Lab_6_FA23

ATMS-120 Fall 2023 © 2023 Dept. of Atmospheric Sciences, University of Illinois-Urbana Champaign 1 Name and NetID: Abigail Pritz Lab #6: Severe Thunderstorms Due 11:59PM Tuesday October 17th, 2023 Part #1 – Wind Shear Strong wind shear is a crucial ingredient for severe thunderstorm development. Wind shear is defined as the change in wind speed and direction with height. Supercell thunderstorms form most frequently on the Great Plains of the central US. The most common supercell wind profile exhibits southeasterly winds at the surface that strengthen and become westerly with altitude. The most common wind profile for a squall line (a.k.a. mesoscale convective system) is unidirectional, meaning that the wind speeds increase with height but the direction of the wind does not change with height. Five different wind profiles are below. Place a triangle around the letter beneath the most common supercell wind profile and place a circle around the letter beneath the most common MCS wind profile (1 point each) . Bulk wind shear is calculated by finding the vector difference between the winds at two different heights. Using the supercell wind profile you identified, calculate the 0-1 km and 0-6 km bulk wind shear values. This means we will find the difference between the surface wind (lowest wind barb on the sounding) and the speed of the wind at 1 km and 5 km. The atmospheric pressure at 1 km above sea level is typically very close to 850 mb. The pressure at 6 km above sea level is very close to 500 mb. Please calculate the 0-1 km and 0-6 km wind shear values in knots (kts). For simplicity, assume that the surface winds are due southeasterly, the 850 mb winds are due southerly, and the 500 mb winds are due westerly. Show your work. (2.5 points each)

ATMS-120 Fall 2023 © 2023 Dept. of Atmospheric Sciences, University of Illinois-Urbana Champaign 2 0-1 km shear value: 30 kts 45 kts (air pressure at 1 km) – 15 kts (air pressure at sea level) 0-6 km shear value: 55 kts 70 kts (air pressure at 6 km) – 15 kts (air pressure at sea level) Part #2 Assessing Thunderstorm Ingredients – 0.5 points each 1. Define vertical wind shear. A change in wind speed and direction with height. 2. Is relative humidity or dewpoint a better variable to examine to determine if there is sufficient moisture for thunderstorms to form? Relative humidity is better to determine if there is sufficient moisture for thunderstorms to form. 3. Imagine a large complex of thunderstorms form in Champaign, IL. The main source of moisture is the Gulf of Mexico. What is the name of the process that is a secondary source of moisture? Evapotranspiration 4. Why is the stratosphere considered a stable layer in the atmosphere? The stratosphere is considered a stable layer because the warmer, less dense air is constantly sitting on top of the colder, denser air, and therefore there isn’t much movement. 5. If the Lifted Index is positive, is the parcel warmer than, the same temperature as, or cooler than its environment? If the LI is positive, then the parcel is cooler than the environment.

ATMS-120 Fall 2023 © 2023 Dept. of Atmospheric Sciences, University of Illinois-Urbana Champaign 4 Part #4 Hail Damage A storm on August 7 th , 2023 dropped giant hail just south of Fort A.P. Hill in Virginia. A storm was found with a circumference of 14.9 inches. Using the circumference and the information below, answer the following questions. What you need to know… Ø Kinetic Energy = 0.5 * Mass * Velocity 2 (Units: Mass à kg, Velocity à m/s, yielding Joules J) Ø Density of the ice in the hail stones = 31 lbs/ft 3 Ø The volume of a sphere is ( 4 / 3 ) × π × r 3 Ø The fall velocity of a hailstone in still air can be approximated using the equation where k = 20 if d (the diameter of the stone) is given in centimeters. V will have units of meters per second (m/s) as long as d is in cm. Ø Density = Mass/Volume Ø Assume the hailstone was spherical What was the diameter of the hailstone in inches? !".$ % = 4.74 ࠵?࠵? 1 pt What was the volume of the hailstone in cubic inches? ( " & ) (࠵?)((4.74/2) & ) = 55.76 ࠵?࠵? & 1 pt What was the fall velocity of this hailstone in m/s? 20√12.0396 = 1.66 ’ ) 2 pts What was the total mass of the hailstone in kg? 31 = * ++.,- , ࠵? = .51 ࠵?࠵? 2 pts What was the total kinetic energy of this hailstone in Joules? 2 pts 0.5 * .51 * (1.66)^2 = 0.703 Joules Part #5: 2019 Hail Data Download the US hail reports for 2022 from January through May from the Lab #6 Link on Canvas. Using these data complete the following tasks. 1. How many people were injured in this period of 2022 by hail (1 pt) ? 4 people 2. How many people were killed by hail in this period of 2022 (1 pt) ? 0 people 3. What was the total economic loss of all hail events combined in this period of 2022 (property+crop) (1 pt) ? i. $59466380 4. What is the total number of hail reports in this period of 2022 (1 pt) ? 3806 reports 5. Create a bar graph that shows the hail reports for each day of the year through May (2 pts) . 6. Create a line graph that shows the cumulative hail reports throughout this period of 2022 (2 pts) . 7. Which day had the most hail reports in this period of 2022 (1 pt) ? April 14 th and the 19 th of May had 253 reports. d k V =

ATMS-120 Fall 2023 © 2023 Dept. of Atmospheric Sciences, University of Illinois-Urbana Champaign 5 8. What was the largest hailstone diameter in this period of 2022 and in which state(s) did it fall (1 pt) ? The largest hailstone diameter was 6 in and it fell in NE. Part #6: Wind Force on a Tree – 4 points In one of your instructor’s neighbor’s backyards is a huge pin oak tree. How fast would the wind need to be in a thunderstorm to uproot this tree and knock it over? The force required to uproot this tree is approximately 640,000 N (640,000 kg m/s 2 ). How fast would the wind have to be (in mph) to generate this force against this tree? Use this information below to solve this problem. (Show all of your work and be careful with units!) F = Force applied by the wind = 640000 kg m/s 2 ρ = Air Density = 1.22 kg/m 3 C D = Drag Coefficient = 0.78 (this is a ratio and therefore has no units) A = Area of the leaves on the tree. Each leaf is 12 in 2 and there are 62,500 leaves on this tree. Consider the area of the branches and trunk to be negligible. v = wind speed in m/s 640,000 = ( !...∗0.,1∗!$,0+0 . ) ࠵? . à 640,000 = 9,063.99࠵?2 à ࠵? . = 70.61, ࠵? = 8.403 ’ ) à 18.797 mph A= 62,500*12*0.0254= 19,050 m^2 F = ρ C D A 2 ! " # $ % & v 2