Solution to Global Warming (Part 7)

ARGON
Discovered in 1894 by British Chemist William Ramsay & Baron William Strutt Rayleigh in 1894. It has been commercially manufactured and used in Welding; Argon derives its name from Greek word “Argos” meaning inert. It is also used in advertising lights which exhibit a blue color.

KRYPTON
Discovered by Sir William Ramsay. It is one of the elements among inert gases that forms compound with fluorine. Krypton is derived from the Greek word kryptos “hidden”. Krypton is present in the atmosphere to the extent of 1 ppm and its liquid form has a specific gravity of 2.41. It is used alone or with Argon and Neon to exhibit a bright orange red color in incandescent bulbs. It is used in lighting air fields because red light is more visible in long distances and penetrates fogs and haze to a greater extent.

XENON
Discovered by British Chemists William Ramsay & Morris Travers in 1898; It was found in 1962 to form compound with fluorine and oxygen; It is a colorless, tasteless element that is insoluble in water; It is used principally in such lighting devices as high speed photographic tubes. Its density is 5.9 Gm/liter at 00C and 1 atmosphere pressure.

RADON Discovered by Madame Marie Curie in 1898. Radon 222 which is isotopes of Radon 220 is the decaying product of Radium which highly cancerous. There portions of the earth which emit Radon that made several thousand afflicted and died of Cancer.

2.6 GREENHOUSE GASES

3.3.1 The Three Types Of Greenhouse Gases

3.3.2 Effects Of Greenhouse Gases

POLLUTANT ppm of Atm Km3 of H2O Eq.
Carbon Dioxide “CO2” 375.00000000 1,553.14987500 Km3
Methane “CH4” 10.60000000 43.90237642 Km3
Ozone “O3” 2.00000000 8.31147898 Km3
Nitrogen Oxide “N2O” 0.60000000 2.48504018 Km3
Freon 11 “CFCl3” 0.00173913 0.00720301 Km3
Freon 12 “CF2Cl2” 0.00144927 0.00600249 Km3
CHF2Cl 0.00130435 0.00540227 Km3
CF3CH2F 0.00115942 0.00480201 Km3
Per-fluorocarbon “PFC” 0.00038673 0.00160173 Km3
Sulfur Hexafluoride”SF6” 0.00024155 0.00100044 Km3


PERCENT INCREASE SINCE 1750 TO 2000

GREENHOUSE GASES INCREASE 1750 2000
Carbon Dioxide “CO2” 31.0% 280.000000 ppm 265.000000 ppm
Methane “CH4” 151.0% 0.701986 ppm 1.761986 ppm
Nitrogen Oxide “N2O” 17.0% 0.270588 ppm 0.316588 ppm


THE COMPONENT OF THE ATMOSPHERE

What is the total weight of the Atmosphere and how much Carbon Dioxide does it hold? To find out this quantity, we have to start from the diameter of the earth and come up with the total earth’s surface area.

Equatorial Diameter 12,756.300 Km
Polar Diameter 12,713.822 Km
Mean Diameter of Earth 12,735.061 Km
Mean Radius 6,367.5305 Km

Area of Earth’s Surface: A = r2 Pi2

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

TOTAL WEIGHT OF THE ATMOSPHERE:

Atmosphere = 10,350,000 Tons/Km2 (400,167,499.2481663018 Km2)
= 4,141,733,617,218,521.22363 Tons
= 4.1417336172 x 1015 Tons
= 4,141,733,6172 Gegatons

We can have a better idea what the above figure is all about if we transform the total weight of the Atmosphere into the weight of one cubic kilometers of water as a unit of measurement. This unit is known as Gegaton.

4.14173361721852122363 x 1015 Tons
Atmosphere = ----------------------------------------------
1.0 X 109 Tons/Km3

= 4,141,733. 61721852122363 Km3 weight of water equivalent
= 4.1417336172 x 106 KM3 of water weight equivalent.

III. HEAT ENERGY EQUILIBRIUM

3.1 SOLAR RADIATION

3.1.1 QUANTITY OF SOLAR RADIATION

Facial Area of Earth: = 127,377,271.1235 Km2
Surface Area of Earth: = 400,167,499.2482 Km2

Radiation = 127,377,271.1235 Km2 (1 x 106 M2) (1,003,900.476 CalM2/Hr (24Hr/day)
= 3.068978474699104914864 x 1021 Calories

3.068978474699104914864 x 1021 Calories
Equivalent Gasoline = -----------------------------------------
8,517,042 Cal/L x 1012 L/Km3

= 360333842982000.66588255 Liters/day
= 360.333842 Km3 of Gasoline/day

“The total amount of Solar Radiation that the earth receives from the sun is equivalent to the heat that can be generated if and when 360 cubic kilometers of gasoline is burned every day.”

127,377,271.1235
Cal/M2 = ---------------------- ( 1,003,900.476)/Hr) = 319,551.4462137316 Cal/M2/Hr
400, 167,499.2481

3.2 OTHER NATURAL SOURCES OF HEAT
SOURCES OF GASOLINE
HEAT INPUTS HEAT EQUIVALENT
• SOLAR RADIATION 360.333842982 Km3/day.
• HEAT FROM THE INTERIOR OF THE EARTH 0.4059445980 Km3/day
• COMBUSTION OF AGRICULTURAL WASTE 0.0523035448 Km3/day
• HUMAN AND ANIMAL BODIES 0.018401107 Km3/day
• COMBUSTION OF CRUDE OIL & DERIVATIVES 0.015768119 Km3/day
• COMBUSTION OF FIREWOODS & CHARCOAL 0.014970056 Km3 /day
• COMBUSTION OF NATURAL GAS 0.013033549 Km3/day
• COMBUSTION OF COAL 0.009789455 Km3/day
• HEAT FROM LIGHT BULBS 0.006269445 Km3/day
• RUNNING ELECTRIC HEATING ELEMENTS 0.005355450 Km3/day
• FERMENTATION OF ORGANIC COMPOUNDS 0.002607567 Km3/day
• REACTIONS OF CHEMICAL COMPOUNDS 0.000117312 Km3/day
• HEAT FROM FRICTIONS & MACHINERIES 0.000112526 Km3/day
• OPERATIONS OF GEOTHERMAL PLANTS 0.000011650 Km3/day
• LIGHTNING & THUNDERSTORMS 0.000004712 Km3/day
• HEAT OF FORMATION OF WATER 0.000003367 Km3/day
• HEAT FROM EXPLOSIVES 0.000000063 Km3/day
• FROM GLACIERS AND ACTIVE VOLCANOES 0.000000011 Km3/day
TOTAL 360.997218433 Km3/day

TEMPORARY HEAT STORAGE DURING DAYTIME AND RELEASED DURING NIGHTTIME
• STORED HEAT FROM THE GROUND/WATER 50.615455093 Km3/day
• HEAT STORED IN THE ATMOPSERE 154.72542329 Km3/day
TOTAL 205.746822981 Km3/day





The Sun
The Sun outshines everything else in the sky. Without the Sun, there would be no life on Earth.
D. Boone /Corbis
Microsoft ® Encarta ® 2007. © 1993-2006 Microsoft Corporation. All rights reserved.




PAGE 60




3.3 SOLAR RADIATION

Solar radiation that reached the ground is 60.0% infrared, 1.0% ultraviolet and 39.0% visible radiant energy.
RADIANT RAYS INFRARED VISIBLE RAYS ULTRAVIOLET
Sunlight 60.0 % 39.0 % 1.0%
Incandescent Lamp 95.0 % 4.8 % 0.2%
Carbon Arc 80.0 % 15.0 % 5.0 %
Resistance Wire 99.0 % 0.5 % 0.5 %
The average solar heat radiated on the surface of the earth amounts to 402.48955 BTU/ft2/hr or 1,091,196.17 Calories/M2/Hr.This is also equal to 1.267 kilowatt/M2/Hr. About 9.0% of this magnitude is invisible radiation in the form of ultra-violet “UV”, about 60.0 % is infra red radiation and the rest is that of the visible light. Out of the 114.03 watts equivalent of UV light that the earth receives per M2/hr, only 12.67 watts/M2/Hr reached the ground surface because about 8.0% of the total UV radiation is blocked and being neutralized by the ozone layer located between the altitude of 40 Km and 50 Km from the ground surface.

By the action of 101.36 watts of ultra-violet radiation which comprises about 8.0% of the solar radiation, three oxygen molecules dissociate and recombine to form two molecules of ozone. This chemical reaction results to the reduction of the quantity of molecules by one third (33.33%) within a given space in the atmosphere and subsequently there is a need for other gases in the vicinity to expand. The expansion of atmospheric gases translate into mechanical loses of energy and eventually enhances the process of cooling. This could be the reason why the atmosphere immediately above the ozone layer gradually cools down from 200C at 50 Km altitude to -82.50C at 82.5 Km in altitude.

TRANSFORMATION OF OXYGEN TO OZONE: 3 O2 + UV = 2 O3
The constant formation of ozone between the altitudes of 40 km to 50 km creates a given density of ozone that absorbs much more heat which is radiated back towards the earth’s surface as well as into the outer space although these are very minimal in magnitude. The space within the ozone layer must be much denser than the space immediately above it as well as that which is immediately below it by virtue of the fact that ozone has a much higher molecular weight which is 48 Gm per mole compared to that of air at 28.86 Gm per mole. The absorption of more heat makes the ozone layer much hotter and in fact the hottest part of the atmosphere from the altitude of 10 km to 120 km. The prevailing temperature at the ozone layer is from 0 0C to 20 0C while temperature is registered at -55 0C at 10,000 meters in altitude and -82.25 0C at 82,500 meters in altitude.

While the formation of ozone raises the temperature of the atmosphere particularly at the altitudes from 40 km to 50 km, such rise in temperature does not have significant effect upon the stratosphere at the ground level and may not be considered as a threat.

The total heat that the ground surface of the earth receives from the sun that amounts to 1,003,900.4764 Cal/M2/hr is equivalent to the heat that can be generated if and when 331.51 Km3 of gasoline is burned every day. This amount of radiant energy referred to is the amount of heat energy that actually reached the ground surface during a sunny day more especially experienced near the equator. This means that heat energy being absorbed by the ozone layer and that of a mantle of gases is excluded in this quantity. To include the heat absorbed in the formation of ozone, it is 1,091,196.17 Calories/M2/Hr. to include the solar radiation absorbed by the ozone layer the total radiant energy from the sun amounts to the equivalent to the heat generated if and when 360.333842 Km3 of Gasoline is burned per day.

Air, which is 21.0 % oxygen, 78.0 % nitrogen and approximately 1.0 % of other components of the atmosphere by volume allow radiant energy or heat to penetrate through despite of the fact that the total mass of the atmosphere amounts to 10.35 metric tons per square meter of the earth’s surface. If the total amount of atmosphere is transformed to the density of water, it will fill the earth’s surface up to 10.35 meters. If we try to scrutinize and figure out deeper, we would realize that the mantles of gases that comprise the atmosphere are particles that are very far apart from each other.

Even at ground level where the atmosphere has its highest density, there exist but only 1.160829 kilograms per cubic meter of space at the temperature of 300C. Barely one part per thousand is occupied by matter and the rest is empty space. We can then imagine that each particle of photon is simply refracted once it hits the particle of oxygen or nitrogen atom in the atmosphere and consequently proceeds to its course until it reaches the ground level where it is eventually either absorbed or reflected.

In view of the fact that 80.0% of the mass of the atmosphere is found within the altitude of 12 kilometer or 50.0% is found within the altitude of 6.7 kilometers, we may therefore deduce that if ever there is absorption of heat in the atmosphere, it takes place within this stratum of atmospheric gases. From the altitude of 12 kilometers and beyond gaseous particles are extremely far apart one from another that heat absorption by atmospheric gases is insignificant. Even at the altitude of 10 kilometers above the ground level, atmospheric temperature is likewise not affected because it remains at -55 0C even during daytime.

In the absence of hard facts and figures to recon with, we may suffice ourselves with the analysis of atmospheric absorption of heat being confined up to the altitude of three kilometers where the temperature is at 0 0C which will allow the condensation of water vapor into droplets of water and where the prevailing atmospheric pressure is at about 0.700 Atmosphere. Referring to the climate in Metro Manila for the month of December where the temperature during the daytime averages at 30 0C and drops to 20 0C at nighttime, the 10 degree Celsius drop during the night at the ground level and 2 degrees drop at the two kilometer level gives us the average temperature change of 6.0 0C. The heat absorbed by 235.70 Gm/Cm2 that is confined from the ground level up to the height of 2,100 meters of the atmosphere for it to change its temperature by 6.0 0C

Specific Heat of Air = 0.25 Cal. /Gm
Temperature Change = 6.0 0C
Mass of Air Affected = 235.70 Gm/Cm2
Calories = 235.70 Gm/Cm2 (6.0 C0) (0.25 Cal. /Gm)
= 353.55 Cal. /Cm2.
10,000 Cm2
Calories = 353.55 Cal. /Cm2 (---------------) = = 3,535,500 Calories/M2/day
M2

This also mean that 2.357 tons of atmospheric gases contained within an area of 1.0 M2 and a height of 2,000 meters (2,000 M3) would need the addition of 3,535,500 Calories/M2/day to change the ground level temperature from 20 0C to 30 0C. To include the heat absorbed by the moisture content of the atmosphere, 3,712,125 Calories/M2/day is required.

Specific Heat of H2O Vapor = 0.50 Cal. /Gm
Temperature Change = 6.0 0C
Mass of Air Affected = 235.500 Gm/Cm2
Average Moisture Content = 2.5%
Moisture Content = 235.5 Gm/Cm2 (2.5%)
= 5.8875 Gm/Cm2
Heat Absorbed = 5.8875 Gm/Cm2 (0.50 Cal. /Gm) (6.0)
= 17.6625. Cal. /Cm2 10,000 Cm2
Calories = 17.6625 Cal. /Cm2 (-------------------)
M2
= 176,625 Calories/M2

SUMMATION OF ABSORPTION:
Absorption by Air = 3,535,500 Calories/M2/day
Absorption by Moisture = 176,625 Calories/M2/day
Total 3,712,125 Calories/M2/day


During the daytime additional heat in the amount of 3,712,125 Calories/M2/day, is absorbed by the atmosphere which it releases or radiates into outer space during the night. The initial temperature of 200C slowly and gradually builds up to 300C because while the atmosphere absorbs heat it is at the same time releases heat.

The absence of solar radiation during the night brings down the temperature back to about 200C because 115.97424 watts (99,882.11835 Cal/M2/Hr) or 1,198,585.420254 Cal/M2 per 12 hours is radiated into outer space. During the daytime, 124.02576 watts (1,281,797.30005 Cal/M2/12Hr is radiated into outer space making the total of 240 watts or 2,480,382.72059 Cal/M2/day radiated into outer space by virtue of the nature of the earth as a black body.

3.3.1 COMPUTATION OF AVERAGE SOLAR RADIATION THAT IS ABSORBED BY EARTH’S GROUND SURFACE

Diameter of Earth: = 12,735.061 Km
FACIAL AREA OF EARTH = (6,367.5305)2 (3.141592654)
= 127,377,271.1235 Km2
AREA OF EARTH’S GROUND SURFACE = (6,367.5305)2 (3.141592654)2/2
= 400,167,549.924 Km2

127,377,271.1235 Km2
PERCENT = (----------------------------) (100) = 31.83098458 %
400,167,549.924 Km2

EQUIVALENT HOURS = 0,3183098458 (24 hours/day)
= 7.6394 Hr/day
IN METRO MANILA:
AVERAGE RADIATION = 1,003,900.4764 CAL./M2/Hr (12) (0.634968729416)
= 7,649,344.919518 Cal/M2/Hr (12 Hr/day)

The forest area has been observed to have the capacity to reflect 40% of solar radiation into outer space. A large portion is temporarily absorb by the plant leaves, branches and stems and eventually pass it into the atmosphere. The visible light that is absorbed by the plants is being transformed into infrared radiation before passing it into the atmosphere. A very small portion, approximately 0.02 percent or 200 Cal/M2 is absorbed by the plant for its daily growth.

3.3.2 SOLAR RADIATION INTENSITY DISTRIBUTION
LATITUDE MEAN
000 - 050 = Cos 00 - Cos 50 = 1.0000000000 - 0.9961946981 0.998097349046
050 - 100 = Cos 050 - Cos100 = 0.9961946981 – 0.9848077530 0.990501225552
100 - 150 = Cos 100 - Cos150 = 0.9848077530 - 0.9659258263 0.975366789656
150 - 200 = Cos 150 – Cos 200 = 0.9659258263 – 0.9396926207 0.952809223495
200 - 250 = Cos 200 – Cos 250 = 0.9396926207 – 0.9063077870 0.923000203911
250 - 300 = Cos 250 – Cos 300 = 0.9063077870 – 0.8660254037 0.886166595411
300 – 350 = Cos 300 – Cos 350 = 0.8660254037 – 0.8191152044 0.842588724037
350 - 400 = Cos 350 – Cos 400 = 0.8191152044 – 0.7660444431 0.792598243703
400 - 450 = Cos 400 – Cos 450 = 0.7660444431 – 0.7071067811 0.736575612153
450 - 500 = Cos 450 – Cos 500 = 0.7071067811 – 0.6427876097 0.674947195437
500 - 550 = Cos 500 – Cos 550 = 0.6427876097 – 0.5735764364 0.608182023019
550 - 600 = Cos 550 – Cos 600 = 0.5735764364 – 0.5000000000 0.536788218176
600 - 650 = Cos 600 – Cos 650 = 0.5000000000 – 0.4226182617 0.461309130870
650 - 700 = Cos 650 – Cos 700 = 0.4226182617 – 0.3420201433 0.382319202533
700 - 750 = Cos 700 – Cos 750 = 0.3420201433 – 0.3420201433 0.300419594214
750 - 800 = Cos 750 – Cos 800 = 0.3420201433 – 0.1736481777 0.213233611385
800 - 850 = Cos 800 – Cos 850 = 0.1736481777 – 0 0871557427 0.130401960207
850 - 900 = Cos 850 – Cos 900 = 0 0871557427 – 0.0000000000 0.0 43577871374
TOTAL 11.44887696749

11.44887696749
PERCENT = ------------------- (100) = 63.6048720416 %
18 Sections

At a given area/spot near the equator, only 63.6048720416 % of radiation is received. Considering that the earth rotates around its axis to produce nighttime, the percentage of solar radiation that a given spot near the equator received 31.8024360208 % during the 24 hour period. Theoretically, this figure is equal to the reciprocal of “Pi” or 1/3.141592654 which incur minor discrepancy in sectional distribution using the Cosine formula.

Cosine Value 0.318024360208
Reciprocal Value 0.318309886142
Discrepancy 0.000285525934
Percent Discrepancy 0.08972%

The reciprocal of “Pi” is exactly the value of the Great Circle divided by the total surface area of the earth.


CIRCUMFERENCIAL
LATITUDE LENGTH AREA KM2
000 - 050 = (0.998097349046)2 = 0.9961983181 39,853.134780 22,145,271.1095
050 - 100 = (0.990501225552) 2 = 0.9810926778 39,249.286238 21,809,729.3824
100 - 150 = (0.975366789656) 2 = 0.9513403744 38,061.581500 21,149,755.1153
150 - 200 = (0.952809223495) 2 = 0.9078454164 36,321.418895 20,182,795.4802
200 - 250 = (0.9230002039112 = 0.8519293764 34,084.309113 18,939,696.2134
250 - 300 = (0.886166595411) 2 = 0.7852912348 31,418.225421 17,458,228.1561
300 – 350 = (0.842588724037) 2 = 0.7099557579 28,404.175486 15,783,405.0007
350 - 400 = (0.792598243703) 2 = 0.6282119653 25,133.739259 13,966,115.3116
400 - 450 = (0.736575612153) 2 = 0.5425436324 21,706.288557 12,061,576.7750
450 - 500 = (0.674947195437) 2 = 0.4555537166 18,225.963471 10,127,657.5739
500 - 550 = (0.608182023019) 2 = 0.3821490136 15,289.160660 8,495,758.4823
550 - 600 = (0.536788218176) 2 = 0.2882411591 11,532.060095 6,408,042.8971
600 - 650 = (0.461309130870) 2 = 0.2128061214 8,514.026895 4,731,006.3528
650 - 700 = (0.382319202533) 2 = 0.1461479862 5,847.143291 3,249,093..8068
700 - 750 = (0.300419594214) 2 = 0.0902519326 3,610.833074 2,006,438.8359
750 - 800 = (0.213233611385) 2 = 0.0454685730 1,819.123674 1.010,836.0902
800 - 850 = (0.130401960207) 2 = 0.0170045778 680.325508 378,037.8357
850 - 900 = (0.0 43577871374) 2 = 0.0018990309 75.977138 42,218.3682
TOTAL 8.9939308566 359,826.77305 199,945,662.8066

AREA COVERED BY SUNLIGHT = 199,945,662.8066 Km2 (2)
= 399,891,325.6132 Km2

ACTUAL AREA = 400,167,549.924 Km2
LESS 399,891,325.1326 Km2
DISCREPANCY = 276,224.3108 Km2
PERCENT DISCREPANCY = 0.069 %

40,008.37408584 Km2
AREA DISTRIBUTED = 359,826.773055 Km2 (-------------------------) (50)
3600
= 199,945,613.790071 Km2

WHOLE EARTH SURFACE = 199,945,613.790071 Km2 (2)
= 399,891,226.189141 Km2

COMPUTED AREA = 400,167,549.924 Km2
DIATRIBUITED AREA = 399,891,226.189 Km2
DISCREPANCY = 276,323.735 Km2
= 0.06905%

3.3.3 SOLAR RADIATION ON EARTH’S SURFACE AREA PER SECTION OF 50 LATITUDE

LATITUDE SECTION AREA INTENSITY Cal./M2/Hr Km3 Gasoline
00 – 050 22,145,271.1095 Km2 0.998097349040 1,001,990.4038 22.94457301 Km3
050 – 100 21,809,729.3824 Km2 0.990501225552 994,364.6518 22.42494421 Km3
100 – 150 21,149,755.1153 Km2 0.97.5366789656 979,171.1813 21.41407814 Km3
150 – 20 0 20,182,795.4802 Km2 0.952809223495 956,525.6030 19.96242675 Km3
200 – 250 18,939,696.2134 Km2 0.923000203911 926,600.3441 18.14683469 Km3
250 – 300 17,458,228.1561 Km2 0.886166595411 889,623.0669 16.05985490 Km3
300 – 350 15,783,405.0007 Km2 0.842588724037 845,875.2211 13.80519103 Km3
350 – 400 13,966,115.3116 Km2 0.792598243703 795,689.7541 11.49092014 Km3
400 – 450 12,061,576.7750 Km2 0.736575612153 739,448.6077 9.22247524 Km3
450 – 500 10,127,657.5739 Km2 0.674947195437 677,579.8108 6.15519826 Km3
500 – 550 8,495,758.4823 Km2 0.608182023019. 610,554.2224 5.36366708 Km3
550 – 600 6,408,042.8971 Km2 0.536788218176 538,881.8477 3.57070836 Km3
600 – 650 4,731,006.3528 Km2 0.461309130870. 463,108.4561 2.26553960 Km3
650 – 700 3,249,093.8068 Km2 0.382319202533 383,810.4294 1.28947948 Km3
700 – 750 2,006,438.8359 Km2 0.300419594214 301,591.3736 0.54277198 Km3
750 – 800 1.010,836.0902 Km2 . 0.213233611385 214,065.3239 0.22374970 Km3
800 – 850 378,037.8357 Km2 0.130401960207 130,910.5899 0.05117352 Km3
850 – 900 42,218.3682 Km2 0.043577871374 43,747.8458 0.00190982 Km3
TOTAL 199,945,662.8066 Km2 11.448900415966 11,036,740.7334 180.16692100 Km3


(1,001,990.4038) (106) )22,145,271.1095 Km2)
Km3 = --------------------------------------------------------) ((24) (0.3183098861)
8,517,042 (1012)

= (2.60528821 Km3) (24) (0.3183098861)
= 19.90293584 Km3

The above data is limited to half of the great sphere and therefore to obtain the total heat radiation, 180.166921 Km3 is multiplied by 2.0 to obtain the total which is 360.333842 Km3 of Gasoline/day. The discrepancy due interpolations appears so large and we remedy the figure by ratio and proportion.

With the above tabulation you would know how much solar radiation you have in your area.

3.3.4 SOLAR RADIATION ON EARTH’S SURFACE AREA PER SECTION OF 50 LATITUDE

LATITUDE SECTION AREA INTENSITY Cal./M2/Hr Km3 Gasoline
00 – 050 22,145,271.1095 Km2 0.998097349040 1,001,990.4038 19.90293584 Km3
050 – 100 21,809,729.3824 Km2 0.990501225552 994,364.6518 19.45219140 Km3
100 – 150 21,149,755.1153 Km2 0.97.5366789656 979,171.1813 18.57533035 Km3
150 – 20 0 20,182,795.4802 Km2 0.952809223495 956,525.6030 17.31611648 Km3
200 – 250 18,939,696.2134 Km2 0.923000203911 926,600.3441 15.74120758 Km3
250 – 300 17,458,228.1561 Km2 0.886166595411 889,623.0669 13.93088734 Km3
300 – 350 15,783,405.0007 Km2 0.842588724037 845,875.2211 11.97511199 Km3
350 – 400 13,966,115.3116 Km2 0.792598243703 795,689.7541 9.96763139 Km3
400 – 450 12,061,576.7750 Km2 0.736575612153 739,448.6077 7.99990189 Km3
450 – 500 10,127,657.5739 Km2 0.674947195437 677,579.8108 6.15519826 Km3
500 – 550 8,495,758.4823 Km2 0.608182023019. 610,554.2224 4.65263493 Km3
550 – 600 6,408,042.8971 Km2 0.536788218176 538,881.8477 3.09735899 Km3
600 – 650 4,731,006.3528 Km2 0.461309130870. 463,108.4561 1.96520935 Km3
650 – 700 3,249,093.8068 Km2 0.382319202533 383,810.4294 1.11854021 Km3
700 – 750 2,006,438.8359 Km2 0.300419594214 301,591.3736 0.54277198 Km3
750 – 800 1.010,836.0902 Km2 . 0.213233611385 214,065.3239 0.19408842 Km3
800 – 850 378,037.8357 Km2 0.130401960207 130,910.5899 0.04438973 Km3
850 – 900 42,218.3682 Km2 0.043577871374 43,747.8458 0.00165665 Km3
TOTAL 199,945,662.8066 Km2 11.448900415966 11,036,740.7334 152.63316278 Km3


(1,001,990.4038) (106) )22,145,271.1095 Km2)
Km3 = --------------------------------------------------------) ((24) (0.3183098861)
8,517,042 (1012)

= (2.60528821 Km3) (24) (0.3183098861)
= 19.90293584 Km3

THEORETICAL TOTAL = 180.16692100 Km3
DESTRIBUTED TOTAL = 152.63316278 Km3
DISCREPANCY = 27.53374722 Km3

PERCENT DISCREPANCY = (27.53374722 Km3/180.16692100 Km3) (100)
= 15.28%


The resultant discrepancy which turned out to so large could be attributed to some interpolations made in between each sections and therefore needs adjustments.