The total volume of the five Great Lakes of North America is 0.0227 × 1015 m3. Together they contain about 18% of the world's surface fresh water.
From table 4.1 and the formula M = ρV, where ρ is the density, we easily compute the total mass of water in the world:
TABLE 4.1
Volume of water in the world
We now summarize the results of this brief analysis. The ratio of the mass of water to the mass of air is 1.435 × 1021 kg/(5.116 × 1018 kg) = 280. So the answer to the question at the beginning of this chapter is that all of the water in the world weighs 280 times more than all the air in the world. This is an interesting result. However, an even more interesting aspect of this short problem is that we were able to determine the two total weights, water and air, and their ratio, without much uncertainty and with relatively little difficulty.
Suppose that All the Ice Were to Melt
Here is an interesting question to examine. What would happen if all the ice presently locked up mostly in the polar ice caps and glaciers were to melt? How much would the surface of the ocean rise? How much of the present land areas of the world would be flooded?
From table 4.1 it is seen that the volume of water contained in the ice caps and glaciers is Vice = 29 × 1015 m3. The surface area of the oceans is A0 = 3.61 × 1014 m2. Dividing the first of these quantities by the second gives the increase in depth, z = 80.3 m.
FIG. 4.1
The hypsometric curve: A plot showing the earth's solid surface area above any given elevation or depth. (From Sverdrup et al. 1942.)
However, as the ocean depth increases, due to melting ice, the ocean area will enlarge as water begins to flood the low-elevation regions of the world. Consequently, the increase in ocean depth will be something less than 80 meters.
To answer the questions we need the so-called hypsometric (or hypsographic) curve shown in figure 4.1. This curve shows the percentage of the earth's solid surface above any given depth or elevation. For example, 29.2% of the earth's land surface is above z = 0 (sea level), 7.9% is above 1,000 meters, and so on.
We now construct a simple mathematical model to guide our analysis. A moment ago we established that the increase in ocean depth due to the melted ice will be something less than 80 meters; how much less is the problem. To solve the problem, it is clear that we need an equation of the hypsometric curve shown in figure 4.1 for values of the elevation z less than, say, 100 meters or so.
As precisely as we can, we match or fit the hypsometric curve of figure 4.1 with a simple straight line relationship applicable for small values of z. Doing so yields the following result:
in which land Pland is the area of the earth's land surface as a percentage of the earth's total area. Thus, if z = 0 (i.e., present sea level) in equation (4.2), we obtain Pland = 29.2%, a result we already have. If z = 100 m, for example, then Pland = 23.4%.
Next, using the relationship Pland + Pocean = 100.0%, we alter equation (4.2) to the form,
where Pocean is the area of the earth's ocean surface as a percentage of the earth's total area.
Finally, knowing that the earth's total area is AT = 5.10 × 1014 m2, we obtain the expression,
in which A is the area of the ocean surface when the surface is at elevation z and A0 = 3.61 × 1014 m2 is the area when z = 0. Solving this equation for A, we obtain
Now we are ready for the final step. From table 4.1, the volume of the ice caps and glaciers is V = 29 × 1015 m3. Since all this ice melts into the ocean, we have
29 × 1015 = z[(3.61 × 1014) + (0.0030 × 1014)z]
or
This is a quadratic equation in z. Recall that if the quadratic equation is written in the general form,
then its solution is given by
Using this expression, the solution to equation (4.6) is z = 75 m. So we have our answer. If all the world's ice were to melt, the oceans would rise by about 75 meters (246 feet).
Finally, we note that the quantity A – A0 represents the area of land surface flooded by the rise of the ocean. We easily compute from equation (4.5) that the flooded area is 0.225 × 1014 m2 or approximately 22.5 million km2. The area of the contiguous 48 states of the United States is 7.62 million km2. So we conclude that the total area flooded by the melted ice, worldwide, would be about three times larger than the 48 states of the United States.
A SUGGESTION FOR A TERM PAPER. Are you looking for an interesting subject for a term paper in your course in geography, science, or mathematics? Well, you might want to use this topic of ice melting in the oceans, but explore the matter to a much greater extent. A good place to begin your analysis is the interesting book by de Blij and Muller (1996).
For example, let's assume that the ocean level rises by 75 meters due to the melted ice. In your term paper, you could examine and respond to the following questions:
What would the substantially altered map of North America look like? What would be the new area of the forty-eight states of the United States? What major cities would be submerged?
Most of Florida vanishes. What about the other low-elevation states, such as Delaware, Louisiana, Mississippi, and Rhode Island? How much of Hawaii disappears?
How are the Saint Lawrence River Valley, Niagara Falls, and the Great Lakes affected? What happens to the Panama Canal?
How are the Central Valley, Death Valley, and Imperial Valley in California changed? Where are the mouths of the Colorado, Mississippi, Columbia, Rio Grande, and Hudson Rivers?
In the United States, approximately how many people have to move to higher ground?
Some Items Involving the Hydrologic Cycle
Most of us have heard about the so-called water cycle or hydrologic cycle. This is the process featuring evaporation of water from ocean and land surfaces, retention of this water vapor in the atmosphere for a relatively short period of time, precipitation of this water, usually in the form of rain, onto oceans and lands, and retention of the water in the ocean, ground, lakes, and rivers for a relatively long period of time. An important component of the cycle is the drainage or runoff of water from land surfaces to the oceans. A diagram illustrating the hydrologic cycle is shown in figure 4.2.
First, we consider evaporation. As indicated in the figure, the total rate of evaporation is Q = 456 + 62 = 518 × 1012 m3/yr. The rate from the surface of the oceans is given by the equation q = Q/A0 = (456 × 1012)/(3.61 × 1014) = 1.263 m/yr, or about 49.7 inches per year. From land surfaces the rate is 0.416 m/yr or 16.4 inches per year. Worldwide, the average rate of evaporation is q = 1.016 m/yr = 40.0 in/yr.
FIG. 4.2
The hydrologic cycle. Units of the quantities are (a) V: 1015 m3, (b) A: 1014 m2, and (c) Q: 1012 m3/yr.
Next, the matter of average retention time or residence time in the atmosphere is examined. Suppose we have a tank containing V = 30 gallons and that water flows into and out of the tank at the rate Q = 5 gallons per minute. Then the average residence time of a water particle in the tank is T = V/Q = 6 minutes.
By the same token, from figure 4.2 it is noted that the amount of water stored in the atmosphere is V = 0.013 × 1015 m3, and the rates of inflow (evaporation) and outflow (precipitation) are Q = 518 × 1012 m3/yr. Accordingly, the average residence time of a water molecule in the atmosphere is T = 0.025 yr = 9.2 days.
Now we go to the topic of precipitation or rainfall. From the figure we see that the total rate of precipitation is Q = 410 + 108 = 518 × 1012 m3/yr. The average rainfall rate over the oceans is q = 1.136 m/yr = 44.7 inches per year and over land surfaces q = 0.725 m/yr = 28.5 inches per year.
On a worldwide basis, the average rainfall is 1.016 m/yr or 40.0 inches per year. Needless to say, there is a wide range of rainfall rates. Rainfalls of nearly 500 inches per year have been recorded in Hawaii and India. At the other extreme, in desert regions in various parts of the world there is virtually no rainfall whatsoever. The Atacama desert in northern Chile holds the record, with an average annual rainfall of 0.02 inches.
Ret
urning to the matter of residence time, we saw that a water molecule is retained in the atmosphere for an average of only 9.2 days between the time it is evaporated and the time it is precipitated. We make the same kind of computation of average residence time for a water molecule on the earth's surface. That is, T = V/Q = (1,387.5 × 1015)/(518 × 1012) = 2,679 yr. This means that, on the average, a water molecule falling in 680 BC. was finally evaporated in 1999 in order to spend about nine days in the atmosphere before falling back to earth once again. Of course, the residence time of 2,679 years is the average. The residence time of a particular molecule could be a few minutes or it could be many thousands of years.
Finally, a word about runoff. The water balance diagram of figure 2.4 shows a drainage of Q = 46 × 1012 m3/yr from the lands to the oceans. This represents the combined discharges of all the world's rivers, streams, and underground aquifers into the Pacific, Atlantic, Indian, and Arctic oceans. The annual flow indicated above is equivalent to Q = 1,460,000 m3/s. To give a scale to this quantity, we note that the world's mightest river, the Amazon, has an average discharge at its mouth of about 180,000 m3/s. So the Amazon alone handles around 12% of all the world's runoff to the seas.
Our thrilling saga will continue after a brief pause.
5
Raindrops and Other Goodies Revisited
Runoff of Rivers
Our saga now continues. Readers will recall that the incredible Amazon river in South America delivers 180,000 cubic meters of water per second to the Atlantic ocean. This is about thirty times more than the flow over Niagara Falls. The drainage area of the Amazon is about 7 million square kilometers—about ten times the size of Texas. Much of the Amazon basin receives heavy rainfall to sustain the well-known rain forests.
On a smaller scale, the same things can be said about the world's second river: the Congo (or Zaire) in Africa. It collects water from a drainage area of 3.5 million square kilometers and empties over 41,000 cubic meters per second into the Atlantic. The Congo basin is also a region of tropical rain forests and extensive rainfall.
And so it goes. The fifteen major rivers of the world (scaled on the basis of discharge, not length), collectively drain about 37 million square kilometers (24.7% of the world's land area). Together, they deliver 460,000 cubic meters of water per second to the oceans (31.5% of total runoff). Eight of these rivers are in Asia, three each in North America and South America, and one in Africa. They are listed in table 5.1.
TABLE 5.1
Kinetic Energy and Power of Rainfalls
Enough of rivers; back to rainfalls. Not surprisingly, in rainfalls there are various sizes of raindrops. For example, in a light shower (r = 5 mm/hr), raindrop diameters range from about D = 0.1 mm to D = 1.5 mm. In a moderate rainfall (r = 25 mm/hr), the maximum diameter is around D = 3.5 mm. And in a heavy deluge (r = 100 mm/hr), drop diameters can be as much as D = 6.0 mm.
Maximum raindrop diameter is about 6 millimeters. Larger sizes tend to be unstable due to shearing forces caused by their motion through the air. Consequently, they rupture and form smaller drops.
What is the velocity of a raindrop? As we shall see, it depends on the diameter of the drop. The raindrop velocity also depends on the elevation of the rainfall above sea level. It is logical that raindrops fall faster at higher elevations because the air is less dense. If you are interested in the complexities of raindrops, and indeed there are a great many, texts dealing with atmospheric science and cloud physics are good places to start. Suggested references are the books on fluid mechanics by Richardson (1952) and on environmental aerodynamics by Scorer (1978).
For our present purpose, a simple equation to compute the terminal velocity of a raindrop is the expression provided by Dingle and Lee (1972):
where D is the drop diameter in millimeters and U is the velocity in meters per second. Thus, if D = 2.5 mm, we obtain U = 7.44 m/s.
We have now set the stage for some very exciting calculations. As we saw before, the earth receives a total of 518 × 1012 cubic meters of precipitation each year, mostly as rain but some as snow and hail. Here's one for people who like big numbers: how many raindrops are contained in this 518 million million cubic meters of annual rainfall?
To make things easy, suppose all the annual rainfall were composed entirely of raindrops with diameter D = 3 mm. The volume of a sphere is or So the number of drops falling each year is
This is quite a few raindrops. To carry this kind of numerical computation a step further, please verify that with this many drops, every square millimeter of the earth is hit, on the average, by 72 raindrops every year. A good thing to know.
The kinetic energy of a single raindrop is
Since the diameter of our drop is D = 3 mm, its velocity, from equation (5.1), is U = 8.08 m/s. With density ρ = 1,000 kg/m3 and volume we compute from equation (5.2) that
For a single raindrop this is certainly not much energy. But we have a lot of drops. The total power produced by all of the raindrops is
This is the result we are after. Now if somehow the kinetic energy of the world's rainfalls could be captured, we would be able to generate 536 million kilowatts of power. To provide a scale to this number: in 1990 the total power-generating capacity of the United States (hydroelectric and steam) was 698 million kilowatts.
Of course, it is not feasible to harness the kinetic energy of raindrops. However, we can appreciate how much power there is in rainfalls. Admittedly, rainfalls do help to alleviate pollution in rivers, lakes, and estuaries by (a) direct addition of oxygen from the oxygen-saturated drops, (b) increasing the rate of oxygen transfer from the atmosphere to the body of water, and (c) outright dilution (“the solution to pollution is dilution”). Otherwise, the enormous kinetic energies of rainfalls do little beyond causing extremely serious problems of soil erosion.
At this point, you might be interested in confirming that rainfalls comprised only of D = 2 mm drops will generate 354 million kW and rainfalls with D = 4 mm drops will produce 637 million kW.
How Many Times Has the World's Water Been Consumed by Humans?
As we have seen, there is an enormous amount of water in the world. Furthermore, we know that a great many people live and have lived on our planet in the past. So a reasonable and interesting question is the following: how many times have all these humans consumed all this water during the past million years or so?
First we turn to the extensive Scientific Tables prepared by Ciba-Geigy and edited by Diem and Lentner (1970). They indicate that per capita daily water intake in all forms (liquids and food) is approximately 2.5 liters. Accordingly, water consumption is per person.
In chapter 15, “How Many People Have Ever Lived?,” we establish that about 80 billion people lived on earth during the period from 1,000,000 B.C. to 1990. With an assumed average life span of 25 years, we are dealing with a total of 2,000 billion person-years. Consequently, the total amount of water consumed by humans during the past million years or so is M(water consumed) = (912.5)(2,000 × 109) = 1.825 × 1015 kg. The amount of water in the world is M(water available) = 1.435 × 1021 kg. Dividing the first quantity by the second gives 1.27 × 10 –6. This means that humans have consumed only about one cup of water per million cups available. Or, stated in a more direct fashion, over the past million years humans have consumed only about one-twelfth of the volume of the Great Lakes; in other words, about the same amount as Lake Ontario.
How Many Times Has the World's Air Been Breathed by Humans?
Our final question has a rather startling answer. The Scientific Tables of Diem and Lentner report that human breathing rates range from about 1.0 liters per minute for infants to around 40 liters per minute for men engaged in heavy work. A reasonable average value is Q = 121/min = 0.2 × 10–3 m3/s. Next, we want to calculate the mass of air we breathe per second. Now the density of air (sea-level pressure; 15°C) is ρ = 1.225 kg/m3. Using the formula M = ρQ, we determine that M = 0.245 × 10–3 kg/s = 7,725 kg/yr per person.
As be
fore, from the year 1,000,000 B.C. to 1990, there were 2,000 billion person-years of human life. So the computed amount of air breathed by humans during that very long period of time is M(air breathed) = 1.56 × 1016 kg.
The amount of air in the world is M(air available) = 5.116 × 1018 kg. The ratio of the amount of air breathed to the amount available is 0.00305 or 0.305%. This answer says that only about 0.3% of the air in the world has been breathed even once by humans. Could this be true? It seems to be a surprisingly small amount. Still, we have defined our computation and that is the outcome.
We conclude our chapter with the following interesting item: Annually, a person breathes about 7,700 kilograms of air and consumes around 900 kilograms of water, or about 8.5 times more air than water.
6
Which Major Rivers Flow Uphill?
As all of us know, jellied biscuits fall to the rug (frequently upside down), baseballs fall to the center fielder (frequently with two outs and bases loaded), and car keys fall to the pavement (frequently through sewer gratings). And rivers fall—or as we usually say—flow downhill. So how can a river flow uphill? Well, be patient. That matter will be explained in a page or two.
Slicing Pizzas, Racing Turtles, and Further Adventures in Applied Mathematics (Princeton Paperbacks) Page 4
