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homework3_answers - ATOC/ASEN 5235 Homework#3 Due Thursday...

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ATOC/ASEN 5235 Homework #3 Due Thursday, 23 September Please staple all pages together !!! Each problem is worth 25 points 1. Petty 2.3. The atmospheric boundary layer is that region near the surface that is “well-mixed” by mechanical and/or convective turbulence originating at the surface. Its thickness may range from a few meters to several kilometers. Heat added by conduction from the surface is typically distributed quickly throughout the boundary layer. At a certain location in the tropics, the sun rises at 06 Local Solar Time (LST), is directly overhead at noon, and sets at 18 LST (=6 PM). Assume that, during the twelve hours that the sun is up, the net flux of solar energy absorbed by dry vegetation and immediately transferred to the overlying air is F(t) = F 0 cos[ π (t-12)/12], where t is the time of day (in hours), and F 0 =500 W m -2 . (a) Ignoring other heating and cooling terms, compute the total solar energy (in J m -2 ) added to the boundary layer over one 24-hour period. (b) If the boundary layer depth Δ z = 1 km, its average air density is ρ a = 1 kg m -3 , and the heat capacity at constant pressure is c p = 1004 J kg -1 K -1 , compute the temperature increase Δ T implied by your answer to (a). (c) If, instead, the boundary layer depth started out at sunrise only 10 m deep and remained at that depth throughout the day, what would be the corresponding change of temperature? Why is it much more likely that the boundary layer would deepen quickly after sunrise? Answer: (a) F(t) = F o cos[ π (t-12)/12], t=local time (hrs), F o = 500 W m -2 Let Q = Energy per unit area (J m -2 ): ( ) [ ] = = 18 6 18 6 dt 12 / 12 cos dt ) ( t F t F Q o π Solve by substitution: u = π (t-12)/12 du/dt = π /12, so dt = 12 du / π : 7 . 3819 24 2 sin 2 sin 12 | sin 12 du cos 12 18 6 18 6 = = = = = = = = = π π π π π π o o t t o t t o F F u F u F Q The units on Q as expressed above: W m -2 hr = J s -1 m -2 hr To convert to J m -2 : Q = 3819.7 J s -1 m -2 hr × 3600 s hr -1 = 1.38 × 10 7 J m -2 (b) Δ z = 1 km, ρ a = 1 kg m -3 , c p = 1004 J kg -1 K -1
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