Hi, I am your Meteorology lab tutor. Please refer to my notes and/or the course instructor when you have any questions. You may NOT use the instructor to check final answers but you may use the instructor for specific help on specific problems. FIRST QUARTER EXERCISE LAB 1, Page 1 - 7 Problem #1: The problem uses the assumption that the atmospheric mass will be cut in half with each 5.6 km rise above the surface. In each row, the % of atmosphere above + % of atmosphere below will equal 100%. Below are how the first 4 bottom rows should look. Height (km) % above % below %above + % below 16.8 12.5% 87.5% 100% 11.2 25% 75% 100% 5.6 50% 50% 100% sfc 100% 0% 100%Problem #2: The y-axis is the vertical axis on the page while the x-axis is the horizontal axis on the page. Plot increments of 10% along the bottom x-axis (% above value) for each box and plot increments of 2 km along the far left y-axis for each box. You will then plot 7 dots. The first dot will be at (100%, 0 km), the second dot will be at (50%, 5.6 km), the third dot will be at (25%, 11.2 km) and so forth. After plotting the 7 dots, connect them with a smooth curve. It should look like a slide when you are done. Problem #3: Use table in problem #1. Read off percent above value at a height of 11.2 km Problem #4: You need to use the graph in #2. Go along the y-axis on the far left side until you come to 6.19 km. Next, move horizontally until you intersect the graph. At this intersection point draw a line straight down to the bottom. The % value where this line intersects the bottom is the % above. The question asks for percent BELOW. 100 minus % above equals percent below. Problem #5: Same method at #4 but this time the question wants the % above Problem #6: Change 10% to 100mb, change 20% to 200 mb, change 30% to 300 mb, and so forth Problem #7: Use information given in numbers 3 through 5. Using new scale from #6, determine the pressure for each height. Problem #8: 23% of the partial pressure is from Oxygen. At sea level, 1000 mb times 0.23 = 230 mb of partial pressure from oxygen. A jet air craft travels at 11.2 km as given in #3. At 11.2 km there is 250 mb of pressure (25% mass of total atmosphere at this height); 250 times 0.23 = 57.5 mb of pressure from Oxygen at 11.2 km Problem #9: Explain what happens to the temperature lapse rate when moving from the upper troposphere and into the lower stratosphere and why the temperature lapse rate changes Problem #10: Plot each point from Table 1-2 for each city. Temperature is on the x-axis while height is on the y-axis on the graph. For Key West, the first point will be (24, 0), the second point will be (18, 800), and so forth. Once all the Key West points are done then connect them. Next, do the same process for Fairbanks. Problem #11: Read temperatures off graph Problem #12: Notice if the temperature warms or cools with height from the surface to 1 km. What is the term for a warming with height? Problem #13: Record values Problem #14: Very important concept in meteorology. Cold air is more dense than warm air. Look at the graph and look for the relationship between average temperature and thickness Problem #15: There are a total of three possible that can be circled. Review questions: Since pressure decreases with height, then air density will also decrease with height. Figure 1-2 shows how temperature changes with height on average. The thickness is determined by the average density of the air- what influences most the average density of the air? This was answered in problem #14 Compare the pressure change between traveling 600 km along the earth's surface to traveling 600 km above the earth's surface LAB 2, Page 9 - 20 Problem #1: Be sure to examine the example problem on page 11. Here are some notes about the example problem: The sun's direct rays are striking 23.5 degrees South, thus the sun is directly overhead at solar noon at that location. For each degree of latitudes moving North or South from 23.5 degrees S, take a degree off the sun angle. For example, the equator is 23.5 degrees away from 23.5 degrees South. Thus, the solar noon sun angle at the equator is 90 - 23.5 = 66.5 degrees on the winter solstice. Problem #2: Keep in mind there are 90 degrees between the equator and pole and that the Tropic of Capricorn is located 23.5 degrees South of the equator. Problem #3: Take answer from #2 and subtract from 90. Another method is to take the difference in degrees between the Arctic circle and where the person is standing. Problem #4: Determine at what latitude the sun is directly overhead on each of those dates Problem #5: New Orleans example: New Orleans latitude is 30 degrees N. On March 21 the solar declination is at the equator. Find the difference in degrees between solar declination and the city's latitude and subtract this number from 90. March 21: difference from equator to New Orleans is 30 degrees. 90 - 30 = 60 degrees Problem #6: Example: June 8 in Starkville, MS. The number of days to the nearest equinox is 78 and N is positive since it is between March and September. June 8 solar declination = 23.5 * SIN (78) = 22.99 degrees N. Starkville latitude = 34 degrees North noon solar elevation= 90 - (34 - 22.99) = 79 degrees noon zenith angle = 90 - 79 degrees = 11 degrees Problem #7 - #8: Optional problem, not required Problem #9: An important piece of information is that the latitude of reference is 30 degrees North. First, calculate the noon sun angle at 30 N on those dates. The sun angle plus the zenith angle equal 90. To find the beam spreading you will need a calculator that can handle the SIN function (scientific calculator). Surface area (beam spreading) = 1 / SIN(sun angle) Problem #10: text on previous page explains why Problem #11: First, measure the length along a latitude in millimeters. Second, record the length in millimeters of the lighted portion along the same latitude. Divided length of lighted portion by total length. Multiple this number by 24 to get the number of daylight hours. Problem #12: Keep in mind there are 24 hours in a day and the hemisphere's are mirror images of each other in size and shape. Problem #13: Determine by looking at the data in the two previous problems. Earth latitudes are often referenced as low (equatorial) latitude, middle latitude and high (polar) latitude. Problem #14: Example for March 21. 1186 units at 30 N and 685 units at 60 N. Difference = 1186 - 685 = 501 Wm^-2 Problem #15: Range is high value minus low value. For 30 N, the range would be 1381 - 815 = 546. The range for 60 N is found by the same type method. This is actually a similar question to #13. Problem #16: The amount of radiation received at the surface decreases at an increasing rate as the pole is approached. Best way to answer question to show this mathematically. Below is the SIN of sun angles from 90 degrees (sun directly overhead at solar noon) to 0 degrees (sun on or below horizon at solar noon). Since the earth is a curved surface on a sphere, solar receipt will follow a SIN function. SIN 0 = 0 SIN 10 = 0.17 SIN 20 = 0.34 SIN 30 = 0.50 SIN 40 = 0.64 SIN 50 = 0.77 SIN 60 = 0.86 SIN 70 = 0.94 SIN 80 = 0.98 SIN 90 = 1.00 From the values above, notice the SIN of the sun angle decreases at an increasing rate when going from the sun being directly overhead to the sun being on the horizon. Also notice, the difference between SIN 90 and SIN 60 is less than the difference between SIN 30 and SIN 0 even though both change by 30 degrees. With this background information you should be able to answer the question. Problem #17: Optional problem, not required Problem #18 - #24: CD exercise, Optional problems, not required Review Questions: a. Same principle as problem #16 b. Keep in mind that locations with lower sun angles will have cooler seasons and cooler climates than locations with higher sun angles, all else being equal. Also, keep in mind that the sun is not able to warm the surface if it is below the horizon and the longer it is below the horizon the less warmth it can bring. LAB 4, Page 27 - 34 Problem #1a: plug 6,000 K into Stefan-Boltzman equation and then solve Example Problem: What is E when 5,000 K is plugged into Stefan-Boltzman equation: E = (5.67 * 10^-8) * (5,000 K)^4 = 35,437,500 W/m^2 Problem #1b: plug 300 K into Stefan-Boltzman equation and then solve Problem #2a: divide sun temperature by earth temperature from problem #1 Problem #2b: 4th power means to multiply the same number by itself 4 times. i.e. 6^4 = 6 * 6 * 6 * 6 = 1,296 Problem #2c: Divide the answer from 1a by 1b and see if this is equal to the answer from 2b. Problem #3: The problem is already done for you in the textbook. You will notice the constants are slightly off. The true value of Wien's constant is the 2898 umK value, but either constant will not throw answer off enough to hurt you in testing. Example problem: What is wavelength of maximum emission using a temperature of 1,000 K? max wavelength of emission = 2898 micrometers*K / 1,000 K = 2.898 micrometers wavelength Problem #4: Take wavelength of maximum emission for the sun found in #3. Use table 4-1 to see the type of energy that corresponds to that wavelength Problem #5: Example for July 24 on 10 am SW in = 700 SW out = -100 LW in = 380 LW out = -450 Net = 700 - 100 + 380 - 450 = +530 W/m^2 Problem #6: There is something between the earth's surface and outer space that is responsible. This something changes the amount of radiation that makes it to the surface, explain why. Problem #7: 1 pm would be 13th hour of the day. Divided SW outgoing by SW incoming and multiply by 100%. Problem #8: Remember that Energy emission magnitude is a function of the temperature of a body. See Stefan-Boltzmann law to see this relationship. Longwave outgoing is earth's emitted radiation. Problem #9: Look at the dates of each location and think of the temperature climate. Problem #10: Think about where shortwave incoming and longwave incoming radiation is primarily absorbed at. Problem #11: The cloudier, the more shortwave incoming is blocked from reaching the surface Problem #12: The warmer a body, the more radiation it emits Problem #13: see graph. Keep in mind clouds help promote an atmospheric greenhouse effect Problems 14-25: Not required Review Questions: Not required SECOND QUARTER EXERCISE LAB 6, Page 45 - 55 Problem 1 through 9: Not required Problem #10: Hint: pick the latent heat process that is removing the additional heat Problem #11: Remember that this is a sunny day. Think of a latent heat process that would result in the playing fields having different temperatures. Problem #12: See figure 6-1. Multiply mass by value in figure 6-1 to get value in calories. The figure 6-1 should be in units of cal/g instead of just cal (i.e. condensation is 600 cal/g). Problem #13: If surrounding air cools then latent heat is consumed. If surrounding air warms then latent heat is released. Problem #14: Use table 6-2 to get the saturation mixing ratio. 14 C = 10.14 g/kg. The relative humidity is the mixing ratio divided by the saturation mixing ratio time 100% (i.e. (5 / 10.14) * 100% = 49% Problem #15: Think about how changing the saturation mixing ratio will change the relative humidity Problem #16: Think about how changing the saturation mixing ratio will change the relative humidity Problem #17: First, fill in the saturation mixing ratio columns. The actual mixing ratio equals the saturation mixing ratio times the decimal form of the relative humidity. For example if the saturation mixing ratio is 10 g/kg and the relative humidity is 50%, then the actual mixing ratio is 10 * 0.5 = 5 g/kg. Next, rank each row. A rank of 1 will have the highest value of actual mixing ratio Problem #18: Can the relative humidity value by itself (not knowing mixing ratio or saturation mixing ratio) tell you how much moisture is in the air? why or why not? Problem #19: The wet-bulb depression is equal to the dry bulb temperature (actual temperature) minus the wet bulb temperature. Use tables to find relative humidity. For example, if the temperature is 14 C and the wet-bulb depression is 2 Celsius degrees change, then the RH = 79% Problem #20-21: Not required Problem #22: Use same method as described in the paragraph above the problem. Problem #23: Use table 6-2 to find saturation mixing ratio. Mixing ratio = decimal form of RH times the saturation mixing ratio. For last part of question perform the same method as in #22 Problems #24-30: Not required Review questions: Answer the review questions on page 55. LAB 7, Page 57 - 66 Problem #1: As you did on the previous lab, look at the table to determine the saturation mixing ratio at each temperature Problem #2: Actual mixing ratio is decimal form of relative humidity multiplied by the saturation mixing ratio Problem #3: plot the two points. Problem #4, #5, #7: resulting temperature is (10 + 38) / 2. resulting mixing ratio is (actual mixing ratio 1 + actual mixing ratio 2) / 2 Problem #6: Use table 7-1 to determine saturation mixing ratio for the resultant temperature from #5 Problem #8: The RH will be greater than 100% for the answer. This is a condition of supersaturation. In the real atmosphere, the excess moisture would condense out of the air and the relative humidity would be 100% Problem #9: Think of fog over a lake and how that forms Problem #10: Since the dry adiabatic lapse rate (called unsaturated lapse rate also) is 10 C/km, the temperature will decrease by 1 C for each 100 meters the parcel rises Problem #11: Here is the answer for parcel A. PARCEL A Temperature height (km) -4.5 C 5.0 -2.0 C 4.5 0.5 C 4.0 3.0 C 3.5 5.5 C 3.0 8.0 C 2.5 10.5 C 2.0 13 C 1.5 (height of LCL, saturated above this height) 18 C 1.0 23 C 0.5 28 C surfaceProblem #12: Same process as parcel A except the directions give you a different lapse rate for the wet adiabatic lapse rate Problem #13: The answer is given to you on the previous page Problem #14: For the earth's surface, the temperature is 30 C with a dewpoint of 14 C on the table. The temperature will decrease by 5 C each 0.5 km (500 m) rise, while the dewpoint will decrease by 1 C each 0.5 km rise. At 2 km, the temperature will be equal to the dewpoint. Problem #15: Plug in numbers into the equation on the previous page. As an example we can use problem #14. T = 30, Td = 14, T - Td = 16, 16/8 = 2 km Problem #16: The question is assuming you are holding surface temperature constant. At a constant temperature, what happens to the height of the cloud base as dewpoint changes? Problem #17: not required Problem #18: When air is unsaturated, rising air will cool at 5 C per 500 meters (DALR) and the dewpoint will cool by 1 C per 500 meters. Cloud bases form at the point at which the temperature first becomes equal to the dewpoint. When the air is saturated, the dewpoint will fall with the temperature. The wet adiabatic lapse rate in this problem is 2.5 C per 500 meters. At the top of the mountain, the temperature and dewpoint will both be 5 C. On the leeside of the mountain the temperature will warm at the dry adiabatic lapse rate as the air sinks and the dewpoint will increase by 2 C per kilometer. Problem #19: Cloud bases form at the point at which the temperature first becomes equal to the dewpoint. Problem #20: The latent heat released on the windward side of the mountain due to condensation will effect the temperature of the air on the leeward side of the mountain Problem #21: Take note of which side of the mountain that air is rising and which side air is sinking Problem #22: It is the level at which the temperature first equals the dewpoint. Problem #23: Dewpoint in unsaturated air decreases by 1 C per 500m (0.5 km) rise while temperature decreases 5 C per 0.5 km. Above the LCL for this problem, temperature and dewpoint drop together at a lapse rate of 2.5 C per 0.5 km. Problem #24: Difference is equal to environmental temperature minus parcel temperature. For example, at 8 km the difference is -28 - (-14) = -14. A negative number indicates an unstable layer. Problem #25: If there is a deep layer of negative numbers then you would forecast thunderstorms. Review Problem: Think about the movement of air in the vertical and how that changes the temperature of the air. A stable atmosphere occurs when there is a deep layer of positive numbers using difference between environment and parcel LAB 8, Page 67 - 72 Problem #1: RH = (vapor pressure / saturation vapor pressure) * 100% Problem #2: pour a little water on your table and see what happens to the water after time Problem #3: Same generally process as #1. This time the RH will be greater than 100% (respect to ice) Problem #4: Study the Bergeron process in the hardback textbook Problem #5: V = 4/3 * PI * 0.025 * 0.025 * 0.025 Example problem, what is V when r is 2 mm? V = 4/3 * 3.1416 * 2 * 2 * 2 = 33.51 mm^3 Problem #6: The volume is a function of the radius to the third power. If the radius is increased by 100, this will increase the volume by 100^3 Problem #7: Area = PI * (0.5 mm)^2 Problem #8: Area = PI * (2.5 mm)^2; how many times greater is = 2.5^2 / 0.5^2 Problem #9: A cubic relationship increases more quickly than a squared relationship Problem #10: Use table 8-1 to get fall speed. Distance = Rate * Time; thus Time = Distance / Rate Problem #11: Time = Distance / Rate Problem #12: Use Table 8-1 to get radius of drizzle, then use table 8-2 to get distance it falls before evaporating Problem #13: Use figure 8-4. draw line through 1 km that intersects the 0-700 m cloud base, then read value using vertical axis Problem #14: Same as #13 except use a 5.5 km thickness Problem #15: Read up in the textbook on precipitation processes Problem #16: Same type problem as #13 and #14 Problem #17: Remember that the air below cloud base is generally unsaturated THIRD QUARTER EXERCISE LAB 9, Page 73 - 94 Problem #1: The pressure gradient is stronger where the isobars are closer together. Go ahead and change the question to include 3 or 4 areas where there is a strong pressure gradient. Flow is cyclonic around lows and anticyclonic around highs (in the real atmosphere, wind arrows will cross from high toward low pressure at about a 25 degrees angle across the isobars). Problem #2: Not required Problem #3- #4: follow directions Problem #5: The solid line should be curved and deflecting to the right of the path of motion Problem #6: The solid line should be curved and deflecting to the left of the path of motion Problem #7: You should get a curved line deflecting to the right of the path of motion Problem #8: error in figure 9-6c, rocket should be on 120 W longitude and A should be at 60 W Problem #9: They have already done it for you Problem #10: Compare figure 9-7d to 9-5d Problem #11: Answer to question is given (inferred) on the previous page Problem #12: gravity is a constant, thus dP/dz = density*constant, Thus, it is the density of the air that determines the magnitude of dP/dz. The way that density changes with height will directly determine how pressure changes with height Problem #13: You should plot 9 dots on figure 9-11 (each dot represents the height above the letter given on the bottom on the figure). After connecting dots you should get three slopping lines. The top line is the 300 mb pressure surface, the middle the 500 mb pressure surface and the bottom the 850 mb pressure surface Problem #14: dP/dz = density*constant; the height of a column of air depends on its average temperature. A warmer column of air will have a higher height since warmer air is less dense. If a column of air is higher, dP/dz will be smaller. If a column of air is shorter, dP/dz will be higher since pressure levels will be encountered more quickly as one samples the vertical pressure levels compared to a warmer column of air Problem #15: Compare the relationship between 850-mb temperature and the height of the column. Problem #16: On the 300 mb pressure surface, higher pressures are below this pressure surface and lower pressures are above this pressure surface. PGF flows from high toward lower pressure on a constant height surface Problem #17: The more a pressure surface is sloped, the stronger the PGF Problem #18: Winds will be stronger when the PGF is stronger Problem #19: PGF always points from higher toward lower heights. The Coriolis acts to the left of the path of motion in the Southern Hemisphere and to the right of the path of motion in the Northern Hemisphere (when in geostrophic balance the Coriolis points in the opposite direction from the PGF). Problem #20: The PGF always points from high to low. A change in direction over time causes an acceleration. This acceleration is called centripetal acceleration. The force directed toward the center of the rotation (either PGF or Coriolis depending on if it is a high or low) must be GREATER than the force directed away from the center of rotation in order for the gradient balance to hold Problem #21: Draw the PGF pointing directly toward the L for the top diagram and directly away from the H on the bottom diagram. The Coriolis will be to the right of the path of motion Problem #22: (Figure 9-16) The drawing that has the PGF most parallel and in the same direction as the resultant wind will have the higher speed Problem #23: See figure 9-15. Use the length of the Coriolis force to determine wind speed. When the length of the Coriolis force increases, then the wind speed increases. Problem #24: See following website: http://weather.ou.edu/~metr4424/Files/CurvedFlow.pdf Problem #25: The magnitude of the Coriolis is a function of the wind speed: Coriolis = 2*velocity*constant*SIN(latitude) Problem #26: Friction points in opposite direction from path of motion. PGF is directly from high toward lower pressure. Coriolis is to right of path of motion Problem #27: Northern Hemisphere: flow around low is cyclonic (counterclockwise), flow around high is anticyclonic (clockwise). Southern Hemisphere: flow around low is cyclonic (clockwise), flow around high is anticyclonic (counterclockwise). The term cyclonic means spinning in the same direction as low pressure in the hemisphere of interest. Problem #28: Do as instructed Problem #29: Upward vertical motion is required to produce clouds and precipitation Review Questions: Refer to textbook and lab manual to help answer questions FOURTH QUARTER EXERCISE LAB 10, Page 95 - 110 Problem #1: Use figure 10-1 as a guide. The R with arrow symbol is a thunderstorm Problem #2: For any isopleth, the values on one side should be higher than that isopleth and the values on the other side should be lower. Isopleths can make circles but they should never cross or merge into each other Problem #3/#4: Use a pencil so you can erase. It takes some trial and error Problem #5: Streamlines are parallel to wind barbs Problem #6: Use textbook for reference Problem #7: Do as instructed Problem #8: Range = highest height minus lowest height Problem #9: Warm air advection is warmer air moving toward the forecast area on a flat plane. Heights will increase over time when low level warm air advection occurs Problem #10: Follow as instructed Problem #11: Range = highest height minus lowest height Problem #12: Take two adjacent height contours and find the difference between the two Problem #13: Like a boat in a river current, storm systems move in the downstream direction Problem #14: Take two adjacent height contours and find the difference between the two Problem #15: 60 knots is a triangle and a long bar together LAB 11, Page 111 - 124 Problem #1: Air masses are dry or humid and are cold or warm. Air mass types are (continental or maritime) and (tropical or polar). Problem #2: Label wind directions Problem #3: The blanks at bottom are for each column Problem #4: Use value of nearest isobar encircling center of low. Compare the first panel's pressure to that of the last panel's pressure Problem #5: Notice spacing of isobars over time and notice changes in the wind barbs over time Problem #6: Compare surface map to Norwegian model Problem #7: Maritime airmass is between the cold and warm fronts in the warm and humid sector Problem #8,#9,#10: see chart Problem #11 - #17: see meteograms Problem #18,#19, #20: If the wind is blowing lower valued height contours into a region then cold air advection is occurring. If the wind is blowing higher valued height contours into a region then warm air advection is occurring Problem #21: Compare surface map to 850 mb map. Make sure to look at same time period Problem #22: Diffluence occurs where height contours spread out over distance Problem #23: Compare maps Problems #24-26: Follow instructions END OF LAB TUTOR |