Solution to Global Warming (Part 13)

3.11.5.3 SUMMATION OF CARBON DIOXIDE SEQUESTERING CAPACITY OF THE EARTH.

Inland Rate of Growth = 15.000 Gm/M2 (70%) = 10..5 Gm/M2
Ocean Rate of Growth = 23.145 Gm/M2 (70%) = 16.2 Gm/M2
Total = 18.0 Gm/M2
Average = 9.0 Gm/M2

342 384 528 198
C12H22O11 + 12 O2 = 12 CO2 + 11 H2O

C12 = 144 528
H22 = 22 CO2 = --------- (9.0 Gm) = 1.3895 Gm
O11 = 176 342
Total 342

The above computation shows that 9.0 Gram of Cellulose that is produced out of plant growth would be able to sequester 13.895 Gm of CO2/M2 of the earth surface. The total amount of carbon dioxide in the atmosphere is 0.377775 Gm/Cm2 or 3.77775 Kg/M2 of the surface of the earth. This means that one part for every 272 parts or 0.3678% of CO2 undergoes chemical exchange every day.

The observation that carbon dioxide increases by 1.5 ppm per year in recent years could be understood that 15.525 grams of CO2 is added per M2/year. On a daily basis, what is added is 42.53 mlligram/day/M2.
.
HEAT ABSORPTION
144 14,000 BTU 2.2 Lb 252 Cal.
C = (--------) (---------------) (------------) (---------) = 3,268,042.105263 Calories
342 Lb Kg BTU

22 60,000 BTU 2.2 Lb 252 Cal.
H = (--------) (---------------) (------------) (---------) = 2,139,789.47368 Calories
342 Lb Kg BTU

Total 5,407,831.578943 Calories/Kg

Heat = (862,310,525.2605 Tons/day) (1,000 Kg/Ton) (5,407,831.578943 Cal/Kg)
= 4,663,230,089,358,657,401.3896515 Cal,

4663230089358657401.3896515
Km3 = ---------------------------------------------- = 0.54751756 Km3 of Gasoline
8,517,042 (1012)
3.12.0 HEAT BALANCE AND DISTRIBUTION

The total amount of radiant heat the earth receives from the sun for ten hours a day at 402.48955 BTU/Ft2/hr is 4,024.8955 Btu/ft2day or 10,911,961.70 Calories/M2 per day. It follows therefore that the heat absorbed by the atmosphere is 45.59% of the total radiant energy that the earth receives from the sun.

It must be borne in mind that all the heat absorbed by the atmosphere is not directly taken from the incoming radiant energy of the sun because part of it is gained from the ground level through reflection. Reflection of radiant energy varies to a wide range depending upon the nature of the surface on the ground level. In the absence of hard facts and figures, we may however have a random guess that 37.65% of what reached the ground surface is reflected back and absorbed by the atmosphere. Pine Woods have been recorded to reflect 40.0% of the sunlight while the ice sheet or snow reflects 78.0 of the solar radiation.

Out of 10,911,961.70 Calories/M2 that the earth receives from the sun per day, 872,956.94 Cal./M2/day or 8, 0% is absorbed by the ozone layer located at about 50 kilometers in altitude. Out of the remaining 10,039,004.76 Cal./M2/day, about 1,363,431 Cal./M2/day (12.49%) is absorbed directly by the atmosphere and allowing 8,675,574 Cal./M2/day (79.50%) to reach the ground level. The total radiant energy that reached the ground directly from the sun in the Philippines amounts to 8,675,574 Cal./M2/day or 320 BTU/Hr. These are the available heat we used for drying palay, corn, fish etc.

The reason why the environment is maintained within a certain range of temperature over a period of time during summer or winter is because everything that is added during the day is being lost during the 24 hour period. Every day, 4,960,765.55 Cal./M2 (45.46%) is lost by being radiated back into the outer space which is inherent in earth’s characteristic as a black body.

SUMMARY OF HEAT INPUTS
2.2.1 SOLAR RADIATION 360.333842982Km3/day
2.2.2 COMBUSTION OF COAL 0.815738571 Km3/day
2.2.3 HEAT FROM THE INTERIOR OF THE EARTH 0.405944598 Km3/Day
2.2.4 HUMAN AND ANIMAL BODIES 0.018401107 Km3/day
2.2.5 COMBUSTION OF CRUDE OIL & DERIVATIVES 0.015768119 Km3/day
2.2.6 COMBUSTION OF FIREWOODS & CHARCOAL 0.014970056 Km3 /day
2.2.7 COMBUSTION OF NATURAL GAS 0.013033549 Km3/day
2.2.8 RUNNING ELECTRIC HEATING ELEMENTS 0.00535545 Km3/day
2.2.9 FERMENTATION OF ORGANIC COMPOUNDS 0.002607567 Km3/day
2.2.10 REACTIONS OF CHEMICAL COMPOUNDS 0.000117312 Km3/day
2.2.11 HEAT FROM FRICTIONS & MACHINERIES 0.0001125260 Km3/day
2.2.12 OPERATIONS OF GEOTHERMAL PLANTS 0.000011650 Km3/day
2.2.13 HEAT FROM LIGHTNING AND THUNDERSTORM 0.000004712 Km3/day
2.2.14 HEAT OF FORMATION OF WATER 0.000003367 Km3/day
2.2.15HEAT FROM EXPLOSIVES 0.0000000629Km3/day
2.2.16 FROM GLACIERS AND ACTIVE VOLCANOES 0.0000000110Km3/day
TOTAL 360.997218433 Km3/day



2.3.1 BLACK BODY RADIATION TO OUTER SPACE 74.19110748 Km3/day
2.3.2 HEAT LOSSES ATTRIBUTED TO OZONE LAYER 28.826707438 Km3/day
2.3.3 LOSSES ATTRIBUTED TO WATER EVAPORATION 23.20819100 Kim3/day
2.3.4 PRECIPITATION OF WATER INTO CLOUDS & RAINFALL 23.20819100 Kim3/day
2.3.5 ATTRIBUTED TO ABSORPTION BY PLANTS 0.68225700 Km3/day
2.3.6 ABSORPTION BY OCEAN PLANTS 0.54751756 Km3/day
2.3.7 EXPANSION OF ATMOSPHERE 210.333246955 Km3/day
TOTAL 360.997218433 Km3/day


3.12.1 HEAT LOSS IN EXPANSION OF THE ATMOSPHERE

Having 872,956.94 Cal./M2/day or 8,0% been absorbed by the ozone layer and 4,960,765.55 Cal./M2 (45.46%) having been lost into outer space, the rest in the amount of 5,078,239.21 Cal./M2/day (46.54%) is being lost as follows:

• Mechanical Energy loss due to air expansion1, 1,795,792 Cal/M2/day
• Precipitation of H2O vapor into clouds 2,600,000 Cal/M2/day
• Utilized in H2O evaporation 650,000 Cal/M2/day.
• Utilized in growth of plants 32,447Cal./M2/day.

COMPUTATION OF AVERAGE SOLAR RADIATION
Diameter of Earth: = 12,735.061 KM
FACIAL AREA OF EARTH = (6,367.5305)2 (3.141592654)
= 127,377,271.1235 KM2

RADIATED AREA OF EARTH = (6,367.5305)2 (3.141592654)2/2
= 200,083,7749.62 KM2

200,083,7749.62 KM2
RATIO = ----------------------------
127,377,271.1235 KM2

= 1.570796327

1,091,196.17 CAL./M2/HR
AVERAGE SOLAR RADIATION = -------------------------------
1.570796327

= 694,677.057263 CAL./M2/HR


RADIATION IN METRO MANILA TO OZONE LAYER
Ground Level Temperature at night = 25 0C
Temperature at Ozone Layer = 20 0C

Radiation = 5.67032 x 10-8 (298 0K)4 - (293 0K)4
= 5.67032 x 10-8 (7,886,150,416 – 7,370,050,801)
= 5.67032 x 10-8 (576,099,615)
= 29.264499 Cal/M2/Sec.
= 2,528,452.77 Cal/M2/day
122.3256 Watts

HEAT LOSS DUE TO EXPANSION OF GASES

Volume of 1.0 M2 and Length of 3 Km = 300 M3

6.0 0C
Volume of Expansion = -------- (3,000 M3) = 65.934066 M3.
273 0K

14.7 lb
Pressure = ----------- 0.700 (1,550.0031 In2 = 15,949.5319 lb
In2

Distance = 65.934066 M (3.28 ft/M)
= 216.2637 ft

Work = 15,949.5319 lb (216.2637 ft) = 3,449.305.3638 ft-lb
= 3,449.305.3638 ft-lb/778 ft-lb/Btu (252 Cal/Btu)
= 1,117,225.7219 Calories/M2/day

Heat absorbed by Plants: Ex. Rice Plantation:
Grain = 0.6 kg/M2/120 days = 0.005 kg/day
Stalk = 3.0 kg/M2/120 days = 0.025 kg/day
Total = 0.030 kg/day = 30 Gms/day/M2
Heat Absorbed =

The total amount of Calories needed to raise the temperature of

We, who are at the earth’s surface, move around the circumference of the earth at the velocity of 464 meters per second due to earth’s rotation around its axis. The oxygen together with the ozone molecule that is continuously formed at 50 km above the ground moves at the velocity of 467.64 meters per second around the rigid component of the earth.
Relative momentum between the two layers:
Momentum, F1 = m1v12
F2 = m2v22

Where m1 = m2, the ratio is expressed by the square of their velocities:

F2 m2v22 v22 (467.64)2
Ratio = ------- = (--------) = ------ = --------------
F1 m1v12 v12 (464.00)2

218,687.17
= -------------- = 1.0157511983
215,296.00

PAGE 125
3.13.1 DIAGRAM OF SOLAR RADIATION DISTRIBUTION IN-LAND


OUTER SPACE------------------------------------------------------------------ OUTER SPACE
+10,911,961.70 Calories/M2day - 4,960,765.55 Cal./M2/day
(Total Solar Radiation) Radiated to Outer Space

- 872,956.94 Cal./M2/day TEMPERATURE
(UV absorbed by Ozone layer) MAINTAINED AT 20.0 0C
50,000 M ALTITUDE -------------------------------------------------------50,000 M ALTITUDE

+ 2,172,438 Cal/M2/day + 5,872,901 Calories/M2/day
(Directly absorbed by Reflected from Ground)
Atmosphere from the Sun)

- 1,117,225.72 Calories/M2/day
(Heat loss in Air Expansion by 6.0 0C rise in temp.)

- 1,055,212.28 Calories/M2/day - 2,172,438 Calories/M2/day
(Heat loss in additional Air Expansion ) (Total Loss in Air Expansion)

3,000 M ALTITUDE----------------------------------------------------------------------3,000 M ALTITUDE
+ 4,975,110Cal/M2/day + 3,270,459.6 Cal/M2/day
(Residual heat of atmosphere) (Absorbed from Ground)
+ 755,655.55 Cal/M2/day
(Reflected from Ground)

+ 1,243,778 Cal/M2/day - 4,960,765.55 Cal./M2/day
(Reflected back to ground) (Radiated to Outer Space)
¬-650,000 Cal/M2/day -120,000 Cal/M2/day - 3,270,459.6 Cal/M2/day
(H20 Vapor transformation) Plant Growth Absorption (Reflected to Atmosphere)
GROUND LEVEL --------------------------------------------------------------- GROUND LEVEL
+ 8,686,107.65 Calories/M2/day_ + 1,243,778 Cal.M2/day + 9, 929,885.15Cal.M2/day
(Absorbed by Ground Surface) (Heat Reflected from atmosphere) (Total GroundAbsorption
4,954,775.15
650,000 + 120,000 + 3,270,459.6 = 4,040,459.6


3.13.2 PROPOSED DIAGRAM OF HEAT DISTRIBUTION AT THE OCEAN NEAR THE EQUATOR.
OUTER SPACE------------------------------------------------------------------ OUTER SPACE
+10,911,961.70 Calories/M2day - 4,960,765.55 Cal./M2/day
(Total Solar Radiation) Radiated to Outer Space

- 982,076.55 Cal./M2/day
(UV absorbed by Ozone layer)
5,872,901 Calories/M2/day (Reflected from Ground)

+ 2,172,438 Cal/M2/day (Directly absorbed by 1,558,894 Cal/M2/day
Atmosphere from the Sun) (Reflected by Atmosphere
To Outer Space)
- 1,117,225.72 Cal./M2/day
(Heat loss in Air Expansion by 6.0 0C rise in temp.)
- 1,055,212.28 Cal./M2/day - 2,172,438 Cal./M2/day
(Heat loss in additional Air Expansion ) (Total Loss in Air Expansion)

3,000 M ALTITUDE---------------------------------------------------3,000 M ALTITUDE
+ 4,975,110Cal/M2/day - 1,243,778 Cal.M2/day + 3,270,459.6 Cal/M2/day
( Heat Absorbed by atmosphere) (Heat Reflected by Atm. (Absorbed from Water Surface)
+ 2,802,672 Cal/M2/day to Water Surface)
(Reflected from Water Surface)

+ 1,243,778 Cal/M2/day - 196,681.6 Cal./M2/day - 4,960,765.55 Cal./M2/day
(Reflected back (Cloud precipitation) (Radiated to Outer Space)
to Water Surface
¬-2,600,000 Cal/M2/day -10,000 Cal/M2/day - 3,270,459.6 Cal/M2/day
(H20 to Vapor transformation) Plant Growth Absorption (Absorbed by Atmosphere)
SEA WATER LEVEL -------------------------------------------------- SEA WATER LEVEL
+ 7,757,447.15 Calories/M2/day_ + 1,243,778 Cal.M2/day + 9,001,225.15 Cal.M2/day
(Absorbed by Water Surface) (Heat Reflected from atmosphere) (Total Water Surface Heat)
Absorption 4,960,765.55 Cal./M2/day

IV. NATURE AND ROLE OF OZONE LAYER

4.1 DEFINITION OF OZONE

Ozone is a substance formed by three atoms of oxygen contrary to two atoms that form a diatomic molecule of oxygen. It exists as a faint blue gaseous substance having a molecular weight of 48 which is 1.5 times that of oxygen. It has the density of 1.658 that of the air which boils at -110 0C (63 0K) and it decomposes at 270 0C. It is soluble in water and more reactive than oxygen gas. It is produced in nature by silent electrical discharge and commercially manufactured for industrial use as bactericide, oxidizing agent, and for bleaching fats, oils, textiles and sugar solutions.

Oxozone is an industrial product resembling a filter paper that is impregnated with Potassium Iodide “KI2.” and starch or indigo solution which turns blue upon exposure to ozone. It is used as Ozone detector.

At high concentration, Ozone is a pollutant that creates photochemical smog. The World Health Organization has proposed an air quality air standard of 60 Parts Per Billion “ppb” at hourly average while Environment Protection Administration “EPA” sets the standard at 120 ppb for human health and 80 ppb for welfare (visibility and avoidance to plant damage. Ozone that exist in the Ozone Layer Traps 8.0% of Solar Radiation being directed towards the earth’s surface.

Shield Ozone or Ozone Layer refers to highly concentrated layer of Ozone that exists at the altitude from 40 kilometers to 50 kilometers. The concentration of Ozone is at least 12 Parts Per Million “ppm” along with undetermined amount of ions of oxygen as well as nitrogen. The density of the Ozone is surprisingly much higher than the density of matters/gases below it as well as the density of materials/gases above it.

The idea of having a dense particle of gaseous materials being suspended in space at such altitude without exerting weight upon the thin air below it was at first a big challenge that this author encountered in his research. He made the attempt to quantify the effect of distance upon gravitation and also the effect of higher velocity and momentum but was disappointed to discover that these effects are minimal with respect to reduction in the quantity of gravitational force.

This author eventually stumbled upon an article which identified oxygen as a paramagnetic material which magnetic strength in inversely proportional with temperature. In a certain experiment, oxygen molecules were found to be suspended in space where there is strong magnetic flux. The interior of the earth is made up of liquefied iron that rotates along with the rotation of the earth. This circular movement of ions within the interior of the earth produces a magnetic flux that extend and establish a close loop with one portion in the interior of the earth and another portion outside the earth some 40 Km to 50 Km in altitude. The flux at the outer portion traps the materials in the Ozone Layer.

Paramagnetic behavior results when the applied magnetic field lines up all the existing magnetic moments of the individual atoms or molecules that make up the material. This results in an overall magnetic moment that adds to the magnetic field. Paramagnetic materials usually contain transition metals or rare earth elements that possess unpaired electrons. Paramagnetism in nonmetallic substances is usually characterized by temperature dependence; that is, the size of an induced magnetic moment varies inversely to the temperature. This is a result of the increasing difficulty of ordering the magnetic moments of the individual atoms along the direction of the magnetic field as the temperature is raised.

4.2 EFFECTS OF ALTITUDE ON WEIGHTS OF PARTICLES

4.2.1 COMPARATIVE ANALYSIS OF PROTON

CENTRIPETAL FORCE

m v2
F = --------
r

Where: F = the centripetal force in lb
m = mass of a molecule of Proton in lb
v = velocity of oxygen molecule in feet/second around the earth’s circumference. (467.6 M/sec)
r = radius of circular motion which is earth’s radius in ft (21,049,500)

m v2
F = ------
r

Proton Mass = 1.67264 x 10-27 Gm
= 0.003679808 x 10-27 Lb
Velocity = 467.64 M/sec
= 1,533.73 ft/sec

Earth’s Radius = 6,367,530,5 M + 50,000 M
= 20,885,500.04 ft + 164,000 ft
= 21,049,500 ft

m v2 0.003679808 X 10-27 Lb (1,533.73 ft/sec)2
F = -------- = ----------------------------------- = 0.000411226601 x 10-27 lb
r 21,049,500 ft