Purpose: The purpose of this exercise is to help the student become familiar with various ways of measuring atmospheric moisture, with the ways of

Transcription

1 Purpose: The purpose of this exercise is to help the student become familiar with various ways of measuring atmospheric moisture, with the ways of expressing atmospheric moisture content, and with the relationships among the expressions of atmospheric moisture. Background: To the meteorologist, one of the most important gases in the atmosphere is water vapor. It not only provides the moisture supply for clouds and precipitation, but it also plays a vital role in atmospheric dynamics because of the latent heat exchanges involved in phase changes of water, and because of its role in atmospheric radiation processes; such as, absorption and emission of some wavelengths of radiation while allowing other wavelengths to pass affects the earth's energy budget. The five elements which are most often used to describe the water vapor content of the atmosphere are: mixing ratio, vapor pressure, dew point, wet-bulb depression, and relative humidity. Others exist, such as specific humidity, but these are the expressions most commonly used. Since all of these expressions describe the moisture of the atmosphere, certain relationships exist among them. The thermodynamic diagram, among other methods, can be used to obtain these relationships. Instruments that are used to measure atmospheric humidity are collectively called hygrometers. Part I. Temperature Water Vapor Relationships The relationship between temperature and the amount of water existing in a vapor state is usually determined under a specific set of conditions. Assume that you have a container in which a pan of liquid water is placed. The surface of the liquid water in the pan is flat, not curved. The water molecules will begin to evaporate from the pan. As time goes on, some of the water vapor, molecules that had evaporated will condense back onto the liquid water surface. There will be a back and forth evaporationcondensation process occurring at the surface of the liquid water in the pan. Eventually, there will occur an equilibrium condition in which the amount of evaporation equals the amount of condensation that is occurring. The number of water molecules in a vapor state above the liquid water surface is sufficient to cause the equilibrium (evaporation = condensation) condition to exist. This is the state of saturation and it is dependent, in part, on the temperature of the environment in which this evaporation condensation process is occurring. Notice, nothing was said about any other gas molecules above the liquid water surface. They play no role in the process. These other gas molecules; e.g. nitrogen, oxygen, etc. can exist above the liquid water surface or not and the amount of water vapor molecules necessary to achieve equilibrium conditions will be the same. These water molecules in a vapor state will exert a pressure, just as nitrogen and oxygen exert a pressure. This pressure exerted by the water vapor molecules in a vapor state is called the water vapor pressure and it contributes a partial pressure to the total pressure of the atmosphere. A partial pressure is the pressure exerted by a particular gas in the atmosphere. Thus there is a pressure that is exerted by only the nitrogen molecules, a pressure exerted by just the oxygen molecules, and a pressure exerted by each of the other gas molecules in the atmosphere. Put together, they add up to the total pressure of the atmosphere. If equilibrium conditions (evaporation = condensation) exist, then the partial pressure exerted by the water molecules in a vapor state is called the saturation vapor pressure (e s ). If equilibrium conditions do not exist, then the pressure being exerted by the water vapor 95

2 molecules is called water vapor pressure (e). The relationship between temperature and the number of molecules required to be in a vapor state above the flat, liquid water surface in order to achieve equilibrium (saturation) is not a linear relationship, however, but an exponential one, in that the amount of water necessary to exist as vapor to achieve equilibrium increases at an increasing rate as the temperature increases. To show this graphically, plot the data from table 1 on graph 1. The independent variable is normally plotted on the abscissa (x-axis) and the dependent variable is normally plotted on the ordinate (y-axis). Thus, plot the temperature along the x-axis and the water vapor pressure along the y-axis. Table 1. Saturation Vapor Pressure C hpa C hpa C hpa C hpa Water molecules in vapor form are not bonded together, but move freely from each other. In a liquid state, the water molecules are bonded together in long chains by ionic bonds holding the molecules in the chain form. Table 1 shows the pressure that water vapor molecules will exert, at the indicated temperature, when saturated conditions exist under the above conditions; i.e., in relation to the flat water surface. For curved surface, such as with cloud droplets and rain drops; then factors such as curvature of the drops, and the purity of the drops play a role. Also, the saturation conditions; amount of water vapor molecules necessary to achieve equilibrium (evaporation = condensation) changes if the molecules are evaporating from and condensing onto a ice surface, such as an ice crystal (snowflake). 96

3 Some other things to understand. In order for molecules to leave the pan as liquid water and become water vapor (evaporation), heat must be added to the molecules. For each gram of liquid water that becomes water vapor, 540 calories of heat must be added to the molecules when the temperature is at 100 o C (212 o F). If the temperature were at 0 o C, 600 cal/gm of heat energy must be added. This heat is called the Latent Heat of Evaporation. When molecules change from water vapor back to liquid water (condensation), the same amount of heat (between 540 calories per gram and 600 calories per gram) is released. We then call it the Latent Heat of Condensation. It is called latent because it is, in a sense, hidden; adding or removing this heat does not change the temperature, but only serves to change the phase, of the molecule (liquid phase or vapor phase). When water freezes, the liquid water molecules release 80 cal/gm of energy and when ice melts, 80 cal/gm of heat is added to the ice to change the phase of the solid water back to liquid. This is called the Latent Heat of Freezing and Latent Heat of Melting. Problem 1. Problem 2. Boiling Plot the values of saturation vapor pressure corresponding to the temperature values given on graph 1 on the answer sheet (every 5 degrees) Draw a line connecting the dots you have plotted on graph 1. Does the graph show a relationship in which the values of the ordinate (y-axis) increase at an increasing rate as the abscissa (x-axis) values increase? Record you answer on the answer sheet. If this is not apparent, calculate the differences between successive ordinate values. What does this say about the ability of water to exist as a vapor as the temperature increases? Record your answer on the answer sheet. Boiling of a liquid occurs when the atmospheric pressure equals the saturation (equilibrium) vapor pressure. In other words, P (atmospheric pressure) = e s. The following formula expresses the temperature at which boiling of water would occur at a particular elevation assuming the atmospheric pressure changes according to the standard atmosphere. T Boiling = L v Rv ( z 7.29km) Problem 3. At sea level where the average pressure is millibars, water boils at 100 o C (212 o F). So, an egg being boiled is cooking in water of 100 o C when the water is boiling. Denver has been called the mile high city because its official elevation which is marked on the steps of the Colorado State Capitol is exactly one mile (1.609 kilometers). The term L v is the Latent Heat of Vaporization which varies with temperature. The term R v is the gas constant for water vapor. As an example, we will use typical values for these so that the term L v /R v is equal to 5423 o K. Determine at what temperature (in o C and o F) would boiling occur, on the average, at Denver, Colorado. Since the temperature is cooler, you can see why it takes longer to boil an egg there. 97

4 Part II. Dew point (Td) Recall the definition of dew point: the temperature to which air must be cooled at constant pressure to produce saturation. You may have noticed that on a humid day condensation will form on a glass of any cold beverage. The cool liquid cools the glass and the air in contact with the outside of the glass. If the liquid is cold enough, the glass will cool to the dew point of the air and then the air in contact with the glass will also cool to the dew point. Further cooling will produce condensation on the outside of the glass since the water vapor condition of the volume next to the glass is one of saturation, so further cooling will cause some of the water vapor to convert to liquid water. An important point here is that water vapor more easily condenses on an object, so you see the liquid condensing on the glass. In the atmosphere, water vapor usually condenses on dust particles, pollen grains, smoke ash, and such. These are called condensation nuclei. In normal circumstances, the dew point cannot exceed the air temperature. If the air cools to the dew point and continued cooling occurs, the dew point will drop as water condenses. Thus, the dew point drops at the same rate as the air temperature and water condenses at a rate fast enough to keep the dew point the same as the air temperature. If the air temperature increases, the dew point is not affected, since the dew point changes only with changes in the amount of actual water vapor in the air. A large difference between the dew point and the air temperature (T- Td)(dew point depression) means conditions are far from saturation, the volume is relatively dry in relation to the amount of water vapor that could be in the volume as determined by its temperature. A small dew point depression means the volume is close to saturation. Part III. Psychrometer The most widely used instrument for determining the water vapor content of the atmosphere is the psychrometer. The word psychros is the Greek word for to cool. It consists of a wet-bulb thermometer and a dry-bulb thermometer. The dry-bulb thermometer measures the ambient air temperature. The wet-bulb thermometer measures the temperature resulting from evaporation of water. If only a small amount of water exists in vapor form in a volume, evaporation of water will occur relatively rapidly and the wet-bulb thermometer will cool by a large amount. There will be a large difference between the dry-bulb thermometer reading, (T), and the wet-bulb thermometer reading, (Tw). This difference is called the wet-bulb depression (T-Tw). If the amount of water vapor in a volume is close to saturation conditions, very little water will evaporate, the evaporation rate will be slow, and the thermometer will not be cooled very much. The wet-bulb depression will be small. The wet-bulb thermometer becomes cooler with evaporation because for every gram of liquid water, approximately 540 calories of energy is required to convert that gram of liquid water to water vapor. You notice the cooling effects of evaporation when alcohol is rubbed on your skin. As the alcohol evaporates, your skin will become cooler. Thus, a large difference between the wet-bulb and dry-bulb thermometer readings means that much evaporation occurred and only a small amount of water existed as vapor in the volume compared to the amount that could have existed as vapor (which is based on the temperature). The kind of psychrometer you will use in this part of the exercise is a sling psychrometer. The sling psychrometer consists of a bracket to which two thermometer are attached. The wet-bulb thermometer has a piece of cotton wicking attached to the bulb. A short handle extends from the bracket so the thermometers can be whirled allowing air to flow across the thermometers. When whirling the psychrometer, be certain you do not strike anything. Following the procedure below, determine the dry-bulb temperature, the wet-bulb temperature and the relative humidity in this classroom, the hallway on the first floor, and outside on the lawn. 98

5 Procedure: 1. Using the bottle of distilled water, saturate the wick with water. 2. Whirl the psychrometer with a steady motion for about 20 seconds and then read both thermometers. Whirl the psychrometer again and make another reading. The dry-bulb temperature should be the same each time, but the wet-bulb temperature will approach some minimum value. Continue whirling and then reading the thermometers until the wet-bulb thermometer reading ceases to fall. You may assume that this wet-bulb reading has stabilized when you get the same reading twice. If the wet-bulb temperature goes up instead of down, re- soak the wick and start again. Record these last dry-bulb and wetbulb thermometer readings in o C in table 5 on your answer sheet. 3. From table 2, determine the relative humidity using the dry- bulb and wet-bulb temperature readings in o C. 4. Use table 3 to determine the dew-points. 5. Repeat steps 1 to 4 for the first floor of this building and outside on the lawn. 99

6 Table 2. Relative Humidity (%) Air Wet-Bulb Depression (Dry-bulb temperature minus Wet-bulb temperature)( o C) Temp ( o C) Continued on next page 100

7 Table 2. Relative Humidity (%) Air Wet-Bulb Depression (Dry-bulb temperature minus Wet-bulb temperature)( o C) Temp ( o C)

8 Table 3. Dew Point ( o C) Air Wet-Bulb Depression (Dry-bulb temperature minus Wet-bulb temperature)( o C) Temp ( o C) Continued on next page 102

9 Table 3. Dew Point ( o C), continued Air Temp Wet-Bulb Depression (Dry-bulb temperature minus Wet-bulb temperature)( o C) ( o C)

10 Part IV. Mixing Ratio (w) Another way to express the amount of moisture is to look at the actual mass of water molecules existing in vapor form. The mixing ratio (ω) is defined as the mass of water vapor in a volume of air (Mv) divided by the mass of the dry air (Md) in the same volume. Or, " = M v M d... (1) The mixing ratio may also be obtained from the weight of water in the air since, ( )( g) ( M d )( g) Weight of water in a volume of air " = Weight of dry air in the volume = M v where, g = the acceleration of gravity. The density of water vapor and dry air can also be used to obtain the mixing ratio since, " = Density of water in a volume of air Density of dry air in the volume = # M v v = Vol = # d M d Vol ( M v) M d The units used for mixing ratio are usually grams of water vapor per kilogram of dry air. Thus, 3 grams of water vapor to 1000 grams of dry air (1 kilogram) would be expressed as 3 gm/kg. For a certain temperature and pressure, there is a maximum amount of water that can exist as vapor. This maximum amount of water vapor can be expressed as the saturation mixing ratio (ωs). When saturated conditions exist, ω = ωs. Values of the saturation mixing ratio for selected temperatures and pressures are given in table 4. The dew point is also related to the mixing ratio. If the ambient temperature were the same as the dew point, saturated conditions would exist. Then the actual mixing ratio would be the same as the saturation mixing ratio and the relative humidity would be 100%. Thus, we can say that the dew point is the temperature to which a parcel of air would have to be cooled so that the actual mixing ratio is also the saturation mixing ratio. If the actual mixing ratio is known and if the pressure is known, one can use table 4 to find the temperature at which the actual mixing ratio is the same as the saturation mixing ratio. This temperature is the dew point. Table 4 shows the saturation mixing ratio values for various temperatures expressed in units of grams per kilogram (g/kg). ( ) As mentioned above, dew point relates to the actual amount of water in vapor form. Thus, if we look at the temperature values of table 4, or table 1, as dew points, then the moisture value in g/kg (table 4) or in hpa (table 1) represents the actual amount of water vapor molecules in the vapor state in a kilogram of air. Problem 4. How many grams of water vapor are added to a kilogram of air when the dew point is increased from -15 o C to -10 o C? If the air temperature were 25 o C, it would require about Joules of energy added to each gram of water just to get all of it to evaporate. This energy is contained in the water vapor molecules as latent heat energy. Latent heat does not cause a change in temperature, only a change in state of the water molecules. How much energy, then, was added to the volume of air by evaporating 104

11 water into it to change its dew point from -15 o C to -10 o C, assuming Joules of energy were used to evaporate each gram of water. Record your answer on the answer sheet. Problem 5. How many grams of water vapor are added to a kilogram of air when the dew point is increased from 15 o C to 20 o C? How much energy, then, was added to the volume of air by evaporating water into the volume to change its dew point from 15 o C to 20 o C, again assuming the air temperature were still 25 o C and Joules of energy was needed for each gram of water to get it to evaporate. Record your answer on the answer sheet. Table 4. Saturation Mixing Ratio For a pressure of 1000 mb C g/kg C g/kg C g/kg C g/kg From the values you should have obtained from the answer to problems 4 and 5, you can see that considerably more moisture must be added to a kilogram of warm air than to cold air to raise the dew point by 5 o C. Part V. Relative Humidity (RH) Relative humidity is probably the expression most widely used by the general public to indicate the amount of moisture in the air. It is a ratio (expressed in percent) of the actual amount of moisture in vapor form to the amount of moisture at saturated conditions for a given temperature and pressure. Relative humidity can be obtained easily from several of the expressions we have already covered. 105

12 The relative humidity is related to the mixing ratio and the saturation mixing ratio by: Relative humidity = Actual mixing ratio Saturation mixing ratio X 100. Or, RH = ω ωs X (2) Relative humidity may also be obtained from pressure considerations, as: Relative humidity = Vapor pressure Saturation vapor pressure X 100 Or, RH = e es X (3) From the above expressions, if the relative humidity and the saturation mixing ratio are known, then the actual mixing ratio can be determined; or if the relative humidity and the saturation vapor pressure are known, then the actual vapor pressure can be determined. Relative humidity is related to temperature because the saturation mixing ratio and the saturation vapor pressure are both related to temperature. As temperature changes, the amount of moisture required for saturated conditions will change, so both saturation mixing ratio and saturation vapor pressure change. We see this on a daily basis. At night when cooling occurs, the relative humidity normally increases; not because there is an increase of water vapor, but simply because the temperature is cooler so the amount of moisture required for saturation has decreased. In the above equations, the denominator becomes smaller so the value of relative humidity must increase. During the day, when the warming occurs, the relative humidity decreases because now more vapor is required for saturation. In neither case does the actual amount of vapor change, only the amount that is required for saturation changes. Problem 6. Problem 7. Problem 8. Problem 9. In light of the above discussion, for the three locations where you have psychrometer data, use the temperature (T), wet-bulb temperature (Tw) and relative humidity (RH) to find the saturation mixing ratio, the actual mixing ratio, vapor pressure, saturation vapor pressure, and the dew point and complete the table on the answer sheet. Although the saturation mixing ratio changes with atmospheric pressure as well as with temperature, we will assume the atmospheric pressure equals 1000 hpa. List at least three errors that were discussed in class that may have resulted from the sling psychrometer method to cause the value you obtained for the dew point to be higher or lower than the true dew point value. Record your answers on the answer sheet. Which of these elements of the air (temperature, wind speed, water vapor content, cloud cover, pressure) would cause the dew point to be different between the three locations? Record your answer on the answer sheet. Which of these elements of the air (temperature, wind speed, water vapor content, cloud cover, pressure) would cause the relative humidity to be different between the three locations? Record your answer on the answer sheet. 106

13 Problem 10. Problem 11. As you compare the actual mixing ratio values going from classroom air to first floor air to outside air, the values either increase or decrease. Do the actual vapor pressure values respond in the same manner as the actual mixing ratio values? Changes in what element of the atmosphere is causing the actual values to change? Record your answer on the answer sheet. Does the saturation mixing ratio and saturation vapor pressure values respond in the same manner as the actual mixing ratio values and actual vapor pressure values when you compared them in problem 10? To what element of the atmosphere does the saturation mixing ratio values and saturation vapor pressure values respond? Record your answers on the answer sheet. The following shows some examples of using tables 1 and 4 and the equations for relative humidity to determine various elements of the atmosphere. Example 1. Given: ω = 8 g/kg, ωs = 20 g/kg Find: RH Solution: RH = = ω ωs 8 20 X 100 X 100 =40% Example 2. Given: RH = 60%, e = hpa Find: es Solution: 60% = hpa es X % = = hpa es es hpa (100)(27.24 hpa) 60 = es = 2724 hpa 60 = hpa Find: T, Td Solution: To find T, if es (saturation vapor pressure) = hpa, then from Table 1, the air temperature must be between 31.0 o C and 32.0 o C. We can set up the following table to interpolate the value for T. 107

14 On the left, the difference between the temperature of 31.0 o C and the temperature we are looking for, T, is an unknown value (T 31.0 o C). We ll put a question mark there. The difference between 32.0 o C and 31.0 o C is 1.0 o C. On the right, the difference between 44.9 hpa and the value (45.4 hpa) associated with the temperature we are looking for is 0.5 The difference between and 44.9 is We can then write the following:? 1 = Meaning that our unknown value is to 1 (bears the same relationship to) as 0.5 is to Then, we cross multiply to determine the value of out unknown (?). This value is the difference between 31.0 o C and the temperature value we are looking for. It turns out to be Then, since as one goes from 31.0 o C to 32.0 o C the values are increasing, we should add 31.0 o C and 0.19 o C to get the value for T, or o C. To find Td, refer to table 1. If e (actual vapor pressure) = hpa, half way between HPa and hpa, the dew point must be 22.5 o C, half way between 22.0 o C and 23 o C. In other words, if the temperature were 22.5 o C, the air would be saturated and the water vapor would exert a pressure of hpa. Problem 12. Given: e = 8.71 hpa, es = hpa. Find: RH, Td, T, Tw Record your answer and computations on the answer sheet. Problem 13. Problem 14. Given: es = hpa, Td = 9.5 o C Find: RH Record your answer and computations on the answer sheet. Given RH = 32%, e = hpa Find: T Record your answer and computations on the answer sheet. Problem 15. Given ω = 7.21 gm/kg, RH = 39% Find: ωs, T, Td Record your answer and computations on the answer sheet. Problem 16. Can Td ever equal Tw? Explain. Record your answer and computations on the answer sheet. 108

15 Figures 1 and 2. The two figures show the dew points at 00Z on August 1, 2006, for the northwestern United States and for the southeastern United States. 109

16 Problem 17. Problem 18. Problem 19. Draw isodrosotherms on both maps at multiples-of-five with isodrosotherm values ending in 0 or 5. In general, how does the water vapor content of the air depend on distance from the nearby large body of water? Explain why the dew points are so different in these two regions, even though both are near large bodies of water. Part VI. Temperature, Moisture and Humans Human beings, like all organisms on the earth, can exist only within certain environmental limits; and comfortably only in more restrictive environmental limits. A certain range exists for oxygen content in the atmosphere, atmospheric pressure, temperature, moisture vapor content, etc. Some, like oxygen content, are more restrictive while others, like the amount of water vapor content are much less restrictive. Problems often arise when a combination of the factors serve to place stress on the human body. Two of these are temperature and moisture content. The human body is homeothermic (heat regulating and producing) and must maintain its body temperature in a rather narrow range. To do this, heat gain must equal heat loss. It is a common misconception that humans feel comfortable or uncomfortable due to the temperature of the air around them. Actually, the comfort or discomfort arises from the body's ability to maintain a balance of heat gain to heat loss. Humans feel a cold discomfort when heat loss is greater than heat gain, and they feel a hot discomfort when heat gain is greater than heat loss. Factors, which account for heat gain are: metabolic heat generated within the body, and radiation, convection and conduction received from sources external to the body. Factors, which account for heat loss to the body are: radiation, convection, conduction, evaporation. This can be written as: metabolic heat (M) + radiation (R+) + convection (C+) + conduction (P+) = radiation (R-) + convection (C-) + conduction (P-) + evaporation (E) where, those terms on the left side of the equation refer to processes by which heat is added to the body and those on the right side are processes by which heat is removed from the body. Figure 3 shows the various aspects, which play a role. Figure 3. The body heat balance of human beings 110

17 Humans adjust the effects of the above factors by various methods: increasing or decreasing activity to change metabolic heating, moving indoors or outdoors to change radiation effects, moving towards or away from regions of warm convection and conduction, sweating to help cool the body or putting on more clothing to warm the body. Additionally, humans change the environment in which they live by heating and air conditioning of homes, buildings, and cars. When one of these mechanisms is hindered, the body has greater difficulty in maintaining its internal temperature. Discomfort and harm to the body may be the results. When the body begins to overheat, it can cool itself by simply losing water. When a gram (1 cm 3 ) of water is evaporated completely (converted from liquid water to water vapor), approximately 2400 Joule of energy is added to the water and removed from the body as the water evaporates. The body loses water by evaporation of liquid water by sweating and by the moisture exhaled from the lungs and upper respiratory tract. At temperatures above 30 o C (86 o F) and with humidity above 50%, the breathing loss is much less than the sweating loss. If the air is saturated and at temperatures above 33 o C (91.4 o F), only sweating loss is available. At these high temperatures and humiditys, the amount of moisture taken in by breathing equals that exhaled. Exhaled breath is at about 80-90% relative humidity. At temperatures above 37 o C (98.6 o F) at high relative humiditys, evaporation from the skin is so minimal that the body cannot cool itself even by sweating. Several indexes have been developed which attempt to quantify the effects of high temperature and high atmospheric moisture content on the human body. The National Weather Service has adopted an index that has become known as the Heat Index. The heat index, we might say, is an attempt to quantify the sultriness of a set of environmental conditions and represents the extent to which humidity aggravates physiological effects of high temperature. The body reacts to the actual air temperaturerelative humidity conditions the same way it would react to the higher heat index apparent temperature - low humidity conditions. The heat index considers several factors such as; ambient air temperature, water vapor content of the air, area of skin which exchanges radiant heat with its surroundings, the average body volume for conduction transfer, clothing cover, internal body temperature, skin temperature, average activity level, etc. The results are shown in table 6, which gives the apparent temperature for a set of environmental conditions. The apparent temperature is that ambient air temperature which by itself would produce the same physiological reduction in magnitude of the body's ability to cool itself as the environmental temperature and humidity conditions together produce. Note: It is an apparent temperature, not the true air temperature. Also included in table 6 are danger categories. Due to the often misunderstanding many people have concerning true air temperature and an index giving an apparent temperature which relates how a body responds to a set of environmental conditions (principally temperature, humidity and wind), a better method of expressing the level of danger or discomfort which one may experience under a certain set of environmental conditions should be used, such as simply using the danger categories I, II, III, IV and not referring to an apparent temperature. The heat index can be calculated at the following site: 111

18 Problem 20. The average daily maximum temperature that Bryan/College Station experiences during the summer months is 95 o F (35 o C). The average daily relative humidity is 68%. What Danger Category does this put Bryan/College Station in? Record your answer on the answer sheet. Table 6. Heat Index AIR TEMPERATURE o F / o C 120/ / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / / /21.1 RELATIVE HUMIDITY (%) II I III IV DANGER CATEGORY IV EXTREME DANGER III DANGER II EXTREME CAUTION I CAUTION HEAT SYNDROME HEATSTROKE OR SUNSTROKE IMMINENT SUNSTROKE, HEAT CRAMPS, OR HEAT EXHAUSTION LIKELY; HEATSTROKE POSSIBLE WITH PROLONGED EXPOSURE & PHYSICAL ACTIVITY SUNSTROKE, HEAT CRAMPS & HEAT EXHAUSTION POSSIBLE WITH PROLONGED EXPOSURE & PHYSICAL ACTIVITY FATIGUE POSSIBLE WITH PROLONGED EXPOSURE & PHYSICAL ACTIVITY Problem 21. The relative humidity of 68% given in problem 20 is a daily average value. Highest relative humidity values occur in the early morning hours when the temperature is lowest and lowest relative humidity values normally occur in the afternoon when temperature is highest. Normal afternoon relative humidity values average about 30-35%. Assume that the afternoon relative humidity was 33%. What value would the temperature have to reach to put the Bryan/College Station area in the level III danger category? Record your answer on the answer sheet. 112

19 Problem 22. The maximum temperature recorded for Bryan/College Station was 110 o F in July. Consider figure 1 and the discussion above, list five actions mentioned in this exercise that a person could take to prevent the heat syndrome dangers. Record your answers on the answer sheet. The following two maps show the weather observations recorded for the southern plains region for 1200Z on March 22, 1994, and 1200Z on March 23, On March 22, the dew points for most of the area were relatively low. Just 24 hours later, the dew point temperatures were considerably higher. The amount of moisture in the air determines, in part, whether a person feels comfortable or uncomfortable. Generally, if the dew point temperatures are below 55 o F, a person usually feels comfortable. This, of course, also depends on the temperature of the air. If the dew point temperatures are above 65 o F, a person will generally feel uncomfortable. The air is considered muggy and unpleasant. Answer the following two questions concerning the below maps. Notice also the shift in wind direction for stations near the gulf coast and the change in total amount of cloud cover at many of the stations in Texas and Louisiana. Figure 4. Surface Observations for 12Z 22 March

20 Figure 5. Surface observations for 12Z 23 March 1994 Problem 23. Problem 24. On the March 23, 1994, map on your answer sheet, circle those stations at which the dew point temperatures on the morning of March 22, 1994, were below 55 o F (air felt comfortable) and then on March 23, 1994, the dew point temperatures were above 65 o F (people woke up to a muggy, sultry, uncomfortable morning). What type of pressure system, high pressure or low pressure, is probably located in Nebraska based on the changes in pressures that are shown on the two maps? 114

21 Part VII. Temperature, Wind and Humans Another danger to humans, although not related to moisture in the air, results from low temperatures. Humans, being homeothermic, must maintain the internal body temperature above a certain value as well as keeping it below a certain value. Wind moving past exposed skin during cold weather increases the body's heat loss. The body pumps warm blood to the extremities in an attempt to maintain the proper body temperature. However, if the temperature is low, and the wind is strong, the body often cannot keep up with heat loss, and the skin temperature decreases. The term "wind chill" was first used by the Antarctic explorer Paul A. Siple in a 1939 dissertation, "Adaptation of the Explorer to the Climate of Antarctica," to describe the cooling power of various combinations of wind and low temperature. He conducted experiments to determine how quickly water in plastic containers would freeze and the factors, which determined the rate at which freezing, would occur. He discovered that the initial temperature of the water, the air temperature to which the water was exposed (outside air temperature), and the wind speed determined the rate. His results implied that if there is no air motion then cooling does not occur, which is not the case. Also, the work was done using water and a plastic container with the assumption that the results can be extended to exposed skin. This concept has been adapted to the effects of temperature and wind on humans. The wind chill compares the effects of a particular combination of temperature and wind on the rate of cooling of the human body to the same effects resulting from temperature and a 3 mph wind blowing against a standing persons uncovered skin, such as the face; or a person walking through calm air at 3 mph. It does not relate rate of cooling by air temperature and wind for that part of the body that is clothed. An early version of the wind chill formula is written as: Where, # WC = T skin " V + V & 0 % ( $ ' V T skin = the skin temperature in o F, T air = the air temperature in o F, * ( T skin " T air ) V o = 3 mph, the average speed that people walk, V = the wind speed. When the wind speed is below 3 mph, the wind chill temperature may be higher than the actual temperature. When wind velocities are near zero, and you are standing still, your body heat warms the air near your body. This warm air near the body provides some insulation from the colder environment. The air is not moving fast enough to carry it away. As a result, it may actually feel warmer than the true air temperature. There is nothing exact about wind chill or, more appropriately, Equivalent Chill Temperature. It is an approximation, or estimate, relating how rapidly your skin is losing heat energy to the environment. The skin cools at a rate proportional to the difference between the skin temperature and the air temperature. If the skin temperature and the air temperature are the same, no further cooling will occur. The U.S. National Weather Service across the southern states computes the wind chill factor whenever temperatures dip below 40 o F and wind speeds rise above 10 mph. 115

22 The formula the National Weather Service uses to compute wind chill is: WC( o F) = T 35.75(V 0.16 ) T(V 0.16 ) where, WC = wind chill in o F, V = wind speed in statute miles per hour, T = Air temperature in o F. Again, the wind chill is an apparent temperature expressing the rate of heat loss by exposed human skin, not the actual air temperature. Table 7 below gives the wind chill equivalent temperature values. The wind chill may also be calculated at the following site. Problem 25. Problem 26. Determine the Wind Chill Equivalent Temperature if the air temperature is 20 o F and the wind speed is 25 mph. Record your answer on the answer sheet. If the air temperature is 38 o F and the wind speed is 20 mph, what is the Wind Chill Equivalent Temperature? Will a person's exposed skin eventually freeze if they are exposed to these environmental conditions for several hours? Explain why exposed skin will, or will not freeze, with these conditions. Table

23 ATMOSPHERIC MOISTURE ANSWER SHEET Section: Name Problem 1. Graph 1. Problem 2. c. Graph relationship? Ability of air to hold water vapor as air temperature increases? 117

24 Problem 3. Problem 4. Temperature that water boils. o C, o F Grams of water vapor. Energy added to air Problem 5. Grams of water vapor Problem 6. Classroom T d from beaker Energy added to air Psychrometer Use of Equations T d T T RH From w Table 3!! s e es First Floor Hall Outside on lawn Problem Table 5. Three errors of the psychrometer method Problem 8. Problem 9. Problem 10. Problem 11. Element of the air that would cause the dew point to be different. Elements of the air that would cause the relative humidity to be different. Response of actual mixing ratio and vapor pressure. Yes / No Response of saturation mixing ratio and saturation vapor pressure. Yes / No Element to which saturation mixing ratio and saturation vapor pressure responds. 118

25 Problem 12. Find RH, Td, T, Tw RH % Td o C T o C Tw o C Problem 13. Find RH RH % Problem 14. Find T T o C Problem 15. Find T ωs g/kg T o C T d o C Problem 16. Can Td ever equal Tw? Explain! 119

26 Problem 17. Analyze the surface dewpoint temperatures. 120

27 Problem 18.In general, how does the water vapor content of the air depend on distance from the nearby large body of water? Problem 19. Why are the dewpoint temperatures so different in the two regions, even though both are near large bodies of water? Problem 20. Danger Category is Problem 21. Temperature would have to be o F. Problem 22. Five actions to take

28 Problem 23. Problem 24. Type of pressure system: low pressure center or high pressure center. Problem 25. Problem 26. Wind Chill value. Wind Chill value. Yes / No Explanation 122