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Solutions to Chapter 3

This document presents the solutions to various assignments concerning environmental science and atmospheric stability. It includes calculations of wind velocity at different elevations, considering the effects of lapse rates in rural areas. It provides examples using given temperatures, initial wind speeds, and stability classes to determine the maximum mixing depth under specified conditions. Through mathematical modeling, we examine how temperature profiles and environmental factors influence air parcel movements.

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Solutions to Chapter 3

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  1. Solutions to Chapter 3 Assignment 3 CE/AE/EnvSci 4/524

  2. 3.7: The wind has a velocity of 10 m/h at 10 m in a rural area. When lapse rate is 0.2°C/100, what is the velocity at an elevation of 100 m in miles per hour? • u/u1 = (z/z1)p • Table 3-2, ΔT/ΔZ = 0.2C/100, Stability Class = E • U = u1 (z/z1)p = 10 mph * (100m/10m)0.35 = 22.4 mph

  3. 3.9: Consider that the wind speed u is 2 m/s at a height of 10 m. Estimate wind speed at a) 200 m and b) 300 m for the 6 stability classes in Table 3-3 for rural conditions. • U = u1 (z/z1)p = 2 m/s (200/10)^P • A: U = 2 (200/10)0.07 = 2.47 m/s • B: U = 2 (200/10)0.07 = 2.47 m/s • C: U = 2 (200/10)0.10 = 2.70 m/s • D: U = 2 (200/10)0.15 = 3.13 m/s • E: U = 2 (200/10)0.35 = 5.71 m/s • F: U = 2 (200/10)0.55 = 10.39 m/s • U = u1 (z/z1)p = 2 m/s (300/10)^P • A: U = 2 (300/10)0.07 = 2.54 m/s • B: U = 2 (300/10)0.07 = 2.54 m/s • C: U = 2 (300/10)0.10 = 2.81 m/s • D: U = 2 (300/10)0.15 = 3.33 m/s • E: U = 2 (300/10)0.35 = 6.58 m/s • F: U = 2 (300/10)0.55 = 12.9 m/s

  4. 3.14: In a given situation the ground-level air temperature is 15° C, while the normal maximum surface temperature for that month is a) 26°C and b) 24°C. At an elevation of 300 m, the temperature is found to be 21°C. What is the maximum mixing depth in meters for the two cases? • Air parcel will rise until its temperature T` = local atmospheric temperature, neutral equilibrium. • Temperature profile = DT/dz, solid line • Dry adiabatic is dashed MMD

  5. 3.14: In a given situation the ground-level air temperature is 15° C, while the normal maximum surface temperature for that month is a) 26°C and b) 24°C. At an elevation of 300 m, the temperature is found to be 21°C. What is the maximum mixing depth in meters for the two cases? • Air parcel will rises until intersection of lines. Slope for existing condition is: (21° – 15°) = 0.02°C/m • 300m • Adiabatic lapse rate (slope) is -0.0098°C/m (equation 3-5) • MMD is point where slopes intersect (Z)

  6. 3.14: for condition A • existing condition is: 0.02°C/m • Adiabatic lapse rate (slope) is -0.0098°C/m (equation 3-5) • MMD is point where slopes intersect (Z) • 15 + Z (0.02°C/m) = 26 + Z(-0.0098°C/m) = • Z = _______26°C - 15°C_____ =369.13 m • (0.02°C/m + 0.0098°C/m) MMD

  7. 3.14: for condition B • existing condition is: 0.02°C/m • Adiabatic lapse rate (slope) is -0.0098°C/m (equation 3-5) • MMD is point where slopes intersect (Z) • 15 + Z (0.02°C/m) = 24 + Z(-0.0098°C/m) = • Z = _______24°C - 15°C_____ =302 m • (0.02°C/m + 0.0098°C/m) MMD

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