Date Due: Wednesday, 17 September 1997
The total maximum points were 50. Point distribution for each question noted below.
1. What is the difference between the concept of weather and the concept of climate? Give an example of each.
(4 pts.)
| Weather - A description of the current state of the atmosphere
Weather maps, today's high or low temperature, tomorrow's weather forecast. Climate - A longer term state of the atmosphere and its interactions with the other components of the earth-atmosphere-hydrosphere system. The "normal" climate for Madison; continental versus maritime climates or polar versus tropical. |
2The barometric pressure associated with one standard atmosphere at mean sea level is [you may round to nearest whole number]:
(5 pts - 1 each)
| 1 atmosphere of pressure is equivalent to:
29.92 (or 30) inches of mercury 76.0 centimeters of mercury 14.7 (or 15) pounds per square inch (psi) 1013.25 (1000) millibar (mb) 34 feet of water |
3. The highest recorded sea level corrected pressure in the world was ---. [Please include units!]
| Highest pressure: 1083.8 mb = 32.01 inches of mercury |
The lowest recorded sea level corrected pressure in the world was ---. [Please include units!]
| Lowest pressure: 870 mb = 25.68 inches of mercury |
The range between the record lowest and highest sea level corrected pressure (above) is approximately ---.
| The range: Range = (High - Low) = (1083.8 - 870) mb = 214 mb =(32.01 - 25.68) in = 6.33 in of mercury |
(4 pts - 1,1,2)
1d. What is the weight exerted by the atmosphere upon the flat, horizontal roof of a 25 foot by 60 foot building? [Assume standard sea level conditions; English units may be used here]. Clearly show your work for partial credit!
| Weight = Pressure x area
Pressure = 15 pounds per square inch (approx.) Area = 25 ft x 60 ft = 1500 sq. ft. Since 1 sq. ft. = 144 sq. in. (count them - since 12 inches on each side of the square), then 1500 sq. ft. = 1500 x 144 = 216,000 sq. inches. Then: Weight = 15 lb per sq. in x 216,000 sq. in. = 3,240,000 lb or 1620 ton (If 14.7 psi were used, the weight would be 3,175,200 lb or 1588 ton) Note that units check too! While this answer may sound large, the roof does not collapse due to the weight exerted by the atmosphere since the air pressure is pushing on the roof in all directions. |
(5 pts)
5. A football fan brought an aneroid barometer to Mile High Stadium in Denver (elevation of 1 mile) and made a reading of 840 mb. What would be the approximate sea level corrected pressure if we assumed that the pressure decreases at approximately 1 mb per 10 meters ascent through the atmosphere?
| Mile High Stadium in Denver is 5280 ft above mean sea level (MSL),
or 1600 m MSL.
Since the air pressure is assumed to decrease at a rate of 1 mb per 10 m, the pressure at the stadium should be 160 mb less than at the mean sea level directly below the stadium. Because the observed station (or in this case, stadium) pressure was 840 mb, by descending to sea level, the pressure would increase, or at the stadium would be [840 + 160] mb or 1000 mb. |
(4 pts)
How does this sea level pressure that you calculated compare with the standard sea level pressure?
| The sea level pressure below Denver on this particular day (1000 mb)
is roughly equal to a typical value of sea level pressure and it was 13 mb less than standard sea level pressure (1013 mb). |
(3 pts)
6. Convert the following temperature readings:
| 59 deg F = 25 deg C = 298 K
-40 deg C = -40 deg F = 233 K 263 K = -20 deg C = - 4 deg F Note: Be careful of signs! If the negative sign does not appear in your answer where appropriate, the answer is not correct. |
(6 pts)
7. The record high temperature for Madison, WI was 107°F (41.7°C) on 14 July 1936, while the record low was -37°F(-38.3°C) on 30 Jan 1951. What is the range of extreme temperatures?
| Range = (High - Low) = 107 deg F - (-37) deg F = 144 deg F. |
(1 pt)
The following values were obtained from Appendix IV of your textbook.
| Record High | Record Low | Range = (High - Low) | |
| United States | 134 deg F or 56.7 deg C |
79.8 deg F or -62.1 deg C (includes Alaska) -69.7 deg F or |
213.8 deg F or 118.8 deg C |
| World | 136 deg F or 57.8 deg C |
-129 deg F or -89.4 deg C |
265 deg F or 147.3 deg C |
(6 pts)
9. The National Weather Service at Madison reported the following information for individual days during this past January. The "normal" high and low temperatures for these days are also included and represent the 30 year averages for the 1961-1990 climatological interval.
|
DAY |
Observed |
Normal |
| 17 Jan 1997 | [5 + (-11)]/2 = -3 deg F | (24 + 7)/2 = 16 deg F |
| 20 Jan 1997 | (33 + 11)/2 = 22 deg F | (24 + 7)/2 = 16 deg F |
(12 pts.)
a.) Actual Heating Degree Day Units:
| HDDU = (65 deg F - Average daily temperature) |
| 17 Jan 1997: 65 deg F - (-3 deg F) = 68 HDDU |
| 20 Jan 1997: 65 deg F - 22 deg F = 43 HDDU |
b.) Normal HDDU
| 17 Jan: 65 deg F - 16 deg F = 49 HDDU |
| 20 Jan: 65 deg F - 16 deg F = 49 HDDU |
c.. Qualitatively, how would the amount of energy required for space heating on each of those dates compare with that of the climatological (or "normal") average for the corresponding dates? Explain your reasoning.
| Since 17 Jan 1997 was a day that was cold (daily average was -3 degrees F) or "below normal" temperature-wise, 68 HDDU were accumulated as compared to the "normal" of 49 HDDU, more energy than normal would have to be consumed to heat your house. However, three days later on 20 Jan 1997 the daily average of 22 deg was above "normal" meaning less heat than normal would be needed, since 43 HDDU were accumulated as compared with the typical 49 HDDU. |
(4 pts.)
Last revision 20 September 1997
Produced by Edward J. Hopkins, Ph.D. Department of Atmospheric and Oceanic Sciences University of Wisconsin-Madison Madison, WI 53706 hopkins@meteor.wisc.edu
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