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Materials: Calculator, Ruler, Colored Pencils Section 1-Introduction (Pre Lab) Recall that a sounding is a set of observations measured by a radiosonde, which is tethered to a weather balloon and launched through the atmosphere to obtain vertical profiles of temperature, dew point temperature, and winds. These observations are often plotted on a skew-T diagram, which is especially designed to facilitate calculations of how air parcels' temperature and moisture content change with height as the air parcels ascend and descend in the atmosphere. below. Figure 2 below shows a skew-T that records a small dataset of temperature and dew point observations, which The skew-T derives its name from the temperature axis, which is skewed to the right at a 45° angle as shown in Figure 1 are given in Table 1 below. 100 400 Below 500 800 700 Abov 800 900 1000 m O 10 20 30 40 ℃ Figure 1. Pressure and temperature lines on a skew-T diagram. 200 300
8 Figure 1. W Jun 200 300 400 500 500 Dew Point 700 800 900 1000 mb 40 -30-20-10 0 10 20 30 40℃ Figure 2. Sample temperature and dew point temperature observations plotted on a skew-T diagram. Table 1. Selected observations from a Greensboro, NC sounding from 12UTC on 4 December 2002 Pressure. Height above sea level Dew point temperature (mb) (m) Temperature (°C) (C) 999 270 -2.9 -18.9 925 877 -6.9 -34.9 850 1545 -0.5 -4.2 700 3094 -5.3 -5.6 500 5700 -13.7 -15.1 NOTE: In saturated air parcels, the temperature equals the dew point temperature, while in unsaturated air parcels, the temperature is greater than the dew point temperature. An inversion is a layer where temperature increases with height 1. Plot the observations in Table 2 below onto the blank skew-T diagram in Figure 3 below. Use red colored pencil for temperature and green colored pencil for dew point temperature. Table 2. Selected observations from a Greensboro, NC sounding from 12UTC on 5 December 2002 Pressure Height above sea level Temperature Dew point temperature (mb) (m) ("C) ("C) 985 270 -1.1 -1.1 925 774 -2.7 -2.7 850 1459 6.2 3.1 700 3030 -0.3 -0.3 500 5660 -13.5 -14.2
N tr structure pare() Figure 3. Blank skew-T diagram to plot the observations from Table 2. 2. On Figure 3 above, which levels are saturated, and which levels are unsaturated? Saturated Levels: Unsaturated Levels: 3. On Figure 3 above, which layer(s) have inversions? Layer(s) with inversions: 4. Comparing Figure 2 to Figure 3, how did the surface pressure AND surface temperature change from 4 December to 5 December 2002 in Greensboro? 5. Given the surface temperature in Figure 3, what type of precipitation was likely at Greensboro on 5 December 2002? Explain your answer. 6. How critical was the surface temperature forecast to citizens of Greensboro, NC on 5 December 2002? Would a forecast error of 3°C, either cooler or warmer, have a significant impact on the weather forecast? If so, how? 46
Temper 20 40 section 2-Forming Clouds Through Mixing (in Lab) One way to bring the air to saturation and, thus, to form clouds, is to mix two different air masses. On a typical day, the troposphere contains multiple vertical layers of air that can be traced back to different locations of origin. For example, summer days producing large hall and extensive wind damage in the Northeast U.S. often contain a layer of very dry air in the middle to upper troposphere that was originally lifted over the Rocky Mountains and then advected across the continental U.S. The injection of this dry air into an otherwise moist environment increases the instability of the atmosphere and can produce large thunderstorms with strong vertical motion In a more ordinary sense, the mixing of different layers of air-one warm and moist, and the other cool and dry-can saturate the air and produce clouds. To understand this process, we need to introduce a new measure of moisture in the atmosphere known as the mixing ratio. The mixing ratio is simply the mass of water vapor in the air in grams divided by the mass of the rest of the air (also known as the mass of dry air) in kilograms. Thus, the mixing ratio is a measure of the actual water content of the air. One can easily use the mixing ratio, along with the temperature, to find the relative humidity-the relative water content of the air. From the temperature of the air, we can find the saturation mixing ratio, which is the mixing ratio value at which the air will become saturated. The saturation mixing ratio is roughly the "upper limit" of water vapor content in the air. If the mixing ratio is greater than the saturation mixing ratio, the air is called supersaturated. If the mixing ratio is less than the saturation mixing ration, the air is unsaturated. Table 3 and Figure 4 below show the values of saturation mixing ratio given the temperature. Table 3. Saturation Mixing Ratio (g/kg) at Sea-Level Pressure as a Function of Temperature ("C). Bán khong tone Dyb 10140 16 11.500 18 13962 20 14956 22 MMI 24 19.210 26 21.734 24357 Unsaturated 30 27.094 31213 15.134 30.500 10 20 30 Temperature (C) Figure 4. Saturation Mixing Ratio (curved line between Supersaturated and Unsaturated regions) -35 -10 21 30 -10 -10 44 42 10 * 4 12 14 GIIN 419 0318 0310 0.7M 0911 1.100 1.300 1329 178 1.009 2,450 2012 130 1419 400 5320 5.80 6771 7.763 682 Mixing Ratio (g/kg) $ Low Vertical struct 6 • 10,000 47
C Mixing (kg) 1. Now we will consider two air masses, A and B, which you can think of as different, uniform layers in the atmosphere. Given the temperature values below for these air masses, use Table 3 above to find their saturation mixing ratios (m.). Air Mass A 8 Temperature 10°C Relative Humidity 75% A: m₂ = ./kg m₂ = a/k The relative humidity is given by the mixing ratio divided by the saturation mixing ratio times 100%, that is RM- mixing ratio saturation mixing ratio -100% 2. Re-arrange the equation above to find the mixing ratio for air masses A and B given above. A: ma g/kg B: m- e/kg 3. Plot a point representing each of these air masses from question 2 on Figure 5 below. 294013 20 Temperature (C Figure 5. Mixing Ratio vs. Temperature. The curved line represents the saturation mixing ratio. M. Figure 1 Vertical struck. Lowerten. 38°C 75% -30 10,000 24.0 48
Let's assume that equal parts of these two air masses mix together. Therefore, the resulting temperature and the mixing ratio will be halfway between the values from above. This will result in a new relative humidity value. Find these values below. 4. What is the new temperature? (Include units!). 5. What is the new saturation mixing ratio? (Use Table 3.) (include units!) 6. What is the new mixing ratio? (Average the mixing ratios in question 2.) (include units!) 7. What is the new relative humidity? (Use equation above.) (Include units!) 8. Plot the new mixed air sample on Figure 5 above. 9. Is the new sample supersaturated or unsaturated? 10. One other example of air mixing to produce clouds is contrails-cirrus-level clouds produced by the mixing of water vapor from warm, moist jet exhaust with cold, dry surrounding air. Some contrails disappear quickly, while others linger. Given the exercise you just completed, what do you think determines the lifetime of a contrail? Section 3-Adiabatic cooling More common than air masses mixing, the lifting of warm, moist air is the primary cause of cloud formation- particularly cumulus clouds- in the troposphere. An air parcel expands and cools as is rises. As an unsaturated air parcel rises, its temperature cools at a constant rate of about 10°C per kilometer. (Recall that 1 km 1000 m). This rate of cooling due to lifting is known as the dry adiabatic lapse rate (DALR). 1. Calculate the temperature of an unsaturated air parcel at 100-m increments as it is forced to rise from the earth's surface. Assume that the parcel's surface temperature is 35°C. Start at the bottom and work your way up. Height (m) Temperature (°C) 1000 m (1 km) 900 m 800 m 700 m 600 m 500 m 400 m 300 m 200 m 100 m Surface 35°C 49
just as the temperature decreases as an unsaturated air parcel rises in the troposphere, the dew point also decreases at a steady rate of 2°C per kilometer as an unsaturated air parcel rises in the troposphere. Because the temperature is decreasing at a much faster rate than the dew point as the parcel rises, at some height, the temperature will be equal to the dew point, and the air parcel will become saturated. The elevation at which saturation occurs is known as the lifted condensation level (LCL). The lapse rate (in kilometers) can be found by dividing the difference between the surface temperature and dew point by the difference in lapse rates, which is 8"C/km, or: LCL (km)= T(C)-T₂(C) 8°C/km In meters, the LCL is: LCL (m) - 125 [T (°C)-T₂(°C)] 2. Calculate the LCL for the two examples below. Surface Temp (T) Surface Dewpoint (Ta) LCL Example A 32°C 20°C 10°C Example 8 32°C 3. What do the above calculations tell you about the relationship between surface dew point and cloud height, provided the temperature is constant? 4. Fill in Table 4 below to show the temperature and dew point of a rising air parcel up to the LCL. For the temperature, you should use a DALR of 10°C/km, and for the dew point, use a dew point lapse rate of 2°C/km. Table 4. Temperature and dew point of an ascending air parcel. Work up from bottom to top. Height (m) Temperature (°C) Dew Point Temperature ("C) 2000 1500 1000 500 Surface 30 °C 14 "C Above the lifting condensation level, the air parcel will be saturated. As the parcel continues to rise, it will continue to expand and cool, which will cause water vapor to condense within the parcel. This condensation process releases heat to the atmosphere, which offsets some of the cooling. As a result, rising saturated air parcels obey a different lapse rate, known as the moist (or wet) adiabatic lapse rate (MALR). The MALR ranges from 5°C to 9°C per kilometer, depending on atmospheric conditions. This variation occurs because the amount of condensation depends on both the amount of water vapor in the atmosphere and the atmospheric pressure. Putting together several of the above ideas, air parcels rise at the DALR until they reach the LCL, because the parcel remains unsaturated from the ground to the cloud base. If the parcel rises above the LCL, the lapse rate changes to MALR above this level, as the parcel is now saturated. the 5. Fill in the left column of Table 5 below. Assume the surface temperature is 28°C and the LCL is at 1.5 km. The parcel should rise at the DALR (10°C/km) up to the LCL Above the LCL, you should use a MALR of 5°C/km. 6. Fill in the right column of Table 5 below. Now assume that Parcel B's surface temperature is 10°C and the LCL is again 1.5 km. Because the air is cooler, let's assume a larger MALR of 7 "C/km above the LCL. 50
t structur er temp Table 5. Temperature of two parcels that rise above the LCL. Work from bottom to top. Parcel A Temperature ("C) Height (km) 5.0 Parcel B Temperature ("C) 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 0.5 28 °C Surface 10°C Section 4-A Mountain Example When air encounters a mountain, the air is forced to rise over the mountain. This scenario of forced lifting often creates condensation and clouds along the mountain slopes. Once the ascending air parcel reaches the top of the mountain, if it is denser than the surrounding environmental air, it will sink back down on the other side of the mountain. As you will see this, lifting process over mountains makes dramatic changes to the properties of the surface air on either side of the mountain. Let's now consider an air parcel forced to rise over a tall mountain range. We'll assume the mountains rise from sea level to an elevation of about 2500 meters (about 8200 feet). The temperature at the base of the windward side of the mountain is 25°C and the dew point is 13°C. The temperature will initially decrease at the DALR and the dew point will decrease at the dew point lapse rate. Fill in one level on the windward side of the mountain at a time, until the temperature and dew point are equal. This is the LCL. Above the LCL, the temperature will fall at a MALR of 5°C/km, and the dew point will decrease at the same rate as the temperature to the top of the mountain. As for the other, sinking air side of the mountain (the leeward side), the air parcel will contract and warm as it sinks, so the air parcel remains unsaturated. Going down the leeward side from the peak of the mountain, the temperature increases at the DALR and the dew point increases at the dew point lapse rate. 1. Use the above paragraph and Figure 6 below to fill in the spaces for temperature and dew point as the air rises over the mountain and sinks on the leeward side. Assume a MALR of 5°C/km. Start at the windward side bottom, work your way up to the top, then go to the leeward side and work your way down. 51
2500m 2000 1500 m 1000m 500 m Windward Side To Figure 1. Vertical str Lowe 2000 1000m Leward Side T T 25° 130 Sele T T T₂ Figure 6. Temperature and dew point changes as air is forced over a mountain. 2. At what height would cloud bases form? meters 3. How do sea-level temperature and dew point on the leeward side compare to their values on the windward side of the mountain? (( 4. Which side of the mountain is more often cloudy and which side is more often clear? Section 5-Atmospheric Stability (Extra Credit) The reason we use the moist and dry adiabatic lapse rates is ultimately to understand the behavior of air parcels in the troposphere-will they rise or sink? In other words, we are looking to determine the stability of the atmosphere with respect to vertical motion. To determine the stability of the atmosphere at any given time, we must measure the observed lapse rate in the atmosphere-the environmental lapse rate (ELR)- which is most frequently achieved with a weather balloon but can be estimated with measurements from satellites as well. We can compare the ELR to the DALR and the MALR to determine the stability of the air. Essentially, what we are doing is comparing the density of a lifted air parcel with the surrounding air to determine whether that lifted air parcel will rise, sink, or remain in place. 1. Circle the correct answers. If we find that the temperature of a lifted air parcel is warmer than the surrounding air, the air parcel is (more/less) dense than the surrounding air, and it will (rise/sink). 2. Circle the correct answers. If we find that the temperature of a lifted air parcel is colder than the surrounding air, the air parcel is (more/less) dense than the surrounding air, and it will (rise/sink). 52
Lower Vertical stru 10,00 New no enpiper ing the principles from the above two questions, we can determine that if the lapse rate of the air is larger than the DALR of 10°C/km, the air parcel will ALWAYS cool at a slower rate as it rises than the surrounding air, and so it will ALWAYS be warmer than its surrounding environment, and the air parcel will continue to rise until the environmental lapse rate decreases below the DALR. This environment, where the EUR is greater than the DALR, is called absolutely unstable or superadiabatic. Superadiabatic lapse rates are rare in the atmosphere, but when they occur, they are usually found in a shallow layer from the surface up to maybe 1 km. Similarly, if the lapse rate is less than the MALR, which varies from 5°C/km to 9°C/km (a typical value is about 6"C/km), then a lifted air parcel will ALWAYS be cooler than the surrounding air and will sink. This environment, where the ELR is less than the MALR, is called absolutely stable. Inversions-layers where the temperature increases with height-are very stable layers in the atmosphere that strongly resist air rising above them. The stronger the inversion, the more stable the air. While it is possible for the air to be absolutely stable or absolutely unstable, most of the atmosphere usually has lapse rates that lie between the DALR and the MALR. This is known as conditionally unstable. If the air is saturated, it will be unstable compared to the MALR, but if the air is unsaturated, it will be stable compared to the DALR...hence the term "conditionally unstable". Earlier this semester, we discussed radiation inversions, which occur at the surface overnight as the surface cools. Another type of inversion is known as a subsidence inversion, which is produced by the sinking of very dry air aloft. The subsidence inversion occurs above the surface, usually quite close to the height of the cloud base. This is often somewhere around 850 mb, but subsidence inversions can occur higher in the troposphere as well. Subsidence inversions are rarely found below 900 mb-near-surface inversions are typically either radiation inversions or frontal inversions (which we will discuss later this semester). A key to identifying subsidence inversions is that the temperature and dew point lines suddenly split very far apart just above the inversion, indicating dry air. C Top temp-bottom Figure 7, which is shown on page 12 in color, shows a typical subsidence inversion, which is located around 775 mb. First, we are going to determine the lapse rates of several different layers in the atmosphere. The observed lapse rate is and has units of "C/km. Note that the heights given by the following equation: Lapse rate- for some pressure levels are given in black to the right of the pressure numbers in blue. You will find the temperatures using the blue lines on the diagram which are slanted to the right at a 45° angle. Top height-Bottom height 14°C 25°C 1554-153 IC 1.401 km -7.85°C/km. Assuming a MALR of 1. The lapse rate in the bottom layer is: LR- 6°C/km, is this air absolutely stable, conditionally unstable, or absolutely unstable? Circle: Stable Conditionally unstable Unstable 2. Calculate the lapse rates and determine the stability condition of each of the following layers. Show your work for each calculation. Assume a MALR of 6°C/km for each layer. 850-700 mb. Show your work: Lapse Rate Circle: Stable Conditionally unstable Unstable 53
oor ao to on and on to as a sa dewao SAGIRA RODINOUJO. 500 mb. Show your work: Lapse Rate= Circle: Stable Conditionally unstable Unstable 500-400 mb. Show your work: Lapse Rate = Circle: Stable Conditionally unstable Unstable 3. Rank these four layers from most stable to least stable: Most stable: Least stable: 4. The word "subsidence" means "sinking". Which layer(s) are likely to have sinking air, based on your lapse rate calculations? *
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