evaporation of water in the ocean together the other portions of the earth would be 2,600,000 Cal/M2/day at the surface of the ocean at the equator and 1,551,810 Cal/M2/day for global average at ocean covered areas.. The total amount of heat that is lost to evaporation in the global scale is equivalent to the heat that can be produced if 72.910682 Km3 of gasoline is burned every day.

650 Cal 1,000 Gm 1012 Kg
HEAT = (-----------) (--------------) (-----------) (304.10021356 Km3)
Gm Kg Km3

= 197,665,138,814,015,064,856.973020 Cal.

EQUIVALENT GASOLINE HEAT CONTENT :

197,665,138,814,015,064,856.973020 Cal.
Km3 = -------------------------------------------------
8,517,042 Cal/Li x 1012

= 23.208191 Km3


EVAPORATION IN THE OCEAN
DAYTIME = (214,369, 729.35 Km2) (106) (4.0 Liters) (0.3183)
= 272.9442 Km3.

EVAPORATION INLAND
DAYTIME = (95,760,082.57 Km2) (106) (1.0 Liter) (0.3183)
= 30.4804 Km3.

Total Evaporation = 302.4246 Km3./day

Heat expended = (302.4246 Km3./day) (109 M3/Km3) (1000)(650,000 Cal/Liter)
= 197,226,012,283,320,150,000 Cal
= 1.97226012283320150000 x 1020 Cal

197226012283320150000 Cal
Km3 Gasoline Equivalent = -------------------------------- = 23.15663 Km3
8,517043 (1012)

The amount of water evaporation in-land may be approximated at the average of 1.0 millimeter or 1.0 Liter/M2/day which is equivalent to 650,000 Cal/M2/day. It must be noted that there is practically no evaporation that takes place in the ice lands which occupies about 15.0% of the total surface of the earth and likewise no evaporation in the deserts which comprise about 7.5% of the total surface of the earth. The amount of water that the atmosphere holds may be computed from the average humidity of 2.5 % of the atmospheric layer existing within the altitude of 4,000 meters. The total amount of water in the atmosphere is 41,400 Km3.

A = (6,367.5305 KM) 2 (3.141592654)2
= 400,167,499.2481663018 Km2
= 4.0016749925 x 108 Km2

TTOTAL WEIGHT OF THE ATMOSPHERE:
Atmosphere = 10,350,000 Tons/Km2 (400,167,499.2481663018 Km2)
= 4,141,733,617,218,521.22363 Billion Tons
= 4.1417336172 x 1015 Tons
Equivalent Weight = 4,141,733.6172 Km3 of water.

In view of the fact that the layer of the atmosphere from the ground to 4,000 meter in elevation which contains water vapor is 62.0569% of the weight of the atmosphere, the to
total water it holds computed at 2.5% humidity would be 68,832.1657 gegatons.

A Summary of World Weather Records
RainfallHighest amount of rainfall in the northern hemisphere in 24 hours Paishih, Taiwan 124 cm/49 in.
Highest amount of rainfall in 24 hours (not induced by the presence of mountains) Dharampuri, India 99 cm/39 in.
Highest amount of rainfall in 24 hours Cilaos, Réunion 188 cm/74 in.
Highest amount of rainfall over 5 days Cilaos, Réunion 386 cm/152 in.
Highest amount of rainfall in 12 hours Belouve, Réunion 135 cm/53 in.
Highest yearly average rainfall in Africa Debundscha, Cameroon 1,029 cm/405 in. (with an average
variability of 191 cm/75 in.)
Lowest yearly average rainfall in Africa Wadt Halfa, Sudan 3 mm/0.1 in.
Lowest yearly average rainfall in Asia Aden, South Yemen 5 cm/1.8 in.
Highest yearly average rainfall in Europe Crkvine, Serbia and Montenegro 465 cm/183 in.
Lowest yearly average rainfall in Europe Astrakhan, Russia 16 cm/6.4 in.
Highest amount of rainfall in 20 minutes Curtea-de-Arge, Romania 21 cm/8.1 in.
Highest amount of rainfall in Australia in 24 hours Crohamhurst, Queensland 91 cm/36 in.
Highest yearly average rainfall in Australia Tully, Queensland 455 cm/179 in.

Lowest yearly average rainfall in Australia
Mulka, SA
10 cm/4.1 in.
Highest yearly average rainfall in South America Quibdo, Colombia 899 cm/354 in.



Lowest yearly average rainfall in South America
Arica, Chile
0.7 mm/0.03 in.
Highest yearly average rainfall in North America Henderson Lake, British Columbia, Canada 665 cm/262 in.
Lowest yearly average rainfall in North America Bataques, Mexico 3 cm/1.2 in.
Highest yearly number of days of rainfall Bahia Felix, Chile 325 days
Longest period without rainfall Arica, Chile 14 years
Highest yearly average period of thunderstorms Kampala, Uganda 242 days
Highest sustained yearly average period of thunderstorms Bogor, Indonesia 322 days per year from 1916 to 1919













Longest snowfall














Bessans, France





PAGE 110





19 hours with 173 cm/68 in. of snow

Source: National Weather Service, National Oceanic and Atmospheric Administration

Highest temperature ever recorded in the world Al'Aziziyah, Libya 58° C/136° F


Lowest temperature ever recorded in the world

Vostok, Antarctica
-88° C/-127° F
Highest temperature in Antarctica N/A near 16° C/60° F
Lowest temperature in Antarctica South Pole -77° C/-107° F
Lowest temperature in Africa Ifrane, Morocco -24° C/-11° F
Highest temperature in Australia Cloncurry, Queensland 53° C/128° F
Lowest temperature in Australia Charlotte Press -22° C/-8° F
Lowest temperature in Greenland Northice -66° C/-87° F
Highest temperature in Europe Seville, Spain 50° C/122° F

Lowest temperature in Europe
t 'Shchugor, Russia
-55° C/-67° F
Lowest temperature in northern hemisphere Oimekon, Russia; Verkhoyansk, Russia -68° C/-90° F
Highest temperature in western hemisphere Death Valley (CA) 57° C/134° F

Lowest temperature in North America (excluding Greenland)
Snag, Yukon Territory, Canada
-63° C/-81° F
Highest yearly average temperature in world Dallol, Ethiopia 34° C/94° F
Highest temperature in South America Rivadavia, Argentina 49° C/120° F
Lowest temperature in South America Sarmiento, Argentina -33° C/-27° F
Highest temperature in Asia Tirat Tsvi, Israel 54° C/129° F
Highest yearly average temperature range Eastern Sayan Region, Russia through 63° C/146° F
Highest average temperature sustained over a long period Marble Head, Australia 38° C/100° F for 162 consecutive days




Fastest temperature rise in short period Edinburgh, Scotland through -6° C/21° F




Wind
Record Location Details
Highest peak wind Thule Air Base, Greenland 333 kph/207 mph
Highest average wind speed in 24 hours Port Martin, Antarctica 174 kph/108 mph
Highest peak wind gust Mt. Washington (NH) 372 kph/231 mph
Highest monthly average wind speed Port Martin, Antarctica 105 kph/65 mph
Source: National Weather Service, National Oceanic and Atmospheric Administration

3.11.4 HEAT LOSSES ATTRIBUTED TO PRECIPITATION OF WATER INTO CLOUDS AND RAINFALL

In evaporation the molecule of water absorbs heat for its transformation into vapor. In precipitation the molecule of water in vapor state must shed off its heat content to gain transformation into liquid state. This happens when the upward current of air brings along with it the water molecule that upon reaching the lesser density of atmosphere eventually expands cools and precipitates. In this process 650 Calories per Gram must be mechanically lost.

Since that whatever is evaporated must be condensed, 2,600,000 Cal/M2/day is being lost every day in precipitation with reference to evaporation near the equator in order to put the humid content of atmosphere in equilibrium

EVAPORATION IN THE OCEAN
DAYTIME = (214,369, 729.35 Km2) (106) (4.0 Liters) (0.3183)
= 272.9442 Km3.

EVAPORATION INLAND
DAYTIME = (95,760,082.57 Km2) (106) (1.0 Liter) (0.3183)
= 30.4804 Km3.

Total Evaporation = 302.4246 Km3./day

Heat expended = (302.4246 Km3./day) (109 M3/Km3) (1000)(650,000 Cal/Liter)
= 197,226,012,283,320,150,000 Cal
= 1.97226012283320150000 x 1020 Cal

197226012283320150000 Cal
Km3 Gasoline Equivalent = -------------------------------- = 23.15663 Km3
8,517043 (1012)



3.11.5 HEAT LOSSES ATTRIBUTED TO ABSORPTION BY PLANTS

3.11.5.1 HEAT LOSSES ATTRIBUTED TO INLAND PLANT GROWTH.

In in-land areas, the average rate of growth of plantation including rice. corn, sugar cane, wheat, and wild grasses and shrubs amounts to an average of 15 Gm per day per square meter. On dry basis, this would sum up to approximately to 6.0 Gm/M2/day. Combustion of organic matter that is similar to that of fire woods showed that for every kilogram of fire woods that is burned, 5,407,831 Calories is produced. Therefore 6.0 Gm of dried plant materials would produce 32,447 Calories of heat. Since that the heat of combustion is equal to the heat of formation, the plants take in 32,447 Cal./M2/day.

It would take approximately four hours and for the solar radiation to change the temperature from 200C to 300C of the ground atmosphere along the equator or an average change of 6.00C for the first 3,000 M thick lower atmosphere. When air is heated, it expands and when expanded becomes lighter. Being lighter, it is buoyed upward or displaced by cooler air. In the morning, the land is cooler than the sea and therefore making wind blow from the sea towards the land producing the sea breeze. In the afternoon, the land becomes warmer than the sea and therefore making the wind blow from the sea toward the land producing the land breeze.

Inside a thick forest, the temperature is much lower even during daytime and the difference with that of nighttime is barely 3.0 0C. In the Philippines, the prevailing temperature inside a thick forest is about 18 0C while immediately outside it is 30.0 0C. This indicates how important the presence of rainforest in Global Warming is.

For the rest of the day during daytime or sunny day, absorption of heat by the atmosphere undergoes a continuous process and at the same time expansion of atmospheric material also undergoes a continuous process. As the heated or warmed air rises or pushed upward it ultimately reach certain elevation where the atmospheric pressure is lower and allowing it to expand further. In the process of expansion, there is a corresponding loss of heat energy that is equivalent to the acting pressure multiplied by the added volume of expansion.

When the air becomes cooler and denser or heavier, it moves downward toward the ground by the gravitational pull of the earth to displace the warmer air. The atmosphere therefore serves as the temperature regulator of the environment. If air is contained within a transparent and air tight glass container and allowed to be exposed to sunlight, the temperature and pressure of air within the glass will rise to a point of possibly breaking the glass container.

General Sherman Tree
This giant sequoia, known as the General Sherman Tree, reigns as the world's largest living tree, rising to a height of 84 m (275 ft).
Ron Sanford/ALLSTOCK, INC.
Microsoft ® Encarta ® 2007. © 1993-2006Microsoft Corporation. All rights reserved.


The rate of growth of plants differs in wide range in vegetative cover. The Sequoia tree in California which is the biggest tree in the world is 8.0 Meters in diameter and its height is measured at 84 meters. It must have a canopy diameter of at least 40 meters or an area of 1,256 M2. It is estimated to have lived for 1,600 years. Its body to include roots and leaves, which is estimated at 1,800 M3 would reveal that its rate of growth is 1,125 kg per year or 3.08 kg/day. If it grows at 2.45 Gm/M2/day for it’s body and three times a much for its leaves annually, its actual rate of growth is 9.81 Gm/M2/day. This shows that when the big trees reaches the age of more than a thousand years its rate of growth becomes much lower compared trees in their prime ages.

Normally, the growth of timber trees in side the rainforest is at the average of 35 grams per square meter per day but in some special cases the rate of growth ranges from 40 grams to 50 grams per day per square meter. In fact there was a certain experiment undertaken by a scientist whereby the quantity of soil was measured and placed in a confined concrete bed before planting and was measured after the tree grew to a total of about 80.45 kg to include leaves, roots and everything. In that experiment he discovered that the reduction in the amount of soil that goes to the growth of the trees is 60.5 grams.

Since it was not made clear whether measurements were quantified in dry basis or not we cannot ascertained with what the figures are revealed. However, one kilogram of earth material will produce 800 kilograms of wood on dry basis is the nearest approximation.

In the absence of the proximate analysis of the wooden materials from different species, we left with the ashes from burned wooden materials as the basis. On the average, the ashes represent about 2.0% to 3.5% of the wooden materials that is burned.


Composition of Wood Ash
Compounds Mol. Wt
Potassium Carbonate “K2CO3” 138
Sodium Carbonate “Na2CO3” 96
Phosphorus Pentoxide “P2O5” 142
Magnesium Oxide “MgO” 40
Magnesium Carbonate “MgCO3” 84
Calcium Oxide “CaO” 56
Calcium Carbonate “CaCO3” 100

The biggest tree this author ever saw in the Philippines is an acacia tree at the Cemetery of Dinalupihan, Bataan. Its diameter is approximately 4.0 meters and the diameter of its canopy at 80 meters and its height is approximately 40 meters.

In the case of farm lands planted with rice, the recent variety being planted can grow, mature and harvested in 120 days cycle. The yield per harvest is estimated at 150 sacks of 50 kilogram for a total of 7,500 kilograms on a hectare basis. If the weight of leaves, roots and stalks is placed at seven times the weight of the grains, the total vegetative growth would be 60,000 kilograms or 6.00 kilograms per square meter in 120 days or 50 grams per square meter per day. The sequestration of CO2 ability of rice plantation is about five times more than the forest but the problem is that it returns almost everything it sequestered back into the atmosphere within the same year. The rainforest stores the CO2 it sequestered for hundreds if not thousand of years.

On the global scale, the average vegetative growth is about 15 grams per square meter, or 10.5 grams/M2 on dry basis. The total amount of CO2 being sequestered from 95,768,085.6807 Km2 that constitutes the land portion of the earth is 1,005,564,899.65 tons/day which is only about 1.883642% greater than that of the ocean.

NATURE PERCENT AREA Gm/M2/Day GEGATONS
Forested 6.50 % 6,240,000 Km2 10.50 0.065520
Orchard Plantation 8.50% 8,160,000 Km2 3.75 0.030600
Shrubs & Bushes 30.750% 29,520,000 Km2 2.55 0.077760
Grassland/Grazing 32.25% 30,960,000 Km2 2.25 0.069660
Grain Production 6.50% 6,240,000 Km2 15.00 0.009360
Vegetable Garden 2.25% 2,160,000 Km2 4.50 0.009720
Root Crops 1.75% 1,680,000 Km2 1.95 0.003276
River and Lakes 2.50% 2,400,000 Km2 -------- -----------
Urban Housing&Roads 5.750% 5,520,000 Km2 0.10 0.000552
Open Space/Arid 1.250% 1,200,000 Km2 0.50 0.000600
Total 100.000% 96,000,000 Km2 0.351288

Every day, approximately 0.351288 Gegaton of the body of plants including their roots and leaves is added into the bulk of vegetative growth all over the world. This is equivalent to 0.1053864 gegatons on dry basis. At the conversion rate of 1.54385965 Kg of CO2 per Kg of wood on dry basis, all the existing land base plants is able to sequester as much as 0.16270181061876 gegatons of CO2 everyday or 59.42683632850209 gegatons per year.
This annual CO2 sequestration capacity of existing plants is equivalent to 14.348299968 parts per million “ppm” of the Atmosphere on a yearly basis. This means that every year, 3.6885% of the total available CO2 in the Atmosphere is sequestered by Land Plants alone and plants such as planktons and algae in the ocean is not yet accounted for.

The problem however is that 80% of the CO2 that is sequestered gets back into the atmosphere during the same year because farmers either burn agricultural wastes or allow it to disintegrate into CO2 and Methane gas. The total amount of heat that is absorbed by plants in relation to its process of growth is similar to the quantities of heat that can be realized if a given dry wood is burned. The heat content of firewood is 5,407,831.5789 Cal/Kg.

HEAT ASORBED BY LAND BASED PLANTS:
HEAT = 5,407,831.5789 Cal/Kg (1.074518Gegatons) (1012 Kg/Gegaton)
= 5,810,812,372,496,470,200 Calories
= 5.810812372496470200 x 1018 Calories

CUBIC KILOMETER OF GASOLINE EQUIVALENT;
5.810812372496470200 x 1018 Calories
Km3 = ---------------------------------------------------------
8,517,042 x 1012
= 0.682257 Km3 of Gasoline/day


3.11.5.2 HEAT LOSS ATTRIBUTED TO OCEAN PLANT GROWTH

Many scientists claimed that 90% of CO2 that is absorbed by plants in the process of photosynthesis come from the bodies of water such as pond, rivers, lakes, seas and oceans. However, the claim of 90% CO2 sequestration is just a wild guess which is not possible upon quantitative verification. They further claim that such magnitude of oxygen emission from the ocean is attributed to high density of growth of phytoplankton that serve as the major supply of food for zooplanktons, various types of fishes and even that of sharks and whales. Large sea mammals such sharks are equipped with gills that filter out planktons that thrive at the surface or upper layer of sea water for their food.

On the average, there are as many as five hundred million planktons in a liter of water that comes from the sea. The zooplanktons likewise feed on phytoplankton and they themselves are eaten by small as well as large fishes and mammals.

If 90% of total photosynthesis takes place at roughly 70% of the total surface area of the earth holds true, it follows that photosynthesis that takes place on the ocean is 3.8575 times greater than what it take on the land for the same unit area... Except for the growth of green algae and under water grasses, the ocean does not appear to produce much more oxygen than the land because planktons are almost microscopic in their sizes to escape notices.

If this is the case, the ocean is much more effective medium in sequestering carbon dioxide from the atmosphere and plankton growth and propagation must be much faster than the way corn and rice plantation grow.

Water Area 90%/70%
Ratio of Growth = ------------- = ------------ = 3.8575
Inland Area 10%/30%

Further scrutiny to verify the above claim that water in the ocean produces more photosynthetic outputs leads us to consider that CO2 in the atmosphere has a density of 389 ppm against the dissolved CO2 at 18 ppm somewhere near the surface in the hydrosphere. This likewise leads to scrutinize as well as to find out what is the mass of 500 million planktons present in every liter of sea water and to what depth of the ocean do planktons thrive.

At the solubility of 18 ppm by volume, there could be 18 Cm3 of CO2 in every cubic meter of sea water in the ocean. If plankton thrives up to the depth of for instance 60 meters, then there would be 1,080 Cm3 of CO2 available per square meter of the water surface. .At the mean pressure of 2.0 Atmospheres, there would be 4.021875 grams of available CO2 in every square meter up to the depth of 60 meter from the surface of the ocean.

We do not have the exact data on how fast is the rate by which CO2 is replenished once being consumed by Planktons but the 24 hour period would probably be safe. With this assumption, 4.021875 grams of available CO2 in every square meter up to the depth of 60 meters would be able to nourish or provide about 2.605508 grams additional weight of planktons on dry basis. The supply of additional 2.605508 grams of planktons on dry basis a day is equivalent to 6.51 gram on wet basis which can feed the equivalent body weight of 130.25 grams of fish in every square meter per day. This means that for every square kilometer in the ocean can take a load of 130.25 tons of fish that could feed on 6.51 tons of planktons that consume as much as 4.0225 tons of CO2.in every period of 24 hours for it maximum sequestration performance.

The total maximum amount of CO2 that the Planktons in the ocean can sequester is as follows:

4,021875 Gm 106 M2 Ton
Available CO2 = (----------------) (----------) (----------) (214,369,729.35 Km2)
M2 Km2 106 Gm

= 862,310,525.2605 Tons/day

This value is the maximum that the Ocean can perform but for actual figure, we may assume that only 2.5% of the amount is being sequestered by the Planktons and Green Algae in the Ocean.

Ocean Sequestering Performance = 862,310,525.2605 Tons/day (0.025)
= 21,557,763.13151 Tons/day
Ocean = 21,557,763.13151 Tons/day = 11.70%
Land = 162,701,810.61876 Tons/day = 88.30%
Total = 184,259, 573.75027Tons/day

Annual CO2 Sequestration = 184,259,573.75027 Tons (365.25 days)
= 67,300,809,312.2861175Tons
= 67.300809 Gegatons

67.300809312 Gegaton
Percent Sequestered Annually = -------------------------------------- (100)
1,552.5 Gegatons
= 4.3349% Annually

Therefore: The sequestering ability of the Ocean is 7.547 times smaller than that of the Land. If we could fertilize the Ocean with potassium using ammonium phosphate “(NH4)3PO4” we might maximize the Sequestering Performance of the Ocean.


The ability of the ocean to sequester 0.75 tons of CO2 in every square kilometer a day would amount to 184 million tons a day or 67.217 gegatons a year. The CO2 sequestering ability of the Land is therefore 5.84 times more than that of the ocean contrary to the claims of some scientists. . The total amount of CO2 by the combined sequestering capacity of land and water is therefore 1.2585 gegatons per day or 459.68904 gegatons per year. This is equivalent to almost 111 ppm or 29.59% of the total weight of the atmosphere.


Ocean 241,092,914.9470 Km2 60.24800 %
Green Land 88,001,075.8889 Km2 21.99106 %
Ice 60,025,124.8872 Km2 15.00000 %
Dessert 11,048,383.5250 Km2 2.76094 %
Total 400,167,499.2481 Km2 100.0000 %

Since that rice plantation grows at the rate 15.0 grams per M2 per day, plankton must be growing at 23.145 grams per day to maintain the ratio of 3.8575 times. To make this rate of growth possible, sea water must have the ability to replenish the store of dissolved CO2 every period of 18 hours.If the size of planktons averages at 10 microns in diameter, 500 million planktons in every liter of sea water would yield 3.927 Kg per M2 of sea water.

Volume = (0.01 mm/2)2 (pi) (0.01) ( 500,000,000 ) = 392.7 mm3/liter
= 392.7 Gm/M3

Load per M2 = 392.7 Gm/M3 x 10 M3
= 3.927 kg/M2
23.145 Gm
Rate of increase in weight of plankton per day = ------------- = 0.589%/day
3,927 Gm

In the absence of sufficient research data, it appears in the above illustrations that the ocean indeed has the capacity to generate oxygen ions that is 3.875 times greater than that of the inland farm areas which essentially justifies the statements of other scientists.