GEOG 2013 \u2013 Weather & Water
\nLab Assignment 4 \u2013 Humidity & Atmospheric Stability.
\nLabs due by midnight of night before your lab on BrightSpace, week of Nov. 14 – 18.
\nRequired: A scientific calculator, color pencils. Lab marked out of 92, value 8% of course mark
\nObjective: to become familiar with the concepts of humidity, vapour pressure and atmospheric stability in the atmosphere. You will be asked to graph, calculate some of these values for different air masses in a number of changing situations. YOU MUST SHOW YOUR WORK FOR ALL QUESTIONS IN THIS LAB.
\nReference: Ross, Chapter 7 & 8
\nGEOG 2013 Lectures 8, 9 & 10 (Atmospheric moisture & Atmospheric Stability 1 & 2)<\/p>\n
Part 1: Humidity (total marks \u2013 20)
\nHumidity is the amount of water vapour in the air, and can be described in a number of ways.
\n\u2022\tWater vapour pressure is the part of the total atmospheric pressure due to the amount of water vapour molecules in the air, and is expressed in mb (millibars). This is dependent on temperature and density of water molecules in the air. In terms of vapour pressure, the maximum water vapour capacity of air is expressed as the saturation vapour pressure.
\n\u2022\tMixing ratio describes the actual amount of water vapour in the air, and is expressed as the mass of water vapour in a given mass of dry air, described as grams of water vapour per kilogram of air (g\/kg). The saturation mixing ratio is the water vapour capacity of air at a given temperature.
\n\u2022\tSpecific humidity is similar to the mixing ratio, except that it describes the number of grams of water vapour per kilogram of air, including water vapour.
\n\u2022\tRelative humidity describes how close air is to saturation with water vapour. It is a ratio that compares the actual amount of water vapour in the air (the mixing ratio) to the maximum capacity of the air at a given temperature (saturation mixing ratio). Cold air has a low water vapour capacity, while warm air has a high water vapour capacity. Relative humidity is calculated with a simple formula, and can be calculated using water vapour content (mb) or mixing ration (g\/kg):
\nRelative humidity =\t Actual water vapour content in air \t x 100\t\t\t[Eq. 1]
\nMaximum water vapour capacity of the air<\/p>\n
Relative humidity = \t\tMixing ratio \t\tx 100\t\t\t[Eq. 2]
\nSaturation mixing ratio
\nYou can work with either the vapour pressure or the amount of water (mixing ratio) to calculate the Relative Humidity. The relative humidity (RH) expresses this degree of saturation as a percentage. For example, 50% RH means that the air contains half of the water vapour necessary for saturation; 75% RH means that the air has \u00be of the water vapour necessary for saturation; 100% RH means that the air is saturated. Relative humidity varies due to evaporation, condensation, or temperature changes, all of which affect both the content and the capacity of the air to hold water vapour. Saturation indicates that any further addition of water vapour (change in content) or any decrease in temperature (change in capacity) will result in active condensation; therefore, cloud formation can take place.
\nExample 1: An air mass at 15\uf0b0C has a water vapour pressure content of 10.2 mb. The saturation vapour pressure (max capacity) of the air to hold water vapour at that temperature is 17.0 mb. Therefore, the relative humidity is:
\nRelative humidity = 10.2 mb = .60 x 100 = 60%\t\t\t\t\t\t\t\t\t\t 17.0 mb<\/p>\n
Example 2: If the mixing ratio is 13.5 g\/kg, and the saturation mixing ratio is 22.5 g\/kg, the relative humidity is \t\t\t\t\t\t13.5 g\/kg x 100 = 60% RH
\n22.5 g\/kg
\n1.\tUsing Table 5-2 (below), determine the saturation mixing ratio of the following air samples, and calculate the relative humidity of each. [5 marks]<\/p>\n
Temp.\tSaturation mixing ratio (g\/kg)\tMixing ratio (g\/kg)\tRelative Humidity
\n14\uf0b0C\t\t5\t %
\n14\uf0b0C\t\t9\t %
\n24\uf0b0C\t\t5\t %
\n24\uf0b0C\t\t2\t %
\n34\uf0b0C\t\t7\t %<\/p>\n
2.\tIn the winter, cold air is brought into homes and heated. How does this change the relative humidity of the air? [1 mark]<\/p>\n
3.\tExplain why the basement of a house often has high relative humidity in the summer. [1 mark]<\/p>\n
4.\tUse Table 5-2 (below), determine the saturation mixing ratio of the following air samples and calculate the actual mixing ratio of each. Rank each of the following air samples from 1 (highest) to 5 (lowest) in order of water vapour content. [5 marks]<\/p>\n
Temp\tSaturation Mixing Ratio\tRH\tActual Mixing Ratio\tRank
\nA\t14\uf0b0C\t\t90%
\nB\t20\uf0b0C\t\t60%
\nC\t24\uf0b0C\t\t40%
\nD\t30\uf0b0C\t\t40%
\nE\t34\uf0b0C\t\t30%<\/p>\n
Using the graph provided above (Fig 6-8), answer the following questions.
\n5.\tThe difference in saturation vapour pressure between -25\uf0b0C and -5\uf0b0C is ___________ mb, whereas the difference between +5\uf0b0C and +25\uf0b0C is _______________ mb. [2 marks]<\/p>\n
6.\tAir at 25\uf0b0C has a saturation vapour pressure of ________ mb (capacity). An air mass at 25\uf0b0C actually contains water vapour exerting 14 mb vapour pressure (content). Using Eq. 1, what is the present relative humidity of this air mass? ___________________% [2 marks]<\/p>\n
7.\tWhat is the approximate dew-point temperature of the air mass in Q 6? ______\u00b0C. (What is the temperature for which 14 mb of vapour pressure is the maximum capacity?) In other words, this air mass must cool down from its present temperature to _______\uf0b0C to achieve saturated conditions at the dew point temperature. The present temperature must cool down to the dew point temperature for its water content to equal the air\u2019s water vapour capacity). [2 marks]<\/p>\n
8.\tUsing the same air mass, assume that it warms up to 35\uf0b0C on a hot afternoon and maintains the same vapour pressure content of 14 mb. What is the relative humidity of the air mass at this time? _________% (Hint: determine the capacity of 35\uf0b0C air to hold water vapour, expressed as pressure. Using Eq.1, calculate the value.) [1 mark]<\/p>\n
9.\tAssuming a different air mass from the above that has a relative humidity of 45% at a temperature of 25\uf0b0C, what is the actual humidity content expressed as vapour pressure? _______ mb.
\nThus, by knowing the relative humidity and temperature, relative humidity provides us with an indirect method of determining the actual water vapour content of the air. [1 mark]
\nPart 2: Dew point temperature (total marks \u2013 15)
\nThe temperature at which a given mass of air becomes saturated is termed the dew point temperature (DPT). This is the temperature at which the water vapour capacity of the air (saturation mixing ratio) is the same as the actual water vapour content of the air (mixing ratio). The dew point is determined by the mixing ratio. Air is saturated when the dew point temperature and the air temperature are the same. Further cooling of an air mass once it reaches the dew point will result in condensation, since the air mass can no longer hold the extra water vapour. The amount of condensation must be equal to the amount of extra moisture beyond the maximum saturation capacity for that parcel of air. If the temperature is known, the water vapour capacity can be determined. If the mixing ratio is known, you can determine the dew point (where the mixing ratio & the saturation mixing ratio are the same). If the dew point is known, the mixing ratio can be determined (will be the saturation mixing ratio at that T).
\nExample: the dew point temperature of a parcel of air with a mixing ratio of 11.1 g\/kg is always 15.6\uf0b0C. Conversely, a parcel of air with a dew point of 15.6\uf0b0C has a mixing ratio of 11.1 g\/kg.
\n1.\tUsing the table below (fig. 13-1), complete the following chart (round off relative humidity to the nearest percent) [8 marks]
\nMixing ratio (g\/kg)\tAir Temp (\uf0b0C)\tSaturation mixing ratio (g\/kg)\tRelative humidity (%)
\n2.2\t-1.1 C
\n2.2\t32.2 C
\n13.2\t\t15.6
\n26.2\t\t36.5<\/p>\n
2.\tThe air inside a room is at a temperature of 21.1\uf0b0C and has a mixing ratio of 5.2 g\/kg:
\na.\tWhat is the relative humidity? __________% [1 mark]
\nb.\tWhat is the dew point?____\uf0b0C [1 mark]
\nc.\tIf the mixing ratio remains the same, but the Temperature of the room increases to 29.4\uf0b0C, what is the new relative humidity? ___% [1 mark]<\/p>\n
3.\tThe air inside a room is at a temperature of 35\uf0b0C and has a mixing ratio of 7.6 g\/kg:
\na.\tWhat is the relative humidity? ___% [1 mark]
\nb.\tWhat is the dew point? ___\uf0b0C [1 mark]
\nc.\tIf the room temperature decreases by 5\uf0b0C per hour, how many hours will it take for the air to reach saturation? ____hours [1 mark]<\/p>\n
d.\tAfter reaching saturation, if the temperature of the room continues to decrease for one more hour, approximately how many grams of water vapour (per kg of air) will have to condense out of the air to maintain a relative humidity of 100% _______g\/kg [1 mark]<\/p>\n
Part 3: Adiabatic cooling & warming: (total marks \u2013 30)
\nIn nature, cooling of air masses to their dew point temperature often occurs when air rises. Consider a parcel of air that is forced to rise and does not mix with the surrounding environment. As the parcel rises, atmospheric pressure around it decreases, allowing the parcel to expand. Since the parcel has the same number of molecules but occupies more volume, its average internal temperature decreases. This decrease in temperature happens at the dry adiabatic lapse rate (DALR) of approximately 10\uf0b0C per kilometer.<\/p>\n
1.\tCalculate the air temperature of an unsaturated air parcel at 100 m increments as it is forced to rise from Earth\u2019s surface, where its temperature is 35\uf0b0C. Complete the table below. [5 marks]<\/p>\n
Heights (M)\tTemp. (\uf0b0C)
\n1000 (1 km)
\n900
\n800
\n700
\n600
\n500
\n400
\n300
\n200
\n100
\nSurface\t35\uf0b0C<\/p>\n
It is possible that rising air parcels will cool to the dew point temperature (DPT). When this happens, a parcel becomes saturated and condensation or clouds form. The height at which this occurs is called the lifting condensation level (LCL). If the air parcel rises above the LCL, it cools as a slower rate referred to as the saturated adiabatic rate (SALR). The saturated adiabatic lapse rate ranges between 5\uf0b0C & 9\uf0b0C per kilometer (often averaged at 6\uf0b0C\/km). It is slower than the dry adiabatic lapse rate (DALR) because latent heat is released into the parcel as water vapour condenses, a warming that partially offsets adiabatic cooling. The SALR varies because the amount of condensation depends both on the amount of water vapour in the parcel and on atmospheric pressure.<\/p>\n
Parcel A Temp. (\uf0b0C)\tHeight (km)\tParcel B Temp. (\uf0b0C)
\n5.0
\n4.5
\n4.0
\n3.5
\n3.0
\n2.5
\n2.0
\n1.5
\n1.0
\n0.5
\n28\uf0b0C\tsurface\t10\uf0b0C<\/p>\n
2.\tConsider an air parcel A, which is 28\uf0b0C at the surface. It is forced to rise to 5 km. The lifting condensation level is 1.5 km, above which the parcel cools at an average SALR of 5\uf0b0C per kilometer. Fill in the left column of the table above indicating the parcel\u2019s changing temperature. [5 marks]<\/p>\n
3.\tNow consider air parcel B with a surface temperature of 10\uf0b0C that is forced to rise. It too reaches the LCL at 1.5 km, but the average SALR in this case is 7\uf0b0C \/km. Fill in the right column of the table above indicating temperature change in the lower 5 km. [5 marks]<\/p>\n
4.\tWhy would the warmer parcel cool at a slower rate between 1.5 and 5 km? [2 marks]<\/p>\n
If air is rising up over a mountain, and continues to rise beyond the LCL, it cools at the SALR. As the air is rising, the Dew Point temperature is also dropping at the Dew Point Temperature lapse rate of -1.8\u00b0C\/km. Once the LCL has been reached, and because water vapour condenses out of the rising air, temperature and dew point temperature both decrease at the SALR to the top of the mountain, and relative humidity remains at 100%. As air sinks on the leeward side of the mountain, assume that its temperature increases at the dry adiabatic lapse rate (DALR) and its dew point temperature increases at 2\uf0b0C per kilometer. Consider an air parcel forced over a mountain range (fig 6-4 below). At the base of the windward side of the mountain the temperature is 25\uf0b0C and the dew point temperature is 13\uf0b0C.<\/p>\n
5.\tFill in the spaces for temperature and dew point temperature at various heights on the windward and leeward sides of the mountain in Figure 6-4 above. Assume that the SALR is 5\uf0b0C \/km. (Note that temperature and dew point drop together when a saturated air parcel rises and water vapour condenses.) [8 marks]<\/p>\n
6.\tAt what height would cloud bases form? [1 mark]<\/p>\n
7.\tHow do the sea level temperature and dew point on the leeward side compare with the sea level temperature and dew point on the windward side? [2 marks]<\/p>\n
8.\tWhich side of the mountain is more often cloudy and which side is more often clear? Why? [2 marks]<\/p>\n
Part 4: Stability, Adiabatic processes & Lapse rates: (total marks \u2013 27)
\nThe stability of air is an important characteristic of the atmosphere. Air is unstable if it rises on its own, and stable if it resists vertical motion and rises only when forced. The temperature of a parcel of air, relative to the temperature of the surrounding air, determines stability for the most part. A parcel of air will be unstable if it is warmer than the surrounding air. A parcel of air will be stable if it is the same temperature, or cooler, than the surrounding air.
\nIn order to understand stability, we must distinguish between the \u201cenvironmental\u201d lapse rate (ELR) and the adiabatic lapse rate. The ELR reflects the temperature of the atmosphere at different altitudes \u2013 this is the vertical temperature gradient. The ELR averages about 6.4\uf0b0C per 1000m within the troposphere. This rate is called the average lapse rate. However, from day to day and from place to place, the ELR frequently deviates from this average rate. Changes in the ELR are also often observed from day to night. Further, there are times when the temperature will actually increase as we move up through the atmosphere \u2013 a situation known as a temperature inversion.
\nBoth ascending & descending temperature changes are measured with one of 2 dynamic rates, depending on moisture conditions in the parcel, the DALR or the SALR. In the context of stability, we can think of the ELR as showing the temperature change of the surrounding air through which a parcel of air is moving and changing its temperature adiabatically (following the DALR or SALR).<\/p>\n
Determining the degree of stability or instability involves measuring and comparing simple temperature relationships between conditions inside an air parcel and the outside surrounding air. This difference between an air parcel and the surrounding environment can produce buoyancy that can contribute to further lifting. The basic temperature relationships that determine stability conditions in the atmosphere between the ELR and dry and saturated adiabatic lapse rates are shown in Figure 8.2 above. The 3 possible conditions are: stable (if the rising air is cooler than or the same temperature as the surrounding air), unstable (if the rising air is warmer than the surrounding air), and conditionally unstable (if, in rising, it becomes warmer than the surrounding air only after it begins cooling at the SALR).
\n1.\tCity A is at mean sea level with an air temperature of 20C. Air flow from City A to City B, at an elevation of 500 m, must cross the crest elevation of the intervening Surprise Mountain at 2500 m. Use graph 6-8, Table 5-2 & Fig 13-1 from Parts 1 and 2 of this lab to Help you.
\na.\tWhat is the relative humidity at City A if the parcel holds 8 g\/kg? [1 mark]
\nb.\tWhat is the dew point at City A? [1 mark]
\nc.\tAt what elevation would saturation occur as air rises above City A? [1 mark]
\nd.\tIf the SALR is 6.0 C\/1000 m, what is the temperature of the parcel at the crest? [1 mark]
\ne.\tWhat is the relative humidity at the crest? [1 mark]
\nf.\tWhat is the temperature of the parcel upon arrival at City B? [1 mark]
\ng.\tAt City B, what is the approximate relative humidity? [1 mark]
\n2.\tCreate a graph with elevation on the y axis and temperature on the x-axis. Use the following sets of hypothetical data to plot the vertical temperature profile (the ELR) of the atmosphere in 2 locations. With a straight edge, connect the temperature points for Location A with a blue line, and for Location B with a green line. [10 points for graph with 2 lines here + 2 more lines in questions 3 + 4 below]<\/p>\n
Remember, the vertical temperature profiles show the temperature of the surrounding air through which the parcels of air can move. For the following questions, assume the DALR is 10\uf0b0C\/1000 m, and the SALR is 6\uf0b0C\/ 1000m. Use the graph you just created to answer the following questions.<\/p>\n
3.\tA parcel of air with an initial T of 15\uf0b0C begins to rise at location A. The LCL of the parcel is 500 m. With a red line, carefully draw the temperature decrease of this parcel of air as it rises to 4000m on the graph you just created. Be sure to consider the LCL, and the DALR and SALR.
\na.\tDescribe the stability pattern of this parcel of air. [1 mark]
\nb.\tWhat is the general name for the change observed in the vertical temperature profile between 1000 and 1500 m? [1 mark]<\/p>\n
c.\tDoes the parcel of rising air become highly stable or highly unstable between 1000 and 1500 m? Why? [2 marks]<\/p>\n
d.\tAt what elevation above the LCL does the air come closest to the temperature of the surrounding air? [1 mark]<\/p>\n
4.\tA parcel of air with an initial temperature of 35\uf0b0C begins to rise in Location B. The LCL of the parcel is 1500 m. With a red line, carefully draw the temperature decrease of this parcel of air as it rises to 5000 m on the graph you just created in Q2. Be sure to consider the LCL, and the DALR and SALR.
\na.\tWill this parcel of air begin to rise from the surface on its own? Why? [1 mark]
\nb.\tDoes the stability of this parcel change with increased elevation? Why? If so, at what elevation does this change occur? [2 marks]
\nc.\tHow would the pattern of stability below 5000 m be different if the lifting condensation level was not reached until 4000 m? [2 marks]<\/p>\n","protected":false},"excerpt":{"rendered":"
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