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Unformatted text preview: P156 A deliver}r vehicle carries a long sign on top, as in Fig. P156. If the sign is very thin and the vehicle moves at 65 mith, (a) estimate the force on the sign with no crosswind.
{b} Discuss the effect of a crosswind. Fig. FILEE Solution: For air at 2119C, talte p: 1.2 kgx’m3 and n: LEE—5 leg/“ms. Convert 65 mifh = 29.06 nits. [a] If there is no crosswind, we ma}r estimate the drag force by ﬂat—plate
theory: 1.2[2Q.DE) [3} [1.1181 _ [1.03] Re = =1.55ET torﬁulent , {2' = _ — a [1.20291
1 ise—s E } D self [1 .555)”
PM = mg] Watts sides] = D.UD291[ %] (cans)E (as) (a) (2 sides] =14 1v Ans. [a] {b} A crosswind will cause a large aide force on the sign, greater than the ﬂat—plate drag. The sign will act like an airfoil. For example, if the 29 mfs wind is at an angle of only 5“
with respect to the sign. from Eq. (Till), CL n: 2;:rsin [5“]!{1 + 3.025} a [1112. The lift on the sign is then about
Lift = CL Lora) vzat .1. [ooa (twanzsosﬁosna) a so N Ans. {b} Test] a heavy sphere attached in a string
shculd hang at an angle 6' when immersed
in a stream at velccity U. as in Fig. Pisa. Derive an eapresaicn for H as a functicn cf
the sphere and ﬂew preperties. What is E if the sphere is steel {56 = 136} cf diameter
3 cm and the ﬂaw is sealevel standard air
at U = 40 rats? Neglect the string drag. Sciatica: 3Fer sealevel air take p: I 225 kgr‘rn3 and Jat: l. TEE—5 kgfm s The
sphere shculd hang sc that string tensien
balances the resultant cf drag and net weight: wast
Bras rand: , er H=tan'l[ ta— pigtm'ﬁlﬂ?’
{aﬁlﬂn atria:1 J Symbolic answers Fer the given numerical data. first check Re and the drag ceefiicieat. then ﬁnd the angle: 1 223%}{03‘3} ReD=—=B3ﬂl]ﬂ. Fig. alas: cu =a.s l .TSE—ﬁ F=%{c.5xt.225}(4cﬁe.e3f  {1.346 N: w: [asstaasi— 1.2251(931 igteaaii  the N a= tan'lﬂ .D‘JMJ‘lﬁ}  '12 Arts. 100 A piekug truck has a clean dragarea
Cy! of 35 ft. Estimate the hersepewer
requiredtudrive the lruekil 55 miﬂnia} clean
and [b] with the 3 by 60 Sign in Fig. P103
installed if the rolling resistance is l50 lhf
at sea IeveL Solution: For; sealevel air. lake p:
0. 00233 slugfﬂg’ and y: 17213—1 aluyﬂ 5.
Carmen V—  55 mii'h: 30. Ir" W5. Calculate the drag witlmut the sign: F= Fm“; “snag v1 =15e+ aﬁemzamnﬂﬂﬁf =421 Ihf Fig. em Hemmer = (421K30.T}+550  62 hp {clean} Arts. (a) With a Sign added. 11111 = 2.0. read CD = 1.19 from Table 13. Then
0.00233 F = 421...," +1 . Ie[ Jar.1}: {61(3 1 = 53? lhf. Fewer = FV = 00 hp Arts. {In} 114 Show that the ontodimensional
laminarthatr pattern with dpttttt = t} u= Uﬂti —eC‘—") u= uIt a u is an exact solution to the bottomlayer equations {7.19}. Find the value of the
constant C in terms of the floutr parameters. Are the bmndary cmditions satisﬁed? 1What Flg. F114
might this ﬂowr represent? Solution: Substitute these {our} into the :Itmomcntum equation (1 19!?) with é‘ufﬂx = D: or: C: pvnftt = constant c I] If the constant is aagathc, 1.1 does not go to m and the solution represents laminar
boundarylayer ﬂour past a ﬂat plate 1with wall suction, tr“ 5 U {see ﬁgure). It satisﬁes at 3 = t]: u = {1 {no slip} and it = cu {suction}: as 1.: —} an, o —} LL, {freestrcam} The thickness :1 where u = H.99Um is deﬁned by cspwvnﬁ'ﬂ}: 0.01,, or H: 4.6,tn'pvﬂ. ...
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 Spring '08
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