hmwk 10 solutions

hmwk 10 solutions - 9.1? An atmospheric boundary layer is...

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Unformatted text preview: 9.1? An atmospheric boundary layer is formed when the wind blows over the earth‘s surface. Typically. such velocity profiles can be written as a power law: 1.: = ay". where the constants a and I: depend on the roughness of the terrain. As is indicated in Fig. P917, typical values arem = 0.40 for urban areas, n = 0.28 for woodland or suburban areas, and n = 0.16 for flat open boun- try (Ref. 23). (a) [f the velocity is 20 ftis at the 150 bottom of the sail on your boat (y = 4 ft), what is the Velocity at the top of the mast (y = 30 ft)? ' (b) If the average velocity is 10 mph on the tenth FIGURE 1:917 floor of an urban building. what is the average ' ' V velocity on the sixtieth floor? E_ 300 W“. (a) u: Cit/0'” 777113, , wile/'3 Cis a combat 0:15 0.”: u ‘ ——3'"(—X2—) or e.’.(‘,_,=20—:--1E 13371:) 11,“)4 I where C is a constant 0.40 or u2=/0mp}1 (45%) 9.22 An airplane flies at a speed of 400 mph at an altitude of l0.0f}0 {L If the boundary layers on the wing surfaces be- havn as those on a flat plate. estimate the extent of laminar boundary layer flow along the wing. Assume a transitional Reynolds number of Rejltr = 5 x 105. [f the airplrine maintains its 400-mph speed but descends to sea level elevation, will the portion of the wing covered by a laminar bnt'mdary layer increase or decrease compared with its value at I0.000 ft? Explain, Al 10,000 H= UX - lb szearl = £1 (a) Rex”: VG?! w/yer‘e U"¥00mpb(3606’:§)(, m; ) 451753 and From 735/3 0.!) V = “T:- = m 1.755 “0'3 -s#% 4 F41 ‘ l 5 52.0/X/0 T file/Ice, wn‘l: Rah: 5mm 1 , 2. X _ VRexcr: (2.0lxla”§i)(5x105) #0171191 "F U 587 ' ‘ '. Hr sea— let/cl -‘ (5) Rex“: va”, where U=400mplt fi)(fl) = 537E _ 2 and Z”=/.57x/0 41—y— HBHCEJ Z X =l/ieifi__ a or U '_ _£_ The laminar boundary layer occupies Nye flit-sf 0. 136‘1‘1‘ of 71/29 wing all .580 level and “ram paN' '(GJ above) #76 19/371 0. I 7f ff af an a/r‘it‘ade of M000 H- 777]: is clue [mfg/y la {/79 lower dam?) ( larger lrr'nemafi‘e wircoJHy ). 7799 afi/Mmic viscosHies are approx/M07219 Hlfi‘ same. 9.47 9.47 A 22 in. by 34 in. speed limit sign is supponqd on a 3-ln. wide, S-ft—Iong pole. Estimate the bending moment in tho pole at ground level when a 30~mph wind blows agalnst the. sign. (See Video V9.6.) List any assumplions uscd in your calculations. For 87Va/I'6PI‘VNJ Z My =0 9/" MB 2,5117%? + (5+ ,iEJfi' 94?; Were J \ m' 0.5;, = day an Me [pa/6‘ and Mi. =‘c/My 0/: 7%? 3322/; T From Fig. 9.23 LII/7% [/13 <0J for #1“:pr D ODS = M’ 1 From F117, W? if 7% pm1 407': 01534 qua/‘9 rod '4'"? WW: slump comm Cop = 2.2 77m: wif/I V= 30mph = 45‘?" DD} =%9U (:1,5 A, = f(0.0023d’§%’%{) f¥¥¢fi)2(/.9)(if;§#)flz)=22.7fi and ._ “‘9'” 22917155») 4° ‘ filo-00138 Wyn-2% m1) = my; r 771% from 557.17.)" mg 2.519! (5.3%,) +(5+fr§)fl{22.7/6) = A42 {HA ml}: .— 9f 5‘5 9.52 How much more power is required to peddle a bicycle al 15 mph into a 20-mph head- wind than at 15 mph through still air? Assume a frontal area of 3.9 fiJ and a drag coefficient of CD = 0.88. 70*pawer- = W and fl=jCa—%9U‘ZI9 J Myers Ug=3peed of Me bike _ - fl _ fl . ~15 0.". 88 .= —22 wind speed relafiva fa blA’E. " 5022i) '5 hr 7721/5, HE '* ’ £1195—1.1='. 12—72 w P =(22 §)(0.33)(g)(o.oozsa H5 JV (3916') Doggazifi Q m; 38.121 E (,3 WW; 5 20mp23 beaa’wde 0:05 +zojfi;(—gO-717) = $53 5 '75? Wins, ' 2 . 7&0 = 0.0396 (5153) = 23; fl and U: 1' 1,; MW; 3221/ my” U= 15mpfi=32 £— 7710: , ‘ H -£ 53 =0.oa¢a(22.)1=4¢3.5 «p— .— , M #20055 flared an aa’dhiiana/ [miller of F “ =(23b "53-5) 3 ‘ = 0.35049 ...
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hmwk 10 solutions - 9.1? An atmospheric boundary layer is...

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