We present here some calculations based on information from references readily available
in local libraries: Encyclopedia Britannica, McGraw-Hill Encyclopedia of Science and
Technology, and Americana 2000 (Grolier). Some of the quantities are very large, and we
express them as some number times ten to some power. For example, 2.5E6 means
2.5 x 10 x 10 x 10 x 10 x 10 x 10, or 2.5 million. The units of measurement are defined
and explained in the following table, their abbreviations are given, and their
equivalents in other units are given. We put the units into the calculations, multiply
them together, and divide by them, as if they were numbers. When we include the units in
the calculation, the answer automatically comes out with the correct units.
kg = kilogram = unit of mass.
m = meter = 39.37 inches.
km = kilometer = 1E3 m.
m3 = cubic meters.
km2 = square kilometers.
1 km2 = the area of a square whose side measures 1 km.
C = degrees Celsius = Temperature on the Celsius scale, or it could be the difference
between two temperature readings on the Celsius scale.
1C = One degree Celsius. Note that there is no space between the number and the C.
When expressing all quantities other than temperature, we put a space between the number
and the abbreviation for the unit name.
J = Joule, a unit of energy (kinetic, potential, electrical).
kJ = kiloJoule = 1000 Joules.
kcal = kilocalorie = a unit of heat.
An expression enclosed in parentheses ( ) is treated as a single entity.
(12) (34) (56) indicates that 12, 34, and 56 are multiplied together.
((12) / (34)) (56) means that 12 is to be divided by 34, and the result of that division
is to be multiplied by 56.
(12) / ((34) (56)) means that 12 is to be divided by the product of 34 and 56.
1 kcal is the heat required to increase or decrease the temperature of 1 kg of water by
1C.
1 kcal = (1 kg) (1C).
4.186 kJ of kinetic, potential, or electrical energy can be dissipated to produce 1 kcal
of heat.
Heat is the energy of random, disordered motion. A heat engine is required to convert
heat into orderly, accessible kinetic, potential, or electrical energy. The efficiency of
a heat engine is always less than 100%, so 1 kcal of heat energy cannot be upgraded back
into 4.186 kJ of organized (high class) energy.
s = second.
/s = per second.
1 kW = 1 kiloWatt of power, which is a flow of energy, or a conversion of one form of
energy into another at the rate of 1 kJ/s.
m3/s = cubic meters per second.
80 kcal = heat required to melt 1 kg of ice.
1E3 kg/m3 = mass density of fresh water and the approximate mass density of salt water.
The Arctic Ocean and its many bordering seas are almost completely surrounded by land.
Water flows in from the Atlantic and the Pacific, and exits to the Atlantic at the rate
of 5.9E6 m3/s. The average temperature of the entering water is about 10C, and the
exiting temperature is about 0C, so the water temperature is reduced by about 10C.
The cooling rate is
(5.9E6 m3/s) (10C) (1E3 kg/m3) (1 kcal) / ((1 kg) (1C)) = 5.9E10 kcal/s
Multiplying this cooling rate by (3600) (24) (182) tells us that, in a half year,
9.28E17 kcal is removed from the water that flows through the Arctic Ocean and
surrounding seas. This heat is removed by melting the ice that is floating on the
surface of the Arctic Ocean and its surrounding seas. Divide by 80 kcal/kg to get
the mass of ice that must be melted in a half year.
(9.28E17 kcal) / (80 kcal/kg) = 1.16E16 kg
The density of ice depends on the amount of voids. We use a value of 8.8E2 kg/m3.
Dividing by this density, we find that 1.32E13 m3 of ice must be melted in a
half year.
The area covered by ice at the end of winter is 12E6 km2. At the end of summer it is
9E6 km2. In a half year, 3E6 km2 melts, or 3E12 m2. The average height of the pack ice
that melts would have to be 1.32E13 / 3E12 = 4.39 m.
The average thickness of ice that is more than 1 year old is 12 ft, or 3.66 m. Ice less
than one year old averages 6 ft. The 4.39 m thickness calculated from the cooling rate
is at least 20% too great. The explanation for the discrepancy might be that some of
the cooling is done directly by the cold air and the cold water discharged by the rivers.
The foregoing calculation assures us that we have a fairly good understanding of how the
Arctic area cools the water that passes through it. The cooling rate, 5.9E10 kcal/s, is equal
to the heat transport rate from the lower latitudes to the Arctic Ocean and
its surrounding seas. Most of
it comes up via the North Atlantic Current.
We want to examine the possibility of reducing the heat load on the Arctic ice by
reducing the flow of the Gulf Stream. The Gulf Stream has a well-defined path and is
flowing at a useful speed from Florida to North Carolina. The following is a preliminary
attempt to estimate how much heat it transports and guess at how much of it reaches the
Arctic. In future work, we will access the relevant information provided by
oceanographers on this subject and refine these estimates.
For the present, we ignore the effects of recirculation and other heat losses along
the path. The summer surface temperature near the Florida Straits is 28C. The Florida
Current transports 2.6E7 m3/s, and we assume that this water somehow reaches the region
of the North Atlantic where the Gulf Stream slows down, meanders, and becomes the North
Atlantic Current that crosses the ocean. The summer surface temperature is 20C at this
point. We assume that the 28C water from the Florida Straits is dumped into this 20C
sink, so its temperature is reduced by 8C. The rate at which heat is given up is
(2.6E7 m3/s) (8C) (1E3 kg/m3) = 2.08E11 kcal/s
We assume that this is the heat transported by the water coming from the Florida Straits.
This is about 3.5 times the Arctic cooling load. When the heat transport of the Antilles
Current is added to the Florida Current, it would seem that the Gulf Stream makes a
significant contribution to the heat load on the Arctic ice, even if only a small
fraction of the heat transported by the Gulf Stream actually reaches the Norwegian Sea.
According to the Americana 2000 Encyclopedia, the Florida Current velocity can be as
high as 1.79 m/s. If all of the 2.6E7 m3/s transported by the Florida Current were
moving at this velocity, the kinetic energy per unit volume would be
(.5) (1.79 m/s) (1.79 m/s) (1E3 kg/m3) = 1.6E3 J/m3
Multiplying this by the volume transport rate of 2.6E7 m3/s gives a kinetic energy
transport rate of 4.17E10 J/s. This is a power flow of 41.7E9 W, or 41.7 million kW.
This has a heat equivalent of 9.95E6 kcal/s. The heat transported by the Florida Current
was calculated in the preceding section to be 2.09E4 times greater. This tells us that
the heat equivalent of the power that we need to extract from the Florida Current,
although large, is a small fraction of the heat transport that we are attempting to
control.
How much power do we need to extract from the Florida Current and other currents to
accomplish the goal of preventing the sea level from rising? We are a long way from
answering this question. But we might guess that converting 10% of the kinetic energy
transport of the Florida Current into electricity ought to make an observable change
in the Arctic heat load. This would be an electrical output of about 4000 Megawatts.
This is somewhat less than the power that the Long Island Power Authority has to
deliver to Long Island, NY on the hottest day of the year. And no radioactive waste,
carbon dioxide, or pollutants would be produced.
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