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HW 13 Solutions

# HW 13 Solutions - 53:050 Natural Environmental Systems...

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53:050 Natural Environmental Systems Solutions to Homework No. 13 – Spring 2011 1. Problem 12-5 in Davis and Masten. The prevailing lapse rate (PLR) is calculated as: PLR = 1 2 1 2 Z Z T - T - . Atmospheric stability is determined by comparing PLR to ALR as described on p. 557-559 and Example 12-2. (a) PLR = 2 - 50 5.00 - 4.52 = -0.010 o C/m = - 1.00 o C/100 m Neutral (= ALR) (b) PLR = 2 - 50 5 - 5 = 0 o C/m = 0 o C/100 m Stable (and Isothermal) (c) PLR = 2 - 50 (-21.01) - 25.17 - = -0.087 o C/m = -8.7 o C/100 m Unstable (PLR > ALR) 2. Problem 12-7 in Davis and Masten. The following are taken from the authors’ Solutions Manual. Here we compare the given conditions to information contained in Table 12-10. (a) At 1:00 pm on a clear, summer afternoon one would expect the solar radiation to be strong. With a wind speed of 5.6 m/s, we select stability Class C . (b) A clear summer night means “≤ 3/8 cloud”. With a wind speed of 2.1 m/s, select Stability Class F . (c) Stability Class D . The neutral class (D) is selected for all overcast conditions – see footnote in Table 12-10. (d) See the notes below Table 12-10. For broken clouds, “strong” solar insolation (as expected on a clear summer afternoon) is reduced to “moderate”, and with a wind speed of 5.2 m/s, we select stability Class C or D . 3. Problem 12-12 in Davis and Masten. First we need to determine the stability class. Since it is a clear sky at night with a wind speed of 2.5 m/s, Table 12-10 indicates we should use Stability Class F. Now, we check to see if Eq. 12-18 or Eq. 12-25 should be used. s z = 0.47(185 – 85) = 47 m 1

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Using Figure 12-24 or Equation 12-22, we can find x L . We demonstrated how to use Figure 12- 24 in the class notes – in this case, we get x L ≈ 11 km, so 2x L = 22 km. We could also use Equation 12-22 for stability Class F: s z = cx d + f From Table 12-11:
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HW 13 Solutions - 53:050 Natural Environmental Systems...

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