Solutions to Chapter 3

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Solutions to Chapter 3. Assignment 3 CE/AE/ EnvSci 4/524. 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/u 1 = ( z/z 1 ) p Table 3-2, Δ T/ Δ Z = 0.2C/100, Stability Class = E

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### Solutions to Chapter 3

Assignment 3 CE/AE/EnvSci 4/524

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

3.9: 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

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

MMD

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)

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

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